electromagnetism - Energy conservation in Electrodynamics

Let us suppose that we have a known electromagnetic wave-train of finite size propagating in a certain direction. There is a probe charge on its way. This EMW is an external field for the charge. The EMW has a certain energy-momentum (integral over the whole space). After acting on the probe charge the wave continues its way away. In the end we have the energy of the initial wave (displaced somewhere father), the kinetic energy of the charge (hopefully it starts moving), and the energy of the radiated EMF propagating in other directions. Thus the total energy may become different from the initial one. How to show that the total energy is conserved in this case?

It is not a Compton scattering. Just a regular electrodynamics problem. How EM energy can change appropriately? Via destructive interference? How to show it if the incident field is a known function of space-time?

EDIT: I can emit a half-period long wave from a radio-transmitter:$E(t)=E_0 sin(\Omega t), 0 < t < \pi/\Omega $. Then the final charge velocity will be clearly different from zero:

$ma=F(t), v(t>\pi/\Omega)=\int_{0}^{t}F(t')dt'=\frac{2qE_0}{m\Omega}$.

In addition, the charge itself radiates some new wave during acceleration period. The radiated energy is only a small fraction of $\frac{mv^2}{2}$. What can guarantee that the total energy remains the same?

EDIT 2: OK, let us simplify the task. I wonder if there is a simplest problem in CED where the total energy with a radiating charge is conserved explicitly?

kinematics - How to remove gravity component from accelerometer $X$, $Y$ readings?

So I have an accelerometer which I am wanting to use in an IMU.

When the device is tilted but stationary I want the x, y values to be 0, so effectively negate the effect of gravity along the x and y axes of the accelerometer.

I have found lots of conflicting advice on the internet regarding rotation matrices and was wondering if anyone here can provide some clarity.

I understand accelerometers can only measure pitch and roll from their x, y, z so how can I use these values to remove the gravity vector component on the x and y axes?

Answer

This is a hard thing to do. If you know the magnitude of $g$ accurately, you can look at the actual total acceleration observed by your accelerometer and subtract the "known" g. What you are left with is the difference vector. The problem is that the difference between two large vectors that point almost in the same direction is a small vector with a large error on it.

This is why it is preferable to have some independent information about the rotation of the sensor: if you have rotation sensing (not just linear acceleration measurement) you can integrate that to get the angular position; this helps improve the estimate of the orientation, and then it's easier to subtract the gravity vector.

For example this answer assumes you know the rotation (orientation of the sensor) after which things are simple. But when you don't, the problem is very ill posed unless the acceleration is large compared to $g$ (or at least "not small").

Details of the calculation can be found in this paper. If that doesn't answer your question, can you please be more specific in your question about the concept you are stuck on?

update

Here is the reason that your problem is not easy to solve. For simplicity I will do this in 2D - it should be easy to see how to extend it to 3D (but that doesn't make it better-behaved).

I show two situations where the $a_x$ sensor records the same acceleration, but the $a_y$ sensor records a slightly different value. When the sensor is rotated and stationary, and the values are perfectly known, then

$$a_x^2+a_y^2=g^2\\ a_y = g\sqrt{1-\left(\frac{a_x}{g}\right)^2}$$

Now in the second case (not rotated, and accelerating), the reading from the $a_y$ sensor will be exactly equal to $g$. In the first case, we can only estimate the rotation from the fact that it is not quite g: from trigonometry, we see that

$$\cos\theta = \frac{a_y}{g}$$

so we expect the measured horizontal component of acceleration to be

$$a_x = g \sin\theta = g \sin\cos^{-1}\frac{a_y}{g}$$

if we want to do error propagation, we can start by looking at the derivative of the $\cos^{-1}(x)$ function, which is $-\frac{1}{\sqrt{1-x^2}}$.

For small angles of rotation, when $a_y\approx g$, that error becomes exponentially bigger - in other words, even with an accurate accelerometer it's hard to estimate $\theta$ when the angle is small. And that means that you can't tell the difference between the two cases I drew very well - the uncertainty in $a_x$ will be very large. The problem becomes less important when the acceleration is very large - but for small (compared to $g$) accelerations this really won't work very well.

This is why it is necessary to have some independent indication of orientation / rotation if you want to use an accelerometer for an IMU.

quantum field theory - What is negative energy and how does it predict antimatter?

According to multiple online sources, antimatter was discovered through the Dirac equation because there were multiple solutions; a positive energy solution, to be expected and a negative energy solution. What does this mean. Also I read something about this giving rise to a symmetry called CPT symmetry, however that is not the main focus of this question.

I would just like an explanation of what negative energy is and why it results in antimatter.

Answer

In brief :

From relativity, the Dirac equation gives the following relation : \begin{equation}\tag{1} E^2 = p^2 c^2 + m_0^2 \, c^4, \end{equation} which is a second order algebraic equation, with two roots : \begin{equation}\tag{2} E = \pm \, \sqrt{p^2 c^2 + m_0^2 \, c^4}. \end{equation} Now, quantum mechanics requires that all solutions be considered, since any superposition of solutions is another solution to the linear Dirac equation. Thus, you have negative solutions to consider. You can't just throw them away just because they don't have "physical" sense to you. Now, the Dirac equation admits an operation (complex conjugate and a matrix multiplication) which can convert a negative solution to a positive solution traveling in the opposite direction, and reversed spin and electric charge. This operation gives another solution of the Dirac equation : \begin{equation}\tag{3} \psi_{\text{c}}(t, x) = \gamma^2 \, \psi^{\ast}(t, x). \end{equation} This suggest that the negative solution may be interpreted as an "anti-particle", i.e. one with a reversed spin and electric charge.

wavefunction - Help me to visualize this wave equation in time, to which direction it moves?

The wave is

$\bar{E} = E_{0} sin(\frac{2\pi z}{\lambda} + wt) \bar{i} + E_{0} cos(\frac{2 \pi z}{\lambda}+wt) \bar{j}$

Let's simplify with $z = 1$. Now the xy-axis is defined by parametrization $(sin(\frac{2\pi }{\lambda}+wt), cos(\frac{2\pi }{\lambda} + wt)$ where $t$ is time and $\lambda$ is wavelength. This parametrization satisfy the equation $1^2=x^{2}+y^{2}$, a circle.

Now, let's variate the value of $z$. We know now that it cannot move into x or y coordinates or do we? Not really, the latter simplification is naive -- $x-y$ parametrization depends on the dimension $z$ -- but can we see something from it? If so, how to proceed now?

The solution is that the wave moves along the $z$ -axis to the negative direction as $t$ increases, a thing I cannot see.

The way I am trying to solve this kind of problems is:

1. Parametrize the equation


2. suppose other things constant and change one dimension, observe


3. check other variable

...now however I find it hard to parametrize the $z$ so a bit lost. So how can I visualize the wave with pen-and-paper?

Answer

Would you agree that $\vec{E}$ depends only on $\frac{2 \pi z}{\lambda} + \omega t$ (taking $E_0$ to be a constant)?

If so, we can imagine picking some spot it space and time, taking note of the value of $\vec{E}$ at that point and looking to see how we have to move to keep the value constant in time

$$ \frac{2 \pi z}{\lambda} + \omega t = C $$

where C is determined entirely by our initial choice of space--time location. So:

$$ z = z(t) = \frac{ \lambda }{2 \pi} \left( C - \omega t \right) $$

represents a locus of $z$-positions as a function of time where $\vec{E}$ continues to have the same value it had at our starting point. And those positions move in the negative $z$ direction as time increases.

Question for the studuent: how fast do they move?

You should be able to answer by inspection.

energy - Fusion vs. Fission

I understand why fission generates large amounts of energy when the nucleus is split, but then why does fusion generate such large amounts of energy. If fission releases energy when some mass is lost as energy, then shouldn't the fusion process absorb energy to fuse nuclei together?

I also am curious as to where the energy released from fusion comes from, while fission releases some of the energy of the strong nuclear force.

Answer

Fission is exothermic only for heavy elements, while fusion is exothermic only for light elements. Intermediate nuclei, in the iron/nickel range, are the most tightly bound, and so you generally release energy moving in that direction.

Fusing stable elements into uranium would consume energy, as would trying to break helium into hydrogen.

For a more thorough background, see for instance this post.

general relativity - Why is $p_phi$ conserved in a Schwarzschild orbit?

This arises from the question What is the relationship between $a$ and $m$, which I'm afraid I answered just by looking it up in Schutz's book. However Schutz (as he frequently does) glosses over details he thinks are irrelevant or too simple to be worth explaining, and I have realised I don't understand an assumption he makes.

Schutz states without proof that if we have an equatorial orbit in a Schwartzschild metric then:

Independence of the metric of the angle $\phi$ about the axis implies that $p_\phi$ is constant.

In the non-relativistic world I assume this corresponds to angular momentum being constant in a central potential. So far so good. But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? The component $p^\phi$ is presumably not constant since (in this case) $p^\phi = p_\phi/r^2$.

Bonus points for also explaining his similar claim that time independance means that $p_t$ is constant rather than $p^t$.

I fear that Schutz didn't explain because it's an insultingly simple question, but if someone can provide a nice intuitive explanation I would be very pleased to read it.

Answer

But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$?

From the bottom of page 189:

The geodesic equation can thus, in complete generality, be written

$$m \frac{dp_\beta}{d\tau} = \frac{1}{2}g_{\nu \alpha,\beta}\;p^\nu p^\alpha$$

We therefore have the following important result: if all of the components $g_{\mu \nu}$ are independent of $x^\beta$ for some fixed index $\beta$, then $p_\beta$ is a constant along any particle's trajectory

Also, be aware that, in the relevant section on equatorial orbits in the Schwarzschild geometry, Schutz is working in a coordinate basis and not a unit basis.

