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Einstein's relativity of simultaneity - train Einstein's thought experiment has two lightning flashes at either end of a moving train as seen by an observer on the train, and a platform observer. They disagree on the simultaneity of the flashes.. But if we alter the experiment as follows they will agree on simultaneity.
Have the flashes located in front of the train on the rails, front left and front right, with the on train observer equidistant from the flashes. He will see them as simultaneous whether the train is moving or stationary.
The platform observer is positioned on a footbridge in front of the train also equidistant from the flashes, so he also sees the flashes as simultaneous.
Both observers agree on simultaneity. How is this possible ?
DAC
| Your interpretation of your modification is correct, and it's also entirely consistent with Special Relativity. What appears to be incorrect here is your assumption that two observers in different frames must disagree on simultaneity of events. This is not actually true (as you have demonstrated), nor does Special Relativity say it is true.
What Special Relativity actually claims, on the other hand, is that observers in different frames can disagree on simultaneity (i.e. it is possible, but not guaranteed). The original thought experiment was designed to illustrate a case where observers in different frames disagreed (which is usually the non-intuitive case). Your modification is a case where they agree.
| {
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How does a Galilean telescope form an enlarged image even though it has a diverging lens? I have been reading about Galilean telescope and the picture in the book is something like this:
After rays pass through the converging lens, there is a real image formed which is intercepted by the diverging lens but as I learnt before, diverging lens cannot form an enlarged image. So, is the ray diagram inaccurate?
| I think that perhaps the ray diagram in this article
https://thesciencegeek.org/2018/03/13/galileo-and-the-telescope/
is perhaps easier to follow. It shows clearly how the angular magnification is achieved. The article provides useful background
| {
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$SU(2)_L$ Gauge theory and particle-antiparticle annihilation I have a problem understanding how to reconcile the particle antiparticle annihilation vertex with the $SU(2)$ gauge theory, in the context of the weak interaction. Let me explain better :
Invoking $SU(2)$ gauge invariance we deduce there must be three gauge bosons, associated to the three Pauli matrices. We take, as usual, the linear combination yielding $\sigma_+, \sigma_-$ and $\sigma_z$ that are respectively associated to $W^+, W^-$ and $Z$. I am aware that I should be considering $U(1)_Y\times SU(2)_L$, but in the context of this question I believe it is irrelevant.
Now consider the SU(2) doublets,$\begin{pmatrix}l^+\\ l^- \end{pmatrix}$, where $l^+$ has weak isospin $1/2$ and $l^-$ has isospin $-1/2$. Let's take $\begin{pmatrix}v_e\\ e^- \end{pmatrix}$, we find that the weak current by coupling to the $Z$ boson is:
$$j^{\mu}_Z \propto \begin{pmatrix}\overline{v}_e & \overline{e}^- \end{pmatrix}\gamma^{\mu}\sigma_z \begin{pmatrix}v_e\\ e^- \end{pmatrix}$$
Where $\overline{u} = u^{\dagger}\gamma^0$. Expanding this, we find that :
$$j^\mu_Z=\frac{1}{2}\overline{v}_e\gamma^{\mu}v_e-\frac{1}{2}\overline{e}^-\gamma^{\mu}e^-$$
Where, $v_e$ and $\overline{v_e}$ stands for the spinors of the neutrino, and likewise for the electron. As we can see from this, it seems that the Z-boson couples particles of same weak isospin. However, we can have an annihilation vertex where $e^-$ and $e^+$ annihilate into a Z boson, despite the fact that $e^-$ has $I_w^{(3)} = -1/2$ while $e^+$ has $I_w^{(3)} = 1/2$. How can this reconciled with the representation of Z as $\sigma_z$ ?
I know that there is some problem with my current, since obviously an $e^-$ cannot annihilate with an $e^-$, in a vertex such as : , but only in a vertex such as : .
However, in my derivation, there does not seem to be a distinction in which one of these vertex I'm considering, so I'm confident that there lies my mistake, but I am unable to figure it out. I think somehow, in an annihilation vertex, particles of opposite weak isospin should interact while in a scattering vertex particle of same weak isospin should interact. This is also consistent with conservation of weak isospin, but I am unable to understand how to make this distinction in the currents using $\sigma_Z$ as the Z boson coupling.
|
How can this reconciled with the representation of $Z$ as
$\sigma_{z}$?
Don't confuse: the isospin $T$ is not the isospin projection $T_{3}$. The $Z$ and $W^{\pm}$-bosons, as well as fermions, correspond to the definite isospin projections. The difference is that the fermions are in the fundamental 2-dimensional representation of the $SU(2)$ group (the so-called $2$ representation), with isospin projections $\frac{1}{2}, -\frac{1}{2}$, while $W^{\pm}, Z$-bosons are in adjoint 3-dimensional representation (the so-called $3$ representation), with isospin projections $1,0,-1$. The isospin projection $T_{3}$ of the fundamental representation (say, just for electron) is defined as
$$
\frac{1}{2}\sigma_{3}\begin{pmatrix} 0 \\ e\end{pmatrix} = -\frac{1}{2}\begin{pmatrix} 0 \\ e\end{pmatrix} \equiv T_{3}\begin{pmatrix} 0 \\ e\end{pmatrix}
$$
The isospin projection of weak bosons is determined in a way
$$
[\sigma_{3}, \sigma_{i}] \equiv T_{3}\sigma_{i},
$$
where $\sigma_{i}$ is the generator associated with the given boson; see also the question. For $Z-$boson, $i = 3$, and $T_{3} = 0$; for linear combinations of $W_{1,2}$ bosons, namely, $W_{+} = W_{1} - iW_{2}$, $W_{-} = W_{1} + iW_{2}$, the projections are $T_{3} = + 1, T_{3} = - 1$ correspondingly.
Therefore the operator $\bar{e}\gamma^{\mu}(c_{V} - c_{A}\gamma_{5})eZ_{\mu}$ has zero total isospin projection.
| {
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If velocity is relative, how can mass (inertia) be coordinate independent? Would inertia grow differently if we switch the reference frame?
LHC accelerator makes protons x7000 heavier at record speeds and I just wonder if we could change the reference frame (keeping velocity), would its mass be different? (eg. in vacuum, far from any gravitational field). If velocity is relative, we can say that protons are in rest and any other frames are moving at that velocity in opposite direction...
| Your main problem here is that you are mixing ideas from two different ways of formalizing relativity.
If you are using the (old fashioned, unnecessary, and easily misapplied) notion of "relativistic mass" then mass is not invariant.
If you are using the modern nomenclature then 'LHC accelerator makes protons x7000 heavier' is incorrect, in that framework the accelrator gives them an energy 7000 times their mass energy, but their mass remains exactly the same.
Now, you add 'inertia' to your title, but I'd like to talk about why that doesn't actually make relativistic mass a good idea.
Let's say that you have a mass moving a speed $v$ with respect to you and you apply a force $F$ to it. From Newton's second law you would expect that the ratio of the force to how fast the velocity changes to be the mass.
If you apply the force perpendicularly to the direction of current direction you'll get
$$ \left(\frac{F}{a}\right)_\text{transverse} = \gamma m \;,$$
where $m$ is the mass you measured for the object at rest and $\gamma = [1 - (v/c)]^{-1/2}$ is the Lorentz factor. This is the usual "relativistic mass"
But if you apply the force along the same direction as the current motion you
$$ \left(\frac{F}{a}\right)_\text{longitudinal} = \gamma^3 m \;.$$
So the relativistic mass isn't the inertia in any general way.
In the modern parlance, this difference is understood to be a consequence of the way velocity composition works, because the mass remains the same.
| {
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How does the sun really produce light in terms of waves? Electromagnetic waves are caused by changing electric and magnetic fields, and these are caused by a charge possible oscillating like an antenna or a varying current etc.
My question is, with the sun, where is this source that causes the electric and magnetic fields to oscillate.
Everywhere I've read stated that it was due to the energy released from nuclear fusion, but when looking at the process of nuclear fusion there are no charges produced. How is this 'energy' supposedly producing the same effect as an electron oscillating?
| The visible, UV, and IR radiation which reaches the earth from the sun is due to blackbody radiation from the surface (photosphere and chromosphere) of the sun. You can see more about blackbody radiation here and in the answer by @annav here. Basically, temperature indicates molecular or atomic vibration, which means moving charges. Higher temperature means more movement which means more EM radiation. Anna's link gives a fuller picture.
The fusion process, deep in the core of the sun, transfers mass-energy into both thermal (aka, atomic kinetic energy) and photons (which are EM energy bundles emitted from nuclei when the nuclei re-arrange during decays and reactions). Fusion involves strong, weak, and electromagnetic interactions, so the emission of photons is not surprising. All of this energy takes a long time to transition to the surface due to the high densities of the inner layers of the sun, plus it gets spread out over a much larger volume, so the temperature at the surface (~6000 K) is much lower than the temperature in the core (~15 MK).
| {
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Are all electromagnetic waves transverse in free space? I understand that in vacuum, monochromatic plane waves are transverse. I suppose that a non-monochromatic plane wave would also have this property, as it can be decomposed using the Fourier transform and each of its components would be transverse.
However, what happens if the waves are not plane? Are they still transverse? The only demonstrations I've found are used for plane waves, so I don't know if this is a necessary condition or is just done for simplicity but can be generalized.
| As discussed in the comments, the plane wave approximation is mostly irrelevant. The only important ingredient is the fact that the homogeneous vacuum Maxwell equations read
$$
\nabla\cdot\boldsymbol E=\nabla\cdot\boldsymbol B= 0
$$
and therefore, after a Fourier transform $\boldsymbol x\to\boldsymbol k$ we get
$$
\boldsymbol k\cdot\boldsymbol E=\boldsymbol k\cdot\boldsymbol B= 0
$$
which means that the fields are transverse.
In practice, the most simple treatment of electromagnetic waves is through the vector potential. The equations are simplest in the Lorenz gauge
$$
\partial_\mu A^\mu=0
$$
which, again, in Fourier space becomes $k\cdot A=0$, i.e., the vector potential is transverse. Other very useful gauge is the radiation gauge
$$
\nabla\cdot \boldsymbol A=0
$$
which, again, implies that $\boldsymbol A$ is transverse:
$$
\boldsymbol k\cdot\boldsymbol A=0
$$
| {
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Why are the windows of bridges of ships always inclined?
What is the reason that the windows of ships' bridges are always inclined as shown in the above picture?
| Look at CandiedOrange's answer
This answer was accepted, but CandiedOrange has the right answer. See this document page 21:
The second way in which reflection can interfer e with controller’s vision is light sources within the cab (or direct sunlight that enters the cab), which can cause disturbing reflections during either day or night operations. The effects of these reflections can be a loss of contrast of the image being viewed, a masking effect of a competing image, or
glare. The two ways to mitigate these effects are to reduce the reflection coefficient or to design the ATCT cab to reduce or eliminate the probability that any light source (artificial or natural, direct or indirect) can produce a reflection in the pathway of a
controller’s view out of the cab windows.
It controls glare. Whenever the sun hits a window, it reflects off of it. If the windows are vertical, its pretty hard to control where that glint could go. When the sun is near the horizon, it could even be seen by other ships, but at the very least it can blind workers on your own ship.
Angling them doesn't prevent this from happening entirely, but it does substantially limit the places on the ship which can be hit by this glint to a small region around the bridge itself.
This requirement appears in specifications such as these regulations from the UK:
1.9 Windows shall meet the following requirements:
1.9.1 To help avoid reflections, the bridge front windows shall be inclined from the vertical plane top out, at an angle of not less than 10° and not more than 25°.
...
These same rules are also applied to air traffic control towers at airports:
| {
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Why does torque produce a force on the axis of rotation? If a door is rotated about its fixed axis in (outer) space, a force parallel to the door on the hinges will arise due to centripetal force on the centre of mass and conservation of momentum (Newton's third law).
But any torque on the door will create a force on the hinges which is equal to $t/r$ or torque divided by radius. I'm looking both an intuitive and mathematically based explanation for this fact. I can sort of 'see' why, but my understanding is vague and uncertain.
| "Any" torque will not create a force. Take the simple example of a force $F$ (and the only force in this problem) perpendicular to the door at a distance $r$ from the hinge. The torque's magnitude is $|\tau|=|rF|$ and there is no force on the hinge by the problem statement.
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Does Bayer demosaicing from RGB-filter sensors work *because* the color filters are imprecise? I have an ongoing friendly dispute with another member over on Photo Stack Exchange about the fundamentals of how RAW works, and I'm hoping you can settle it.
My understanding is that Bayer demosaicing works basically entirely on the assumption that one can deduce likely color information from monochromatic pixels. That is, for a pixel with a blue filter, you can assume that the correct green value is close to the average of neighboring green-filtered pixels, and the red value close to the average of neighboring red-filtered pixels — even if those filters were to be theoretically perfect.
My site-colleague argues that, as shown in this graph:
the filters have a lot of overlap, and describes the demosiacing algorithms as working because they recover that overlap.
Who is correct here? Would Bayer demosaicing work if the filters had no overlap?
(Bonus questions: would this be an improvement, or actually a downside? Presumably you'd be letting in less light overall; would you get more accurate color for that price? I know that simple averaging "works" for demosiacing; are there more complicated algorithms which do take overlap into effect explicitly?)
| Most situations, you would probably not notice the difference if there was no overlap. There are a few situations where we see monochromatic light in the real world. Rainbows, lasers, polarized reflections among other things. A rainbow is good example because you have the whole visible spectrum in one shot. If there was not overlap in the detectors sensitivity bands, you would have weak sensitivity zones in the colors in between the sensitivity peaks since the yellow and cyan in the rainbow are not actually made from a composition of either neighboring color. It's just pure yellow and there is no yellow detector.
| {
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Partiton function of lattice gas Suppose I have a single particle on a cubic lattice with lattice site length $a$ and $N$ lattice cells. The particle is feeling a potential $v(\boldsymbol{r})$ which is a continuous spatially dependent potential and a kinetic energy $t(\boldsymbol{p})=-\frac{p^2}{2m}$. One can then formulate the partition function as
$Z=\frac{1}{h^3}\int\mathrm{d}\boldsymbol{p} \mathrm{e}^{-\beta t(\boldsymbol{p})} \int \mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta v(\boldsymbol{r})}=\frac{1}{\Lambda^3}\int \mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta v(\boldsymbol{r})}=\frac{a^3}{\Lambda^3} \sum_i^N \mathrm{e}^{-\beta v(\boldsymbol{r_i})}$
Is there something wrong with the last step? Taking the limit $a\rightarrow0$ should give the partition function of a free particle feeling the potential $v$,i.e.
$Z=\frac{1}{\Lambda^3}\int \mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta v(\boldsymbol{r})}$
but here $a\rightarrow0$ gives just zero...
| You are forgetting a -1, should be subtracted from your exponential and then evaluate the integral
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Wave equation: $y=A \sin(\omega t-kx)$ or $y=A\sin(kx-\omega t)$? What is correct wave equation:
$y=A \sin(\omega t-kx)$ or $y=A\sin(kx-\omega t)$?
How are these wave equations used in the positive $x$-direction
and negative $x$-direction?
| One of the problems of illustrating a wave is that the displacement of the particle $y$ is a function of two variables the equilibrium position of the particle $x$ relative to some origin and time $t$.
So to graph a wave one might draw.
However is is more practical to compare the displacement with time of particles at fixed positions or the displacement with position of particles at fixed times.
The graph above shows the motion of a particle as a function of time at a fixed position, $x=0$ in this case.
If one now adds the graph of displacement against time for a particle at $x = \Delta x$, ie at an equilibrium position further away from the origin in the positive direction.
What one notes from these two graphs is that the red graph lags behind the grey graph.
This means that the wave is travelling in the positive x-direction.
You can think of it as the particle at $x=0$ telling the particle at $x = \Delta x$ what to do but that information is delayed because the information (wave) travels at a finite speed.
With $\omega t$ and $kx$ reversed you will see that it is still a wave travelling in the positive x-direction but $\pi$ out of phase with the other one.
If you look at my attempted representation of the 3D graph you are essentially looking at a series of sections of the graph which are parallel to the $y-t$ plane ie looking left from the right hand side.
The other wave of representing a wave is to draw a wave profile which is essentially an instantaneous photograph of the wave.
Both the graphs illustrate quite clearly that the waves are travelling in the positive x-direction.
If you look at my attempted representation of the 3D graph you are essentially looking at a series of sections of the graph which are parallel to the $y-x$ plane ie looking from the front into the screen.
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How can a railcar withstand high pressure but fail under a vacuum? A 25,000 gallon (95,000 litre) bulk chemical storage railcar can store products with vapor pressures in excess of 200 psia (1.38 Mpa). The same railcar can not withstand a vacuum when being unloaded. I want to understand why.
