Coordinate and proper time, null geodesic

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I have a question which asks show that a null geodesic to get to r> R , r some constant, given the space time metric etc, takes infinite coordinate time but finite proper time. ( It may be vice versa ).

I just want to confirm that, ofc there is no affine parameter for a null geodesic and so you could take coordinate time and proper time to be the same, and so that this question is referring to the time as observed by a timelike observer.

So if I denote the Lagrangian by L , you would proceed as follows :
1)set L=1 to find the relationship between proper time and coordinate time for the timelike observer ( only ##\dot{t} \neq 0 ## since at rest to get proper time, all other coordinates 0 )
2) set L=0 solve for the relationship between r(t) and (t) - assume radial for simplicity and then
a) not substituting in the expression from 1) will give you coordinate time
b) substituting the expression in from 1) will give you the proper time.

Can I confirm these ideas are correct - basically the question does not specify, but it should be the proper time of a timelike observer ?

Many thanks
 

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  • #2
PeterDonis
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ofc there is no affine parameter for a null geodesic
This is not correct. It is perfectly possible to find a affine parameter for a null geodesic. (You can always find an affine parameter for any curve.) What you can't do is use arc length as the affine parameter for a null geodesic.

and so you could take coordinate time and proper time to be the same
Certainly not.
 
  • #3
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This is not correct. It is perfectly possible to find a affine parameter for a null geodesic. (You can always find an affine parameter for any curve.) What you can't do is use arc length as the affine parameter for a null geodesic.



Certainly not.
Proper time : timelike
1=g_00 (dt/ds)^2

Seperate variables : dt/ds=\sqrt{g_00}

Proper time null :
dt/ds= 0 , set the constant of integration to zero ?
 
  • #4
Ibix
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Proper time : timelike
1=g_00 (dt/ds)^2
This is only correct for a massive object at rest in this coordinate system. Otherwise the spatial derivatives are non zero and you have a collection of ##dtdx^i## and ##dx^idx^j## terms that you have dropped.
Proper time null :
dt/ds= 0 , set the constant of integration to zero ?
Again, you appear to be neglecting the spatial coordinate differentials. Only this time doing so is self-contradictory - you are trying to describe a null path that's time-like (unless you were intending t to be a null coordinate, which would be an odd choice of notation). Furthermore, the defining characteristic of a null path is that ##ds=0##, so ##dt/ds## is undefined. In fact ##d\mathrm{anything}/ds## is undefined.

The question as you've written it in the OP makes no sense to me. There is no notion of proper time for a null path, so the answer is mu (as the Chinese philosopher would say). What exactly are you asked?
 
  • #5
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This is only correct for a massive object at rest in this coordinate system. Otherwise the spatial derivatives are non zero and you have a collection of ##dtdx^i## and ##dx^idx^j## terms that you have dropped.
Again, you appear to be neglecting the spatial coordinate differentials. Only this time doing so is self-contradictory - you are trying to describe a null path that's time-like (unless you were intending t to be a null coordinate, which would be an odd choice of notation). Furthermore, the defining characteristic of a null path is that ##ds=0##, so ##dt/ds## is undefined. In fact ##d\mathrm{anything}/ds## is undefined.

The question as you've written it in the OP makes no sense to me. There is no notion of proper time for a null path, so the answer is mu (as the Chinese philosopher would say). What exactly are you asked?
Proper time is defined as when the observer is at rest ??
That is what I was asked, which is what my whole post is about, is it instead asking for a comparison of proper time of a timelike observer ..
 
  • #6
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Question word for word

Show that it takes infinite proper time for any null geodesic to go from r =
ri > 1 to r → ∞, but the same trajectory is covered in finite coordinate time
t. What happens when ri → 1?

Is proper time referring to a non-geodesic observer ? So timelike ?
 
  • #7
stevendaryl
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Proper time is defined as when the observer is at rest ??
That is what I was asked, which is what my whole post is about, is it instead asking for a comparison of proper time of a timelike observer ..
No, proper time is defined for any trajectory, whatsoever. But in the special case in which an object is at rest in some coordinate system, then the proper time for that object's trajectory is given by: ##1 = g_{00} (\frac{dt}{d\tau})^2##.

In general, proper time for any trajectory is given in a coordinate system by: ## 1 = \sum_{\mu \nu} g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}##
 
  • #8
Ibix
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Proper time is defined as when the observer is at rest ??
Proper time is the interval along a timelike worldline. The particular formula you were proposing is only valid for a timelike path that is at rest in your coordinate system, since you've left out all the other coordinate differentials.