In the case that $\theta = \frac{\pi}{2}$ (as in this example), we have

$$\vec e_\phi \cdot \vec e_\phi = r^2$$

which is why, I believe, $p^\phi$ is $r$ dependent.

electromagnetism - Why is a stellarator-type nuclear fusion reactor so oddly-shaped?

My first impression: It's a mess.

Why is it shaped like that? I can't find any info about its shape other than it's a special arrangement of magnetic coils.

Answer

Suppose you start with a linear solenoid. Due to the Lorentz force charge particles travel in circles (or helices) inside the solenoid so they can't reach the walls of the solenoid. But obviously the trouble is that they will leak out of the ends.

Now we curve the solenoid round and join its ends together to make a torus so now the particles can't leak out of the ends. The trouble is that by bending the originally straight solenoid we have made the spacing between the turns of wire smaller on the inside that on the outside, and as a result the magnetic field inside the solenoid is higher on the inside than on the outside. This means the charged particles will no longer move in circles but will instead spiral towards the outside wall until they hit it.

A stellarator gets round this by deliberately making the field inside the stellarator uneven, but then rotating the unevenness round the ring. So at one point in the ring the charged particles tend to drift to the outside but a bit farther along the ring they tend to drift to the inside. If you can get the particles moving around the ring then this unevenness cancels out and on average the particles stay in the middle of the ring.

If you look at the picture of a stellarator from Wikipedia:

you can see how each coil is rotated slightly compared to the ones immediately before and after it.

This sounds a nice simple idea, but containing plasma makes herding cats seem easy because the plasma generates its own fields as it flows and these interact with the stellarator fields in complicated and non-linear ways. To make the idea work in practice requires a design with the fearsome complexity shown in your picture.

newtonian mechanics - Is centripetal acceleration mandatory for circular motion?

If we consider a case where a closed circular wall is present, and inside the boundary, just adjacent to the inner wall, two particles are placed touching the surface of inner boundary such that the distance between both particles tends to 0. Particle A and B have identical masses, shapes and physical properties. Particle B applies a constant force on particle A and particle B is in uniform circular motion. Thus, particle A also moves in a circle. And here, centripetal force is not required to make particle A move in a circle.

What's going on?

Answer

Let me try to add a missing intuitive piece in the explanation to the discussion to @Gert's answer of why there must always be radial acceleration for any circular motion.

Remember what acceleration is: Change in velocity. $\vec a=d\vec v/dt$. More mathematically it is change in the velocity vector.

If velocity is changed - either magnetude or direction - then we define this rate of change as acceleration.

• If the velocity magnetude is changed, then acceleration is pointing in the same direction as the velocity (and it goes faster or slower). We call this tangential acceleration.


• If the velocity direction is changed, then acceleration is pointing to the side and not parallel to the velocity direction. If it is exactly perpendicular to the direction we call it radial acceleration.

Now imagine what happens if an object has no tangential and only radial acceleration. Then it keeps its magnitude (so the speed doesn't change) and only changes direction. Thinking more about this it should be clear that such a situation is a circular motion. If there both is tangential and radial acceleration, then the we have an elliptical motion.

Naturally, for any path that is not straight, there must be a change of the velocity direction. And we call such a change acceleration - for a circular motion that acceleration happens to be perpendicular and towards the center of the circular path.

Any forces that act on a particle in a circular motion therefor must result in such a sideways acceleration. If they do not, then the motion cannot be circular because of the above explanation.

quantum field theory - How to calculate the charges of W and Z bosons?

In the Standard Model, electroweak unification is based on the symmetry breaking of $ \text{SU(2)}\times \text{U(1)}_Y \rightarrow \text{U(1)}_{EM}$ by the VEV of a complex doublet $H$ with hypercharge $\frac{1}{2}$.

The Lagrangian is

$$\mathcal{L}=-\frac{1}{4}(W^a_{\mu\nu})^2-\frac{1}{4}B^2_{\mu\nu}+(D_\mu H)^\dagger(D_\mu H)+m^2H^\dagger H-\lambda(H^\dagger H)^2,$$

where $W^a_\mu$ are the $\text{SU(2)}$ gauge bosons and $W^a_{\mu\nu}$ are given by $W^a_{\mu\nu}=\partial_\mu W^a_\nu-\partial_\nu W^a_\mu+gf^{abc}W^b_\mu W^c_\nu$, and $B_\mu$ is the $\text{U(1)}$ gauge bosons and $B_{\mu \nu}=\partial_\mu B_\nu-\partial_\nu B_\mu$.

The covariant derivative is $$D_\mu H=\partial_{\mu}H-igW^a_\mu \tau^aH-\frac{1}{2}ig'B_\mu H.$$

After $H$ gets its VEV $$H_0=\begin{pmatrix} 0 \\ \frac{v}{\sqrt{2}} \end{pmatrix},$$ the mass term of the gauge bosons comes from $$(D_\mu H_0)^\dagger(D_\mu H_0)=\frac{v^2}{8}\left[ g^2(W^1_\mu)^2+g^2(W^2_\mu)^2+g'^2(g'B_\mu-gW^3_\mu)^2 \right].$$

My questions are why we define W bosons as $W^\pm_\mu=\frac{1}{\sqrt{2}}(W^1_\mu\mp iW^2_\mu)$ rather than just $W^1_\mu$ and $W^2_\mu$ respectively, and how to see the charges of $W^\pm$ and $Z$ bosons?

energy conservation - Would Portal-style portals transmit gravity?

In the video game Portal, there are often puzzles which must be solved by gaining a large amount of momentum. Typically, this is accomplished by putting one portal on the ground and another directly above it on the ceiling, then jump into the portal in the ground, and the person would appear at the ceiling and started to fall off. When the person fall off, he/she gain kinetic energy and reach a certain speed and jump into the portal in the ground again. After that, he/she appear at the ceiling again. Thus you create an infinite loop of falling.

Although the idea of a portal is absurd to common sense, let's discuss its possibility to existence.

The portal would seem to be creating potential energy. Typically when an item falls, its potential energy is converted to kinetic energy by virtue of it losing height, and additional energy would need to be expended to raise it up again before it could fall again.

What is the possible source of the potential energy, if portal would exists? Does a portal provide potential energy in order to push people to higher position?

And this has lead me to wonder another point, though, whether the portals should also transmit gravity. This might remove the whole problem of conservation of energy; if the gravity is transmitted through the portal, then the object would be pulled back up with equal force, and would actually reach an equilibrium floating within the portal, rather than falling continuously.

Interpretation of the Dirac-measure property

First and foremost, apologies in advance for using an abuse of notation by placing the Dirac measure inside an integral. But given the circumstances, I have no choice.

This is essentially a word by word copy of an interpretation given on page 1 of these Berkeley notes:

The important property of the delta function is the following relation $$\displaystyle\int f(t) \delta(t) \, \mathrm{d}t = f(0)$$ for any function $f(t)$. This is easy to see. First of all, $\delta(t)$ vanishes everywhere except when $t = 0$. Therefore, it does not matter what values the function $f(t)$ takes except at $t = 0$. You can then say $f(t)\delta(t) = f(0)\delta(t)$. Then $f(0)$ can be pulled outside the integral because it does not depend on $t$, and you obtain the r.h.s.

Here's the problem, it was my understanding that $$\delta(t) = \begin{cases} 0 & \space \mathrm{for} \space t \ne 0 \\\infty&\ \mathrm{for} \space t = 0 \end{cases} $$ So by my logic this means that $\delta(0)=\infty$ and therefore undefined; which implies that when $t=0$ $$\displaystyle\int f(0) \delta(0) \, \mathrm{d}t = \displaystyle\int f(0) ~\infty \, \mathrm{d}t$$ which is manifestly not true and certainly not equal to $f(0)$.

Clearly I am missing the point of this argument, so if someone would be kind enough to explain it to me I would be most grateful.

Answer

A distribution is not a function, it is a functional acting on (a suitable space of) functions.

In particular, let's consider the functions of rapid decrease $\mathscr{S}(\mathbb{R}^d)$. Its topological dual, $\mathscr{S}'(\mathbb{R}^d)$, is the space of continuous linear functionals of $\mathscr{S}$, and it is called the space of tempered distributions.

How does a distribution work? It is a map that associates to each $f\in\mathscr{S}$, a complex number. Its action is usually denoted by $(\phi,\cdot)$, where $\phi\in\mathscr{S}'$. The Dirac delta distribution $\delta\in\mathscr{S}'$, is the distribution defined by: $$(\delta,f)=f(0) \quad, \quad f\in\mathscr{S}(\mathbb{R}^d)\; .$$

Now the rapid decrease functions are dense in the tempered distributions, in a suitable topology (the $\sigma(\mathscr{S}',\mathscr{S})$ one). And there is a natural identification of $f\in\mathscr{S}$ with the corresponding element $\tilde{f}\in\mathscr{S}'$: $\tilde{f}$ is the distribution whose action is defined by $$(\tilde{f},g)=\int_{\mathbb{R}^d}f(x)g(x)dx\quad ,\quad g\in\mathscr{S}(\mathbb{R}^d)\; .$$ So if the $\delta$ distribution was a real function $\delta(x)$ (but it is not!), we could write $$(\delta,f)=\int_{\mathbb{R}^d}\delta(x)f(x)dx=f(0)\; .$$ This is what it is usually done by physicists, but it is an abuse of notation. Another abuse of notation, is to write $\int_{\mathbb{R}^d}\delta(x)dx=1$. In principle, it is not possible to define the integral of a distribution. However this abuse of notation may be justified as follows:

• It is possible to approximate the delta distribution by rapid decrease functions, since $\mathscr{S}$ is dense in $\mathscr{S}'$. Indeed, given an integrable function $\eta$, such that $\int\eta(x)dx=1$, then $$\delta_h(x)=\frac{1}{h^d}\eta\bigl(\tfrac{x}{h}\bigr)$$ is an approximation of $\delta$, in the sense that $$\lim_{h\to 0}\int_{\mathbb{R}^d}\delta_h(x)f(x)dx=f(0)\quad ,\quad f\in\mathscr{S}(\mathbb{R}^d)\; .$$ Since, in addition, $$\int_{\mathbb{R}^d}\delta_h(x)dx=1$$ uniformly in $h$, it is tempting to conclude that $\int_{\mathbb{R}^d}\delta(x)dx=1$ (but it is not true, since- very roughly speaking -you are not allowed to take the limit inside the integral).

