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General Relativity

  • #1
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Hi.

I'm reading these notes:
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf
and have a few questions.

(i) On page 11, why is equation (2.1) true? Where does this come from? I guess the timelike feature manifests itself as the minus sign at the front but I could really just use some help on understanding this equation.

(ii) Why does [itex]d \tau = \sqrt{ - ds^2}[/itex] as is written in (2.2)?

(iii) Where does (2.6) come from? What does he mean by [itex]D_{(\lambda)}[/itex]?
I can see he's tried to define it in (2.7) but that's not really helping much!

Thanks.
 

Answers and Replies

  • #2
dextercioby
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Proper time is invariant wrt the Lorentz transformations. Since the action integral must be invariant wrt the Lorentz transformations (just like the nonrelativistic one is invariant wrt the Galilei transformations), d\tau must be there alongside the invariant mass and the invariant speed of light. (mass times speed squared times time = action)

The minus under the sqrt in 2.2 comes from the convention he chose for the metric, mostly plus.

As for the <mysterious> derivative,it's the derivative along the geodesic. See here http://en.wikipedia.org/wiki/Geodesic and at the bottom of here http://en.wikipedia.org/wiki/Covariant_derivative (<derivative along a curve> section).
 
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  • #3
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Proper time is invariant wrt the Lorentz transformations. Since the action integral must be invariant wrt the Lorentz transformations (just like the nonrelativistic one is invariant wrt the Galilei transformations), d\tau must be there alongside the invariant mass and the invariant speed of light. (mass times speed squared times time = action)
And the minus at the front? That's because it's for a timelike curve, right?

The minus under the sqrt in 2.2 comes from the convention he chose for the metric, mostly plus.
But if we expand this out, we get [itex]ds^2=-d \tau^2[/itex] what happened to the rest of it? Surely, since we have [itex]ds^2=g_{\mu \nu} dx^\mu dx^\nu[/itex], we should have spatial terms as well? Why are they set to zero?

As for the <mysterious> derivative,it's the derivative along the geodesic. See here http://en.wikipedia.org/wiki/Geodesic and at the bottom of here http://en.wikipedia.org/wiki/Covariant_derivative (<derivative along a curve> section).
So does [itex]D_{( \lambda)} V^\mu(\lambda)= \nabla_{\dot{\lambda}} V^\mu(\lambda)[/itex] ?
Is that correct definition of this "covariant derivative along a smooth curve"?

If so, I still can't manage to prove (2.6)?

Thanks again.
 
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  • #4
dextercioby
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And the minus at the front? That's because it's for a timelike curve, right?
Not really. It's there for convenience. It's only when the particle is coupled to the e-m field, the utility of the minus will be seen.

The minus under the square root comes from:

wikipedia said:
Vectors are classified according to the sign of η(v,v). When the standard signature (−,+,+,+) is used, a vector v is:

Timelike if η(v,v) < 0
Spacelike if η(v,v) > 0
Null (or lightlike) if η(v,v) = 0
.

latentcorpse said:
But if we expand this out, we get [itex]ds^2=-d \tau^2[/itex] what happened to the rest of it? Surely, since we have [itex]ds^2=g_{\mu \nu} dx^\mu dx^\nu[/itex], we should have spatial terms as well? Why are they set to zero?
Nothing is set to 0. You're confusing d\tau with dt, the former is the proper time, the latter the <normal> time.

latentcorpse said:
So does [itex]D_{( \lambda)} V^\mu(\lambda)= \nabla_{\dot{\lambda}} V^\mu(\lambda)[/itex] ?
Is that correct definition of this "covariant derivative along a smooth curve"?
Yes.
 
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  • #5
dextercioby
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If so, I still can't manage to prove (2.6)?
Well, the Christoffel symbols are computed wrt the flat spacetime metric, so they're zero. What are the Euler-Lagrange equations for the action with einbein ? (2.4)
 
  • #6
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Well, the Christoffel symbols are computed wrt the flat spacetime metric, so they're zero. What are the Euler-Lagrange equations for the action with einbein ? (2.4)
Why are the Christoffel symbols computed wrt the flat metric? We haven't said anywhere that it was flat space, have we?

