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What does "commuting with the Hamiltonian" mean? In quantum mechanics an observable or an attribute to a particle (like spin) is conserved if and only if it commutes with the Hamiltonian. What does this mean? What observables do not commute with the Hamiltonian?
| In quantum mechanics, observables are represented by Hermitian operators. Mathematically, two operators $\hat A$ and $\hat B$ commute if $$\tag 1 [ \hat A, \hat B]=\hat A \hat B- \hat B \hat A = 0$$
Hermitian operators which satisfy (1) are also called compatible observables meaning that both can be measured simultaneously.
Hermitian operators which do not commute and do not satisy (1) are called incompatible observables, and a good example of this in quantum mechanics are the canonical commutation relations.
Now:
Operators or observables that commute with the Hamiltonian of the system are conserved quantities, e.g. angular momentum or spin. This means that these quantities do not change with time. Those that do not commute with the Hamiltonian, are not conserved quantities.
To summarise, if $\hat H$ is the Hamiltonian of the system then if
$$[\hat H, \hat O] = 0$$ then $\hat O$ is conserved and $$\frac{d \hat O}{dt}=0$$ but if
$$[\hat H, \hat O] \ne 0$$ then $\hat O$ is not conserved and $$\frac{d \hat O}{dt} \ne 0$$
Commutation relations also do not depend on our choice of basis meaning they work just as well in coordinate space as they do in momentum space. This is a powerful result used in quantum mechanics.
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If quantum entanglement causes the violations of Bell's inequality why don't we see perfect correlation? From wikipedia:
For example, if a pair of entangled particles is generated such that
their total spin is known to be zero, and one particle is found to
have clockwise spin on a first axis, then the spin of the other
particle, measured on the same axis, is found to be counterclockwise.
Yet the Bell tests all observe a lesser correlation:
If S is numerically greater than 2 it has infringed the CHSH
inequality. The experiment is declared to have supported the QM
prediction and ruled out all local hidden variable theories.
Is there any way to run an experiment that could observe (near) perfect correlation?
| No. Tsirelson's bound states that, e.g., the value of $2\sqrt{2}$ obtained for the CHSH inequality is optimal within quantum theory, while the maximum value obtainable within all theories which cannot communicate instantly (no-signalling theories) would be 4.
| {
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Zener effect - how does the probability of tunnelling increase under an increasing potential barrier? From Wikipedia (the basis of my lectures):
Under a high reverse-bias voltage, the p-n junction's depletion region widens which leads to a high-strength electric field across the junction.
Sufficiently strong electric fields enable tunneling of electrons across the depletion region of a semiconductor, leading to numerous free charge carriers.
I'd expect the widening of the depletion zone and the increasing potential barrier to reduce the chance of tunnelling, not to increase it. How is the opposite possible?
| I don't think that the Wikipedia article is correct. The widening doesn't cause tunneling; the shifting of the bands does. Maybe this figure helps (also from wikipedia):
(Zener tunneling is the right-most subfigure.)
In fact, I've never heard of this widening. I guess that it could happen, but I've never seen it in any models of Zener tunneling, so I don't think widening is important if it does happen.
I should add that the hyperphysics link conflates the Zener effect and avalanche breakdown. The two are different things altho they have a similar effect and can happen in the same device. (In fact, many "zener" diodes that you can buy at electronics suppliers don't really rely on Zener tunneling; they use avalanche breakdown.) Zener tunneling is in fact quantum tunneling.
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Physics experiment with sound I am doing an experiment for a school project which consists of blowing into 6 different bottles to create different notes/harmonics. Each bottle is filled with different volumes of water to create a different sound. I have to calculate the theoretical frequency of each, but I do not know how.
| I want to add to the answer given about Helmholtz resonator.
What you are expected to calculate (in principle) is the resonance of sound waves in a tube which has one side closed (your bottle). In this simple case the frequency of the different nodes depends on (among others) the length of the tube. As you fill water to your bottle, you decrease the length of the tube, that is, the empty region (air) in your bottle where sound waves interfere. This will cause the resonance frequency, therefore, the pitch of the sound to change.
I suggest you to look at
https://openstax.org/books/university-physics-volume-1/pages/17-4-normal-modes-of-a-standing-sound-wave
and scroll down to section called Resonance in a Tube Closed at one End.
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What trigger a capacitor to discharge when I connect a wire externally if $E$ is 0 outside? I know that there is a $Vab$ between the plates but I know also that the charges attract each other in the inside side of plates. So, the questions are:
what happens when I just connect a wire on both plates before connecting the leads? Do the charges move on the wire? Why?
Is probably because I don't have anymore two infinite plates?
I think that this question can justify the overall question.
| Yes some charge from the plate will go into wire. Electric charge of same sign will expand to all available conductor surface due to repulsive force of the charge being stronger than the attractive force of the opposite charge (because that is more distant).
| {
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What is meant by "spectral resolution of 0.5 GHz"? I sometimes see spectral resolution uses unit Hz. But spectral resolution is defined here and its unit is given as the smallest difference in wavelengths(Δλ) that can be distinguished at a wavelength.
What is the equivalent way of saying "0.5 GHz spectral resolution" in terms of wavelength(Δλ)?
| The dispersion relation can be written as $c=\lambda f$. Differentiate both sides using the product rule and the fact that $c$ is a constant:
$$0 = d\lambda f + \lambda df$$
Solve for $d\lambda$:
$$d\lambda = -\lambda \frac{df}{f}$$
Usually people consider the absolute value and express $\lambda$ through $f$:
$$|d\lambda|=\frac{c}{f^2}|df|$$
| {
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Intuition behind field transfomations Consider a real field $V^{\mu}(x)$ defined on a 4-dimensional Minkowski space. Acted by a transformation $\Lambda = \Lambda^{\mu}{}_{\nu} $ it transforms like
$$V^{\mu}(x) \to V^{'\mu}(x) = \Lambda^{\mu}{}_{\nu} V^{\nu}(\Lambda^{-1}x)$$
My question is: what is the intuition behind this transformation? I can't wrap my head around it.
| The components of the vector field transforms the same way as the components of the position vector.
For a scalar field with the value of 7 at point (3,0), for another reference frame rotated $90 ^\circ$, the rotated point (0,-3) has the same value.
But for a vectorial field with value (0,1) at point (3,0), for the new frame, the rotated point (0,-3) has now the value (1,0).
The same rotation matrix
$\begin {bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$
is applied to the vector position and to the vector field value itself.
It is easy to draw a picture and see what happens in this toy model 2-D.
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Lattice spacing in lattice QCD It is known that the lattice spacing in lattice QCD is not an external parameter and needs to be calculated, also the lattice beta parameter scales the lattice spacing ($a$) and goes as a function of the coupling constant as $$β=\frac{2*N}{g_0^2}$$
Where $N$ is a number of colors.
So the question is, how exactly is the lattice spacing calculated in lattice QCD?
I would also like to know how this lattice spacing is translated into physical units (GeV,fm).
| The lattice spacing can be obtained in a procedure usually referred to as scale setting for which one requires a physical quantity computed on the lattice. Take as an example the mass of the proton $am_P$ computed at a given value of the bare gauge coupling $g_0^2$ from the corresponding two-point correlation function. Given knowledge of the experimental value of the proton mass $m_P^{phys}$, the value of the lattice spacing can now be obtained via
$$a(g_0^2)=\frac{am_P(g_0^2)}{m_P^{phys}}\,.$$
A result in fm can then be directly derived using $\hbar c=197.3269788(12)$ MeV fm.
This scale setting is of course only valid up to lattice artifacts and depends on the quantity used to set the scale.
In practice people usually do not use the proton mass but rather the $\Omega$ or $\Xi$ masses or the decay constants of the pion and/or kaon. Intermediate scales like $r_0$ or $t_0$ which do not have a direct physical counterpart but can be easily and precisely computed on the lattice are also very common.
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Why is the radial velocity considered zero? I had recently come across a question which is stated as below:
A disc placed on a large horizontal floor is connected from a vertical cylinder of radius $r$ fixed on the floor with the help of a light inextensible cord of length $l$ as shown in the figure. Coefficient of friction between the disc and the floor is $\mu$. The disc is given a velocity $v$ parallel to the floor and perpendicular to the cord. How long will the disc slide on the floor before it hits the cylinder?
I thought hard for a few days but I couldn't solve it as the mathematics was terrible. Finally, while trying many other things, I tried considering the radial component of velocity to be zero and it worked! I got the answer.
But I am not able to understand the logic behind considering the radial velocity zero. Would someone please help me to understand it?
Edit: The figure is given as below:
| Some commentators also seem to have fallen into the ambiguity of Radial velocity. Radial normally means toward a centre. and a fixed centre. as the string wraps around the cylinder clearly there is no obvious centre as the motion isn’t circular but some weird spiral and the straight part of string doesn’t continually point back toward a constant common centre. There’s no centre to have radii properly. If the string is unstretchable and doesn’t go slack then however we can say the ball motion is perpendicular to the string along its momentary straight length(which we may try to call radius even though it’s changing centre. Anyway the motion is sort of circular instantaneously about the contact point of string with cylinder which must lead to answer.
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Perturbative treatment of infrared divergencies in QED of arbitrary order It is often claimed in QFT literature that infrared divergences of cross-sections of various processes can and should be compensated by taking into account processes of emission of soft photons in any order of perturbation theory.
However in the sources I have seen this is proven only in lowest orders for some concrete situations. An exception is Ch. 13 of Weinberg's book "The quantum theory of fields". However the discussion there is not detailed enough for me. Is there an alternative place to read about it? The case of QED would be sufficient for me.
| In the comments, OP clarifies that the issue is with equation (13.2.3). I was confused about this part as well when I first learnt it. The proper way to perform the calculation is as follows.
*
*First, do the $q^0$ integral by contour integration. We assume that the numerator of the integrand (which involves the lower point amplitude, the numerator of propagators, vertex factors, etc.) does not contain any poles in $q^0$. Thus, all the poles arise from the zeros of the denominator so there are four poles in $q^0$.
*Note that some poles are $q^0 = {\cal O}(|\vec{q}|)$ and some poles are $q^0 = {\cal O}(1)$ in the small $|\vec{q}|$ limit.
*Show that the ${\cal O}(1)$ poles do not contribute to the infrared divergence. More precisely, the ${\cal O}(1)$ poles give an integral which is ${\cal O}(|\vec{q}|^{-1})$. The integral over $\vec{q}$ then gives a term that is ${\cal O}(\Lambda^2)$ which vanishes in the limit $\Lambda \to 0$.
*With the foresight that only poles which are ${\cal O}(|\vec{q}|)$ contribute to the infrared divergence, we can expand the entire integrand in small $q^\mu$ (not just small $\vec{q}$) and therefore arrive at eq. (13.2.3).
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If we have a net negative charge in a spherical conductor, why negative charges goes to the surface? So imagine that we have a set of net negative charges, in physics books they said that this set of charges goes to the surface because they repel each other in such a way that this reach to an electrostatics equilibrium when this set of negative charges sits on the surface. This must create a negative potential energy in the surface of the sphere, so imagine that we have a negative charge in the middle or anywhere inside the sphere. By definition, negative charges tends to move to higher potential energy position, so why they are still going to the surface?
| The question relates only to a charged conductor without current flow. Many students have observed the effects of static electricity generated in the classroom. A test subject stands on an insulator touching a conductive ball which becomes charged from a moving carpet belt and an electrode. Hair standing on end demonstrates that accumulating static electric charge does in fact move to points on the surface of the conductor, in this case a person. The hair remains standing until the subject dismounts the insulator and POP, the charge is discharged.
Gauss's law advises if there is net free charge in a conductor an electric field is produced within the conductor. Excess charge in a conductor is free to move and so move to the surface, spreading out as far as possible. At equillibrium q(inside) = 0 since E=0. All the charges inside the Gaussian surface sum to zero so any charge on the conductor is in fact on the surface.
A useful analogy is that the force of gravity at the center of the earth is zero since it is surrounded by mass and we feel gravity is strongest at the surface of the earth. Likewise the static charge in the center of the conductive sphere is zero.
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What does this Euler-Lagrange equation represent? I have the following Lagrangian:
$$L=\frac{\mu}{2}\left(\dot{r}^2+r^2\dot\phi^2\right)-U(r),$$
The Euler-Lagrange equations are thus:
$$\frac{d}{dt}\left(\mu r^2\dot\phi\right)=0$$
$$\frac{d}{dt}(\mu \dot r)=\mu r\dot{\phi}^2-\frac{\partial}{\partial r}U(r).$$
I am trouble understanding what each Euler-Lagrange equation represent:
For example the first one: $\frac{d}{dt}\left(\mu r^2\dot\phi\right)=0$
What does $\mu r^2\dot\phi\ $ means?
Would it be the angular momentum?
| Yes, the term $\mu r^2\dot{\phi}$ represents the angular momentum, and the first Euler equation telling you that the angular momentum is the constant of motion for the given problem. The second Euler equation is just the radial force equation. If you can try to write newton's equation of motion like
$$F_r=m(\ddot{r}-r\dot{\theta}^2)\ \ \text{and}\ \ \ F_\theta=m(r\ddot{\theta}+2\dot{r}\dot{\theta})$$
Then you can compare it to the Euler equation to get the meaning of the terms.
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How do you incorporate friction into kinematics? A small disc of mass $m$ is pushed to $v_0$ on a surface with kinetic coefficient of $\mu$.
Is $v_f = v_0 -{\mu mg} \text{ t} $ ?
(Just based on the idea that increasing $\mu$ , $m$ or $g$ would increase the slow down.)
| Since disc is a continous body the magnitude of friction causing on each part will be different just like in a rope having mass tension force is different at different points
You can solve this by considering torque of friction
$f_r = {\mu } (dm) g$
$f_r = \frac{M}{\pi R^2}2pirdr $
$\int T.dt= f_rr$
After finding expression for torque you can equate it to $I{\omega}$
To find final velocity I leave the calculations to you
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Calculate height of a projectile in rotating spaceship reference frame In a rotating reference frame, or in a rotating spaceship, the apparent gravity felt by an object is the centrifugal force. Any moving object also experiences a Coriolis force. When I want to do calculations with a projectile (an object thrown across the circle in such a rotating frame), how do I take into account that the Coriolis acceleration constantly changes? I have been doing some calculations using the projectile equation and changing the acceleration variable to the Coriolis and centrifugal acceleration. But obviously, this is not correct. Lastly, how can I derive the maximum height of such projectile in this rotating frame of reference, given launch angle, angular velocity, and the radius of the circle?
