Recent content by Tomsk

Interesting things happen when you have a charged field that interacts with the EM field. I'll try to describe scalar electrodynamics, describing electrons requires spinors which is just extra complication at this stage. The main object of study is the action functional, which is the integral...

The kets |+\rangle and |d+\rangle refer to different systems, so the product |+\rangle |d+\rangle is a tensor product, it could be written |+\rangle \otimes |d+\rangle. Sometimes you will see |+d+\rangle meaning the same thing. The outer product of tensor products (|+\rangle \otimes...

I think your latter conclusion is right. If you have two entangled particles A and B, and you measure A, nothing actually travels from B to you. So there isn't really any sort of speed involved. You gain some information from A, which tells you about B, but only because you already knew A and B...

Sorry but I don't understand how you got that expression. I may have messed up, I don't know. I got the standard Maxwell stress energy tensor from the general equation for a Noether current. If you have some fields \phi_a (x) and a lagrangian \mathcal{L}(\phi_a (x),\partial_\mu \phi_a (x)), and...

OK that sounds right You'd sum over realizations for one volume. The probabilities are probabilities of finding a particular realization of the volume V, i.e. they all correspond to the same volume. A microstate might be a list of positions of all the particles: {x_1 ... x_N}, then the Gibbs...

I'm not sure that's right. The index i in Gibb's entropy formula indexes microstates equivalent to a macrostate, not macrostates themselves.

A neat way of visualizing spin is with the Bloch Sphere. In fact this let's you visualize any two state quantum system, such as the ground and excited state of an atom, photon polarization and so on. The state space can be visualized as a unit sphere in three dimensions (i.e. the surface of a...

Can anyone explain why the effect propagates at the speed of sound in the bar, and not the speed of light? The interactions between adjacent atoms is electromagnetic, so I would have thought it would be closer to c. You can imagine giving the bar a very strong tug so that the end near you is...

If you see your brother off in the distance 2ls away, and you want to meet up at a clock half way between you, you can get there in a time less than 2s (according to that clock) no problem, because it is only 1ls away. The only way to BOTH meet up there is to BOTH accelerate towards the clock...

They have to BOTH have been stationary in the ground frame in order to make that initial measurement. In order to reach 0 distance in 1.333 seconds, they both have to have traveled towards each other. This means they BOTH ACCELERATED. Acceleration is absolute in special relativity. The only way...

You have taken a distance measurement and time measurement (2ls and 1.333s) in one frame (the ground frame, which you're now trying to forget) and assuming those distances and times are the same in other frames (the traveler's frame and his brother's). This is not true. Having no ground...

I think it goes something like this. If \sigma_x , \sigma_y , \sigma_z are the Pauli matrices, then \sigma_0 , i\sigma_x , i\sigma_y , i\sigma_z act like the unit quaternions, where \sigma_0 is the 2x2 identity matrix and i=\sqrt{-1}. A vector v can be written v = v_x\sigma_x + v_y\sigma_y +...

My original dumb idea came about by thinking about whether integral operators had a nice geometric interpretation, but I just wonder if I've found it. I found the following definition of the exterior derivative d\alpha (v_1 ... v_k) = \lim_{h \rightarrow 0} \frac{1}{h^k} \int_{\partial P(hv_1...

You're right that there's a problem with the normal derivation from Noether's theorem, the tensor T_1{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4 is not gauge invariant. But this can be solved with a couple of tricks. One is to perform a gauge transformation when you vary A...

It can also be derived from Noether's theorem, it is the conserved current of translations x \rightarrow x+a.