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Does the Lorentz transformation necessary follow from the two postulates of relativity? The two postulates of special relativity are:
The choice of what inertial frame to use is arbitrary: all laws of physics are invariant. (the principle of relativity)
The metric $$(\Delta s)^2 = (\displaystyle\sum_{\mu=1}^3 \Delta x^\mu)^2 - c^2 (\Delta x^0)^2 = 0$$ for any light-like geodesic. (the invariance of the speed of light)
Now, my question is whether the Lorentz transformation necessary follow from these two postulates, or if the Lorentz transformation is just one possible solution that satisfies the postulates. Why does it has to be of the form $\gamma(x^1-v x^0)$?
You could, of course, have an empirical argument that the Lorentz group is preserved in experiments but that's not really what I'm searching for .
| In some sense it is not the only possibility. And that's ignoring the obvious typos like missing signs or inconsistent use of superscripts and subscripts, and the fact that your equations require velocities to be dimensionless to even be dimensionally correct.
In particular the first principle tells you next to nothing if you haven't specified which particular statements you label as laws of physics.
And the second one all by itself is pretty weak as well. Basically you want to say that if two events can be traversed by an object going at speed $c$ then $(\Delta (ct))^2-(\Delta x^1)^2=0$ but that's pretty much a tautology all by itself. The real content should be that different frames agree on whether the object is going at $c$ but you haven't actually stated that $x^0=ct$ or that $c$ is actually a constant that doesn't depend on frame. So even if you find that $ct$ transforms a certain way, this might not give us Lorentz transformations. If $c$ can depend on frame then $t$ can change differently so long as $ct$ transforms correctly.
And your laws of physics might be the same, for instance Maxwell in Gaussian units really only has $ct$ appear.
As you can tell, a lot of this boils down to your statements being vague about what $c$ is. And there are ways to derive the Lorentz transformations by assuming a linear transformation, a group structure, an isotropy, a homogeneity, and then getting a frame invariant speed that is determined empirically (an infinite invariant speed giving Galilean Relativity and an invariant speed equal to $c$ giving Einsteinian Relativity). When you just set it equal to one, you exclude Galilean Relativity, and you make it hard to distinguish all the other Relativities that were possible.
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What happens to a body if it rotates extremely fast? I am thinking on a object, e.g. ball or planet that starts rotating with increasing speed. Let's assume that his speed get's closer to the speed of light, what happens to this object? There are several forces acting. But I always get caught thinking that it will get heavier and heavier because of the additional energy which is needed to accelerate it. Is that all? Or anything else interesting happens?
| Note that "c" means linear velocity, not angular. Then, you would refer to tangential velocity.
If an object rotated at relativistic (tangential) velocity, then each "shell" will have a different space and time compression. By contraction of the tangent lengthes, the perceived length of large circles would be smaller than $2\pi r$, leaning to 0 lenght at equator as the tangential speed approach c.
Appart from that, yes you would need more and more energy to accelerate it, and also the material coherency of the object is unlikely to resist to centrifugal force !
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Prospects for detection of gravitons? With the announcement of the detection of gravitational waves, questions about the implications proliferate. Some relate to the possible existence of gravitons. The analogous relationship between gravitons/gravitational waves and photons/electromagnetic waves is frequently mentioned.
The detection of individual photons required experiments of very low intensity light, yet their existence was inferred (prior to their actual detection) by Planck and Einstein (among others) using the properties of experimental black body radiation and the photo-electric effect.
If the prospects for detection of gravitons requires similar study of very low intensity gravitational waves, then those prospects are very dim indeed. My question: are there similar indirect "experimental" methods for inferring the existence of gravitons?
| Well, phenomenologists never give up. Here is a paper peer reviewed which explores the discovery of gravitons in future, but not too far future, colliders.
All one needs is large extra dimensions in a string theoretical model, to give predictions.
Two birds with one stone, graviton and extra dimensions.
Edit with new information:
In the paper, "Using cosmology to establish the quantization of gravity," published in Physical Review D (Feb. 20, 2014), Lawrence Krauss, a cosmologist at Arizona State University, and Frank Wilczek, a Nobel-prize winning physicist with MIT and ASU, have proposed that measuring minute changes in the cosmic background radiation of the universe could be a pathway of detecting the telltale effects of gravitons.
from the abstract :
We argue here, however, that measurement of polarization of the Cosmic Microwave Background due to a long wavelength stochastic background of gravitational waves from Inflation in the Early Universe would firmly establish the quantization of gravity.
The BICEP2 experiment measures polarization, and if the future measurements show gravitational waves the quantized nature of gravity, and therefore gravitons, will have been detected.
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What actually is the event that we think we have detected with gravitational waves? This answer shows the "event" that is creating excitement. It looks to the untrained eye like a single "blip" on a detector. It appears to last less than a second.
It is, later in the answer, referred to as a "black hole merger".
Are we seriously saying that two black holes merged in under a second?
| The signal was fit to that produced by two black holes merging, with each having a mass of roughly 30 times that of our sun. For a solar mass the Schwarzschild radius is about 3 km. So that means the black holes, if off by themselves, would have an event horizon radius of about 90 km.
Only right near the end of their fate during the merger was the gravitational wave strong enough to make it above the detector noise threshold. Yes gravitational waves were emitted during the entire process, but we can only detect the violent end which releases the strongest waves.
In the process of merging, the black holes were moving relativistically. So moving on the order of 100 km many times in that short chirp recorded is quite reasonable.
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Would you hear a gravitational wave, if its amplitude and frequency were suitable? If there was a source of a continuous gravitational wave at (say) 50hz, and amplitude of say a micrometer (a typical sound wave displacement, I think), and you were nearby (standing happily on a planet in an atmosphere), with your ear pointing to the source, would you hear it?
It seems to me that since the gravitational wave is reducing and increasing the distance between points in the atmosphere right at your eardrum, surely the density and pressure of the air there is likewise increasing and decreasing, so you might expect to hear it. What I can't "intuit" is whether you would actually hear it due to the fact that you yourself are also being distorted.
My tentative conclusion is that you would hear it. At any given time, there appears to be a pressure differential across your eardrum due to this distortion in space pressurising the materials - so ... deflection?
(note: I know that in the recent LIGO announcement they talked about "hearing" the waves, but this is something completely different: an electro-acoustic rendition of the waveform. I'm asking about direct physical sensing.)
| I don't see how you could hear gravity waves even if they were of the appropriate frequency and amplitude. Hearing depends on the motion of hairs in the vestibular system, specifically the relative motion between the hair and its attachment in the cochlea. If a gravitational wave were to pass, all parts of the organ would move together and the hairs would not transmit a signal to the nervous system. Therefore you could not hear a gravitational wave. That's the way I see it.
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Is the Landauer limit reversible As I understand it, the Landauer limit, $kTln(2)$, is the minimum amount of energy to erase a bit. Is it also the minimum amount to create a bit? I'm asking statistical, like Avogadro's number of bits, i.e., $RTln(2)$.
My thinking: $RTln(2)$ is the work required, probably both ways (create and erase), but I don't know whether that is the same as the energy. By "same" I mean classical like heat, like something that can be used in $E=mc^{2}$.
| To create a bit in some defined state like 0 takes energy $kT\ln(2)+W$, where $W$ is the work required to create whatever the bit is made of (a flip-flop, a particle with spin, etc.). The first factor is just the Landauer cost of setting it in a definite state. As you say for a mole of bits you will have to pay at least $RT\ln(2)$ Joules.
$W$ will be problem dependent, and could in principle be zero: we can just name existing particles with spin as our bits without doing anything to them ("That electron over there will be bit 1, the other one bit 2, and the one over on Mercury bit 3..."). That kind of arbitrary naming still has the problem that we need to remember which electron is which bit, paying a Landauer cost for the naming, but often the indexing of bits can be done implicitly like when they are all spins along a crystal lattice we just found and decided to make our memory.
Since the work $W$ of making a bit typically involves irreversibly changing multiple degrees of freedom while the Landauer limit is about just one degree of freedom, most of the time $W \gg kT\ln(2)$.
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Somewhat unusual projectile motion question
A stone must fly over two walls of height $h_1$ and $h_2$ $(h_2~ > ~h_1)$ from the side of the lower wall. The distance between the upper points of the two walls near which the stone's trajectory lies is $L$. Find the minimum velocity of the stone. (source: AN Matveev's Mechanics and Relativity).
Answer:$\sqrt{g(h_1+h_2+L)}$, where $g$ is acceleration due to gravity
The rather terse nature of the problem statement is very typical of Russian texts. I'm an intermediate/advanced self-learner and I love to wrangle with these kinds of problems. This problem seemed like a basic projectile motion problem but this has got me in knots - If the limiting height is $h_2$, why does the problem need $h_1$? And, there is no angle to use either. What's the significance of the requirement for minimum velocity? Clearly, it has to cross the walls, and based on what I see, $h_1$ likely lies below the trajectory and $h_2$ must just touch the trajectory and this can potentially give the velocity. But, I'm unable to find the approach to solve this. Can anyone provide a way to think about this problem?
| The problem is kind of hard to solve if one is not careful. One can write the satisfying equations but still might not be able to find the answer. I will outline a way to this.
The idea is that at the minimum required velocity the projectile just touches the top of the walls. Let the velocity and angle of projection at the origin be $v_0$ and $\theta _0$. From the origin the projectile goes and touches the top of the first wall. You can view the rest of its motion as a projectile projected from the top of the first wall with velocity $v_1$ at an angle $\theta_1$(both are unknown at this point). The following relation holds $$ v_1 ^2 = v_0 ^2 -2gh_1.$$
Define $\sin(\phi) = \frac{h_2 - h_1}{L}$.
The projectile should now go and touch the top of the second wall which means w.r.t the new origin it must pass through the point $(L cos(\phi), h_2 - h_1)$. You can plug this in to the equation of projectile parabola to get.
$$h_2 - h_1 = \tan(\theta_1)[L\cos(\phi)] - \frac{gL^2\cos^2(\phi)}{2(v_0^2 -2gh_1)\cos^2(\theta_1)}$$.
Now rearrange this and write $v_0$ in terms of $\theta_1$. Everything else are parameters given in the question. Use calculus to minimize the value of $v_0$. You'll get the answer. Let me know if anything is not clear.
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Vacuum is not really empty Vacuum should contain something in it. Because nothing is perfectly empty that's what I feel, but what is there left in it? Is there any matter or its just enegry. Can energy be pulled out of some space?
| In quantum field theory, the vacuum is the state containing exactly zero particles anywhere in space and at all times. Since it is an eigenstate of the number operator, there is no uncertainty at all about this.
On the other hand, empty space between matter (i.e., what is informally called a vacuum) is never completely empty; it is still filled with the quantum fields emanating from the matter. Just like the space between the sun and the planets is not empty but filled with the gravitational field.
If this field is strong enough one can extract energy from it. For example, a ball falling in a conventional vacuum gains kinetic energy from the gravitational field.
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Homogenuous Maxwell Equations in the Language of Differential Forms
*
*I understand that if I define electric field to be $E=E_i dx^i$, magnetic field to be
$B=B_1 dx^2 \wedge dx^3 + B_2 dx^3 \wedge dx^1 + B_3 dx^1 \wedge dx^2 $, and field strength to be $F= dx^0 \wedge E + B$, I would get the two homogenuous Maxwell equations from $dF=0$. The last equation is a nontrivial equation.
*However I've read somewhere else that if I define the vector potential to be the one form $A=A_\mu dx^{\mu}$, and the field strength to be $F=dA$ I would get the two homogenuous Maxwell equations.
My questions:
*
*First, with this second definition the equation $dF=0$ is trivially true since $d^2=0$ and I don't understand how this would give me anything non-trivial.
*Secondly, from the second definition if I write $F$ in components I would have [*]:
$$
F=dA=\partial_{\beta}A_{\mu} dx^{\beta} \wedge dx^{\mu}\\
= \partial_0 A_i dx^0 \wedge dx^i + \partial_i A_0 dx^i \wedge dx^0 + \partial_j A_i dx^j \wedge dx^i\\
= \partial_0 A_i dx^0 \wedge dx^i + x^0 \wedge E + B
$$
which has the first term extra compared to the first definition, and I don't have any reason that this term is zero. So, what am I missing here? Which of the the two above approaches are correct?
[*] I use Greek letter super/subscripts for 4 space-time components and small english letters for 3 space componenets.
| *
*Yes, written in terms of the gauge potential $A_{\mu}$, the source-free Maxwell equations become trivially satisfied.
*It seems OP is using the electrostatic definition of $E_i$. In full electromagnetism, besides the $\partial_i A_0$ term, there is also a $\partial_0 A_i$ term in the definition of $E_i$.
See also e.g. my Phys.SE answer here.
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If pencil tip is heated why doesn't it write? Why doesn't a pencil write if its tip is heated in a candle flame?
| The reason must come from the hardening it experiments under fire.
Pencils' leads (the writing core) are made today of a mixture that contains clay (see for example this patent) which hardens under the heat, but mose importantly, the compound will loose the softness because of the separation of salts under the heat.
The mixed results come from the fact that there is no unique definition of the mixture and different manufacturers will use different ones. But you should expect that soft leads, which seem to have larger content of clay, will be more affected by heat.
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What is the dielectric constant of a pure conductor? Dielectric constant is the ratio of permittivity of a medium to the permittivity of free space. How to find dielectric constant of a conductor?
| The permittivity of a conductor is infinite.
Let the value of an external electric field in free space (relative permittivity = 1) be $E$.
If this is applied to a material of relative permittivity $\epsilon_r$ then the electric field in the material is $\dfrac {E}{\epsilon_r}$
Inside a conductor the electric field is zero hence its relative permittivity is infinite.
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Why absorption spectum is not identical to emission spectrum? Hydrogen emission and absorption spectral lines are typically depicted as the same:
(source)
However, in more complex systems, the emission and absorption spectra are significantly different. For example:
Absorption and photoluminescence spectra of DCJTB doped in a PS film.
(source)
What causes this shift of $\lambda_{\text{max}}$ etc. between absorption and emission?
| As @MikaelKuisma mentioned, the reason for this difference is the contribution of nuclear vibrational overlap to the transition moment. When an electron is excited from a ground state $S_0$ to the first excited state $S_1$, the bond is stretched and the internuclear separation increases:
(source: Martin Pope and Charles E. Swenberg, Electronic Processes in Organic Crystals and Polymers, 1999)
Assuming harmonic oscillations, the overlap between each two vibrational wavefunctions (different energy level for different frequencies/number of nodes) is defined as the nuclear vibrational overlap, which in turn contributes to the total transition moment.
The intensity of absorption/emission is proportional to the transition moment squared, $d_{nm}^2$ (following Beer-Lambert's law):
$$I = I_0 10^{-\varepsilon c l}$$
$$A = \log \frac{I_0}{I} = \varepsilon c l \propto d_{nm}^2$$
which means that factors contributing to the transition moment are expressed in the absorption/emission spectra.
For more information see Franck-Condon principle.
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Practical time travel: time dilation above the event horizon Imagine there is a huge black hole with very small gravity gradient so that one doesn't get killed by spaghettification after even nearing the event horizon.
Now imagine a very curious creature wanting to know how the universe would evolve. Even though it isn't sure it will be able to return, it decided to travel just close enough to the event horizon to survive, and stay there until for as long as it can make.
For how far would the creature see the future? Is this even possible? What should one expect?
This question was inspired by Miller's planet from the movie Interstellar.
| time travel between horizon and sigularity = $\pi G M/c^3$.
i.e. a 100th of second for M = 1000 solar mass.
| {
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Confusion on Time and Ensemble Averages of Classical Harmonic Oscillator Assume we have a classical harmonic oscillator
$$ \ddot{x} = -k^2x.$$
Then the general solutions are of the form
$x(t) = x_0cos(kt) + \frac{v_0}{k}sin(kt)$ where $x_0$ and $v_0$ are initial conditions. Lets assume that $x_0=0$. The time average of any quantity $f$ is given by
$$
\frac{1}{T}\int_0^Tf(\frac{v_0}{k}sin(kt))dt.
$$
For instance average oscillation amplitude as $T\rightarrow \infty$ is
$$
<x^2>=\lim_{T\rightarrow \infty}\frac{1}{T}\int_0^Tv^2_0sin^2(kt)dt = \frac{v_0}{2k^2}.
$$
However if you consider the ensemble average using Boltzmann density $e^{-\beta(\frac{1}{2}mv^2 + k^2x)}$ you get that the average value is
$$
<x^2>=\frac{1}{\beta k^2}.
$$
Now this is a system near equilibrium (i.e the measure above is ergodic for this dynamics is what I am assuming) so shouldnt one get that the space average is equal to time average for all most all initial conditions.I could say that I am just choosing bad initial conditions but I can not see how one would generally produce a term like $\beta$ by choosing initial conditions correctly.
| First, your Hamiltonian is wrong; you should be getting $(\beta~k)^{-1}$ for the force $k x$.