Any advice would be appreciated. Thank you.
| The efficient way to calculate that is to use the non-rotating coordinate system.
Given is:
1 Launch velocity with respect to the rotating frame
2 Angle with respect to the circular perimeter of the rotating frame
3 Angular velocity of rotating frame with respect to non-rotating frame
The velocity with respect to the non-rotating frame is the vector sum of:
1 The velocity with respect to the rotating frame
2 The instantaneous velocity of the point of launch with respect to the non-rotating frame
The motion of the projectile with respect to the non-rotating frame is along a straight line. So you can, for example, calculate when that straightline motion will intersect the perimeter again.
By contrast, if you would insist on doing the calculation exclusively in the rotating frame then I think numerical analysis is the only way. The motion with respect to the rotating frame is not some nice function, such as a parabola, or (semi)circle, or hyperbola, etc. The simplest algorithm for plotting an arbitrary trajectory is Euler's method.
Clearly, numerical analysis is an option only if you can set up a computer to do the calculation.
| {
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Baryon to photon ratio In Dodelson's "Modern Cosmology", the current baryon-to-photon ratio is defined as $$ \eta_b \equiv \dfrac{n_b}{n_\gamma} = 5.5 \times 10^{-10} \left(\dfrac{\Omega_b h^2}{0.020}\right). $$
I have tried to found a discussion of the evolution of this parameter, but all I could find was its present value. Why, it seems, is its evolution never mentioned? What approximate value would it have for the early universe?
| The baryon number is conserved. Dating from the baryogensis, the baryonic number density is $n_B=N_B/a^3$, where $a$ is the scale factor and $N_B=\text{const}$.
The number of photons isn't strictly conserved. However, it is approximately $10^{10}$ times larger than the number of baryons. Even if photons are sometimes produced in processes involving baryons, it changes the total number of photons only by a tiny fraction. Hence, the photon number density is $n_\gamma=N_\gamma/a^3$ with $N_\gamma\approx\text{conts}$.
Both $n_B$ and $n_\gamma$ contain $a^3$ since we are talking about number density, not energy density (usually denoted by $\rho$). Therefore, their ratio, $\eta$, remains approximately constant.
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Divergence of Green function expansion at the origin? The expansion of Green's function in spherical coordinate is given in Eq. (3.125) in Jackson by
$$G\left(\mathbf{x}, \mathbf{x}^{\prime}\right)=4 \pi \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{Y_{l m}^{*}\left(\theta^{\prime}, \phi^{\prime}\right) Y_{l m}(\theta, \phi)}{(2 l+1)\left[1-\left(\frac{a}{b}\right)^{2 l+1}\right]}\left(r_{<}^{l}-\frac{a^{2 l+1}}{r_{<}^{l+1}}\right)\left(\frac{1}{r_{>}^{l+1}}-\frac{r_{>}^{l}}{b^{2 l+1}}\right),$$
where $r_<$ ($r_>$) is the lesser (greater) of $r$ and $r'$.
The electric potential can be found using Eq. (1.42)
$$\Phi(\mathbf{x})=\frac{1}{4 \pi \epsilon_{0}} \int_{V} \rho\left(\mathbf{x}^{\prime}\right) G\left(\mathbf{x}, \mathbf{x}^{\prime}\right) d^{3} x^{\prime} +\frac{1}{4 \pi} \oint_{S} [...]$$
If we have a single charge and it is located at the origin, then, $r_< = r'$, and the term $\frac{a^{2l+1}}{r_<^{l+1}}$ in (3.125) diverges.
What am I missing? Expression (3.125) should allow us to compute the potential due to a charge at the origin without diverging.
| Equation (3.125) is the Green function for a spherical shell bounded by $r=a$ and $r=b$, where it vanishes. If you want to place a charge at the origin, you should first set $a=0$, and the problem will disappear.
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What does 3D buckling reconstruction mean? In condensed matter/material science papers, I often encountered phrases like structural reconstruction, 3D reconstruction, 3D buckling reconstruction, etc. What do these phrases mean (especially the last one)?
For example, this paper (and a non-paywalled link).
| Surfaces of materials don't look like the inside of the bulk, otherwise you'd have dangling bonds and a ton of energy. So the dangling bonds combine in a certain way to minimize that energy. This is what's called a reconstruction. In silicon, the 111 face usually reconstructs into the 7x7 pattern, which is flat and has steps. A buckling reconstruction is one where some of the atoms pop out of the plane, so the bonds form "ripples" on the surface.
Transition metal dichalcogenides, which you see in that paper, were novel for existing in 2D structures like graphene. I suppose the 3D is emphasizing that the buckling made the 3rd dimension important.
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Significance of the fact the the final expression for acceleration in rolling motion over a slope is independent of friction In accelerated rolling motion on an incline , once we simplfy the equations we find that:
$$ f_s = \frac{I_{cm}}{r^2} \left(\frac{g \sin \theta}{ 1 + \frac{I_{cm} }{mr^2} }\right) \tag{1}$$
What is strange to me is that here the expression seems completely independent of the normal force applied. Usually we have we speak of friction, we find that the frictional force is proportional to the weight which is applied on the surface and it takes on the form:
$$f_s = \mu N$$
But why is it different for rolling on an incline?
Refer
| I think that as long as $$ f_s \le \mu N$$
there is no sliding and therefore no energy in the system is lost to heat. Under this condition friction can be ignored in the conservation dynamics of the description.
The first example in this link describes the situation in some detail:
rolling motion
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What does the concept of an "infinite universe" actually mean? When physicists talk about the universe being infinite, or wondering whether it is or not, what do these two options actually mean? I am not interested whether the universe is infinite or not, I am interested in what are the two options actually looking like.
Does an "infinite universe" mean infinite space, but when you go far enough there isn't any matter anymore, just more of infinite empty space in that direction?
Or does an "infinite universe" mean infinite matter, where it doesn't matter how far you go, you will find infinite stars all along the way?
In short, is the "infiniteness" meant to apply to space or to matter (which then would also include space of course)?
| Generally when physicists talk about the universe being finite, they are talking about the existence of an upper bound $R$ on the distance between any two points in space. Such an upper bound could arise in several ways - perhaps the universe has an edge - a boundary which cannot be crossed - or perhaps the universe has the topology of a 3-sphere, and so if one travels sufficiently far in any direction they would eventually return to their starting point. A spatially infinite universe is one which does not have this feature - given any real number $M$, there exist two points in the universe which are separated by a distance which is greater than $M$.
A finite universe of course can presumably only host a finite amount of matter. An infinite universe could in principle host either a finite or infinite amount of matter. Mainstream cosmology generally assumes the cosmological principle, which states that on sufficiently large scales the distribution of matter in the universe is homogeneous; in such cases, an infinite universe would have an infinite amount of matter in it, but of course this principle may not be accurate.
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Why is electric field across a resistor constant, or voltage gradient linear? Consider this complex circuit :
*
*Consider an ideal resistor with uniform cross section and made of isotropic material with constant resistivity.
*Consider an ideal battery giving a voltage difference of $V$ across terminal.
*Consider that the wires are superconductors with $0$ resistance.
*The voltage drop across the length of current flow is linear, that is $\frac{\partial V}{\partial L}$ is constant.
*This implies that Electrostatic force in the direction of current flow is constant throughout the resistor, since $\vec E = \frac{\partial V}{\partial L}$.
** Q: Is there a way to prove that the Electric Field ($\vec E$) is constant all along the length of the resistor in the direction of current flow, derive this mathematically using fundamental laws of electrostatics, or prove why the $\frac{\partial V}{\partial L}$ is linear across the length of the wire in the direction of the current flow?
How do the charges in the wire (at the edges of the resistor) exert a constant electrostatic force/electric field throughout the resistor?
**Note:
*
*Most other texts give the analogy of a capacitor with 2 charged plates, but I feel those examples are wrong, since they assume plates of infinite surface area which is clearly not analogous to the above circuit.
*Don't tell $E = \frac{V}{L}$ is constant at all points assuming $V = IR$ (ohm's law), because its actually the other way.
$V=IR$ is probably derived from $E = \frac{V}{L}$ being constant.
| As to your last question: There will be an excess of electrons on the negative end of the resistor, and a deficiency of electrons at the positive end. A uniform field inside requires a gradient in the charge density.
| {
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Peskin & Schroeder, section 12.2, page 409, analytic continuing from $d=2$ to $d=4$? In Peskin & Schroeder section 12.2，page 409, consider the one-loop correction to the propagator in Yukawa theory, it has the form
The authors first spot the pole at $d=2$, and find out that it could be completely absorbed by counterterm $\delta_m$, then they analytically continuing this equation from $d=2$ to $d=4$, and claim that we don't need to worry about the $\delta_m$ anymore, since in the limit $d\to4$, this equation has the form Eq(12.33)
My question is, why couldn't we just stick with the situation $d=4$ and claims that in the massless limit of scalar field, there will be no contribution to the mass shift, why should we bother to consider the pole at $d=2$?
Additionally, even if we absorb the mass shift at $d=2$ with $\delta_m$, can we justify the analytic continuation? I think $d=2$ and $d=4$ are two completely different and independent situation, how could we just analytically continue one to the other?
| Probably there are some important details that I do not fully understand. But on a high-level, we want to do this because we don't want $\delta_m$ to depend on $M^2$.
If we did it directly in $d=4$, (12.32) would have a pole depending on $p^2$, and therefore we would have to make $\delta_m$ depend on $M^2$. This would mess up all the following derivations.
| {
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When is $a_c = v^2/r$ valid? All the derivations I've seen of the formula in the title, namely $a_c = \frac{v^2}{r}$ rely on the tangential speed $v$ being constant. However, I watched this video:
https://www.khanacademy.org/science/physics/centripetal-force-and-gravitation/centripetal-forces/v/yo-yo-in-vertical-circle-example
and in it he happily uses the formula even though the tangential speed is not constant. For example, it will be much lower at the top of the circle than at the bottom. Is it true that the formula holds even for non-constant tangential speeds?
| I treat 2D first and then comment on 3D.
The complete formula for acceleration in a plane,
treated in plane polar coordinates, is
$$
{\bf a} = (\ddot{r} - r \dot{\theta}^2) \hat{\bf r}
+ (r \ddot{\theta} + 2 \dot{r} \dot{\theta} ) \hat{\boldsymbol \theta}.
$$
Using now that the velocity is
$$
{\bf v} = v_r \hat{\bf r} + v_\theta \hat{\boldsymbol \theta}
=
v_r \hat{\bf r} + r \dot{\theta} \hat{\boldsymbol \theta}
$$
we can write the acceleration as
$$
{\bf a} = \left(\ddot{r} - \frac{v_\theta^2}{r} \right)\hat{\bf r}
+ \left(r \ddot{\theta} + 2 \dot{r} \dot{\theta} \right) \hat{\boldsymbol \theta}
$$
The four terms in the expression for acceleration may be understood or named as follows
*
*$\ddot{r} \hat{\bf r}$ radial acceleration owing to motion in the radial direction
*$- \frac{v_\theta^2}{r} \hat{\bf r}$ the centripetal acceleration which causes any transverse velocity to change direction (and thus follow a circle if the other terms were absent)
*$r \ddot{\theta} \hat{\boldsymbol \theta}$ acceleration around a circle associated with second derivative of the angle.
*$2 \dot{r} \dot{\theta} \hat{\boldsymbol \theta}$
the Coriolis acceleration, which is present only when both $r$ and $\theta$ are changing.
Three dimensions
At any point on a trajectory you can define a plane singled out by the velocity and acceleration vectors. In that plane you can set up rectangular coordinates and polar coordinates in two dimensions, and then the above formulae apply.
| {
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Can internal conversion happen at (close to) absolute zero? My understanding is that, if an atom finds itself in an excited state, for instance following electron capture and gamma emission, the atom relaxes to the ground electronic state because of the perturbation caused by the room temperature. The atom may temporarily borrow energy in order to relax, if you like.
So in the absence of the kick provided by the temperature of the room, would excited electronic states be stable?
Radioactive decay is spontaneous because essentially, everything is at absolute zero for nuclear relaxations, because of the large energy changes involved in transitioning between nuclear states.
This is of course not true in stars.
Thanks in advance, I look forward to hearing the answer.
| Usually, radioactive decay has nothing to do with temperature and does not at all depend on it. (*)
Radioactive transitions are probabilistic as they are best modeled just like the tunneling effect. There is some potential barrier, but the (quantum) state of the nucleus can tunnel through that and relax to another state.
This has nothing to do with thermal energy at all. Thermal motion is kinetic motion of the entire nucleus, and thus does not provide any additional energy for the tunneling process. In the OP's picture, thermal motion kicks the entire nucleus, not just individual nucleons.
Also, typical kinetic energies at room temperature are k$_B$T $\sim 10^{-4}\mathrm{eV/K} \cdot 300\mathrm{K}\sim 10^{-2}$eV which is orders of magnitude below typical decay energies $\gg$keV. From this you also get the caveat to the initial statement: (*) ...Unless you go to crazy high temperatures and densities, or highly ionized states of atoms.
| {
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Combining spins of two particles of spin |1,0> and |1/2, 1/2> I am trying to combine the spin of two particles.
Their individual spins are:
$|1,0\rangle $ and $\left|\frac{1}{2},\frac{1}{2} \right>$
Now I am told that they combine to give a total spin state of:
$$\left|\frac{1}{2},\frac{1}{2} \right>$$
However, I am confused as to how this works. Thinking physically I am confused as:
I understand that this doesn't really make sense as spin is quantized, however, I am confused as to how the states combine if this is not the case.
| $
\newcommand{\ket}[1]{{\textstyle\left|{#1}\right>}}
\newcommand{\sqrtfrac}[2]{{\color{lightblue}{\textstyle\sqrt{\frac{#1}{#2}}}}}
%%%
$There is no way to combine $\ket{1,0}$ and $\ket{\frac12, \frac12}$ to get a pure $j=\frac12$ state. The only possible value for the $z$-axis projection $m$ is $0 + \frac12 = \frac12$, but the total angular momentum can take either of $\frac12$ or $\frac32$.
