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I am struggling to understand how to calculate gravitational time dilation using four-vectors, based on my understanding of them. Specifically, I would appreciate if someone could point out the misconception in the following steps, to answer the question:Alice is sitting on the Earth's surface at a radius $r$, and Bob is hovering in a hot air balloon at a radius $r+\Delta r$. If $\Delta T$ passes for Alice, how much time passes for Bob?

  1. Let Alice hold an atomic clock, with two consecutive radioactive decays being the events that she measures a time $\Delta T$ between.

    1. Then for Alice, the spacetime vector between the events is $(c\Delta T , 0,0,0)$ and for Bob $(c\Delta T_B ,0,0,0)$

    2. (I think this is the part where is goes wrong). Thinking in terms of the spacetime Manifold, Alice and Bob each go along their worldlines. Alice needs only to measure the time elapsed on her own clock, while for Bob the relationship between the time on his own clock and the time at which he thinks the events take place is not so simple.

    3. What I thought one could do in this case is use the invariant spacetime interval like so: Alice and Bob both agree on the invariant spacetime interval between the events. For Alice, this is $g_{\mu\nu} dx^\mu dx^\nu \approx (1-\frac{2GM}{rc^2})c^2\Delta T^2$ and since Bob is in a region where the spacetime metric is different, the spactime interval for him is $ds^2 \approx (1-\frac{2GM}{(r+\Delta r)c^2})c^2\Delta T_B^2$. The I would be inclined to equate these, but then I get $T_B \approx \sqrt{\frac{1-\frac{2GM}{rc^2}}{1-\frac{2GM}{(r+\Delta r)c^2}}}T$. Apparently, the answer is the reciprocal of this! (page 25)

I think it will help me to understand the problem mathemtically. What is the nature of the vector that Bob is measuring, and how he determines its length? I think it comes down to constructing a chart on the spacetime manifold, for which the metric is Minkowski along Bob's worldline. Now we have proper spacetime coordinates for both events, according to Bob. I suspect the 'equating the spactime intervals' that I tried to do above, is something akin to finding the transition function relating the Alice-frame coordinate chart and the Bob-frame coordinate chart.

Even if the above is correct, I am struggling to see to what the metric corresponds. The metric I am using above seems to be that for the frame of an observer at infinity...

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  • $\begingroup$ The way the question is phrased is based on a couple of misconceptions about general relativity. (1) We don't have spacetime displacement vectors in GR. (2) The metric I am using above seems to be that for the frame of an observer at infinity. GR doesn't have global frames of reference, and coordinate systems do not correspond to frames of reference. See physics.stackexchange.com/questions/458854/… $\endgroup$
    – user4552
    Oct 13, 2019 at 19:46
  • $\begingroup$ @BenCrowell Thank you four your reply. I am aware of the first point- which is why I was careful to speak of Bob's coordinates in the 'mathemtical' section! Though I agree the rest of what I wrote sounds like I think he is measuring the length of a vector... The second point is new, however. The link is helpful. $\endgroup$
    – Meep
    Oct 13, 2019 at 20:04

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I whould approach it like this: \begin{equation} ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu} = c^2 d\tau^2 \end{equation} \begin{equation} \Rightarrow d\tau = \sqrt{g_{00}} dt \end{equation} Where I used that the clocks are resting. \begin{equation} d\tau_B = \sqrt{\frac{g_{00,B}}{g_{00,A}}} d\tau_A \end{equation} You used the approximation $g_{00} = 1 + 2\frac{\Phi}{c^2}$ for small gravitational fields. And you obtain: \begin{equation} \Delta T_B = \sqrt{\frac{1 - \frac{2GM}{(r+\Delta r)c^2}}{1 - \frac{2GM}{rc^2}}} \Delta T_A \end{equation} Corresponding to your book. So I think you might have confused $dt$ and $d\tau$. $dt$ is the same in bouth Systems. It is the time passed at $r\rightarrow \infty$. While $d\tau$ is the time the clock is showing in the corresponding frame.

I hope I was able to help.

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  • $\begingroup$ Hi tomtom1-4, thank you for your reply! I see how the answer comes about when dt is taken to be the interval measured by an observer at infinitym however I am not quite sure why this is the case. Could you give a reference or more explanation as to why the quantities dt/dtau refer to the given observers? $\endgroup$
    – Meep
    Oct 13, 2019 at 20:28
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So $dt$ is the time at $r \rightarrow \infty$ because in this case $g_{\mu \nu} = \eta_{\mu \nu}$ hence $d\tau = dt$. $d \tau $ is the propper time in a given frame and is pretty much defined in the way of the first equation of my original answer. This should be known from special reletivity. From time dilation you get \begin{equation} d\tau = \sqrt{1 - \frac{v^2}{c^2}} dt = \sqrt{\eta_{\mu \nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}} \frac{dt}{c} = \sqrt{\eta_{\mu \nu} dx^{\mu} dx^{\nu}}/c \end{equation} $d\tau$ is always the time shown on the clock in a given frame. In general reletivity you have the same laws of special reletivity but only in the lokal frame of refrence. So $\eta_{\mu \nu} \rightarrow g_{\mu \nu}$.

I hope this has become a little clearer.

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