# Proper time of an accelerated frame in a external gravitational field

1. Feb 1, 2014

### npnacho

hi everyone. i'm having trouble understanding the concept of proper time in general relativity.

suppose we have some metric given by a fixed mass distribution, say schwarzschild or something (it's not important) and a test particle go over some path between two events A and B.

if we want to compute the proper time measured by this particle in its travel, we have to consider the metric due its acceleration besides the metric given by the external gravitational field, right? is this doable? how would do you calculate the resulting metric in a problem like this?

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i think i also have problems with the idea of coordinate time... i'm looking at a derivation of the gravitational redshift and it's something like this: suppose you have an uniform gravitational field in the z direction and two observers A and B with positions zA and zB respectively. if a electromagnetic wave travels from A to B, the periods of the wave measured by them are related by:

$\frac{T_{A}}{T_{B}]}=\frac{\Delta\tau_{A}}{\Delta\tau_{B}}=\frac{\sqrt{g_{00}(z_{A})}}{ \sqrt{g_{00}(z_{B})}}\cdot\frac{\Delta t_{A}}{\Delta t_{B}}=\frac{\sqrt{g_{00}(z_{A})}}{\sqrt{g_{00}(z_{B})}}$

(T goes for period, tau goes for proper time and g00 is the obvious component of the metric tensor)

i can't get the last step, they say that the periods in t (the coordinate time) are the same for both observers so $\frac{\Delta t_{A}}{\Delta t_{B}}=1$. i could understand it in the framework of special relativity where both observers are at rest with respect to each other and the coordinate time becomes the proper time for both (and, obviously, as the metric remains unchanged, you don't get any redshift), but here? what is the frame that measures that coordinate time and why is the same for both observers?

i'm sorry if my english sounds a little funny, i'm not used to write it.

thanks!

2. Feb 1, 2014

### Staff: Mentor

No; there is only one metric, the one "given by the external gravitational field". I don't understand what you mean by "the metric due to its acceleration", but whatever it is, you don't need it to compute proper time.

Let's take a simple example: an object "hovering" at a constant altitude in the Schwarzschild geometry. The only part of the metric that comes into play is this...

$$d\tau^2 = \left( 1 - \frac{2M}{r} \right) dt^2$$

...because the object's worldline only changes its $t$ coordinate; the other coordinates $r$, $\theta$, $\phi$ stay constant. So the object's proper time is just

$$\tau = \int_{A}^{B} \sqrt{1 - \frac{2M}{r}} dt = \sqrt{1 - \frac{2M}{r}} \left( t_B - t_A \right)$$

where $t_A$ and $t_B$ are the coordinate times in Schwarzschild coordinates of the endpoints A and B.

It looks like you are looking at an argument using flat spacetime, not curved spacetime; the redshift that appears here is not really "gravitational" because the spacetime is flat. However, the equivalence principle allows us to model the local effects of gravity using accelerated frames; see below.

The coordinates being used are Rindler coordinates:

http://en.wikipedia.org/wiki/Rindler_coordinates

It looks like the derivation you saw (do you have a link?) used $z$ instead of $X$, which is what the Wikipedia page uses for the spatial coordinate in the direction of the acceleration. However, the formula that you wrote works for Schwarzschild coordinates as well: you just have

$$\sqrt{g_{00}} = \sqrt{1 - \frac{2M}{r}}$$

so the redshift of light climbing from radius $r_1$ to radius $r_2 > r_1$ is just

$$\frac{\sqrt{g_{00} ( r_1 )}}{\sqrt{g_{00} ( r_2 )}} = \frac{\sqrt{1 - \frac{2M}{r_1}}}{\sqrt{1 - \frac{2M}{r_2}}}$$

3. Feb 1, 2014

### WannabeNewton

There's only one metric for a given space-time by definition. The proper time along a given world-line $\gamma$ is simply the arc-length of $\gamma$ as given by this metric.

If the space-time is asymptotically flat and stationary (the latter of which your formula relies on) then the coordinate time $t$ corresponds to the proper time read by the clock of a observer at spatial infinity at rest in the gravitational field. Otherwise it's just a global time coordinate, that's all.