Second, your definitions of $k$ are going to be slightly off; the Hamiltonian $\frac12 m v^2 + \frac12 k x^2$ corresponds to the dynamics $\ddot x = -\omega^2 x$ only for $\omega^2 = k/m.$ Best then to write $\frac12 m (v^2 + \omega^2 x^2)$ with $\langle x^2\rangle = (\beta m \omega^2)^{-1}.$
Third, your interpretation is wrong; the Boltzmann factor comes from the canonical ensemble which is derived from the microcanonical ensemble in the limit where it is connected to a large thermodynamic system maintained at a constant temperature; your spring in the first case is not connected to any such system but rather is undergoing simple harmonic motion as if undisturbed by anything.
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Why does LIGO have an arm length of few kilometers? Is the distance dependent on Gravitational Wave wavelength? Antennas for capturing radio waves need to have $\frac{\lambda}{2}$ length for optimum reception of signal. Does it imply LIGO arm length is $\frac{\lambda}{2}$ of Gravitational Wave it is trying to capture?
| Pretty close. The effective LIGO arm length is 1600km (the light beam is reflected forth and back 400 times). LIGO is most sensitive at approx. 150Hz (advancedligo.mit.edu/summary.html), which would be a wavelength of 2000km... so the LIGO arms are approx. $\lambda /2$. The noise minimum depends on the noise spectrum, of course, so the sensitivity max. won't be exactly where one would expect it.
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How does gravitational wave compress space time? My question came from the talk of how gravitational wave stretches and compresses space time.
Say there are two protons that are 1 centimeters apart, as a G-wave passes through them, would the electrostatic force experienced by the protons change?
What about Plank's constant? If two particles are x number of Plank distances apart, is the new x smaller than the old x?
| Suppose we choose our coordinates so both protons are on the $x$ axis, at $x = -0.5$cm and $x = +0.5$cm:
The distance $d$ between the protons is obviously 1cm - well, that may seem obvious but actually it's only true in flat spacetime. More generally the geometry of spacetime is described by a quantity called the metric tensor, $g_{\alpha\beta}$, and the proper distance between the two protons is given by:
$$ d = \int_\text{x=-0.5cm}^\text{x=0.5cm} \sqrt{g_\text{xx}}\,dx $$
In ordinary flat spacetime the value of $g_\text{xx}$ is constant at one, and the integral turns into:
$$ d = \int_\text{x=-0.5cm}^\text{x=0.5cm} dx = 1 \,\text{cm} $$
as we expect. Now suppose we have a gravitational wave coming out of the screen towards you. This causes an oscillating change in the spacetime geometry that looks like this (picture from Wikipedia):
Whatb this is supposed to illustrate is that the value of $g_\text{xx}$ (and $g_\text{yy}$) oscillates with time, so it alternately becomes greater than one and less that one. In that case our integral becomes:
$$ d(t) = \int_\text{x=-0.5cm}^\text{x=0.5cm} \sqrt{g_\text{xx}(t)}\,dx $$
and our distance $d(t)$ is no longer a constant but oscillates above and below $d= 1\text{cm}$ as the gravitational wave passes through.
This isn't some mathematical trick, the gravitational wave really does cause the distance $d$ to change with time. If you shone a light ray between the protons and timed how long it took you'd find that time oscillated as well. Assuming the oscillation is slow compared to the time the light rays takes, the time the light ray would take to travel between the protons, $T$ would be:
$$ T(t) = \frac{d(t)}{c} $$
The electrostatic force between the protons would also change with time:
$$ F(t) = \frac{ke^2}{d^2(t)} $$
However Planck's constant is just a constant and this wouldn't be changed by the gravitational wave.
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Is it wrong to say that an electron can be a wave? In QM it is sometimes said that electrons are not waves but they behave like waves or that waves are a property of electrons. Perhaps it is better to speak of a wave function representing a particular quantum state.
But in the slit experiment it is obvious to see that electrons really are a (interfered) wave. So can you say that an electron is a wave? And is that valid for other particles, like photons? Or is it wrong to say an electron is a wave because it can be also a particle, and because something can't be both (a behaviour and a property)?
| The rules of how electrons move are analogous to waves because an internal state is cyclic and different possible paths are summed showing an interference pattern.
That's not the same as saying that electrons themselves are waves. The formulas for waves are used to explain where to find an electron.
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Force on current carrying square loop I'm asked to find force on square loop (side a) carrying current $I$, flowing counter clockwise, when we look down x-axis, lying in yz plane. the loop is centered at the origin. The magnetic field is given as:
$\vec{B} = kz\hat{x}$
Its solution states that force on left an right cancel each other .The force on top is $IaB=iak(a/2)$ pointing upward and the force on bottom is$IaB=-iak(a/2)$ also pointing upward .How the force on bottom is upward? (From where minus sign came?). By R.H.R it should be downward.
| You are correct, and the "solution" is apparently in error. The force on the top leg is equal and opposite to the force on the bottom leg. They cancel just as the forces on the left and right legs do. There is no net force on the loop.
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Question about source terms in scalar quantum field theory I'm having a bit of a mental block when trying to interpret the inhomogeneous Klein-Gordon equation $$(\Box +m^{2})\phi(x,t)=j(x,t)$$
In particular, how does one interpret the term on the right-hand side of the equation, $j(x,t)$ as a source term for the scalar field $\phi(x,t)$?
Is it analogous to the situation in classical electromagnetism? For example, the flux of an electric field through an closed surface, $S$ $\bigl(\oint_{S}(\mathbf{E}\cdot\mathbf{n})\,dS$, where $\mathbf{n}$ is the unit normal vector to the surface $S\bigr)$ is proportional to the total charge density $\rho$ in the enclosed volume $V$. This can be written in differential form as $$\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_{0}}$$ and so the charge density $\rho$ is interpreted as a source for a spatially varying electric field.
| Generally, source terms are what stands in the rhs of the differential equation $$\hat{\Theta}_x f(x) = j(x),$$ where $\hat{\Theta}_x$ is some linear differential operator (in your case $\hat{\Theta}_x = \Box_x + m^2$).
A generic solution of this equation is a sum of any solution of the homogeneous equation $\hat{\Theta}_x f(x) = 0$ (which usually is a superposition of plane waves) and any particular solution of the nonhomogeneous equation. Moreover, this particular solution is given by the Green's function:
$$ f(x) = f_0 (x) + \int d^4 y \, G(x, y) j(y) $$
where $G(x, y)$ is the Green's function of the differential operator $\hat{\Theta}_x$ (which means that it satisfies $\hat{\Theta}_x G(x, y) = \delta^{(4)}(x - y)$ as a distribution), and $f_0 (x)$ is any solution of the homogeneous equation (a superposition of plane waves).
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Is the local Lorentz transformation a general coordinate transformation? There is a saying in Nakahara's Geometry, Topology and Physics P371 about principal bundles and associated vector bundles:
In general relativity, the right action corresponds to the local Lorentz transformation while the left action corresponds to the general coordinate transformation.
Because the structure group right acts on Principal bundles and left acts on associated vector bundles.
But I don't think that the local Lorentz transformation is general coordinate transformation. Since for local Lorentz transformation, the structure group is $O^{+}_{\uparrow}(1,3)$ while for general coordinate transformation, the structure group is $GL(4,\mathbb{R})$.
So is the book wrong? Or I didn't understand correctly.
| Here I answer on the question from the title.
There is some feature which is related to the spinor representation of the proper Lorentz group. In fact, there is homomorphism $SL(2, C) \to SO(3,1)$, whose core contains two elements - unity and minus unity. The second one corresponds to representation
$$
T(N)= -T(-N), \quad N\in SL(2,C)
$$
and gives rise to applications of spinors in physics. For this representation we must introduce Lorentz group representation for which
$$
T(\Lambda)=\pm T(N), \quad \Lambda (N)\in SO(3,1)
$$
We can't throw out one of signs, since this breaks the continuity and the group property $T(\Lambda_{2})T(\Lambda_{1}) = T(\Lambda_{2}\Lambda_{1})$. So if we talk about the "usual" representations of the proper Lorentz group, we don't deal with spinor representations. If you mean the "usual" (unique) representations of the Lorentz group, then the statement that they are $GL(4,R)$ elements is correct. If you, however, talk about projective representations, which are realized in QFT and include "two-sign" representations, then the statement is incorrect: the group $GL(4, R)$ of general coordinate transformations doesn't contain representations similar to the spinor representation of the Lorentz group.
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Gravitational waves induce changes in the $h_{00}$ (time) component of the metric? I'm rather stumped by a subtle point regarding metric perturbations of GW. I'm well aware the GW are able to produce changes in the flat space metric, They are transverse and have planes of polarizations (namely $\times$ and $+$). However a friend and I are arguing over if GW can also produce warps in time, His argument is that in the quadruple approximation there is no $h_{00}$ component, $h_{00}$ is not a solution of the wave equation.
However he says once you leave the quadrupole approx regime such assumption can be valid. I believe that it is possible for a GW to induce both length variations and time variations just like what goes on in special relativity. Who is right? And can anyone provide a more solid explanation?
Thanks.
| Group theoretically you can split the graviton $h_{\mu \nu}$, where as usual $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$, in 3 irreducible representations of $SO(3)$:
$$h_{00} \rightarrow Spin \, 0 \; (Scalar)$$
$$h_{i0}\rightarrow Spin \, 1 \; (Vector)$$
$$h_{\mu \nu}\rightarrow Spin \, 2\; (Tensor)$$
Before choosing a gauge, look at the equations of motion: the equation for the spatial components is something like $\partial_i \partial_j h_{00}= \dots$, where the derivatives are spatial. So there aren't time derivatives acting on $h_{00}$. This means that $h_{00}$ is not a true degree of freedom of the gravitational field, but can be determined from the other components. It turns out that the only propagating degrees of freedom are the tensorial ones. In four dimension this means 6 d.o.f, but after gauge fixing only 2 survive.
Notice that this a feature of General Relativity. In alternatives theories of gravitation (for instance with additional terms $R^2, R^4, \dots$ in the lagrangian, or with more fields $\phi, \psi, \dots..$) the scalar d.o.f can became dynamical.
This discussion is parallel to the one in electrodynamics, with the vector potential $A^{\mu}$.
Reference: S. Carroll, Spacetime and Geometry
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Is there a theory where there are (recursively) infinitely smaller particles? So I read that electrons are just points, with no mass, and furthermore, protons look like they have some "size" but that's really 3 "point-like" quarks. We first thought atoms were the smallest possible particle. Then we thought the same about protons, electrons, and neutrons, I think, until we found quarks.
Is there a popular theory that assumes this pattern continues infinitely, with quarks being made of of some smaller things, and so on? Or otherwise, is there a popular theory that specifically predicts that this isn't the case?
| There is no such popular theory now, but I understand a somewhat similar idea was kicked around in the 70's, before the Standard Model was worked out. This was associated with the "Bootstrap model" of Geoffrey Chew. I am both too young and in the wrong subfield to give you a very detailed account of how this works, but here's a quick Wikipedia quote:
Geoffrey Chew and others went so far as to question the distinction between composite and elementary particles, advocating a "nuclear democracy" in which the idea that some particles were more elementary than others was discarded. Instead, they sought to derive as much information as possible about the strong interaction from plausible assumptions about the S-matrix, which describes what happens when particles of any sort collide, an approach advocated by Werner Heisenberg two decades earlier.
(from Bootstrap model)
However, this idea fell out of favor after QCD was worked out. Aside from practical issues, I think the relatively simple symmetries and few ingredients seen in the Standard model seemed to be at odds with this kind of idea. Unless those many lower-level particles somehow only combine to create fewer composite ones, something that is not seen at any other level of reality, we're just running out of room for simplification.
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Why do charged particles follow magnetic field lines? This may seem like a dumb question, but I can't think of the answer. The context I am curious about is the solar wind. Specifically particles flowing out of coronal holes and following the magnetic field lines arching out into space. Why do both the positive and negative particles follow these field lines and flow into space?
| Background
In the absence of an electric field, a charged particle experiences a force that is perpendicular to the magnetic field and its velocity relative to that field, called the Lorentz force. This is given by:
$$
\mathbf{F}_{s} = q_{s} \ \mathbf{v}_{s} \times \mathbf{B} \tag{1}
$$
where $q_{s}$ is the charge of species $s$, $\mathbf{v}_{s}$ is the velocity of species $s$ with respect to the magnetic field, $\mathbf{B}$. We can approximate the plasma as a fluid, i.e., magnetohydrodynamics, which will help simplify some things.
In the fluid limit, we can show that Ohm's law, in a frame of reference moving relative to the fluid at velocity $\mathbf{U}$ (or equivalently, one can say we are at rest and the fluid moves at $\mathbf{U}$), is given by:
$$
\mathbf{j} = \sigma \left( \mathbf{E} + \mathbf{U} \times \mathbf{B} \right) \tag{2}
$$
where $\mathbf{j}$ is the current density and $\sigma$ is the electrical conductivity. Note that this is in the non-relativistic limit, where we can approximate by using gallilean transformations (e.g., $\mathbf{E}' = \mathbf{E} + \mathbf{U} \times \mathbf{B}$) instead of Lorentz transformations. We can then use Ampère's law combined with Ohm's law (e.g., Equation 2 above) to show:
$$
\nabla \times \mathbf{B} = \mu_{o} \ \sigma \left( \mathbf{E} + \mathbf{U} \times \mathbf{B} \right) \tag{3}
$$
where $\mu_{o}$ is the permeability of free space. We can take the curl of Equation 3 and combine with Faraday's law to show:
$$
\partial_{t} \mathbf{B} = \nabla \times \left( \mathbf{U} \times \mathbf{B} \right) + \frac{1}{\mu_{o} \ \sigma} \nabla^{2} \mathbf{B} \tag{4}
$$
In the limit as $\sigma \rightarrow \infty$, one can further show that:
$$
\frac{d \Phi_{B}}{dt} = \int_{S} \ dA \ \left[ \partial_{t} \mathbf{B} - \nabla \times \left( \mathbf{U} \times \mathbf{B} \right) \right] \cdot \hat{\mathbf{n}} = 0 \tag{5}
$$
where $\Phi_{B}$ is the magnetic flux and $dA$ is an arbitrary surface with unit normal vector, $\hat{\mathbf{n}}$. Equation 5 is known as the frozen-in condition because it states that the magnetic fields are tied to the fluid.
Answer
Why do both the positive and negative particles follow these field lines and flow into space?
The answer is two fold relating to the frozen-in condition being partially satisfied and the Lorentz force (Equation 1 above). Equation 1 shows that if a particle tries to move orthogonal to a magnetic field, the field will turn it back towards the field resulting in roughly circular motion (when considering only magnetic fields and only the perpendicular components of the velocity). There are intuitive reasons for why this should be so, as I previously stated in this answer.
If, however, a particle's velocity ($\mathbf{v}_{s}$) is parallel to the magnetic field, it will experience no force in the absence of electric and gravitational fields. Thus, it is very easy for particles to move along the magnetic field but difficult to move across it.
The frozen-in condition also shows that the field and particles are tied to each other, such that if one changes the other changes to compensate (assuming the changes are slow enough).
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How does the drift velocity of electrons in a conductor depend on the temperature? How does the drift velocity of electrons in a conductor depend on the temperature?
I have two contradicting views for this.
*
*First, we can say that increasing the temperature of the conductor will increase the kinetic energy of the electrons. Hence, their drift velocity should increase with increase in temperature.
*Or, from the relation $v_d = \frac{eE}{m}T$ ($T$ is the relaxation time) we can say that the drift velocity is directly proportional to the relaxation time. Increasing the temperature will obviously decrease the relaxation time - as collisions will become more frequent - and thus decrease the drift velocity. Hence, an increase in the temperature will cause a decrease in the drift velocity.
So which view is correct?
| Let us consider the following equation:
$$i=ev_dAn$$
where $i$ is the current passing through the material, $e$ the amount of charge on the charge carrier (here it's the charge on an electron), $v_d$ is the drift velocity, $A$ is the area of cross section of the material and $n$ is the number of charge carriers per unit volume of the material.
For the sake of simplicity, let's assume the material's dimension doesn't change on increasing the temperature. So $A$ is a constant. Also $e$ is invariant of temperature. In a conductor, we know that on increasing the temperature, the current passing through it decreases when the terminals are maintained at a constant potential difference. This is because its resistance increases.