A bulk chemical storage car in this example (assume refrigerant on a warm day) is unique in it's design in the sense that it is a pressure vessel contains well over 200 psig (1.38 MPa) internal pressure whereas many metal tanks are rated for far lower pressures thus the pressure differential between atmosphere and the inside of the metal tank may be typically be close to the difference found in this example, 14.7 psi (101 kPa) or less.
The definition of a metal tanks can be open to interpretation not only with regard to pressure ratings but also wall thicknesses. One of the answers posed refers to plastic or aluminum soda containers. The material properties are far different than those typically found in the types of rail cars I have presented here.
| The sealing of a port in a pressure vessel might use the pressure in the
vessel to compress a seal (push an O-ring firmly against a door); such
a seal will not be properly compressed if the vessel is under vacuum.
Many pressure vessels, with thin walls, use pressure-induced tension to hold
the shape (like a balloon); under negative pressure, like a balloon, that kind
of vessel will collapse.
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Rewriting the equation for the virial mass of a cluster I have a question that I don't know how to solve, any help is appreciated.
Here it is:
Rewrite the equation for the virial mass of a cluster so that if velocities are entered in km/sec and dimensions in Mpc, the resulting mass is in solar mass
Well the equation is: $M=\frac{5(v_r)^2R}{G}$
I think I have to use dimensional analysis but I really don't know how to get there
$[M]=\frac{(\frac{km}{s})^2km}{\frac{m^3}{kg*s^2}}$
with one Mpc being $3.09*10^{17}$ km
Thanks in advance :)
| This is a very common (and quite subtle) thing done, especially in astrophysics, that is extremely useful to master. It can be a little tricky at first, but it isn't bad. The idea is just to convert standard equations into units that are a little more convenient. The key is to remember that units (like grams, years, or solar-masses) are quantities that can be canceled out in equations just like variables or other numbers.
Let's work with a simpler example: kinetic energy.
$$E = \frac{1}{2} m v^2$$
If we want to calculate energy from the above equations, we put in values on the right, and get an answer out. If we want our answer in erg (the standard CGS unit for energy, s.t. $1 \, \rm{erg} = 10^{-7} \, \rm{Joule}$), then we need to put our mass in grams, and our velocity in cm/s. That's because $1 \, \rm{erg} \equiv g \, \rm{cm}^2/\rm{s}^2$. We could build this into our equation, by rewriting it as:
$$\frac{E}{1 \, \rm{erg}} = \frac{1}{2} \left(\frac{m}{1 \, \rm{g}}\right) \left( \frac{v}{1 \, \rm{cm/s}} \right)^2$$
or equivalently (just multiplying both sides by $(1 \, \rm{erg})$ as,
$$E = 1 \, \rm{erg} \cdot \frac{1}{2} \left(\frac{m}{1 \, \rm{g}}\right) \left( \frac{v}{1 \, \rm{cm/s}} \right)^2.$$
This makes it very clear that if I plug in the mass in grams, and the velocity in cm/s, then I get out a number of ergs.
But what if we want to use different units? For example, what if I want to use (US "short") tons for mass? That's simple. We know (or can google search) that, $1 \rm{ton} = 907,185 \rm{grams} \approx 9 \times 10^{5} \rm{grams}$. That means that $1 = (9 \times 10^{5} \rm{grams}) / (1 \, \rm{ton})$. So, we just need to multiply the initial equation by that conversion factor:
$$E = 1 \, \rm{erg} \cdot \frac{1}{2} \left(\frac{m}{1 \, \rm{g}}\right) \left(\frac{9 \times 10^{5} \rm{grams}}{1 \, \rm{ton}}\right) \left( \frac{v}{1 \, \rm{cm/s}} \right)^2.$$
The grams cancel on the top and bottom, and we're left with:
$$E = 9 \times 10^{5} \, \rm{erg} \cdot \frac{1}{2} \left(\frac{m}{1 \, \rm{ton}}\right) \left( \frac{v}{1 \, \rm{cm/s}} \right)^2.$$
But what if we also want to use miles-per-hour for the velocity? We know that, $1 \rm{mph} \approx 45 \rm{cm/s}$, so:
$$E = 9 \times 10^{5} \, \rm{erg} \cdot \frac{1}{2} \left(\frac{m}{1 \, \rm{ton}}\right) \left( \frac{v}{1 \, \rm{cm/s}} \right)^2 \left( \frac{45 \, \rm{cm/s}}{1 \rm{mph}} \right)^2.$$
Now, the $(\rm{cm/s})^2$ cancels, and we use: $45^2 \cdot 9\times 10^5 \approx 2000 \cdot 9 \times 10^5 \approx 2 \times 10^9$...
$$E = 10^{9} \, \rm{erg} \cdot \left(\frac{m}{1 \, \rm{ton}}\right) \left( \frac{v}{1 \, \rm{mph}} \right)^2.$$
(Note that I canceled out the factor of 1/2).
This equation is much more convenient for calculating the energy of a speeding car (for example) because the units are much nearer to unity (1.0). For example, it's easy to see that a 2 ton car, going at 10 mph, has ($2 \times 10^2 \times 10^9 = $) 200 billion ergs of kinetic energy!
Try using these methods on your particular problem. You can edit your question to show your work. If you're still having trouble just comment on this answer to let me know.
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Why is voltage inversely related to capacitance? $C=\frac{Q}{\Delta V}$ More like charge on each capacitor plate increases $Q$, which increases $C$. However, more like charge on each plate also increases $\Delta V$ right? Because compared to before the capacitor was charged, there was an equal amount of each type of charge and everything cancelled to $0$. But now when $C$ is charged, not only do lots of like charges exist in the same place, but right over on the other plate is a bunch of the other type of charge. Doesn't this create a potential difference between two points in space between which a strong electric field is generated? So as a capacitor accumulates charge, it increases in capacitance; yet simultaneously it gains voltage, and still manages to increase in capacitance despite the inverse relationship.
What's wrong with this logic?
I've just read a thread about how even though $\Delta V=IR$, voltage can actually exist without current. So even though the capacitor may be disconnected, it still has a voltage. My question applies to all possible states of a capacitor; including charging, discharging or charged capacitors.
| What you have written about more charge means a larger capacitance is not true.
The capacitance of a conductor depends on its dimensions and its composition.
For an ideal parallel plate capacitor the capacitance is $\dfrac{\epsilon A}{d}$ where $A$ is the area of the plates, $d$ the separation of the plates and $\epsilon$ the permittivity of dielectric between the plates.
For such a capacitor it is found that the charge stored (on one of the plates) $Q$ is proportional the potential difference between the plates $\Delta V$, so $Q\propto \Delta V \rightarrow Q = C\; \Delta V$ where the constant of proportionality $C$ is called the capacitance.
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Symmetry of two-particle system of electrons Consider a two particle system consisting of two electrons. The complete state of the electron includes its position wave function and also a spinor describing the orientation of its spin: $$\psi(r) \otimes \chi(s).$$ Why does it follows that for the the two particle system that if we have an anti-symmetric spin state of the two electrons such as the singlet state $$\frac{1}{\sqrt{2}}(| {\uparrow} \rangle \otimes | {\downarrow} \rangle - | {\downarrow} \rangle \otimes | {\uparrow} \rangle)$$ then this has to be joined with a symmetric spatial function (and similarly if we have a symmetric state of two electrons such as $| {\uparrow} \rangle \otimes |{\uparrow} \rangle$ then this has to be joined by an anti-symmetric spatial wave function?
Also if two electrons occupy the singlet spin state then the spatial wave function describing the two particle state would be symmetric, but I thought that for identical particles which are fermions (such as electrons), the spatial wave function is always antisymmetric?
Thanks for any assistance.
| For fermions, the total wave function, including both the spatial wave function and the spin state, must be antisymmetric under exchange. Since the product of two antisymmetric functions is symmetric (as is the product of two symmetric functions), it is necessary that either the spin is antisymmetric or the spatial wave function is antisymmetric but not both.
| {
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Why is a nonzero VEV for a spinor field said to break Lorentz invariance? Consider a spinor field $\psi(x)$. Its vacuum expectation value is given by $$v=\langle 0|\psi(x)|0\rangle.$$ Using the fact that the vaccum is invariant under Lorentz transformation, we get, $$v=\langle 0|\psi(0)|0\rangle.$$ Why is it that, if $v\neq 0$, the Lorentz invariance is broken?
| To make ACM's argument more explicit, consider
\begin{align}
v&=\langle 0|\psi|0\rangle\\
&=\langle 0|\overbrace{UU^\dagger}^1\psi\overbrace{UU^\dagger}^1|0\rangle\\
&=\overbrace{\langle 0|U}^{\langle 0|}\overbrace{U^\dagger\psi U}^{D_\Lambda \psi}\overbrace{U^\dagger|0\rangle}^{|0\rangle}\\
&=D_\Lambda v
\end{align}
where $U=U(\Lambda)$ is the unitary operator that corresponds to Lorentz transformations in the Hilbert space, and $D_\Lambda$ its representation in the space of spinors.
Considering $\Lambda$ to be, say, a rotation around the $z$ axis with angle $\theta$, and expanding to first order in $\theta$, we get
$$
S^zv=0
$$
which is impossible for representations of the Lorentz Group with half-integer spin, as $S^z$ has eigenvalues
$$
-j,-j+1,\cdots,+j
$$
none of which is zero.
Therefore, we must conclude that $U(\Lambda)$ doesn't exist, that is, the Lorentz symmetry is broken.
| {
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Did inflation stop because of energy density drop or some other reason? If I am understanding big bang correctly... During big bang
Approximately $10^{−37}$ seconds into the expansion, a phase transition caused a cosmic inflation,
Which was 'free lunch' of energy and it continued for some time but then suddenly 'free lunch' was over and energy conservation started...
The universe continued to decrease in density and fall in temperature, hence the typical energy of each particle was decreasing.
and quarks and other particles formed....
Did inflation stop because of energy density drop or some other reason?
| The standard Big Bang model meets at the singularity proposed by the original Big Bang model, a classical General Relativity model, which was proposed when it was observed that all clusters of galaxies are retreating from each other as if coming from a cosmic explosion.
Diagram of evolution of the (observable part) of the universe from the Big Bang (left) - to the present.
The inflationary model assumes an effective quantization of gravity during the period of rapid inflation. This was deemed necessary because the homogeneity of the cosmic microwave background radiation (the blue plane in the diagram) could not be explained . At the time of separation of the electromagnetic radiation there were regions in the observable universe that could not come into thermodynamic equilibrium .
The inflaton field was proposed so that it would create the homogeneity from the quantum mechanical indeterminacy, generating an enormous expansion. At some point the expansion stopped because the energy density dropped and the elementary particles in the table were created out of the available energy.
The inflaton field is a hypothetical scalar field that is theorized to drive cosmic inflation in the very early universe.The field provides a mechanism by which a period of rapid expansion from 10^−35 to 10^−34 seconds after the initial expansion can be generated, forming a universe consistent with observed spatial isotropy and homogeneity.
If there is another reason than energy density, it is dependent on the specific quantum mechanical modeling. The gross reason is the drop in the energy density, the same for determining all periods of the Big Bang expansion.
| {
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Why are light rays able to cross each other? See the image first:
Why are light rays able to cross each other? Air isn't able to.
| Note: this answer was in response to the original question:
My question is that Why the light rays able to cross each other weather water waves and air could not cross each other
Other waves pass through each other just as with light. This is easy to test. Place four people at the corner of a large room. Have two of them, at adjacent corners talk to the person at a diagonal corner. Use a cone such as a cheerleader might use to somewhat channel the sound. You may be a bit distracted by the other voice but you will clearly hear the voice from the opposite corner.
Here's a standard demo in a high school science class. Have two students hold each end of a moderately stretched slinky resting on a smooth floor. Have each student give the slinky a sharp snap to their right. Since the students are facing each other, the pulses will be opposite one another as they travel toward opposite ends. When the two pulses meet in the middle the slinky will appear relatively straight but only for an instant. The two pulses will continue to travel past one another as if they never had met.
Waves of the same kind traveling through one another maintain their original identity after the encounter. This is a basic property of waves, you can read about it in any introductory Physics text.
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Is there any advantage in stacking multiple images vs a single long exposure? Suppose I have a source object that is not time varying, to be concrete let's say it's a galaxy. Is there anything additional that can be learned or done with multiple short exposure images of exactly the same field as compared to a single long exposure, given that the total integration time is identical? I'm thinking of things along the lines of noise suppression, background removal, image processing magic...
So far the only thing I can think of is that a long exposure could saturate the detector (I'm thinking CCD here). Short exposures could avoid this, allowing for accurate photometry across the entire image. I've tagged this [astronomy] since that's the area of application I'm most familiar with, but perspectives from other fields are welcome.
| A significant increase in dynamic range can be achieved by taking exposures with different exposure settings. Wind,vibration, periodic errors etc. Can be combined with all other techniques. It is used in spectrum analyzers under the name of video averaging. Can also be used in radio astronomy. I wonder if it could have been adapted to address the reciprocity failure in film?
| {
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contravariant components of electromagnetic field tensor under lorentz transformation I have to show, how the contravariant components of the electromagnetic field tensor behave under Lorentz transformation.
I guess the answer should look something like this
$$F'^{\mu\nu}=\frac{\partial x'^\alpha}{\partial x^\gamma}\frac{\partial x'^\beta}{\partial x^\delta}=\Lambda^\alpha{}_\gamma\Lambda^\beta{}_\delta F^{\gamma\delta}$$
That's not a big deal, one can see this immediately (although I don't really get the message of this exercise, since $F^{\mu\nu}$ just transforms as a 2nd rank tensor has to by definition)
However, the exercise wants me to derive this from the inhomogeneous Maxwell equations in covariant formulation:
$$\partial_\mu F^{\mu\nu}=\mu_0j^\nu$$
I know how $\partial_\mu$ transforms:
$$\partial'_\mu=\frac{\partial x^\nu}{\partial x'^\nu}\frac{\partial}{\partial x^\nu}=\Lambda_\mu{}^\nu \partial_\nu=(\Lambda^{-1})^\nu{}_\mu \partial_\nu$$
as well as $j^\nu$:
$$j'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}j^\nu=\Lambda^\mu{}_\nu j^\nu$$
But how do I link these equations to the 3d one and arrive at the 1st ?
Any hints?
| What you're maybe missing is the requirement that the covariant equation should still be valid in the new coordinate system:
$$ \partial'_\mu F^{\mu\nu} = \mu_0 j^{\prime\nu} $$
You can now plug in your three equations and solve for $F'$.
| {
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How can Entropy Distinguish Reversible and Irreversible Process if It is a State Variable? If entropy is a state variable, it means that the change in entropy should not depend on the path between two equilibrium states. But the second law of thermodynamics states that, in an isolated system the entropy change due to irreversible process must be greater than zero, while the entropy change due to reversible process is always zero. If both processes start from state A and end in state B, how can the entropy change be zero?
| A similar question has been asked before, although I cant find it now. In an isolated system, between two given states $A$ and $B$ there can be either a reversible process or an irreversible process, but not both. If entropy of isolated system at $A$ and $B$ are same then the process must be reversible, otherwise not.
| {
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Single photon, double slit question If photons are fired conitnuosly one at a time through the double slits is the statistical outcome going through each slit 50%?
| The key feature of the double slit experiment is that the light passes through both slits at the same time. It does not go through one slit or the other but instead passes through both.
This happens because unlike macroscopic objects like baseballs quantum objects do not have a position. They are fuzzy objects that are spread out over a region of space. The double slit experiment only works when the photons are delocalised enough that their spatial extent covers both slits.
| {
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How are club-style weapons effective? The First Law of Thermodynamics states that I can't swing an object held in one hand with more energy than I can swing my arm, and the Second Law says that the total energy would probably even end up being somewhat less. And yet, a person who might not be afraid of getting punched by me would certainly be more cautious if I had some sort of blunt object, such as a club, baseball bat, or crowbar.
If it were a blade, I could understand that: the edge focuses the impact down to a much more concentrated line. But a blunt weapon doesn't do that, so how is it able to be an effective weapon, hitting harder than you can hit with an unarmed blow?
| The answer: It's less soft.
It is not as important how much energy is transferred, as how fast. Something more soft makes the impact last longer. The soft material dissipates some energy during its deformation.
You can see this from the momentum version of Newton's 2nd law:
$$\sum F=\frac{dp}{dt}$$
You need to make a certain change in momentum $dp$. If you can do that very faster (small $dt$), the force is huge.
| {
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What is the correct way to estimate the work done by a climber? My teacher gave us a worksheet with word problems and their solutions. It is in German, so I have tried my best to translate it to English:
A 26 year old man climbs Mount Everest (8848 m) in only 8 hours 10 mins from the base camp at 5300 m. Estimate the "lifting work" (Hubarbeit) that the man exerted in the climb.