Proper time is not defined for a null path because the interval is zero along such paths.
Question word for word

Show that it takes infinite proper time for any null geodesic to go from r =
ri > 1 to r → ∞, but the same trajectory is covered in finite coordinate time
t. What happens when ri → 1?
I'm sorry, this question still makes no sense to me.

Is there any context to it? What metric are you working in? What coordinates are you supposed to use? What's the significance of an r coordinate of 1? Critically, whose proper time are we talking about? It can't be the light's because that's undefined. But no other observer is specified, so I'm stumped by the question as written.
 
  • #9
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No, proper time is defined for any trajectory, whatsoever. But in the special case in which an object is at rest in some coordinate system, then the proper time for that object's trajectory is given by: ##1 = g_{00} (\frac{dt}{d\tau})^2##.

In general, proper time for any trajectory is given in a coordinate system by: ## 1 = \sum_{\mu \nu} g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}##
ah okay , apologies my bad
 
  • #10
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Proper time is the interval along a timelike worldline. The particular formula you were proposing is only valid for a timelike path that is at rest in your coordinate system, since you've left out all the other coordinate differentials.

Proper time is not defined for a null path because the interval is zero along such paths.
I'm sorry, this question still makes no sense to me.

Is there any context to it? What metric are you working in? What coordinates are you supposed to use? What's the significance of an r coordinate of 1? Critically, whose proper time are we talking about? It can't be the light's because that's undefined. But no other observer is specified, so I'm stumped by the question as written.
will post the rest of the question in a second, on my phone not computer, in terms of the question, may you assume a radial trajectory, as you do for example, in the derivations of red-shifts, setting ##d\phi## and ##d\theta=0## for simplicity? Also, if the question seems to be referring to a time-like observer- physically observers are time-like so perhaps this is why it should be assumed, although very vague- since you can find a frame in which the observer is at rest, may you take this frame to get the relation between coordinate time and proper time, and assume the observer is at rest? (like again we do in red-shift derivations?)
 
  • #11
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Apologies for the delay, full question attached.
Many thanks for your help, it is greatly
fullquestion.png
appreciated.
 

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  • #12
stevendaryl
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My approach for timelike geodesics is to let ##\tau## be your "time" parameter, and then you have corresponding "velocities": ##V^t \equiv \frac{dt}{d\tau}##, ##V^r \equiv \frac{dr}{d\tau}##, ##V^x \equiv \frac{dx}{d\tau}## and ##V^y \equiv \frac{dy}{d\tau}##. Then the quantity

##\mathcal{L} \equiv \frac{1}{2} g_{\mu \nu} V^\mu V^\nu##

acts like an effective Lagrangian. (The 1/2 in front is irrelevant, but it makes it look a little more like the kinetic energy from classical physics). Then there are corresponding "momenta":

##P_\mu = \frac{\partial \mathcal{L}}{\partial V^\mu}##

They obey a Lagrangian equation of motion:

##\frac{d P_\mu}{d\tau} = \frac{\partial \mathcal{L}}{\partial x^\mu}##

So you get that ##P_\mu## is constant whenever ##\mathcal{L}## does not depend on ##x^\mu##. So for your case, ##\mathcal{L}## depends only on ##r## (and the velocities) so ##P_x, P_y, P_t## are all constants.

Then there is one more constant of the motion, because since ##d \tau = \sqrt{ds^2}##, it follows that ##\mathcal{L}## itself is a constant (equal to 1/2, just because I put a 1/2 in front---it would be 1 otherwise).
 
  • #13
stevendaryl
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My approach for timelike geodesics is to let ##\tau## be your "time" parameter, and then you have corresponding "velocities": ##V^t \equiv \frac{dt}{d\tau}##, ##V^r \equiv \frac{dr}{d\tau}##, ##V^x \equiv \frac{dx}{d\tau}## and ##V^y \equiv \frac{dy}{d\tau}##. Then the quantity

##\mathcal{L} \equiv \frac{1}{2} g_{\mu \nu} V^\mu V^\nu##

acts like an effective Lagrangian. (The 1/2 in front is irrelevant, but it makes it look a little more like the kinetic energy from classical physics). Then there are corresponding "momenta":

##P_\mu = \frac{\partial \mathcal{L}}{\partial V^\mu}##

They obey a Lagrangian equation of motion:

##\frac{d P_\mu}{d\tau} = \frac{\partial \mathcal{L}}{\partial x^\mu}##

So you get that ##P_\mu## is constant whenever ##\mathcal{L}## does not depend on ##x^\mu##. So for your case, ##\mathcal{L}## depends only on ##r## (and the velocities) so ##P_x, P_y, P_t## are all constants.

