Is the speed of light constant even for an object which is accelerating in free space?
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2$\begingroup$ Related: physics.stackexchange.com/a/232137/520 $\endgroup$– dmckee --- ex-moderator kittenCommented Dec 23, 2019 at 17:29
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2 Answers
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No, as the object would be in a non-inertial frame of reference.
answered Dec 23, 2019 at 17:10
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$\begingroup$ This is wrong, and in any case the question is a duplicate. $\endgroup$– user4552Commented Dec 23, 2019 at 23:05
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1$\begingroup$ I'm happy to delete it if it is wrong, but a couple of words saying why would be helpful. $\endgroup$ Commented Dec 24, 2019 at 10:06
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Yes, but only instantaneously. If at any instant the acceleration were to stop, the speed of light would be $c$ relative to the observer who was accelerating. Acceleration means to continually change inertial frames. In each of those frames the speed of light is $c$ relative to the frame. Mathematically this may be stated $\frac{dc}{d\tau}=0$.
answered Dec 23, 2019 at 17:18
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3$\begingroup$ This is not correct. Firstly because it doesn't make much sense to say "the speed is constant but only instantaneously" but second and more importantly, because it just isn't so. See, for instance, the Rindler coordinates for an observer with constant acceleration. Such an observer only observes the speed of light as being $c$ for light rays immediately adjacent to them. Light rays ahead of them go faster, and light rays behind go slower, and at some point the speed of light becomes zero and the Rindler Horizon is formed. $\endgroup$– dA-VeCommented Dec 23, 2019 at 17:29
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$\begingroup$ It really depends on what one considers the frame of reference of the accelerated observer. Or even what one considered an observer. It is a fact that $\frac{dc}{d\tau}=0$ for an accelerated observer. Arbitrarily extending an accelerated observer's frame of reference leads to degenerate coordinate system, and other problems such as reference points moving faster than light, in the case of a rotating system. See MTW Chapter 6 and Chapter 13 section 6. $\endgroup$ Commented Dec 23, 2019 at 18:29