Let's replace the constant terms in the above equation by $k$:
$$i=kv_dn$$
As $i$ decreases, the product $v_dn$ must also decrease to maintain the equality. In a conductor, the value of $n$ remains fairly constant and hence the decrease in current is due to decrease in the drift velocity. And hence the drift velocity decreases with increase in temperature. This is just a reverse reasoning based on our observations from experiments.
An interesting note on semiconductors: As we increase the temperature, the value of $n$ increases in a semiconductor. Even if $v_d$ decreases with increase in temperature, the increase in $n$ outshines it and hence the product $v_dn$ increases on the right hand side and finally the current increases with increase in temperature.
For further reading: Doris Jeanne Wagner and Rensselaer Polytechnic Institute
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Ammeter range and shunt resistance Its said that for an ammeter to give good reading, the full current in the circuit must pass through it. But if I am right, the ammeter is basically a galvanometer connected parallel to a very low resistance called a shunt. I am aware that connecting a low resistance in parallel will reduce effective resistance to a value lesser than the least resistance.
But in an ammeter, if the shunt is a low resistance (lesser than galvanometer's resistance), then most of the current would pass through the shunt than the galvanometer. Thus, the reading given by galvanometer would decrease (as its the component which gives deflection in an ammeter), which means that the reading of ammeter would decrease.
Is my interpretation correct? If its wrong please explain me where I have gone wrong.
Also, how will range and sensitivity of a an ammeter change if we increase or decrease shunt resistance?
| An Ammeter which is a current measuring device has some limitations. If the instrument is designed as a moving coil galvanometer -it can measure small currents and the maximum limit may be about 30 micro amperes.
Therefore one uses a shunt resistor to share larger part of the current and limited(within range ) current is allowed to pass through the ammeter.
one can select a range of shunt resistance for utilizing them in measuring current of different ranges.
the ammeter can be calibrated along with different shunts and the same instrument can have switching device for different ranges.
Regarding sensitivity of an ammeter usually its defined as amount of current required for a full scale deflection- and it varies from 10 micro-Ampere to 30 micro-Ampere .
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In solids, is it phonons, or is it the oscillations of electrons in bands, that emit most of the blackbody radiation? In solids (most any object we see), which tends to emit most of the blackbody radiation: phonons (atomic, or molecular dipole, lattice vibrations) or oscillating electrons in their energy bands?
| Excellent question. The excitations involved are from occupied conduction (metals) or valence states (insulators, semiconductors) to unoccupied conduction band states. These form a continuum so can emit a continuous black body spectrum. These excitations are also responsible for the dielectric response.
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How is a human voice unique? Well, I am quite new to concepts of vocal sounds. From the physics point of view I believe a sound has two basic parameters i.e, frequency and amplitude.
Considering the end sound wave produced by human voice it must have frequency and amplitude as parameters. Well, when a human can speak in multiple frequencies (multiple pitches) and amplitudes (multiple volumes) I was wondering what makes every human voice unique?
Even if two persons produce a sustained note(say a same music note) their voices can be easily distinguished.So why do the voices seem different?
Are there any others parameters that distinguishes or do I have a misconception?
| A "pure tone" is a sound that has a single sine function as its pressure profile. The human voice is not a pure tone; it is a superposition of many different sine waves with different frequencies and different amplitudes. Here is an image illustrating how many sine waves of different frequencies can combine to make a more complicated waveform like the human voice:
(image credit)
Thus a human voice has many more parameters than just a single amplitude and frequency. It has many amplitudes, one for each of many different frequencies (along with a phase for each as well). Furthermore, these amplitudes change over time as the human voice makes different sounds.
This picture, for example, is a "spectrogram" of a human voice.
(image credit: By Dvortygirl, Mysid - FFT'd in baudline; original sound by DvortygirlThis file was derived from:En-us-it's all Greek to me.ogg, CC BY-SA 3.0 )
The x-axis is time, the y-axis is frequency, and the intensity indicates the amplitude of each frequency component at each point in time. A pure tone would show up as a single solid horizontal line. You can see that the human voice is made of many many frequency components of various amplitudes.
This is the same reason a violin, oboe, and piano sound different even when they play "the same note". The musical terminology for the specific balance of different frequency components is known as "timbre".
See the Wikipedia article for further reading.
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Holonomic and Topological Quantum Computing In topological quantum computation, anyons are braided in spacetime, performing non-trivial evolutions of some degenerate groundstate.
In holonomic quantum computation, the system is braided in parameter space, performing non-trivial evolutions of some degenerate groundstate (via the holonomy of a non-abelian connection).
What are the key differences between the two?
| In brief, the holonomy depends on the area enclosed by the path in parameter space, and is therefore sensitive to perturbations of the path, e.g. its length. In contrast, a topological evolution is largely independent of the path geometry, and depends only on topological data, specifically homotopy (e.g. how many times a path encircles a singular point, which for a pair of anyons is $r=0$, with $r$ the relative coordinate). In particular, two homotopically equivalent paths lead to the same evolution. This makes topological quantum computing much more robust in principle.
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A counterexample to the law of stable equilibrium? This is the law of stable equilibrium, according to Hatsopoulos and Keenan:
A system having specified allowed states and an upper bound in volume can reach from any given state one and only one stable state and leave no net effect on its environment.
Consider the following system: two sealed containers of gas. Container A contains a weight on a raised platform. Container B contains a flywheel. A string-pulley system connects the flywheel to container A. Let our system be both containers combined.
Here are two options to reach stable equilibrium
*
*We slide the weight in A off the platform; it crashes to the bottom, raising the temperature of A.
*Instead, we slide the weight onto the string's hook, such that, as the weight gently falls, the flywheel spins and raises the temperature of B.
It seems we have reached two different stable states despite making no effect on the environment, contrary to the law. What am I missing here?
| The system is not in a stable state as long as the temperature in the two containers is different.
It might take some time for the heat to flow through the string.
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What is the difference between gravitational force and gravitational field? I see two different formulas describing gravity:
$$F=\frac{GMm}{r^{2}}$$
$$g=\frac{F}{m}$$
But I don't understand the difference between gravity as a force and its field as a vector.
| Gravitational force depends on both the source mass and the test mass while the field is purely a property of the source mass.
For concreteness, consider the gravitational force acting on a person of mass $m$ due to the Earth having mass $M$. The gravitational force on the person is then given by \begin{equation}F=\frac{GMm}{r^{2}}\end{equation} (I am ignoring the direction now). It says that in general, gravitational force magnitude depends on the source mass, test mass and their separation.
But there is more fruitful way of seeing this: Suppose the person does not exist. Then the Earth will, due to its mass $M$, spread a gravitational field everywhere. The gravitational field strength is given by \begin{equation}g=\frac{GM}{r^{2}}\end{equation}
This vector field clearly depends only on the Earth; it measures how much force the Earth will exert on a unit mass, hence $g$ is sometimes called gravitational force per unit mass. If you now put the person at distance $r$ away from the centre of the Earth, this person will fall under the influence of the field and experience a force equal to $F=mg$, and this gives back the first formula.
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Voltage Drop and Current Flow Assume we have a source @100 DC volts in series with a 200 ohm resistor. According to Ohm's Law, current flow in 0.5 amps. Voltage drop across this resistor is 100 volts and equal to the source voltage. If voltage drop is 100 volts(meaning none is left on the "outflow" or neg. side of the resistor), then how can current continue to flow to the neg. term ? There is no electrical pressure(voltage) left to push the electrons.
| This problem is an idealization; real batteries have an internal resistance, and so do the wires.
In this idealized circuit the wires have no resistance, nor do the connections. You can logically collapse them to points -- and this brings the battery terminal into contact with the resistor; the current has obeyed Kirchoff's rules, and has arrived at the battery.
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How can we define a velocity for quantum objects? I have a question about quantum mechanics: I know that velocity is defined as the change of position with time, $v = \frac{\mathrm{d}x}{\mathrm{d}t}$.
In quantum mechanics, the position of a particle is not certain, but of a statistical nature. How can we define a velocity for quantum objects?
| In the Heisenberg picture of quantum mechanics, the position operator is itself time-dependent, and you may just define the "velocity operator" $\dot{x}$ as in classical mechanics. However, the Heisenberg equation of motion says
$$ \dot{x} = \mathrm{i}[H,x]$$
and e.g. for a free particle with $H= \frac{p^2}{2m}$, we have $[H,x] \propto p$, so this velocity operator is just proportional to the momentum operator (as one would classical expect). In particular, it does not commute with the position operator, you cannot know the position and the velocity of a particle simultaneously to arbitrary precision.
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Rayleigh equation as explanation for sky being blue I've been reading up on the internet as to why the sky is blue. The answer usually cites Rayleigh scattering that I've checked on wikipedia: https://en.wikipedia.org/wiki/Rayleigh_scattering:
$$
I=I_0 \frac{1+\cos^2\theta}{2R^2}\left(\frac{2\pi}{\lambda}\right)^4\left(\frac{n^2-1}{n^2+2}\right)^2\left(\frac{d}{2}\right)^6
$$
This answer raises more questions in my mind, that I hope some people can help to answer.
First of all, I can't understand the $\lambda^{-4}$ dependence in that equation. It means that the scattered intensity goes to infinity as $\lambda\to 0$. It also means that the observed intensity $I$ can be greater than the incident intensity $I_0$.
On several web pages, the $\lambda^{-4}$ dependence has been cited to explain why the sky is blue, but that doesn't make sense either. According to this reasoning the sky should be purple or indigo which has a higher frequency than blue. I've seen another explanation online that says that the sunlight impinging on our atmosphere has less indigo frequency than blue. However I can see the indigo part of a rainbow; it doesn't seem significantly dimmer than the blue part, so sunlight must have a decent indigo frequency content and as mentioned, the 4th power is a very strong dependency. This argument says that the sky ought to be indigo.
A second question I have is concerning the angle dependency. $1+\cos^2 \theta$ has a maximum at zero and at 180, and a minimum at 90 degrees. According to this dependency the sky should look brightest when looking at 0 degrees (towards the sun), and 180 degrees (with the sun to your back), but it should have half the intensity at 90 degrees. This doesn't match with our experience of the sky. Given the hand-wavy nature of the explanations, I wonder if Rayleigh scattering truly is the explanation for why the sky is blue.
| Rayleigh Scattering was calculated using classical theories of EM radiation.
Perhaps you should also look at Quantum Spectroscopy explanations. Scattering is different from absorption, at dawn and dusk water vapour has a greater effect, hence "Red sky at Morning/Night'.
Also consider the sky colour on Mars, where very little water vapour, CO².
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Semiconductors, Solid-State Physics We know, that conductors, conduct because their valence energy band is "half" full, and k ("wave vector") can increase and therefore the electrons under the influence of a electric field can "move", and similarly insulators wont conduct, because ther valence band is full, and we have a "big" energy gap between the valence and conductivity band.
My question: Why can't under the infulence of a electric field electrons "jump" onto the next energy band (that's the conductivity band) , where their k ("wave vector") can increase "more freely" and therefore conduct electricity?
| You said it here:
we have a "big" energy gap between the valence and conductivity band.
If you supply enough energy, electrons will jump to the conduction band (become excited). Semiconductors and insulators do not have many differences, it is simply called insulators when the gab is big, and semiconductors when small.
Semiconducting materials may react with sunlight (photovoltaic effect in solar cells), while inductors (like air) has become conducting when you see a lightning flash.
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What is this optical phenomenon, as looked from outer space? I just saw a photo from NASA. This is it.
This is beautiful. But i am not being able to understand, what are those red and green lines above the earth surface? These cannot be due to Rayleigh scattering of course as there are no particles to do that.
Moreover, just above the surface of the earth, we see a blue colour( which is very faint ) . Is this due to Rayleigh scattering?
PS- I can assure you, that these are not photographic effects, so please don’t take it into account.
| This optical phenomenon is called "airglow". A couple of other images, both also taken from the ISS:
(source: nasa.gov)
The blue band just next to the Earth in the second image is caused by Rayleigh scattering, which occurs primarily in the troposphere. The blue band is just barely visible in the first image, which was taken much later at night. This blue band is not airglow. The airglow starts in the stratosphere and continues for hundreds of kilometers up.
You can see airglow from the surface of the Earth. It gets in the way of astronomy, even in the darkest of places. All together, airglow is ten times brighter than is the light from all of the stars in the night sky. Airglow is one of the key motivators for space-based telescopes.
Airglow is, as the name suggests, glowing air. The orangish/yellowish band closest to Earth is sodium glowing at 589nm. The sodium comes from micrometeoroids that disintegrate into tiny particles 80 to 105 kilometers above the surface. Just above is a greenish band caused by excited oxygen atoms glowing at 558nm. Sometimes a thin blue band appears above this greenish band, caused by molecular oxygen glowing at 464 to 467 nm (multiple spectral lines). High above, atomic oxygen glowing at ~630 nm (multiple spectral lines) forms the faint red part of the airglow.
Aside: The images are clickable. Hover your cursor over the images to see a description of the image and of the link.
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Can a coffee cup really jump off the table? I've been reading Nassim Nicholas Taleb's book The Black Swan which largely concerns the uncertainty of social and economic systems (and the futility of predicting outcome); the Black Swan being the rare event.
But in one of the chapters Taleb talks about the motion of atoms and probabilities of the trajectories of atoms. He illustrates using a coffee cup saying that the most likely atomic trajectories are such that they all cancel one another and the cup remains motionless on the table. But he cites there is the very rare possibility that all the atoms in the cup move in the same direction at the same time - and the coffee cup jumps off the table.
I have a basic understanding of entropy and Maxwell's Demon, but question that if Taleb's cited event were to ever happen would there be enough momentum, energy to actually cause the cup to fly off the table? I know atoms can move very fast, but then they don't tend to move very far for very long.
Is Taleb's illustration possible?
| I have seen some wet plastic coffee cups with very hot coffee poured inside moves horizontally. This happens not during the coffee is being poured but after a minute or two.
I theorized this is possibly due to trapped air between cup base and smooth table surface get expanded and create air cushion on which the cup slides until the air finds its way out from the water-seal.
But no way in this universe the cup can take off vertically because of all molecules strike. Its nonsense.
| {
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What is time dilation really? Please will someone explain what time dilation really is and how it occurs? There are lots of questions and answers going into how to calculate time dilation, but none that give an intuitive feel for how it happens.
|
Asked by lucas:
I know nothing about relativity but I cannot accept that there is a phenomenon called time dilation. However I have no problem with it because of mathematics behind it. I have no problem if time is dilated, because I don't know what time is. But I wonder when they say a clock will work slowly with respect to the other same clock if its speed is higher.
*
*Which kind of clocks they mean? Analog clock, digital clock, etc.
*As far as I know some mechanical clocks work by a torsion spring inside them. So, how does the material of the spring know that it must unroll slowly at higher speed? Does higher speed change chemical structure or physical properties of the material of the spring?
Answered by Gennaro Tedesco:
The clock obviously neither slows nor speeds. That is only unfortunate terminology to mean that time intervals depend on the reference frame and different observers in different reference frames may measure different time intervals if in relative motion with respect to each other.
| {
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How much energy does proton - carbon 12 fusion produce? Page 25 of this document from the California Institute of Technology says that proton-carbon 12 fusion releases 7.54 MeV, while Wikipedia says it releases 1.95 MeV. Which one is correct?
| OK, you are referring to the fusion reaction
$^{12}$C + $^{1}$H $\rightarrow$ $^{13}$N + $\gamma$.
Let's neglect the $\gamma$ and insert the atomic masses
$12\cdot u + 1.00782503223\cdot u = 13.005738609\cdot u + x$,
where $x$ is an upper boundary for the released energy (since we neglected the $\gamma$). The data for the atomic masses is from the NIST website and from wikipedia.
Solving that equation yields
$x=0.00208642323\cdot u$
which corresponds to ($1\cdot u$ corresponds approximately to $931.5$ MeV)
$x\approx 1.94 $ MeV.
Therefore, the wikipedia value seems to be right.
| {
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Impulse Equations A solid sphere of mass $m$ rolls without slipping on a horizontal surface and collides with a vertical wall, elastically. The coefficient of friction between the sphere and wall is $\mu$. After the collision, the sphere follows a parabolic trajectory, with range $R$. What is the value of $\mu$ to maximize $R$?
Since the collision is elastic, we can say impulse normal impulse $J = \Delta P = 2mv$.
As frictional impulse is $\mu$ times normal impulse, $J'=2mv\mu$ (upwards).
Therefore, sphere acts like projectile with horizontal velocity $v$ and vertical velocity $2v\mu$. To maximize $R = 4\mu v/g$, $\mu$ should be maximum i.e.$1$.