The above type of approximations $\delta_h$, also suggest the "pictorial" representation of $\delta$ as a function (but it is not!) that is zero everywhere, and infinity in zero. However, this is just a pictorial representation (that may be useful, but is not rigorous); so it cannot be used to argue pro or against the notation $\int_{\mathbb{R}^d}\delta(x)f(x)dx$ (that again is just an abusive notation, and not a rigorously defined integral).

electromagnetism - How to derive the expression for the electric field in terms of the potential?

How can I derive that $$\vec{E}=-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t}$$ where $\phi$ is the scalar potential and $\vec{A}$ the vector potential?

Answer

$\def\vA{{\vec{A}}}$ $\def\vB{{\vec{B}}}$ $\def\vD{{\vec{D}}}$ $\def\vE{{\vec{E}}}$ $\def\vH{{\vec{H}}}$ $\def\vS{{\vec{S}}}$ $\def\eps{\varepsilon}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$

Faraday's law: $$\rot(\vE)+\dot\vB=0$$ Source-less B-field: $$\div(\vB)=0$$ From this in a simply connected domain there follows the existence of a vector potential $\vA$ with $$\vB = \rot\vA$$ Therewith, Faraday's law reads $$\rot(\vE+\dot{\vA})=0$$ The curl-freeness of the vector field $\vE+\dot{\vA}$ (in a simply connected domain) implies the existence of a scalar potential with $$\vE+\dot{\vA} = -\grad\varphi$$ and that is your formula. You see $\varphi$ is just defined in the way you wrote it down.

The purpose of introducing $\vA$ is to solve $\div\vB=0$ and the purpose of introducing $\varphi$ is to solve Faraday's law. In most cases the only equation which remains to be solved is Ampre's law $$ \rot\vH = \vS + \dot\vD $$ which reads with our new independent variables $\varphi$ and $\vA$ as $$ \rot(\mu\rot \vA) = (\kappa + \eps\partial_t)(-\grad\varphi - \dot\vA). $$ Maybe, one has also a pre-defined space-charge density $\rho$. That would imply the equation $$ \rho = \div \vD = \div (\eps(-\grad\varphi - \dot\vA)) $$ For constant $\eps$ you have $$ -\frac{\rho}{\eps} = \Delta\varphi + \div \dot\vA $$ Now, you have some degrees of freedom in the choice of $\vA$. If $\vA$ is a vector potential for $\vB$ then for any smooth scalar function $\varphi'$ also $\vA':=\vA+\grad\varphi'$ is a vector potential for $\vB$ since $\rot\grad=0$. One possible choice is $\div\vA = 0$. Therewith, the equation for the space charge reads just $$ -\frac{\rho}{\eps} = \Delta\varphi $$ which we know from electro-statics. The nice thing of $\div\vA=0$ is that the above equation decouples from the magnetics. (But only if $\rho$ assumed to be predefined.) So one can solve the problem staggered.

The condition $\div\vA=0$ can always be satisfied. If we initially have a vector potential $\vA'$ with $\div\vA'\neq 0$ then we define $\vA := \vA'+\grad\varphi'$ such that $$ 0 = \div \vA = \div(\vA' + \grad\varphi')= \div\vA' + \Delta \varphi' $$ To find the function $\varphi'$ for the modification of $\vA'$ we just have to solve the poisson equation $$ \Delta\varphi' = -\div\vA' $$ for $\varphi'$. The coice $\div A=0$ is the so called Coulomb gauging.

A field is determined by its curl and sources. Under the assumption that we have already the required boundary conditions for $\varphi$ and $\vA$ then the fixation of the divergence for the vector potential (i.e., the gauging) determines the potentials uniquely. Beside the Coulomb gauging there are other forms of useful gauging (e.g., Lorentz-gauging).

electromagnetism - Maxwell's equation for non inertial observer

Applying Maxwell's equation we can prove that light will move at the speed of light for every inertial frame, is it true as well for non-inertial frames? How light moves slowly near a black hole??

Answer

These are two separate questions.

1) Does light move at c in non-inertial frames?

First to get a bit pedantic, I assume you mean the coordinate velocity. As light does not have a well defined proper velocity, this seems to be the most reasonable way to interpret your question here.

While rewriting Maxwell's equations in a non-inertial frame, and then solving this potential mess can be difficult, we can easily take a solution in an inertial frame and just do the coordinate transformation to get the solution in some non-inertial frame. In particular, if we have the position of some light wave packet vs time in an inertial frame, we can just apply the coordinate transformation to find the new coordinate labels for the points on its path.

Consider the inertial frame with coordinates $t,x,y,z$. A simple example of a non-inertial coordinate system $t',x',y',z'$ is be given by the transformation: $$t'=t,\ x'=x+vt,\ y'=y,\ z'=z$$ This Galilean transformation gives us a non-inertial frame, and the coordinate velocities will just change by $v$ in the $x$ direction as is usual with Galilean transformations. So if in the inertial frame the velocity of the light packet was $c$ in the $x$ direction, the velocity in this new frame would be $c+v$. This is probably the easiest counter example to see.

2) How light moves slowly near a black hole?

Now you are talking about curved spacetime as well. In curved spacetime, there is no choice of coordinate system in which the coordinate speed of light will everywhere be c.

However, locally we can still choose a coordinate system in which that is the case. A free falling frame is such an example, and so if you fall into a black hole nothing strange will be seen with your local measurements of the speed of light. When people say light "slows" while approaching a blackhole, they are again talking about the coordinate speed of light, and they are implicitly assuming a coordinate system (usually the Schwarzschild coordinates). Light slowing to a stop in the limit of approaching the event horizon is just an artifact of a coordinate singularity in this particular choice of coordinate system. This singularity is only an issue with the coordinate choice and can be completely removed if another coordinate system is used. A very similar coordinate singularity occurs even in flat space-time in a common choice of coordinate system with the spatial origin having a constant proper acceleration (Rindler coordinates).

quantum mechanics - What does it mean "not to have a definite trajectory"?

In a comment to my question someone stated the following:

"photons do not travel at some definite number of oscillations per second. In fact, they do not "travel" at all, no more than electrons or other quanta do, as by the Uncertainty Principle they don't have a definite speed and/or trajectory"

Nobody objected or denied it, can someone explain what that actually means?

• Does it mean that they do not have a definite/straight/regular trajectory and they wander erratically or that they have none at all? Can you try to graphically describe their motion ?

It is generally thought that QM describes weird things, laws and phenomena that are quite different from macroscopic world, can you be precise about one feature, please, i.e if it respects the basic old tenet "Natura non facit saltus":

• Does QM allow a particle to disappear from a point and reappear in another point that is not continuous to it? If so, what is the explanation?

newtonian mechanics - Throwing a ball upwards in an accelerating train

If I throw a ball upwards to a certain height in an accelerating train, will it end up in my hand? At the moment I release the ball, it will have a velocity equal to that of the train at that instant. But because the train is accelerating, the train will have a greater velocity and thus the ball will travel lesser distance and should fall behind me. Is my reasoning correct?

Answer

User Sahil Chadha has already answered the question, but here's the math and a pretty picture for anyone who is unconvinced that you're right.

Since the train is accelerating, from the perspective of an observer on the train, the ball will experience a (fictitious) force in the direction opposite the train's travel having magnitude $ma$ where $m$ is the mass of the ball and $a$ is the magnitude of the acceleration of the train. If we call the direction of travel the positive $x$-direction, and if we call the "up" direction the positive $y$-direction, then the equations of motion in the $x$- and $y$-directions will therefore be as follows: \begin{align} \ddot x &= -a \\ \ddot y &= -g. \end{align} The general solution is \begin{align} x(t) &= x_0 + v_{x,0} t - \frac{1}{2}a t^2 \\ y(t) &= y_0 + v_{y,0} t - \frac{1}{2}g t^2 \end{align} Now, let's say that the origin of our coordinate system lies at the point from which the ball is thrown so that $x_0 = y_0 = 0$ and that the ball is thrown up at time $t=0$ with velocity $v_{y,0} = v$ and $v_{x,0} = 0$ in the positive $y$-direction, then the solutions becomes \begin{align} x(t) &= -\frac{1}{2}a t^2\\ y(t) &= vt - \frac{1}{2} gt^2 \end{align} So what does this trajectory look like? By solving the first equation for $t$, and plugging this back into the equation for $y$, we obtain the following expression for the $y$ coordinate of the particle as a function of its $x$ coordinate along the trajectory: \begin{align} y(x) = v\sqrt{-\frac{2x}{a}} +\frac{g}{a} x \end{align} Here's a Mathematica plot of what this trajectory looks like for $v = 1.0\,\mathrm m/\mathrm s$ and the list $a = 9.8,5.0,2.5,1,0.1\,\mathrm m/\mathrm s^2$ of values for the train's acceleration

From the point of view of someone on the train, the ball flies backward in a sort of deformed parabola, but the smaller the acceleration is, the more it simply looks as it would if you were to throw a ball vertically in an un-accelerated train.

quantum mechanics - Is the wave function objective or subjective?