I thought to use the formula: $W = F \cdot \Delta S$, but I don't know what $F$ would be. I think the $\Delta S$ is $8848 - 5300 = 3548$. But then I looked at the solutions and my teacher used this formula: $W = m g h$. He guessed the man's weight and also used $g = 10\ \mathrm{N}/\mathrm{kg}$. I don't understand this.
Could someone maybe help me? Anyone have an idea?
| $W=mgh$ is the potential energy the climber (of mass $m$) gains by changing his altitude by $h=8848~\textrm{m} - 5300~\textrm{m} = 3548~\textrm{m}$.
This will be a lower bound for the work done by the climber. In reality work/energy will be lost in friction and any other activities (e.g. setting up a tent...) the climber does which are not related to raising his altitude.
| {
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What causes a circle of light to appear opposite the sun when looking from an airplane? Today I saw a circle of light outside my plane window on the clouds, as if someone was shining a bright, tightly focused flashlight, or perhaps like the halo that sometimes appears around the sun. I think it was approximately where I would expect our shadow to be (at least the sun was shining on the opposite side of the plane).
I took a video with my phone. I move the camera around a bit to show that it doesn't seem to be an artifact of the window.
| It is an optical phenomenon called a "glory" caused by a diffraction in very small water drops. In fact, the radius of the glory depends on the size of the drops of water and can change in a dynamic fashion as you fly above different clouds with different size of drops. It is quite different than a rainbow though.
Professor Lewin gave a fantastic lecture at MIT about the rainbow and other optical phenomenon, where he mentions glories: https://www.youtube.com/watch?v=pj0wXRLXai8&list=PLyQSN7X0ro2314mKyUiOILaOC2hk6Pc3j&index=32.
However, as pointed out in the comments, he does not provide a thorough explanation of the glory which is extremely difficult, there is still no complete explanation that is agreed upon. Still the lecture as a whole is very instructive and is worth watching.
| {
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Why does each plate receive a charge exactly the same magnitude in parallel plate capacitor? When a parallel plate capacitor is connected through a cell, each plate of the capacitor receives a charge with the same magnitude, but with the opposite sign. Is it because of the battery or the area of the plates?
| The battery creates a potential difference between its terminals. Because, in a steady state situation, the total voltage through a closed circuit is zero, this will create, when the capacitor is charged, the same voltage over this component.
Then, if the charges on the capacitor plates wouldn't be equal in magnitude (!), the electric field outside (and inside as well) the capacitor would not be zero. However, in steady state the electric field inside a (perfect) conductor, the wires in this case, is always zero.
| {
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How to calculate an object breaking point or force require to break it? for example, i have a 10 ft rope that has a strength to hold 2000lbs.
when I attach the rope to an object of 1500lb and pull it toward me, (keep it simple with no friction apply) the rope should not be damaged
Now I wrap the rope around the object and have 2 persons identity with the same amount of force pulling toward to themselves. does each person share equal amount of 2000lbs ? which law of physics define this?
| You have asked at least two different questions here. I will answer the one in the title of your post.
To solve for when the rope will break, you need to know three things: 1) the magnitude of the load, 2) the cross-sectional area of the rope, and 3) the yield strength of the rope material.
The yield strength is quoted in terms of pounds of load per square inch of area at the breaking point. So you solve for (magnitude of load)/(cross sectional area of rope). If this number is greater than the yield strength, the rope breaks.
| {
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Is hermiticity a basis-dependent concept? I have looked in wikipedia:
Hermitian matrix
and
Self-adjoint operator,
but I still am confused about this.
Is the equation:
$$ \langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A.$$
independent of basis?
| There is a definite sense in which self-adjointness is indeed a basis-dependent concept. It really depends on what you start with. If you have a particular inner product in mind, then the truth of your equation is entirely independent of what basis you choose to represent $x, y$, and $A$ with respect to. However, if you do not have a particular inner product in mind, then the choice of inner product itself plays a role.
The idea is that choosing an inner product makes some bases special: namely, those which are orthonormal with respect to it. It is not so restricting as choosing a single basis, but it still means that we can make a choice that determines whether or not the matrix (or rather the linear transformation) is self-adjoint.
More precisely, we can find a matrix $A$ and two inner products $\left< \cdot, \cdot \right>_{1}$ and $\left< \cdot, \cdot \right>_{2}$ such that $A$ is self-adjoint with respect to the first but not the second. For instance, take
$$ A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$
and consider the inner product $\left< \cdot, \cdot \right>_{1}$ such that
$$\beta_{1} = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$$
is orthonormal
(i.e. the standard dot product), and the inner product $\left< \cdot, \cdot \right>_{2}$ such that
$$\beta_{2} = \left\{ \begin{pmatrix} 1 \\ 2 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$$
is orthonormal.
Then $A$ is self-adjoint with respect to the first (since it is hermitian in that basis) but not with respect to the second. To see this, note that with respect to the second basis, $A$ takes the following form.
$$ [A]_{\beta_{2}} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -3 & -1 \end{pmatrix}$$
The main point is that the notion of being basis-independent is a vague concept. When only talking about vector spaces (i.e. without inner products), we can speak precisely about what it means to be basis-independent since we can distinguish between things that require us to pick a specific basis and those that do not. However, when we add the structure of an inner product, we cannot really talk about something being basis-independent in the same absolute terms since the choice of inner product is basis-dependent. However, we can talk in a weaker sense about something being basis-independent assuming we already have an inner product. In this way, self-adjointness is basis-independent.
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Direction of normal force on stick on box
What is the direction of the normal force on the stick in this case, assuming gravity? Is it right angled with the stick? Or is it upwards? Or is it impossible to determine?
| I think the problem is not the stick, it is the edge of the box. For an ideal box the edge changes direction discontinuously : one side is vertical, the other is horizontal. So how can the normal to the surface of the box at an edge be anything other than horizontal or vertical?
For a real box the edges cannot change direction discontinuously. If we look on a small enough scale an edge changes direction gradually, and can be approximated by a section of a continuous curve, such as a circle. Then it is easy to identify a point of contact at which the surfaces of the stick and box are parallel, and a common direction which is normal to both.
However a continuously changing surface is really yet another idealisation. On a microscopic scale most real contacting surfaces are jagged an irregular, and deform in response to the forces between them. The result is that there are many individual points of contact at different angles, and many different contact forces. The macroscopic contact force is the sum of these, and its resolution into parallel and perpendicular components is a convenience for mathematical analysis of the situation.
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Do we suspect that any 2 seemingly identical experiments actually change under the passage of time? For example, let's say that I set up 2 consecutive identical experiments where I know that the conditions are exactly the same (go through whatever difficulties you need to). The only thing I can't control is the passage of time, of course.
We currently accept that if 2 experiments are set up and carried out in precisely the same manner, then there is no difference in the results. We even accept the converse because we see that the results are exactly the same and claim that the setups must have been exactly the same as well. But can anything be said about a possible difference simply due to the passage of time? Because even if you try to set up completely identical conditions and procedures, the reality is that for the first experiment you did it some time ago and the other experiment you did it some time after that. There is always that difference, even though we try to mimic identical conditions. Is there any inquiry into this question?
|
I set up 2 consecutive identical experiments where I know that the
conditions are exactly the same (go through whatever difficulties you
need to). The only thing I can't control is the passage of time.
Totally impossible.
Perhaps the OP was asking as to whether some of the fundamental constants of Physics (speed of light, gravitational constant, fine structure constant etc) are really time invariant?
A change in these constants would produce different results in similar experiments done at different times and a lot of work for theoretical Physicists.
The best that has been done is to say that over such and such a time scale no detectable change has been found for a lot of these constants and this is expressed as an upper bound for the fractional change over a number of years.
This Wikipedia article states:
For the fine-structure constant, an upper bound on time variation of
$10^{−17}$ per year has been published in 2008
and
an upper bound of less than $10^{−10}$ per year for the gravitational
constant over the last nine billion years
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Why don't these two methods of finding the electric potential in a semicircle agree? I was given the following problem:
A wire of finite length that has a uniform linear charge density $λ = 5.51\times 10^{-9}\ \mathrm{C/m}$ is bent into a semi-circle. Find the electric potential from the center of the semi-circle.
After trying it myself and seeing several examples on the internet, I've seen that the typical way to approach this problem is from the formula
$$V = \int \frac{k\,\mathrm{d}q}{r}\tag{1}$$
where $λr\,\mathrm{d}\theta$ is substituted for $\mathrm{d}q$. The integral is bounded from $0$ to $\pi$, resulting in $kλ\pi$.
However, when I first solved this problem, my instinct was to use another approach:
$$V = -\int \vec{E}\,\mathrm{d}s\tag{2}$$
I knew from previous exercises that the electric field of a semi-circle is given by $-2kλ/R$, which when inputted into the electric potential formula in terms of electric field, gives a $V$ of $2kλ\pi$.
This second value is off from the first by a factor of two, when the two values should be the same. Why don't these methods agree?
| Actually you've got that latter formula wrong. It's
$$\Delta V = -\int \vec{E}\cdot\mathrm{d}\vec{s}$$
The $\Delta$ is important. It reflects the fact that you are calculating a change (or difference) in potential between two points, not the potential at a point, as you are being asked for in this problem. So you can't use that equation. It simply does not apply to the physical situation you're dealing with.
The integral in this equation is a path integral, too, which means you integrate along some path running between the aforementioned two points, not along the path where the charge is.
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Canonical quantization of bosons During my studies on QFT a fundamental question occurred concerning the canonical quantization. In our course, we mentioned that:
"The canonical quantization of a field with values in the complex numbers can lead only commutation relations, as opposed to anticommutation relations."
How can I interpret this statement? Is there any justification?
| The product of two complex numbers is commutative and that's why you can't have anticommutation relations. If you want to have anticommutative fields you need Grassmann numbers that are numbers whose product is defined to be anticommutative.
This is only a problem for classical fields though. Once you impose the canonical commutation/anticommutation relations you are replacing classical fields (i.e. functions of spacetime) with operators on a Hilbert space. When you are working with classical fields, before replacing them with operators, and you need classical fields to be anticommutative (for example because they are fermion fields) you need to use Grassmann numbers. This is mostly used when working with path integrals.
I hope I was clear enough.
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Spin of cobalt-60 At many places in beta decay of cobalt-60, the ground state spin of the isotope is given as 5+. However that's not what is predicted by shell model and applying nordheim's strong rule. Is the data given purely experimental in such cases and outside purview of shell-model? Similarly, daughter nucleus Ni has 4+ spin-parity.
Is there a theoretical way to determine the nuclear spins in ground state?
| Shell Model arguments usually work well for nuclei close to magic neutron and proton numbers, but sometimes fail when that is not the case. Co-60 is just not magic enough. Here is a link to an old paper that might help: http://ist-socrates.berkeley.edu/~schwrtz/PhysicsPapers/01PhysRev_94_95.pdf The data in the nuclides chart is experiemntal and there is no theoretical model that explains all the measured gs spins, especially for the most challenging odd-odd isotopes.
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Why the electrons below the Fermi level do not conduct electricity? Physically, why is it that the electrons need to excited above the Fermi level to conduct electricity? In other words, why is the current zero when the electrons lie below the Fermi level? Does Pauli exclusion principle play any role here?
| I think of the TV commercial where the escalator stops and all the people are trapped. The steps are energy levels, and there are two people on each step (spin up and spin down). No third person can occupy a step (Pauli exclusion). Only the ones at the top (Fermi surface) are free to move (gain energy). When they do, the steps are free and more can move up.
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"question_score": "16",
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Binding energy per nucleon error I am given that the mass deficit of 114-Cd is -90.01 MeV, and asked to calculate the binding energy, which should simply be 90.01 MeV then, due to the use of natural units (i.e. where $c=1$). I'm then asked to calculate the binding energy per nucleon which should simply be $\frac{90.01}{114}$. However, according to the solutions manual, the binding energy per nucleon is 8.54 MeV instead. I'm not entirely sure where the discrepancy arises. Any help would be appreciated.
| The mass deficit is related to, but different from, the binding energy. The bare neutron and bare proton have zero binding energy, while the canonical source for mass deficits defines $\Delta[{}^{12}\rm C] = 0$.
Note also that the NNDC mass deficits are for neutral atoms, and therefore include electron masses.
| {
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Will a contiguous, low-Re, low-Ca, liquid body always become a sphere at zero gravity? Let's assume zero gravity, zero initial speed everywhere, $Re \ll 1$ and $Ca \ll 1$
Will such a liquid body always become a sphere or will it sometimes split?
Formally speaking, I'm talking about
$$
\lim_{viscosity -> \infty} \lim_{t -> \infty} ShapeAtTime(t)
$$
(Sufficiently high viscosity will also limit $Ca$, even though it is not directly in the expression)
I think it helps to think about this kind of experiment, but with an hourglass-like shape: Will its neck widen or expand at zero gravity?
| If the object is infinitely viscous (and non-volatile, so that it can't evaporate), it will never change shape.
| {
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Why doesn't Young's modulus change with length and diameter? In this question:
The Young modulus of steel is determined using a length of steel wire and is found to have the value $E$.
Another experiment is carried out using a wire of the same steel, but of half the length and half the diameter.
What value is obtained for the Young modulus in the 2nd experiment?
I know that the Young's modulus is an intrinsic property of a object. But what I found confusing is that, when I calculated the Young's modulus for the 2nd experiment, I got $2E$. But the answer was $E$, instead of $2E$.
However, my thought kept lying with the equation:
$$\text{Young's modulus} = \frac{\text{force}\times\text{length}}{\text{extension}\times\text{area}}$$
Doesn't the change in length and diameter affect the Young's modulus value? How can it be an intrinsic value for a object?
| This question was asked about 3 years ago and by now, you might have passed the A Levels too so this answer might seem irrelevant to you now. But as it turns out, I was having the same problem with the exact question and was seeking the same answer as you did.
I finally found it and wanted to share with you and I hope this will help anyone else who is facing the same problem.
This is just a very stupid question that only has one explanation. It is a fact. The Young Modulus of Steel is between 190 to 215 GPa according to Google. No matter how you change the length and diameter and force, it will always stay constant. The reason is because if you increase force or diameter, the value of stress will increase and so will the expansion and the strain. No matter what variable you change (except for the material), the ratio will always stay the same. To prove this you will need experimental values and mathematics beyond the A Level syllabus.
| {
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How does $I = \mathrm{d}q/\mathrm{d}t$ work for a capacitor? When the capacitor is charging in a circuit consisting of a resistor, a capacitor and an alternating sinusoidal generator at $t=0$, the charge across the capacitor is 0 and the current is $I =\mathrm{d}q/\mathrm{d}t$. Does this make the current zero too? While it is max across the resistor in the same circuit and they are connected in series which means that the current should be the same in all the components of the circuit.
| For a capacitor $Q=CV \Rightarrow \dfrac{dQ}{dt} = I = C \dfrac{dV}{dt}$.
When the rate of change of voltage $\dfrac{dV}{dt}$ is a maximum, ie $V=0$ the current $I$ is a maximum.
When the rate of change of voltage $\dfrac{dV}{dt}$ is zero, ie $V= \pm V_{\rm max}$ the current $I$ is zero.
There is a $90^\circ$ phase difference between the current and the voltage with the voltage leading the current.
| {
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What happens to the temperature of the container during phase change of the substance contained inside I know that as the phase of a substance changes the temperature of the substance remains constant. Now, I was wondering what would happen to the temperature of the container (assuming during the phase change of the substance contained in the container, the container itself doesn't undergo phase change) when the substance contained in the container was undergoing phase change.
I searched the web for any clue but could not find anything regarding the temperature of the container.
Would the temperature of the container go up or remain constant if heat is being supplied at a constant rate to the system(system containing the container and the substance undergoing phase change).
I think the temperature of the container depends upon the conductivity of the substance undergoing phase change. If the conductivity of the substance undergoing phase change is more than the container then the temperature of the container would remain same and if the conductivity of the substance is less than that of the container then the temperature of the container rises.
I am not sure if my line of reasoning is correct, hence would like to have some conformation regarding the same and do support your statement by a sound reasoning.
| Elementary thermodynamics tells us that during phase change a system can absorb or release heat without changing its temperature. You have to give to (or take from) the system a certain quantity of energy to have it complete phase transition: this energy must come from (or go to) somewhere.
Think about a block of ice at temperature 0 Celsius, immersed in water. To melt this quantity of ice, water must give heat to the block until fusion is complete. In this heat exchange, water's temperature decreases.