However, this is not correct. What am I missing here?
|
only thing you left was the fact that friction stops acting when slipping stops. Range is maximised when upward velocity is max as horizontal is fixed
| {
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Gauss law in gravitation Is it possible to use Gauss's law of electromagnetism, (The net electric flux through any closed surface is equal to $1⁄\epsilon$ times the net electric charge enclosed within that surface.) to calculate the gravitational field at point by making certain changes, i.e, by replacing electric flux with gravitational flux, $1⁄\epsilon$ with $1/(4\pi\,G)$, and charge with mass?
| Yes, you can use Gauss's law for gravity.
$$\nabla\cdot\vec{g} = 4 \pi\, G\, \rho $$
or
$$ \oint \vec{g}\cdot\mathrm{d}\vec{a} = 4 \pi\, G\, M_\mathrm{enc} $$
where $\vec{g}$ is the gravitational field (equivalently, acceleration due to gravity), $\rho$ is mass density, and $M_\mathrm{enc}$ is the total mass enclosed by the Gaussian surface.
When you make the comparison to Gauss's law for electric fields, you can see how the constants work out the way that they do:
$$E = \frac{1}{4\pi\, \epsilon_0} \frac{Q}{r^2}, \quad\quad g = G\, \frac{M}{r^2}, $$
so $1/\epsilon_0\rightarrow 4\pi\, G$.
One common use for Gauss's law for gravity is to determine the gravitational field strength at a given depth inside the Earth. It is very similar to the calculation for the electric field inside a charged, insulating sphere.
| {
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Can a qubit have an imaginary component? My knowledge of linear algebra is limited and my physics knowledge mostly comes from high school and Youtube so please bear with me.
In the equation $$|x\rangle = a|0\rangle+b|1\rangle,$$ I read that $a$ and $b$ can be complex. Does that mean that they can be imaginary? If so, does something like that have any sort of non-abstract meaning? Would it be some sort of imaginary vector that can't actually be visualized?
| A complex number is equivalent to a vector in $\mathbb{R}^2$. Therefore you can visualize these vectors very easily. Of course, once you want to visualize a two-dimensional vector of complex numbers, you'd need four real dimensions, which is hard to "visualize" in three dimensions. However, for qubits, you don't really have four dimensions: A global phase leaves the state invariant and the state needs to be normalized, which means that any pure state is given by only two real dimensions. This leads naturally to the visualisation of a qubit on the Bloch sphere.
The term "imaginary" does not mean that it can't be visualised - it is a bit of an unfortunate name, because there is nothing imaginary about it. Do you need complex numbers in physics? Yes and no. See this question and answer for more details: QM without complex numbers
One thing you need to understand though is that the mathematical desciption of an object is always an abstraction. After all, the vector lies in a "Hilbert space", a purely mathematical concept that you cannot "see" in the real world.
| {
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Merging and Splitting of Black Holes We know that two black holes can merge to form a single black hole with the release of vast amounts of energy. Can the reverse happen? That is can it happen that large energy is supplied to a black hole and it splits into two?
| Given that as far as anyone knows there is only one black hole in the centre of every galaxy, if it were to split into two it would split the galaxy. We know that galaxies often collide and can see one feeding into another, but surely there is no evidence of a split galaxy. It would look rather like a cell dividing wouldn't it?
| {
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Integral of a velocity profile? Part of my fluid mechanics homework asks me to solve:
$${\partial u\over \partial x} = 0$$
Which represents how the velocity profile, u, changes in the x. I'm not sure whether you can integrate partial derivatives but my intuition says that you just integrate and get:
$$u=C$$
Is this logic correct?
| Velocity u could be a a function of time AND space in general. From the first relation mentioned in the problem, all you can infer is that velocity does not depend on space. All remains would be its time-dependency. You can also show all this in partial-derivative notation.
| {
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Is there a maximum energy for a relativistic particle? So I was told today that the Standard Model breaks down at really, really high energies. The lecturer mentioned particles such as electrons hypothetically having energies equivalent to that of entire stars and it got me thinking, surely the maximum theoretical energy any particle can have is limited by the speed of light. I understand I'm talking about only kinetic energy here, but I fail to see how any other form of energy is relevant at those sorts of speeds. I did wonder if massive particles at the speed of light have infinite energy (which satisfies my question if this is the case), but I don't see that sort of behaviour from Einstein's relativistic mass-energy relation.
So my question is- is there a maximum theoretical energy particles can have?
| There is no maximum energy of a freely moving massive particle in special relativity.
The relativistic energy of a particle of rest mass $m$ moving in your frame at speed $v$ is given by $E=\gamma m c^2$ where $\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$. If you look closely at $\gamma$ you will see that it is not defined at $v=c$ ($c$ is the speed of light), and that $\underset{v\rightarrow c}{\lim}\gamma = \infty$. From this you conclude that the energy of a particle increases without bounds as its speed approaches that of light.
From this you see that: 1) you can always increase the energy of a massive particle by accelerating it, 2) you need more and more energy to approach $c$, so no massive particle can travel at the speed of light.
| {
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Does increasing the width of the slit increase the intensity of the light passing through it considering the slit as a source I did an experiment in which I tried to show that the visibility of the interference fringes is related to the relative slit width in a double slit interferometer. In other words if one slit had a width that was twice the width of the other, would that change the visibility of the interference fringes. I thought it would because I assumed that by changing the width of the slits the intensity of the light passing through the slit would change, but now I am not so sure about that. All of the equations I have seen regarding intensity in a double slit interferometer do not indicate a linear relationship between intensity and slit width.
| The fringe pattern is simply the fourier transform of the slit aperture, in 1 or two dimensions.
The fourier transform of a single slit ~ $sinc(x/a)$ where a is the width of the slit.
The fourier transform of a double (equal) slit ~ $cos(x/a)$ eqn.(1)
The fourier transform of two slits , one twice the width of the other is something like $a*cos(x/a) + b*sin(x/a)$
Which, using well known trigonometric identities is simply eqn. (1) displaced slightly along the x direction
there is no change in intensity
| {
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Why fuse needs to be slightly more able to bear current than the appliance? if I have a bulb with rated current of $4 A$ then why I need to connect an $5A$ fuse?( which has slightly more ability to bear current.)
| The actual amount of current that will cause a fuse with a given rating to open is a random variable that depends on random variables in the manufacturing process. Same goes for the light bulb: The actual amount of current that the light bulb will draw from a given supply voltage is variable.
In order to prevent blown fuses, you want the nominal rating of the fuse to be somewhat higher than the nominal rating of the light bulb, and in order to prevent house fires, you want the nominal capacity of the wires in the walls to be a bit higher than the rating of the fuse.
| {
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Does the rotational form of Newton's second law always hold? Does the equation,
$$\vec \tau=I\vec \alpha$$
(rotational form of the second law) always hold? If not, in what conditions does it hold true?
|
Does the equation,$$\vec \tau = \mathrm I \vec\alpha$$(rotational form of the second law) always hold?
No. This assumes a rigid body, that the body is rotating about one of the principal axes of the rigid body, and that the external torque is parallel (or anti-parallel) to the body's rotation axis.
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Putting a capacitor into a strong magnetic field, will this change the capacity? I'm wondering, does a magnetic field change the number of electrons, placed and displaced on the two plates of a capacitor. To prove or disprove this, I think the capacitor could be connected to an other capacitor outside the magnetic field and it has to be measured the current flowing between the capacitors during the increase and decrease of the magnetic field.
Edit: Was such an experiment carried out?
|
does a magnetic field change the number of electrons, stored on a capacitor.
No, because ...
*
*The purpose of a capacitor is not to store electrons but to store energy. A "charged" capacitor contains the same number of electrons as an "uncharged" capacitor.
*Electrons don't easily disappear or appear, they have to be moved somewhere. If you move the electrons around, you change the amount of stored energy, you don't change the capacitance.
*The capacitance depends on factors like plate-area, separation-distance and permittivity of separator. These are not normally affected by a magnetic field.
$$C = \epsilon \frac{A}{d}$$
| {
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Why so much electric charge? I'm working on the Millikan experiment using the following apparatus:
http://www.unitedsci.com/product-catalog/millikan-oil-drop-apparatus-0
when I compute the value of charge of the oil drops, the vast majority is around 200 times the charge of the electron, but according to the manual and many websites with just 10 times is bad data. So my question is what could I been doing wrong?
At first I thought it was because I was using the mercury lamp all the time during the experiment, but after turning it off the oil drops become neutral again. Also I used the data in the manual to make sure my equations were correct. So if someone has experienced this before let me know.
| I'm not sure how this manufactured device looks from the inside. During my undergraduate my friend and I built our own device. But I can give you some possible explanations.
First of all there is a slight chance that the mercury lamp causes this due to the emission of UV. Is it build in? Because it also generates a lot of heat (not responsible for the charge). Second, what source for ionization do you use? Some radiation sources emit a lot of radiation, this causes the oil drops to become charged with a huge amount of electrons. Another possibility is more mathematical. The best statistical analysis is minimizing $\chi^2$ distribution of the measured electron charges. More info on https://www.physics.uci.edu/~advanlab/millikan.pdf
| {
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Invariant polynomials of the Landau theory of phase transitions (crystal symmetry?) I'm convinced I'm missing something so obvious but here goes
Typically, one can define something like a "general" expansion of an order parameter, ${\boldsymbol \Gamma}$, up to 6th order as follows
$$F = \frac{1}{2} A_{ij} \Gamma_i \Gamma_j + \frac{1}{3} B_{ijk} \Gamma_i \Gamma_j \Gamma_k + \frac{1}{4} C_{ijkl} \Gamma_i \Gamma_j \Gamma_k \Gamma_l + \frac{1}{5} D_{ijklm} \Gamma_i \Gamma_j \Gamma_k \Gamma_l \Gamma_m + \frac{1}{6} E_{ijklmn} \Gamma_i \Gamma_j \Gamma_k \Gamma_l \Gamma_m \Gamma_n$$
Then, a symmetry consideration from a lattice can be applied, to reduce the above expression to a simpler form. Usually this is done by looking at a parent phase and then lowering the symmetry through an irreducible representation of a soft mode and getting a symmetry invariant polynomial.
For example, if the parent phase is cubic, sometimes defined as $O_h$, one can find, in three dimensions (${\boldsymbol \Gamma} = \left\{ \Gamma_x, \Gamma_y \Gamma_z \right\}$), that
$$F = \alpha_1 \left(\Gamma_x^2 + \Gamma_y^2 + \Gamma_z^2 \right) + \alpha_{11} \left(\Gamma_x^4 + \Gamma_y^4 + \Gamma_z^4 \right) + \alpha_{12} \left(\Gamma_x^2 \Gamma_y^2 + \Gamma_y^2 \Gamma_z^2 + \Gamma_x^2 \Gamma_z^2 \right)
+ \alpha_{111} \left(\Gamma_x^6 + \Gamma_y^6 + \Gamma_z^6 \right) + \alpha_{112} \left[\Gamma_x^4 \left(\Gamma_y^2 + \Gamma_z^2 \right) + \Gamma_y^4 \left(\Gamma_x^2 + \Gamma_z^2 \right) + \Gamma_z^4 \left(\Gamma_x^2 + \Gamma_y^2 \right) \right] + \alpha_{123} \left(\Gamma_x^2 \Gamma_y^2 \Gamma_z^2 \right) $$
is the free energy for a system whos order parameter obeys tetragonal symmetry. (I'm sure there is some Voight transformation on notation between the general expression and this example).
What are the exact operations one would have to do if say we wanted the above expression? How about a system where the order parameter obeys rhombohedral symmetry? Hexagonal symmetry? What if the parent phase is one of the above symmetries and we want to lower further?
Thanks!
| The brute-force method to find the symmetric polynomials of a point group is to start with the generic polynomial and perform a symmetrization over group elements. This can be done for small finite groups and does not require much knowledge about the representation theory.
Let $G$ be the point group (can be $O_h$, $T_d$, $D_{6h}$ etc) in consideration. For any element $g\in G$, first write down its three-dimensional representation $R(g)$, as a $3\times 3$ matrix acting on the vector $\Gamma=(\Gamma_x,\Gamma_y,\Gamma_z)$, s.t.
$$(g\Gamma)_i=\sum_{j}R(g)_{ij}\Gamma_j.$$
Then take a generic polynomial expansion of the free energy:
$$F = A^{(0)}_{ijk\cdots}\Gamma_{i}\Gamma_{j}\Gamma_{k}\cdots,$$
symmetrize the coefficient tensor $A^{(0)}_{ijk\cdots}$ by
$$A^{(0)}_{ijk\cdots}\to A^{(1)}_{ijk\cdots}=\frac{1}{|G|}\sum_{g\in G}A^{(0)}_{i'j'k'\cdots}R(g)_{ii'}R(g)_{jj'}R(g)_{kk'}\cdots ,$$
where $|G|$ is the order of the group $G$ (the number of elements in $G$). Then the symmetrized coefficients $A^{(1)}_{ijk\cdots}$ will produce all (and only) symmetric polynomials to the desired order.
Example: Let us consider a simple example of the tetrahedral group $T_d$, which has 24 group elements. They are generated by two generators:
*
*the two-fold reflection (denoted as $a$) about the mirror plane normal to the $(1,1,0)$ vector,
*the three-fold roation (denoted as $b$) around the $(1,1,1)$ axis by $2\pi/3$.
With these, the tetrahedral group has the presentation $T_d=\langle a, b|a^2=b^3=(ab)^4=1\rangle$. To obtain the matrix representations $R(g)$ for all group elements $g\in T_d$, we first find the representations for the generators $a$ and $b$, which can be done in Mathematica quite easily:
a = ReflectionMatrix[{1, 1, 0}];
b = RotationMatrix[2 Pi/3, {1, 1, 1}];
Then the whole $T_d$ group can be generated from $a$ and $b$ by consecutive multiplications until closure, which can be formulated as a problem of finding the fixed-point under right multiplication of the generators.
G = FixedPoint[DeleteDuplicates@Join[#, #.a, #.b] &, {IdentityMatrix[3]}];
$G$ now contains the 24 elements of $T_d$ group. Now we can use it to symmetrize the coefficient tensors $A^{(0)}_{ijk\cdots}$. Let us start from rank-2 tensors (corresponding to quadratic polynomials). How should we initialize the tensor $A^{(0)}_{ij}$? One nice trick is to initialize $A^{(0)}_{ij}$ by random numbers!
A0 = RandomReal[{}, {3, 3}];
d = ArrayDepth[A0];
The tensor rank (depth) is kept as $d$, which will be useful later. Given the group $G$ and the tensor rank $d$, we can construct the symmetrization operator (also known as the projection operator) $P_G$ for that rank:
PG = Sum[TensorProduct @@ ConstantArray[SparseArray@g, d], {g, G}];
The final step is to apply the symmetrization operator $P_G$ to the random tensor $A^{(0)}$ by tensor contractions:
A1 = TensorContract[TensorProduct[PG, A0], {2 # - 1, 2 d + #} & /@ Range@d];
To read out the symmetric polynomials from the symmetrized tensor $A^{(1)}$, we create an order parameter $r$ to contract with the coefficient tensor.
r = {x, y, z};
Dot @@ Prepend[A1]@ConstantArray[r, d] // Expand
The result will be something like
11.6381 x^2 + 11.6381 y^2 + 11.6381 z^2
After factoring out the common coefficient (which is random), we know that $x^2+y^2+z^2$ is the only $T_d$ symmetric polynomial on the quadratic level. The above procedure can be extended to higher order polynomials by increasing the rank of the initial random tensor $A^{(0)}$. For example, to obtain symmetric quartic polynomials, we need to set
A0 = RandomReal[{}, {3, 3, 3, 3}];
and rerun the program. The result will look like
18.2274 x^4 + 66.6814 x^2 y^2 + 18.2274 y^4 + 66.6814 x^2 z^2
+ 66.6814 y^2 z^2 + 18.2274 z^4
Different symmetric polynomials can be identified by different coefficients. So we know there are two $T_d$ symmetric polynomials on the quartic level: $x^4+y^4+z^4$ and $x^2 y^2+x^2 z^2+y^2 z^2$. In this way, one can find the $T_d$ symmetric polynomials for the lowest few orders:
$$\begin{array}{cc}\text{order} & \text{symmetric polynomials}\\ 2 & x^2+y^2+z^2\\ 3 & xyz\\ 4 & x^4+y^4+z^4\\ & x^2 y^2+x^2 z^2+y^2 z^2\\ 5 & xyz(x^2+y^2+z^2)\\6 & x^6+y^6+z^6\\ & x^4(y^2+z^2)+y^4(x^2+z^2)+z^4(x^2+y^2)\\ & x^2 y^2 z^2\\ \cdots & \cdots \end{array}$$
However, it worth mention that the computation complexity grows exponentially with the polynomial order. So very quickly we will get stuck and not able to proceed to higher orders.
