Relationship between Coordinate Time and Proper Time

Question

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?

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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!

Answer 1

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 $ds^2=d\tau^2 = -g_{tt}dt^2+...$

Answer 2

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.

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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).

Answer 3

The answer provided by Rumplestillskin is correct $$d\tau = \sqrt{1-\frac{r_g}{r}} dt$$ but it is only valid for the clock statically residing in the gravity at coordinate distance $r$ from the center. For moving clocks in the Schwarzschild coordinates, for example in radial free fall, the relation between the proper time and coordinate time is different $$d\tau=dt+\sqrt{\frac{r_g}{r}}\left(1-\frac{r_g}{r}\right)$$