4. Feb 1, 2014

### npnacho

what i mean is that, in virtue of the equivalence principle, an accelerated frame in vacuum must have a non flat metric, so i suppose that the metric "observed" by a particle at rest in front of the sun it's not the same that the metric observed by another particle that is accelerated with respect to the sun.

another formulation of my question could be: suppose we are in a free falling frame of reference, say a free falling elevator in the gravitational field given by the earth... the metric that we observe is obviously minkowski... my question is: if some external observer (that knows that we are in a free falling elevator towards the earth) wants to calculate our metric... what is the computation that he does to arrive to the minkowski metric? how is that the metric due to our own acceleration "cancels out" the metric given by the external field to give the metric of flat spacetime?

sorry, i don't have a link. the derivation is in my class notes...

like you said, the derivation uses the equivalence principle and compute the redshift in an uniformly accelerated frame... but the equations i've wrote are valid (i think) in the GR case... that's why appears the g00

so, another formulation for this question could be (i think)

when we say that some metric is given by

dS2 = g00(x,y,z,t) dt2 + another terms...

that "t" is the coordinate time in the frame that observes this metric, right? then, what is the frame that measures the coordinate time when i compare the proper time measured by two observers at rest in different places, like in this case? i don't know if i'm being clear... i hope so.

5. Feb 1, 2014

### Staff: Mentor

No; you're confusing the metric, i.e., the spacetime geometry, with the coordinate charts that can be used to describe it. If you are in flat spacetime and you accelerate, spacetime is still flat, and the expression for the metric in any coordinate chart you use will still show it to be flat. For example, Rindler coordinates are an alternate way of expressing the flat metric of Minkowski spacetime, in a coordinate chart in which an observer with a constant acceleration is at rest (which is, I assume, what you mean by "an accelerated frame").

Conversely, if you are in curved spacetime (because gravity is present, for example due to a large mass like the Earth--see below), the expression for the metric in any coordinate chart will show it to be curved. The metric will, however, be *locally* flat, in the sense that you can set up a local coordinate chart centered on a given event in which the metric will be the standard Minkowski metric, *within the confines of the chart*. The region of spacetime covered by the chart must be small enough that deviations from flatness are not measurable. See further comments below.

No; both observe the same metric, i.e., the same geometry of spacetime. They may describe it differently, but it's still the same geometry.

No, it isn't. It's *locally* flat, but that "locally" is a crucial qualifier (see above). Globally, spacetime is curved because of the presence of the Earth.

There is no "metric due to our own acceleration". The metric is a property of spacetime. If gravity is present (due to a large mass being present, like the Earth), spacetime is not flat; it is only locally flat (as above).

As long as you use the correct $g_{00}$, yes, the equation is valid for Schwarzschild spacetime; I wrote down what the formula looks like for that case in my prevous post.

A clarification here: if the concept of "gravitational redshift" is to be well-defined, $g_{00}$ can't be a function of $t$. Another way of saying this is that the spacetime must be "stationary".

Also, you should be using the term "coordinate chart" here instead of "metric". See below.

It's the coordinate time in the *coordinate chart* in which the expression for the metric is as above. Changing coordinate charts doesn't change the metric (the underlying geometry); it only changes how the metric is described in terms of the coordinates.

If the observers are both at rest, the coordinate chart must be one in which they are both at rest--i.e., both their worldlines must have constant values for the spatial coordinates. The coordinate time in that chart is what you use in the computation of proper time.

6. Feb 2, 2014

### A.T.

No, it's the other way around: The space-time in a gravitational field is locally flat, just like the space-time in vacuum. Locally means, small enough so that tidal effects are negligible, because that's what requires intrinsic curvature of space time.

The local coordinate acceleration of free fallers (observed in the hovering frame in a gravitational field and the accelerated frame in vacuum) is just a coordinate effect, that doesn't require intrinsic curvature of space time.

Have a look at this post with a nice diagram by DrGreg:

7. Feb 2, 2014

### WannabeNewton

Oh boy. You're falling into the unfortunate but common trap of confusing frames and coordinate systems. When it comes to SR and GR, a slew of archetypal misunderstandings and confusions can often be solved simply by realizing the difference between a frame and a coordinate system; even more can be solved by working past the sloppy language in standard GR texts when it comes to terms such as "locally inertial frames" and "local flatness". Actually based on your posts there are quite a few deeper misunderstandings at play as well. What textbook is your class using?

8. Feb 2, 2014

### npnacho

the class was a couple of years ago, now i'm starting to study for the final exam... but between the strange education system we have here in argentina and my little memory... it's like i wasn't studied GR in my entire life...

this couple of things were part of the first classes (the subject started with a little review of concepts of SR, and a couple examples of why SR isn't suitable to explain gravitational phenomena -like the redshift-)

but i've not started the important part yet (differential geometry, the class used the bernard shutz's book -geometrical methods of mathematical physics-), so i'm a little (very) slow with concepts like charts and coordinate systems in a rigorous way. so i'll start with this mathematical part to see if it starts to ring a bell to me.

but definitely your answers have made some light here, i'm starting to think i know where the problem is... i'm going to re-read carefully all this later and i'll post my doubts.

thank you all!