A more clever and more systematic construction will require some knowledge of group representation theory. The idea is to start with the polynomial representation of each irreducible representation of the group and build up higher order polynomials by tensoring the irreducible representations. The symmetric polynomials will be found as the trivial representation. Useful tables for these can be found at this website: http://www.webqc.org/symmetrypointgroup-td.html.
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Can the quantum mechanical current density be imaginary? I am dealing with a situation where I get an imaginary transmission current density.
Is this possible?
Does it imply a zero transmission probability?
| The formula for the probability current is
$$
\boldsymbol j=\frac{1}{2m}(\psi^*\hat{\boldsymbol p} \psi-\psi\hat{\boldsymbol p}\psi^*) \tag{1}
$$
which is manifestly real. As a cross-check, if you write $\psi=R(x)\mathrm e^{iS(x)}$, then the current is given by
$$
\boldsymbol j=\frac{R^2}{m}\nabla S \tag{2}
$$
yet again, a real quantity.
Anyway, note that the transmission coefficient is defined as a quotient of currents$^1$, which means that if you get an imaginary current (because you use a different definition rather than $(1)$, which is fine), the quotient will be real, as one would expect. In the end, the global phase of $\boldsymbol j$ is unphysical, as mentioned in the comments.
$^1$: $$T=\frac{j_\text{out}}{j_\text{in}}$$
All the formulas are taken from the Wikipedia article probability current.
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Lead shielding very close to the Sun I understand that eventually lead would melt when it nears the sun. In a liquid state how effective would lead be in blocking radiation? Would it still be as effective as solid state of lead? What about lead in a gas state?
| All that really matters is the column density of the lead - how much mass per unit area you have. What this means is that if you change the state or density of lead then so long as you arrange it so that $\rho x$ is the same, where $x$ is the path traversed by the radiation, then the absorption due to the lead will be the same.
Everything might change if you ionise the lead, but not if you merely change it into an atomic gas.
| {
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Vierbeins in General Relativty; degrees of freedom? I am self-learning GR.
I want to ask if vierbeins $e^b_{\ \ \nu}$ need to satisfy any relations or if I am free to choose any type of vierbein I like
So I have been looking into tetrads again. I understand I can calculate $g_{\mu \nu}=e^a_{\ \ \mu}e^b_{\ \ \nu}\eta_{a b}$ and use it to transform into a local tetrad frame, hence simplifying some equations.
The great thing I learned about this is that when I transform into a locally minkowskian frame, I can use special relativistic formulas to calculate some quantities. Yet I can not find a source telling me how to choose the vierbein. Am I free to choose any vierbein which satisfies $g_{\mu \nu}=e^a_{\ \ \mu}e^b_{\ \ \nu}\eta_{a b}$?
Small note: I have not studied differential geometry :)
| Yes. You are free to choose any such vierbein. Note that any two choices of vierbeins are related by a local Lorentz transformation. This appears then as a gauge symmetry in tetrad formalism. By choosing a particular veilbein, you have essentially fixed the gauge.
| {
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When does the principle of superposition apply? I assumed from my general physics courses that the principle of superposition was just an empirical fact about forces. Then I could understand that derived quantities like the $E$ and $B$ fields would also obey it because, for instance: $$F_1 + F_2 = qE_1 + qE_2 = q(E_1+E_2) = F_{total} \\ \implies E_1 + E_2 = E_{total}$$But yesterday I saw that the Wikipedia page on gravitational potential stated that "the potential associated with a mass distribution is the superposition of the potentials of point masses." So apparently gravitational potential energy also obeys the superposition principle.
This leads me to wonder what all are the quantities that obey superposition. Do all types of energy obey it, for example? Better yet, is there some way of determining whether a given quantity (number/ vector/ etc) will obey the principle of superposition theoretically or do we need an empirical law for each?
Looking at the Wikipedia page on the superposition principle didn't help as it stated that all linear systems obeyed it. But how do we know whether a system is linear? I know how to determine whether a function is linear, but let's take for example gravitational potential energy: $$U_g = - \frac{GMm}{r}$$ This law has $3$ independent variables. It is linear in $M$ and $m$ but not in $r$. So how would I determine which of those variables needs to be linear for the gravitational potential energy to obey the principle of superposition?
| The principle of superposition is not obvious in any sense. However, it is an experimentally verified fact to a certain accuracy. If you look at Newton's law or Coulomb's law, it does not say anything about the fact that the net force is the sum of individual forces as if all other particles were absent. There is no reason for the net force not to be related non-linearly.
It is one of those fundamental principles that is taken for granted, similar to the Equivalence principle or Hamilton's principle.
| {
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Help! An 8 year old asked me how to build a nuclear power plant I would really like to give an explanation similar to this one.
Here's my current recipe:
(i) Mine uranium, for example take a rock from here (picture of uranium mine in Kazakhstan).
(ii) Put the rock in water. Then the water gets hot.
(iii) [Efficient way to explain that now we are done with the question]
This seems wrong, or the uranium mine would explode whenever there is a rainfall. Does one need to modify the rock first? Do I need some neutron source other than the rock itself to get the reaction started?
As soon as I have a concrete and correct description of how one actually does I think I can fill in with details about chain reactions et.c. if the child would still be interested to know more.
| You may need to explain the concept of a chain reaction, at least enough to explain why the rocks don't explode. Every fission event in Uranium generates neutrons. This occurs naturally at a slow rate, or can also be triggered by a Uranium atom getting hit by a neutron. The more densely packed the Uranium is, the more fission occurs (this is drastically simplified, skipping the issues of neutron speeds, but good enough to expand on later). Rocks with a lower Uranium density do heat up water, but not to the extreme that concentrated Uranium does. The larger the volume of Uranium, the more neutrons hit other Uranium atoms. The more Uranium there is per volume, the more neutrons hit other Uranium atoms. This is why we enrich Uranium for use in reactors.
Then you can show them this example of what chain reactions look like.
| {
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Neutron population growth in a reactor In nuclear reactors physics, the effective constant $k_{eff}$ is defined as
$$k_\text{eff} \equiv {\text{number of neutrons in one generation}\over \text{number of neutrons in the preceding genereation}}$$
And the equation that describes the change of the neutron population, the point-kinetics equation, can be obtained as
Consider a core in which the neutron cycle takes $l$ seconds to
complete. The change $\Delta n$ in the total number of thermal
neutrons in one cycle at time t is $(k_\text{eff}-1)n(t)$, where $n(t)$ is
the number of neutrons at the beginning of the cycle. Thus $$\Delta
n(t) \equiv l {dn(t) \over dt} = (k_\text{eff}-1) n(t) $$
(Fundamentals of Nuclear Science & Engineering by John K. Shultis, Richard E. Faw, Chapter 6.1)
which solution is
$$n(t) = n_0 \exp{\left[{k_\text{eff}-1\over l} t\right]}$$
so if we now calculate the rate between the population in one generation and the preceding we have
$${n(2l) \over n(l)} = e^{k_\text{eff}-1} $$
that only coincides with the first definition of $k_\text{eff}$ at first order.
Where is the condition $k_\text{eff} \rightarrow 1$ in the derivation of the point-kinetics equation, so both results are equal?
| It's okay if the two results are equal only at first order. The implicit assumption is that the length $\ell$ of the "neutron cycle time" is short enough that the higher-order terms are negligible. This is the standard calculus technique for deriving the exponential function; it shows up all over the place.
Note that if $k_\text{eff} \lesssim 1$ then the reaction dies away, and if $k_\text{eff} \gtrsim 1$ the reactor explodes, so $k_\text{eff} \approx 1$ is the case of interest for nuclear engineering anyway.
| {
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Could an asteroid land slowly on Earth's surface? The concept in my mind is that an asteroid is on a vector similar to Earth's, but slightly slower (e.g., 50kmh slower). As Earth passes it, it enters the atmosphere at a sharp angle, and since Earth was passing it, it just barely touches down due to Earth's gravity and atmospheric drag.
Given a large asteroid (e.g., 500 meters wide), is there any reason something like this couldn't happen? And, is there any evidence that it has happened?
| If the asteroid is in parallel to the orbit of the earth and at rest it will feel the gravitational attraction and will fall with velocity growing as $g\cdot t^2.$ This force will be there whatever the angle and velocity of the asteroid, centrifugal forces may make it miss the earth in a parabolic orbit, or be caught in an elliptical as the path of the satellites. To avoid falling on the earth with great velocity it would need not only to have a small velocity relative to earth but also an acceleration equal or larger and opposite to the acceleration of gravity.
| {
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Units don't match in the torsional spring energy! According to Wikipedia's description of torsion springs and according to my understanding of physics the energy of a torsional spring can be written as
$$U=\frac{1}{2}k \varphi^2$$
where $k$ is a constant with units of $\rm N\,m/rad$.
I am freaking here because if the energy of a torsional spring is really $k \varphi^2$ than the units are
$\rm (N\,m/rad) \cdot rad^2=Joule\cdot rad$. ??
What on earth am I missing here?
| Let's see how the units work out if we convert a linear spring (where we know everything) to a torsion spring, by attaching our linear spring to a stiff rod some distance $R$ from a pivot:
The (linear) force due to the spring is $\vec F = -k\Delta \vec x$, for spring constant $k$ having units of newtons per meter. The torque is
$$
\tau = RF = -R \cdot k (R \Delta\theta) \equiv -\kappa \Delta\theta
$$
So apparently the torsion spring constant $\kappa = kR^2$ has units of newton-meters, which is equivalent to newton-meters per radian, because the radian is a dimensionless ratio. If you mistrust the apparatus of calculus and would like to do a lot more work you could use the appropriate trig function $\sin\Delta\theta = \Delta x /R$; in that case only follow my argument in the small-angle approximation $\lim_{|x|\ll1} \sin x = x$.
The energy stored in the linear spring is
\begin{align}
U = -\int_0^{\Delta x} \vec F \cdot d\vec x
= \frac12 k(\Delta x)^2 = \frac12 k(R\Delta\theta)^2 = \frac12 \kappa (\Delta\theta)^2
\end{align}
which is exactly the same thing you get if you integrate the torque
$$
U = -\int_0^{\Delta\theta} \vec\tau \cdot d\vec\theta = \frac12 \kappa (\Delta\theta)^2
$$
As another answer says much more succinctly: it all works because the radian, a ratio between two lengths, is dimensionless.
lemon asks in a comment elsewhere how you would convert $\kappa$ in each case if you were tied to a log in a sawmill and commanded either to use degrees or die a hideous bloody death. (My paraphrase; I would probably choose death, myself.) In that case you might grudgingly admit that the relevant constant in the torque equation has units of foot-pounds per degree, while in the energy equation you have picked up another angular factor so that the unit is b.t.u. per degree squared. I don't think there's anything profound about the coincidence that torque has units of energy; I do think there's something profound about the fact that we have invented the SI units to make these pointless problems go away.
| {
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Why does the kinetic energy work theorem apply to inclined surfaces?
Work is supposed to be force x distance, how did we end up with work equaling the difference of hights x gravitational acceleration ?
Sorry for my poor understanding :D
| Look at the forces in your system:
*
*In the y-direction, you have gravitation force.
*In the x-direction, there is no force.
Edit:
Suppose you have vectors for the force and the displacement:
$F=(F_x,F_y)$ and $d=(d_x,d_y)$, now to get the work W, you calculate the dot product* of these two:
$W=F*d=F_x*d_x + F_y*d_y$
With
$F_x=0$ (because there is no force on your ideal frictionless inclined plane) and
$F_y= -m*g$ (gravitational force; m=mass, g=gravitation constant=$9.87 m/s^2$)
above equation will give you:
$W=F_y*d_y = -m*g*d_y$.
So we see, since no force acts in the x-direction, no work is done in the x-direction.
*=in genereal this would be a line integral, but it boils down to the same idea.
| {
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How does one calculate how big something has to be, to be seen at a given distance? Ignoring curvature of the Earth.
How do I calculate the size an object would need to be in order to appear to be approx 1cm tall at a given distance?
| Let $w$ be the actual size of the object, $d$ be the distance to the object, $w_r$ be the size of a reference object, and $d_r$ be the distance to the reference object.
If you want your object to appear to be the same size at some distance $d$ as a reference object at a reference distance $d_r$, then using the properties of similar triangles...
$$\frac{w}{d} = \frac{w_r}{d_r}$$
Solving for $w$, the size of your object,
$$w = w_r \frac{d}{d_r}$$
For example, if you want an object at 20 meters to appear to be the same size as a 1 cm object at 3 meters, you would plug in like so:
$$w = 0.01 ~\rm{m} \frac{20 ~\rm m}{3 ~\rm m} = 0.067 ~\rm{m} = 6.7 ~\rm{cm}$$
| {
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Is acceleration relative in relativity? Suppose a box A is moving relative to a Box B, then by time dilation equation if I take 1 sec passed for an observer in A then for an observer in B will be little longer. Now if I suppose that the box B is moving while A is stationary under the same condition, then by the time dilation equation time passed in B must be shorter than A. How is this issue resolved? In the twin paradox problem my book says that it's due to acceleration, but in my opinion acceleration is relative (please correct me if I'm wrong), but here it is not the case.
| Are you familiar with the concept of the relativity of simultaneity? Because simultaneity is relative, when they move apart they will disagree about which pairs of readings on each of their clocks are simultaneous. For example, suppose they move apart inertially at 0.6c, with each clock set to read a time of 0 seconds at the moment they were right next to one another at the same position coordinate. Then in the inertial frame where Box A is at rest, the event of Box A's clock reading 20 seconds is simultaneous with the event of Box B's clock reading 16 seconds, so in this frame Box B's clock has ticked less time. But in the inertial frame where Box B is at rest, the event of Box A's clock reading 20 seconds is simultaneous with the event of Box B's clock reading 25 seconds, so in this frame Box A's clock has ticked less time.
And as Conifeld said, acceleration is not relative--you can determine whether you're accelerating using an accelerometer that measures G-forces, for example. In the twin paradox, one of the twins accelerates so that they can compare clocks at the same location in space--this means there is no longer any problem with disagreements about simultaneity, but the acceleration breaks the symmetry, and the one who accelerated will have aged less than the one who moved inertially. There's a good discussion of a number of different ways to think about the twin paradox on this page.
| {
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Why does the temperature of the gas in a container moving with constant velocity not change?
Systematic changes do not affect thermodynamic equilibrium.
What does this mean? And what kind of systematic changes are allowed?
The container with gas is stationary till some time then it's given a constant velocity and the final temperature is asked; the answer says that systematic changes don't affect thermodynamic equilibrium and temperature remains constant.
| This means that changes that happen to the whole of the gas won't affect its temperature.
This is because the temperature of a gas is proportional to the average kinetic energy of the particles.
$ T \alpha KE_{average}$
Since in an ideal gas all particles move in straight random paths, an increase in velocity in one direction for all particles will not actually change their average kinetic energies.
Therefore the thermodynamic equilibrium remains unchanged.
| {
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How to write BdG Hamiltonian in graphene? In Beenakker's paper:Specular Andreev Reflection in Graphene, the BdG Hamiltonian is written as:
$$
H_{BdG}=\begin{pmatrix}H-E_F&\Delta\\ \Delta^*& E_F-H\end{pmatrix}
$$
from equation (1).
Where $H$ is the Hamiltonian of pure graphene and it is:
$$
H=\begin{pmatrix}H_+&0\\ 0& H_-\end{pmatrix}
$$
Where $\pm$ denotes different valleys and:
$$
H_{\pm}=-i\hbar v(\sigma_x\partial_x\pm\sigma_y\partial_y)
$$
Moreover, $H$ is written in the basis of four dimensional spinor $(\Psi_{A+},\Psi_{B+},\Psi_{A-},\Psi_{B-})$
Question is, what basis is $H_{BdG}$ written in? What does the $4\times4$ matrix of $\Delta$ looks like?
Finnally, why the original $8\times8$ BdG equation can be valley decoupled like this (Equ.7 in the paper):
$$
\begin{pmatrix}H_\pm-E_F&\Delta\\ \Delta^*& E_F-H_\pm\end{pmatrix}{u\choose v}=\epsilon {u\choose v}
$$
It is a bit strange because paring is bewteen two valleys, how can we decouple the equation into two seperate valleys?
| According to @Bercioux answer, if we choose the basis:
$$
\phi_1=(\Psi_{A+},\Psi_{B+},\Psi_{A-}，\Psi_{B-},\Psi_{A+}^\dagger,\Psi_{B+}^\dagger,\Psi_{A-}^\dagger,\Psi_{B-}^\dagger)
$$
The BdG Hamiltonian should be written like this:
$$
H_{BdG}^1=\begin{pmatrix}H_+-E_F&0&0&\Delta_2\\0&H_--E_F&-\Delta_2&0\\
0&-\Delta^*_2&E_F-H_+&0\\ \Delta_2^*&0&0&E_F-H_-
\end{pmatrix}
$$
where $\Delta_2=\Delta I_2$, and note the that the minus sign is crucial.
