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Feb 9, 2019 at 16:35 comment added Philip Wood This is SO useful. Thank you. Notation is a challenge, is it not? I like $\vec u$ for $\frac{d \vec r}{dt}$ and $\vec U$ for the 3-vector proper velocity $\frac {d\vec r}{d \tau} $ and $\mathbf{U}$ for the four vector $(\gamma c,\ \vec U)$ but the accelerations are a nightmare...
Jan 8, 2018 at 17:12 comment added inya @Pulsar is this proper acceleration, as you have defined it, also the acceleration a particle will 'feel' in it's own instantaneous rest frame?
Jun 17, 2013 at 10:50 comment added Pulsar @diffeomorphism I've written an explanation in this post: physics.stackexchange.com/a/68331/24142
Jun 15, 2013 at 8:53 comment added Calmarius So does this mean the acceleration felt by an accelerating observer is $du/dt$, and not $du/d\tau$? Isn't the local accelerometer measures $du/d\tau$?
Jun 3, 2013 at 0:05 comment added diffeomorphism shouldn't the module of the non-inertial forces in the accelerated frame be defined as a Lorentz-invariant scalar? if the frame feels one-G, that one-G should be agreed by all observers, instead you are saying that the proper acceleration is a derivative of coordinate time, not proper time, so a different coordinate system will measure a different G-force on that frame. How is that possible?
Jun 2, 2013 at 20:51 history edited Pulsar CC BY-SA 3.0
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Jun 2, 2013 at 20:24 history edited Pulsar CC BY-SA 3.0
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Jun 2, 2013 at 20:16 comment added Pulsar His mistake is that he confuses proper acceleration with four-vector acceleration. I explained that these are two different quantities.
Jun 2, 2013 at 20:13 comment added Larry Harson This doesn't answer the question. He wants to know where he's made a mistake in his working
Jun 2, 2013 at 20:11 history edited Pulsar CC BY-SA 3.0
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Jun 2, 2013 at 20:04 history answered Pulsar CC BY-SA 3.0