The orthonormal combinations are given by the Clebsch-Gordan coefficients, which are usually presented in horrible tables like
The way to read this horrible table is that, if you wanted to construct the composite state like $\ket{\frac32,\frac12}$ or $\ket{\frac12,\frac12}$, you would read down the columns of the second table:
\begin{align}
\ket{\frac32,\frac12} &=
\sqrtfrac13\ \ket{1,1}\ket{\frac12,{-\frac12}}
+ \sqrtfrac23\ \ket{1,0}\ket{\frac12,{+\frac12}}
\\
\ket{\frac12,\frac12} &=
\sqrtfrac23\ \ket{1,1}\ket{\frac12,{-\frac12}}
- \sqrtfrac13\ \ket{1,0}\ket{\frac12,{+\frac12}}
\end{align}
If your constituent particles are in a pure state $\ket{1,0}\ket{\frac12,\frac12}$, your composite system is in a superposition of $\ket{\frac12,\frac12}$ and $\ket{\frac32,\frac12}$. You should convince yourself that you can find its coefficients either by solving the system of equations above for $\ket{1,0}\ket{\frac12, \frac12}$, or by reading across the Clebsch-Gordan table.
| {
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What are the Newton's third law pairs in this diagram? In the diagram below, block $C$ accelerates downwards when released. I'm having some trouble identifying the third law (action $-$ reaction) force pairs in my free-body diagrams. Please provide any insights to determining the force pairs.
| *
*B and A - apply force on each other using friction (because they are sliding one over the other).
While it seems that B and C apply force on each other using the rope they are not count as third law because the force do not apply directly on each other - block C apply force on the rope (and vice versa so the rope and block C count as a third law pair) and then the rope apply force on block B (and vice versa so the rope and block B count as a third law pair) .
The only remaining pair, A and C do not have any direct interaction between them and they are not applying any force on each other.
| {
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Do resistance have to heat? Not a physicist.
Is there a way to build resistances that do not heat when opposing current?
More generally, is it necessary to waste energy to resistance?
If I cut a wire, there will be an almost infinite resistance (unless the tension is large enough to break the air), but no energy cost.
If I had a switch with a duty cycle of 50 pc, I would oppose the flow of current similarly to a resistance without heating, maybe. Is that an actual method for loss free resistance?
| If there is resistance then there is momentum being passed from electrons to other electrons and atomic nuclei which we call heat. In superconductors electrons pass some momentum to positive charges but also get it back because superconductors have charges that move in a non random coordinated manner.
| {
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A light problem: What happens when light completely destructively interferes? So here's the setup. We have a spherical source. It emits a pulse of light in all directions with some wavelength $\lambda$. It reflects off of a spherical mirror that is centered around this source.
Now, when the light comes back, it bounces off of the source again (or some percentage, whatever). The source emits another pulse of light at the same time with exactly the same energy as the light that bounced off, but shifted back $\lambda$/$2$, so all the crests line up with all the troughs and the light completely destructively interferes.
I can't see a way out of this. I've spent energy - but where did it go?
| The energy went back into your spherical source. For the first pulse the source had to provide energy. For the second pulse the source had to absorb energy. The mistake is simply assuming that the energy required to drive the source is independent of the external fields acting on the source.
| {
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What factors does magnetic field due to a circular loop depend on? I know of three.
*
*Strength of current
*number of loops ( a cheeky way to 'increase' current?)
*radius
I'm having doubts about radius of circular loop affecting the magnitude of magnetic field ta centre of loop. As if magnetic field due to an infinitesimal current carrying element is inversely proportional to radius, and number of such elements would be proportional to the radius, the magnetic field at centre of loop would be independent of radius
| The magnitude of the field at the centre is-
$$B=\dfrac{\mu_oNI}{2r}$$
andis thus inversely proportional to radius.
| {
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Confusion in Signs when formulation mathematical equations for physics problems One of the common confusion I face when writing and solving mathematical equations for physical problems is regarding signs. Consider this problem;
A ball is dropped from rest from top of the building,it falls down subjected to air resistance which is directly proportional to velocity. Find Velocity as function of time?
Let $+y$ be along downward direction
Drawing free body diagram of the body at any instant gives two forces acting on it gravity downward and air drag upwards
Assuming velocity and acceleration is downwards,writing Newton equation gives;$$mg-c(v)=m(a)$$ where $g$,$m$,$c$ are positive constants. Since this is a scalar equation since $v$ and $a$ are magnitude of the vectors. But mathematically $v$ and $a$ are variables hence can be positive or negative. Since $a={d|v|\over dt}$,the fact that speed is decreasing or increasing will mess up the sign of acceleration.
Now consider this problem;
A ball is dropped from a top of a building at a velocity greater than the terminal velocity,moving under air drag. Find $v(t)$?
Here since the ball is dropped at a velocity greater than the terminal velocity we may expect that the net acceleration is upward and velocity is downward so that it's speed decreases and it obtains terminal speed. By using same coordinate system as above and writing force equation $$mg-c(v)=m(-a)$$ but this equation is wrong in terms of sign( $(-a)$ because acceleration was in upwards direction which is negative).
So how do we deal with the sign of variables in mathematical equations in physics problems?
| We must always pick a direction to be positive and stick to it throughout.
A good tip to reduce confusion is to pick the positive direction to be the way you expect the object to accelerate.
So in your first example pick positive to be down. In the second pick positive to be up.
After calculating $a$, you'll have your answer. If it comes out positive that's fine, it's in the direction you expected. Occasionally it might come out negative, this means that your initial guess for the direction of the acceleration wasn't right and it's accelerating the other way.
| {
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Where from is the distance calculated in Newton's law of universal gravitation? In the equation for universal gravitation $(1)$ between two objects, where from is $r$ calculated? From the surface, from the center? Also, are the objects assumed to be particles in this equation or could the be multiple particles like molecules?
$$F=G \frac{m_{1} m_{2}}{r^{2}}\tag{1}$$
| The object in this law assumed to be points (so the distance is just distance between points). Any two object that are very small compare to the distance between them (no matter from which points on the objects you measure it) can be considered as points.
So, for example if you have a two rigid molecule that are small compare to the distance between them, then you can use this equation to calculate the force between them.
| {
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Is conduction band discrete or continuous? My question is very simple. I just want to know that is conduction band discrete or continuous?
| As other have noted, it's discrete but with fine enough spacing to treat as continuous.
However I disagree that quantum mechanics is the reason. You see the exact same thing in a classical 1-D chain of masses connected by springs.
The allowed wave vectors $\vec{k}$ in the bands are reciprocal lattice vectors, and the number of reciprocal lattice vectors is equal to the number of real-space (aka direct) lattice vectors. That in turn is equal to the number of atoms. So, if you have an infinite number of atoms in your lattice, then there are an infinite number of reciprocal lattice vectors, and your band is continuous. (I guess it's countably infinite which is a little different that continuous, but let's not go there.)
Real materials have a finite number of atoms, so the bands are, strictly speaking, not continuous. However, most crystals that people deal with are large enough that the number of atoms is huge --- effectively infinite.
| {
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How does Dodelson calculate the number density of a particle species? In chapter three of Modern Cosmology, Dodelson models the evolution of a particle plasma as the universe expands. On page 61, the author gives the formula for the species-dependent equilibrium density as:$$n_i=g_i\space e^{\mu_i/T}\int\frac{d^3p}{(2\pi)^3}e^{-E_1/T}\tag1$$For equilibrium (when a species is created as often as it's annihilated), we have:$$n_i^{(0)}=g_i\space\int\frac{d^3p}{(2\pi)^3}e^{-E_1/T}\tag2$$Now this is the part I don't follow. How does Dodelson go from Eq. (2) to this:$$n_i^{(0)}=g_i\left(\frac{m_iT}{2\pi}\right)^{3/2} e^{-m_1/T}\tag3$$I get that $\int d^3 p=\frac{p^3}{({2\pi})^3\space6}$, and I get that $E_1=m_1$, and I get that $p=\sqrt{E^2-m^2}$, but I can't put them together. How do we go from Eq. (2) to Eq. (3)?
| I'm afraid your primary error is one of elementary calculus. $\int d^3p$ is not equal to $\frac{p^3}{(2\pi)^3 6}$. Secondly, $E\neq m$ (how could it, since as you correctly say in your next words, $p=\sqrt{E^2-m^2}$??).
To evaluate this integral we go to spherical coordinates, in which $\int d^3 p = \int_0^\infty p^2 dp \int_{0}^{\pi} sin(\theta)\space d\theta \int _0^{2\pi} d\phi = 4\pi \int_0^\infty p^2 dp$, where we've used the spherical symmetry of the problem to evaluate the angular integrals. From there, we have
$$n_i^{(0)} = g_i \frac{4\pi}{(2\pi)^3} \int_0^\infty p^2 e^{-\sqrt{p^2+m_i^2}/T} dp$$
This integral doesn't have a nice form. However, if we assume that $m_i \gg p$, we can approximate $\sqrt{p^2+m_i^2} \approx m_i + \frac{p^2}{2m_i}$ (a non-relativistic approximation), which allows us to simplify to
$$n_i^{(0)} \approx g_i \frac{4\pi}{(2\pi)^3}e^{-m_i/T} \int_0^\infty p^2 e^{-p^2/2m_iT} dp$$
$$ = g_i\frac{4\pi}{(2\pi)^3} e^{-m_i/T} \frac{\sqrt{\pi}}{4}(2m_iT)^{3/2}$$
which simplifies after a bit of algebra to
$$n_i^{(0)} \approx g_i e^{-m_i/T} \left(\frac{m_iT}{2\pi}\right)^{3/2}$$
The non-relativistic approximation is valid as long as $m\gg T$ or, in SI units, $m\gg \frac{k_B T}{c^2}$.
| {
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Why black holes lose energy? It is often said that black holes lose energy throughout the process called Hawking radiation. My question is, since the process of creating particles is random, why do the black holes captures more particles with negative energy than particles with positive energy?
| As Wikipedia notes, when one particle escapes the event horizon but a partner particle doesn't, the latter isn't the issue (as its energy neither enters nor leaves), but the former has more energy due to a gravitational redshift. This amplification produces a partner wave. It's this wave that actually returns negative energy to the black hole.
| {
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Is it possible to lift an object from rest with constant power? This is inspired by the following question.
Consider some object which I want to lift from rest with a constant power throughout the whole process; the power I apply when lifting the object from rest is the same power I apply to keep lifting it. The force $F$ and the speed $v$ may change, but power may not.
The power is $P=Fv$. If we want constant power, then $dP/dt=0$.
First, differentiate wrt time, $$\dfrac{dP}{dt} = F \dfrac{dv}{dt}+v \dfrac{dF}{dt}.$$
Set equal to zero, this guarantees constant power, which implies:
$$F \dfrac{dv}{dt} = -v\dfrac{dF}{dt}.$$
From there, I use $F=\dfrac{d(mv)}{dt}\implies F = m\dfrac{dv}{dt}$ and get $$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0.$$
How may I solve this non-linear ODE?
EDIT: Initially, I got to the wrong conclusion that $F(v-v_0) = -v(F-F_0)$ is a solution. That's what the answer by Vilvanesh addresses.
| $$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0$$
$$\frac{dv}{dt}\times\Big[\frac{dv}{dt}+v\Big]=0$$
When $\displaystyle{\frac{dv}{dt}=0}$, $v$ has to be a constant. But that isn't possible because initially it was at rest but while being lifted, it is not.
$$\frac{dv}{dt}+v=0$$
$$\frac{dv}{v}=-dt$$
$$\int\limits_{v_o}^v\frac{dv}{v}=-\int\limits_0^tdt$$
$$\ln{v}-\ln{v_o}=-t$$
But since $v_o=0$, $\ln{v_o}$ won't be defined.
Hence a solution doesn't exist for the above differential equation.
| {
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How does the temperature affect the diameter of a brass ball, and the hole it has to go through (hole is in a thin steel plate)? A brass ball has a diameter of 25.232 mm and a hole in a thin steel plate has a diameter of 25.220 mm at 21.5 C. Both the steel and the brass ball have the same temperature all the time. At which temperature does the brass ball go through the hole?
First I calculated the volume of the brass ball
$V=\frac{4}{3}\cdot\pi\cdot\left(12.616\right)^3=8411.1165379117\ mm^3$
Then the volume needed for the brass ball to go through the hole.
$V=\frac{4}{3}\cdot \pi \cdot \left(12.61\right)^3 =8398.1225396174\ mm^3$
$V=\left(57\cdot\frac{10^{-6}}{C}\right)\cdot\left(8411...mm^3\right)\cdot\left(x\right)=8398mm^3$ (1)
$x=17516.8C$ (1)
But then I realised that the diameter of the hole gets also effected by the temperature, and now I'm stuck. Does anyone know how to solve this kind of a problem?
edit (1): added calculations that I've calculated
| The coefficient of thermal expansion describes the increase in a diameter or a circumference, since one is a linear function of the other. You don't need to work with the volume.
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How does the growing up of trees works with respect to energy conservation? We know that the energy (and hence mass) can neither be created nor destroyed. In the light of above statement, I would like to understand views on growing up of fauna and flora on earth.
A small seed is planted in earth. The only visible input for the seed is amount of manure, fertilizer and water added to it. Agreed, it receives a good amount of energy from sun and other minerals from the earth.
The output: The tree grows up amounting into hundred of tons of wood mass, thousands of flowers and fruits.
Considering that this has been happening for billions of year, Is it fair to assume that the above output is produced from the above set of inputs only and there has been no increase of mass in the process.
| You are exactly right. A plant like a tree is basically a chemical machine for storing chemical energy by converting water and carbon dioxide into cellulose and other hydrocarbons, using solar energy to do the work. Mass is conserved at every step of the process.
Animals then chemically convert plant matter into muscle, bone and brains (in some cases). Mass is strictly conserved here too.
Now since we know that energy has mass via E = mc^2, the solar energy in principle adds mass to any object that stores it up, but the amount of mass increase in a tree via E = mc^2 is millions of times smaller than the mass increase of the tree as it manufactures cellulose and other hydrocarbons, and hence completely negligible in practical terms.
| {
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By how much does the gravitational redshift change a neutron star emission spectrum, thus disturbing the measurement of its surface temperature? How much does the gravitational redshift change a neutron star emission spectra disturbing so the measurement of its surface temperature? I remember some tv-shows talking about Italian pasta when talking about neutron stars....does this imply neutron stars are not all of same density so gravitational redshift differs depending on differences in densities between neutron stars?
| The standard gravitational redshift formula applies to neutron stars.