9. Feb 2, 2014

### npnacho

ah! and for the GR part i think i'm going to use "gravity" by james hartle... but another recommendations will be welcome.

10. Feb 2, 2014

### WannabeNewton

That's one of the best GR books in existence. Read chapter 3 in detail, it will clear up a lot of things for you.

11. Feb 7, 2014

### npnacho

well, i have been reading hartle's book and it's clarified many of my doubts... but i still got one:

suppose someone gave us some metric, a diagonal one to simplify, say:

$dS^2=g_{00}(cdx^0)^2-g_{11}(dx^1)^2-g_{22}(dx^2)^2-g_{33}(dx^3)^2$

where the components of the metric tensor, g, could depend on the coordinates.

i've learned in this thread and with the book that this isn't enough to know the spacetime geometry, you also have to say me what the coordinates ($x^0,x^1$,etc.) are.

now, suppose you give me that, i.e: you say me: i have selected this coordinate system, and the geometry of the whole universe descripted in that system is given by that $dS^2$ above.

we have something we call the coordinate time, in this case represented by $x^0$. my question is: this coordinate time is only a mathematical parameter or it has physical meaning?

what i guess is that if, for example, that metric is the way i see the spacetime geometry of the universe when i'm sitting in my bedroom, and i observe a fly going from the spatial point $(x^1_{A},x^2_{A},x^3_{A})$ to the spatial point $(x^1_{B},x^2_{B},x^3_{B})$, then $(x^0_{B}-x^0_{A})$ matches the duration of the trip measured by my own clock (at rest with me).

is this correct? in that case there is my problem understanding the meaning of the coordinate time in "singular metrics" like scharzschild, because in that one, for example, the spatial part of the metric looks like your coordinate system is at rest on the singularity, so i have no choice to think, in this kind of cases, that the coordinate time (the factor which multiplies the "timelike part" in the metric, i.e: the part with the opposite sign to the rest) is simply a parameter without a definite physical meaning... some of you said something like "it's the time measured by an observer at infinity", but i didn't catch that.

thanks!

Last edited: Feb 7, 2014
12. Feb 7, 2014

### WannabeNewton

In general it is only a mathematical parameter and has no physical meaning. However if the coordinate time is a time parameter adapted to a family of observers then we can give it physical meaning as the time function that defines the surfaces of simultaneity for the entire family of observers. As a trivial example go back to Schwarzschild space-time. We have the privileged family of observers who are at rest with respect to the asymptotic Lorentz frame (the distant stars) and these are just the observers who hover in place outside the self-gravitating source (static spherically symmetric star) using say rockets. For them the surfaces $t = \text{const.}$ are simultaneity surfaces as is very easy to see., so in this sense the coordinate time has physical meaning. This amounts to having the hovering observers synchronize all their clocks by readjusting individual clock rates in light of gravitational time-dilation.

On the other hand if we consider the coordinate time in the coordinates of a rotating disk there is no such physical meaning for $t$ for obvious reasons (obstruction of global synchronization).

13. Feb 7, 2014

### Staff: Mentor

Just a note, the quantities inside the parentheses should be coordinate differentials, i.e., $cdx^0$, $dx^1$, $dx^2$, $dx^3$. That's because, as you note, the components of $g$ can depend on the coordinates, so in order to evaluate the path length along a finite curve, you have to do an integral.

In general it's only a mathematical parameter. In certain spacetimes, you can choose coordinates such that coordinate time has a direct physical interpretation, but you can't always do that, and even if you can, those coordinates won't always be the best ones to use to solve problems.

If you are in a spacetime where you can choose coordinates that way, yes. But not all spacetimes will allow you to do this.

Also, in any curved spacetime, you have to be careful interpreting "duration" along curves that are spatially separated from you. See further comments below.

I'm not sure what you mean by this. If by "the singularity" you mean $r = 0$, that's not a spatial point in this spacetime. If you're outside the horizon of the black hole, you can view Schwarzschild coordinates as coordinates in which the hole is "at rest", but you have to be careful with that since it can lead to misunderstandings if you push it too far.

Locally, this is true; the coordinate time itself doesn't in general represent any local observable. (Note that the coordinate usually called $t$ is not even timelike at or inside the horizon, so even the indirect interpretation of it as "time measured by an observer at infinity"--see below--doesn't work at or inside the horizon.)