Here is a question I am curious about.

Is the wave function objective or subjective, or is such a question meaningless?

Conventionally, subjectivity is as follows: if a quantity is subjective then it is possible for two different people to legitimately give it different values. For example, in Bayesian probability theory, probabilities are considered subjective, because two agents with access to different data will have different posteriors.

So suppose two scientists, A and B, have access to different information about the same quantum system. If A believes it has one wavefunction and B believes it has another, is one of them necessarily "right" and the other "wrong"? If so then the wavefunction is objective, but otherwise it must contain some subjective element.

gravity - Identify binary stars in nbody simulation

I'm doing a nbody simulation, and I'm interested in the formation of binary systems in the temporal evolution. I can identify them by eye, but I don't know an algoritmic criteria that can say me if two stars are a binary system or not.

Is there a deterministic criteria that can say me if two arbitrary stars in the simulation are a binary system?

Answer

This is really a difficult problem, but possibly not for the reason you imagine. The following naif criterion seems at first highly appropriate:

take any pair of stars, subtract the center-of-mass motion, compute the kinetic ($T$) and gravitational ($W$) energies, and check whether

$T + W < 0$.

If so, the pair is bound, otherwise it is not.

How can this criterion fail? I can think of three ways:

1. 
   

   Even in an uncrowded field, there may be three or more stars bound. It is well-known that multiple systems are unstable, so establishing which of the stars will remain bounded requires more analysis. Normally, the two heaviest stars are those that remain bounded, because the lighter ones gain energy at the expense of the heavy double, to escape the bound system.

   
   


2. 
   

   A supposedly unbound pair may later turn out to be bound after all, because of loss of angular momentum. It is well-know that the centrifugal barrier, under angular momentum conservation from the original cloud in which the stars form, is so high that nearly no stars should form. Since gravity conserves the overall amount of angular momentum, there must be non-gravitational forces at work that remove it from the collapsing gas. No one knows for sure what is responsible for these torques, even though everyone claims that they must be magnetic in origin.

   
   

   Thus it can happen that a given pair has, initially, so much angular momentum as to violate the above condition, but it can also happen that angular momentum removal is still occurring (at the time your simulations end) so that more binary pairs will form than is apparent from your simulations. Since this involves processes which are, as of now, largely speculative, no one can claim to know the exact time-evolution of these torques. This is a fundamental element of ignorance, on our part.

   
   


3. 
   

   Lastly, and least likely, in crowded fields, interactions between the (transient) pair and passers-by may unbind the binary, at the expense of the kinetic energy of the passer-by. In the limit of fully-developed stars, this has been studied especially in connection with the evolution of binaries in globular clusters (GCs, see Lyman Spitzer's Dynamical Evolution of Globular Clusters, especially Ch. 6). But you should keep in mind that in GCs two circumstances enhance the relevance of these interactions: the high stellar densities (higher than in Open Clusters), and the long time scales available, of the order of a Hubble time as compared to the few million years available before newly-formed stars walk away from their formation sites.

   
   
   

   Still, one circumstance makes many-body encounters more effective in Open Clusters, the large size of forming stars: when a binary encounters a passer-by, energy may be dumped from the enterloper's kinetic motion into tidal heating of either star in the binary, to be later dissipated thru radiative mechanisms. This tidal heating slows down the interloper, hence it makes the contact last longer, and the impulse transferred to the stars in the binary larger. In other words, it facilitates unbinding the binary.

   
   

Of the above, objections 1 and 3 are easily kept under control if your simulations do not involve unrealistically crowded fields (after all, Open Clusters are barely bound, and do no last long!). But objection 2 is more fundamental, and harder to keep under control. Keep in mind that, in Nature, 1 out of two stellar systems observed from Earth is a binary, so just about any simulation resulting in many fewer binaries than this is likely to be tainted by an improper accounting of electromagnetic torques.

general relativity - What does this depiction of a black hole in the movie Interstellar mean?

I was expecting a whirlpool in 3D and the matter glowing from friction as it nears the center, as I expected a event horizon to be negligible visually.

How does this depiction work? How big is the central sphere? I am puzzled by the perpendicular circles. Are they the event horizons if they are both visible? What would be a path of a particle as it gets swallowed into the singularity?

Answer

First note that this is a fictional movie and the image is an artist's impression, not a detailed simulation. The public seems to think the movie is some sort of fictionalized documentary, which it never claimed to be. That said, the image is qualitatively conveying some of what happens near a black hole.

The diagonal disk is the accretion disk -- this is where matter is spiraling inward due to gravity, friction, and electromagnetic forces too. It glows because it is very hot.

The circular ring is the result of gravity bending the light emitted from the far side of the accretion disk into our line of sight. A similar effect happens when the light source is much farther behind the black hole, as seen in this Wikimedia image:

As for event horizons, no, you would only see the outer one by definition. (An event horizon is nothing more or less than a surface which delineates which regions of spacetime can communicate with each other.) Since the image in question is made for primarily artistic purposes, I wouldn't try to read too much into details that you see.

homework and exercises - Shape of a rotating rope with one free-end

One end of a uniform rope (with total mass $M$) is fixed on the edge of a cylinder. The cylinder has a radius $R$ and rotates with angular velocity $\omega$. The axis is vertical in a gravitational field. Air drag is neglected. What is the shape of the rope?

——————

I tried to use the Euler-Lagrange equation to solve this problem but I don't know how to handle free-end boundary condition. Maybe some other method?

homework and exercises - Question about tensor form of Maxwell equation

By variating the Maxwell Lagrangian we get the equation of motion. The remaining two Maxwell equations can be written as $$\epsilon_{\mu\nu\rho\sigma}\partial^{\rho} F^{\mu\nu} = 0.$$ I have also seen it written as the Bianchi identity: $$\partial_{[\lambda}F_{\mu\nu]} = 0.$$ Why are these two forms equivalent?

Answer

It's basically just a duality relation analogous to the cross product in three dimensions. But if you want to do some work to show the equivalence, then:

Going from the second equation to the first is easy, just hit it with $\epsilon_{\mu\nu\rho\sigma}$.

Going from the first to the second equation, is a little trickier and relies on knowing how to evaluate the products of Levi-Civita symbols. The basic idea is that you should contact the first equation with $\epsilon^{\mu'\nu'\lambda'\sigma}$ and compare the resulting antisymmetric combination of $\delta$s with the antisymmetrization of the indices in the second equation.

homework and exercises - Movie Interstellar - Question about Escape Velocity

The movie Interstellar shows people on a water planet where time is dilated so much that 1 hour is equal to 7 years back on Earth. Even though they lift off from Earth using a Saturn-V two stage rocket, they leave this water planet in a shuttle craft.

Can someone provide the escape velocity for a place where time is dilated by 60,000 times? The stated mass of the black hole creating the gravity well is 100 million solar masses, but I'm not sure that would work into the calculation.

Answer

The gravity from the black hole (BH) will have no effect on their ability to take off from the water planet itself. Objects in orbit feel weightless (think astronauts in the ISS). If you're only worried about getting off the water planet then there should be no problem.

However, if they were to try to put some distance between themselves and the BH then they'd find it a much more difficult task.

Assuming the BH isn't rotating very fast, the time dilation factor for a body in orbit relative to a stationary observer at infinity is:

$$\frac{d \tau}{dt} = \sqrt{1-\frac{3GM}{c^2 r}}$$

The escape velocity of a BH looks the same as it does in Newtonian mechanics:

$$v_0^2 = \frac{2GM}{r}$$

So if the mass of the BH is ~100 million solar masses, the radius at which the time dilation is 60,000 times normal is at $r \approx$ 275 million miles $\approx$ 3x average Earth-Sun distance. The event horizon itself is at $r \approx$ 2x the average Earth-Sun distance. Meanwhile the escape velocity at the planet radius is about 82% the speed of light, or $v_0 \approx$ 250 million meters/second.

experimental physics - How the nucleon structure has been identified experimentally?

It is known that nucleons (proton, neutron) are composed of partons (quarks, etc.). How was this identified experimentally? In particular, how it has been identified that nucleons comprise of more than one constituent?

Answer

Matt Strassler goes into detail with LHC data here:

http://profmattstrassler.com/articles-and-posts/largehadroncolliderfaq/whats-a-proton-anyway/checking-whats-inside-a-proton/

Is there a magnetic line that is a Eucliden straight line?

In my (though few) experiences, magnetic lines are always curved (with respect to a Euclidean plane). I have never seen there is any magnetic line that is Euclidean straight.

If there is a magnetic line that is Euclidean straight, could you please give me an example? If not, then the existence of a Euclidean straight magnetic line contradicts what physics theories?

Answer

Magnetic field lines are not straight because we have not found a mono magnetic pole yet. magnetic lines by a single pole can only be straight like electric field lines by a single positive charge or single negative charge. But unlike electric charge ,magnetic charge always found is pairs(as far as we know there exist only magnetic dipole). Thats why magnetic lines are not straight.

string theory - Applications of Algebraic Topology to physics

I have always wondered about applications of Algebraic Topology to Physics, seeing as am I studying algebraic topology and physics is cool and pretty. My initial thoughts would be that since most invariants and constructions in algebraic topology can not tell the difference between a line and a point and $\mathbb{R}^4$ so how could we get anything physically useful?

Of course we know this is wrong. Or at least I am told it is wrong since several people tell me that both are used. I would love to see some examples of applications of topology or algebraic topology to getting actual results or concepts clarified in physics. One example I always here is "K-theory is the proper receptacle for charge" and maybe someone could start by elaborating on that.