That's the same case as yours, when you consider the water as the container. If you provide heat to the system, the situation can change wildly, but the basic rule is to use the first principle and think in terms of energy exchanges.
| {
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Covariant Riemann tensor indices Trying to follow a calculation through a paper where, I think, something strange is happening with the indices in the product terms:
How does $\Gamma_{isl}\Gamma^s_{jk}$ in the second line become $-\Gamma_{jks}g^{st}\Gamma_{ilt}$ in the third line? This amounts to setting $\Gamma_{isl} = -\Gamma_{ils}$, given the definition. But the first two indices are the symmetric ones and the author is swapping one with the last index (as in, $[is,l]=-[il,s]$). This shouldn't be anti-symmetric in general.
| Got it. Had to look at the derivative terms. Use:
$$g_{ij,k}=\Gamma_{kij}+\Gamma_{kji}$$
and
\begin{align}
0&=(\delta_i^{\,j})_{,k}\\
&=(g_{is}g^{sj})_{,k}\\
&=g_{is,k}g^{sj}+g_{is}g^{sj}_{\,\,\,,k}\\
\therefore\,\,&g_{is,k}g^{sj}=-g_{is}g^{sj}_{\,\,\,,k}
\end{align}
Then:
\begin{align}
R_{ijkl}
&=g_{sl}(\partial_i\Gamma^{\,\,\,s}_{jk}-\partial_j\Gamma^{\,\,\,s}_{ik})+\Gamma_{isl}\Gamma^{\,\,\,s}_{jk}-\Gamma_{jsl}\Gamma^{\,\,\,s}_{ik}\\
&=g_{sl}[\partial_i(g^{st}\Gamma_{jkt})-\partial_j(g^{st}\Gamma_{ikt})]+\Gamma_{isl}\Gamma^{\,\,\,s}_{jk}-\Gamma_{jsl}\Gamma^{\,\,\,s}_{ik}\\
&=g_{sl}[(g^{st}_{\,\,\,,i}\Gamma_{jkt}+g^{st}\Gamma_{jkt,i})-(g^{st}_{\,\,\,,j}\Gamma_{ikt}+g^{st}\Gamma_{ikt,j})]+\Gamma_{isl}\Gamma^{\,\,\,s}_{jk}-\Gamma_{jsl}\Gamma^{\,\,\,s}_{ik}\\
&=[(-g_{sl,i}g^{st}\Gamma_{jkt}+\delta_l^{\,t}\Gamma_{jkt,i})-(-g_{sl,j}g^{st}\Gamma_{ikt}+\delta_l^{\,t}\Gamma_{ikt,j})]+g^{st}(\Gamma_{isl}\Gamma_{jkt}-\Gamma_{jsl}\Gamma_{ikt})\\
&=\Gamma_{jkl,i}-\Gamma_{ikl,j}+g^{st}(g_{sl,j}\Gamma_{ikt}-g_{sl,i}\Gamma_{jkt}+\Gamma_{isl}\Gamma_{jkt}-\Gamma_{jsl}\Gamma_{ikt})\\
&=\Gamma_{jkl,i}-\Gamma_{ikl,j}+g^{st}[\Gamma_{ikt}(g_{sl,j}-\Gamma_{jsl})-\Gamma_{jkt}(g_{sl,i}-\Gamma_{isl})]\\
&=\Gamma_{jkl,i}-\Gamma_{ikl,j}+g^{st}(\Gamma_{ikt}\Gamma_{jls}-\Gamma_{jkt}\Gamma_{ils})\\
&=\Gamma_{jkl,i}-\Gamma_{ikl,j}+g^{st}(\Gamma_{iks}\Gamma_{jlt}-\Gamma_{jks}\Gamma_{ilt})
\end{align}
| {
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Why buckets have rims curled outwards? My brother told me that household buckets have rims curled outwards to increase their polar moment of inertia.
But for what? This increased moment of inertia is needed to counteract what?
|
Why have buckets with curved rims?
The polar moment of inertia is used to gauge the capacity of an object to resist twisting, or torsion, (in circular cross sectioned objects). In other words it is a measure of the angular displacement of the bucket when undergoing torque.
So if we filled the bucket with water, and then used the handle to lift it vertically, any slight horizontal motion would cause the bucket walls to twist, and reinforcement of the top by means of a curved rim would possibly reduce this twist, if the handle pivot points were located there on appropriately reinforced areas.
Extract and image from Polar moment of Interia
A schematic showing how the polar moment of area is calculated for an arbitrary shape about an axis $o$.
$ρ$ is the radial distance to the element $dA$.
Definition of polar moment of inertia,
$${\displaystyle I_{z}=\int _{A}\rho ^{2}\,dA}$$
$I_z$ = the polar moment of area about the axis $z$
$dA$ = an elemental area
$ρ$ = the radial distance to the element $dA$ from the axis $z$
For a circular section with radius $r$:
$${\displaystyle I_{z}=\int _{0}^{2\pi }\int _{0}^{r}\rho ^{2}\rho \,d\rho \,d\phi ={\frac {\pi r^{4}}{2}}} $$
Two other advantages , with respect to plastic and other thin walled buckets, should also be mentioned .
The curved rim strengthens the plastic bucket and also increases the mass at the top, raising the center of gravity, making it easier to tip it over, (but only slightly).
The curved rim may reduce drips whilst pouring.
| {
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Reconcile three-point function with OPE in CFT Three-point function of primary operators in conformal field theory can be fixed up to a constant by symmetry considerations
$$\left\langle V_{h_1}(z_1)V_{h_2}(z_2)V_{h_3}(z_3)\right\rangle=\frac{C_{h_1h_2h_3}}{(z_1-z_2)^{h_1+h_2-h_3}(z_1-z_3)^{h_1+h_3-h_2}(z_2-z_3)^{h_2+h_3-h_1}}$$
On the other hand, it should be possible to derive this result using the operator product expansion
$$V_{h_1}(z_1)V_{h_2}(z_2)=\sum_{h,K}C^{h,K}_{h_1h_2}(z_1-z_2)^{h+|K|-h_1-h_2}L_{-K}V_h(z_2)$$
here the sum is performed over all primary dimensions $h$ and their descendants parametrized by multi-index $K=\{k_1,...,k_2\}, |K|=k_1+\dots+k_n$. The OPE expansion is valid inside correlation function given that there are no other fields between points $z_1$ and $z_2$. For the three-point function this implies that we must have $|z_3-z_2|>|z_1-z_2|$. With this assumption we can rewrite the three-point function as
$$\left\langle V_{h_1}(z_1)V_{h_2}(z_2)V_{h_3}(z_3)\right\rangle=\sum_{h,K}C^{h,K}_{h_1h_2}(z_1-z_2)^{h+|K|-h_1-h_2}\left\langle L_{-K}V_h(z_2)V_{h_3}(z_3)\right\rangle$$
Now, I believe that all the two-point functions inside the r.h.s. vanish except for $h=h_3$ and $K=\{\}$, i.e. for a primary field of the weight $h_3$ (this might be the wrong assumption leading to the inconsistency). Then, using the two-point correlator $V_{h}(z_1)V_h(z_2)=C(h)(z_1-z_2)^{-2h}$ one arrives at the following expression
$$\left\langle V_{h_1}(z_1)V_{h_2}(z_2)V_{h_3}(z_3)\right\rangle=(z_1-z_2)^{h_3-h_1-h_2}(z_2-z_3)^{-2h_3} C(h_3)C^{h_3,\{\}}_{h_1h_2}$$
which clearly has a different coordinate dependence. So, how does one derive the correct three-point function using the OPE?
| You should sum over all descendents $K$. In general their contributions do not vanish, although they are subleading in the limit $z_1\to z_2$. For example the two-point function
$$
\langle L_{-1}V_{h_3}(z_2)V_{h_3}(z_3) \rangle = \frac{\partial}{\partial z_2}\langle V_{h_3}(z_2)V_{h_3}(z_3) \rangle
$$
is clearly nonzero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Question about simple harmonic motion
A massive bird lands on a taut light horizontal wire. You observe that after landing the bird oscillates up and down with approximately simple harmonic motion with period $T$.
You also notice that during the initial oscillations the maximum height the bird reaches above its position at landing is equal to the distance below its landing position at which the bird finally comes to rest, when all oscillations have stopped.
In terms of only $T$ and $g$, the acceleration due to gravity, estimate the magnitude of the vertical velocity $v_0$ that the bird had at the instant of landing on the wire.
So, to attack this problem by energy conservation I use
\begin{equation}
\frac{1}{2}mv_0^2=mgA
\end{equation}
Where now $v_0 = \omega A$. So
\begin{equation}
\frac{1}{2}\omega^2 A^2=gA \Rightarrow A=\frac{2g}{\omega^2}=\frac{2g}{\frac{4\pi^2}{T^2}}=\frac{2gT^2}{4\pi^2}
\end{equation}
So
\begin{equation}
v_0 = \omega A = \frac{2\pi}{T} \frac{2gT^2}{4\pi^2} = \frac{gT}{\pi}
\end{equation}
But the answer is
\begin{equation}
v_0 = \frac{\sqrt{3}gT}{2\pi}
\end{equation}
What am I doing wrong?
| The second line is the clincher in the problem. Once you understand its significance, the problem becomes very easy. When the bird comes to rest permanently on the wire, it reaches the equilibrium position which is the mean position of SHM. If this is A units below landing position, and the maximum height reached is A units above landing position, then amplitude is 2A. This means that at the landing position the displacement is half of maximum amplitude and hence, the velocity at this point( which is required in the problem) is √3/2 times the maximum velocity.( This comes from the fact that velocity and amplitude are cosine and sine functions respectively in case of SHM).
Now, you can get the maximum velocity by equating $mω^2A=mg$ since the bird is in equilibrium at A units below the landing position. Remember, the amplitude is 2A. The maximum velocity that you get from this is the same as the answer that you have obtained in your own attempt, $gT/π$. The initial velocity occurs at half amplitude and is √3/2 times the maximum velocity i.e √3/2 $gT/π$.
| {
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Why is a sleeping bag so cold when you first get in? I sometimes put a sweater on when I first get up on a cold morning. The sweater isn't so cold against my skin, but the zipper is. I get this. Even though they are both at the same temperature, the zipper is a better conductor of heat. Heat flows more easily from warm skin into the cold zipper, cooling the skin more effectively.
The reason a zipper is a better thermal conductor is that it is a metal. Electrons in the conduction band conduct heat as easily as electricity. On the other hand, wool is hair - an insulator.
When I get into a sleeping bag, it is cold like a zipper, even though the nylon and feathers are both insulators. Why is this?
I saw Stephan Bishof's answer to Could a sleeping bag be warmer if you are naked inside?. He says evaporation plays a role. I am not sure I buy this. It is cold even against dry skin, when you get static electricity from sliding in.
Is it simply that the sleeping bag is smooth and a sweater is fuzzy? Better contact makes better heat transport?
In summer you can avoid the whole by unpacking your bag when you first get to camp. It fills with warm air, and keeps it warm until night.
| i have a potential solution but i'm not sure if it's accurate
say you get up on your cold morning, and your sleeping bag, as rightly stated is a good insulator, but because its been there all night, it will have slowly gotten cold overnight, but because it is such a good insulator, it will prevent a lot of heat getting inside, which should keep it cold
| {
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Why doesn't increasing resistance increase brightness if $P=I^2\cdot R$ Light bulb brightness increases with power, $P$.
So why doesn't increasing $R$ increase $P$ and hence increase brightness as $P=I^2\cdot R$ due to $P=I\cdot V$ and $V=I\cdot R$?
I read increasing $R$ decreases brightness.
| The two equations of relevance are
${\rm power} (P) = \,{\rm voltage} \,(V) \times {\rm current} \,(I)$
${\rm resistance}\, (R) = \dfrac{{\rm voltage}\,(V)}{{\rm current}\,(I)} $
From those two equations you can get $P = I^2R$ and $P = \dfrac {V^2}{R}$
Suppose that it is assumed that the resistance of the light bulb does not vary with the voltage across it / the current through it.
In your room you have a light stand with a $240 \, \rm V,\;60\, \rm W$ light bulb in it and you want to replace it with a brighter $240 \, \rm V,\;100\, \rm W$ light bulb.
Using $R = \dfrac {V^2}{P}$ he working resistance of the $60\, \rm W$ bulb is $694\, \Omega$ and that of the $100\, \rm W$ bulb is $576\, \Omega$.
So decreasing the resistance increases the brightness.
The problem with using $P=I^2R$ is that you might get the impression that because the resistance $R$ goes down the power also decreases but in doing that you have assumed that the current $I$ stays constant.
The current is not constant but actually increases by the same fractional amount as the resistance decreases.
But that is not all because in the equation $P=I^2R$ the current is squared and so the fractional increase in the current squared $I^2$ is double the fractional decrease in the resistance $R$.
So overall the power dissipated increase as the current decreases which leads to increased brightness.
| {
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"timestamp": "2023-03-29T00:00:00",
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Invasive blood pressure measuring and hydrostatics I don't understand why the solution-manometric liquid must be at the catheter level.
I'm studying biophysics, in particular fluid mechanics.
A link to the image
Here is another image
Why A' has to be in the same level of A?
| I'm unable to see your picture but believe I know what you are asking. An arterial line is placed to directly measure arterial blood pressure by accessing an artery with a catheter and measuring the pressure in a continuous manner with a pressure transducer. Access to the artery requires that the catheter remain patent, that is that the line does not clot, so some means is required for low flow rate, continuous flushing of saline into the artery. This requires a saline bag with a pressurized cuff that is inflated to roughly match the mean arterial pressure and a metering valve to adjust the infusion rate. The pressure transducer is 'T'eed off this line thus measuring the mean pressure and the dynamic excursions.
Without a cuff, the saline bag would have to be raised to a much higher level than the heart to at least balance the pressure that the heart maintains (about 65 - 110 mm Hg or about 50 to 60 inches of water pressure) and additionally higher to maintain the flush rate.
So the manometric fluid must be kept to at least the catheter (pressure) level to prevent arterial blood from flowing out of the patient, and marginally higher to maintain a flush rate to keep the catheter patent. Additionally heparin is often used in the saline to prevent clotting.
| {
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Conservative force and its Potential Energy Function We are given that $\vec{F}=k\left<y,x,0\right>$, and asked whether $\vec{F}$ is a conservative force. If yes, we are asked to find $U(x,y,z)$ and then find $\vec{F}$ back from $U$ and show it matches the original form.
Given $\vec{\nabla} \times\vec{F}=\vec{0}$, force is conservative.
Therefore, $U(x,y,z)=-\int_{r_0}^r \vec{F} \cdot d\vec{r}=-\int_0^x kydx-\int_0^y kxdy=-2xyk.$ (Note that the $z$ component is $0$).
Such a potential function yields $\vec{F}=-\vec{\nabla}U=2k\left<y,x,0\right>.$
Something must be wrong or I must be missing something because I get an extra factor of 2 and I do not understand why.
| Well you integrated it the wrong way.
$$\int ky \, dx + \int kx \, dy = \int k \,d(xy) $$
X and y are not independent.
| {
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What is the significance of the phase constant in the Simple Harmonic Motion equation? The displacement of a particle performing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. What is the significance of $\phi$. How is it used? Please explain the meaning of the phase constant
| Based on a point raised by @docscience this answer addresses the phase in terms of "initial conditions" introduced by driving forces. In fact one can think of this as answering how the SHO was set in motion in the first place.
The position of a simple harmonic oscillator at time $t$ that experienced force at time $t'$ and that was at rest in the far past
$$
\lim_{t\to -\infty} x(t)=0 \\
\lim_{t \to -\infty} \dot x(t)=0
$$
is given by
$$
x(t) = \int_{-\infty}^t \frac{1}{\omega} \sin (\omega (t-t')) f(t')
$$
This has been obtained by using retarded Green's function for the SHO details of which can be found elsewhere but one can check that this satisfies the SHO equation of motion.
(1) For the simplest case lets take the case of a pulse of force at time $t'=t_0$ then we get
$$
x(t) = \frac{1}{\omega} \sin(\omega( t- t_0)) \Theta(t-t_0)
$$
where $\Theta(t-t_0)$ is the Heavyside step function. Thus we see that the oscillator is at rest for $t<t_0$ and after that the 'phase' is $-\omega t_0$.
(2) Now lets take the case of two pulses at times $t_0$ and $t_1$ with amplitude $f_0$ and $f_1$ i.e.
$$
f(t)=f_0 \delta(t-t_0) + f_1 \delta(t-t_1)
$$
with $t_1>t_0$. Its easy to see the solution is
$$
x(t)=\frac{f_0}{\omega} \sin(\omega( t- t_0)) \Theta(t-t_0) + \frac{f_1}{\omega} \sin(\omega( t- t_1)) \Theta(t-t_1)
$$
Here is where we see the meaning of phase clearly: If we take $f_1 = f_2$ then we see it is possible to choose $t_0$ and $t_1$ such that the two pulses are "in-phase" and the amplitude doubles or "out-of-phase" such that the amplitude cancels and the second pulse just stops the SHO. These correspond to $\omega(t_1- t_0)=2 n \pi$ and $\omega(t_1-t_0)=n\pi$ for $n$ an odd integer.
| {
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What is the infinity that strikes quantum field theory? I'm confused, on a book I recently read, it talked about renormalization, the creation of renormalization is because of the infinity problem.