This means that if a distant observer observes a spectral feature that was emitted at a rest wavelength of $\lambda_0$ from the NS surface, then it will be observed at a wavelength
$$\lambda = \lambda_0\left(1 -\frac{2GM}{Rc^2}\right)^{-1/2}\ ,$$
where $M$ and $R$ are the neutron star mass and radius (actually the Schwarzschild $r$ coordinate of its surface).
Since establishing a temperature from a blackbody spectrum involves finding the peak wavelength, then the observed temperature is reduced by
$$T = T_0\left(1-\frac{2GM}{Rc^2}\right)^{1/2}\ .$$
The ratio $M/R$ probably does vary a little bit (up to a factor of 2) between the most and least massive neutron stars. Roughly, $M/R \propto M$, since the equilibrium radii of neutron stars are insensitive to their mass (in most models).
"Nuclear pasta" occurs in neutron star interiors and doesn't affect radiation from their surfaces.
| {
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How to explain smooth absorption spectrum of nature (plants, etc.) while the atoms have discrete absorption lines? From quantum mechanics, a photon of energy exactly equal $h\nu=E_2-E_1$ could be captured by an atom that has the energetic level levels $E_1$ and $E_2$ (with $E_1<E_2$). This corresponds to "absorption lines" for the photons in the material made of a single type of atoms.
How to explain that in nature, for example in plants, we see absorption spectrum : that is : there is no "Dirac-like" missing lines, but we have really a smooth curve of absorption spectrum.
Example :
(Source)
| I'm going to go with an intentionally provocative answer. Atoms don't have discrete absorption lines. Nothing has discrete absorption lines. Any state that can decay or be excited necessarily has a finite lifetime $\Delta t$, and thus its energy is uncertain by an amount $\Delta E$ that can be estimated from a simple Fourier argument to be
$$ \Delta E = \hbar\Delta \omega \sim \frac{\hbar}{\Delta t}.$$
In the case of atoms, the typical line widths are $\Delta \omega \lesssim$ GHz and the typical frequencies can be hundreds of THz, meaning that the lines look very sharp on a coarse-grained resolution. But "Dirac delta" peaks just don't occur in Nature. Anything that can be measured couples to the outside world and therefore has a finite lifetime/linewidth.
In reality, the broad spectra of typical materials are of course explained by the points raised in Vadim's answer: you have "classical" broadening by mixing the spectra from different sources, as well as the truly "quantum" broadening that arises from coupling between systems with the same spectra. Note that the latter is essentially the same physics as I described above, since localised states of a composite system have a finite lifetime due to the coupling between subsystems.
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How do I find the approximate surface area of a chicken? I'm working on building a chicken army and I'm trying to find out how much metal or kevlar (still deciding) I need to make armor for the chickens. this measurement does not need to be exact I'm just trying to get an estimate for how much I will need. You will be spared when my chickens take over the world if you give me a working answer.
| Use the ancient Japanese art of Gyotaku. This method was used to record the catches of Japanese fishermen. They would paint the fish with ink and then roll the fish onto rice paper to get a print.
*
*Dye the chicken in harmless food coloring.
*Carefully roll one side of the dyed chicken onto a large piece of paper.
*Wash the chicken before returning it to the coop.
*Estimate the area of the image on the paper using polygons. Multiply the result by 2 (to account for the other side of the chicken).
An advantage to this method is that it will also provide a template for stamping steel sheet to be formed on a rigid chicken armory dummy, or to cut the Kevlar prior to sewing.
I embrace my chicken overlords and submit to their rule!
| {
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Is no acceleration a cause or consequence of no net force? If a body is moving with constant velocity, or is at rest, then the net force on it must be $0$. If the net force on a body is $0$, then it must be moving with constant velocity or must be at rest.
Is $0$ net force a consequence of being at rest or moving with constant velocity or is moving at constant velocity or being at rest a consequence of $0$ net force?
| The latter.
Every frame is at rest for observers on it. If they note that the objects on the surroundings are passing by at increasing velocities, how they can decide who is really accelerating, their own frame or the objects on the surroundings?
There is a (fictitious) force in the accelerated frame that can be identified by observers on it. Based on that force, they know that they are accelerating.
On the other hand, if there is no net force in the frame, they know that their own frame is inertial. And in this case, if the surroundings seems accelerating, they are really accelerating.
| {
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Mass-mismatch in the low-lying states of the bosonic string $\newcommand{\ket}[1]{\left|#1\right>}$I've recently been studying the quantisation of the bosonic string (using GSW as my main text). However, I have some issues which I believe should be pretty straightforward, but I can't seem to work out!
We derive the spectrum of states of (open, for example) strings. The masses of these states are given by the mass operator $M^2 = 2(N-a) = 2(N-1)$ setting $\ell = 1$, $\alpha' = \frac{1}{2}$, and with $N$ the 'level/number' operator. We act with this on the space of states generated by the sequential application of any of $\alpha^i_{-n}$ onto the ground state $\ket{0;k^\mu}$ (which we understand to exist from something to do with the Stone-Weierstrass theorem). This is the tachyonic ground state.
This implies that (in light-cone gauge):
$M^2\ket{0;k^\mu} = -2\ket{0;k^\mu}$
This means that $k^\mu k_\mu=2$! That would be fine - but now I apply $\alpha^i_{-1}$ to my (string) vacuum state and create a level $N=1$ massless photon-like particle. The momentum operator commutes with the above creation operator, and so the state still has the same 26-momentum. However, it should also now have mass $M^2=0$. I don't see how these two statements can be coherent. Is it simply that if one attempts to create a state with the incorrect 26-momentum for its mass, the result is unphysical and so decouples? I have found neither hide nor hair of such a result or argument in GSW, Tong, etc.
What am I missing? Thanks very much...
| You've demonstrated that the creation operator $\alpha^i_{-1}$ can map a physical state to a non-physical one. There is no contradiction. If you've seen the BRST approach yet, a succinct way to phrase this is that the action of the oscillators does not preserve the subspace of BRST-cohomology classes (i.e. physical states).
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Can speed of image in mirror exceed the speed of light? Suppose I'm moving toward the mirror at the speed of v, then relative speed btw me and mirror would be v.
And I thought that we can think like there exist world that is the same as our world inside the mirror,
so I guessed we will measure speed of the image as 2v/(1+v^2), according to relativistic velocity addtion.
But if we think that I'm stationary and mirror is moving toward me, then it looks obvious that speed of image should be 2v because distance btw me and mirror and mirror and image must be always same.
So Im very confused about this, and honestly i have no idea about how can we define speed of image because it doesn't actually exist.
(If it existed as normal object, we shoot the light and by measuring reflected light and getting average of two instance, we might be able to specify its position and time)
| Prepare to have your mind blown: hold up a mirror to the night sky, and point it at a quasar. Flip it 45 degrees. You've now made the mirror image of that quasar travel several billion light years in less than a second.
The motion of mirror images is virtual, just a mathematical construct, and relativity does not apply to it.
| {
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Characteristic timescale of a can of beer? Modelling a 440 ml can of beer by Newton's law of cooling, the temperature difference between the can and its surroundings (e.g. a fridge) decays exponentially in time at a rate determined by the e-folding time or "characteristic timescale".
Ideally, this characteristic timescale would be determined by experimentation to give the best approximation for a particular environment (fridge). In the absence of necessary equipment to perform the experiment, can the characteristic timescale be estimated to reasonable accuracy from first principles?
| No, because this timescale is dependent on the material inside and outside the can.
The can will lose heat to the environment by radiation and conduction from its surface. The rate at which this occurs is dependent on the colour of the surface (dark surfaces are more effective radiators) and the rate at which heat is transferred from the body of the can to the surface. In the case of a real can, the outside edges will start to cool first, which will set up convection currents within the beer, transferring heat from the interior to the surface.
Finally the effects of the air in conducting and convecting heat away from the can need to be considered.
For the case of a spherical perfectly black body in a vacuum with perfect interior heat conduction, the rate of heat loss can be calculated. For real cans of beer, an empirical approach is best. (45 minutes is enough to chill most drinks to close to fridge temperatures https://www.omnicalculator.com/food/chilled-drink)
| {
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What is the significance of the sign of the velocity for a particle executing SHM? So while deriving equation for the velocity of particle executing SHM at any point, I noticed a difference in the result depending on what wave (sine or cosine) you chose.
For $x=A\cos\omega t$:
$\quad \ \,v=-\omega\sqrt{A^2-x^2}$
For $x=A\sin\omega t$:
$\quad \ \,v=\omega\sqrt{A^2-x^2}$
Can anyone explain to me why the difference is there and what it means, since both equation are basically the same with only a phase difference?
| Neither of those expressions are right. If $x(t)=A\cos(\omega t)$, then $v(t) = -A\omega \sin(\omega t)$. You wrote
$$v(t) = -\omega \sqrt{A^2-x^2}=-A\omega \big|\sin(\omega t)\big|$$
If we write $x(t) = A\cos(\omega t - \delta)$, then your two examples correspond to $\delta = 0$ and $\delta = \pi/2$. In the general case,
$$v(t) = -A\omega \sin(\omega t- \delta)= A\omega \cos(\omega t - \delta -\pi/2)$$
so we see that in general, the velocity function is related to the position function by a $\pi/2$ phase shift and scaling by $\omega$.
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Distribution of charge when 2 metallic spheres are connected It is given in my textbook that when 2 charged conducting sphere of different radius are connected by finite wire the redistribution of charges takes place such that the potential just outside of both spheres become equal.
But why potential is the necessary condition?
Like if net electric field just outside one sphere is 0 then even if there is some potential difference charge will not flow ,so why doesn't electeic field is necessary condition ??
And what should be the relation between the charges of two sphere after equillibrium is established when they are just touched?
| When we connect two spherical conductors with a conducting wire, essentially the potential on surface of both of the conductors become the same. Basically they are equipotential surfaces. Now given $E = dV/dl$, and in this case, $dV =0$, $E=0$. Now you are telling that electric field is zero at only one point between the two conductors, but given what I just mentioned, shouldn't it be $0$ everywhere between the surface of the two conductors?
| {
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Can I push something standing on a frictionless floor? Suppose I stand on a friction less floor, and another object of finite nonzero mass stand in front of me. Can I push the mass, so that it has a nonzero acceleration? Also, where does this force arise, since I cannot push the ground, since that would make me slip on the floor.
| Absolutely you can apply a force on some other object and make it accelerate. What will happen is that you too will accelerate - in the opposite direction.
You push the thing, the thing pushes you, it's classic Newton's Third Law. This is akin to rocket propulsion.
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How do I go from Planck units to string units? I am not sure exactly what "string units" are. I know that $\alpha' = l_{string}^2 = \frac{1}{M_{string}^2}.$ But these are not units per see, I still need values for $l_{string}$ and $M_{string}$ that could be in whatever units?
My question is, how do I go between string units and Planck units? Alos, what is the unit of $g_s$ the string coupling constant?
| *
*The closed-string string coupling constant $g_s$ in both bosonic and superstring theory is the vacuum expectation value of $e^\Phi$, where $\Phi$ is the dilaton field. It is dimensionless: indeed, all dimensionless parameters in string theory arise as VEV's of scalar fields.
*In the bosonic case, the 26D Planck length $\ell_p$ is set via $\ell_p^{24}=\ell_s^{24}g_s^2$. This is seen via the low-energy bosonic effective action. The 26D Planck mass is related as $M_p=\ell_p^{-1}$
| {
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In a sealed chamber, can a vapor be heated enough, that the pressure increases enough, to condense to liquid? Assuming a sealed chamber, containing a vapor, where volume is constant:
Given that the ideal-gas law states $PV = nRT$, if heat is added to an ideal-gas, the pressure will increase as a function of temperature according to:
$$ P = \frac{nR}{V}T$$
So if this constant $ nr/V $ is great enough, then I'm assuming that water vapor would increase enough for a vapor to enter the liquid phase.
So theoretically you could heat steam into condensing?
Or would there be issues at the boundary due to the reduction in latent heat needed to condense?
|
So theoretically you could heat steam into condensing?
Try it (graphically)! Choose an amount of water vapor $n$ and calculate $V$ for a gas-phase $T_0$ and $P_0$ combination on the phase diagram:
Now draw the line $P=nRT/V$ that passes through ($T_0$, $P_0$). (This will appear as a curve on this log-linear plot.) Can you enter the liquid phase through raising the temperature?
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Why is a large cavity with a small hole at constant temperature approximated as a black body? I can't understand how a large cavity maintained at a constant inner surface temperature T, with a small opening on its surface behaves like a black body.
How is it a perfect emitter and a perfect absorber?
Please explain in as simple terms as possible.
| The hole is the black body.
The term black body is to do with an object that doesn't reflect anything, and so appears black.
Any light falling on the hole doesn't get reflected, but bounces around inside the box until absorbed. The chance of it re-emerging from the small hole is zero, so it's a perfect absorber.
As there is no reflected light, the only light leaving the hole is due to thermal radiation from the walls of the box, depending only on the temperature.
| {
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Buoyant force on partially submerged bodies When a cylinder is placed on a fluid, it is partially submerged and attains equilibrium at that state.
In equilibrium,
$$F_B=mg$$
where $F_B$ is the buoyant force and $mg$ is the weight of the cylinder.
But we know that $F_B=m_{\text{liquid displaced}}\ g$. Thus, $ m_{\text{liquid displaced}}\ g=mg$
But in the above case, $m_{\text{liquid displaced}}\neq m$ as only a part of $m$ is inside the fluid.
During partial submersion, how can we write the equation for equilibrium state?
| There are two equal and opposite forces on the cylinder, $w = m_{cyl}g$, and $F_B$.
$F_B$ is the force from pressure that water exerts on the cylinder. Since the pressure increases with depth, forces on the bottom are greater, and these are directed upward.
These bouyancy forces are the same, no matter what object with the same size and shape is subject to them. In particular, just enough water to fill the hole is subject to the same bouyancy force.
$$F_B = m_{water}g$$
So $m_{cyl}g = m_{water}g$. We know the cylinder has a larger volume because some of it stick up above the water. So the cylinder is less dense.