I am sure there are other common examples I am missing.

quantum mechanics - What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-omega t)}$?

I've read that plane wave equations can be represented in various forms, like sine or cosine curves, etc. What is the part of the imaginary unit $i$ when plane waves are represented in the form $$f(x) = Ae^{i (kx - \omega t)},$$ using complex exponentials?

Answer

It doesn't really play a role (in a way), or at least not as far as physical results go. Whenever someone says

we consider a plane wave of the form $f(x) = Ae^{i(kx-\omega t)}$,

what they are really saying is something like

we consider an oscillatory function of the form $f_\mathrm{re}(x) = |A|\cos(kx-\omega t +\varphi)$, but:

• we can represent that in the form $f_\mathrm{re}(x) = \mathrm{Re}(A e^{i(kx-\omega t)})=\frac12(A e^{i(kx-\omega t)}+A^* e^{-i(kx-\omega t)})$, because of Euler's formula;


• everything that follows in our analysis works equally well for the two components $A e^{i(kx-\omega t)}$ and $A^* e^{-i(kx-\omega t)}$;


• everything in our analysis is linear, so it will automatically work for sums like the sum of $A e^{i(kx-\omega t)}$ and its conjugate in $f_\mathrm{re}(x)$;


• plus, everything is just really, really damn convenient if we use complex exponentials, compared to the trigonometric hoop-jumping we'd need to do if we kept the explicit cosines;



• so, in fact, we're just going to pretend that the real quantity of interest is $f(x) = Ae^{i(kx-\omega t)}$, in the understanding that you obtain the physical results by taking the real part (i.e. adding the conjugate and dividing by two) once everything is done;


• and, actually, we might even forget to take the real part at the end, because it's boring, but we'll trust you to keep it in the back of your mind that it's only the real part that physically matters.

This looks a bit like the authors are trying to cheat you, or at least like they are abusing the notation, but in practice it works really well, and using exponentials really does save you a lot of pain.

That said, if you are careful with your writing it's plenty possible to avoid implying that $f(x) = Ae^{i(kx-\omega t)}$ is a physical quantity, but many authors are pretty lazy and they are not as careful with those distinctions as they might.

---

(As an important caveat, though: this answer applies to quantities which must be real to make physical sense. It does not apply to quantum-mechanical wavefunctions, which must be complex-valued, and where saying $\Psi(x,t) = e^{i(kx-\omega t)}$ really does specify a complex-valued wavefuntion.)

kinematics - Why do we equate an indefinite integral to a specific value?

Suppose we want to obtain a displacement vector defined as $\mathbf s(t) = x(t)\mathbf i + y(t)\mathbf j + z(t)\mathbf k$ from the components of a velocity vector $\mathbf v(t) = \dot x(t)\mathbf i + \dot y(t)\mathbf j + \dot z(t)\mathbf k=\mathbf 0$. According to my notes, this can be done by equating each scalar component of the displacement vector to the indefinite integral of the corresponding scalars of the velocity vector, i.e. $$ \mathbf s(t)=\begin{pmatrix} x(t)=\int \dot x\ dt \\ y(t)=\int \dot y\ dt \\ z(t)=\int \dot z\ dt \end{pmatrix} $$ But, as $\int f(x)\ dx=\{F(x): \frac {dF}{dx}=f(x)\}$, this should be syntactically wrong, because we're implying that a number is equal to an infinite set of numbers, or am I missing something?

Moreover, this also leads to a weird equation when solving the integral; for example, by taking into consideration the $x$-component of $\mathbf s$, we would have that $$ x(t)=\int \dot x\ dt=c_1 $$

Which is correct, but it would also mean that the $x$-component of the velocity could be equal to any value belonging to $\mathbb R$. Because of that, we substitute $c_1$ with the initial condition and we equate it to zero, giving it a specific value: $c_1=0$. But, to me, this sounds like a break of the definition of indefinite integrals, as $\int \dot x\ dt=c_1=0$ would basically mean that an indefinite integral is one, specific function.

I know this may be a very stupid question, and maybe it has to do with the same shortcuts that make us not specify "$\forall c \in \mathbb R$" when adding the constant $c$ in the solutions of an indefinite integral, but this doubt is really challenging me and I still don't understand whether I'm missing some point or it should actually be written $x(t)=c_1=0 \in \int \dot x\ dt$. Thanks a lot in advance!

Answer

In physics we frequently leave off the limits of the integral when the limits can be figured out from the context. So, in the first case, the actual relation is: $$x(t) = x(0) + \int_0^t \dot{x} \operatorname{d}t'.$$

Most often, though, when the limits are left off the implied limits are over all possible values of the dummy variable. For example: $$ Q = \int \rho(\mathbf{x}) \operatorname{d}x^3$$ is understood to be the integral over all of space.

classical mechanics - How to find a particle's dynamics in general relativity?

About a year ago, I took a course on general relativity. It isn't until now that I realize that, given a metric, I am unsure how to find a particle's dynamics. What I mean by that is, normally I would set up either a Lagrangian and Hamiltonian and try to find dynamics the usual ways from these methods. But how do these methods change when I am in curved space time? During the relativity course, we spent a very large amount of time studying tensor analysis, covariant derivatives, geodesics, etc, but not finding the dynamics of a physical system. I am sure we learned this somewhere, maybe I just forgot.

I thought to myself that maybe I could use the geodesic equation, but I don't think this equation would be valid in the presence of a potential. Can I just use the normal methods but just replace the normal derivative with the covariant derivative?

EDIT:

Thanks for all the answers and the bounty put up. I will wait a couple more days to pick the best answer. I thought I would add this for future readers of this question (and possibly for more discussion).

I found a section in my relativity book (Spacetime and Geometry by Sean Carroll) in the beginning of Chapter 4, he says in order to do the "physics of space time" do the following:

1. Take a law of physics, valid in inertial coordinates in flat spacetime



2. Write it in a coordinate invariant (tensorial) form.


3. Assert that the resulting law remains true in curved spacetime.

His example is then to do these steps with a free particle to obtain the geodesic equation, so according to this, these steps could be followed for any dynamical theory. The second step being the most complicated part as (for example) second derivatives need modifications to be considered a tensorial object (which he covers in the book).

Answer

As in classical mechanics, there are several approaches to determining the dynamics of a particle in general relativity. Crucially we should observe that, in general relativity, the effect of a gravitational field is entirely captured by the behaviour of the metric. Providing the particle in question experiences no other external influences, we can determine its motion using the geodesic equation as you suppose. This is one of the central axioms of general relativity:

A particle moving only under the influence of gravity follows geodesics in curved spacetime. These geodesics are timelike if the particle is massive and null if the particle is massless.

Knowing the above, we can directly write down the equation governing particle motion. In Riemannian geometry, it is natural to define a geodesic as a path that extremises the length between two fixed points. For us, this definition comes with some subtleties, due to the fact that two distinct points can be zero distance apart. For our purposes, there is a better definition:

A geodesic is a path whose tangent vector is parallel transported along it. In wordier terms, if you take the tangent vector at one point, and then carry it along the path without changing the direction it points in, the vector remains tangent to the path at the destination.

In less wordy terms, geodesics are straight lines.

Mathematically this is expressed as

$$ \nabla_U U = 0 \qquad \mathrm{where} \qquad U^\mu = \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}\,.$$ Here $\lambda$ is a parameter along the curve and $U$ is tangent to it. If we expand out this equation using the definition of the covariant derivative, we arrive at the famous geodesic equation: $$ \frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\lambda^2} + \Gamma^\mu{}_{\nu \rho}\frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}\frac{\mathrm{d}x^\rho}{\mathrm{d}\lambda} = 0 \,.$$ This equation is the analogue of Newton's first law: particles under the influence of no external force move in straight lines at constant speed. It is nothing other than

$$\vec{a} = 0\,.$$

This comparison informs us how to modify the geodesic equation when external forces are present. For definiteness, suppose that the particle is moving under the influence of an electromagnetic field, in addition to gravity. We simply take geodesic equation and replace the right hand side with force over mass. Writing the Lorentz force in the covariant form used in special relativity, and choosing $\lambda$ to be proper time $\tau$, we find that a charged particle obeys the equation: $$ \frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau^2} + \Gamma^\mu{}_{\nu \rho}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\rho}{\mathrm{d}\tau} = \frac{q}{m} F^\mu{}_\nu \frac{\mathrm{d}x^\nu}{\mathrm{d}\tau} \,.$$ Note that if we turn off the gravitational field by setting the connection components to zero, we recover the special relativistic form of Newton's second law for a particle moving in an electromagnetic field, as we must do.

One of the advantages of Lagrangian over Newtonian mechanics is that it's often easier to write down some scalar 'energy' that characterises our system than it is to specify the vectorial forces acting on it. With no external influences present, we want an action that, when extremised, gives us geodesics. For massive particles, we can employ the definition of geodesic familiar from Riemannian geometry – action equals length:

$$S = m\int\mathrm{d}\tau = m\int \mathrm{d}\lambda \sqrt{-g_{\mu \nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}\frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}} \,.$$

It's a worthwhile exercise to check that this indeed yields the geodesic equation when extremised (note that one should be careful about the limits on the integral when performing the variation!).