Then there is one more constant of the motion, because since ##d \tau = \sqrt{ds^2}##, it follows that ##\mathcal{L}## itself is a constant (equal to 1/2, just because I put a 1/2 in front---it would be 1 otherwise).
This approach to finding geodesics completely avoids ever computing ##\Gamma^\mu_{\nu \lambda}##, but it's equivalent to doing it that way.
 
  • #14
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This approach to finding geodesics completely avoids ever computing ##\Gamma^\mu_{\nu \lambda}##, but it's equivalent to doing it that way.
Seen all this before ?
Not sure how it addressed my question, other than perhaps a method as to how to approach it if ##dx^i \neq 0 ##, which I see I incorrectly assumed, but I also didn't say I dont know how to approach If this is case.

I'm more interested in some of the questions above such as :
- interpretation of the question
- when it is valid to work in the local rest frame, whcih would let ##dx^i =0 ## and massively simplify the maths
 
  • #15
PeterDonis
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interpretation of the question
I'm not sure what you mean. The question is perfectly clear about what it's asking for; there's no need for any "interpretation". But you do, of course, need to have the requisite background to understand the terms the question is using. For example, do you know what a Killing vector is and why Killing vectors are significant?

Also, it might help to give the source for the question (what textbook/chapter/problem). At least one of the items, item (e), seems to be using language in a rather sloppy way; there is no such thing as "proper time" along a null geodesic. I suspect the intended meaning is infinite affine parameter, but context would help.
 
  • #16
stevendaryl
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Seen all this before ?
Not sure how it addressed my question, other than perhaps a method as to how to approach it if ##dx^i \neq 0 ##, which I see I incorrectly assumed, but I also didn't say I dont know how to approach If this is case.

I'm more interested in some of the questions above such as :
- interpretation of the question
- when it is valid to work in the local rest frame, whcih would let ##dx^i =0 ## and massively simplify the maths
Okay, I guess I don't understand your question. I thought you were trying to answer the questions you posted.
 
  • #17
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I'm not sure what you mean. The question is perfectly clear about what it's asking for; there's no need for any "interpretation". But you do, of course, need to have the requisite background to understand the terms the question is using. For example, do you know what a Killing vector is and why Killing vectors are significant?

Also, it might help to give the source for the question (what textbook/chapter/problem). At least one of the items, item (e), seems to be using language in a rather sloppy way; there is no such thing as "proper time" along a null geodesic. I suspect the intended meaning is infinite affine parameter, but context would help.

' no such thing is proper time ... '
Is what my question was. You say the question is perfectly clear and then say this, which is what my whole question was about, and that some language is sloppy ? Im asking whether the question means proper time and coordinate time for a null observer or a timelike observer, since the proper time for a null observer seems confusing. That is what I mean when I said the question is unclear.
Okay, I guess I don't understand your question. I thought you were trying to answer the questions you posted.
I was asked to post the full question in case it may clarify some of my questions? Doesn't change my initial question to please help me with all of these questions ?
 
  • #18
stevendaryl
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So, I'm confused. Do you know how to answer the questions you posted, or not?

(a) Identify three constants of the motion
(b) Show that the geodesic equation can be reduced to an equation of the form...
(c) Sketch the potential...
(d) Sketch V(r) for timelike geodesics
(e) Show that it takes infinite proper time...

Do you know how to answer those questions, and you're just asking additional questions that came up, or did you need to ask additional question in order to answer the problem questions?
 
  • #19
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So, I'm confused. Do you know how to answer the questions you posted, or not?

(a) Identify three constants of the motion
(b) Show that the geodesic equation can be reduced to an equation of the form...
(c) Sketch the potential...
(d) Sketch V(r) for timelike geodesics
(e) Show that it takes infinite proper time...

Do you know how to answer those questions, and you're just asking additional questions that came up, or did you need to ask additional question in order to answer the problem questions?
I'm genuinely lost
Do I need to repeat myself ?
My question in my op is the only questions that outstands,and still outstands.
I repeat, I posted the other questions because someone asked for the whole question. Would you like me to say that again ?
 
  • #20
stevendaryl
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But could you answer my question: Do you know how to do those problems, or not?

If you don't, then I think it's more worthwhile for you to try to answer those questions, because they are more concrete than your questions, which I don't understand (it doesn't help to repeat them).
 
