4
$\begingroup$

In SR consider two time like separated events - In some frame \begin{equation}ds^2= dt^2 - dx^2\end{equation}

In a frame where the events occur at the same place ( rest frame; $dx' =0$) then according to what I know proper time is the time elapsed in that frame i.e. $d\tau=dt'$.

Hence \begin{equation}ds'^2 = d\tau^2 =dt'^2\end{equation} ( since $d\tau =dt'$ in that frame)and since the interval is invariant; \begin{equation} d\tau^2=dt'^2= dt^2 - dx^2.\end{equation}

Consider the same event in GR, in some frame ( coordinate system) \begin{equation}ds^2= g_{00}dt^2 - g_{11}dx^2....\end{equation} In the frame ( coordinate system) where the events occur at the same place ( $dx'^2=0$) then according to what I know proper time is the time elapsed in that frame i.e. $d\tau=dt'$, and we should have, \begin{equation} ds'^2= g_{00}dt'^2 = g_{00}d\tau^2\end{equation} ( by the same analogy as in SR, $d\tau =dt'$) and since the interval is invariant, we should have, \begin{equation} g_{00}d\tau^2 = g_{00}dt^2 - g_{11}dx^2....\end{equation}.

But from time dilation formula in GR, I know this wrong.

Precisely, according to what I have unsertood, proper time is the time elapsed in the rest frame of the particle, just like in SR, so for the interval\begin{equation} ds^2= g_{00}dt^2 - g_{11}dx^2....\end{equation}, if two events happen at the same place \begin{equation}ds^2 = g_{00}dt'=g_{00}d\tau^2\end{equation} ( by definition just as in the SR case).

  1. Why is this wrong. Is my reasoning that proper time is the time measured in the rest frame wrong

                             Or 
    

2)is that the time coordinate in a general metric not represent time measured by any clock.

$\endgroup$
1
  • $\begingroup$ I have seen a few questions along the similar lines. But the answer wasn't clear to me and I couldn't understand. Hence I have asked this question to understand what is wrong with my reasoning. A detailed explanation pointing out what I am doing wrong and and what is the corresponding correct t thing would be very helpful. $\endgroup$
    – Shashaank
    Dec 24, 2020 at 21:06

2 Answers 2

3
$\begingroup$

This is not a contradiction, but simply a constraint on the allowable coordinate systems that would qualify as a particle’s rest frame. You have discovered that along the worldline of the particle the metric in the particle’s rest frame must have $g_{00}=1$. It is not mandatory to use such coordinates, but only such coordinates will be called the particle’s rest frame.

As an example, consider the standard Schwarzschild metric for a manifold with (-+++) signature and in units where $c=1$ $$g=\left( \begin{array}{cccc} -\left(1-\frac{R}{r}\right) & 0 & 0 & 0 \\ 0 & \left(1-\frac{R}{r}\right)^{-1} & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin (\theta ) \\ \end{array} \right)$$ which is the line element $$ds^2 = -d\tau^2 = -\left(1-\frac{R}{r}\right) dt^2 + \left(1-\frac{R}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 \sin (\theta ) d\phi^2$$

Now, for a particle at rest we have $dr=d\theta=d\phi=0$ so $$ds^2 = -d\tau^2 = -\left(1-\frac{R}{r}\right) dt^2 = g_{00} dt^2$$ Notice that for $r=\infty$ we have $g_{00}=-1$ so the proper time $d\tau$ is equal to the coordinate time $dt$. So these coordinates are a valid rest frame for an observer at rest at $r=\infty$. However, for $r=10R$ we have $ g_{00} = -0.9$ so $d\tau^2 = 0.9 dt^2$ thus the coordinate time is not equal to the proper time. These same coordinates are not a valid rest frame for an observer at rest at any finite $r$.