Anyway, (2.4) is independent of derivatives of the Einbein so the E-L equations are just [itex]\frac{\partial I}{\partial e}=0[/itex] Which is precisely eqn (2.5) right?
 
  • #7
dextercioby
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Yes, about the second part. As for the flat part, yes, you're right. The metric needn't be constant, that's why the Christoffen symbols come in.
 
  • #8
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Yes, about the second part. As for the flat part, yes, you're right. The metric needn't be constant, that's why the Christoffen symbols come in.
Ok.
But you just said that the Christoffel symbols are computed wrt the flat metric and now you're saying it needn't be flat? I don't understand.

Also, going back to (2.6), how do we go from [itex]\frac{\delta I}{\delta x^\mu}=0[/itex] to [itex]D_{(\lambda)} \dot{x}^\mu -(e^{-1}\dot{e})\dot{x}^\mu[/itex]

Cheers.
 
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  • #9
dextercioby
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What's [itex]\frac{\delta I}{\delta x^\mu}[/itex] equal to ?
 
  • #10
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What's [itex]\frac{\delta I}{\delta x^\mu}[/itex] equal to ?
Should I use [itex]\frac{\partial}{\partial x^\mu} = \frac{\partial \lambda}{\partial x^\mu} \frac{\partial}{\partial \lambda}[/itex]?

I'm unsure whether this is correct because this will give us lots of terms:

[itex]-e^{-2} \frac{\partial \lambda}{\partial x^\mu}{ \frac{\partial e}{\partial \lambda} \dot{x}^\sigma \dot{x}^\rho g_{\sigma \rho} + 2e^{-1} \frac{\partial \lambda}{\partial x^\mu} \frac{\partial \dot{x}^\sigma}{\partial \lambda} \dot{x}^\rho g_{\sigma \rho} - m^2 \frac{\partial \lambda}{\partial x^\mu} \frac{\partial e}{\partial \lambda}=0[/itex]

How's that? The only simplification I can see is to divide through by [itex]\frac{\partial \lambda}{\partial x^\mu}[/itex], giving:

[itex]-e^{-2} \frac{\partial e}{\partial \lambda} \dot{x}^\sigma \dot{x}^\rho g_{\sigma \rho} + 2e^{-1} \ \frac{\partial \dot{x}^\sigma}{\partial \lambda} \dot{x}^\rho g_{\sigma \rho} - m^2 \frac{\partial e}{\partial \lambda}=0[/itex]

And I guess dot mean differentiation wrt [itex]\lambda[/itex] so we can write:

[itex]-e^{-2} \dot{e} \dot{x}^\sigma \dot{x}^\rho g_{\sigma \rho} + 2e^{-1} \ddot{x}^\sigma \dot{x}^\rho g_{\sigma \rho} - m^2 \dot{e}=0[/itex]
 
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  • #11
dextercioby
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Should I use [itex]\frac{\partial}{\partial x^\mu} = \frac{\partial \lambda}{\partial x^\mu} \frac{\partial}{\partial \lambda}[/itex]?
No, it's wrong. x is a function of lambda, so [itex] \frac{\partial I}{\partial x^{\mu}} [/itex] means just to differentiate the lagrangian wrt x. No lambda involved.
 
  • #12
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No, it's wrong. x is a function of lambda, so [itex] \frac{\partial I}{\partial x^{\mu}} [/itex] means just to differentiate the lagrangian wrt x. No lambda involved.
I don't understand. How come, we are sometimes able to use the chain rule and sometimes not?