  • #21
Ibix
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Question e doesn't make sense as written. My guess is that the author is using proper time to refer to the affine parameter along the null geodesic. That's highly non-standard terminology as far as I'm aware, but it's the only way I can interpret the question.

Presumably you have expressions for ##dr/d\lambda## and ##dt/d\lambda## where ##\lambda## is an affine parameter (you may have called it ##\tau##). Presumably you can integrate them to get ##r(\lambda)## and ##t(\lambda)##, with initial conditions ##t=0## and ##r=r_i##. Then what do you find about ##\lambda## as ##r\rightarrow\infty##? What about ##t##?
 
  • #22
stevendaryl
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Your first post only has one question in it: "Can I confirm these ideas are correct?" That has been answered: no, they are not exactly correct. A null geodesic can have an affine parameterization.
 
  • #23
stevendaryl
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Question e doesn't make sense as written. My guess is that he's using proper time to refer to the affine parameter along the null geodesic. That's highly non-standard terminology as far as I'm aware, but it's the only way I can interpret the question.

Presumably you have expressions for ##dr/d\lambda## and ##dt/d\lambda## where ##\lambda## is an affine parameter (you may have called it ##\tau##). Presumably you can integrate them to get ##r(\lambda)## and ##t(\lambda)##, with initial conditions ##t=0## and ##r=r_i##. Then what do you find about ##\lambda## as ##r\rightarrow\infty##?
If you take an affine parameter ##\lambda## and let ##\mathcal{L} \equiv g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}## be an effective Lagrangian, then the Lagrangian equations of motion give the geodesics, regardless of whether the geodesics are timelike or null. Or at least, I think that's true.

But I think that the definition of an "affine parametrization" is equivalent to ##\mathcal{L} = ## constant. Choosing that constant to be 1 (or -1, depending on the signature convention) means that ##\lambda## is proper time. In the case of null geodesics, ##\mathcal{L} = 0##.

I'm interpreting question e as asking about the case ##\lambda \rightarrow \infty## for the null geodesic, not ##\tau##.
 
  • #24
Ibix
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I'm interpreting question e as asking about the case λ→∞\lambda \rightarrow \infty for the null geodesic, not τ\tau.
Agreed. But I think a lot of the confusion on this thread stems from us (or me at least) trying to work out whether the use of proper time implied that there was a timelike geodesic in the problem somewhere. There isn't. It's just a badly worded question.

This, incidentally, is why there's a homework template in the homework section that demands that the exact question be posted. We could have had this bit of the conversation on Tuesday.
 
  • #25
vanhees71
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This is not correct. It is perfectly possible to find a affine parameter for a null geodesic. (You can always find an affine parameter for any curve.) What you can't do is use arc length as the affine parameter for a null geodesic.



Certainly not.
Indeed. The trick is instead of using the general parametrization independent action with the square root, i.e., the Lagrangian
$$L_1=-\sqrt{\dot{x}^{\mu} \dot{x}^{\nu} g_{\mu \nu}},$$
where the dot denotes the derivative wrt. an arbitrary world-line parameter. The corresponding action is obviously parameter invariant, i.e., changing from ##\lambda## to ##\lambda'## doesn't change the metric. This is, however, not the most convenient action (already not in SR by the way) but instead you use
$$L_2=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
Then the used parameter is automatically "affine" in the sense that ##L_2## is a constant of the motion, because the Lagrangian does not explicitly depend on ##\lambda##, i.e., the "Hamiltonian" is a constant of the motion:
$$H_2=p_{\mu} \dot{x}^{\mu}-L_2,$$
where the canonical momentum is
$$p_{\mu} =\frac{\partial L}{\partial \dot{x}^{\mu}}=g_{\mu \nu} \dot{x}^{\nu}$$
and thus
$$H_2=L_2=\text{const}$$
along the solution of the Euler-Lagrange equations which are the equations for spacetime geodesics.

The only specialty of the motion of massless particles, i.e., the null geodesics is that then ##L_2=0## along the geodesic, while for massive particles ##L_2>0##. In the latter case the value can be arbitrarily choosen, and a physical choice is ##L_2=c^2##. Then ##\lambda=\tau## the particle's proper time.

If you have also other interactions than the external gravitational field, it's clever to keep the corresponding Lagrangians as homogeneous of 1st order in ##\dot{x}^{\mu}##, because then also in these cases the parameter ##\lambda## still is affine and can be chosen as the proper time for massive particles.

It's clear that for massless particles there's no preferred choice of the affine parameter, simply because there's no scale in the problem. It's of course clear that a massless particle doesn't have something like "proper time", because there's no (local) rest frame of such particles.
 
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