With this free-particle action in place, it's a simple matter to add extra terms that capture the effect of interactions of the particle with external fields or other particles. If an electromagnetic field is present, the action becomes

$$ S = m\int \mathrm{d} \tau - q \int \mathrm{d} \tau A_\mu \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} \,. $$

For massless particles this picture is complicated by the fact that every null path has zero length. To talk about the Lagrangian mechanics of massless particles, we need to introduce a new action which contains a Lagrange multiplier $e(\lambda)$ defined along the worldline. The following action is valid for particles of any mass:

$$ S = \frac{1}{2} \int \mathrm{d} \lambda \left(\frac{1}{e(\lambda)} g_{\mu \nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}\frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda} - m^2 e(\lambda)\right)\,.$$

This is a technical point – more information can be found in this and this stackexchange post. At the end of the day, we're defining these actions so that they return the geodesic equation when extremised. Adding extra terms to the action to account for interactions is then often easier than adding extra terms directly to the geodesic equation.

Perhaps some other time; this answer is already long enough. Let me just note here that it is entirely possible to formulate a Hamiltonian approach to particle dynamics in a curved spacetime, but a little tricky. There are additional subtleties associated with the fact that the conjugate momenta are not independent – they satisfy a mass-shell constraint – which compels us to include a Lagrange multiplier as above. For more information see these lecture notes by Paul Townsend.

thermodynamics - What is entropy really?

On this site, change in entropy is defined as the amount of energy dispersed divided by the absolute temperature. But I want to know: What is the definition of entropy? Here, entropy is defined as average heat capacity averaged over the specific temperature. But I couldn't understand that definition of entropy: $\Delta S$ = $S_\textrm{final} - S_\textrm{initial}$. What is entropy initially (is there any dispersal of energy initially)? Please give the definition of entropy and not its change.

To clarify, I'm interested in the definition of entropy in terms of temperature, not in terms of microstates, but would appreciate explanation from both perspectives.

Answer

There are two definitions of entropy, which physicists believe to be the same (modulo the dimensional Boltzman scaling constant) and a postulate of their sameness has so far yielded agreement between what is theoretically foretold and what is experimentally observed. There are theoretical grounds, namely most of the subject of statistical mechanics, for our believing them to be the same, but ultimately their sameness is an experimental observation.

1. 
   

   (Boltzmann / Shannon): Given a thermodynamic system with a known macrostate, the entropy is the size of the document, in bits, you would need to write down to specify the system's full quantum state. Otherwise put, it is proportional to the logarithm of the number of full quantum states that could prevail and be consistent with the observed macrostate. Yet another version: it is the (negative) conditional Shannon entropy (information content) of the maximum likelihood probability distribution of the system's microstate conditioned on the knowledge of the prevailing macrostate;

   
   


2. 
   

   (Clausius / Carnot): Let a quantity $\delta Q$ of heat be input to a system at temperature $T$. Then the system's entropy change is $\frac{\delta Q}{T}$. This definition requires background, not the least what we mean by temperature; the well-definedness of entropy (i.e. that it is a function of state alone so that changes are independent of path between endpoint states) follows from the definition of temperature, which is made meaningful by the following steps in reasoning: (see my answer here for details). (1) Carnot's theorem shows that all reversible heat engines working between the same two hot and cold reservoirs must work at the same efficiency, for an assertion otherwise leads to a contradiction of the postulate that heat cannot flow spontaneously from the cold to the hot reservoir. (2) Given this universality of reversible engines, we have a way to compare reservoirs: we take a "standard reservoir" and call its temperature unity, by definition. If we have a hotter reservoir, such that a reversible heat engine operating between the two yields $T$ units if work for every 1 unit of heat it dumps to the standard reservoir, then we call its temperature $T$. If we have a colder reservoir and do the same (using the standard as the hot reservoir) and find that the engine yields $T$ units of work for every 1 dumped, we call its temperature $T^{-1}$. It follows from these definitions alone that the quantity $\frac{\delta Q}{T}$ is an exact differential because $\int_a^b \frac{d\,Q}{T}$ between positions $a$ and $b$ in phase space must be independent of path (otherwise one can violate the second law). So we have this new function of state "entropy" definied to increase by the exact differential $\mathrm{d} S = \delta Q / T$ when the a system reversibly absorbs heat $\delta Q$.

   
   

As stated at the outset, it is an experimental observation that these two definitions are the same; we do need a dimensional scaling constant to apply to the quantity in definition 2 to make the two match, because the quantity in definition 2 depends on what reservoir we take to be the "standard". This scaling constant is the Boltzmann constant $k$.

When people postulate that heat flows and allowable system evolutions are governed by probabilistic mechanisms and that a system's evolution is its maximum likelihood one, i.e. when one studies statistical mechanics, the equations of classical thermodynamics are reproduced with the right interpretation of statistical parameters in terms of thermodynamic state variables. For instance, by a simple maximum likelihood argument, justified by the issues discussed in my post here one can demonstrate that an ensemble of particles with allowed energy states $E_i$ of degeneracy $g_i$ at equilibrium (maximum likelihood distribution) has the probability distribution $p_i = \mathcal{Z}^{-1}\, g_i\,\exp(-\beta\,E_i)$ where $\mathcal{Z} = \sum\limits_j g_j\,\exp(-\beta\,E_j)$, where $\beta$ is a Lagrange multiplier. The Shannon entropy of this distribution is then:

$$S = \frac{1}{\mathcal{Z}(\beta)}\,\sum\limits_i \left((\log\mathcal{Z}(\beta) + \beta\,E_i-\log g_i )\,g_i\,\exp(-\beta\,E_i)\right)\tag{1}$$

with heat energy per particle:

$$Q = \frac{1}{\mathcal{Z}(\beta)}\,\sum\limits_i \left(E_i\,g_i\,\exp(-\beta\,E_i)\right)\tag{2}$$

and:

$$\mathcal{Z}(\beta) = \sum\limits_j g_j\,\exp(-\beta\,E_j)\tag{3}$$

Now add a quantity of heat to the system so that the heat per particle rises by $\mathrm{d}Q$ and let the system settle to equilibrium again; from (2) and (3) solve for the change $\mathrm{d}\beta$ in $\beta$ needed to do this and substitute into (1) to find the entropy change arising from this heat addition. It is found that:

$$\mathrm{d} S = \beta\,\mathrm{d} Q\tag{4}$$

and so we match the two definitions of entropy if we postulate that the temperature is given by $T = \beta^{-1}$ (modulo the Boltzmann constant).

Lastly, it is good to note that there is still considerable room for ambiguity in definition 1 above aside from simple cases, e.g. an ensemble of quantum harmonic oscillators, where the quantum states are manifestly discrete and easy to calculate. Often we are forced to continuum approximations, and one then has freedom to define the coarse gaining size, i.e. the size of the discretizing volume in continuous phase space that distinguishes truly different microstates, or one must be content to deal with only relative entropies in truly continuous probability distribution models Therefore, in statistical mechanical analyses one looks for results that are weakly dependent on the exact coarse graining volume used.

Why $-ihbarvecnabla$ for momentum in quantum mechanics, while $mvec{v}$ in classical mechanics?

I am a little bit confused when thinking of the momentum representation in QM and CM.

In QM, momentum is represented as $-i\hbar\vec\nabla$, while in classical, momentum is represented as $m\vec{v}$.

At least, where does the mass $m$ in CM gone when meets QM please?

Once I saw a sentence like "What quantum theory really unites is matter and information" from Prof. Xiao-Gang Wen's PPT file. Though I do not understand this sentence at all for the moment.

Answer

$-i ħ \nabla$ is the momentum operator. You have to apply it to a wave function to get the actual momentum.

Consider the plane wave solution to the Schrödinger equation: $\Psi = e^{i \mathbf{k} \cdot \mathbf{r} - \omega t}$. Applying the momentum operator gives $-i ħ \mathbf{k} \Psi$. You can see the eigenvalue has units of momentum. (If you can't see it, note that $\mathbf{k} \cdot \mathbf{r}$ in the exponent is dimensionless, so clearly $\mathbf{k}$ has units of inverse length. $ħ$ has units of angular momentum, so $ħ \mathbf{k}$ has units of momentum.)

As far as where the mass of the particle factors in, it's in the Schrödinger equation (and thus related to the wave function): $i ħ \frac{\partial}{\partial t} \Psi \left(\mathbf{r}, t \right) =\left[-\frac{ħ^2}{2m}\nabla^2 + V \left(\mathbf{r}, t \right)\right]\Psi \left(\mathbf{r}, t \right)$

In particular, the classical relationship between momentum and kinetic energy is $E = \frac{p^2}{2m}$. (That's the same as your $mv$, for $E = \frac{1}{2}mv^2$.) Note for the free particle in quantum mechanics, it's the same. $E \Psi = - \frac{ħ^2}{2m}\nabla^2 \Psi = \frac{\left(-iħ\nabla\right)^2}{2m} \Psi = \frac{p^2}{2m} \Psi $

observable universe - What's the total Cosmic Microwave Background energy?

This is a part of my wider attempts at accounting matter in the observable universe.

The total light ever emitted by stars and other space objects is capped around 1% of baryonic mass (in energy equivalent).

How much light is there in space and how heavy is it?

How does the CMB compare to that? How would we calculate it?

Answer

The energy density of the Cosmic Microwave Background (CMB) is $4.19 \times 10^{-14} J/m^3$.

This value can be calculated from the CMB temperature being 2.728 K and

$energy$ $density = aT^4$

where $a$ is the Radiation Constant.

You can multiply the energy density by your volume of interest such as that of the observable universe if you want total energy.

electromagnetism - Why does my cell phone still work inside a metal box?

I just did the following experiment:

Using a metal Whitman's chocolate box, I placed my phone it it, closed the lid then called it from another phone. To my surprise, it rang.

The normal parameters of a faraday cage is any gaps have to be small compared to the wavelength of the radiation. The edges of the box overlap by about 1 cm, and the gap is a fraction of a mm.

Phones operate at 1-2 GHz.  Skin depth for 1 GHz is about 2 microns Wavelength of 1 GHz is about 8 inches.