Here's is the problem, what is the infinity problem that needs to be eliminated in quantum mechanics? What is the cause?
| It is all about the perturbative expansion for the interaction amplitudes. These are expressed in Feynman diagrams of increasing order.
in calculating loop Feynman diagrams in QED one finds integrals which diverge Such apparent pathologies are dealt with by the process of renormalization. This procedure can be viewed in several ways. From one perspective it is a formal manipulation, part of the definition of the quantum field theory, which allows one to calculate finite, testable expectation values and scattering amplitudes. From a more physical perspective, one starts by noting that the key divergences come from the high energy limits of the momentum integration.
So concretely it is going to large momenta that loop feynman diagrams diverge. The paper describes the procedure of avoiding the problem, which is the whole project of the renormalization procedure in this case.
| {
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Acceleration of car. One dimensional motion easy problem A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
My attempt at solving the problem:
$$a(x) = \frac{v - u}{t}$$
where
$v =$ final velocity
$u =$ initial velocity
$$$$
I get the answer as $4.05 \space ms^{-2}$
But the correct answer given to the problem is $8.10 \space ms^{-2}$.
They used a different equation to reach that answer.
Did I use the wrong equation? I have the average velocity and not the instantaneous veolcity?
| Yes, I think so. Below is the proper formula for the distance an object accelerating at a constant rate goes over time. I used it to get a formula where you enter distance and time traveled to get the acceleration.
m=0.5at^2
2m/t^2=a
(2*110)/5.21^2=8.10489203915 m/s^2
Hope this helps.
| {
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What causes change in planet's angular velocity? A satellite moving in an elliptical orbit will increase in angular velocity as it nears a planet. I understand that this is consistent with angular momentum. But what causes the increase in angular velocity if there is no torque acting on the satellite?
| Consider an object moving straight past a point that doesn't gravitate at all.
In uniform linear motion, the angular velocity with respect to this point starts out very small, then grows to a finite value, and then decreases towards 0 again -- all without any force at all acting on the moving object.
Change in angular velocity doesn't need a cause other than the relative positions of the involved objects changing.
| {
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An appropriate way to store neodymium magnets Okay so I've bought a few small neodymium magnets to play around with, they're very powerful and I really like them, but I was wondering what's the actual best way of storing those magnets in a way that doesn't affect their magnetic fields or degrades them in any way.
I'm currently storing them stuck to one another, is it a good practice? Thanks a lot!
| I just purchased a large-ish neodymium magnet. It is 2.75 inches in diameter, disk-shaped. It also came with a thin disk of steel sheet metal the same diameter that is called a "striker plate." It can be used in non-metal applications to get the magnet to attach. (e.g. if you want to use the magnet to hold a wooden cabinet door shut, you can attach the magnet to the frame and the striker to the door, which will cause the wooden door to "stick" to the magnet.) The striker was separated from the magnet by a half-inch piece of styrofoam.
It didn't come with much in the way of instructions, however, the box said, "Do not allow the striker to stick to the magnet or magnetic strength will be severely reduced." So, apparently you should not store the magnet near other metals. I'm trying to figure out if this rule also applies to other magnets (which I doubt because I see so many people store their magnets together) or if it is necessary to align the poles in a certain manner if magnets are stored together?
| {
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The equilibrium concentration of vacancies In the derivation of the equilibrium concentration of vacancies by statistical mechanics method, I was stumped by this procedure (marked by "?").
$\textbf{Physical Model}:$
1.Solid viewed as a collection of $N$ atomic sites;
2.Each site may or may not be occupied, and assume now that $N_o$ sites are occupied and $N_v$ sites are vacant;
3.If a site is not occupied then the system has an additional energy, namely the formal energy $E_v$;
$\textbf{Solution}:$
1.multiplicity function
$$\Omega = C_N^{N_o} = \dfrac{N!}{N_o!N_v!}$$
2.entropy:
$$S=k_B\ln \Omega = -N k_B (c\ln c+(1-c)\ln(1-c)) \qquad (c=\dfrac{N_v}{N} \quad ;\quad (1-c) = \dfrac{N_o}{N})$$
3.the internal energy ($\textbf{?}$)
$$U = N c E_v$$
(Why we don't consider the internal energy of the whole system rather than the vacancies ? )
4.The Helmholtz free energy
$$F = U-TS = N(cE_v + k_B T (c \ln c + (1-c) \ln (1-c)))$$
and taking $c \ll 1 $
$$\dfrac{F}{N} = c E_v + k_B T c \ln c$$
5.equilibrium concentration (by minimizing the Helmholtz free energy.)
$$ c \rightarrow e^{-\dfrac{E_v}{k_B T}}$$
So what's the missing points to understand the marked procedure above?
| Calculate the ratio of the number of vacancies in equilibrium at 27 C in aluminum to that produced at 527 C. The energy of the formation of vacancies in aluminum is 68 kJ/mol
| {
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Magnetic behaviour of magnet A very direct question regarding behaviour of a magnet-
When a magnet is heated such that it melts, why does it lose its magnetism?
This is what I was wondering -
When a solid magnet is heated, the heat is able to provide potential energy to the constituent atoms and thus the magnet turns into molten state. So there is comparitively more inter particle space as compared to the earlier solid state. Moreover, the atoms are now comparitively even more mobile so the alignment of the atoms inside the magnet has been altered as the atoms are not in a fixed position. However some amount of particle movement is also present in solid state.
Please provide a detailed explanation to my intuition if it's right. Otherwise provide me with a suitable explanation for my question.
| Ferromagnetism is a quantum mechanical effect that relies on the ability of atoms to achieve "long range ordering". This ordering results in a coupling of the angular momentum of electrons in neighboring atoms, which leads to magnetic domains. When enough domains align, you get net magnetism.
Heating a material above the Curie temperature destroys this long range ordering - and this can happen below the melting point.
It is possible for a liquid or even a gas to become ferromagnetic - when the conditions are right for this long-range ordering to take place. But most of the time, when the material melts you will lose the order.
Also, for magnetic materials that derive their properties from a particular lattice structure (different types of atoms in a specific arrangement), melting the material will destroy that arrangement.
| {
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What is a "thermal" particle? I have seen various uses of the word "thermal" recently (e.g. in black hole physics) that make me wonder what is the exact meaning of "thermal" in particle physics. There are also "thermal neutrons", for example, so my initial understanding of "thermal" as related "thermal radiation", i.e. propagated by photons, seems inadequate.
Can I apply "thermal" to any particle which meets the following criterion?
A thermal particle is a free particle with a kinetic energy corresponding to the most probable velocity at a temperature of 290 K (17 °C or 62 °F), i.e. the mode of the Maxwell–Boltzmann distribution for this temperature.
Or should the criterion be:
A thermal particle is a free particle with a kinetic energy corresponding to the most probable velocity at the temperature of its environment, i.e. the mode of the Maxwell–Boltzmann distribution for this temperature.
| Thermal means a type radiation (i.e photons) which is given out by a body to reduce the temperature of itself. It is generally in microwave range.
| {
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Is it possible for light to fall into a geo-synchronous orbit around a sufficiently massive planet? Is it possible for (visible) light to fall into a geo-synchronous orbit around a sufficiently massive planet?
Can a planet or heavenly body be massive enough for it to happen?
What will that suspended light look like to an observer on that planet?
What role will the mass of light play?
| *
*No, light cannot fall into Geo-synchronous orbit. Because light is emitted from on point and travels in straight line and terminates due to scattering or absorption. Considering the massive Celestial objects where it is seen that light bends are near the black holes.
*It is impossible for any planet to be as massive as a black hole. Considering Black hole by itself, it is the one which gulps the light. There are no proofs of any planets such massive other than stars.
*Assume if the light is being suspended, you cannot see light though u see the particles that are being illuminated by the light.
*By the Energy, mass and velocity equations say mass of light at its speed is infinity theoretically, it stays same as long as the light is terminated.
| {
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Do physicists use particle "energy" to refer to kinetic energy? In 1963, this paper was written about the effects of radiation on solar panels. The paper states that:
When electrons at energies greater than 145 KeV and protons at energies
greater than 98eV bombard a silicon crystal, they can displace an atom from the
crystal lattice, producing a lattice vacancy and a recoil atom which comes to rest
as an interstitial atom.
However, the resting energies of electrons and protons are far greater than this, at roughly 511 KeV and 938 MeV respectively. I concluded that the paper was referring to kinetic energy rather than total energy, and adjusted my calculations based on this conjecture.
So: Was I correct to assume that the paper referred to kinetic energy, or was it instead some other measure of the particles' energy?
More generally, is there a standard meaning for a particle's "energy" when referring to such particles moving at relativistic speeds?
| Yes, in this case what's meant by "energy" is the kinetic energy $K = (\gamma-1)mc^2$ that the incident particle can transfer to the target system. As you point out, it wouldn't make sense to talk about an electron, which has rest energy $E_0=mc^2=511\rm\,keV$, to have a total energy $E=\gamma mc^2 = mc^2 + K$ of only 100 eV.
For ultra-relativistic particles with $E\gg E_0$ it's a reasonable approximation to think of the kinetic energy and the total energy as being identical, and for nonrelativistic systems the meaning is unambiguous from context, so it's only in a fairly narrow energy region around $\gamma\approx2$, or $K\approx E_0$, that you have to be careful to distinguish between kinetic and total energy. That makes us sloppy. Sorry.
| {
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Translation of Vectors I am a bit confused about translation of vectors. In the first class in physics itself we are told that we can translate vectors as we like to the desired position to do whatever that we are trying to do. For example, if someone draws two random vectors then to get the sum, we translate them, make a parallelogram and draw the diagonal as the resultant.
However I have some doubts on this. In the following example, clearly we cannot translate the vectors.
Consider this rigid body. We want to calculate the torque about origin of a force.
Now if we translate the force vector, then we would obtain the following.
Obviously the situation are very different and its not equivalent.
So are we really allowed to translate vectors?
| The definition of a vector as having magnitude and direction is typically used in physics, so the exact location of the vector is not included in the definition. This is also called a "free" vector. As you say the effect of the vector depends on its location. For example, a force (vector) cannot act upon a point mass unless is acts "at" the location of the mass.
Physicists typically regard a vector as a free vector, recognizing the effect of the vector depends on its location. [Davis, Introduction to Vector Analysis] [Symon, Mechanics]
Engineering texts (and some older physics texts) sometimes distinguish among free, bound, and sliding vectors.
A good discussion of a free vector and its use in transforming between coordinate systems (for example, from an inertial system to a rotating and translating system) can be found in the textbook Mechanics by Symon.
| {
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Is the concept of in-compressible fluid valid in special theory of relativity? Is the concept of incompressible fluid valid in special theory of relativity?
Can anybody answer this question without going into speed of sound and fluid dynamics ?
| In an incompressible fluid the density does not change in response to changes in the pressure. This means that the speed of sound is infinite,
$$
c_s^2= c^2\left.\frac{\partial P}{\partial \rho}\right|_s=\infty .
$$
Here, $P$ is the pressure and $\rho$ is is the energy density. In the non-relativistic limit $\rho=mnc^2$, where $m$ is the mass of the particles and $n$ is the particle density.
This is obviously incompatible with relativity, disturbances in the fluid propagate faster than the speed of light.
Of course, non-relativistic fluids are not truly incompressible either, but the approximation is useful if the fluid velocity is much smaller than the speed of sound, $u\ll c_s$. In a relativistic fluid in which $u$ is comparable to $c$ this cannot be true.
Postscript: Note that in a non-relativistic fluid incompressibility means that $n=const$. Then $\rho=mnc^2$ is also constant. In a relativistic fluid we could either mean $\rho=const$, or $n=const$. Note that $\rho=const$ is the more natural generalization, and it is incompatible with $c_s^2<c^2$ as explained above. Constant particle density also not allowed, because $\partial P/\partial n \sim \chi/n$, where $\chi$ is the susceptibility,
| {
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How does one show that the curl of magnetic field is equal to $\mu_0\mathbf J+\mu_0\epsilon_0\frac{\partial \mathbf E}{\partial t}$? Normally, the curl is supposed to be equal to $\mu_0\mathbf J$. However, when checking for invariance for Maxwell's equations under duality transformations, the term $\mu_0\epsilon_0\frac{\partial \mathbf E}{\partial t}$ is introduced. How does it come about?
| I do not see what you mean by "normally, the curl is supposed to be equal to $\mu_0\vec J$". I will assume that you mean: experiments show that $${\rm curl}\ \!\vec B=\mu_0\vec J,$$ i.e. an electric current generates a magnetic field turning around the wire. However, this equation cannot be true: taking the divergence, the l.h.s vanishes $${\rm div}\ \!{\rm curl}\ \!\vec B=0$$ while the r.h.s. is (up to a factor $\mu_0$)
$${\rm div}\ \!\vec J+{\partial\rho\over\partial t}=0$$
(charge continuity equation, i.e. conservation of the total charge). Since the Maxwell-Gauss equation reads
$${\rm div}\ \!\vec E={\rho\over\varepsilon_0}$$
it is easy to see how to modify our original equation:
$${\rm curl}\ \!\vec B=\mu_0\vec J+\mu_0\varepsilon_0
{\partial\vec E\over\partial t}$$
The divergence of the two sides of the equation now vanishes. Physically, you should imagine that when electric wires are finite, there is an accumulation of charge at the two ending points of these wires. These varying charges also contribute to the magnetic charge.
| {
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Can you build a compass that is attracted to the South Pole? Was just curious, since all compasses point to the North Pole.
South is just the opposite polarity of of North, so it seems very likely, but I've never seen an example of this. Is there any videos demonstrating this?
Could a South attractor be added to a standard compass to help confirm the integrity of the North's signal? (For situations where the compass is being affected by another magnetic source).
| Another approach is to remagnetize the needle in the opposite direction using a strong magnetic field. Then the painted end will point south.
| {
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Breit-Wigner Peak - Not quite a Lorentzian? The non-relativistic Breit-Wigner peak is given by:
$$\sigma_{fi}=\frac{\pi\hbar^2}{q^2} \frac{2j+1}{(2S_1+1)(2S_2+1)} \frac{\Gamma_i \Gamma_f}{(E-Mc^2)^2+\Gamma^2/4}$$
where $q$ is the center of mass momentum of one of the particles and $E$ is the center of mass energy. From what I am aware $q$ is a function of $E$ (changing the center of mass energy will change the center of mass momentum) typically $q\propto E$ so we would get:
$$\sigma_{fi}\propto \frac{1}{E^2} \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$
however in every source I can find (e.g. Martin, 2012 pg 27) it is stated that
$$\sigma_{fi}\propto \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$
and is usually compared to been the same as a Lorentzian. My question is therefore what happens to the $E$ dependence of $q$ and why is it not considered?
| You are right that the relativistic Breit-Wigner distribution isn't functionally quite the same as the Lorenzian distribution. However, the approximation that you described and which compares the relativistic distribution to the non-relativistic, Lorentzian one may be extremely accurate.
Well, it is extremely accurate whenever $Mc^2\gg \Gamma$. Note that the distribution is almost entirely concentrated in the interval
$$E=Mc^2 \pm {\rm few}\cdot \Gamma$$
because of the denominator that you kept in your last formula. But when $E$ belongs to this interval, $q$ may be well approximated by substituting $E=Mc^2$.
In other words, the relativistic Breit-Wigner distribution only "changes abruptly" in the same interval where most of the distribution is concentrated and when one accounts for all the abruptly changing factors in this region, the distribution becomes indistinguishable from the non-relativistic, Lorentzian one.
| {
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Can relative velocity be found? Can you find the relative velocity between two cars with constant velocity of U1 and U2=-U1. In other words is there any kind of experiment you can do to understand that you are moving also and not just seeing the other car with 2U1?
| "Can relative velocity be found?"
Yes. More on that in a moment.
"is there any kind of experiment you can do to understand that you are moving also"
Theoretically, no. However, in actual practice, an accelerometer will give you a pretty good idea of whether or not you are "moving" in the sense that you're thinking. Ie, starting with both cars at rest, if your accelerometer never budges off zero, then the other car is moving at 2U. If you read some acceleration for a while, then you can use those readings to calculate your new velocity relative to your original (assumed to be "stationary"), and then compare that to the other car.
"and not just seeing the other car with 2U1"
So getting back to where I started with this, the relative velocity is 2U1. Consider the following three scenarios:
your car velocity | their car velocity
0 | 2U1
2U1 | 0
U1 | U2 = -U1
In all three scenarios, the relative velocity is 2U1. It's simply the difference between the two velocities.
| {
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Does a purely tangential force create centripetal force? Suppose there's a circular disc pivoted at is center and a purely tangential force is applied to change the angular speed of the disc. So as the disc's rotation speed changes constantly, the particles get accelerated towards the center too. Has the tangential force created centripetal force or something ?
| The tangential force somewhat creates the centripetal force.
