general relativity - At black-hole horizon time stops...for whom? The traveller itself or for an observer from the Earth?

1) Time slows-down in strong gravity field. But I cannot understand for whom (in which reference frame)

What about the paradox that time stops for an observer at earth who thus would never see the traveller disappear behind blackhole horizon. But I think that strong gravity happens around the traveller, so it is the traveller whose time shall stop! And for earth observer time is never changed b/x gravity around the observer never changes.

2) And is there time inside black-hole if time stops at the horizon - also in which reference frame we have this problem? If time stops at horizon in one reference frame than we have this problem only in this reference frame.

Answer

This is called gravitational time dilation, and is not caused by mass (contrary to popular belief) but by stress-energy. Even massless photons have gravitational effect, because they have stress-energy. The black hole has a strong gravitational field (stress-energy), but only compared to the stress-energy of Earth. The observer is on Earth, in a much weaker gravitational field (less stress0energy).

Now it is the difference between the stress-energy of the black hole and Earth that causes this time dilation.

Now the traveler sees his own clock as ticking normally. The observer on Earth sees his own clock tick normally too. It is when they try to compare their clocks that they see that the traveler's clock is ticking slower when it closes up to the black hole.

Now the traveler will not see anything unusual happen. He will cross the event horizon, just like normally, he won't even notice the event horizon (other then being spaghettified).

Now the observer will see the traveler's clock tick slower (relative to his own clock), and at the point where the traveler reaches the horizon, the observer will see the traveler's clock stopped, meaning, that the tick on the traveler's clock would be equal to infinite time on the observer's clock.

There, the traveler seems to have frozen on the event horizon (from the observer's view).

quantum electrodynamics - How to get Hamiltonian of QED from lagrangian?

I have the QED lagrangian: $$ L = \bar {\Psi}(i \gamma^{\mu }\partial_{\mu} + q\gamma^{\mu}A_{\mu} - m)\Psi + \frac{1}{16 \pi}F_{\alpha \beta}F^{\alpha \beta} . $$ I tried to get hamiltonian by getting zero component of energy-momentum tensor: $$ T^{\mu}_{\quad \nu} = i\bar {\Psi}\gamma^{\mu}\partial_{\nu}\Psi + \frac{1}{4 \pi}F^{\mu \gamma}\partial_{\nu}A_{\gamma} - \frac{1}{4 \pi}J^{\mu}A_{\nu}\Rightarrow $$ $$ T^{0}_{\quad 0} = i\Psi^{\dagger}\partial_{0}\Psi + \frac{1}{4 \pi}F^{0\gamma}\partial_{0}A_{\gamma} - \frac{1}{4 \pi}J^{0}A_{0} = i\Psi^{\dagger}\partial_{0}\Psi + \frac{1}{4 \pi}F^{0\gamma}\partial_{0}A_{\gamma} - \frac{1}{4 \pi}\Psi^{\dagger}A_{0}\Psi = H_{density}. $$ But it seems that it's incorrect, because I never get by this expression term $\bar {\Psi} \gamma^{\mu}\Psi A_{\mu}$, which refer to interaction part.

So how to find the true hamiltonian?

Thank you.

Added. Hmm, I find the mistake in expression of energy-momentum tensor. Fixed.

rotational dynamics - For a solid sphere rolling (pure roll) up a slope (with friction) does friction play a role in slowing it down?

A solid sphere rolling up a slope is slowed down - is this only due to gravity, or is it also because of friction? I need to know this, to calculate the final translational and angular velocity of a solid sphere, at the top of the slope, initially rolling (pure roll) with velocity v on a plane surface and then comes onto this slope.. I'll do this using work - energy theorem, but I need to know the total work done on the sphere first.

I think that friction is the only force that can affect the rotational motion of the sphere, as only it exerts a torque on the sphere..

Answer

In the case of pure rolling (no slipping or sliding) enough friction is provided so that at all times:

$$v=\omega R$$

where $R$ is the radius of the sphere.

Does friction play a part in the energy balance? Yes.

Let $T$ be the total energy of the sphere, $U$ be the potential energy of the sphere and $K$ its kinetic energy, then because the friction force does no work and we assume no other external forces:

$$\Delta T=\Delta U+\Delta K=0$$

when rolling up the incline.

Of course $\Delta U=mg\Delta h$ (for smallish $\Delta h$).

But for $\Delta K$ the situation is slightly more complicated because the sphere is translating and rotating at once, so that:

$$K=K_{trans}+K_{rot}$$

You'll need to find the expression for $K_{rot}$ and use $v=\omega R$ to make everything about $v$.

Without friction, $\omega$ would remain constant and thus also $K_{rot}$.

electromagnetism - Relation between magnetic moment and angular momentum -- classic theory

How do I prove the relation between the vectors of magnetic moment $\vec\mu$ and angular momentum $\vec L$, $$\vec\mu=\gamma\vec L$$ ?