This interpretation can sometimes be helpful, but it has limitations. As I noted above, in curved spacetime you have to be careful interpreting "duration" along curves that are spatially separated from you.

For example, if you are an observer at infinity, and I am an observer "hovering" at some finite altitude not far above the hole's horizon, then we are at rest relative to each other and can exchange light signals whose round-trip travel time will be unchanging. A given round-trip light signal will take a certain amount of Schwarzschild coordinate time $\Delta t$ to travel back and forth. Since you are at infinity, the proper time you measure for that round-trip light signal will be the same as the coordinate time, i.e., $\Delta t$. But the proper time *I* measure for the same round-trip light signal will be smaller by a factor $\sqrt{1 - 2M / r}$, where $r$ is the finite radius at which I am hovering.

So if you try to say that $\Delta t$ is the "duration" of the round-trip light signal, without qualification, you and I are going to disagree; to me, the "duration" of the signal is the proper time I measure, not the coordinate time. And you can't directly measure "time" along my worldline; your clock only directly measures time along your worldline. So saying that $\Delta t$ is the "time measured by an observer at infinity" along *my* worldline, which is *not* at infinity, is not really correct; the proper time along my worldline is $\Delta t \sqrt{1 - 2M / r}$.

14. Feb 7, 2014

### npnacho

oh, thanks for the differentials, i just forgot. now i've corrected that.

anyway, i'm not sure if my question was totally clear... in the scharzschild case, the coordinate time matches the propper time measured by an observer at spatial infinity because $\sqrt{1 - 2M / r}$ → 1... but what i asked, in this case, becomes: suppose this observer at spatial infinity observes a ship going from some spatial point A to another spatial point B... how do we compute the elapsed time in his watch?

i mean, if you are in the vacuum case, and you are in an inertial frame, and you choice cartesian coordinates so g=diag(1,-1,-1,-1)... i.e: you are an inertial observer in SR... the time measured by you when some other guy goes in a trip from A to B is the variation of the coordinate t (the $x^0$ on your coordinate system) between the moment that you see the guy starting to move and the moment that you see the guy arrives to B... am i correct?

so i can't understand why this becomes false in the GR case

15. Feb 7, 2014

### WannabeNewton

You are confusing proper time with coordinate time. In an inertial frame corresponding to an inertial observer the only reason an expression like $\Delta t = t_2 - t_1$ even makes sense is because we've setup clocks at each point in space and synchronized their proper times with the proper time of the inertial observer's own clock. Therefore if event A in the vicinity of our observer has proper time $t_1$ according to the observer's clock, at which point the observer emits a light beam, and distant event B in the vicinity of the light beam's reception has proper time $t_2$ according to a clock there then the only reason we can say the observer's own clock reads proper time $t_2$ when the light beam reaches B is we've synchronized these two clocks (i.e. we've defined simultaneity of events). Obviously this fails in Schwarzschild space-time for if we place clocks at each point in space there is no way to synchronize their proper times due to gravitational time dilation as I've already explained above: gravitational time dilation makes clocks at different points in space tick at different rates of proper time.

A side note: when I say "space" I mean the $t = \text{const.}$ surfaces.

16. Feb 7, 2014

### Mentz114

The elapsed time along a worldline is given by ( i,j run over 1,2,3)

$\tau=\int_{t_1}^{t_2}\sqrt{g_{00}+g_{ij}\dot{x}^i\dot{x}^j}\ dt$

in GR and SR.

Last edited: Feb 7, 2014
17. Feb 7, 2014

### npnacho

i don't understand what coordinate time is... so i'm hardly confusing it with propper time

i'm still having the same question:

in the SR case i've mentioned above, the time elapsed isn't propper time, because we are talking about the trip of a spaceship which is not at rest in our coordinate system... but in that case the change on the coordinate $x^0$ is the time elapsed in my point of view, right? i'm not sure about this, it's a question too.

that is the propper time measured by the guy who drives the spaceship... but i'm asking which is the elapsed time measured by the guy how is sitting at spatial infinity seeing how the spaceship goes from A to B.

18. Feb 7, 2014

### WannabeNewton

First of all "seeing" is different from distant simultaneity/clock synchronization. "Seeing" only requires a local clock, apart from the obvious necessity of a radar set-the reason we need distant simultaneity/clock synchronization in the first place is because of the finite signal propagation speed between local events corresponding to "seeing" and distant events corresponding to the actual occurrences of whatever it is you "see". Secondly, I've already answered your question multiple times in the thread, twice just in the last two posts of mine. I don't see any point in repeating it again.