Even if it's 30 ga metal that's still 12 thou or about 300 microns.  Now it's steel, not copper, but steel isn't THAT much worse a conductor.

I don't understand what is going on.

Edit: More experiments:

I have phoning over wifi enabled, and I'm sitting 20 feet from the access point. Disabling the wifi on the phone and using the 2 bar local cell signal, no calls.

Re-enable wi-fi and I can reach the phone in the box intermittently. I can ping the phone from my local network which makes for faster tests. Curiously, the phone doesn't respond to pings when it's asleep.

My current working hypothesis is that the electrical connection between the box and the lid isn't reliable.

The box is laquered with red and gold Christmas colours which may be acting as an insulator.

After opening and closing the box several times, while running a ping to the device over my local I get erratic results. Sometimes closing the box gets me 'request timed out.' sometimes not.

Wrapping it with alumininum foil loosely gives more odd results.

At one point I had it loose enough that I could see the phone -- gap of maybe 1 cm, and got device unreachable, and didn't become reachable until it was open to about 2 inches. I suspect that my AP has some kind of adaptive power setting.

Right now it's sitting in it's metal box and I'll get a series of 20 unreachables, 4 pings with variable times, more timeouts.

Answer

The normal parameters of a faraday cage is any gaps have to be small compared to the wavelength of the radiation. The edges of the box overlap by about 1 cm, and the gap is a fraction of a mm.

The important dimension is the longest dimension of the gap. So if you have a seam 0.01 mm wide, but 10 cm long, it's the 10 cm dimension that matters when deciding if the gap is "small" compared to the wavelength.

So if the seal between the lid and the body of your candy box goes all the way around the box, as they typically do, then the critical dimension is the length of the box.

It's also possible the box parts have some coating (paint or a plastic film) that prevents the two metal part making a metal-to-metal contact anywhere in the seam.

The fact that the two parts overlap by 1 cm does mitigate this somewhat, so you would expect some blocking from this arrangement.

The point made by Jon Custer in comments about the high resistivity of the carbon steel used in this kind of box is also likely relevant.

quantum information - How can one get the density operator from the characteristic function?

To solve analytically the master equation of two qubits interacting with a cavity mode through their environment we use the charactristic function,
$$\chi (\beta)=\operatorname{tr}[\rho D(\beta)],$$ with $D(\beta)$ being the displacement operator.

How do we get the $\rho$ from the $\chi$ expression?

Answer

The fundamental reconstruction theorem in phase-space quantum mechanics inverts the Wigner transform of an arbitrary operator $\hat G$ through the Weyl transform, $$ \hat G= \frac {\hbar}{2\pi} \int\!\! d\tau d\sigma~~ e^{i(\tau \hat p + \sigma \hat x)} ~~\operatorname{tr} [\hat G~ e^{-i(\tau \hat p + \sigma \hat x) } ] . $$ (If you blink in a right-brain hemisphere vision, your might dream of a collapsing 2d operator δ-function there, but don't worry if this does not evoke anything. The trivial formal proof is in our Concise Treatise of Quantum Mechanics in Phase Space, Ch 0.18. It first appeared in Groenewold's 1946 breath-taking dissertation.)

Mindful of the connection to optical phase space, $$ \hat D(\beta)= e^{\beta a^\dagger -\beta ^* a}= e^{\hat x (\beta -\beta^*)/\sqrt{2}-i\hat p (\beta + \beta^*)/\sqrt{2}}, $$ and using $\hat ρ$ as your operator, you translate this to a linear combination of translation operators, your building blocks, $$ \hat\rho= \int \frac{d^2\beta }{\pi} ~~ \hat D(-\beta) ~~ \operatorname{tr} [ \hat\rho \hat D(\beta) ]= \int \frac{d^2\beta }{\pi} ~~ \hat D(-\beta) ~~ \chi(\beta) . $$

newtonian mechanics - Relation and difference between work and kinetic energy

I don't really understand the difference and the relation between work and kinetic energy. When you move an object a distance you do work (or does the object work?), what's the object's kinetic energy? Is the kinetic energy the work you've put in to the box all gathered up at the end of the distance given the box keeps moving when you stopped pushing the box? The box can't have kinetic energy if's stationary after I've pushed the box can it?

Sorry if it's unclear.

Answer

The kinetic energy is indeed the work you have done (assuming no change in potential energy), however this is also neglecting the effect of friction. In real life, the kinetic energy of the box is going to be constant if its moving at a constant velocity even if you are exerting a force on the box and hence doing work on the box. This is because in this case, the work you are doing is being converted into heat and sound due to friction. If we neglect friction, your box would keep accelerating as you applied a force and hence the kinetic energy would keep increasing.

general relativity - Metric and Connection

I have read and heard that there are several theories of gravity and Quantum Gravity which treat the metric (defining rods and clocks) and connection (defining free fall equation of geodesic) of the manifold as independent quantities. I know that the Levi-Civita connections are given as, $$ \Gamma_{\mu\nu}^{\alpha}=\frac{1}{2}g^{\alpha\delta}(g_{\mu\delta,\nu}+g_{\nu\delta,\mu}-g_{\mu\nu,\delta}) $$ But in this case, the connection is uniquely determined by the metric tensor. However, as I see it, the converse is not true (that is the metric cannot be uniquely determined by a given Levi-Civita connection).

So, Given that the Levi-Civita connection can be uniquely determined by the metric tensor, does this fact ascribe certain general properties to the manifold? Hence, by violating these properties of the manifold, one can find other mathematical forms of connection which can be uniquely determined by the metric tensor.

Answer

The Levi-Civita connection (or Christoffel symbols) are coming from the metric compatibility constraint :

$$\tag{1} \nabla_{\lambda} \, g_{\mu \nu} = 0,$$

plus the "desire" to get a symetrical connection : $\Gamma_{\mu \nu}^{\lambda} = \Gamma_{\nu \mu}^{\lambda}$. This last commandement is arbitrary, and is only justified to get the simplest theory possible (i.e classical General Relativity). If you remove the symetrical constraint but only imposes equ (1), you then get the Levi-Civita connection and a contorsion tensor :

$$\tag{2} \Gamma_{\mu \nu}^{\lambda} = \frac{1}{2} \, g^{\lambda \kappa} \, (\, \partial_{\mu} \, g_{\nu \kappa} + \partial_{\nu} \, g_{\mu \kappa} - \partial_{\kappa} \, g_{\mu \nu}) + K^{\lambda}_{\mu \nu}. $$ The antisymetrical part $T^{\lambda}_{\mu \nu} \equiv \Gamma_{\mu \nu}^{\lambda} - \Gamma_{\nu \mu}^{\lambda}$ is the torsion tensor, which is assumed to vanish in GR. This guy is not 0 in the Einstein-Cartan theory, which is the natural extension of GR (a bit more general than GR).

Some theories consider that the connection is an independant variable when they don't impose the symetric constraint, or even the compatibility (1) above.

quantum mechanics - Why, for a spin-½ particle, are the possible outcomes of measuring spin projection along any direction the same?

If one measures the projection of spin of a spin half particle along the $x$ axis one will always get $\pm\tfrac12\hbar$.

Measuring it along the $y$ axis one will always get $\pm\tfrac12\hbar$.

Measuring it along the $z$ axis one will always get $\pm\tfrac12\hbar$.

Measuring it along "any" axis one will always get $\pm\tfrac12\hbar$.

Why are the possible outcomes the same in all directions?

What is meant by “combinatorial background” in experimental high energy physics

My guess is that they find a certain tracks coming from a certain source by "combinatorially" selecting all track pairs and finding their invariant mass. If this is true, of which I am not sure, how is the background defined, i.e how far away from the invariant mass peaks? Any references will help too.

Answer

Especially in hadron colliders, often-times you have a lot going on in a single event. There might be multiple jets (which originate from quarks or gluons, which cannot survive alone, cf. confinement etc.).

Many of the particles one would be looking for in such events have a very short life-time, such as a top quark, a W or Z-boson. Thus, as opposed to more obvious things in an event, one has to reconstruct the existence of those particles by correctly identifying their decay product.

For example, a W-boson can decay into two quarks, which will form jets. Now, if you have many jets in an event, which of those two will you pick? Probably the ones whose combined invariant mass is near the W-mass. But still, since unfortunately the objects in the event don't come with labels, we cannot be sure these jets are the correct combination.

Another example would be a top-antitop-event. The top decays, almost to a 100%, into a W boson and a b quark (which will become a jet). Each W can either decay leptonically or hadronically. So in a all-hadronic top-antitop event you have 6(!) jets you have to combine correctly to reconstruct the event. Even if you're looking at a pure sample of top-antitop events (like in a Monte-Carlo simulation), you are bound to get those combinations wrong.

These things are what is called combinatorial background.

homework and exercises - Inclined plane question

An object, mass $m$ is placed on an incline, angle $\theta$. System is at equilibrium. coefficients of static and kinetic frictions are $\mu_s$ and $\mu_k$ respectively. Then:

1) What is the Total contact force on the block by the surface??

2) Which are necessarily true?

a) $tan \theta$ is less than or equal to $\mu_s$

b) $f = \mu_s N$

c)$f= mgsin\theta$

My approach to 1)

total contact force will be $mgcos \theta$ (exerted by the incline as a normal reaction) + $\mu_s mgcos \theta$ (exerted by the incline in the form of friction) which are at right angles, result will be the square root of their squares, giving $mg cos \theta \sqrt{1+ \mu_s^2}$.

But the answer given is just mg!!! And the explanation is that total contact force = $mg cos \theta (= N) + mg sin \theta (= f)$ $\implies$ total force $= \sqrt{N^2+ f^2} = mg \sqrt{cos^2 \theta + sin^2 \theta }$

For 2)

I think all are right, except, $tan \theta = \mu_s$

but answer says b) is false!