Many text books and lecture notes about the principles of magnetism show the relation of $\mu$ and $L$ as scalars only and then just state that the relation holds also for the vectors. An example: http://folk.ntnu.no/ioverbo/TFY4250/til12eng.pdf

$$\mu=I\cdot A = \frac{q}{t}\pi r^2 = \frac{qv}{2\pi rm}m\pi r^2$$ ($I=q/t$: current, $A$: area of a loop, $q$ charge, $t=2\pi r/v$: time of 1 rotation, $v$: velocity of particle, $m$: mass )

The angular momentum is $\vec L = \vec r\times\vec p$ or $L=mrv$ and therefore $$\Rightarrow \mu = {\frac{q}{2m}} L = \gamma L.$$

Why is this also true for the vectors? Is there a general explication by classical physics without the need of quantum theory?

Answer

The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop. However, the more general case is that of a a volume current $\vec{J}$ in some finite region of space. In this case, the the general formula for the magnetic dipole moment of the configuration is $$ \vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \,d^3r. $$ (Showing that this reduces to the above formula for a flat planar loop is left as an exercise to the reader.) If we further assume that the current density is due to a number of particles with number density $n$, charge $q$, velocity $\vec{v}$, and mass $m$, then we have current density $\vec{J} = nq\vec{v}$; thus, $$ \vec{\mu} = \frac{1}{2} \int \vec{r} \times (n q \vec{v}) \, d^3 r = \frac{q}{2 m} \int \vec{r} \times (n m \vec{v}) \, d^3 r. $$ But $n m \vec{v} = \rho \vec{v}$, where $\rho$ is the mass density of the cloud; thus, the above integral can be rewritten as $$ \vec{\mu} = \frac{q}{2 m} \int \rho \vec{r} \times \vec{v} \, d^3 r = \frac{q}{2m} \vec{L}. $$ QED.

The above relationship will hold so long as we can model the object as made out of particles with a definite charge-to-mass ratio, or (which is equivalent) so long as the ratio of charge density to mass density is constant throughout the body. The easiest way for this to happen is, of course, for both densities to be constant.

experimental physics - Cause for spikes in Trinity nuclear bomb test

In Richard Rhodes' book, The Making of the Atomic Bomb, I was reading about the Trinity nuclear test. High speed photos were taken and this one is from <1ms after the detonation. The book mentions the irregular spikes at the bottom of the image, but does not explain them. Is there a specific reason or explanation for these odd spikes in the relatively spherical explosion?

Nuclear explosion photographed less than one millisecond after detonation. From the Tumbler-Snapper test series in Nevada, 1952, showing fireball and "rope trick" effects. The fireball is about 20 meters in diameter in this shot

Answer

The answer is in wikipedia

The photograph on the right shows two unusual phenomena: bright spikes projecting from the bottom of the fireball, and the peculiar mottling of the expanding fireball surface.

The surface of the fireball, with a temperature over 20,000 kelvin, emits huge amounts of visible light radiation (more than 100 times the intensity at the sun's surface). Anything solid in the area absorbs the light and rapidly heats. The "rope tricks" which protrude from the bottom of the fireball are caused by the heating, rapid vaporization and then expansion of mooring cables (or specialized rope trick test cables) which extend from the shot cab (the housing at the top of the tower that contains the explosive device) to the ground. Malik observed that when the rope was painted black, spike formation was enhanced, and if it were painted with reflective paint or wrapped in aluminium foil, no spikes were observed – thus confirming the hypothesis that it is heating and vaporization of the rope, induced by exposure to high-intensity visible light radiation, which causes the effect. Because of the lack of mooring ropes, no "rope trick" effects were observed in surface-detonation tests, free-flying weapons tests, or underground tests.

mathematical physics - Does Clifford algebra depend on the topology of manifold?

We know the greatest feature of Clifford algebra is coordinate-free. One can do vector operations without knowing the representation of vectors. And due to its very characteristc, Clifford or geometric algebra is believed to a reinterpretation of differential geometry suggested mainly by Hestenes and Doran. But as far as I know, many manifold-related theorem depends on the topology of the manifold such as connectedness, compactness, boundaryless or not. I want to know how Clifford algebra behave in different topologies?

thermodynamics - Balloon of ideal gas pops in vacuo - K.E. per atom unchanged, velocities become locally correlated?

My question is: is the following thinking pretty much correct?

I'm interested hearing about anything here that is fundamentally wrong. Of course there is a bit of hand waving and approximation.

If a balloon of ideal gas in equilibrium at a finite temperature T suddenly bursts while in space, the gas cloud will immediately begin to expand.

If the initial average kinetic energy per particle was $\frac{3}{2}{k_b}T$, the initial total kinetic energy would be just $\frac{3}{2}N{k_b}T$. There shouldn't be much loss of energy anywhere, so I'm assuming the final total kinetic energy would be about the same.

The much-larger gas cloud is now cold "due to expansion", but what's really happened is that the current position of each atom is determined by it's velocity at the last scatter event, when the mean free path became larger than the diameter of the cloud. Now, if a group of atoms are still near each other, it's because their velocities have been similar for a while.

That means the relative velocity of a group of nearby atoms with respect to each other is much lower than the average velocity they are moving, and therefore the local temperature is low.

my question: The average kinetic energy per atom hasn't changed. However their velocities are now highly correlated locally, and so we say the temperature has decreased - it is "cold".

Answer

After writing this answer I recalled asking this question two years ago. I'll answer my own question just to wrap this up.

My question is: is the following thinking pretty much correct?

Yes. For a given small region of the expanding gas, the temperature would be defined by the distribution of velocities around the collective center of mass velocity of that region.