The centripetal force is a reaction of the object or system to the tangential force. In the case of a disk, when you apply the tangential force somewhere, if the disk is being held so it can only rotate, then the tangential motion won't be able to accelerate the disk in the direction of force.
Instead, the internal forces holding the disk together create an acceleration inwards so that the place where you apply the force doesn't just break off. Since it can only rotate the tangential force induces the centripetal force.
It's easy to picture with a ball on a string spinning around. The ball wants to move forward but because of the length of the string being fixed, the tension of the string makes a centripetal acceleration.
| {
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Experimentally measure velocity/momentum of a particle in quantum mechanics In the context of quantum mechanics one cannot measure the velocity of a particle by measuring its position at two quick instants of time and dividing by the time interval. That is,
$$ v = \frac{x_2 - x_1}{t_2 - t_1} $$ does not hold as just after the first measurement the wavefunction of the particle "collapses".
So, experimentally how exactly do we measure the veolcity (or say momentum) of a particle?
One way that occurs to me is to measure the particle's de Broglie wavelength $\lambda$ and use $$p = \frac{h}{\lambda}$$ and $$v = \frac{p}{m}$$
to determine the particle's velocity. Is this the way it is done? Is there any other way?
| In condensed matter physics community, one can use the ARPES apparatus. ARPES gives information on the direction, speed and scattering process of valence electrons in the sample being studied (usually a solid). This means that information can be gained on both the energy and momentum of an electron, resulting in detailed information on band dispersion and Fermi surface.
| {
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Interaction terms in the Standard model which cause Higgs decay to electroweak bosons The coupling of the Higgs boson to the electroweak gauge bosons in the Standard model is given by
$$\mathcal{L}_{\text{H-g}} = - \left( \frac{H}{v} + \frac{H^{2}}{2v^{2}} \right) \left(2M_{W}^{2}W_{\mu}^{+}W^{-\mu} + M_{Z}^{2}Z_{\mu}Z^{\mu} \right).$$
However, in Cliff Burgess' textbook 'The Standard Model: A Primer,' the author suppresses the contribution $H^{2}/2v^{2}$ in equation (4.55) on page 146 when he discusses the decay of the Higgs boson to electroweak gauge bosons.
Is there a reason why the contribution $H^{2}/2v^{2}$ is suppressed?
| The reason is very simple -- term of the form $H A_\mu A^\mu$ corresponds to the vertex with one Higgs and two bosons -- directly the Feynman diagram giving the decay. The term $H^2 A_\mu A^\mu$ is the vertex with two Higgs bosons and two gauge bosons. At tree level this is some scattering process (say, $2H\to 2Z$), but not decay. It contributes to decay only at higher order loop diagrams, thus it is significantly suppressed.
$HZ_\mu Z^\mu$: ,
$H^2 Z_\mu Z^\mu$:
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Orbitals and electron jumping Bohr model (tries to) explain how electron goes to higher state(orbit) from lower state(orbit) on photon absorption.But in quantum mechanics, we have orbitals which can at most accommodate 2electrons.
Let's consider an electron that has absorbed a photon sufficient to make it jump from Orbital 1 to Orbital 2.If the Orbital 2 is already filled with 2 electrons where will the excited electron go?
| If the excited electrons can't be stimulated, then the photon simply won't interact with the lower-level electron as the electron's transition is forbidden.
| {
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Calculate which composite will give the lightest beam for a given force and deflection. A uniform rectangular-section beam of fixed width, W, unspecified depth, d, and fixed length,L, rests horizontally on two simple supports at either end of the beam. A concentrated force, F,acts vertically downwards through the centre of the beam. The deflection, $\delta$, of the loaded
point is:
$$\delta = \frac{^{3}}{4_{}^{3}}$$
Where $E_{c}$ is the modulus of the composite material used to produce the beam.
Ignoring the deflection due to self-weight, calculate which of the three composites in Table 1 will give the lightest beam for a given force and deflection.
Table 1
\begin{array}{l|c|r}
Material & Density / Mg m^{-3} & E_{c} / GPa \\
CFRP & 1.53 & 197 \\
GFRP & 1.85 & 37.5 \\
Steel/concrete & 2.51 & 48.1 \\
\end{array}
| The mass of the beam, $M$, is equal to $$M = \rho L W d,$$ where $L$ and $W$ are the same for every beam. For a given force $F$, we can calculate the minimal width of the beam $d$ so that the deflection would not exceed $\delta$: $$d = L\left(\frac{F}{4E_cW\delta}\right)^\frac13.$$
The mass of the beam is then $$M = \rho L^2 W\left(\frac{F}{4E_cW\delta}\right)^\frac13 \propto \rho E_c^{-\frac{1}{3}}$$
In arbitrary units, the $\rho E_c^{-\frac{1}{3}}$ factor is 0.263 for CFRP, 0.553 for GFRP and 0.690 for steel/concrete.
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Current as the time derivative of the charge I have been told that the current $i$ can be defined as
$ i = \displaystyle\frac{dq}{dt} $,
where $q$ is the charge and $t$ is the time.
I do not understand this definition because, if the charges are moving so that the net charge remains constant in an infinitesimally thin cross-section of a wire, $q$ is constant with time and hence $dq/dt = 0$. That result would mean that no constant current can exist unless the charge change has a linear dependence with time (i.e. $q = q(t) \propto t$).
As I assume my reasoning is wrong, where is my mistake?
Thank you.
| Current is not related to the charge "in an infinitesimally thin cross-section of a wire". Electrical current is the amount of charge that passes across that cross-section per unit of time.
If the current is constant, the last expression you show works:
$ I = \displaystyle\frac{q}{t} $,
If the current varies, your first expression is appropriate.
| {
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Heat shields and the atmosphere Space capsules and the Shuttles were equipped with heat shields to protect the spacecraft and those on board from the tremendous heat generated from friction as they reentered earth's atmosphere at thousands of miles per hour. Why weren't heat shields installed on the vehicles to protect them as they rocketed through the atmosphere at thousands of mph heading into space?
| The answer is simple. During launch, much of the acceleration occurs above the bulk of the atmosphere. During entry, almost all of the acceleration (deceleration) occurs within the atmosphere. Launching vehicles carefully balance the need for horizontal velocity versus losses due to drag. Reentering vehicles rely very much on atmospheric drag to reduce the vehicle's speed.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can the free charges of a conductor redistribute to the surface? I have studied that if I put a conductor (perfect or non-perfect) at rest in a place where there is an electrostatic field, the charges of the conductor will distribute so that negative charges will be in the surface near the source of the electric field, leaving positive charges on the other side.
I cannot understand this. If the conductor is idle, e.g. not connected to a battery, I assume there are no 'extra' electrons within its lattice. In that case, the number of free charges (electrons) will be such that the net charge of the metal is zero and if an electron moves from its nuclei to another, another electron will come to replace it, avoiding the creation of 'holes' (positive free charges). Unless there exist natural 'holes' in every metal, I cannot see how an accumulation of negative charge can appear on one side of the conductor generating a positive accumulation on the other side. How is it possible? Should not be 'extra' electrons (from a battery, from a capacitor...) be needed?
EDIT: Even in the case of 'extra' electrons, I cannot see how positive charge could appear on the other side. I understand that, in that situation, a negative accumulation of charge would be in the surface of the metal near the source of the field, but how could the positive zone of charge take place? Where is the space to allocate those 'holes'?
| Think of it a different way.
The electrons from only the atoms on the surface of the metal redistribute themselves.
Part of the surface layer becomes positive net, and part of the surface layer becomes negative net.
The net charge of the entire surface remains zero.
| {
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Can an object falling in vacuum generate electricity by itself? When an object falls through vacuum, gravitational potential energy is converted to kinetic energy. Is there some way to get electrical energy out of the equation by itself (i.e. somehow convert the gravitational potential energy to electrical energy)? Is this physically possible? If so, what properties must this object have?
By by itself, I mean without using any external (possibly stationary) "reference object" (e.g. a stationary coil), so a magnet falling through a coil does not count, i.e. the electricity is generated solely by the object that is falling. Note that the object itself can be arbitrarily complex internally, just that whatever mechanism it has inside must also be falling along with the object.
| If by 'electrical energy' you mean 'an electric current', then no, that won't work. The stationary coil you are excluding is there to provide the supply of electrons to be moved by the magnet - no electrons, no current.
And no, nothing can be generated within the falling object. Current is generated by the relative motion of a magnet and an electron source; an object in free-fall is never in motion relative to itself.
| {
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Can Zener Breakdown be converted to Avalanche breakdown? Wikipedia says:
The Zener effect is distinct from avalanche breakdown. Avalanche breakdown involves minority carrier electrons in the transition region being accelerated, by the electric field, to energies sufficient for freeing electron-hole pairs via collisions with bound electrons. The Zener and the avalanche effect may occur simultaneously or independently of one another. In general, diode junction breakdowns occurring below 5 volts are caused by the Zener effect, whereas breakdowns occurring above 5 volts are caused by the avalanche effect. Breakdowns occurring at voltages close to 5V are usually caused by some combination of the two effects. Zener breakdown voltage is found to occur at electric field intensity of about 3×107 V/m.[1] Zener breakdown occurs in heavily doped junctions (p-type semiconductor moderately doped and n-type heavily doped), which produces a narrow depletion region.[2] The avalanche breakdown occurs in lightly doped junctions, which produce a wider depletion region. Temperature increase in the junction increases the contribution of the Zener effect to breakdown, and decreases the contribution of the avalanche effect.
My question is can in a diode, primarily under Zener breakdown, increasing the reverse-bias voltage cause avalanche breakdown ?
| Zener breakdown (the breakdown mechanism described by Clarence Zener) occurs
in practical semiconductor devices at low voltages. Avalanche breakdown
dominates at high voltages and includes amplification by carrier-generating
processes similar to a small rockfall initiating an avalanche.
Zener breakdown has a negative temperature coefficient of voltage, while
avalanche breakdown has a positive temperature coefficient.
So, in practical Si devices see figure #1 here both processes are
present, but the Zener process dominates low-voltage breakdown behavior, while
avalanche dominates at higher applied voltages.
The temperature coefficient at 5V is very near zero, indicating that both processes contribute nearly equally. This doesn't mean Zener behavior is
'converted', though, it simply means that the flood of extra charge carriers produced by avalanche effects can be more important than the Zener contribution.
| {
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Why Newton's equation of motion is time reversally invariant (TRI)? I am really perplexed by the fact that Newton's equation is time reversal?
Newton's equation of motion is time reversally invariant, evident from the equation itself:
$$m\dfrac{d^2x}{dt^2} = F(x).$$
My question is why?
Is there some deep reason they come out to be time reversally invariant(may be connection to Principle of Least action, which is a global picture by the way, instead of being local which is the case of Newton's equation)? or connection to something else(which is more evident)?
In the equation of motion(eom) because of the acceleration term, instead of velocity or other(which defy TRI). Is there a reason of it coming out to be like this? Links from geometry(variational principle) or where I see clearly that, it has to be like this(very basic and physically intuitive).
A bit detailed explanation will be of great use(origin of such symmetry here).
Forgive me, if question is unclear(make it clear, if asked) or if it has been asked(I checked but not my question), any help is highly appreciable.
| I'm not sure if this helps or even if it addresses your question, but one way to think about this is that Newton's law says that accelerations (not, say, velocities) are proportional to forces. If you accept that as a "non-perplexing" given, then the resulting differential equation can be interpreted as an expression of "space-time kinematics", and the time-reversibility simply follows from the fact that acceleration is the second time derivative of the position. This equation therefore describes invariant curves in space time: Particles are constrained to follow these curves independent of direction.
Of course, as you mention, you can view things from a different perspective as well, and if you start from a variational formulation, you may end up with fundamental equations that do not explicitly contain time, which makes the result (of time reversibility) trivial.
| {
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About effects of torque off center of mass I am trying to deepen my intuition: If in outer space is a rod of length 2 meters standing still. At the both ends it has some heavy wheels of equal mass and the associated motors. If a motor starts spinning one of the wheels ... will the center of mass of the rod stand still or move? I think that the following are possible:
*
*the center of mass will slowly rotate about the end of the rod with the wheel spinning. That end of the rod will stand still. So the c.o.m of the rod moves in a circle
*the center of mass will stand still and the ends will slowly rotate about the center of mass of the rod. So the c.o.m will not move, but the ends of the rod will rotate about it.
*neither ...
The motors are attached (welded) to the rod. Of the motor's rotors are attached the heavy wheels. (One motor at one end of rod)
I used to think that option 1 is the correct one ...
| Farcher has already given a good explanation : the centre of mass of the system does not move; the total angular momentum of the system remains zero unless ther is an external torque.
If the centre of the rod coincides with the centre of mass of the sytem, the centre of the rod will not move. If the centre of the rod does not coincide with the CM of the system, it could move around the CM of the system, depending on how other parts rotate.
If the wheels rotate in the same sense (eg both anti-clockwise, as in Farcher's diagram) then there is a net torque on the rod which turns it clockwise. If the wheels rotate in opposite directions there is zero net torque on the rod so it does not rotate.
Total angular momentum must be conserved, for the same reason that the CM of the whole system does not move : because of Newton's 1st and 3rd Laws. So if the wheels both rotate anticlockwise, then the whole system rotates clockwise with an equal amount of angular momentum in the opposite direction. The total amount of clockwise angular momentum must equal the total amount of anticlockwise angular momentum at all times.
| {
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More light reflected off of double pane windows? In glass windows, you get light passing through and light reflecting off the front and back surfaces. Would you get more light reflecting off of triple or double pane windows? Because you have double or triple the amount of surfaces for the light to reflect off of.
| You hit on half of the idea of optical thin film coatings. The other half is that light is a wave. Multiple reflections can interfere constructively or destructively. This can be used to make either an antireflection coating, or a mirror, or something more complex.
This shows the idea of an antireflection coating. Attribution below. The coating is 1/4 wavelength thick. Light that bounces off the back surface travels an extra 1/2 wavelength. It emerges out of phase with the front reflection. The two reflections cancel.
It doesn't really work on something as thick as a pane of glass. You can't make the front and back surfaces of glass exactly parallel and exactly the right separation.
Also, as WetSavannaAnimal pointed out, ordinary light does not have a perfectly stable wavelength. It is produced by a lot of individual short lived events in many different atoms, and each atom is independent. By the time a beam of light travels through a pane of glass twice, the light arriving at the top surface is from different events. There is no consistent relationship in the phases of light separated that far. So you don't get a cancellation that lasts longer than a fraction of a nanosecond.
A good quality laser has a very stable wavelength. With it, you can get interference patterns from the front and back surfaces of a pane. This is the original of the speckle that a laser produces when you shine it on a rough surface. You get interference from reflections from low and high points.
Illustration by DrBob - Own work (enwiki), CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=12305780
| {
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Looking for a good casual book on quantum physics I'm looking for something that is going to blow my mind without any scientistic ideas (e.g. something that sounds like science, but doesn't have anything in common with science), written by a professional physicist who spent a lot of time considering "what it all means". I'm reasonably proficient in math and stats, but I'd prefer something that I could spend time listening to in my free time.
Any recommendations on good and exciting books on quantum physics written by scientists?
| I would like to recommend this book, the theoretic minimum: quantum mechanics written by Leonard Susskind. It is very nice to talk about many important points in quantum mechanics.
Here is course website for this course.
| {
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What does one second after big bang mean? Consider the following statement:
Hadron Epoch, from $10^{-6}$ seconds to $1$ second: The temperature of the
universe cools to about a trillion degrees, cool enough to allow
quarks to combine to form hadrons (like protons and neutrons).
What does it mean to say "from $10^{-6}$ seconds to $1$ second"?
How is time being measured?
One particle might feel just $10^{-20}\ \mathrm s$ having passed and another could feel $10^{-10}\ \mathrm s$ having passed.
Is saying "1 second after the big bang" a meaningful statement?
| Adding a little more argument about co-moving, we know that momentum is conserved, even in relativity, so we can talk about the center of mass frame, the frame where the total momentum is zero. This is the preferred frame.
| {
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Method of image charges for a point charge and a non-grounded conducting plane I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential $V$.
| I also think that there would be a difference between the two situations. Lets take a non-grounded PEC plane and say a positive charge lies above it. The field created by the charge will induce negative charges to be stacked on the top (towards the positive charge) of the PEC, and since there is no creation nor destruction of charge, the same and opposite charge will be induced on the bottom surface of the PEC. In this way, treating the thickness of the PEC as infinitely small, both induced surface charges cancel each other over large distances, as seen by the positive charge. In this way, nothing changes in respect to the field generated around the positive charge (since the PEC as no total charge).