Please help me!

Answer

Here we need to see that $$f\le N\mu_s= mg \cos\theta\ \mu_s$$

$N\mu_s$is the maximum friction that may be present in the surfaces,whereas here we need equilibrium.

The two forces along the incline must balance each other. So, $$f=ms\sin\theta \le \mu_s mg\cos\theta$$ If this inequality does not hold , ie. $$\tan\theta\le\mu_s$$

then the body will never remain in eqiulibrium. Now you can proceed !

experimental physics - Recommendations for a book that compiles how the theory & mathematical equipment of quantum mechanics was developed from various experiments?

Is there a book or other resource that compiles how the theory & mathematical equipment of quantum mechanics was developed from various experiments? Something that shows clearly the relationships between real experimental data and the mathematical models of QM?

For instance, a chapter on how the Stern-Gerlach experiment is designed/works in high detail, and exactly how the data is measured and represented by theory.

special relativity - How can a photon have no mass and still travel at the speed of light?

I've read a number of the helpful Q&As on photons that mention the mass/mass-less issue. Do I understand correctly that the idea of mass-less (a rest mass of 0) may be just a convention to make the equations work?

From a layperson's view, it's difficult to understand how a particle of light (photon) can be mass-less. A physical object (everyday world-large or quantum-small) must have a mass. Yet, if my understanding is correct, the mass of a moving object/particle increases in proportion to its speed/velocity...so that at the speed of light, its mass would be infinite. A photon travels at the speed of light, but it obviously doesn't have infinite mass, right? Can someone formulate a practical explanation that can be understood by middle-school to high school kids? Much thanks for the help.

---

Wow--your answers to my original Q below clear up much of my confusion. I now have the daunting task of going over these nuggets and working up an equation-less (hopefully) explanation of the mass-less photon for non-physicist types.

Yes, from a layperson's view, it does seem remarkable that an existing piece of matter-- which has to be made of physical substance--could have zero mass at rest (though a photon is never at rest). It would be almost understandable if a piece of matter made of nothing had zero mass, but that seems to be an oxymoron, and "nothing" would equate to nonexistent, right?

In case you might find it interesting: I'm working on a writing project that posits we inhabit a universe that consists of matter (physical stuff) only, and that the NON-physical (aka supernatural) does not (and cannot) exist. For instance, if a purported supernatural phenomenon is found to actually exist, then by definition, its existence is proof that it is mundane/natural. All it would take to disprove this premise is reliable proof that ONE supernatural event has occurred. Despite thousands of such claims, that's never yet happened.

Who else better than physicists to confirm my premise? However, I do wish the TV physicists would explain the terms they throw about, some of which mislead/confuse their lay viewers. Case in point: "The universe is made up of matter and energy" (without properly defining the term "energy" as a property of matter).

The result is that laypersons are left with the impression that energy must therefore be something apart from or independent of matter (ie, nonphysical). Their use of the term "pure energy" without specifying exactly what that means adds to the confusion. (Thanks to your replies on this forum, I now understand that "pure energy" refers to photon particles.) However, "psychics" and other charlatans take advantage of such confusion by hijacking terms like energy (as in "psychic energy"), frequencies, vibrations, etc to give perceived scientific legitimacy to their claims that a supernatural spirit world, etc., exists. As you may realize, the majority of people in the US (per 2009 Harris Poll) and around the world believe in the existence of nonphysical/supernatural stuff such as ghosts and spirits.

My purpose is to give laypersons the information they need to distinguish what's real from what's not.

Thanks so much for help...And, PLEASE, add any further comments you think might be helpful/insightful to better inform laypersons.

optics - Changing transmittivity of polarizing filters for different wavelengths

I suppose, that a polarizing filter works as described here:

The polarizing filter absorbs or reflects some of the photons in the light beam, which is why the energy/power measured after the polarizer is reduced.

Furthermore any filter works only for some range of the EM spectrum, outside this range the filter is not transparent, e.g. no light is going through.

In my understanding between these to states has to be a range of wavelengths, where the transmission for unpolarized light has to be between zero and less than 50%. Rotating such a filter, for a polarized light beam the intensity behind the filter has to be not a sine function. For example, the absolute impermeability of the filter will not only be at 90° and 270°, but from 70° to 110° and from 250° to 290°. Has this situation been observed?

classical mechanics - How does the speed of an incoming pitch affect the speed of a baseball after it's hit?

Which will go further if a batter manages to hit it with a baseball bat: a baseball thrown to the batter at 90 miles per hour or one thrown at 60 miles per hour?

electromagnetism - Is a Perfect/Lossless Mirror possible?

In traditional mirrors, some of the input light is absorbed by atoms in the mirrors surface and are 'lost' as heat, degrading the quality of the reflected image.

Could this loss be compensated by an array of "powered elements" arranged to reflect nearby photons repelling them with some sort of electo-magnetic force, to completely prevent loss?

This would realize the perfect mirror, reflecting all light back.

Answer

There aren't mechanical oscillators that have the same frequencies as visible light. The most common oscillators with the same frequencies as visible light are the electrons orbiting atoms, which is why atomic transitions emit and absorb visible light.

If you're willing to accept a powered system, you could absorb all the light into an image-processing system and then project that image in another direction, possibly with much more light intensity than the original image had. For instance, folks with telescopes sometimes use video cameras to observe objects too faint to see naked-eye in real time.

statistical mechanics - number of states in microcanonical ensemble

I have a problem with the definition of $\Omega(E,V,N)$ — the number of microstates with $V$, $N$ and energy $E$. It starts with the definition of the PDF. If one defines the PDF as follows:

$P(\{q_i,p_i\})=\dfrac{1}{\Omega(E,V,N)}$ if $H(\{q_i,p_i\})=E$ and $P(\{q_i,p_i\})=0$ otherwise.

In this case it follows that in order for $P$ to be normalized we should demand that: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H=E} d\Gamma$$ but this integral is zero because the domain of integration is a set of measure zero. So as I saw, there are two options. the first is to redefine the PDF as: $$P(\{q_i,p_i\})=\frac{1}{\Omega(E,V,N)}\cdot \delta(H(\{q_i,p_i\})-E)$$ and then we get: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int \delta(H(\{q_i,p_i\})-E) d\Gamma$$ which solves the problem of zero integral but now it has the dimension of $\dfrac{1}{[E]}$ which is problematic if I want to consider $\log \Omega(E,V,N)$.

The second option is to redefine the PDF as:

$P(\{q_i,p_i\})=\dfrac{1}{\Omega(E,V,N)}$ if $H(\{q_i,p_i\})\in[E,E+dE]$ and $P(\{q_i,p_i\})=0$ otherwise

and then we get: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{E

can someone please shed some light on which of these 2 definitions are "better"?

Answer

The dimension issue is solved easily by defining the probability density function(PDF) as

$$P(\{q,p\})=\frac{E_0}{h^{3N}} \ \delta (H(\{q,p\})-E)$$

where $E_0$ is an arbitrary constant which will not affect any thermodynamic quantity or equilibrium property.

Actually, this definition is incomplete. We have to take into account the indistinguishability of particles, or we will encounter Gibbs' paradox. The correct PDF is therefore

$$P(\{q,p\})=\frac{E_0}{N! h^{3N}} \ \delta (H(\{q,p\})-E)$$

Also, it can be shown that the two definition of the PDF that you mention are almost equivalent:

Imagine to divide the phase space in hypercubes of volume $\delta x^{3N} \delta p^{3N} = h^{3N}$, and let's say that you can only determine the energy if each microstate to be within $E$ and $E+E_0$. We can then define a coarse-grained partition funcion

$$\tilde \Omega(N,V,E) = \sum_{\text{hypercubes}: E

Now, since we are integrating over a shell, we can approximate this integral as the hypersurface $H=E$ times the thickness $E_0$, obtaining

$$\tilde \Omega(N,V,E) \simeq \frac{E_0}{h^{3N}} \int d^{3N} p \ d^{3N} q \ \delta (H(\{q,p\})-E) = \Omega(N,V,E)$$

(The indistinguishability must be taken into account separately based on combinatorial arguments and is not included in the previous discussion).

The approximation

$$\int_{E

is good when $E_0$ is small. To see that this is plausible, consider the particular case of the integral over an hyperspherical shell:

$$V_{shell}=\int_{R<\| \vec x \|

where $V_D (R)$ is the volume of the hypersphere

$$V_D(R) = \frac{R^D \pi^{D/2}} {\Gamma(D/2+1)}$$

We then have

$$V_{shell} =\int_{R<\| \vec x \|

we do the following approximation for $\epsilon \ll R$

$$\left(1+\frac \epsilon R \right)^D \simeq 1 + D \frac{\epsilon}{R}$$

from which

$$V_{shell} \simeq \frac{ \pi^{D/2} R^{D-1}} {\Gamma(D/2+1)} \ \epsilon D$$

Now, since

$$\Gamma(D/2+1) = \frac D 2 \Gamma\left(\frac D 2 \right)$$

we have

$$V_{shell} \simeq \epsilon \frac{2 \pi^{D/2} R^{D-1}}{\Gamma(D/2)} = \epsilon S_D(R) $$

where $S_D(R)$ is the surface area of the hypersphere.

Fun fact: if we are dealing with non-interacting particles, i.e.

$$H=\sum_i^N \frac{p_i^2}{2m}$$

we have

$$\int_{E<\sum p_i^2/2m

and this integral is exactly an hyperspherical shell.

References: M. E. Tuckerman, Statistical Mechanics: Theory and Molecular Simulation