0
$\begingroup$

In my GR lecture I was given the following equation for the spacetime interval (signature $(+,-,-,-)$):

$$ ds^2=(1+\frac{2\phi}{c^2})c^2 \, dt^2-(1-\frac{2\phi}{c^2})\delta_{ij}\, dx^{i}dx^{j} \tag{1}$$

I'm having trouble understanding what it means physically. I would suggest it means that the interval between two events in spacetime, measured by an observer using the time $t$ on a clock at some position $y^{i}$ (in his hands), and position coordinates $x^{i}$, can be calculated using the above formula by plugging in the gravitational potential at the position of the clock.

Is this correct? I think this interpretation leads to a contradiction:

We have two observers, one outside a gravitational potential (system S) and the other one inside (system S'). The observers are not moving relative to each other. The two events mentioned above are: the state of the clock (the time it shows) of observer S' at two different times $t'$.

(Observer S will measure this interval in his own coordinates. He should measure a different time between the events. The $ds$ should however be invariant. Since the observers are not moving relative to each other, in special relativity we would trivially have $dt = dt'$.)

Since $dx = 0 = dx'$, we have

$$ ds = (1+\frac{2\phi(S)}{c^2})c \, dt = (1+\frac{2\phi(S')}{c^2}) c \, dt' $$

Since the potential of observer S is zero:

$$ dt = (1+\frac{2\phi(S')}{c^2}) dt' $$

But this means the time elapsed for the observer outside the gravitational potential is smaller then for the one inside! (Because $\phi <0$)

Therefore, it is not in agreement with the fact that time moves slower for observers under the influence of a gravitational field.

Can you help me find the correct physical interpretation of formula $(1)$?

EDIT:

This question is not about proper time. I only want to know the connection between formula (1) and the time on clocks of observers. I don't care if some of those times are called proper time.

$\endgroup$
2
  • $\begingroup$ "But this means the time elapsed for the observer outside the gravitational potential is smaller then for the one inside !" - What's wrong with that? $\endgroup$ May 12 '20 at 20:57
  • $\begingroup$ It is not in agreement with experiment. It is known that time moves slower for observers under the influence of a gravitational field. $\endgroup$
    – curio
    May 12 '20 at 20:59
1
$\begingroup$

I would suggest it means that the intervall between two events in spacetime, measured by an observer using the time 𝑡 on a clock at some position 𝑦𝑖 (in his hands), and position coordinates 𝑥𝑖, can be calculated using the above formula by plugging in the gravitational potential at the position of the clock.

You appear to be over complicating the interpretation. $ds^2$ is the interval along a worldline with coordinates $x^i$. That interval can be measured by a clock which follows that same worldline and for a timelike worldline this interval is often called the proper time. This is different from the coordinate time $t$, which you seem to have confused with the time on a given clock.

Here you have two different coordinate systems with different potentials but the same spatial coordinates, each covering the same stationary clock's worldline. Since the proper time read by that clock is an invariant then $ds=\left( 1+\frac{2\phi}{c^2}\right)dt$ means in either coordinate system the clock's proper time $ds$ goes equal or slower than the coordinate time $dt$ since $1+\frac{2\phi}{c^2}\le 1$. This interpretation is in agreement with observation. Note that $\phi$ is evaluated at $x^i$ when calculating $ds^2$

$\endgroup$
13
  • $\begingroup$ Why does it matter that the clocks have different worldlines? Im arguing about the worldline of the clock of one of the observers. The interval along this worldline can be evaluated in different coordinate systems. They use different coordinates, but the intervals should come out equal. Furthermore, I don't know why I should care about the proper time as defined in GR. I care about the time on the clocks of the observers. $\endgroup$
    – curio
    May 12 '20 at 22:07
  • $\begingroup$ Im not equating two intervals of different worldlines but rather the same intervall in different coordinates. $\endgroup$
    – curio
    May 12 '20 at 22:13
  • 1
    $\begingroup$ @curio hmm, that is a lot of additional questions for a comment, many of which don’t make sense given the context of the question. You might be better served at a discussion forum than a Q&A forum. $\endgroup$
    – Dale
    May 13 '20 at 3:22
  • $\begingroup$ I disagree. Im pointing out that your answer misses the point, since you claim that I'm comparing two different intervals. Please substantiate that claim. Also, I didn't bring up proper time, you did. Please show why it is relevant. $\endgroup$
    – curio
    May 13 '20 at 8:42
  • 1
    $\begingroup$ @curio $dt$ is the coordinate time, it is not the proper time. $d\tau$ is the proper time and is not generally equal to the coordinate time. For a quick overview see: en.m.wikipedia.org/wiki/Proper_time “proper time along a timelike world line is defined as the time as measured by a clock following that line” and “proper time is usually represented by the Greek letter τ (tau) to distinguish it from coordinate time represented by t”. A physical interpretation in terms of times on clocks is in terms of proper time. I didn’t recognize that this was confusion so I will update the answer. $\endgroup$
    – Dale
    May 13 '20 at 18:55
0
$\begingroup$

My own take on this, after having put way too much time into it:

The interpretation given in the question is utterly wrong. $dt$ is not the time on the clock of some arbitrary observer, like it was in SRT. Rather it is the time that a clock at $r \rightarrow \infty$ shows. The potential has to be evaluated at the position of a clock, for which the intervall $ds$ is equal to the time difference on that clock (called proper time).

The mistake in the apparent contradiction is thus the idea that $dt$ is the time on the clocks of the observers, aswell as evaluating the potential at the positions of the clocks.

$\endgroup$

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.