Timeline for What's wrong with this application of Thomas Precession to circular motion velocity measurements?
Current License: CC BY-SA 3.0
12 events
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| May 29, 2013 at 20:14 | comment | added | Man | Whoever downvoted this answer, can you please comment? | |
| May 29, 2013 at 20:13 | history | edited | Man | CC BY-SA 3.0 |
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| May 28, 2013 at 16:35 | comment | added | Izzhov | Yeah, I've taken a seminar on special relativity, though we didn't get quite as advanced as learning Thomas Precession. I was trying to teach myself this more advanced material using Jackson's book when I came up with this problem. | |
| May 28, 2013 at 15:54 | comment | added | Man | Have you had a course on relativity? Because that's what is messing up things. | |
| May 28, 2013 at 15:53 | comment | added | Man | (1) Yes, so that's settled. (2) I agree, it still doesn't give the answer. I am trying. | |
| May 28, 2013 at 14:45 | comment | added | Izzhov | For (1), does that mean that, in fact, $\mathbf{a} \neq \frac{d\mathbf{v}}{dt}$, but rather $\mathbf{a} = \gamma\frac{d\mathbf{v}}{dt}$? If so, point taken. However, I believe I've already accounted for (2), as the left hand side is in the lab frame, meaning you don't need to refer to proper time, and the derivative on the right hand side is zero, so multiplying it by a constant won't change it. Furthermore, all this only adds factors of $\gamma$, but the solution is off by a factor of $1 - \gamma$. | |
| May 28, 2013 at 12:46 | history | edited | Man | CC BY-SA 3.0 |
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| May 28, 2013 at 12:13 | history | edited | Man | CC BY-SA 3.0 |
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| May 27, 2013 at 23:49 | comment | added | Man | This is still not giving the correct answer, something else is wrong too. | |
| May 27, 2013 at 23:47 | history | edited | Man | CC BY-SA 3.0 |
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| May 27, 2013 at 23:38 | history | edited | Man | CC BY-SA 3.0 |
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| May 27, 2013 at 23:27 | history | answered | Man | CC BY-SA 3.0 |