This gives two decoupled $4\times4$ Hamiltonian, however the valley is not decoupled as in the paper:
$$
\begin{pmatrix}H_+-E_F&\Delta_2\\ \Delta_2^*&E_F-H_-\end{pmatrix}\text{ and }
\begin{pmatrix}H_--E_F&-\Delta_2\\ -\Delta_2^*&E_F-H_+\end{pmatrix}
$$
If we choose another basis:
$$
\phi_2=(\Psi_{A+},\Psi_{B+},\Psi_{A-}，\Psi_{B-},\Psi_{A-}^\dagger,\Psi_{B-}^\dagger,-\Psi_{A+}^\dagger,-\Psi_{B+}^\dagger)
$$
The BdG Hamiltonian can be simply written as:
$$
H_{BdG}^2=\begin{pmatrix}H-E_F&\Delta_4\\ \Delta_4^*&E_F-H\end{pmatrix}
$$
where $\Delta_4=\Delta I_4$, the Hamiltonian can be easily valley decoupled, resulting equation (7) in the main text.
| {
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In a waveguide, where does the energy in attenuated waves go? In an electromagnetic waveguide, there is generally a "cutoff frequency." Electromagnetic waves with a frequency that is lower than this cutoff frequency will not propagate at all -- i.e., they will be exponentially attenuated.
Suppose that there is a source of electromagnetic radiation at the center of a waveguide (e.g., an oscillating electric dipole inside a rectangular waveguide), and that the emitted waves have a frequency below the cutoff frequency of the waveguide. All of the waves will be exponentially attenuated and thus will not propagate within/outside of the waveguide.
But, before the waves are sufficiently exponentially attenuated, they will be carrying away some energy from the radiating source in the form of Poynting flux. What happens to that radiation energy immediately before and after attenuation -- where does it go? Maybe the radiation is very quickly absorbed by the conducting boundary of the waveguide and thus heats up the waveguide's surface?
Most people would say that the Poynting flux is zero if the frequency of the waves is below the cutoff frequency of the waveguide, but the language of "exponential attenuation" suggests to me that the waves did travel for some short amount of time before being exponentially attenuated (and thus the Poynting flux was not zero before exponential attenuation).
| I believe energy will be reflected from all sides in this case, which will not be much different from a case of a source in a cavity. The amplitude of the electromagnetic field will grow until losses in the walls of the waveguide become comparable to the power radiated by the source.
| {
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Synthetic Photometry - Calculating a colour index I have a theoretical black body spectrum as described by plancks law. I also have the bandpass sensitivity function for various filters. I would like to calculate a colour index from this information, so I can compare it to an experimental result.
My proposed method is to take the black body spectrum and convolve it with the passband. I would then bin the resulting spectrum and convert to photons using the bin's average wavelength. Summing up the photons should give me counts that can be used to calculate a colour index. This method is described at the bottom of the page here http://spiff.rit.edu/classes/phys440/lectures/filters/filters.html
My question is - is this the correct method? Why is the passband convolved with the blackbody spectrum, rather than multiplied together? Which is the correct method?
| In the link, the spectrum is very noisy (possibly by the way of absorption peaks). Then, convolving it by a band-pass filter is simulating the integration - or average - over the band, resulting in a smooth curve, that you can then sample (i.e. pick up discrete values) without suffering aliasing artifact (as you would if the sampling rate of your bining is less than twice the highest frequency of these peaks in the curve).
In your problem (that is not 100% clear to me), you start with a spectrum curve that is already smooth, so you don't need to filter it. (Or you might want to integrate if you want ultra acurate results in non-affine areas, or if you BB is narrow peaked, or your bins are very sparse. But your figure doesn't seem to correspond to any of these cases :-) ).
| {
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Which properties control the strength of strong interactions? Does the strong nuclear force depend on mass, charge or spin? Accordingly, can we order neutron-neutron, proton-proton and neutron-proton pairs by the strength of the strong force between them?
| The picture is basically "upside-down" or rather "inside-out":
According to our most current understanding, mass, charge and spin for the particles you are asking about, are actually the by-product of the elementary constituents of the Proton,Neutron and other baryons/mesons.
All particles mentioned are actually combinations of 3 quarks ( or $qqq$ those would be baryons), and one quark plus one anti-quark (or $q\bar{q}$, those would be mesons).
Quarks are spin 1/2 elementary particles, which have very small masses (not enough to make up the composite particle's mass) and broken charge (-1/3,2/3).
the composites particle's characteristics are the results of the 3(2) quarks that compose them, charges add up, spin is the result of angular momentum addition, and mass is the result of the interaction energy, we perceive that energy as mass.
Again, as per our current understanding, the strong force is nothing but a "Van der Waals" type force for the basic color force.
Obviously all of this is a broad-stroke overview, and I suggest you look into QCD and "the eightfold way".
Since (AFAIK) the strength of the color force is not affected by spin/mass/charge (they do not affect the QCD interaction term), I would have to answer no - these interactions should be similar to an amazing degree...
Even so, it might be beneficial to look into some tables of those interactions. I would look in the Particle Data Group tables (http://pdg.lbl.gov/), to see if indeed there is some asymmetry there.
| {
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Is the electric field strength along an equipotential surface constant? I'm trying to determine whether or not the electric field strength $|\vec{\mathcal{E}}|$ is constant everywhere on an equipotential surface. I know an equipotential surface is defined as
$$
S = \{\mathbf{x}\in\Bbb{R}^3\text{ s.t. } V_{\mathbf{x}} = c\}
$$
where $V_{\mathbf{x}}$ is the electric potential at $\mathbf{x}$ and $c$ is some constant.
At first it seemed incredibly intuitive, but upon trying to prove it I keep going in circles. It seems true for simple cases such as a single charge (in which the equipotential surfaces are circles surrounding the charge) and parallel plates (in which the equipotential surfaces are parallel lines). I'm not sure about other cases though.
Any help is appreciated! :)
| Consider the symmetry plane between two equal magnitude but opposite sign charges. It has the same potential everywhere (zero Volts in the usual gauge). Does it have the same electric field strength everywhere?
For simplicity we'll place a $+1 \,\mathrm{C}$ charge at $+1\,\mathrm{m}$ on the x-axis and a $-1 \,\mathrm{C}$ at $-1\,\mathrm{m}$ on the x-axis.
The symmetry plane is the y-z plane (where x = 0).
Now we compute the field strength at the origin and at $(0,+1)\,\mathrm{m}$ and at $(+1,+1)\,\mathrm{m}$.
*
*At $(0,+1)\,\mathrm{m}$ this is easy because both field contributions point in the -x direction:
\begin{align*}
E
&= 2 * \left( (9\times 10^9 \,\mathrm{N\,m^2/C^2})\frac{1\,\mathrm{C}}
{(1\,\mathrm{m})^2} \right) \\
&= 1.8 \times 10^{10} \,\mathrm{N/C}
\end{align*}
*A $(+1,+1)\,\mathrm{m}$ both field contributions point toward negative x, but one points toward positive y and the other toward negative y. The y-directed components cancel out so we only care about the x-directed parts:
\begin{align*}
E
&= 2 * \frac{1}{\sqrt{2}}\left( (9\times 10^9 \,\mathrm{N\,m^2/C^2})\frac{1\,\mathrm{C}}
{(\sqrt{2}\,\mathrm{m})^2} \right) \\
&= 6.4 \times 10^{9} \,\mathrm{N/C}
\end{align*}
And the field strength continues to drop rapidly as you move away from the origin in the y-z plane.
| {
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Are black holes very dense matter or empty? The popular description of black holes, especially outside the academia, is that they are highly dense objects; so dense that even light (as particle or as waves) cannot escape it once it falls inside the event horizon.
But then we hear things like black holes are really empty, as the matter is no longer there. It was formed due to highly compact matter but now energy of that matter that formed it and whatever fell into it thereafter is converted into the energy of warped space-time. Hence, we cannot speak of extreme matter-density but only of extreme energy density. Black holes are then empty, given that emptiness is absence of matter. Aren't these descriptions contradictory that they are highly dense matter as well as empty?
Also, if this explanation is true, it implies that if enough matter is gathered, matter ceases to exist.
(Sorry! Scientifically and Mathematically immature but curious amateur here)
| (fellow amateur here) I believe that the notion that black holes are empty comes from that fact that all the matter that gets sucked past the event horizon is packed into an infinitival small ball of infinite density, thus it would be the same as if you had a hollow sphere the size of earth, and one particle in the center, one would say that the sphere is empty, and to my knowledge that is why most people would say that a black hole is empty.
| {
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Wave intensity and superposition Let us say we have 2 point sources of sound. My question is how do we consider the intensity to vary according to position?
Let's say both have same amplitude, frequency and speed, just different phase.
Does intensity add up individually or do we calculate the net displacement in pressure due to the superposition and then relate max intensity to regions of max pressure?
Like here, will the intensity at A and B be the same? (Take any 2 points arbitrarily such that constructive interference is happening there)
Our teacher told us this, but I'm not sure about it.
| Any physics problem regarding waves, to my knowledge, will need you to compute the amplitude dependance in time and space before anything.
Waves superposition can be constructive or destructive depending on the sum of the amplitudes of the superposing waves. Intensity is defined as the product of the sound velocity by the sound pressure which is defined as the derivatice of the amplitude regarding space.
Along x axes with the example of a 1D propagating wave:
$$p=\frac{\partial y}{\partial x}$$
where y is the amplitude (say $y=a\sin(\omega t -kx)$ for a plane wave), you get the instantaneous pressure:
$$\textbf{I}=pc$$
where $c$ stands for the wave celerity (usually $c=\frac{\omega}{k}$).
For a sum of waves you have to add $y_1$ and $y_2$ the amplitudes of both wave and do the calculation again. You'll see that intensities just don't add up.
If you want to calculate intensity over a given time $T$, you have to integrate:
$$\textbf{I}_T=\frac{1}{T}\int_0^T pc\ dt$$
I hope this helps.
| {
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What would be the minimum fissile material needed to start fusion within a pulse unit? I am trying to calculate the weight of the pulse unit and composition and the approximate fallout, for a ground launch in regards to Project Orion`s nuclear pulse propulsion for a 0.03 kiloton and 0.35 kiloton detonation. The fusion of Deuterium (or another) would be primary focus so that the lowest possible amount of fallout is generated from the launch.
| Basically, you are asking for the minimum fissile material needed to build a nuclear bomb. This site might help with some of that. However, it seems likely that the real answers are classified. I have heard that a nuke can be created using as little as 1kg of Plutonium. I also saw a reference to a Polish paper during the Cold War that suggested it might be possible to go as low as 100 grams, although it sounds dubious.
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Why is the cut off mass for massive stars 8 solar masses? Why can't it be 10-11 solar masses or so? I know that stars having a mass greater than or equal to 8 solar masses are termed "massive stars". But why is the cut-off 8 solar masses?
| The division is conventionally made at the boundary between where stars end their lives as white dwarf stars and where more massive stars will end their lives in core collapse supernovae.
The boundary is set both empirically, by observations of white dwarfs in star clusters, where their initial masses can be estimated, and also using theoretical models.
The division is not arbitrary, it is of fundamental significance in studying the chemical evolution of a galaxy. The nucleosynthetic products of massive stars are fundamentally different to those of lower mass stars. The products also get recycled into the interstellar medium in a rather different way. Further, massive stars will affect the interstellar medium through supernova explosions in a manner that just doesn't occur in lower mass stars.
The reason for the 8 solar mass division (it is uncertain by about 1 solar mass and also depends to a certain extent on rotation and the initial metallicity of the star, so is not a sharp threshold) is that this is where the carbon/oxygen core (during He shell burning)$^{1}$ becomes hot enough to ignite further fusion. Core burning continues through to iron-peak elements, then there is a core mass collapse, a violent supernova and large quantities of processed material (O, Mg, Ne, Si, r-process elements) are ejected at high speeds. A neutron star or black hole remnant is formed. In lower mass stars, the core becomes degenerate, supported by electron degeneracy pressure, and core nucleosynthesis halts. The star ends its life by expelling the majority of its envelope (mostly H and He, with some enrichment with C, N and s-process elements) at low speeds through stellar winds. The degenerate core becomes a white dwarf.
$^{1}$ Actually it may be possible to go a step further along the fusion ladder and still avoid a supernova. Stars with a mass just a little more than 8 solar masses (and possibly even as high as 10.5 solar masses -Garcia-Berro 2013) may produce Oxygen/Neon white dwarfs as the final outcome.
| {
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DNA breaks with particles or photons (Radiation therapy) When protons (or other particles) or photons are used in radiation therapy to treat cancer patients, the main effect is for it to make DNA breaks that hopefully will make the cancer cell die eventually (At least that is the very easy explanation. There are a lot of biology I do not know).
Now, as I understand, photons interact with matter, and excites electrons from molecules, which then ionize for instance water molecules in order to create free radicals, which can then interact with the cancer cells (and other cells) and create DNA breaks (single or double). At least, that is the idea.
My question is: How does this work for particles ? I know that photons are indirect ionization, and protons for instance doesn't interact that greatly with matter until the Bragg Peak, where it then deposits most of it's energy. But how is this energy transferred in order to make the DNA breaks ? The Coulomb interaction ? If so, wouldn't that just create electrons again, making it indirect as well, or have I misunderstood something ?
Thanks in advance.
| Yes, it is essentially just the Coloumb interaction.
Sometimes, this will strip electrons from atoms. However, those electrons might have been responsible for a bond in a molecule, so this will have effectively destroyed the molecule.
Other mechanisms include having a nucleus recoil from a proton (perhaps imparting enough energy to break a bond), or causing a nuclear reaction to take place (that will change the molecule entirely).
| {
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Driven harmonic oscillator Given the Hamiltonian of a loaded particle
$$\hat H = \frac{\hat p^2}{2m}+eE(t) \hat x + \frac{1}{2}m\omega^2 \hat x^2$$
show that
*
*The time dependent expected values $\langle \hat p\rangle$ and $\langle \hat x\rangle$ satisfy the classical laws of motion
*Discuss the time dependence of the expected values for the case $ E(t) = E_0 \sin(\Omega t)$.
I used Ehrenfest's theorem $(d/dt)\langle \hat A \rangle = (i/h) \langle [\hat H,\hat A]\rangle + \langle (\partial/\partial t)(\hat A)\rangle$ to show 1) but I don't know how to do 2). I know the definition of expected values $\langle \hat A \rangle = \langle \psi|A \psi \rangle$ where $\psi$ is the state of the particle.
| Ehrenfest's theorem proves that expectation values of $\hat{x}$ and $\hat{p}$ obey classical equations of motion in general. This is true regardless of whether the system you are considering is that of a driven harmonic oscillator or not.
The difference in the case of a driven harmonic oscillator, is that solutions to the Schrodinger Equation are proportional to coherent states. The expectation values of $\hat{x}$ and $\hat{p}$ in coherent states obey the equations of motion for a classical harmonic oscillator.
| {
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Why do I get two different answers? Using Calculus and Using Newton for Distance/Velocity Why do I get two different answers?
QUESTION: How long does it take to slow down from 8000 m/s to 6000 m/s with an acceleration of -400 m/s^2 ?
GIVENS
D distance
V velocity
T time
V initial velocity 8000 m/s
V** final velocity 6000 m/s
A acceleration -400 m/s^2
FORMULA 1
D = AT^2 + VT
D = -400T^2 + 8000T
V = D'
V = -800T + 8000 (first derivative)
Setting T=0, we find V=8000, which is correct for initial velocity.
Setting V=V**=6000 for final velocity, we find:
V = -800T + 8000 = 6000
T = 2.5
FORMULA 2
V** = V + AT
6000 = 8000 + -400T
-2000 = -400T
T = 5
I do not understand why these formulas are giving two different answers. Thank you!
| The formula 2 is correct and it gives correct answer t=5s. But formula 1 is wrong. The correct formula is $D=\frac {1}{2}at^2+vt$. If you differentiate $D=at^2+vt$, you will get $v=v_0+2at$, where $v_0$ is initial velocity. That is wrong.