This means if you had a truncated cylinder the same size and shape as the water that filled the hole, it would not be in equilibrium fully submerged in the same position.
| {
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What is light, a wave or a particle or A wave-particle? What is light?
And how do we know that light is an electromagnetic wave?
I asked my teacher and he said that when you place a compass in light's path, the needle of the compass rotates. Which I think is not a valid answer and thats not what actually happens when we place a compass in path of a light.
| This is the simplistic answer that I found useful when I was a kid.
It isn't strictly rigourous, but it was useful. (See also "All models are wrong, but some models are useful".)
Light is something (a "thing", a phenomenom, whatever) which has the property that:
*
*if you do a test that asks "Does this thing behave in this way, which particles behave in?", it will mostly say "Yes".
*and if you do a test that asks "Does this thing behave in that way, which waves behave in" it will also mostly say "Yes".
It isn't meaningfully either thing.
It isn't "a wave" (in any sense that can used to draw analogies with other waves)
And it isn't "a particle" (again, in any sense that you use to draw analogies with other particles).
It's just light. It has properties of both / either / whichever one you ask about.
| {
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Reconstructing wavefunction from the density matrix Say I have a state, $$| \Psi \rangle = \frac{1}{\sqrt 2} \left( | 0 \rangle + \exp( \text{i} \phi ) | 1 \rangle \right) = c_{0} | 0 \rangle + c_{1} | 1 \rangle.$$
Now I construct the density matrix (DM), $$\hat \rho = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left( | 0 \rangle \langle 0 | + \exp( - \text{i} \phi )| 0 \rangle \langle 1 | + \exp( \text{i} \phi ) | 1 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right).$$
So from the DM $\hat \rho$, I can read off $|c_{0}|^{2}$, $|c_{1}|^{2}$, $c_{0}c_{1}^{*}$, and $c_{0}^{*}c_{1}$. Basically $3$ equations and $4$ unknowns.
Is there a way to reconstruct $| \Psi \rangle$ uniquely from the DM, $\hat \rho$?
| If you write $c_0 \equiv |c_0|\, e^{i\phi_0}$ and $c_1 \equiv |c_1|\, e^{i\phi_1}$, then you can write the wave function as
$$|\Psi\rangle = e^{i\phi_0}\left(|c_0| \,|0\rangle + |c_1|\,e^{i(\phi_1-\phi_0)}\,|1\rangle\right)\quad . $$
The associated density operator is then given by $\rho_{\Psi}\equiv |\Psi\rangle\langle \Psi|$. The diagonal elements will yield $|c_0|$ and $|c_1|$ and from the off-diagonal terms you can reconstruct $|c_1| \, e^{i(\phi_1-\phi_0)}$. However, you can only reconstruct the wave function up to the global phase, which is also intuitive, since two wave functions $|\Psi\rangle$ and $|\psi\rangle$ which differ only by a global phase will yield the same density operator.
| {
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My question is about the concept of acceleration What would happen to the velocity of an object if it is given an initial acceleration of 2 meter per second per second in a case if there is no dissipating force such as gravity, air resistance or friction. I think its velocity would increase by 2 meter per second after every second and will go on increasing forever. Am I right? please tell me.
| An object with mass accelerates only if is being pushed by a force. If the force is removed, the acceleration will stop. The problem with a rocket is that you have to accelerate the fuel until you run out.
| {
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What is the exponential (or geometric) rule (or law) for uranium enrichment? Uranium ore starts at about .72% U-235...
At ~20% U-235, it is considered to be about '90% of the way' to weapons-grade uranium, which is about ~90% U-235...
Because uranium enrichment in centrifuges follows a geometric (or exponential) law...
I have read about this repeatedly when hearing about Iran's enrichment program....
Does anybody know what the 'rule' or 'equation' is for uranium enrichment...
(I am not trying to build a bomb, I swear....)
Edit: P.S.:
In the Work equation $W_extract = -T R ln(x)$ , what are T, R, and x?
I can find that equation nowhere else....
| The equation you are looking for is the SWU calculation (Separative Work Units). The calculations are described at:
https://energyeducation.ca/encyclopedia/Separative_work_unit
Figure 1 shows the exponential dependence that you describe. Enriching from natural to 4-5% takes more energy then going from 4-5% to 95%.
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Acceleration of cylinder both rotating and moving under force from a pulley The following states the problem 1.2 from the book Lagrangians And Hamiltonians by Patrick Hamill.
A cylinder of mass $M$ and radius $R$ is set on end on a table at a distance $L$ from the edge, as shown in Figure 1.11. As the string is wound tightly around the cylinder, the free end of the string passes over a friction-less pulley and hangs off the edge of the table. A weight of mass $m$ is attached to the free end of the string. Determine the time required for the spool to reach the edge of the table.
The given answer for acceleration of cylinder across table is $\frac{mg}{M+3m}$ (using which time is calculated). Can we explain intuitively why the term $M+3m$ appears in the solution without resorting to the dynamical equations of motion?
| If the cylinder starts from rest and the acceleration, a, is given, then L = (1/2)a$t^2$.
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According to general relativity planets and Sun bend the spacetime (explaining gravity), but does this hold true for smaller objects? According to general relativity planets and the sun bend spacetime, and that is the explanation of gravity. However, does this hold true for smaller objects, like toys, pens, etc.? Do they also bend spacetime?
| A large body is an accumulation of many small bodies. Its properties — mass, and with it gravitation and inertia, heat capacity etc. — are nothing but the accumulation of the properties of the constituting small bodies. And vice versa :-).
Remember that all small objects you hold in your hands were once parts of very heavy stars which did visibly bend spacetime. Their atoms got expelled when that star went nova, billions of years ago, and later condensed again to form our solar system. The atoms in your toy bend spacetime just like they did when they were part of that large star.
| {
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Muon $g-2$ experiment: is there any theory to explain the results? The nature of the experiment has been discussed here, but my main question is this: is there any theory that has predicted the results of this experiment or are we completely clueless about what's happening? In other words, have we come up with a new hypothetical interaction that could explain the results?
| This answer was already suggested above. This is another source that has some really nice embedded links on the authors and the university that hosted the study. Very nice reading some might enjoy so I thought it would be nice to share. Assuming I am allowed to put in a link. Not sure it is allowed by stack exchange.
https://thenextweb.com/news/did-we-just-discover-new-physics-theoretical-physicists-dont-think-so-syndication
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Why do images not appear inverted when looking directly through a pinhole camera? I understand that the way light takes through a pinhole creates an inverted image on a surface behind the pinhole. I remember this effect from school experiments, it's also described in this wikipedia article. I punctured a piece of paper and looked through it (instead of watching the reflection), the image appeared as normal to me. Why is that? Why doesn't the scene appear upside down when looking through the hole?
| The directly-through-the-pinhole image is upside down on the retina of your eye. But all images on the retina are upside down. When the lens in your eye forms a real image on the retina it is inverted. It only looks the right way up to you because you brain post-processes the retina image in the visual cortex.
By looking with your eye at the image formed on the screen at the back of the camera you see a triply inverted image – inverted once by the pinhole, once by the lens in your eye, and once by you brain – so you see an upside down image.
| {
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What is the bare minimum of air density you need to fly a helicopter? How high will a helicopter be able to fly before the propellers have not enough air particles to achieve lift?
What is the minimum air density needed to achieve flight with a helicopter?
Could you give me the air density from $100\%$ to $1\%$ in meters? Only when there is a significant decrease, otherwise it might be too much work.
Correction :
Let say one is a drone and one is a helicopter.
The helicopter weighs between $500-730 kg$.
The drone weighs between $1-2 kg$.
Which air density is needed for each of them to fly?
At what altitude would there be not enough air particles to achieve lift and what would the air density be?
| The lift generated by a helicopter is linearly proportional to the air density (as it is for planes). To fly we need the lift to be greater than the weight of the aircraft, so as the density reduces there will be a point where the lift generated is less than the weight, so we can't fly.
The minimum air density will depend on the weight and aerodynamic design of a given helicopter.
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What is momentum? Momentum tells you the mass of the object and how fast it is going right? So if I have a 2 kg ball moving at 2 m/s, then the ball has 4 kg⋅m/s of momentum. My question is why do we multiply mass and velocity to get momentum. (From the example above) Why cant we just say the ball is 2 kg moving at a speed of 2 m/s and that is momentum. Why do we have to multiply it?
| I think we should let the Original Gangster speak when talking about intuitive definitions of momentum. From Newton's Principia:
"The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly.
The motion of the whole is the sum of the motions of all the parts; and therefore in a body double in quantity, with equal velocity, the motion is double; with twice the velocity, it is quadruple."
Since conjunctly means as a product. We can paraphrase this:
Momentum is the quantity of motion. Motion can double in quantity by having 2 things of equal size move at the same rate, or by having something go twice as fast. The two are equivalent.
These ideas have been updated by modern theory to some degree, but it is still safe to call momentum a measure of the quantity of motion.
| {
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How to calculate the amount of standards deviations a measure is from another value I have a theoretical value $\gamma_t$ and a measured values $\gamma_m$, with uncertainties $\sigma_{\gamma_t}$ and $\sigma_{\gamma_m}$ respectively. If the errorbars of $\gamma_m$ is inside of $\gamma_t$-'s, as illustrate below, would that mean that $\gamma_m$ is within one standard deviation of $\gamma_t$? Or would the best guess for $\gamma_m$ have to be with the errorbar for $\gamma_t$?
A specific example:
$$ \gamma_t = 4.9\pm 1.1, \quad \gamma_m = 3.1\pm 1.5$$
In this case, the errorbars overlap but $\gamma_m=3.1$ does not lie within one standard deviation of $\gamma_t$. Would $\gamma_t$ be within one $\sigma_{\gamma_t}$ or would it instead be $n$ away, where $n$ is:
$$ \gamma_t = n\sigma_{\gamma_t}+\gamma_m $$
$$ \Rightarrow n \approx 1.6 $$
I hope I made my case clear. Here is the illustration:
| When you want to see if two quantities with uncertainty agree, you should compare their difference with zero. If $\gamma_t-\gamma_m=0$ within the uncertainty, they agree. The difference is the best way to check the degree of (dis)agreement, too. The number of standard deviations, $n_\sigma$, is:
$$n_\sigma = \frac{\gamma_t-\gamma_m}{\sigma},$$
where $\sigma$ is the total uncertainty on the difference.
Assuming the two quantities are independent, then their uncertainties add in quadrature:
$$ \sigma = \sqrt{{\sigma_t}^2 + {\sigma_m}^2}.$$
If the two quantities are not independent, a more detailed error analysis is needed. For example if both $\gamma_t$ and $\gamma_m$ were calculated using the length of some part of the experimental apparatus, $\ell \pm \sigma_\ell$, the uncertainty in that length would contribute to both $\sigma_t$ and $\sigma_m$. In that case the two are not independent.
Assuming your two quantities are independent,
$$ \gamma_t - \gamma_m = (4.9 \pm 1.1) - (3.1\pm 1.5) = 1.8 \pm 1.9,$$
which does agree with zero within the combined "$1\sigma$" uncertainty. In your case $n_\sigma<1$.
| {
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Why are the wavefunctions of the excited states so symmetric? The excitation from a lower level to an excited state happens through absorption of a photon. But the photon comes from a certain direction. I would expect that the wavefunction of the excited state has to mirror that circumstance and has some non-symmetry in direction of the incoming photon. Even if the photon is 'bigger' than the atom it has at least an impulse in some direction and one could think that that impulse should be visible in the pattern of the wave function, which would made it nonsymmetric. Can there be a transient wavefunction which reflects the impulse of the photon and rapidly migrates to the known excited state wavefunction?
| When a particle (for example an atom) absorbs a photon, it gets a very small momentum kick in the direction in which the photon was traveling. Usually this isn't large enough to make an appreciable difference, but if the particle is cold enough then this can be observed experimentally. See How does one account for the momentum of an absorbed photon? for a more detailed description of this.
In terms of the wavefunction, it's important to specify what part of the wavefunction you are looking at, whether that is the position wave function or the atomic orbital. The position wavefunction will be affected by this momentum kick. The orbital transition will only have to do with the frequency of the photon (and in a more complicated picture, the polarization of the light), and therefore should not depend on the direction of the photon.
| {
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Finding the final velocity of a car after 8 seconds with energy I'm having trouble on finding how to solve this question that I received.
A 1500kg race car's engine can produce 150,000 Watts of power. Assuming the race car is able to convert 75% of the energy into kinetic energy, how fast could the car go in 8 seconds?
I understand that I have the mass, work (with efficiency), initial velocity, and time. I tried using the Energy formula (Ei +/- W = Ef). Using that I found the final velocity but how was I supposed to incorporate 8 seconds into it?
| Well, power is just energy over time, one watt is just one joule/second. Hence, convert 150000w to Joules by canceling the seconds. --> 150000j/s * 8s = 1200000j
Now you stated that 75% of the energy is converted to kinetic. so, --> 800000j
now just solve for velocity using the definition of kinetic energy; (1/2)(m)(v^2)
upon doing the algebra one fines that the car would be going 32.65m/s sigfigs(30m/s).
Which is aproximatly 72mph sigfigs(67).
Also, have a great day =)
| {
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WKB application on symmetric potential well I am a little confused how one can find a wave function by using WKB approximation? I do know the oscillation frequency $$\Omega ~=~ {2E\over h}{\rm Re} \langle L|R \rangle~=~ {E\over \pi\hbar}{\rm Re} \langle L|R \rangle, $$
where $L$, $R$ are the eigenstate of the left and right well.