Though, if the PEC is grounded, it acts like an infinite source of charges that can be arranged in any way, dictated by the surrounding charges. So the positive charge induces an electric field on the upper surface of the grounded PEC, BUT no charge on the lower surface since charge must not be conserved in this situation (that is what ground means!). We are then left with a charge that as electric field lines pointing towards the PEC, all coming at right angles with the latter. This situation is exactly equal to a dipole made of the same positive charge, and a equal negative charge situated an equal distance normal to the PEC.
| {
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Density of states of Bogoliubov quasiparticles For a simple fermionic system the formula for calculating the density of states (DOS) is $N(E) = \sum_{n}\delta(E-E_{n})$ where $\{E_{n}\}$ is the set of eigenvalues obtained after diagonalizing the hamiltonian. Now to diagonaloize a hamiltonian with pair correlation terms ($\sum_{k}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}$) Bogoliubov transformation ($c_{k\uparrow}=u_{k}\gamma_{k\uparrow}-v_{k}^{\ast}\gamma_{-k\downarrow}^{\dagger}; c_{-k\downarrow}^{\dagger}=v_k\gamma_{k\uparrow}+u_{k}^{\ast}\gamma_{-k\downarrow}^{\dagger}$) is used. Now after diagonalizing we get a set of eigenvalues in the form:$\{E_n,-E_n\}\forall n$. Now to find the density of states I found a formula like this: $N(E)=\sum_{k}|u_k|^2\delta(E-E_k)+|v_k|^2\delta(E+E_k)$ where $\{E_k\}$ is the set of positive eigenvalues only. I don't understand this particular formula for density of states of bogoliubov quaisparticles. If anyone can explain it that would be very helpful.
| Write BCS ground state (reference) $|\Psi_{BCS}\rangle$:
$$|\Psi_{BCS}\rangle=\prod\limits_{\bf k}(u_{\bf k}+v_{\bf k}c_{\bf k\uparrow}^\dagger c_{\bf -k\downarrow}^\dagger)|0\rangle$$
and calculate the number density expectation:
$$\langle \hat n_{\bf k}\rangle=\langle c^\dagger_{\bf k}c_{\bf k}\rangle$$
| {
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Centrifugal force in inertial systems My textbook gives an example:
There are two systems, $O$ is inertial and $O'$ non inertial. $O'$ is rotating whit $\omega=\mathrm{constant}$ and $O=O'$.
We assume that a disc is rotating whit the same $\omega$ of $O'$, and we put an object on the disc. There isn't friction between the object and the disc, so the object is quiet in system $O$. In system $O'$ the object seems to rotating (in opposite verse than $O'$).
For the second part of the exemple we connect the object whit the system $O'$ whit a rope, so the object starts rotating with $O'$. Now for $O'$ the object is quiet and $O'$ sees that the rope is in tension, so $O'$ hypotizes that there is a force that tenses the rope (centrifugal).
For $O$ the object is rotating, but how $O$ explains the rope's tension? Centrifugal forces does not exist in inertial system, I think. Am I wrong in thinking that? Why? If I'm not wrong, how is the tension explained?
Thank you all, sorry for bad english
| You are correct in suposing centrifugal forces are not the explanation because they do not appear in inertial frames of reference.
In $O$ the object is performing a circular motion, and this motion does not appear when there is no force on the object (it would perform a straight motion).
What the rope does in $O$ is to constrain the motion of the object. The constraint is that the object must remain at a constant distance of the other point the rope is attatched to. For the object to satisfy this constraint it must perform the circular motion, and by the laws of mechanics there is no other way this could happen but the appearence of the tension force.
This tension force is "the rope's way to force a circular motion on the object".
| {
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Would a gas "weigh" less than a liquid if they have the same mass? Thought experiment: I acquired two boxes of the same dimensions and same weight. One box contains $1\ \mathrm{kg}$ of water at room temperature while the other box has $1\ \mathrm{kg}$ of water, but in steam form, because the temperature of the box is above $100^\circ\mathrm{C}$. The volume of the boxes is large relative to the amount of space the $1\ \mathrm{kg}$ of water would take (let's arbitrarily say $10\ \mathrm{L}$). Both boxes contain the same amount of air (at $1\ \mathrm{atm}$) which is why the second box has water in steam form at $100^\circ\mathrm{C}$.
I put each box on a simple electronic scale to measure their respective weights. Unsurprisingly, the box containing water comes out to be $1\ \mathrm{kg}$. But what about the box containing steam?
My guess: Electronic scales measure the amount of force being exerted on it, then divide that force by $g$, to get the mass of the object. I think the box with steam in it will be exerting less force on to the scale and therefore the scale will think its mass is less than $1\ \mathrm{kg}$.
| Weight is proportional to mass, end of story, the two boxes will weigh the same. However to contain the mass of 1kg of steam in a ten litre box will require an enormous pressure. The volume of an ideal gas (which steam is not but neglect that for a moment) is 22.4 litres per mole at STP. One mole of water weighs 18g, so 1kg of steam will have a volume of 1244 litres at STP. At 100C the volume will be higher by a factor of around 373/298 in other words 1558 litres. So you want to put this in a 10L box which means applying a pressure of 156 atmospheres, which is a force of around 15.6 MPa, equivalent to 15.6 newtons per square millimetre. You will need oxygen cylinder type technology which will weigh a huge amount, which will be a tremendous burden on your accurate weighing scale.
Some other objections, a large pressure will elevate the boiling point of water, so you need to heat it up a lot more to ensure it remains gaseous at these huge pressures. Ideally you need to operate above the critical temperature of water which is 374C, resulting in a pressure ratio of 647/298 which is 270 atmospheres which is comfortably above the critical pressure.
| {
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Schrödinger equation and non-Hermitian Hamiltonians Is the Schrödinger equation still valid if we use a non-Hermitian Hamiltonian with it? By this I mean does:
$$\hat{H}\psi(t) = i\hbar\frac{\partial}{\partial t}\psi(t)$$
if $\hat{H}$ is not Hermitian?
| Let's suppose that we have an operator $\hat{\mathcal{O}}$ such that $i \hat{\mathcal{O}} |\psi\rangle = \partial_t |\psi\rangle$. The question is then: What is $\partial_t (\langle \psi | \psi \rangle)$? Well, by definition of the adjoint we have $\partial_t \langle \psi | = \langle \psi| (i \hat{\mathcal{O}})^\dagger = -i \langle \psi| \hat{\mathcal{O}}{}^\dagger$, and so we have
$$
\partial_t \left(\langle \psi | \psi \rangle \right) = \left( \partial_t \langle \psi | \right) | \psi \rangle + \langle \psi |\left( \partial_t | \psi \rangle \right) = -i \langle \psi |\hat{\mathcal{O}}{}^\dagger| \psi \rangle + i \langle \psi |\hat{\mathcal{O}}| \psi \rangle = i \langle \psi |(\hat{\mathcal{O}} - \hat{\mathcal{O}}{}^\dagger)| \psi \rangle
$$
This means that if $\hat{\mathcal{O}}$ is Hermitian, then $\partial_t (\langle \psi | \psi \rangle) = 0$ for all states (and vice versa). But under the usual interpretation of QM, $\langle \psi | \psi \rangle$ is the norm of the state $ | \psi \rangle$, and so we would like it to be constant (and usually equal to 1.) The fancy words for this are that a Hermitian Hamiltonian implies unitary time evolution.
There are some cases where we don't actually want unitary time evolution; the most well-known (which was mentioned by Emilio Pisanty in his answer) are decay processes, which can be modeled quite nicely by a non-Hermitian Hamiltonian. In this case, you want the probability of the particle's existence to decrease (exponentially) with time, so the non-unitarity of the time evolution is actually desirable.
| {
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Conceptually, why is acceleration due to gravity always negative? As the title states, why is acceleration due to gravity always (-). Say you assign "up" as the positive direction. If an projectile is thrown at a 24 degree angle above the horizontal, I get that acceleration due to gravity before the vertex is negative. However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?
What I think is that acceleration due to gravity is always towards the ground. Even if a projectile is going downwards, and we assign downwards as (-), the acceleration due to gravity is still (-), because the object still accelerating downwards. Despite it going upwards or downwards, the net acceleration of the object is downwards.
| Acceleration due to gravity in itself is not negative but it is directed toward center of earth (downward) and we take
Downward direction as negative by convention.
And as force of gravity is pointing to same direction at every point on the trajectory hence acceleration due to gravity is same for before or after the vertex.
| {
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Expectation value of $p^2 (1/r) + (1/r) p^2$ I'm trying to derive $\langle kl | \lbrace 1/r , p^2 \rbrace | nl \rangle$, where the states satisfy the equation of motion (I omit factors of $1/2m$ etc.):
$$(p^2 + V)| n,l \rangle = E_n | n,l \rangle$$
and $\lbrace A,B \rbrace = AB + BA$ is the anticommutator.
I have two solutions at hand that differ. In general I can write:
$$\langle k,l | \lbrace 1/r , p^2 \rbrace | n,l \rangle = \langle k,l | 1/r~p^2 + p^2~1/r | n,l \rangle$$
and I can use that $p^2 \propto -\nabla^2$ and $\nabla^2 1/r \propto -\delta^{(3)}(r)$.
Now I want to let $p^2$ act first on the respective states and let $1/r$ act later on, such that I can use the e.o.m. Thus I have:
$$\langle k,l | 1/r~p^2 + p^2~1/r | n,l \rangle = \langle k,l | 1/r~(E_n - V) + (E_k - V)~1/r | n,l \rangle$$
In this solution I let the first summand act on the ket and I let the second summand act on the bra. However, when I do it the other way around, using the product rule $\nabla^2 (1/r~\psi) = (\nabla^2 1/r) \psi + 1/r (\nabla^2 \psi)$ and use the e.o.m for the second term I obtain:
\begin{eqnarray}
\langle k,l | 1/r~p^2 + p^2~1/r | n,l \rangle &=& \langle k,l | \delta^{(3)}(r) + (E_k - V) + (E_n - V) + \delta^{(3)}(r) | n,l \rangle\\ &=& \langle k,l | 1/r~(E_n - V) + (E_k - V)~1/r | nl \rangle + 2 \langle kl | \delta^{(3)}(r) | n,l \rangle
\end{eqnarray}
These two solutions are clearly different if and only if both of the wave functions do not vanish at the origin. This is to say that these two solutions give different results for s-waves.
Am I missing something essential here? Any input would be appreciated!
| A paradigm is worth a hundred puffs of bloviation.
Exploit your spherical symmetry, $\hat{r}\cdot \vec{\nabla}=\partial_r$, and $\vec{\nabla} f(r) =\hat{r} \partial_r f(r)$; and, since the challenge you are concerned with is the singularity at the origin, r =0, we might as well drop all l>0, which are softer than s-waves, as you noted.
In your $\hbar=1$, $m=2$ units, consider, for simplicity, the s-waves of the Hydrogen hamiltonian, sufficient to illustrate the point,
$$
H=p^2-2/r= -\frac{1}{r^2} \partial_r (r^2\partial_r) -\frac{2}{r}~.\tag1
$$
The Hilbert space integrations will be over $4\pi \int^\infty_0 dr~ r^2 ~~$ .
Thus,
$$
\nabla^2 \frac{1}{r}=-4\pi \delta^{(3)} (\vec{r})~,\tag2
$$
which, in our radial context, amounts to just
$$
-p^2 \frac{1}{r}=\frac{1}{r^2} \partial \left(r^2\partial \frac{1}{r}\right )=- \frac{\delta(r)}{r^2}~.\tag{2'}
$$
The ground eigenstate (n =1) is then
$$
\psi_0= \frac{e^{-r}}{\sqrt{\pi/2}} \qquad \Longrightarrow E_0=-1, \tag3
$$
while the first excited state (n =2) is
$$
\psi_1= \frac{e^{-r/2}}{4\sqrt{2\pi}} (2-r) \qquad \Longrightarrow E_1=-1/4, \tag4
$$
so that $\langle \psi_1| \psi_0\rangle=0$, since $\int_0^\infty \! dx~ e^{-x} x^n =n! ~~$ .
It follows that
$$
\frac{1}{r}~ p^2 ~\psi_0= \left (\frac{2}{r^2}-\frac{1}{r}\right )\psi_0 \tag5 \\
p^2 \left (\frac{1}{r} \psi_0 \right) = \left(\frac{\delta(r)}{r^2}-\frac{1}{r}\right )\psi_0 .
$$
(The second eqn is the screened Poisson eqn -- "massive photon".) Hence
$$
[p^2,1/r]\psi_0 = \left ( \frac{\delta(r)}{r^2}-\frac{2}{r^2} \right) \psi_0. \tag{5'}
$$
(In general, $[p^2,1/r]=\frac{\delta(r)}{r^2} + \frac{2}{r^2} \partial_r ~$ .)
Now note that
$$
\langle \psi_0|[p^2,1/r]| \psi_0\rangle=0= \langle \psi_1|[p^2,1/r]| \psi_1\rangle .\tag6
$$
Consequently, as per your first evaluation path using hermiticity,
$$
\langle \psi_0|\{p^2,1/r\}| \psi_0\rangle=
\langle \psi_0|(H+2/r) \frac{1}{r} + \frac{1}{r} (H+2/r) | \psi_0\rangle=
\langle \psi_0| \frac{2}{r} \left(\frac{2}{r}-1\right) | \psi_0\rangle=12.\tag7
$$
Acting on just the right ket, as per your latter intention, also yields the same,
$$
\langle \psi_0|\left( \frac{\delta(r)}{r^2} -\frac{2}{r} +\frac{2}{r^2} \right ) | \psi_0\rangle=12.\tag{7'}
$$
You never had to cancel the δ : it does not fail to know its place. It had better be, since the expectation of the commutator vanishes.
But this helpful vanishing is not actually necessary for consistency. More generally, as per your off-diagonal example,
$$
\langle \psi_1|\{p^2,1/r\}| \psi_0\rangle= \langle \psi_1|\left(-\frac{1}{4}+ \frac{2}{r}\right)\frac{1}{r} + \frac{1}{r}\left(-1+\frac{2}{r}\right ) | \psi_0\rangle=86/27~. \tag8
$$
Alternatively, simply acting on the right, as done above,
$$
\langle \psi_1|\left( \frac{\delta(r)}{r^2} -\frac{2}{r} +\frac{2}{r^2} \right) | \psi_0\rangle=\int dr e^{-3r/2}( \delta(r) -2r +2 )(2-r) =86/27~.\tag{8'}
$$
Your two approaches are consistent, after all. The linchpin is that the integration by parts implicit in the hermitean maneuver works free of surface terms at the origin.
This is not hard to generalize. For singular potentials (unlike yours, I understand), i.e. with $r^2V(r)$ not vanishing at the origin, take a look at
Khelasvili & Nadareishvili 2010.
| {
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Liquid vs. gas cooling I have an aluminum can that needs cooled. I put it in the refrigerator, where it is cooled by the cold air surrounding the can.
If I were to place the same can in water that had been cooled to the same temperature as the refrigerated air, would the can cool down faster?
I'm inclined to assume the liquid would cool the can faster due to the density of the molecules surrounding the can, but I am interested in what people who know physics have to say.
| The two different methods of cooling the can are completely different and governed by different properties and have different heat transfer coefficients.
Because the can is warmer the direction of heat transfer is from the can.
In the refrigerator the method of heat transfer is radiation. The heat transfer is governed by the radiative property of the can surface termed emissivity. Assuming there is no moment of the air in the refrigerator the heat transfer by forced convection will be zero.
The can in water the method of heat transfer is conduction. The heat transfer from the can is governed by the thermal transport property known as the thermal conductivity. The heat transfer is also due to the fluid thermodynamic properties of the water.
| {
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How to remove gravity component from accelerometer $X$, $Y$ readings? So I have an accelerometer which I am wanting to use in an IMU.
When the device is tilted but stationary I want the x, y values to be 0, so effectively negate the effect of gravity along the x and y axes of the accelerometer.
I have found lots of conflicting advice on the internet regarding rotation matrices and was wondering if anyone here can provide some clarity.
I understand accelerometers can only measure pitch and roll from their x, y, z so how can I use these values to remove the gravity vector component on the x and y axes?
| well as i experienced, you could compute rotation matrix from Azimuth(yaw), pitch & roll. computationally it is simple as mentioned by formulas for kinematics everywhere.
then simple by multiplying 3*3 rotation matrix with 3*1 accelerometer data to convert the readings and compute new values corresponding to fixed global coordinate system.
then you can easily subtract g from newly computed value of z axis.
| {
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Ball inside an accelerating frame Why does a ball inside a moving bus at rest start moving backwards when the bus suddenly accelerates? Also does the moving ball have some acceleration?
