Tag Info

New answers tagged

1

The reason why the commutation relations between a field and its conjugate at equal times are of the form $$ \left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y}) $$ is only to mirror and copy the canonical hamiltonian commutation relations $[q_i,p_j]=i\hbar\,\delta_{ij}$. No causality is involved, rather it is somehow ...


3

One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if $x^0 = y^0$ the separation is spacelike. Alternatively you can work from the definition where two events are spacelike separated if (and ...


1

To reach the Lienard-Wiechert potentials or to prove Feynman's equation (exposed in his lectures without proof), it's necessary to begin with the so-called retarded potentials expressed here conveniently by the following. \begin{equation} \phi\left(\mathbf{r},t\right)=\dfrac{1}{4\pi\varepsilon_{o}}\iiint ...


13

To add to ACuriousMind's answer on the Liénard-Weichert potentials, you can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula from them for the radiation from a moving charge: $$\vec{E} = ...


14

The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.


3

The force is not propagated instantly. It takes time for the information to get from one point to another. You can treat that as an instant if you are working with small enough distances and velocities, but it's not. If you'll ever study field theory you'll meet retarded potentials that are just this: the field propagates at the speed of light and it's no ...


1

In the entangled system we do not have two separate particles. Instead we have a single wavefunction describing a single system. When you interact with the wavefunction you are not interacting with particle $A$ or with particle $B$, you are interacting with a single wavefunction and causing it to change as a result. So the statement measuring $A$ affects ...


0

Let's say you have previously agreed with your friend doing the other measurement that if the combination of spins is up, down you go to the cinema and if it is down, up you go eat a pizza. Now you do the measurement and he does, both just before going out of the lab. You now have excluded two possible future states, i.e., not meeting because one went to the ...


2

There exists no "spooky action at a distance" except in the mind of the bemused. Correlation is not causation, correlation is not causation, correlation is not causation .... I have read the article about entanglement and EPR paradox. The spin of two particles is measured when they are very far apart, and they always make opposite choices. It seems that ...


1

The MWI explains the EPR experiment without invoking any non-local influences. Each observer measures one particle. The measurement affects only the particle being measured and the measurement device. Each measurement device differentiates into two versions, one for each measurement outcome. The correlations are established only after the results are ...


2

Answer to your question The term "quantum non-locality" refers to the fact that quantum mechanics cannot be described by a local classical hidden variable model. This is the content of Bell's theorem. In particular, locality is not violated. The statement of the theorem is that if we assume (wrongly) that quantum mechanics is described by a classical ...



Top 50 recent answers are included