Last edited: Feb 7, 2014
19. Feb 7, 2014

### Mentz114

Whoops. I think what I wrote has a wrong sign ...

But, for the guy at infinity in the Schwarzschild spacetime, this gives $t_2-t_1$ as $r\rightarrow \infty$ so what is your problem ?

20. Feb 7, 2014

### npnacho

first of all, i'm sorry if i'm being repetitive, i know you've answered my question already, the thing is i don't understand that, i''l try to explain my problem with another example:

that was exactly my initial question, i.e: the coordinate time (on my own reference system) has something to do with "what i see", or with "what is really happening" or neither?

so, after the generalized answer "it's only a mathematical parameter", there was my second question:

how do you compute the time measured by an observer between two events he sees happening?

i have this question because i don't understand, for example, what are you doing (physically) when you compute the gravitational redshift for schwarzschild, so i'll try to explain my problem in the context of that example:

suppose you have two observers (1 and 2) at rest with $r_{1}<r_{2}, θ_{1}=θ_{2} , ø_{1}=ø_{2}$ and a light beam goes through both. so the observer 1 measures the time elapsed between two "peaks" of the wave

$$\tau_1 = \int_{A}^{B} \sqrt{1 - \frac{2M}{r_1}} dt = \sqrt{1 - \frac{2M}{r_1}} \left( t_B - t_A \right)$$

where $t_A$ and $t_B$ are the coordinate times when the first and second peaks of the wave reaches the observer 1.

by the other hand, if the observer 2 does the same thing, we have:

$$\tau_2 = \int_{A}^{B} \sqrt{1 - \frac{2M}{r_2}} dt = \sqrt{1 - \frac{2M}{r_2}} \left( t_D - t_C \right)$$

where $t_C$ and $t_D$ are the coordinate times when the first and second peaks of the wave reaches the observer 2.

and then we say that the redshift (relation between "propper periods" measured by each observer) is:

$$\frac{\sqrt{g_{00} ( r_1 )}}{\sqrt{g_{00} ( r_2 )}} = \frac{\sqrt{1 - \frac{2M}{r_1}}}{\sqrt{1 - \frac{2M}{r_2}}}$$

so we obviously have said that $(t_B-t_A)=(t_D-t_C)$

so there is what i don't understand about coordinate times... first of all, there are 4 coordinate times like i wrote? or in each equation i should write only $t_A$ and $t_B$?

my point is: if we take "t" to be the propper time measured by an observer at infinity... then in both equations we have $(t_B-t_A)$ which is the time separation between peaks measured by a third observer at spatial infinity and, in that case, i have no further questions... but if we take t as a simply mathematical parameter, i don't understand how we can say that $(t_B-t_A)=(t_D-t_C)$, i.e: how we know that the variation of this parameter between the first two events (two peaks going through observer 1) is the same that the variation of the parameter between the third and fourth events (the two peaks going through observer 2)?

thanks! (and sorry again)

21. Feb 7, 2014

### npnacho

i wasn't referring to the sign, i hope i made my question clear in my previous post, but what i say is that to get $t_2-t_1$ as $r\rightarrow \infty$, you are measuring the time of events in infinity, that's way the propper time of the observer at infinity matches the coordinate time in scharzschild, i think... but my question was about a observer at infinity looking at something happening in another place, for example away from him, where the spacetime is curved.

thanks!

22. Feb 7, 2014

### WannabeNewton

To start with, do you understand the difference between local time and distant time in the context of inertial frames? We can come back to your example later because it's important that you first understand this difference in the simplest possible case.

23. Feb 7, 2014

### Mentz114

OK, I can see what I've written does not address the question. But 'something happening' is a bit too general for me. Your question is too vague for me to answer ( I probably couldn't anyway ) so I'll leave it.

24. Feb 7, 2014

### npnacho

of course, one way to do that (calculate the elapsed in the observer's clock between two events A and B) is to compute the geodesics of two light rays which come out from A and B and find the times they reach the observer (this is what wannabenewton tried to say me before, i think)

but i was asking if there was another way (more direct) to do that... not because i need to do that calculation, but i was trying to understand what the coordinate time is (now i'm totally convinced that it has nothing to do with this... so now my only question is the one in my 6:19 post, je)

25. Feb 7, 2014

### npnacho

i've never heard "distant time" or "local time" (maybe i know what it is but not in english, i don't know...)

what do you mean with that?