This is my theory:
Initially the bus and the ball are at rest. When the bus starts accelerating, due to inertia of rest, the ball resists change in motion and tends to remain at rest. Since it is an accelerating frame, pseudo forces of unknown origin act on the ball in the direction opposite to the direction of motion of the bus. This pseudo force is responsible for the ball to move in the opposite direction with some acceleration. This explanation with respect to the accelerating frame.
But how do I explain this fact from the frame of reference of a person on the road? I understand that the acceleration of the ball is in the same direction as that of acceleration of bus. There is no other force to balance this force which is in the direction of motion of bus. So the ball should have moved in a direction along the direction of the bus. But the reverse happens. Please help me.
| Take it like this imagine yourself as an by stander watching the bus then you would notice the ball goes in the direction opposite to bus but when we only take the ball the direction of pseudo force is opposite to the motion of the bus when we consider the floor to have friction then this $μN$ force to opposethis pesudo force hence the real motion occurs in the opposite direction of bus.while considering the consequences the motion of ball by the by stander is same as the one in the bus watching the ball.
| {
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How do I find the time evolution of a ket? I have a question which reads:
Let \begin{bmatrix}
{E_0} & 0 & A \\
0 & E_1 & 0 \\
A & 0 & E_0
\end{bmatrix}
be the matrix representation of the Hamiltonian for a three-state system with basis states $|1>, |2> \mbox{and } |3>$.
a. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|2>$ what is $|\psi(t)>$?
b. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|3>$ what is $|\psi(t)>$?
$\textbf{My attempt at a solution:}$
a. For both problems we can use $|\psi(t)> = \hat{U}(t)|\psi(0)>$ where $\hat{U}= e^{\frac{-i\hat{H}t}{\hbar}}$. Since
$$|2> = \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}$$
is an eigenvector with eigenvalue $E_1$ we can simply replace the Hamiltonian in the time evolution operator by $E_1$, so $$|\psi(t)> = e^{\frac{-iE_1t}{\hbar}}|2> $$
Is this correct? I am finding other solutions online which have a different answer, although I can't see how this could possibly be wrong, unless my representation for $|2>$ is wrong.
Assuming this is the correct way of doing this, I am having a hard time doing b. I can find the eigenvalues and eigenvectors of the hamiltonian easily, and can represent |3> = $(0,0,1)^T$ as a linear combination of those vectors, thereby allowing me to operate on it. However, my final answer is in terms of |1> and |3>, which I feel is incorrect somehow.
| Actually I believe both answers are correct. I can't seem to find anything wrong with either. Certainly a. is correct since the hamiltonian in the time operator should just be replaced by the eigenvalue, seen simply if we expand the matrix exponential.
For b, there is nothing wrong with expressing our time dependent state as a linear combination of the initial state and another basis state.
| {
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Michelson interferometer beam splitter phase shift In a Michelson interferometer (image from Optics by E. Hecht ) .
To quote from the same book:
As the figure shows, the optical path difference for these rays is
nearly $2d \cos 0$. There is an additional phase term arising from
the fact that the wave traversing the arm $OM2$ is internally
reflected in the beamsplitter, whereas the $OM1$-wave is externally
reflected at $O$. If the beamsplitter is simply an uncoated glass
plate, the relative phase shift resulting from the two reflections
will be $\pi$ radians.
Consider a beam from $B$ from $S$ towards $O$. At $O$ it splits into two beams:
Beam B1:
*
*Results from refraction of $B$ at the beam splitter $O$ towards mirror $M1$.
*Is reflected in the opposite direction at $M1$.
*Is reflected towards $D$ at $O$. Is it reflected before entering $O$ or after entering $O$ and encountering the air at the other side?
Beam B2:
*
*Results from reflection of $B$ at the beam splitter $O$ towards mirror $M2$. Should there be a phase shift here? This is should be an air/(glass/metal coating) interface.
*Is reflected in the opposite direction at $M2$.
*Goes through $O$ towards detector $D$.
Question:
Could someone tell me at which of these steps a phase shift occurs? It seems to me that it could happen at steps $B1-2$, $B1-3$, $B2-1$ and $B2-2$, but that is probably not right.
| There is an automatic change in the phase when the wave goes ..... bounces off the mirror. Then after all that you start to play with the path lengths to introduce more phase changes.
| {
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What is the difference between mean free path and intermolecular distance? Why is the mean free path not be equal to the intermolecular distance?
A particle moving in a particular direction should strike the object in that direction after the traveling the same distance as the distance between them initially.
| There's a few things to point out. @lemon already pointed out one of them in the comment -- it is possible for a molecule to move and end up going between other molecules and missing them. So even though they start out, say, equally spaced (think like an equilateral triangle), it's possible for a molecule to move between the other two rather than directly at it.
The other is to remember that all the molecules are moving. Consider a very simplified case where you have two molecules on the X axis, initially a distance Y apart. They are both moving at +Z on the axis. Their mean free path is infinite -- they will never collide -- but their intermolecular distance is constant. They are just following each other.
So, all of this is to say that the molecules are all moving at the same time and in random directions. It is not as if you have a single molecule moving and the rest are frozen. Nor are they guaranteed to be moving directly towards one another. The mean free path is how long a molecule travels before it hits something on average, while the intermolecular distance is the mean spacing between the molecules without consideration for their direction of motion.
| {
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Proper volume in different frames of reference (and tetrads) I am interested in the behavior of the proper volume when we switch the frame of reference. For example, I know that the proper volume element for the Schwarzschild and Kerr black holes may be extracted easily from the formulae:
$V=\sqrt{g}dr d\theta d\phi = \frac{r^{5/2} \sin (\theta )}{\sqrt{r-2 m}}$
(Where we only picked the spatial components of the $g_{\mu \nu}$ for the determinant)
However, I am interested in seeing how this proper volume changes when according to different observers. For example, I know that the local frame of reference can be expressed with a tetrad:
$e^m_\mu=\left(
\begin{array}{cccc}
\sqrt{\frac{r-2 m}{r}} & 0 & 0 & 0 \\
0 & \sqrt{\frac{r^2}{r^2-2 m r}} & 0 & 0 \\
0 & 0 & \sqrt{r^2} & 0 \\
0 & 0 & 0 & \sqrt{r^2} \sin (\theta ) \\
\end{array}
\right)$
Which fulfills: $e_{c}{}^{b} e_{e}{}^{d} g_{bd} = \eta_{c e}$ (the tetrad transformation into the local frame of reference)
Now I am interested in finding out what the proper volume $V_{\rm local}$ is in this local frame of reference given by the tetrad. Is this possible?
| Well, to calculate volumes, you need a hypersurface.
Let $\Sigma$ be such a hypersurface, locally described as $$ x^\mu=\Phi^\mu(\xi), $$ where $\xi=(\xi^1,\xi^2,\xi^3)$ are local coordinates on the hypersurface. Let us assume $\Sigma$ is spacelike, since that's what you seem to be going for.
The induced metric is then positive definite and is given by $$\gamma_{ij}=g_{\mu\nu}\frac{\partial\Phi^\mu}{\partial\xi^i}\frac{\partial\Phi^\nu}{\partial\xi^j},$$
the volume element is then $$d\Sigma=\sqrt{\det\gamma}d^3\xi.$$
Let $e_{(a)}=\{e_{1},e_{2},e_{3}\}$ be an orthonormal frame on $\Sigma$, so we have $$ \delta_{ab}=\gamma_{ij}e^i_{(a)}e^j_{(b)}, $$taking the determinant of both sides gives $$ 1=\det\gamma\cdot(\det e)^2, \\ \sqrt{\det\gamma}=\det(e^{-1}), $$ is we denote the dual frame with $\theta^{(a)}=\theta^{(a)}_id\xi^i$ (we have $e^i_{(a)}\theta^{(b)}_i=\delta^b_a$), we have $$\sqrt{\det\gamma}=\det\theta,$$ so the volume element is $$ d\Sigma=\det\theta d^3\xi. $$
| {
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When can one omit a total time derivative in the Lagrangian formulation? I am studying Lagrangian and Hamiltonian mechanics and i am using Landau & Lifshitz and Goldstein books. Both of them state that a modified lagrangian $$L'=L+\frac{df}{dt}$$ gives the same solutions than $L$ wich i kind of understand but its not the main problem.
In landau there is a problem in which a pendulum whose attachment point is oscillating. setting up the equations is not a problem for me, but when he gives the solutions he states that he omits total time derivatives as if it was the most obvious thing to do, i guess that this omission is related to the "invariance" of the Lagrangian but i fail to se the direct relation with this. How do you know from which function are you supposed to omit the total time derivarive?
How do you identify this function?... etc etc
The example that i'm talking about is in Landau & Lifshitz book page 11 exercise 3)b.
| To understand this f you need to be aware that f can only be a function of q and t, not of q dot.
I got tripped up by the same question a long time ago as an undergrad. When you get $a^2 \omega^2 sin^2(\omega t)$ then this is a function of t only (a and omega are not dynamical variables, right?) We know any function of one variable (t) can be written as the total derivative w.r.t t of another function, so thisnterm you can drop.
For the other terms in your solution, you are probably using a different zero point/angle for theta. There is then still the factor of 2 and a factor of l missing, and if i dont remember totally wrong these were typos in L.L.
About your question how to identify: a) just drop any function only depending on t
b) if you see a function like $g(q) \dot q$ you can also drop it (total derivative of G(q) with G stemfunction of g
I think no other rule which would be obscure is used in the book.
| {
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Spinor decomposition I just read in the book "Covariant loop quantum gravity" of Rovelli about spinors (section 1.7.1) I'm confused about the decomposition of an spinor of two indices that it is done:
$$z^{AB}= z_0 \epsilon^{AB} + z^{(AB)} $$
where $z^{(AB)}$ is the symmetric part, $\epsilon$ is the antisymmetric matrix and $z_0 = (1/2)\epsilon_{AB}z^{AB}$.
I understand the symmetric part but I'm rather confused with the anti symmetric part that should be $(1/2)(z^{AB}-z^{BA})$. How can I obtain the expression given by Rovelli?
| \begin{eqnarray}
z_0 \epsilon^{AB} &=& \frac 1 2 \epsilon _{CD}\,z^{CD} \times \epsilon^{AB}\;,\\
&=& \frac 1 2 \delta^{AB}_{CD} \,z^{CD} \;,\\
&=& \frac 1 2 \big( z^{AB}-z^{BA} \big)\;,
\end{eqnarray}
where
$$
\epsilon _{CD}\,\epsilon^{AB}= \delta^{AB}_{CD} \equiv \det \begin{bmatrix} \delta^A_C & \delta^A_D \\ \delta^B_C & \delta^B_D\end{bmatrix}\;,
$$
| {
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LIGO Gravitational wave discovery - how did they know the cause of spike? I understand how unbelievably lucky the discoverers were to catch the wave produced billions of years ago by an event that happens so rarely one hour into a test run of their equipment. But one thing is still not clear to me – how did they know what exactly caused the spike? Was that merely a conclusion that this must have been two black holes colliding or there was any additional astronomical observation done that spotted the collision?
EDIT:
This question is similar, and one of the answers mentions that LIGO observation can not be interpreted as evidence of WHAT the actual event was, but it also never directly tells me if any proven-to-work ways of observation parallely detected the black hole merger.
Because if only LIGO was involved in detection, than how do we know WHEN and WHERE the merger happen? As far as I understand, LIGO technology does not communicate any space-time coordinates of the event. Or am I wrong?
This and other articles insist that event took place 1.3 billion years ago.
| People ran computer simulations that told them what wave pattern would be observed for various cosmological events.
Each even has its own "fingerprint" than can be used to distinguish the various events.
As far as I know there were no additional astronomical observations that gave definite results on the origin of the wave. This is due to the bad resolution of ligo. People don't know where exactly they should search for the origin of the wave.
REQUESTED EDIT:
I can answer your "where" question but I don't know enough about the subject to answer the "when" question.
Gravitational waves stretch spacetime, which is exactly what has been observed via the interferometer. The way in which this stretching happens is not arbitrary and tells you something about the direction in which the signal is traveling. As I mentioned before the angular resolution is not very good such that an exact pinpoint of the origin is impossible (for now)
| {
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Conservation of momentum in refraction Light, when passing through a boundary refracts. How is momentum conserved here? There can't be an impulse, the energy doesn't change.
| In your question, you are assuming that you cannot have a change in momentum without a change in energy. Consider, however, what happens if a force is applied perpendicularly to the trajectory of an object: no work is done but the direction of the momentum is changed. An example of that is a planet in orbit of its star.
I am not sure that this explanation addresses specifically your question about light refraction, but it corrects a misunderstanding that motivated your question.
| {
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If two particles are entangled and you collapse the wave function of one of the particles. Does the other particle collapse as well? Let's suppose you entangled two photons, you separate the photons, and then you measure the polarization of one the photons collapsing its wave function. The wave function of the other photon collapses also?
| If you have two spins in an entangled state they define a wave function
$$
|\psi\rangle~=~\frac{1}{\sqrt{2}}\left(|+\rangle|-\rangle~+~e^{-i\phi}|-\rangle|+\rangle\right)
$$
in a singlet state of entanglement. What exists is the entangled state. In effect the individual spin states do not exist. A measurement of one spin state does mean that the total qubit information of the entangled state is now in spins, which means the other spin appears to the. So if Alice measure a spin, then Bob necessarily has the opposite spin.
We can think of this as a mutual collapse. the so called collapse of a wave just means the observables of some system becomes localized in a way that does not obey Schroedinger or any quantum dynamics. This is how we identify states with particles. In the case of entanglement it is the case that if one spin state is localized "here," then it is also localized "there."
| {
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If free quarks can't exist, how did the universe form? As I understand, the Big Bang started with a photon gas that then created the other particles. Thus obviously there would be some free quarks in the early Universe unless quarks are always created in pairs for some reason. How does physics resolve this?
| "Free quarks can't exist" is simply an oversimplification of the actual situation in quantum chromodynamics (QCD). A better statement is "free quarks cannot exist at low energies", where "low energy" means below the deconfinement scale.
Confinement is precisely the phenomenon that says that the force between two quarks rises linearly with distance, meaning you can never separate two quark, since doing so would require infinite energy. Now, we unfortunately do not have a full theoretical understanding of confinement in continuum QCD, but what we do know - both from heuristic arguments and from lattice computations - is that QCD exhibits a phase transition between a confining phase and a deconfining phase as the energy scale increases.
This is not due to the running coupling as such, but to the expectation value of the order parameter of this phase transition, the Polyakov loop, a variant of the Wilson loop, becoming non-zero. As long as the Polyakov loop is zero, the free energy of a two-quark system is infinite, meaning they cannot be separated. Lattice calculations indeed show that the phase transition to a non-zero Polyakov loop happens as the energy scale increases, so in the early hot universe, free quarks could exist without contradiction to our current situation.
The symmetry which is broken by the Polyakov loop is the so-called "center symmetry" of the gauge theory on the lattice, see this question.
| {
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Elasticity of solid A rod $1m$ long is $10cm^2$ in area for a portion of its length and $5cm^2$ in area for the remaining. The strain energy of this stepped bar is $40$% of that a bar $10cm^2$ in area and $1m$ long under the same maximum stress. What is the length of the portion $10cm^2$ in area.
My attempt:
$$strain-energy = \frac{1}{2}(stress)^2 × \frac{volume}{Y}$$
where $Y$ is Young's modulus.
Since the material of both the rods is the same so $Y$ is same and stress is also same so $40$% of the volume of the uniform rod = volume of the stepped rod. But this expression is not yielding the result. I am confused as what do the line " under same maximum stress" mean in the question as for same force stress cannot be same. Where I am wrong.
| The tension is same same all along the length of the composite bar, as it is in the uniform bar. So the stress will be higher where the cross-section is lower.
The question is saying that the stress in the narrow section of the composite bar is the same as in the uniform bar with which it is being compared.
| {
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Low power loss in electricity transmission lines To reduce the heat lost during transmission of electricity, we say we increase the voltage of transmission, taking the formula $I^2R$ in consideration. Couldn't I consider $V^2/R$? If I consider the second form, increasing voltage will increase the power dissipated. No?
| No, I believe you are misinterpreting something.
For same power to be transmitted in simple terms,
$$VI=constant$$
$V$ = voltage between source and sink.
So higher the votage drop, lesser would be the current.
The tricky part is that when we talk about current, it's passing through the transmission line and being a long wire it has a good amount of resistance so some voltage drop would also be present over it and some votage drop across the town.
Say $R$ is the resistance of the wire and $r$ be of the town.
power loss in wire $= I^2R$
Voltage drop across wire = $IR$
So here you see that some voltage drop across wire is getting wasted.
But by making the voltage of transmission very large, we make the current small and voltage drop across the wire decreases and hence on using
$(V')^2/R$, we get a smaller power consumption.
Note that $V'$ is the voltage across the wire.
You would get similar result as the voltage across the wire is in the term and not the voltage across source and sink.
| {
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