Ok. Well then the second term vanishes. And we are left with

[itex]e^{-1} \frac{\partial}{\partial x^\mu} (\frac{\partial x^\sigma}{\partial \lambda} \frac{\partial x^\rho}{\partial \lambda} g_{\sigma \rho}) = 2 \frac{\partial}{\partial \lambda} \frac{\partial x^\sigma}{\partial x^\mu} \dot{x}^\rho g_{\sigma \rho} + \dot{x}^\sigma \dot{x}^\rho g_{\sigma \rho, \mu}[/itex]
I'm fairly sure I've gone wrong again though. Aargh!
 
  • #13
dextercioby
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Just like in the other thread, you make things so complicated. Jesus and your notation is a killer.

[tex] \frac{\partial I}{\partial x^{\sigma}} = e^{-1} \dot{x}^{\mu} \dot{x}^{\nu} \frac{\partial g_{\mu\nu}}{\partial x^{\sigma}} [/tex]

[tex] \frac{\partial I}{\partial \dot{x}^{\sigma}} = 2 e^{-1} \dot{x}^{\nu} g_{\nu\sigma} [/tex]

Now complete the dots

[tex] \frac{d}{d\lambda} \frac{\partial I}{\partial \dot{x}^{\sigma}} = ... [/tex]

EDIT: the 1/2 factor in front of the action integral is unimportant and can be discarded when computing the E-L equations.
 
  • #14
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Just like in the other thread, you make things so complicated. Jesus and your notation is a killer.

[tex] \frac{\partial I}{\partial x^{\sigma}} = e^{-1} \dot{x}^{\mu} \dot{x}^{\nu} \frac{\partial g_{\mu\nu}}{\partial x^{\sigma}} [/tex]

[tex] \frac{\partial I}{\partial \dot{x}^{\sigma}} = 2 e^{-1} \dot{x}^{\nu} g_{\nu\sigma} [/tex]

Now complete the dots

[tex] \frac{d}{d\lambda} \frac{\partial I}{\partial \dot{x}^{\sigma}} = ... [/tex]

EDIT: the 1/2 factor in front of the action integral is unimportant and can be discarded when computing the E-L equations.
[itex]\frac{d}{d \lambda} \frac{\partial I}{\partial \dot{x}^\sigma}=-2e^{-2} \dot{e} \dot{x}^\nu g_{\sigma \nu} + 2 e^{-1} \ddot{x}^\nu g_{\sigma \nu}[/itex]

Ok. Then I think I'm getting somewhere. The E-L eqns give:

[itex]e^{-1} \dot{x}^\mu \dot{x}^\nu \frac{\partial g_{\mu \nu}}{\partial x^\sigma} = - 2 e^{-2} \dot{e} \dot{x}^\nu g_{\sigma \nu} + 2 e^{-1} \ddot{x}^\nu g_{\sigma \nu}[/itex]
[itex]\Rightarrow \ddot{x}^\nu g_{\sigma \nu} - \frac{1}{2} \dot{x}^\mu \dot{x}^\nu g_{\mu \nu, \sigma} = (e^{-1} \dot{e} ) \dot{x}^\nu g_{\sigma \nu}[/itex]
Multiplying through by [itex]g^{\sigma \rho}[/itex] we get
[itex]\ddot{x}^\rho - \frac{1}{2} \dot{x}^\mu \dot{x}^\nu g_{\mu \nu , \sigma} g^{\sigma \rho} = ( e^{-1} \dot{e} ) \dot{x}^\rho[/itex]

So now it's a case of getting a Christoffel symbol out of that second term somehow?
 
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  • #15
dextercioby
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You missed one term from the chain rule.

[itex]\frac{d}{d \lambda} \frac{\partial I}{\partial \dot{x}^\sigma}=-2e^{-2} \dot{e} \dot{x}^\nu g_{\sigma \nu} + 2 e^{-1} \ddot{x}^\nu g_{\sigma \nu}+ 2e^{-1} \dot{x}^{\nu} \frac{\partial g_{\sigma\nu}}{\partial x^{\mu}} \dot{x}^{\mu} [/itex]

Now can you complete your calculations ?
 
