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While I was reading Ta-Pei Cheng's book on relativity, I was unable to derive the correct relationship between coordinate time $dt$ (the book defined it as the time measured by a clock located at $r=\infty$ from the source of gravity) and proper time $d\tau$ from the definition of metric.

The book states that for a weak and static gravitational field, $g_{00}(r)=-\left(1+\frac{2\Phi(r)}{c^2}\right)$ (with the metric signature $(-1,1,1,1)$ and $\Phi(r)$ is the gravitational potential) and the proper time $d\tau=\sqrt{-g_{00}}\,dt$.

From the gravitational redshift result I know that the above result is correct (in a more unambiguous form $d\tau=\sqrt{-g_{00}(r_\tau)}\,dt$).

However, if I simply use the formula for spacetime interval $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ (assuming two clocks that measure proper time and coordinate time are at rest relative to each other), I have

$$ ds^2=g_{00}(r_\tau)c^2d\tau^2=g_{00}(r_t)c^2dt^2=-c^2dt^2\\ \implies \sqrt{-g_{00}(r_\tau)}\,d\tau=dt$$ This suggests that time flows faster with a lower gravitational potential which is incorrect.

I'm not sure why the above method lead to a wrong conclusion, did I misunderstood the the definition of proper time, coordinate time or spacetime interval?


Update:

  1. One mistake I've made is letting $ds^2=g_{00}(r_\tau)c^2d\tau^2$, which should be $ds^2=-c^2d\tau^2$ by definition. However, I'm confused about two definitions of $ds^2$ now. $ds^2=-c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$, this suggests that $g_{00}$ is always $-1$ for the frame that measures proper time, but in my problem $g_{00}$ is a function of $r$ which is only equals to $-1$ if $r=\infty$, how could two both be true at the same time?
  2. Assuming $ds^2=-c^2d\tau^2$ is true, as all the answers pointed out that $d\tau=\sqrt{-g_{00}}\,dt$. But by the definition of $g_{00}$ and $ds^2$ the $g_{00}$ used here must be $-(1+2\Phi(r_t)/c^2)=-1$, but I want $g_{00}$ here to be $-(1+2\Phi(r_\tau)/c^2)$ so that $$d\tau=\sqrt{-g_{00}}\,dt=\sqrt{1+2\Phi(r_\tau)/c^2}\,dt\approx (1+\Phi(r_\tau)/c^2)\,dt\\ \implies \frac{d\tau-dt}{dt}=\frac{\Phi(r_\tau)}{c^2}=\frac{\Phi(r_\tau)-\Phi(r_t)}{c^2}$$

Please correct me if I've made any mistakes!

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Everytime I try and think of time dilation, length contraction or any other strange phenomenon predicted by this strangely beautiful theory I get confused!! Luckily we have a metric to do all the thinking for us. In coordinates $x^\mu=(ct,x,y,z)$ with spacetime signature $(+1,-1,-1,-1)$ the metric is given by \begin{equation} c^2d\tau^2 = ds^2 = c^2dt^2 - d\vec{r}^2 \end{equation} where $d\vec{r}^2=dx^2+dy^2+dz^2$. If the coordinates are parametrised by $\tau$ so that $t=t(\tau), x=x(\tau), y=y(\tau)$ and $z=z(\tau)$ then we may write the above equations as \begin{equation} d\tau = \sqrt{dt^2-d\vec{r}^2} \end{equation} which is equivalent to \begin{equation} d\tau = dt\sqrt{1-v^2} \end{equation} where we adopt a timescale for which $c=1$ and $d\vec{r}/d\tau$ is equvialent to the velocity and hence the relation between coordinate time and proper time between two events at $t_1$ and $t_2$ is \begin{equation} \tau = \int_{t_1}^{t_2}\sqrt{1-v^2}dt \end{equation}

To answer your question, a spacetime interval $ds^2=d\tau^2=-g_{tt}dt^2$, can be expressed as \begin{equation} d\tau=\sqrt{-g_{tt}}dt, \end{equation} by definition. Your definition of the spacetime interval $ds^2$ is slightly off, it should read $ds^2=d\tau^2 = -g_{tt}dt^2+...$

