Recent content by K^2

The best way to understand Stress Energy Tensor is to first understand current density. Say you have a bunch of particles with some number density ##\small \rho##. Now, let's say you go to a different coordinate system, one moving at some speed ##\small \vec{u}## with respect to the original...

Apparently, some theories predict that in addition to a massless photon, there should be a massive one as well. I really don't know any details on that, but I know there are some experiments over at JLab looking for a massive photon.

You can't connect a resistor so that it's in parallel with internal resistance of battery with respect to the voltage drop of the battery. So yeah, you get maximum current if you just short the battery out.

In general, we write expectation of operator A as ##\displaystyle \langle \psi |A|\psi \rangle = \int \psi^*(r') A \psi(r') dr'##. Substituting ##A = \delta(r-r')## gives you the correct expression for expectation of ##\rho(r)##. So what's the problem?

From perspective of modern theory, all forces arise due to local gauge symmetries. There are gauge fields corresponding to every gauge symmetry, and these fields can be quantized. Electric field is one of the gauge fields of the U(1)xSU(2) symmetry group. Second quantization of electric field...

It's worth noting that basic QM and QFT treat this a little different. In QM, photon is a wave, which has uncertainty in momentum and position. Momentum, of course, being directly linked to its frequency. Quantum Field Theory would say that no, a photon has an exact frequency and an exact...

Basically, because flow is inviscid, vorticity is zero within the flow. But you can still have circulation. Imagine that the cylinder rotates. In that case, the flow around it will have circulation. However, because vorticity is zero in the fluid, that circulation is exactly the same around any...

Keep in mind that Kepler's Second Law is just another way of saying that angular momentum is conserved. In other words, ##\omega r^2## is a constant. Keeping in mind that for a small section h = r, this is in agreement with what you wrote.

Of course. I specifically chose a function that violates B. If distance fulfills these criteria, you have a metric space. We are specifically talking about manifolds that are not metric spaces. Therefore, it does not have a distance that fulfills all of these axioms. On the other hand, a...

Having a distance that is one in the sense of a metric space is the definition of having a metric space. Not every distance is a metric. That's kind of what the whole discussion so far has been about. You asked for an example of a distance that's not a metric, and I gave you one. And now...

Sure. x,y in ℝ. d²(x,y) = Sin²((x-y)(x+y)). Works basically like a distance locally; let's you have an element ds² = 4x² dx². Fails some of the metric axioms, and is not invariant under coordinate transformation x' = x+c. Absolutely irrelevant to GR, because GR manifolds are always...

You can have distances defined, but still not have a metric. You have to have distances follow certain axioms, such as the triangle inequality, in order for it to be a metric space. So yes, you can have a manifold with ds² defined, but which is not a metric manifold, and where ds'² ≠ ds².

1) IR is a broad range. There are electron transitions that correspond to near-visible IR. As you get to thermal ranges IR, the reason you don't get electron transitions that correspond to that is because any electrons that could pick up that much energy are going to be stripped off due to...

Yes. There exists a plane containing both vectors, and in that plane, all the planar geometry you know applies. So θ and θ' are complimentary angles, and therefore, their sum is 180°.

It's the same thing as in the other thread. The voltage across the primary is going to be the sum of the induced voltage and the voltage needed to drive the given current through the wire. The induced voltage works exactly the same way in both coils, but in primary it's the back EMF, while in...