Proper time of an accelerated frame in a external gravitational field

[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?

---------

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 z_A and z_B 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!

 

[PeterDonis]

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?

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.

how would do you calculate the resulting metric in a problem like this?

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.

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 z_A and z_B respectively.

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.

what is the frame that measures that coordinate time and why is the same for both observers?

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

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

 

[WannabeNewton]

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?

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.

what is the frame that measures that coordinate time and why is the same for both observers?

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.

 

[npnacho]

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.

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?

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.

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.

thanks for the answer!

 

[PeterDonis]

what i mean is that, in virtue of the equivalence principle, an accelerated frame in vacuum must have a non flat metric

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.

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.

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.

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...

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.

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?

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).

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

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.

when we say that some metric is given by

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

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.

that "t" is the coordinate time in the frame that observes this metric, right?

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.

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?

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.

 

[A.T.]

what i mean is that, in virtue of the equivalence principle, an accelerated frame in vacuum must have a non flat metric

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:
https://www.physicsforums.com/showthread.php?p=4281670&postcount=20

 

[WannabeNewton]

that "t" is the coordinate time in the frame that observes this metric, right?

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?

 

[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!

 

[npnacho]

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

 

[WannabeNewton]

ah! and for the GR part i think I'm going to use "gravity" by james hartle...

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.

 

[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!

 

[WannabeNewton]

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?

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).

 

[PeterDonis]

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

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.

this coordinate time is only a mathematical parameter or it has physical meaning?

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.

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).

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.

the spatial part of the metric looks like your coordinate system is at rest on the singularity

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.

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

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.)

some of you said something like "it's the time measured by an observer at infinity", but i didn't catch that.

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 $\Delta t \sqrt{1 - 2M / r}$.

 

[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

thanks for the answers!

 

[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.

 

[Mentz114]

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

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.

 

[npnacho]

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.

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

i'm still having the same question:

suppose (schwarzschild case) 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?

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.

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.

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.

 

[WannabeNewton]

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.
...
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.

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.

 

[Mentz114]

..
..
that is the proper 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.

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 ?

 

[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:

First of all "seeing" is different from distant simultaneity/clock synchronization. "Seeing" only requires a local clock

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)

 

[npnacho]

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 ?

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!

 

[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.

 

[Mentz114]

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!

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.

 

[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)

 

[npnacho]

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.

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?

 

[PeterDonis]

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

Yes, if you want to find out at what times, by the distant observer's clock, he actually *sees* (as in, receives light rays) what happens at events A and B, this is how you do it. If this is what you mean by "the elapsed time by the distant observer's clock", then yes, this will tell you what it is.

but i was asking if there was another way (more direct) to do that

To do what? Basically it looks like you're confused about what meaning you want to assign to "the elapsed time by the distant observer's clock". Sometimes you seem to mean the time between his actually *seeing* (receiving light rays) events A and B, in which case the above calculation is how you get the answer. But sometimes you seem to mean something else, only it isn't quite clear what. Part of the problem there may be that, other than the time between actually seeing the light rays, there is no other "elapsed time for the distant observer" that is invariant; so if you're looking for anything other than that, what you're looking for may not be well-defined, as I noted before.

 

[WannabeNewton]

what do you mean with that?

Here's a simple mundane example. You're at home and you want to watch TV. You turn on the TV and you look at your watch (which is set to read proper time) and you say the TV turned on at time $t_1$ according to your watch reading. But what the event $t_1$ really represents is the moment when the initial light from the TV screen reaches your eyes, not the moment when the TV "actually" turns on. You can't use only your watch to figure out at what time the TV "actually" turned on. You need something extra like, for example, a clock sitting by the TV that's been synchronized with your watch. Why is this sufficient? Well now if a person holding the clock says the TV turned on at time $t_2$ according to said clock (this is a local event for this person since they're standing right next to the TV) then you know that your watch said $t_2$ when the clock said $t_2$ because the watch and clock are synchronous, meaning they tick in unison. In other words you now know what event in your vicinity is simultaneous with the "TV turning on" event in the vicinity of the clock and the person holding the clock-this is called distant simultaneity. So if we want to compare the proper times of distant events we need synchronized clocks or equivalently a notion of distant simultaneity.