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  • $\begingroup$ Hi there, perhaps I have not stated my question clear enough, sorry for that. I want to know why my derivation of the relationship between coordinate time and proper time from the definitions of the metric (not the Minkowski flat metric) and spacetime interval is wrong. $\endgroup$ – Taptic Jan 13 '17 at 3:18
  • $\begingroup$ Please see above, I've edited my answer. $\endgroup$ – Rumplestillskin Jan 13 '17 at 3:26
  • $\begingroup$ I'm assuming that in your answer $g_{tt}=g_{00}$. The problem is that given the definition of $g_{00}(r)$, $g_{00}(r_t)=-1$ which is not the desired coefficient. Maybe you could further expand what do you mean by $g_{tt}$ with its explicit expression. Also, I'm not sure what the ellipsis represent in the end, please give a further explanation, thank you! $\endgroup$ – Taptic Jan 13 '17 at 3:45
  • $\begingroup$ Yep, $g_{tt}=g_{00}$. I alternate between notation to keep myself sane :) Hmmmm I am confused myself now. Why is $g_{00}(r)=g_{00}(r_t)=-1$? In your question you state $ds^2 = g_{00} d\tau^2 = g_{00}dt^2$. This is incorrect. This statement is equating proper time and coordinate time. The line element should read $$ds^2 = d\tau^2 = g_{00}dt^2 - d\vec{r}^2$$ where $c=1$. Do you follow me? $\endgroup$ – Rumplestillskin Jan 13 '17 at 6:09
  • $\begingroup$ Yes, but $g_{00}$ is a function of the gravitational potential ($g_{00}(r)=-\left(1+\frac{2\Phi(r)}{c^2}\right)$) which depends on the radial position. So the two $g_{00}$ that used have different values (I indicated that through the use of $g_{00}(r_\tau)$ and $g_{00}(r_t)$ instead of just $g_{00}$). According to the book, the coordinate time $dt$ is measured at $r=\infty$, put that into the formula for $g_{00}$ we have $-1$.($g_{00}(r)\neq g_{00}(r_t)=-1$) Also, in this question the assumption that two clocks that are at rest relative to each other implies $d\mathbf{r}=0$. $\endgroup$ – Taptic Jan 13 '17 at 14:08
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Your mistake is where you just use the formula for the space time interval - I think you have just confused your $dt$'s and $d\tau$'s.

Set the metric signature to be $(-1,1,1,1,)$. Then for $dr=d\theta=d\phi = 0$ the spacetime interval can be written,

$$ ds^2 = -c^2 d\tau^2 = g_{00} c^2 dt^2$$

and so

$$ d\tau = \sqrt{-g_{00}} dt$$

which is the original result you got before.


Clarification as requested in comments:

Be clear that the metric signature $(-1,1,1,1)$ is not equivalent to the value of the metric components $g_{\mu \nu}$. Let me be explicit:

If we take then metric to have signature $(-1,1,1,1)$ then we can write the temporal components of the schwarzchild metric as:

$$ g_{00} = -\left( 1 - \frac{r_s}{r}\right)$$

where $$ r_s = \frac{2GM}{c^2}$$

Now, given that we can approximate GR by newtonian gravity in the weak field regime, we can say that the potential is,

$$ \Phi = - \frac{GM}{r}$$

which substitutes into $g_{00}$ to give,

$$ g_{00} = - \left( 1+ \frac{2 \Phi}{c^2}\right)$$

but this does not equal -1, which is merely the signature of the metric component. But by using $g_{00} = -(1+2\Phi/c^2)$, you have chosen to use the metric signature (-1,1,1,1).

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  • $\begingroup$ Could you give an explicit expression for the $g_{00}$ you used above? Is it $-\left(1+\frac{2\Phi(r_\tau)}{c^2}\right)$ or $-\left(1+\frac{2\Phi(r_t)}{c^2}\right)=-1$? The correct result should be the first one, however, I got the latter one from $ds^2=g_{\mu\nu}dx^\mu dx^{\nu}$. $\endgroup$ – Taptic Jan 13 '17 at 16:47
  • $\begingroup$ Ahhhhh i see where you are going wrong!! They metric tensor components are not equal to $(-1,1,1,1)$. That is their sign. The $g_{00}$ is equal to $1-2\Phi/c^2$ $\endgroup$ – Rumplestillskin Jan 13 '17 at 22:52
  • $\begingroup$ I have not assumed the metric tensor's components as $(-1,1,1,1)$, the only thing I've used is $g_{00}=-(1+2\Phi/c^2)$. If I use $g_{00}=1-2\Phi/c^2$ then $-g_{00}$ is always negative (since $\Phi\leq 0$), which is definitely incorrect. $\endgroup$ – Taptic Jan 13 '17 at 23:39
  • $\begingroup$ You are just mixing up your coordinate time and proper time that's all there is to it. $\endgroup$ – Rumplestillskin Jan 14 '17 at 4:05
  • $\begingroup$ Maybe I'm not understanding. I actually have a copy of the book what page and section are you referring to? $\endgroup$ – Rumplestillskin Jan 14 '17 at 7:46

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