$\endgroup$
11
  • $\begingroup$ Umm, sorry, I am a bit too confused. I couldn't get why it's not a contradiction. Would be really great if you could please elaborate a bit with an example or so. What I am getting is like the time coordinate appearing in general metric g, need not be the time measured by any clock and so we can't just put $dt=d\tau$ when the events happen at the same place because they wasn't initially depicting time. Could you please explain through an example why this is not a contradiction but a constraint $\endgroup$
    – Shashaank
    Dec 25, 2020 at 6:18
  • $\begingroup$ To be exact why does in $ds^2 = g_{00}d\tau^2 =g_{00}dt^2$, the $g_{00}$ has to be equal to 1. And are the other components of the metric $g_{ij}$ identically 0 in the rest frame of the particle. Also will the above fact hold JUST LOCALLY or globally. $\endgroup$
    – Shashaank
    Dec 25, 2020 at 9:04
  • $\begingroup$ @Shashaank no problem. See the edited answer. $\endgroup$
    – Dale
    Dec 25, 2020 at 17:34
  • $\begingroup$ Thanks, it looks like it's making sense now. Just 2 small issues-1) In SR dt was always a physical time interval. It looks to me that dt in GR doesn't need to be always a physical time interval. Is that true isn't that strange that dt isn't physical time in GR 2) why g_00 need to be 1 for d(tau) to be equal to dt, that is why don't you write the last equation as - $ds^2=(-1-R/r)dt^2=-(1-R/r) d\tau^2$ by the SR reasoning that when the events happen at the same position then dt is the proper time itself. Why you have used the other reasoning that ds=d(tau). $\endgroup$
    – Shashaank
    Dec 25, 2020 at 18:15
  • $\begingroup$ Isn't d(tau) Always equal to dt for simultaneous events. What is the rigorous definition of proper time. Is it equal to ds or is it equal to dt (whether or not g_{00} = -1 or anything else) for simultaneous events. Kindly help mee in these 2 points as well. I think it will be more then. $\endgroup$
    – Shashaank
    Dec 25, 2020 at 18:19
1
$\begingroup$

Your second statement is correct. In GR, one can always change to a new set of coordinates $x'=x'(x)$, so it's clear that any particular choice of time coordinate doesn't have any absolute physical significance. What is invariant is the interval $s$. An observer moving along some timelike path will experience time elapsing according to the invariant interval along the path, and this is independent of the choice of coordinates. Now, if you are that observer, one thing you could do is watch your clock, and "draw a tick mark" on the spacetime once per second according to your clock. By doing this you would have constructed a time coordinate which has the simple property that $ds = dt$. But this relation only holds for that choice of time coordinate.

Edit to follow up on some comments: You are wondering whether/why in SR $g_{00}$ must be equal to $-1$. The answer is that it doesn't matter. Or more precisely, that's a meaningless statement. The reason is that time is a dimensionful quantity. If I measure time in seconds, and I have $g_{00}=-1$, then I could also keep the same time coordinate but set $g_{00}=-1/{3600}$ and then I would get the time measured in minutes. So it's really a moot point. The difference with GR is that in GR, the metric varies over spacetime. So the worldline of an observer can pass through points where the metric is different, and those relative changes will produce interesting effects. In GR, you would make some choice of units at one point on your worldline, and you could use that to set $g_{00}=-1$ at that point, but then as you continue on the worldline, you would experience different values of the metric, and thus the value of coordinate time and your accumulated invariant interval will start to differ.

$\endgroup$
7
  • $\begingroup$ In SR the dt was ALWAYS a time interval. So isn't it a bit awkward that the dt in GR isn't a physical time interval. More precisely do you mean that in the Schwarzschild Metric the dt doesn't represent a physical time interval... Also in your 2nd last line is then ds= dt = d(tau), i.e dt in your 2nd last line is the proper time measured by the observer. Also then is the following true-" In the rest frame of the observer, the interval is just ds=d(tau) and the Metric has just the 1st component which would be 1 (or c^2) and all other components of the Metric are 0"..Is this true and correct. $\endgroup$
    – Shashaank
    Dec 25, 2020 at 8:59
  • $\begingroup$ To be exact why does in $ds^2 = g_{00}d\tau^2 =g_{00}dt^2$, the $g_{00}$ has to be equal to 1. And are the other components of the metric $g_{ij}$ identically 0 in the rest frame of the particle. Also will the above fact hold JUST LOCALLY or globally. Could you please add a bit about these queries as well in the answer. $\endgroup$
    – Shashaank
    Dec 25, 2020 at 9:05
  • $\begingroup$ Let me know if in SR dt always represents a physical time and only in GR it doesn't necessarily $\endgroup$
    – Shashaank
    Dec 26, 2020 at 12:41
  • $\begingroup$ @Shashaank added something that I hope will be helpful. $\endgroup$
    – kaylimekay
    Dec 26, 2020 at 14:57
  • 1
    $\begingroup$ @Shashaank yes that sounds right. You always need the metric to define a physical distance. $\endgroup$
    – kaylimekay
    Dec 28, 2020 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.