So the correct answer is t=5s.
| {
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How do temporary dipole-dipole interactions work in quantum mechanics? The standard presentation of temporary dipole-dipole interactions (in high school at least) is classical: the electrons in an atom/molecule 'orbit' around its nucleus/nuclei. As a direct result of this orbital motion, at any particular time there will be a higher charge density in one region of the atom/molecule than in another. This creates a temporary dipole in the atom/molecule and can lead to interactions with other atoms/molecules.
However when we model atoms using quantum mechanics the electrons no longer have classical 'orbits'. How then do these interactions arise?
| Consider two atoms at some distance $R$ from each other. The Hamiltonian of this system is then the sum of the Hamiltonians of the two atoms plus interaction terms involving the electrostatic interaction between the electrons of one atom with the electron and nucleus of the other atom. You can then calculate what the shift in the ground state energy of the system is due to the interaction term.
To first order in perturbation theory the shift is zero, there is nonzero shift at second order. This means that the effect is due to the interaction changing the wavefunction of the system which in turn has an effect on the energy. This gives you the force between the atoms. Due to conservation of energy, if the energy changes as a function of distance then the kinetic energy of free moving atoms will also have to change. Here you assume that the two atom system will remain in the instantaneous ground states, which is approximately correct for low velocities, this follows from the Adiabatic Theorem.
So, the total energy of atoms placed a distance $R$ from each other can be interpreted as an effective potential energy, the effective force is then minus the derivative of the effective potential energy.
| {
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Why don't all the gluon's get converted into energy in an uranium atom undergoing fission? I have following questions and arguments.
*
*why do only gluons get converted into energy in an uranium atom undergoing fission?
*why don't all the gluons get converted into energy?
*which conditions lead to such process?
| In general an atom undergoing fission breaks up into other, smaller nuclei and stable particles: photons (x rays and gamma rays), electrons(beta decays) , alpha particles an other lower mass nuclei.
Gluons are never free, because of QCD, and always inside a proton or a neutron. Their spill over attractive force is due to virtual gluons which will never materialize except as a transfer of an energy momentum vector.
The excess energy from the mass energy fourvector balance appears as kinetic energy of the products, which finally becomes thermal energy after secondary interactions.
Here we see the neutron inducing a chain decay in uranium238. No gluons.
| {
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Why does force cause acceleration? Consider a ball placed on a smooth plane. If you take a plank made out of wood and whack it (apply a force on it), does it accelerate because
*
*First, the force compresses the part of the ball in contact and hence the ball is now like a compressed spring with potential energy
2.Next, the ball pushes on the plank using the potential energy and pushes off in the direction of force.
Is this how a force on a body at rest causes motion? If so do perfectly rigid bodies accelerate on the application of force?
| I am not sure that i understand your question 100%
If your question is - as mentioned in the title 'why does force cause acceleration '- then the answer is
simple: the Newton's second law of motion (the net force on an object is equal to the rate of change
of its linear momentum p in an inertial reference frame) which is an axiom.
But then you continue with a description of how a force is transmited when two object collide each other.
If this is your question then the answer is : the electrostatic repulsion as very nicely is explained here:
https://www.quora.com/What-does-it-mean-for-two-things-to-be-touching
In general, each type of force is considered to have a particle 'carrier'. You can find informations
about 'force carrier' here : https://en.wikipedia.org/wiki/Force_carrier
| {
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Difference between scattering and refraction? I while back I learnt that when light is incident on a dipole the dipole will scatter the light, and when it is incident on a material of a different refractive index then the light refracts. From the Ewald-Oseen extinction theorem, it seems that refraction is caused by scattering. So what is the distinction between scattering and refraction (i.e. when would we call something scattering and something refraction)? (sources would be helpful if you have them)
| Refraction occurs when a large number of dipoles scatter coherently. Each individual dipole scatters light in response to the incident radiation in (almost) all directions, but when you have a large collection of scatterers, each one scattering in many directions, you have to sum the contributions of each one in order to arrive at the total field. Each contribution interferes with every other contribution. When you do this at an abrupt interface, the result is reflection and refraction (and cancellation of the incident light, ala the Ewald-Oseen thm).
So the main difference is that scattering generally refers to small scatterers (having a size on the order of the wavelength), and refraction requires a large number of scatterers, and a clean interface.
| {
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Link between energy per unit frequency and derivative of the energy in regard of frequency I take a simple example to illustrate my question. I consider $\frac{du}{d \omega}$.
It physically represents the volumic energy per unit interval of frequencies (imagine a system with electromagnetic fields inside).
If I take $\frac{du}{d \omega}(\omega_0)$, it will physically give me the volumic energy of all the waves at the frequency $\omega_0$. If I had a filter that would select only $\omega_0$ and delete all other frequencies, it would be the total energy of my system.
Well ok. But what is the link with the derivative ? Indeed this quantity is written in the same way than the derivative of the volumic energy. But I don't see the link between those quantities...
Could you help me ?
Thank you
| They are the same thing. That is the reason "energy per unit frequency" is written the same was as the "derivative of energy with respect to frequency."
When we ask how energy changes with respect to frequency (the derivative) it is the same as asking "for every unit change in frequency, how does the energy change".
To put it more generally, these word expressions mean the same thing:
(1) "derivative of A with respect to B"
(2) "quantity A per unit of B"
In general people tend to think of (1) as a function, and (2) as that function evaluated at a specific point, or as an average. However both can and should be considered as functions. Hope that helps.
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How Are Quantum Computers Able to Store Any Data at all? So if qubits can have more than two states, and according to this video,
https://www.youtube.com/watch?v=T2DXrs0OpHU
you don't know what you get until you actually "open the box", if its all randomness and probability,then how can it store anything?
Like, if you tried opening your word document, won't it show up differently everytime you opened it?
Sorry if some of my questions seem stupid, I'm a high school student who has just recently gotten interested in quantum computing.
| With quantum computers, as with classical computers, the initial states (variables) are set at the beginning of the process.
Suppose you write an ordinary computer program to find the largest prime number less than $N$; you might use the Sieve of Eratosthenes. At the end of the process your computer program has many numbers in its working memory, and you must select the result and share it with the outside world through some output channel, such as a display or a message.
There are currently only a few quantum computing algorithms, and each has a method for extraction of the solution. Quantum information and cryptography are related fields which use the same quantum techniques.
It is a very rich field, and requires a background in mathematics and computer science; if you are interested in building such devices, it requires a very strong background in physics and engineering.
Just as in the early days of computing, there are no programming languages. Instead you work with a series of quantum logic gates, each of which can be represented by a matrix. This means that the mathematics required is linear algebra, which includes the study of matrix algebra.
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What is uppercase ${\cal O}$ in electrodynamics? I'm a bit puzzled as to what the symbol ${\cal O}$ means in electrodynamics, I'm reading this paper here http://arxiv.org/abs/astro-ph/0404512.
See equation 43 which is in page 12, what is this uppercase ${\cal O}$? I can't find its definition anywhere.
| This symbol means that there's an error of aproximation, an example is the equation you mentioned:
$f(r) \sim \frac{1}{4r} + \mathcal{O}(\frac{ln(r)}{r^2})$
Here, the function $f(r)$ is being aproximated by the first term, it means that you can aproximate the actual value of the function $f(r)$ by computing the term $\frac{1}{4r}$, but by doing so, you'll have an error in the same order of magnitude as $\frac{ln(r)}{r^2}$. You can see that the expression for the error goes to zero as $r \to \infty$ meaning that the function $f(r)$ is better and better aproximated by $\frac{1}{4r}$ as $r$ increases.
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What is the difference between emu and esu? My textbook contains the following two statements:
*
*In the CGS system the unit of charge is electrostatic unit of charge (E.S.U). It is also called Stat Coulomb (StatC).
*In the CGS system, the unit of charge is electromagnetic unit (E.M.U).
How can e.m.u and e.s.u both be the units of charge in the same system?
| Quoting from the Wikipedia page on the CGS system:
*
*The e.s.u of charge, also called the franklin or statcoulomb, is the charge such that two equal $q=1\:\mathrm{statC}$ charges at a distance of $1\:\mathrm{cm}$ from each other exert an electrostatic force of $1\:\mathrm{dyn}$ on each other.
*The e.m.u. of current, also called the biot or abampere, is the current such that two infinitely-long straight, parallel conductors carrying $1\:\mathrm{abA}$ of current and separated by $1\:\mathrm{cm}$ exert a magnetostatic force of $2\:\mathrm{dyn}$ on each other.
*The relations between these units are such that
$$\frac{1\:\mathrm{statC}}{1\:\mathrm{abA\times 1\:s}}=\frac{1\:\mathrm{statC}}{1\:\mathrm{abC}}=\frac{1}{c}=\frac{1\:\mathrm{statA}}{1\:\mathrm{abA}}=\frac{1\:\mathrm{statC/s}}{1\:\mathrm{abA}},$$
where $c$ is the speed of light.
The ESU and EMU systems of electromagnetic units are different systems and they should generally be considered as separate and independent (if relatively similar), and they do not coincide with the gaussian set of electromagnetic units. For example, the electric displacement vector $\mathbf D$ is defined as $\mathbf E+4\pi\mathbf P$ in the ESU system and $\frac{1}{c^2}\mathbf E+4\pi\mathbf P$ in the EMU system, so you cannot interchangeably use formulas for one system in another without the use of a formula dictionary like the one at the end of Jackson's Classical Electrodynamics:
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Bifundamental representations Can someone give me explicit examples (in matrix form) of bifundamental representations? Illustrative would be for instance:
a) SU(3) x SU(2)
b) SO(4) x U(1)
c) E6 x U(1)
but other you may have ready would also work. A reference would also be great.
| I will talk about $SU(3) \times SU(2)$.
First, a matrix $T_3 \in SU(3)$ acts in the fundamental representation on $\mathbb C^3$ in the following way: A vector $\vec v \in \mathbb C^3$ with components $v_i$ is mapped to $v'_i = (T_3)_{ij} v_j$. Similarly, a $T_2 \in SU(2)$ acts on $\vec w \in \mathbb C^2$ as $w'k = (T_2)_{kl} w_l$.
The bifundamental representation of $SU(3) \times SU(2)$ acts on $\mathbb C^6$, the only difficulty is the choice of basis. It is easiest to describe the action if we label the components of $\vec v \in \mathbb C^6$ as $v_{ik}$ for $i \in \{1,2,3\}$ and $k \in \{1,2\}$.
Then the element $(T_3, T_2) \in SU(3) \times SU(2)$ acts as
$$ v'_{ik} = (T_3)_{ij} (T_2)_{kl} v_{jl} \;. $$
If you really want to write this in matrix form, you first have to fix some order of the basis of $\mathbb C^6$. For example, we could write the components in as a vector
$$ (v_{11}, v_{12}, v_{21}, v_{22}, v_{31}, v_{32})^T \;. $$
After some thinking, it turns out that the group action can then be written as the following matrix:
$$ \begin{pmatrix}
(T_3)_{11} (T_2)_{11} & (T_3)_{11} (T_2)_{12} & (T_3)_{12} (T_2)_{11} & \cdots & (T_3)_{13} (T_2)_{12} \\
(T_3)_{11} (T_2)_{21} & (T_3)_{11} (T_2)_{22} & \ddots && \vdots \\
\vdots &&&& \vdots \\
(T_3)_{31} (T_2)_{21} & \cdots &&& (T_3)_{33} (T_2)_{22}
\end{pmatrix} \;. $$
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Question about aircraft/rockets Lets say that you're sitting an inverted airplane. How do you determine how fast the plane must accelerate in order for you to not fall out?
| That depends on two things: The coefficient of friction between the pilot and his seat and the direction of acceleration.
First case: The aircraft accelerates along its flight path. The pilot is pressed against the seat by the acceleration, and if that pressure is sufficient, friction will keep him in place. Since the coefficient of static friction $\mu_s$ is equivalent to the tangent of the inner frictional angle, and the ratio between gravity and acceleration is also a tangent, the acceleration $a$ along the flight path must be
$$a > g\cdot \mu_s$$
assuming a horizontal flight path and a vertical backrest. For different flight path and backrest angles correct accordingly.
Second case: The aircraft flies a parabola such that the pilot is pressed into his seat by centrifugal forces. If the angular velocity of the pitch motion is $q$ and the centrifugal force has to be greater than the pilot's weight, the condition is
$$q > \frac{g}{v}$$
The higher the speed $v$ you accelerate to is, the smaller the minimum pitch rate becomes.
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How Fermi momentum and Fermi energy are related? The Fermi momentum is defined as $\sqrt{2mE_F}$, where $E_F$ is the Fermi energy. Does this equation mean that the Fermi momentum depends only on the kinetic part of the Fermi energy?
Is it correct to say that turning on an external field for a spin-half system affects only the Fermi energy and not the Fermi momentum?
| The energy and momentum are related by the "dispersion relation." For a classical free particle, the kinetic energy of a particle is the total energy and so
$$ E=\frac{p^2}{2m} \rightarrow p = \sqrt{2mE}.$$
That is the origin of the relation you mentioned. For a relativistic free particle, you have the Einstein dispersion
$$ E^2 = p^2 + m^2,$$
and you would get a different relationship between the energy and momentum. So in general, translating from the fermi momentum to the fermi energy requires knowledge of the dispersion.
So the answer is yes, the Fermi energy to Fermi momentum translation is different for a free particle and one which is in some potential.
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What is the moment of inertia really? Is moment of inertia or second moment of inertia, simply the resistance of a body to rotate it over an axis? What is radius of gyration? What if the axis is via the center of mass or somewhere different? can you give me please an overview of these issues with SIMPLE words, and without nonsense, like maths who nobody will ever remember. I need the SENSE how the brain comprehends these stuff in simple terms.
| The radius of gyration is roughly the proportion of the moment of inertia of a body to its mass. A higher gyradius means that one body will have a larger moment of inertia than a body with a lower gyradius, if both bodies have equal mass.
| {
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Which coordinate is to be considered for the energy of simple pendulum? For an simple harmonic oscillator energy can be represented as in picture. Consider in particular picture (b) with the energy as a function of the coordinate $x$.
Consider now a simple pendulum. The coordinate $x$ in (b) is the coordinate of an horizontal axis (as in picture 1) or the coordinate or the circular trajectory, as in picture 2.
The motion of the pendulum is indeed a one dimensional simple armonic motion, but the path followed is circular, so I guess that the "$x$ coordinate" of the graph (b) is the one in picture 2. Is that correct?
| The equation of motion for a simple pendulum is actually not the same as for a simple harmonic oscillator. In fact, the pendulum motion can is described by the differential equation
$$\frac{d^2\theta}{dt^2}+\frac{g}{l} \sin{\theta}=0$$
It is only in the small angle approximation (where $\sin{\theta} \approx \theta$) that this equates to a simple harmonic oscillator
$$\frac{d^2\theta}{dt^2}+\frac{g}{l} \theta=0$$
This means that a simple pendulum can not be accurately described as a simple harmonic oscillator for large swing amplitudes. One noticable effect is that for large angles, the oscillation period will depend on the deflection angle (this is not the case for a cycloidal pendulum). For a deflection angle of 10 degrees, the error will be 0.26% (see this site for a calculation).
Now, to answer your question: because we are only looking at small deflection angles, the motion of the end of the pendulum can be approximated as purely horizontal. Thus, you can ignore any curvature of the path of the pendulum and write its deflection as its $x$-coordinate only.
| {
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Why does a ceiling fan start up slowly? I think it probably has something to do with the capacitor inside but I don't get it why doesn't it just Start spinning instantly when we push the button why does it slowly start to spin and gradually gets faster?
| The fan motor provides a torque $\tau$ which has to accelerate $\alpha$ the fan blades whose moment of inertia is $I$:
$$\tau=I\alpha$$
Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the motor applied maximum torque when starting from rest, it would still take a noticeable period of time for the blades to reach their final rotational speed.
The capacitors are there to make sure the motor rotates in the right direction when switched on and/or to allow the speed of the motor and hence the blades to be controlled.
| {
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Physical meaning of partition function in QFT When we have the generating functional $Z$ for a scalar field
\begin{equation}
Z(J,J^{\dagger}) = \int{D\phi^{\dagger}D\phi \; \exp\left[{\int L+\phi^{\dagger}J(x)+J^{\dagger}(x)}\phi\right]},
\end{equation}
the partition function is $Z(0,0)$. We know that the derivatives of the generating functional give the propagator for the system, and it is often said that $Z(0,0)$ relates to the vacuum energy, and it is formally given by
\begin{equation}
Z(0,0) = \langle 0,t_f|0,t_i \rangle.