However, the key of finding this inner product is a pain to me, can someone teach me through it?
| Hints:
*
*In Ref. 1 it is claimed that
$$ \frac{2\pi}{\tau}~=~\Omega~=~\frac{E_n^--E_n^+}{\hbar}~=~\frac{\omega}{\pi}e^{-\phi},\tag{8.63/8.64}$$
where
$$ \phi~\equiv~ \int_{-x_1}^{x_1} \!dx |k(x)|, \tag{8.60}$$
so that
$$ \phi \sim~ \alpha a^2 \quad \text{for} \quad V(0)\gg E \quad \text{where} \quad \alpha~\equiv~ \frac{m\omega}{\hbar}.$$
*Now let's for simplicity assume $n=0$. If we define $$\begin{align}\psi_{R/L}(x)~=~&A\exp\left(-\frac{\alpha}{2}(x\mp a)^2\right)~=~\psi_{L/R}(-x), \cr A~\equiv~& \left(\frac{\alpha}{\pi}\right)^{1/4},\end{align}$$
to be the $E_0=\frac{\hbar\omega}{2}$ ground state in the right/left well
$$ V_{R/L}(x)~=~\frac{1}{2}m\omega^2(x\mp a)^2, $$
then indeed
$$\begin{align}\langle L | R\rangle~=~& \int_{\mathbb{R}}\! dx~\psi_L(x)\psi_R(x)\cr
~=~&\sqrt{\frac{\alpha}{\pi}}\int_{\mathbb{R}}\! dx~\exp\left(-\alpha(x^2+ a^2)\right)\cr
~=~&e^{-\alpha a^2}~\sim~e^{-\phi}. \end{align}$$
References:
*
*D. Griffiths, Intro to QM, 1995; problem 8.15.
*L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S50$ problem 3.
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Kinetic Energy and Inductor energy violates conservation of energy? Suppose we have an ideal LC circuit (no resistance) and an open switch where the capacitor has an initial voltage $V_o$. Initially, the energy stored in the capacitor at $t=0$ is $\frac{1}{2}CV_o^2$ and the energy in the magnetic field of the inductor is zero because no current is flowing. Now at time $t=0+dt$ we close the switch and current slowly begins to build up. When the current is a maximum, the energy stored in the magnetic field of the inductor is $\frac{1}{2}LI^2$ but now the energy stored in the capacitor is zero. Thus we must have that $\frac{1}{2}LI^2=\frac{1}{2}CV_o^2$ because no energy is dissipated since there is no resistance.
But there seems to be something very wrong here at a fundamental level. The charge (the electrons) traveling through the inductor at the instant that the current is a maximum have a non-zero kinetic energy (denote this kinetic energy $K_{charge}$). They have to have non-zero kinetic energy since they constitute a current. But if they do posses this energy in addition to the magnetic field energy $\frac{1}{2}LI^2$, then the total energy at the moment the current is a maximum will equal $E_{tot}=\frac{1}{2}LI^2+K_{charge} >E_{initial}=1/2CV_o^2$. So its seems we have created energy in this process?
The only way I can work around this issue is to assume that the kinetic energy is already somehow factored into the magnetic field energy but I am not sure.
Any help on this issue would be most appreciated!
| Kinetic energy of electrons due to electric current $I$ in an inductor is much smaller than magnetic energy $\frac{1}{2}LI^2$ (provided the inductor has large enough $L$, which is usually the case).
So yes, strictly speaking total energy stored in the capacitor is transformed into magnetic energy and kinetic energy of current-carrying charges, but the latter energy is so small compared to magnetic energy it is customary to ignore it.
| {
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Which types of strain tensor are positive definite? I am taking a look at different types of strain tensor. Specifically, I am thinking about if the infinitesimal strain tensor
\begin{align*}
\epsilon_{ij} = \frac{1}{2} (\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})
\end{align*}
is positive-definite. I have Google searched some resources, and one of them says it is positive-definite. However, I think that it is not always positive-definite, as in the one-dimensional trivial case, if $\partial u/\partial x$ is negative, then it will not be positive-definite. Other sources say that other strain tensors, like the Lagrangian strain tensor, are positive-definite. I am not sure which types of strain tensor are positive-definite and also the implications if so. (I am thinking about the strain surface being an ellipsoid or not.) Any ideas will be greatly appreciated!
| The strain tensor does not have to be positive definite, as mentioned before.
On the other hand, you are asking what type of strain tensor are positive definite. A positive tensor would have positive principal strains. That implies a tensor where you have elongation in 3 different directions.
| {
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Why experiments of electrostatic induction are best done in dry weather? Basic experiments which demonstrate the presence of electric charges and forces..
For instance, when we rub a comb on hairs...it gets negatively charged..and can even suspend bits of paper due to this charge.
And many such experiments of rubbing glass rod, ebonite rod or rubber to silk, wool or paper which lead of their electrification.
They work best on dry days..why so?
I have read in a book that due to excessive moisture in environment the charges induced get leaked from the body to the earth.
How so?
| Electrostatic demonstrations work best when materials classed as insulators do not allow the flow (leakage) of charge. However if even a very thin layer of water adheres to the surface of so called insulators then although the bulk of the material does not allow the passage of electric currents the thin layer of water does.
So instead of using glass and ebonite in electrostatic experiments plastics like acetate and polythene are used because they a less prone to have water adhering to them and heating them up in a stream of hot air from a fan heater usually removes any water present on the surface.
So a high humidity means that there is a greater chance of there being a very thin layer of water on the surface of an "insulator" and that layer in effect makes the material a conductor and charges can leak away to earth when you do not want them to.
In electrostatic experiments the human body is an excellent conductor as is a piece of wood where although the cellulose of which the wood is made is a good insulator water on the pores of the wood provide a conducting path. Carefully heating a piece of wood does improve its insulating properties but if left in the air it soon becomes a conductor and more quickly in more humid conditions..
| {
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Isotopy class of spacetime We know that spacetime is an orientable manifold:
Can spacetime be non-orientable?
But supposing that spacetime is an orientable closed 2D surface, one might envision a variety of non-equivalent solutions in the following sense:
Given a 2D strip, by one rotation(twist), one can create a Moebius strip (it's non-orientable so discarded), but by another rotation (360 degrees) one finds an orientable 2D surface.
Suppose one can repeat this for arbitrary many times(integer multiples of 360 degrees), then one has a countable set of possible orientable spacetimes
Is there any way to determine which spacetime relates to ours(2D), given the fact that Einstein's Field equations are pretty much open-minded regarding the topology of spacetime?
Can I find any physical observable in QFTs on such spacetime that is related to the number of turns in general?
If not, can one hypothetically say that the real spacetime is a superposition of all these possibilities?
Is it possible to extend the idea of twist to 3D hypersurfaces?
| The orientability of space (rather than spacetime) was first raised by Kant in his 1768 paper, Concerning the Ultimate Foundations in the Differentiation of Regions in Space.
We generally take space to be orientable because experiment shows that it is and Kant alluded to this in his paper, but his paper was really about highlighting this property of space.
In a non-orientable space we can't define the Levi-Civita tensor that is used to integrate volumes in the usual presentation of GR. So GR itself requires an orientable spacetime.
Mobius strips, intrinsically speaking, are the same as a twisted line bundles over a circle. In fact, intrisically speaking, there are only two such line bundles, classifed by no twists or just a single half-twist. Extrinsically speaking, and by this I mean the line bundle is embedded in a vector space, we can have any number of half-twists.
We can generalise the intrinsic picture to twisted vector bundle over spacetime. Instantons, which are certain solutions to the Yang Mills equations are described by these and are used to describe vacuum condensates, which are a superposition of vacua, say in QCD.
| {
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Why do my white shoelaces get blue when exposed to UV? My shoes and shoelaces are both white. But when there is a UV light only my shoelaces appeared in a different colour. What is the reason?
| The shoes and shoelaces are made from different materials. The shoelaces contain a substance that fluoresces: it converts the UV light to blue light so that it becomes visible to you. The remainder of the shoes does not contain this substance, so it just absorbs or reflects the UV light, still invisible to you.
There are many blue fluorescent substances, so we don't know which one it is from looking at the picture. The intensity of the glow makes clear, that the shoelaces were especially designed to glow in UV.
Btw, fluorescent substances are also contained in some detergents (but in smaller amounts), where they appear as "optical brighteners" on the ingredients list. They too convert UV into visible light and make, for example, white shirts look bright & shiny in sunlight (which contains UV) and night clubs.
| {
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How does a copper ring levitates above a AC supply primary coil? Doesn't AC current gives decreasing current and so decreasing magnetic fields or flux. This should then attract the copper ring above primary coil. When magnitude of current increases in AC then due to Lentz law as we know ring will get repelled. But what about the decrease of current. Dosen't current decreases like it increases in AC. What is the reason of that lift. Even when anyone blocks ring at a place it still try to lift. Why??? I am very confused on that topic. I am trying to find it hard from everywhere and thinking myself too. But it seems like how?why? Why only lift above?
| Besides the bad English, I think what you're referring to is the jumping ring experiment, which is a demonstration of Faraday's laws of electromagnetic induction and also of Lenz's law. You can read more about it here.
| {
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When putting a floating object in water, why the displacement of mass of water is not equal to only the part of the object inside the water? In this question Will a bathtub full of water weigh more if I add something that floats in the water? the marked answers mentioned a special case where the bathtub is full to a brim and the boat is put into it, floats and it weighs the same, justifying that the extra mass is the spilled water.
I can't wrap my head around the "The boat will displace a mass of water equal to the mass of the boat", If I imagine the boat as a rectangle where the mass is distributed uniformly and half of the rectangle is inside the water and half of it outside the water. Shouldn't the mass of water displaced be equal to half the mass of the rectangle, the part inside the water? Why is also the amount of mass outside of the water being displaced?
| Imagine lowering your rectangular box into a bowl of water. When the box first comes into contact with the water, the upward force upon it from the pressure of water is less than its weight, so it will continue to sink into the water as you lower it.
As the box sinks, the level of the water in the bowl rises because the sinking box has displaced some water which is forced upwards.
There comes a point at which the weight of water that has been forced to rise above the surface level of the empty bowl equals the weight of the box. At this point the weight of the boat is balanced out by the weight of water that has been displaced upwards, so the boat will not sink any further but will float.
The volume of water that has been raised is equal to the volume of whatever part of the box is under the surface. However, the weight of the water displaced is equal to the total weight of the box.
| {
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Active vs passive transformation in right handed particle People often says that active transformation is equivalent to passive transformation.
Suppose that we have a right handed particle that is, the spin and the momentum are pointing in the same direction, call this direction right. Under a passive parity transformation the the spin continues to point to right while the momentum is now pointing to the left.
My question is, in a real world if we inverse the particle momentum , does it turn in to a left handed particle? If not ,does this mean, that passive transformation and active transformation are not equivalent?
Note I am considering here passive transformation as transformation in a measurer apparatus. People often consider passive transformation as coordinate transformation but coordinates are imagination of our mind ,it does has no effect in physics
| Active & Passive Transformations
Active and passive transformations are not the same. To understand this properly requires a local frame in a space. Now, we can either transform the space itself or the frame. In the first case, the transformation is called an active, in the second case, it is called passive.
It is called active in the first case because the points of space actually move, whilst the frame is left alone. It is called passive in the second case because the points of space don't move, whilst the frame itself changes.
Vectors & Pseudo-vectors
Vectors can be classified by how they behave when space is inverted, that is reflected in a plane. When the vector stays as it is we call it an ordinary vector and when it also inverts, then we say it is a pseudovector. Linear momentum is a vector whilst angular momentum is a pseudovector.
| {
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In the theory of special relativity speed is relative so who decides which observer’s time moves slower? If for example we have 2 people one moving in speed v relative to the other, according to special relativity the time passing for the moving person is slower than for the stationary person. However from the moving person’s prospective he is stationary and his friend is moving so time should move faster for him. What’s going on?
| One of the keys to understanding SR is to remember that all the effects it predicts are reciprocal. If Observer A thinks that Observer B's spaceship has contracted in length, Observer B will consider that Observer A's spaceship has contracted by the same proportion. There are many analogous cases in everyday life. If we stand some distance apart, I will seem smaller to you and you will seem smaller to me. There is no contradiction- we just have different perspectives on the same reality.
| {
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How to determine amplitude of compton scattering using Feynman rules? I am currently studying from Griffiths' introduction to elementary particles.
I started reading about the Feynman rules for caclulating scattering amplitudes, and while I understood them for the cases of electron/muon scattering, I can't understand how to proceed in the case of Compton Scattering.
Particularly, considering the following case:
I know that I should consider going from the "end" of the trajectory back in time. But the fact that the electron has a weird trajectory, I don't know if that is the way to go here. Should I consider $\bar{u}(p_4)$, then the vertex $ig_e\gamma^\mu$, then the propagator $\frac{i(\gamma^\mu q_\mu + mc)}{q^2-m^2c^2}$ then another vertex and then the incoming electron $u(p_1)$? But what do I do with the photons then, since the trajectory is interrupted?
I would really appreciate an explanation as to how to approach these diagrams
| The incoming photon will contribute with a $\epsilon_\mu(p_2)$, and the outgoing one with a $\epsilon_\nu(p_3)^*$. Note that the $\mu$ index of the incoming photon will be contracted with the one of $\gamma^\mu$, and the $\nu$ index of the outgoing photon will be contracted with the index of the other vertex. You can place them wherever you want, since they are just numbers ($\epsilon_\mu$ is the $\mu$-th component of the polarization vector).
By the way, I recommend you use another index in the propagator, since $\mu$ is already in use.
| {
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Were Kepler's Laws of Planetary Motion the first formal definition of an ellipse? It seems to me that Kepler's Laws necessitate some definition of an ellipse in terms of a coordinate system. I am wondering whether Kepler's Laws mathematically defined what an ellipse is, or if he used an already defined shape in his laws.
The reason I got curious about this in the first place was that I realized that Descartes may not have invented the Cartesian coordinate system by the time that Kepler discovered these laws and I have only seen proofs of Kepler's Laws in a Cartesian system.
| Due to the geometric formulation, Kepler's Astronomia Nova is full of elaborate illustrations of ellipses, epicircles and whatnot. It is not necessary to spell out the position of points in Cartesian coordinates, because the relation between points are specified in angles, lengths, and geometric constructions.
Kepler presents the use of ellipses for planetary motion in Astronomia Nova (1609, originally in Latin; translated by William H. Donahue, 1992, Cambridge University Press), when René Descartes was thirteen years old. Paging through introductory words by Kepler himself, translators, and commenters (Max Caspar, Kepler, 1993, Dover Publications) is highly recommended because, oh, do they throw shade.
The definition of an ellipse as a conic section, as known to the Greeks, already is a formal definition. Kepler actually cites the Greek philosophers for many propositions. For an in depth historical view, see chapter 2.1 of The Ellipse: A Historical and Mathematical Journey by Arthur Mazer (2010, Wiley).
