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I am currently taking a course in general relativity and one of the homework problems for the course asks to justify the fact that time dilates in a constant gravitational field using the equivalence principle. Although this might be fairly straightforward to many, this pushed me into questioning everything. My reasoning for this problem goes as follows: We know from the equivalence principle that a constant gravitational field is the same as being accelerated with a proper acceleration '$g$'. So I have tried to use the Rindler coordinate system to describe two observers $A$ and $B$ a certain distance apart in an inertial frame (with coordinates x,t). Although this may be going a little overboard, I just wanted to understand time dilation in its full glory. In the image below, I have shifted the Rindler coordinate system by a constant $a - r_0$ such that an observer with acceleration $1/r_0$ would be located at $a$ when $t=0$. I then derived the coordinates $(r,w)$ in terms of $(x,y)$ and derived the Rindler metric: $ds^2 = dr^2-r^2dw^2$. It is clear that an observer A (with constant acceleration $1/r_0$) will trace out a hyperbola and will experience proper time given by $-ds^2=r_0^2dw^2$. Now, it is possible to relate the proper time and $w$ as follows: $\tau _A/r_0 = w$. So following this, I placed an other observer B that is located a distance $\Delta a$ to the right of Observer A with the same acceleration $1/r_0$. I parameterised the trajectory of Observer B in the $(x,t)$ inertial frame and used the coordinate transformation equations to obtain the trajectory in terms of $(r,w)$ (rindler coordinates of Observer A). Using this, I (if this is correct) derived the relation between the proper times of Observers A and B.

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This seems to predict some sort of time dilation between Observers A and B (although it does not quite agree with the time dilation equations I have seen). However, just so I could understand how different coordinates can affect physics, I considered a different coordinate system defined by the following transformation: $(x,t) = (r_0 cosh(w)+r-r_0, r_0 sinh(w))$. This coordinate system corresponds to observers accelerating with the same acceleration $1/r_0$ starting at different points on the x-axis (they start at $x=r$ when $t=0$). Now, I used the same procedure as before to determine the metric and transformation functions. I considered Observer B to be located at a distance $\Delta a$ ahead of Observer A in the inertial frame just like before and tried to relate their proper times. However, this time, I just got $\tau _A = \tau _B $

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I think this is because of the definition of simultaneity in the second coordinate system - it coincides with the definition of simultaneity of the inertial frame. However, I don't see why that would necessarily be wrong. How can there be time dilation in one coordinate system but not the other? Isn't time dilation supposed to be coordinate independent?

In addition, my equation for the time dilation in normal Rindler coordinates does not match the common equation for time dilation, so I must be doing something wrong. There is either an elementary misunderstanding of coordinate systems and time dilation or I am completely ignorant of something. I am sorry for the sheets of paper but I am not proficient in Latex. I would be grateful to anyone who can help; I am completely lost.

Edit: I am trying to calculate the time dilation between 2 Observers A and B who are accelerating with the same acceleration, not the dilation between Observer A and a free falling observer.

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You have gone way overboard here. Once you have derived the Rindler metric you are nearly done.

From $ds^2= dr^2-r^2 dw^2$ we immediately obtain $d\tau^2=r^2 dw^2-dr^2$. Time dilation is given by $$\frac{1}{\gamma}=\frac{d\tau}{dw }= \sqrt{r^2-dr^2/dw^2}$$ and the gravitational time dilation is obtained for $dr/dw=0$ which gives $\gamma=1/r$.

That is it, all of that extra stuff is not needed.

Isn't time dilation supposed to be coordinate independent?

Definitely not. Time dilation is the ratio of the coordinate time to the proper time on some clock. As such it definitely depends on the coordinate system. This should be clear from the fact that in SR a clock is not time dilated in its rest coordinates, but it is time dilated in other coordinates.

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  • $\begingroup$ That looks way simpler, thanks! However, when Observer A is measuring the coordinates of Observer B in his Rindler coordinates, why would he measure a constant r? I would assume that since they are moving with the same acceleration, the shifted hyperbola of observer B will not fit on any of the constant r lines in Observer A's coordinates (since rindler coordinates have lower accelerations to the right). In addition, isn't the time coordinate w for observer A related to his proper time as follows: dtau = r_0*dw? implying that his time coordinate is not w but r_0*w? $\endgroup$
    – Chandrahas
    Mar 17, 2021 at 15:14
  • $\begingroup$ @Chandrahas in A’s Rindler frame B would not be at rest, so B would have a nonzero $dr/dw$. In other words they would experience a combination of gravitational and velocity time dilation. $d\tau$ is the proper time of any observer. $w$ is the coordinate time. Only an observer at $r=1$ has no gravitational time dilation. $\endgroup$
    – Dale
    Mar 17, 2021 at 18:52
  • $\begingroup$ Right, that makes sense. So when we apply this same method to the second coordinate system I considered, I believe we still see that there is no time dilation. Now, since you've already mentioned that time dilation is a coordinate dependent concept, it makes sense that there could be a coordinate system in which there is no dilation. However, what "physical" mechanism are the two observers using to synchronise their clocks in this case such that there is no time dilation (if my calculation is right and there is really no dilation)? $\endgroup$
    – Chandrahas
    Mar 18, 2021 at 3:17
  • $\begingroup$ Not all coordinate systems have a physical mechanism or interpretation. In the end coordinates are just a way of labeling events. It is really up to the inventor (you) to decide what is the meaning of their coordinates, if any. $\endgroup$
    – Dale
    Mar 18, 2021 at 3:24

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