So let's go back to Schwarzschild space-time. Consider the observer at infinity $O_{\infty}$. If $O_{\infty}$ emits a beam of light at an event $p$ that gets sent to a hovering observer $O_R$ deep in the gravitational field who receives it at an event $q$ and subsequently reflects it back to infinity which arrives to $O_{\infty}$ at event $p'$ then we can say that there has been an elapsed proper time $t_{p'} - t_p$ between events $p$ and $p'$ both of which lie in the vicinity of $O_{\infty}$ (i.e. they are local to $O_{\infty}$) because for $O_{\infty}$, coordinate time $t$ and proper time $\tau$ are the same thing for local events. However $t_q - t_p$ doesn't correspond to the elapsed proper time on $O_{\infty}$'s clock between emission of the light beam and reception of the light beam by $O_R$ because $O_{\infty}$'s clock is not synchronized with $O_R$'s clock and in fact cannot be synchronized if we are adamant about using proper time because of gravitational time dilation.

 

[npnacho]

ohh, man. that's really good.

so that's why we have to resign to the idea of assigning physical meaning to the coordinate time and treat it like a global parameter only, right? (except in extreme cases like this one when we know that it coincides with the propper time of a particular observer, o like the flat spacetime case when we can synchronize clocks in every space point)

thanks for the answers and the patience!

 

[WannabeNewton]

Yep.

The following two threads may be of further help: https://www.physicsforums.com/showthread.php?t=732122, https://www.physicsforums.com/showthread.php?t=732892

 

[npnacho]

thanks! i'll take a look.

 

[WannabeNewton]

thanks! i'll take a look.

Perhaps post #40 of the first thread in particular will help clear the air. I could explain it here in more detail using the geometric language intrinsic to GR but I don't know what your mathematical level is.

 

[pervect]

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 proper 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.

There may or may not be something strictly equating to a "point of view", depending on what you think a "point of view means".

See for instance http://arxiv.org/abs/gr-qc/9508043, "Precis of General Relativity".

A method for making sure that the relativity effects are specified correctly
(according to Einstein’s General Relativity) can be described rather briefly.
It agrees with Ashby’s approach but omits all discussion of how, historically
or logically, this viewpoint was developed. It also omits all the detailed
calculations. It is merely a statement of principles.

One first banishes the idea of an “observer”.

This is basically your idea of "point of view", I think.

This idea aided Einstein in building special relativity but it is confusing and ambig
uous in general relativity. Instead one divides the theoretical landscape
into two categories.

One category is the mathematical/conceptual model of whatever is happen-
ing that merits our attention. The other category is measuring instruments
and the data tables they provide

Misner goes on to describe what the measuring instruments would be. Since this isn't GPS, it's not directly relevant, but you might imainge that measuring instruments would be actual clocks that you carry around with you, and timestamped radar signals, so that you know the time of emission and time of reception (by your local clock) of various radar signals.

You can extend this somewhat to imaginary more clocks and radar emitters that are remotely located (such as the GPS clocks, or the receipt of signals from astronomical reference stars, or perhaps even slide slightly over the line and imagine an infinite array of clocks that don't actually exist, though the later starts to blur the line of what "actual measurements" are, since they are really mental constructs.

The other part of GR, other than the measuring instruments, is the "mathematical conceptual model of whaterver is happening". This is the metric. The metric is sort of a space-time map, it assigns coordinate numbers (which in general are more or less arbitrary) to every event in space-time.

Using the metric, you can compute the path of any desired radar signal, and you can compute the proper time reading of any clock following any specified trajectory (though you need the map-coordiantes to specify the trajectory).

Your question I think relates ultimately on how you assemble a mental picture of what's going on from actual observations. There are some conventions here, but they may or may not match your intuitive idea of what a "point of view" should be. The conventions boil down to the fact that if you have a metric such that it is diag(-1,1,1,1) in some area, that in that area the coordinates used by the metric are a "local" point of view that works while you are in that neighborhood.

For instance, in the Schwarzschild map, the coordinates have this property at infinity. Hence the talk about observers at infinity.

In general, there isn't any unambiguous way to have a "point of view" regarding distant objects. You replace this with the ability to use your map of space-time (the metric) to compute anything you can actually measure (with clocks, and radar signals, as specified eariler).

One final note. Misner notes that this is "A method" to make sure one specifies relativistic corrections correctly. It doesn't necessarily mean it is "the only method". Part of the issues here are philsophical, so you might find people with different philosophies. This particular method is particularly simple and useful, I think, however, and avoids a lot of digressions.