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This is a question that is asked the world over and the answer seems obvious to me, yet everyone deems it to be from perception which for me seems to be the issue apply some 1st grade logic to the situation and the answers reveal themselves quite simply. For instance, travelling faster than light from a single point of perception this is absolutely possible ...


0

@benrg I got it. Follow your first half suggestion. Pick $s\in I^+(r)$, so $r\in I^-(s)$. $I^-(s)\cap I^+(p)\ne\emptyset$. Choose any point $q\in I^-(s)\cap I^+(p)$. There is a trip from $q$ to $s$, and at the same time, there is trip from $p$ to $q$, so glue the 2 trips to get a 3rd one from $p$ to $s$. Therefore, $s\in I^+(p)$. $s$ is an arbitrary point in ...


3

You should not imagine a virtual photon as an individual object wandering from one charged particle to another. This picture is simply inapplicable. Unfortunately, Feynman diagrams mislead people to imagine such things. Actually, Feynman diagrams are good for calculation and bad for imagination. Feynman diagrams have been introduced to help physicists to ...


1

All charged particles emit photons which are uncharged. They may, given the right boundary conditions. So how does the photon "know" that it's leaving one kind of charge and "lands" on another? What you are describing here is a "virtual photon", an interaction between two charged particles. There is the complicated way, i.e. mathematical ...


0

You can think of a photon as a quanta of energy. In that case, it can impart its kinetic energy to a charged particle, or, vice vers, a unit of energy can be released when a charged particles slows down and loses kinetic energy


1

If you construct your metric in such a way that $\phi$ is a killing vector generating an axisymmetry of the spacetime, and there is a value of the other three coordinates where $g_{\phi\phi} < 0$, then you have a timelike killing vector generating an axisymmetry. The curve traced out by this vector will then be a closed timelike curve by construction. ...



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