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What are virtual and apparent forces? Acceleration in a rotating frame can be written as: $$\underline{a}=\frac{\partial^2\underline{r}}{\partial t^2}+\frac{\partial\underline{\omega}}{\partial t}\times\underline{r}+2\underline{\omega}\times\frac{\partial\underline{r}}{\partial t}+\underline{\omega}\times(\underline{\omega}\times\underline{r})$$ My lecturer referred to $\frac{\partial^2\underline{r}}{\partial t^2}$ as apparent acceleration and $\frac{\partial\underline{\omega}}{\partial t}\times\underline{r}+2\underline{\omega}\times\frac{\partial\underline{r}}{\partial t}+\underline{\omega}\times(\underline{\omega}\times\underline{r})$ as virtual acceleration but I am unsure as to what this means and was hoping somebody could help explain.
| Apparent forces are those that show up as a result of an accelerating reference frame, i.e., they are not forces over the system, but an effect of being in a reference system over which a force IS being applied. Therefore, if you accelerate in an accelerating reference frame, you would start already with some "virtual" acceleration.
| {
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What is the formula for ripple voltage in integrating circuit? With a rectangular wave input Vin and a typical RC circuit with output taken across the capacitor, what is the formula to calculate Delta V due to ripple?
| I guess you know how to calculate the output with a sinusoidal input wave. There are several techcnics for that, but the easiest one is probably using phasors.
And thanks to Fourier series you can write your periodic rectangular wave input as a sum of sinusoidal waves. Let's take for example a square wave $V_{in}(t)$ with periodicity T:
$
V_{in}(t) = \left\{
\begin{array}{ll}
û & \mbox{if } 0 \leq t < T/2 \\
-û & \mbox{if } -T/2 < t < 0
\end{array}
\right.$ $\>$ $\>$ with $\>$ $V_{in}(t+T) = V_{in}(t)$
Since it is an odd function, the real Fourier series will only have sinus terms since it is odd too. So we can calculate the coefficients $b_n$ given by:
$b_n=
\frac{1}{T}\int_{-T/2}^{T/2}{V_{in}(t)\sin(nwt)} dt \>$ which satisfy $\>$ $V_{in}(t)=\sum_{n=1}^{\infty} b_n \sin(nwt)$
Solving this integral would give you $b_n = \left\{
\begin{array}{ll}
\frac{4 û}{n \pi} & \mbox{if $n$ is odd } \\
0 & \mbox{if $n$ is even }
\end{array}
\right.$ which implies
$\>$ $V_{in}(t)=\frac{4 û}{\pi}\big[ \,\sin(wt)+ \frac{1}{3}\sin(3wt) + \frac{1}{5}\sin(5wt) \>+ \>... \big] \,$ with $ w = 2 \pi /T$
Then since the RC circuit is linear you can say that the effect of the square wave is equal to the sum of the effects of all these sinus waves, which I assume you know how to compute. Your answer will therefore be an infinit sum, but as you can see the coefficients decrease with a factor of $\frac{1}{n}$ so it is a good approximation to only consider the few first elements of this sum.
| {
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Does the shape of a wire affect the electric current in it in any way? Say we have a straight wire connected across two terminals carrying electricity between them. If everything is kept the same (voltage across the two terminals, resistance of the wire, etc.) will there be ANY difference at all if the wire were to be, let's say, tightly wound into a spiral as opposed to being straight?
I'm imagining there should be at least some difference because the spiral would make the electrons "bump" more along the edges. Is there any logic at all to this reasoning, or am I hopelessly illiterate about electricity?
| Yes, there is a difference.
If you made the wire as you mentioned, into a spiral, like this:
then there is quite a big difference between this and a straight wire. The difference between a straight wire and a coil or spiral wire is that the spiral wire resists changes in current flow.
This is called an inductor or solenoid.
It resists changes in the current flow because when there is a changing current in the coil of wire, there is a back EMF that counters the EMF of the current entering the coil of wire. This makes it hard for current to pass (at least initially) through the coil of wire.
Inductors are characterized by the following equation:
$$v = L \frac{dI}{dt}$$
Where $v$ is the voltage, $L$ is the inductance and $\frac{dI}{dt}$ is the change in current with respect to time or how does current change when time ticks.
When the current is unchanging, the inductor behaves just like a normal wire. But when the current is changing, it behaves differently: the greater the change of the current, the greater the effect of the back EMF will be. At very high frequencies, the current will be mostly blocked, and theoretically, as the frequency approaches infinity, the resistance or impedance of the inductor will be infinite, and no current will pass.
Although straight wires also have inductance, it can be usually ignored because it is so small. Only in certain situations such as power line transmission, it becomes important to know the inductance of the straight wire where it cannot be ignored from the math when designing the power lines.
| {
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How can air pressure be equal in all directions? Lay on the ground in a bright sunny day of spring and look the sky. Feel the air pressure on you.
Now lay on your stomach and your back exposed to sky.
You won't feel any change in air pressure. You will breathe normally. Even though air above you ( below you actually ) is so less that it cannot exert the same force/pressure on you.
I know that at a horizontal plane, pressure in fluids is same in all directions but why is it that way?
Its like someone is applying a force or exerting a pressure on you from above but you feel the same force/pressure from all directions, even from sideways. (Maybe the same pressure from downwards can be explained by Newton's third law but even from sideways, the same thing?)
I think fluid particles' mobility is the cause of it.
| Actually, the weight of the air above you contributes to the pressure that you feel. Because of this, air pressure drops as you go to higher altitude. This change is not so significant on the scale of you turning your head from pointing down to pointing up, so no real change is noticeable.
Why does pressure only depend on altitude and not change when you walk around at the same altitude (in a simplified model)? Since the earth is mostly spherical, the atmosphere is mostly spherical, so at any point on the earth's surface there is the same total weight of air from the atmosphere on top of you.
I think some of your questions are easier to understand when we remember that pressure is force divided by area. To hold up the weight of the atmosphere, the particles at the surface have to push back. The more they push back, the faster they must be moving on average. Now, the direction of their movement can be in three dimensions, so they not only push up by also push horizontally. The average force contributed by these particles is the same in all directions, which is why the pressure is the same in all directions.
| {
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Question about length contraction If an observer A is in motion on a train relative to an observer on the train platform, B, and observer A shoots a laser which hits the wall and comes back, who measures the interval to be the shortest? I don’t quite get it because the train would be contracted according to B, thus distance and time would decrease would be shorter for him, but on the other hand isn’t proper time always the shortest and hence the time interval would be shorter for A?
| The time is indeed shorter for A. Your reasoning about A is correct. If the time between two events is measured by a single inertial clock between the two vs a pair of synchronized clocks, the time will always be shorter for the single clock vs the pair of synchronized clocks. So A measures the shortest time.
the train would be contracted according to B, thus distance and time would decrease would be shorter for him
The error is here. You are correct that the train is contracted according to B, but the “thus distance and time would decrease” is not correct. The problem is that the length of the train is not the same as the distance that the light travels according to B.
When the light travels in the same direction as the train, the path of the light is longer than the length of the train. When the light travels in the opposite direction, the path is shorter than the length of the train. It is the length of this path that determines the time, not the length of the train.
| {
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Why electric field is scaled by gamma? Two opposite charges are in a spaceship and are attracted by the electric field $E_s$
But for an observer on earth the Electric force is $$E_e=\gamma E_s$$
Normally the forces are scaled down by $\gamma$ in the earth frame and here also the total force is scaled down.
But why the Electric component of force is scaled up?
Is it because the Electric fields are now closer together because of length contraction?
But i think this is not the answer because if that was the case,then gravitational fields and hence Forces would have scaled up.But this is not the case in nature.
Can you provide a derivation or something which explains how the Electric field is scaled up?
| Electric field E transforms this way:
$$E'=\gamma E$$
Gravity field G transforms this way:
$$G'=\gamma G$$
Force F, be it electric or gravitational, transforms this way:
$$ F'= F / \gamma $$
| {
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Why does conservation of angular momentum explain why planet orbits lie in a plane? I just learned about sector velocity and how you can manipulate the equation to show that angular momentum $L$ is conserved by $L/2M$. I know that sector velocity is constant and therefore angular momentum is constant. I have made it that far but I still can't see why a constant angular momentum would explain the coplanar concept.
| The angular momentum is given by a vector product $\vec{L}=m[\vec{r}\times\vec{v}]$. One o the properties of the vector product is that the result is orthogonal to both factors. I.e. both $\vec{r}$ and $\vec{v}$ should be orthogonal to $\vec{L}$. Or you can say that both $\vec{r}$ and $\vec{v}$ lie in the plane orthogonal to $\vec{L}$. As $\vec{L}$ is conserved this plane is conserved.
| {
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How do turbine engines improve compression ratio? As I have learnt from the Venturi tube that Pressure is inversely proportional to velocity. When a general aircraft is flying at say Mach 0.8 and at the same time there is a pressure decrease with increasing altitude. How is pressure achieved in the case. Is there stagnation pressure in play here?
| I will answer the question in the body of the post.
The venturi effect is not applicable here. The air intake for a small plane with a piston engine is usually exposed head-on to the incoming air and also to the backwash from the propeller. Both of these add ram air pressure to the carburetor inlet.
In fact, it is possible to magnify this effect and get a few inches of manifold boost by positioning an air scoop on the inlet which is very close to the trailing edge of the propeller in front of it, and then mounting the propeller so that the blade passes across the scoop mouth coincident with the opening of the intake valves on the engine.
Regarding turbine engines, the air inlet to the compressor stage is sized so that at cruise conditions, the velocity of the incoming air is the same as that of the air passing through the first stage of the compressor fan so there is no spillage of excess air nor suction pressure inside the inlet duct.
In this case, the fan and compressor stages of the engine perform the compression work to support the thermodynamic cycle of the engine, and ram air pressure recovery is not important.
| {
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What does it mean for the gravitational force to be "between" two bodies? What is the meaning of the word "between" in the law that the force between two masses at separation $r$ is given by $\frac{GM_1M_2}{r^2}$? I am confused about how can a force be in-between, either it is on body A or on body B, or on both.
Suppose body A exerts force $F$ on Body B, so according to Newton's 3rd law of motion B should also exert a force on A.
Let's consider this case for gravitational force between two bodies. If body A exerts force $g$ on Body B, then B body should also exert a force $g$ on A, but B is also exerting the gravitational force $X$ on A, hence A will also exert force $X$ on B.
So, how are two forces acting?
I have given the representation in this diagram.
| It is not a individual force that exists in the space between them, it is rather saying between in the case that both bodies exert a force on each other, which gradually pulls them to a point between them as they are pulled towards each other, they ultimately simply are counterparts to each other, the equal and opposite reactions.
| {
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In this ray diagram, a plane mirror seems to form a real image In this ray diagram the image formed seems to be real with the given position of the eye. I have learnt that plane mirrors cannot form real images at any circumstance. But at this one it does. Please explain the answer like I'm 5 and how you deduced what you propose.
| Farcher's answer is correct. But it can be elaborated a bit to make it easier to understand.
If you observe the above ray diagram for real images, you can see that the real images are formed when rays from the same point of an object intersect to form an image.
As your image shows, this is not the case. Rays from the top and bottom of the object intersect at the eye. Hence the image formed is not real.
As Farcher's modified image shows, the rays from the same point never intersect, but diverges. Hence the image formed is virtual.
| {
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Which force came first in Newtonian gravity? Force A on B or force B on A? Can a body $X$ apply a gravitational force on a body $Y$ without itself being acted on by a gravitational force from $Y$? Just per intuition, we deduce from Newton's third law that if $A$ applies force on $B$ then $B$ applies force on $A$. But in the case of gravitation, they both apply force on each other. So which force came first as an action, and which came later as a reaction?
| Neither comes first, they are mutual. In the Newtonian model gravity acts instantaneously at-a-distance (unlike Relativity, where gravity propagates at the speed of light).
Which force you consider the action depends on which one grabs your attention first. The other one then has to be the reaction. But the distinction in this case has no physical meaning (unlike the case where some agent initiates an action, such as when someone steps off a dinghy).
| {
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Velocity is relative, which means acceleration is relative, which further implies that forces are relative as well So how would we know whether a force truly exists or not. I can be accelerating a car my 5 meters per second squared but another car accelerating with the same acceleration would think that my car is at rest relative to them. So is there any force on the car? Or are forces just relative and their existence just depends on our reference frame?
| People in an accelerated car know that it is accelerating due to inertial forces that can be observed inside.
In cases like that, besides the inertial forces, there is the reference of the fixed environement, so it is easy for the passengers to know about their acceleration, just looking outside.
More interesting is the case of the Earth. It was not so obvious to decide if all the sky rotates around us, or if we are rotating and the stars can be considered a fixed environment. Here, inertial forces like Coriolis, that explain trade winds and hurricanes rotation are a proof that we are rotating, what means an accelerating frame.
| {
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What's the meaning of potential energy in quantum physics? We know that the definition of potential energy in classic physics. When we start learning about quantum physics and especially the Schrodinger equation we blindly accept the potential energy part. I would like to know what's the meaning of potential energy when it comes to quantum physics.
| Quantum mechanics is based on Hamiltonian mechanics or if you want, analytical mechanics in general. That means it utilizes the same concepts of kinetic energy and potential energy among others that you may have encountered already in Newtonian mechanics. Their meaning is exactly the same in classical and quantum mechanics, since quantization does not modify these definitions.
That is quantum mechanics addresses the commutation of certain conjugate variables, such as position and momentum but leaves the definition of potential energy intact. Specifically, the potential captures the interactions from the environment (including possibly other particles) where a particle/body is "moving". In perhaps traditional terms it represents energy stored via working against some force, may it be a Coulomb force, or gravitational or whatever effective force you might encounter.
| {
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Do partial derivatives of different coordinate systems commute? Consider an arbitrary set of coordinates $x^\mu$ and another set of coordinates $y^{\mu}$, which is a (lorentzian) transformation from $x^\mu$ given by $y^\mu = f(x^\mu)$.
So I want to know whether $\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial y^\beta} = \frac{\partial}{\partial y^\beta}\frac{\partial}{\partial x^\alpha}$ holds true or false?
| No they don't commute in general. At least not with the usual understanding that
$$
\frac{\partial}{\partial x^2}
$$
is the partial derivative with $x^1$, $x^3$ etc held fixed.