\end{equation}
How does this matrix element represent the vacuum energy of the system? Is it to do with the size of the fluctuations between the times $t_i$ and $t_f$? Or what is another interpretation of $Z(0,0)$?
| In terms of Feynman diagrams, the partition is represented by the sum over so-called vacuum bubbles - diagrams with no external legs. In formulae and in terms of the interaction picture and the free vacuum $\lvert 0 \rangle$ and the interacting vacuum $\lvert \Omega \rangle$, we have that
$$ \lvert \Omega\rangle = \lim_{T\to\infty(1-\mathrm{i}\epsilon)} \left(\mathrm{e}^{-\mathrm{i}E_\Omega T}\langle\Omega \vert 0\rangle\right)^{-1}\mathrm{e}^{-\mathrm{i}H T}\lvert 0 \rangle$$
and hence
$$ Z = \langle \Omega \vert \Omega\rangle = \lim_{T\to\infty(1-\mathrm{i}\epsilon)} \lvert \langle \Omega\vert 0 \rangle\rvert^2\mathrm{e}^{\mathrm{i}E_\Omega 2T}$$
Now, if you write $Z$ as $\mathrm{e}^{\sum_i V_i}$ where $V_i$ is the contribution of the vacuum bubbles of order $i$, you see that, schematically, $\sum_i V_i \propto E_\Omega T$, so the partition function is the exponential of the vacuum energy.
Heuristically, it should not be surprising that the logarithm of the partition function is the vacuum energy, since $Z \sim \langle 0 \rvert\mathrm{e}^{-\mathrm{i}\int H} \vert 0 \rangle$ so $\ln(Z) \sim \langle 0\vert T \int H \vert 0 \rangle$.
| {
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Deviation of free falling objects (Coriolis effect) using conservation of angular momentum I read this pdf on non inertial frame, in particular I have a question on the deviation of free falling object due to Coriolis effect.
Consider a ball let go from a tower at height $h$. The displacement due to Coriolis effect, calculated with formulas in Earth system, is $(4.19)$, after it there is explanation of the effect that uses the conservation of the angular momentum of the ball in a inertial frame.
$$x =\frac{2\sqrt{2}ωh^{3/2}}{3g^{1/2}} \tag{4.19}$$ Just before being
dropped, the particle is at radius $(R+h)$ and co-rotating, so it has
speed $(R+h)ω$ and angular momentum per unit mass $(R+h)^2ω$. As it
falls, its angular momentum is conserved (the only force is central),
so its final speed v in the (Eastward) direction of rotation satisfies
$Rv = (R+h)^2ω$, and $v= (R+h)^2ω/R$. Since this is larger than the
speed $Rω$ of the foot of the tower, the particle gets ahead of the
tower. The horizontal velocity relative to the tower is approximately
$2hω$ (ignoring the $h^2$ term), so the average relative speed over
the fall is about $hω$. We now see that the displacement $(4.19)$ can
be expressed in the form (time of flight) times (average relative
velocity) as might be expected.
But $$v_{average} t_{flight}=h \omega \sqrt{\frac{2h}{g}}$$
Which differs by $\frac{2}{3}$ from $(4.19)$. Is that due to the approximation made?
I also don't understand completely why the average relative velocity $v_{average}$ is taken to be half the relative velocity found. Isn't this valid only for constant accelerated linear motions?
| You're right about the wrong average velocity. But you can indeed use the conversation of angular momentum to calculate the right displacement. In order to do that let $\omega_0$ be the angular velocity of the earth and $h_0$ be the initial height of the ball. Then it follows from the conservation of angular momentum that $$(R+h_0)^2\omega_0 = (R+h)^2 \omega(h)\Rightarrow \omega(h)=\omega_0(1+\frac{2}{R}(h_0-h))$$ Let $\Delta \omega$ be the difference of the angular velocities of the ball and the earth and $\Delta \phi$ be the difference of their angles. Then it follows that
$$\Delta \phi = \int_0^T \Delta \omega \,\mathrm{d}t = \frac{2\omega_0}{R}\int_0^T h_0-h(t) \,\mathrm{d}t$$ Plugging in $h_0-h(t) = \frac{g}{2}t^2$ and $T = \sqrt{\frac{2h_0}{g}}$ one obtains
$$\Delta \phi = \frac{\omega_0 g}{R}\int_0^{\sqrt{\frac{2h_0}{g}}}t^2 \, \mathrm{d}t$$ and therefore $$x = R\Delta \phi = \frac{2\omega_0 h_0^{\frac{3}{2}}\sqrt{2}}{3\sqrt{g}}$$
| {
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Atmospheric Pressure inside a closed room
Even though they’re too tiny to see, all the molecules of air in the atmosphere above your head weigh something. And the combined weight of these molecules causes a pressure pressing down on your body of 10,000 kg per square metre. This means that the mass of the air above the 0.1 square metre cross section of your body is 1,000 kg, or a tonne.
I would agree with the argument that the atmospheric pressure is a result of the weight of the air above me were I standing in an open area. I do not understand how, by this model of atmospheric pressure, the reason of atmospheric pressure can be explained in a closed room say.
Sourcehttp://www.physics.org/facts/air-really.asp
| "Closed room" word is ambiguos. If first the room was open , then was closed, without pumping all of air out(creating a vacuum), the pressure conditions would remain same, because there will be still air inside the room.However, had we pumped all air out, our blood capillaries would collapse, since our internal pressure is 1 pascal, and it exceed the external pressure by 100000 Newtons.To understand what i mean is, think about how a lizard can stick to a wall,or how do suction pumps work(in bow and arrow toys).The basic principle is that, there is vacuum created at the surface of contact.From your mechanics course you could easily draw FREE BODY DIAGRAM and notice forces due to pressure, explaining the suction.
Note that forces along horizontal direction dont mater because air pressure would be equal on both sides
| {
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Why is Kinetic Energy = (-) Total Energy and Potential Energy = 2 $\times$ Total Energy? I came across this relation while reading on the Bohr atomic model. Are there any other forces for which these relations hold good?
| This is true for central potential problems. Because in that case, both potential energy and kinetic energy are in (1/r) terms. Also, Potential energy is negative and its magnitude is twice that of Kinetic energy . Thus we conclude that only the mathematics of this problem allows it to be expressed as the formula u proposed and it is general for any central force problems
| {
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How does a tranverse wave propagate? Sound waves can be understood as particles hitting each other and to conserve momentum the vibration travels in air. Each particle transfering it's momentum to the other until it reaches our ears. Atleast we can think of a mental picture of why they propagate. But what about transverse waves? Like for instance when you jerk a rope or a slinky? Can somebody give me an intuitive reason for the propagation of these waves or better (if possible) a simple mathematical model?
| You pull a small piece of a rope up, and as that piece goes up it pulls the piece adjacent to it up and as that piece goes up... When you move your hand back to it's original position you're applying a force to the piece again and it pulls the adjacent piece down, etc... Model of displacement as a function of position and time: $y(x,t) = y_{max} sin(kx-\omega t)$ where $k=2\pi/\lambda$ and $\omega = 2\pi/T$ where $\lambda$ is wavelength and $\omega$ is angular frequency.
| {
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Meaning of a certain value at Fourier Transform Define the Fourier Transform of a certain signal in the time domain FT[$x(t)$]=$X(j\omega)$
$X(j\omega)$ = $\int$ $x(t)$ $e$^($j\omega$$t$)$ $dt
I'd like to ask what is the meaning of the value obtained from $X(j\omega)$ with certain frequency $\omega$
for example if we have a voltage signal of 1 Volt and found that
$$X(j100) = 100$$
- the unit is weird for me which will be voltage*sec- what does that 100 mean here?
Also: Why there is a factor of $\frac{1}{T}$ difference between the units of Fourier series and Fourier transform ?
I've asked at Signals processing/Math stack exchange but no answer
I've read this answer but it says:
The multiplication by $T$ in the limit is to account for the differences in definition between the Fourier series and Fourier transform: the series representation typically has a factor of $\frac{1}{T}$, while the transform does not. I don't know that there is a lot of insight to be gained via this analysis, but it shows that the series and transform representations are intimately related.
which didn't satisfy me.
| *
*Your Fourier transform eq. is incorrect. it should be $e^{-jwt}$ not $e^{jwt}$
*Physical meaning: Split your integral into real and imaginary parts, the Fourier transform equation is split into integral with sine and cosines instead of $e^{jwt}$, but with same frequency.
$F = \int \Re{x(t)} cos(wt) dt + j \int \Im {x(t)} sin(-wt) dt$.
You could think of each term in the sum as dot (or inner) product of your function with a sin and cosine at the frequency. If your function is orthogonal to the $cos(wt)$ and $sin(-wt)$, only then the Fourier transform at that frequency is zero. Else it is a magnitude of the projection of your function on the sine and cosine at that frequency.
| {
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Is power still consumed when we switch off the lights? I was having heated discussion with one of my friends. We were discussing Earth Hour and its role in saving electricity, and creating awareness.
He stated that even if you turn off the lights, the load from the transmission lines is there at you home, so power is still being consumed.
But then I replied that, Power P $= I^2R$. and since there is no current flowing in the household equipments( whose R is much higher than that of the transmission lines), so there is no(or negligible) power loss.
Please tell which logic is correct, and whether switching off the unused electric/electronic equipments helps to save power or not?
| First answer this: does it take more power to run one light bulb or two?
Transmission lines are designed to be low loss, but they run a long way. Lightbulbs are designed to be "lossy" because that's how they work. Let's say for the sake of argument that the whole transmission circuit loses 100 watts and you're talking about a 100 watt bulb: those numbers are likely not anywhere close to accurate, but that's conceptually irrelevant. With the light on, the circuit uses 200 watts; with the light off, it only uses 100 watts. Then if we assume that the realistic scenario is that transmission losses are close to negligible, the extrapolation is that removing the household load saves a lot of energy.
As devil's advocate though, I'd like to point out that lights really don't account for much energy usage in the average home. Major appliances such as the refrigerator, air conditioners and electric ranges or water heaters pull much higher wattages, and tend to run often during the day, while lights are only useful in interior rooms or at night. So perhaps the better option for "Earth Hour" is to improve our overall energy efficiency to reduce usage.
| {
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Submerged water jet I am trying to calculate at what pressure and speed water from a high pressure water jet will hit an object 100 millimeters away from the nozzle opening, but i can't figure it out.
The nozzle will be used at approximately 100 meter below sea level in salt water. Its a circular nozzle with a diameter = 2 millimeters. The water from the water jet is at 400 bar and with a flow of about 60 liters per minute.
Does anyone know what formula to use?
| The formula to use for the nozzle velocity is:
V = (gpm * .321) / A where: gpm = gallons per minute
V = Velocity in ft/sec
A = Area of nozzle in sq. inches.
A 2mm nozzle = 0.0787 inches. Area A = .785(0.0787^2) = 0.00486 in^2
Flow of 60 L/M = 15.7 gpm.
Nozzle velocity V = (gpm * .321) / A
V = (15.7 * .321) / 0.00486 = 1036 ft/sec.
Impact force F = (Pn - Po) * A where: Po = pressure at ocean depth of 100
meters or 330 feet or 147 psi
units.
Pn = nozzle pressure of 400 bar in
Psi = 5880.
F = pounds force
Ocean water pressure Df at 330 feet depth is: Po = (Df * SG) / 2.31
Where: SG = Ocean water specific gravity = 1.03
H = one psi for each 2.31 feet of head.
Thus: Po = (330 * 1.03/ 2.31 = 147 psi.
F = (Pn - Po) * A: F = (5880 - 147) * 0.00486 = 5733 pounds.
Keep in mind that F is an impingement on only .00486 in^2 area.
Because type of nozzle not known, I assumed nozzle eff = 100 percent. In reality the nozzle eff. Will be some where between 60 and 90 percent.
By experience, in a stand-off distance of only 100 mm in subsea environment, the water jet stream will remain tight. Also, water speed velocity deterioration is assumed to be negligible.
| {
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Why do X-rays go through things? I always heard that the smaller the wavelength, the more interactions take place. The sky is blue because the blue light scatters. So why is this not true for X-rays, which go through objects so readily that we need often use lead to absorb it?
| Light is composed out of a large ensemble of photons, and photons are quantum mechanical elementary particles. Matter is composed out of atoms and molecules , which have small dimensions and are in the quantum mechanical range.
The quantum mechanical "size of interaction region" is given by the Heisenberg uncertainty relation. Even though a photon is an elementary zero mass particle it has a momentum given by
$$p = \hslash k= \frac{h\nu}{c}= \frac{h}{\lambda}\,.$$
As the electromagnetic wave impinges on a solid, each individual photon will interact/scatter with an atom or molecule on its path.
The Heisenberg uncertainty principle says that if the photon has momentum p
\begin{align}\sf \Delta x\Delta p &\gt \frac{\hslash}{2}\\\end{align}
its position x is uncertain by a volume bounded by the HUP.
The uncertainty in the position of the photon, is inversely proportional to the wavelength. If $\lambda$ is large the photon has the probability to exist in a large x dimension in order for the HUP to be fulfilled.
One can think of the volume defined by the HUP as the measure of how "large" the photon is. The smaller the wavelength the more "point like" the interactions of the photon will be.
For optical frequencies, large $\lambda$ s, this distance is composed of a huge number of atoms and molecules on its way, and the probability that the photon, and therefore the electromagnetic wave built up by photons, will interact, is practically 1.
For x-rays the (HUP limit) $\sf \Delta x$ becomes smaller than the distances between the lattice distances of atoms and molecules, and the photon will interact only if it meets them on its path, because most of the volume is empty of targets for the x-ray wavelengths of the photon.
| {
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Wavefunction Collapse I believe my Lecturer and the textbook have contradicted one another. My lecturer gave the example that if the spatial part of the wavefunction of a particle is given by
$\psi(x) = c_1\psi_1(x) + c_2\psi_2(x)$
for the infinite square well potential (where $\psi_1$ and $\psi_2$ are the ground and first excited energy eigenfunctions). He stated that if we were to measure an observable, for example the energy of the particle, the wave function will collapse to one of the two energy eigenstates.
Where as in Griffiths the following is stated:
Which one is correct? I have no reasoning for which one I think could be correct, so I can't figure it out...
| Both are correct, actually. If you measure an observable for that wave function you'll either find the eigenvalue corresponding to state 1 with probability $|c_1|^2$ (similarly for state 2), subject to the condition $|c_1|^2 + |c_2|^2 = 1$.
Edit: What Griffiths is saying is that before you perform the measurement, the particle is neither in state 1 or 2, but in a quantum superposition. Only the act of measurement forces the wavefunction to collapse to a particular state (at least according to the orthodox interpretation of quantum physics).
| {
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What is the temperature of compress air entering the atmosphere? Hi to all of you enthusiasts,
I have a 1 liter compressed air tank with around 207 bar of air pressure inside. It has been sitting out for a while and I am assuming its temperature has almost reached the surrounding temperature which is 40 degrees Celsius. I am also assuming that the surrounding is at around 1 bar (atmospheric pressure). I am not sure how to go about calculating the air temperature as it leaks out of the tank so I can predict any potential frosting.
I did use the ideal gas law of $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ where $P_1$, $V_1$ and $T_1$ belong to the tank and $P_2$, $V_2$ and $T_2$ belong to the surrounding. I thought if any volume of air leaks to the surrounding its volume would increase by 207 times (i.e. $207\times V_1 = V_2$), because the tank can contain 207 times more air than the atmosphere (which has roughly 1 bar of pressure).
However, the resulting temperature ends up being 40 degrees Celsius again. Am I doing this right?
Any help or hint is appreciated.
| Your mistake: the volume is not 207 times greater.
This would be only true for constant temperature; so it's no wonder if you get constant temperature :)
The volume increases more than this. It is governed by the laws for adiabatic expansion. See wikipedia. The relevant equation is $$T^\kappa p^{1-\kappa} = const.$$
Here $\kappa$ is a parameter of the gas. It can be measured, and, in some approximations, computed. It changes with temperature, but it's probably sufficient for you to know that it is near 1.4 for air.
| {
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How, if at all, does the behavior of a parallel plate capacitor change in the presence of a magnetic field? I'm an AP physics student curious about magnetism and my teacher couldn't answer this question. We already learned about velocity selectors, so I know that, when the plates of the capacitor are in the XZ plane and the magnetic field is in the Y or Z direction, there is no change in capacitance. What would happen if the magnetic field were in the X direction? Would the plates conduct the magnetic flux? Would there be a change in capacitance or any induced current? Thank you for your response.
| What 'behaviour' are you asking about?
Magnetic fields only affect moving charges. If the charge on the capacitor is not changing, the magnetic field has no effect, regardless of whatever direction it is in. No charges cross the space between the capacitor plates. Perhaps you are thinking of a situation where charged particles are fired into the space between the plates of the capacitor, where there are crossed electric and magnetic fields.
| {
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