Timeline for An electromagnetic twist on Ehrenfest's paradox
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2017 at 0:58 | comment | added | WillO | I'm glad this helped. I remember asking myself exactly the same question some years ago, and being very happy when I got it figured out. | |
| Feb 4, 2017 at 0:56 | comment | added | tparker | That last sentence cleared everything up for me! I was under the impression that the proper charge density (i.e. the local charge density as measured from a comoving frame) would stay the same during the acceleration process. But in fact, it decreases by a factor of $\gamma$, which is exactly canceled by the Lorentz contraction you get from boosting back into the lab frame. So the density observed in the lab frame is the same before and after the acceleration process. | |
| Feb 4, 2017 at 0:53 | vote | accept | tparker | ||
| Feb 3, 2017 at 22:37 | comment | added | WillO | Again, take the simplest case: At time $0$ (in the lab frame) all parts of the stick jump from velocity $0$ to velocity $v$. If the left and right ends of the stick are at points $0$ and $1$ to begin with, then they'll be at points $vt$ and $1+vt$ at time $t$ --- so the stick will still have length $1$ (again, all as measured in the lab frame). A rider on the stick will say the stick has stretched. | |
| Feb 3, 2017 at 22:34 | comment | added | WillO | As for what's special about a circular charge, the answer is nothing. If every point of a straight-line stick accelerates by the amount $a(t)$ at time $t$, then the length of the stick cannot change in the lab frame and therefore the charge density does not change. Of course the charge density can change if different parts of the stick follow different acceleration paths according to a lab-based observer. In general, you can't just imagine that the "stick starts moving" without being more precise about which parts start moving when and according to whom. | |
| Feb 3, 2017 at 22:28 | comment | added | WillO | (CONTINUED), who rides on the disk and is present at event $E$, will say that $F$ took place earlier than $E$. He will still say the analogous thing at every time in the future as long as the disk is moving. It's complicated to show this if you insist on looking at Bob as a non-inertial observer, but easy if you look at Bob, over any short time interval, as an effectively inertial observer. | |
| Feb 3, 2017 at 22:26 | comment | added | WillO | I mean this: At time $0$, in the lab frame, the entire disk starts rotating counterclockwise at speed $v$, and continues to rotate at this speed. Let $E$ be the event that a lab observer says is at time $0$ and location $\theta$ and let $F$ be the event that a lab observer says is at time $0$ and location $\theta+\epsilon$. Let $\epsilon$ be small enough, and look over a small enough time interval, that we can interpret an observer at $\theta$ as moving in a straight line, so we can just use special relativity. Then Bob, (CONTINUED) | |
| Feb 3, 2017 at 22:16 | comment | added | tparker | What do you mean by "the front of his stick will have started accelerating before the back of his stick started accelerating"? Do you mean while the ring is speeding up in the lab frame, or long afterward, when everyone agrees it's moving uniformly? Also, what's special about circular motion? For a straight line of charge, the density does go up by a factor of $gamma$ after it starts moving. | |
| Feb 3, 2017 at 21:38 | history | answered | WillO | CC BY-SA 3.0 |