Here is a counterexample:
Let $x= r \cos \theta$, $y= r \sin \theta$ be cartesian and polar coordinates. Then
$$
\frac{\partial x}{\partial y}=0\Rightarrow \frac{ \partial}{\partial r} \frac{\partial x}{\partial y}=0.
$$
but
$$
\frac{\partial x}{\partial r}\equiv \left(\frac{\partial x}{\partial r}\right)_\theta= \cos \theta
$$
is the derivative with $\theta$ being held fixed. Now
$$
\frac {\partial} {\partial y} \frac{\partial x}{\partial r}= \frac {\partial} {\partial y}\cos \theta =\frac {\partial} {\partial y}\frac{x}{\sqrt{x^2+y^2}}\ne 0.
$$
so
$$
\left(\frac {\partial} {\partial y}\left( \frac{\partial x}{\partial r}\right)_\theta\right)_x \ne
\left(\frac {\partial} {\partial r}\left( \frac{\partial x}{\partial y}\right)_x\right)_\theta,
$$
where I have made it explicit what is being held fixed for each derivative.
It may be that they commute if the transformation is linear, but even then I have doubts. Why don't you try and see if it it's OK in this restricted case?
| {
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Is the Least Squares Fit always the accepted Fit in Physics? Is the Least Squares Fit always the accepted Fit in an experiment? Suppose you have $N$ data points and a function $f$ with some parameters. There is a one Least Square sum, it may be obtained for more than one set of parameters but we can find that most optimal fit. Is it true that sometimes that Least Square fit is discarded because the expected values for the parameters are assumed to be different?
Why is it necessary to give starting values to the algorithms? Why can't the computer find those parameters on its own?
| No, it’s not. Sometimes it’s much more complicated than least squares, and sometimes it’s a by-hand “guide for the eye.” Not everybody has the expertise to apply a fancier data analysis, and they don’t let perfect be enemy of good.
Least squares is just sort of default for a typical situation, where you have a curve and a model, and you know nothing about the fitting parameters a priori. Then you play around with the parameters to give the computer a decent guess and report what least squares tells you. Of course, you also have to make sure you’re not over- or under-fitting.
The key here, and with all of science, is that you honestly and thoroughly report what you did. That will help your readers decide whether whatever choice you made was reasonable.
| {
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Nonlinear Schrödinger equation in a potential I've recently become interested in the integrability of nonlinear PDEs while reading these lecture notes.
Question 1: Would the equation $i\Psi_t + \Psi_{xx} - (2|\Psi|^2 + V) \Psi = 0$ for a potential function $V(x)$ posses any physical significance?
I was curious as to how the dynamics of the system would change if a potential function were to be introduced. Given that $V(x,t)\Psi(x,t)$ are multiplied together in the time-dependent Schrodinger equation, I wondered if a similar principle could generalize to the nonlinear case as well. My background is primarily in mathematics, and I was unsure if such an addendum would reduce the physical significance of the system.
| The equation you've written is known as the Gross-Pitaevskii equation, which is one of the central concepts used to describe Bose-Einstein condensates.
(Indeed, this is mentioned explicitly in the disambiguation text at the top of the Wikipedia page for the NLSE.)
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Why does increasing tension in a string increase the speed of travelling waves? We know that $$ v = \sqrt\frac{T}{\mu} $$ meaning that increase in the tension of a string increases the velocity of the traveling wave. But how exactly does this happen? If we consider that the travelling wave is just a certain amount of energy and momentum ($C$) which is propagating then, I think that increasing tension (along one direction) stretches the string in that direction hence decreasing the density (along the other two directions) therefore for a fixed amount of momentum $C=mv$ to travel less $m$ means more $v$ for a fixed $C$. However, I am not sure of this interpretation.
| Tension determines the vertical force (perpendicular to wave motion) on molecules of string and hence determines the speed of perpendicular motion. Faster the perpendicular motion, faster the wave has passed by.
| {
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Why is momentum not quantised in the photoelectric effect? I know that the momentum transfer mechanism in photoelectric effect is pretty complex etc, but why do we quantise the Energy of the photon but not the momentum ?
Edit: had written conserved instead of quantised
| Just because we don't use the equation $$\vec{p}_2=\vec{p}_1+\int_{t_1}^{t_2}\vec{F}\mathrm d t \tag{1}$$
doesn't mean that we ignore momentum. It's simply that there is a large mass (compared to the photoelectron and the mass-energy added by the photon, and that mass absorbs a lot of the momentum of the photon. That integral at the end of (1) tells us how much momentum is transfered into or out of the system. If the mass of the photosensitive substance is included in your system, then the momentum of the system will remain constant. The problem is that the force is very complicated, and we can't measure the momentum of the big mass after the interaction occurs. So, we don't deal with it. Plus, it wouldn't tell us anything useful for the problem.
We can determine the energy of the incoming photon (by its wavelength), the energy of the photoelectron (indirectly by measuring a stopping potential), and the amount of work needed to free the electron (by looking at a variety of wavelengths of photons). And we assume that the kinetic energy or temperature rise due to the interaction is very small compared to the work function.
Basically, we can account for all the energy changes. We can't account for all the momentum changes. But we trust that both energy and momentum (along with their "currents" of work and impulse, respectively) are conserved.
By the way, (1) is the statement of conservation of momentum. And conservation doesn't require constancy within the defined system. Impulse (that integral) tells us how the system momentum changes.
| {
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$SU(2)$ Symmetry of Hubbard Model I am confused with the $SU(2)$ spin rotation symmetry of the fermion Hubbard Hamiltonian. If the Hubbard model has $SU(2)$ rotational symmetry, it means that the Hubbard Hamiltonian commutes with the global spin operator in all direction:
\begin{equation}
[\vec S, H] = 0 ~~,~~ \vec S = \frac{1}{2}\sum_{i}
\begin{pmatrix}
c^{\dagger}_{i \uparrow} & c^{\dagger}_{i \downarrow}
\end{pmatrix}
\vec{\sigma}
\begin{pmatrix}
c_{i \uparrow} \\
c_{i \downarrow}
\end{pmatrix}
\end{equation}
Where $\vec \sigma$ is the Pauli matrices in vector form. My confusion is that whether $[\vec S, H] = 0$ implies $[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0 $. I have proved that they are equal to zero but I thought that my calculation was wrong. The reason why I think my calculation is wrong since if both $S_{x}, S_{y} , S_{z}$ commutes with $H$, it means that they simultaneously share the same eigenstates and $[S_x, S_y] = 0$. However, we know that from elementary QM course:
\begin{equation}
[S_{i}, S_{j}] = i \epsilon_{ijk} S_{k} ~~,~~ ijk = xyz
\end{equation}
In my opinion, $[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0 $ is not true because they cannot satisfy $SU(2)$ commutation relation if all commutes with $H$. May I know is it true that $[\vec S, H] = 0$ implies $[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0 $?
| From what I understand of the standard notation, the statement that $$[\vec{S}, H] = 0$$ is exactly the same as saying $$[S_j, H] =0$$ for all $j$. Then, to give a proof that a vector operator commutes with some operator simply amounts to proving each component commutes with the given operator.
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What would you see dropping a sufficiently strong chain with substantial length into a black hole? Here's a visual representation of the scenario
Here you can see we have a black hole on the left. The event horizon is the edge of the black hole. You are far away from the event horizon, and a chain is passing you by fast heading toward the black hole. Due to the size of the black hole and the makeup of the chain, the chain will not break before reaching the event horizon.
After a while, the following scene happens:
As the end of the chain approaches the event horizon, the chain slows down due to the immense gravity as it approaches a frozen state.
As I show in the diagram, you can observe this phenomenon as well as observe the chain moving fast by you toward the black hole.
As for my question..
How can the part of the chain near you appear to be moving quickly toward the black hole, while the end near the black hole is frozen (or close to it)? Where does all that chain go?
Let me ask the same question in another way..
If the distance between you and the black hole is 1000 units, and the chain appears to be almost frozen 1000 units away from you, how could you reconcile watching 10000 units of chain speed past you? How does that 10000 units appear to fit within a distance of 1000 units from your perspective?
| The first thing to remember is that at the event horizon of a black hole, time dilation is so extreme that, from our frame of reference here on Earth, time has essentially stopped. So you would see nothing happening at all. You could wait for your lifetime, and for the lifetime of all your decedents and nothing would have happened. Not a single link of the chain would have moved by the smallest amount. Even before the chain got to the horizon, time dilation would have slowed the movement so greatly that movement would be extremely slowed.
| {
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Friction between 2 objects stacked on top of each other Let's say I have a particle on top of a block that is resting on top of a rough plane. Let's also assume that there is friction between block & particle as well as block & surface. In which direction would friction act and if we were given some information such as the frictional force between the particle & block, how would I be able to calculate things like frictional force acting between the block and surface, provided the block is in equilibrium?
I appreciate that these questions are hypothetical but the main purpose of these questions is for me to understand that when 2 objects are stacked on top of one another, how do the forces act on them etc?
Thanks
| Generally speaking, you should always draw each internal force between objects according to the 3rd law of Newton, and then apply the 2nd law of Newton on each object.
Usually one thinks that friction always is in the opposite of movement, but this is not true. For example, put a book on the palm of your hand and slowly move your hand; the book moves too! This is because there is a friction between your skin & book, and technically speaking, the friction causes the book to move in the direction which you move your hand. Duo to 3rd law of Newton, there exists an equal and opposite force (friction) which you feel it as some kind of "resistance" from the book.
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For how long does the force in $τ=rF\sinθ$ exert on the object that it sets into rotation? When calculating torque using formulas such as $τ=rF\sinθ$, is the force $F$ exerted on the object for an extended period of time (i.e. the force is still exerted on the object even when the object is at a different position on its path than its starting position) or does the force leave the object immediately after it sets the object into rotation? How can we tell which is the case from examining the formula $τ=rF\sinθ$? And what is this formula lacking so that it can't be the other case?
| $\boldsymbol\tau=\mathbf r\times \mathbf F$ is just the definition of torque. Its behavior is completely dependent on $\mathbf r$ and $\mathbf F$. So, if you want to know torque as a function of time, look at those values as a function of time: $\boldsymbol\tau (t)=\mathbf r(t)\times \mathbf F(t)$. The definition applies in any situation where a force is being applied.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why can you not use equation of motion $v^2 - u^2 = 2as$ when attempting escape velocity questions? There is the this question:
Imagine you are projecting a ball with $3/4$th the escape velocity from the surface of the earth. What is the farthest distance will it reach it from the centre of earth? ($R=$ Radius of the earth)
The answer given in my workbook is $16R/7$
I could solve this question using the conservation of energy theorem. Why can I not use the third equation of motion
$$v^2 - u^2 = 2as~?$$
$s=$ distance from the surface of the earth
I tried solving through the equation of motion method but my answer is widely different.
| That equation is for constant acceleration motions. The value of $g$ decreases with altitude according to $g=\frac{GM}{r^2}$. In this problem, the decrease in $g$ is significant, as you could expect the projectile to reach a great height.
| {
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Odd or Even symplectic structure in BV formalism I am studying Batalin-Vilkovisky formalism. I am a little bit confused on what an odd (or even) symplectic structure is (i know what the degree of the underlying 2-form is). I can not find a clear definition anywhere. I read that an even bilinear form $B$ is even if $B(x,x)=0$ for all $x$ and odd if it exists $x$ such that $B(x,x)=1$ but this does not seem to me to be the right definition for a symplectic form. Could someone enlighten me on this point?
| *
*An even (odd) symplectic 2-form
$\omega~=~\frac{1}{2}\mathrm{d}z^I ~\omega_{IJ}(z)~ \mathrm{d}z^J $ on a supermanifold means a Grassmann-even (Grassmann-odd) closed non-degenerate 2-form, respectively.
*There is a straightforward generalization to even and odd Poisson structures on supermanifolds.
*The Hamiltonian Batalin-Fradkin-Vilkovisky (BFV) formalism uses an even Poisson bracket, while the Lagrangian Batalin-Vilkovisky (BV) formalism uses an odd Poisson bracket (also known as an antibracket).
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Computing curvature singularities from a metric Suppose I have the metric
$$ds^2 = f(r)(dt^2-dr^2-dz^2) - \frac{1}{f(r)} d\phi^2. $$
How would you calculate the curvature singularities of this metric if we assume that $f(r)$ takes value $0$ for $r_0$?
| You have to compute some curvature scalars such as $R$, ${R_{\mu \nu }}{R^{\mu \nu }}$ etc. for your spacetime in order to find where those curvature invariants diverge. In order to understand whether there is singularity at all or not, it is enough to show that one of those curvature scalars diverges. But, the information related to the spacetime singularity may be lost for some curvature scalars such as $R$ and ${R_{\mu \nu }}{R^{\mu \nu }}$ due to the contraction of Riemann tensor (For example, in the case of the Schwarzschild metric, one obtains $R=0$ while the Kretschmann invariant diverges at the origin). Since the Kretschmann (scalar) invariant, ${R_{\mu \nu \delta \lambda }}{R^{\mu \nu \delta \lambda}}$, is a sum of squares of Riemann tensor components, it can be used to find the true (essential) singularities of a spacetime since, by definition, it preserves all the information about the singularities. So the Kretschmann invariant is preferred.
Considering the metric in your question, i.e.,
$$ds^2 = f(r)(dt^2-dr^2-dz^2) - \frac{1}{f(r)} d\phi^2,$$
the Kretschmann invariant is computed as (using the Maple's GRtensor package)
$${R_{\mu \nu \delta \lambda }}{R^{\mu \nu \delta \lambda }} = \frac{{12f{{(r)}^2}{{\left( {\frac{{{d^2}f(r)}}{{d{r^2}}}} \right)}^2} - 32f(r){{\left( {\frac{{df(r)}}{{dr}}} \right)}^2}\left( {\frac{{{d^2}f(r)}}{{d{r^2}}}} \right) + 27{{\left( {\frac{{df(r)}}{{dr}}} \right)}^4}}}{4{f{{(r)}^6}}}.$$
Next, more information about the metric function $f(r)$ is needed. At first glance, it seems that the Kretschmann invariant diverges at $r=r_0$ since $f(r_0)=0$. But, this is a naïve guess. To be sure, you have to put the explicit form of the metric function $f(r)$ in it and examine different limits such as $r \to 0$, $r \to r_0$ etc.
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