  • #16
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Yeah so using the E-L equations, we get:

[itex]\frac{d}{d \lambda} \frac{\partial I}{\partial \dot{x}^\sigma} = \frac{\partial I}{\partial x^\sigma}[/itex] and leaving the [itex]\frac{1}{2}[/itex] in,

[itex]-e^[-2} \dot{e} \dot{x}^\nu g_{\sigma \nu} + e^{-1} \dot{x} \ddot{x}^\nu g_{\sigma \nu} + e^{-1} \dot{x}^\nu \dot{x}^\mu g_{\sigma \nu, \mu} = \frac{1}{2} e^{-1} \dot{x}^\mu \dot{x}^\nu g_{\mu \nu, \sigma}[/itex]
Then dividing by [itex]e^{-1}[/itex] and multiplying by [itex]g^{\sigma \mu}[/itex] as well as symmetrising the third term on the left, we get

[itex]\ddot{x}^\mu + \frac{1}{2} g^{\sigma \mu} \left( g_{\sigma \nu, \rho} + g_{\nu \sigma, \rho} - g_{\nu \rho, \sigma} \right) \dot{x}^\nu \dot{x}^\rho = (e^{-1} e) \dot{x}^\mu[/itex]
[itex]\ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho = (e^{-1}e) \dot{x}^\mu[/itex]
[itex]\Rightarrow D_{(\lambda)} \dot{x}^\mu = (e^{-1} e) \dot{x}^\mu[/itex] as required. However, this has come from the E-L equation not [itex]\frac{\delta I}{\delta x^\mu}=0[/itex] as required. How do I sort this?
 
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  • #17
dextercioby
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[itex]\frac{\partial I}{\partial x^\mu}=0[/itex]

is wrong, you should have it

[itex]\frac{\delta I}{\delta x^\mu}=0[/itex]

which is the variational derivative of the Lagrangian. Which is what you have computed with my help.
 
  • #18
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You missed one term from the chain rule.

[itex]\frac{d}{d \lambda} \frac{\partial I}{\partial \dot{x}^\sigma}=-2e^{-2} \dot{e} \dot{x}^\nu g_{\sigma \nu} + 2 e^{-1} \ddot{x}^\nu g_{\sigma \nu}+ 2e^{-1} \dot{x}^{\nu} \frac{\partial g_{\sigma\nu}}{\partial x^{\mu}} \dot{x}^{\mu} [/itex]

Now can you complete your calculations ?
Yeah so using the E-L equations, we get:

[itex]\frac{d}{d \lambda} \frac{\partial I}{\partial \dot{x}^\sigma} = \frac{\partial I}{\partial x^\sigma}[/itex] and leaving the [itex]\frac{1}{2}[/itex] in,

[itex]-e^[-2} \dot{e} \dot{x}^\nu g_{\sigma \nu} + e^{-1} \dot{x} \ddot{x}^\nu g_{\sigma \nu} + e^{-1} \dot{x}^\nu \dot{x}^\mu g_{\sigma \nu, \mu} = \frac{1}{2} e^{-1} \dot{x}^\mu \dot{x}^\nu g_{\mu \nu, \sigma}[/itex]
Then dividing by [itex]e^{-1}[/itex] and multiplying by [itex]g^{\sigma \mu}[/itex] as well as symmetrising the third term on the left, we get

[itex]\ddot{x}^\mu + \frac{1}{2} g^{\sigma \mu} \left( g_{\sigma \nu, \rho} + g_{\nu \sigma, \rho} - g_{\nu \rho, \sigma} \right) \dot{x}^\nu \dot{x}^\rho = (e^{-1} e) \dot{x}^\mu[/itex]
[itex]\ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho = (e^{-1}e) \dot{x}^\mu[/itex]
[itex]\Rightarrow D_{(\lambda)} \dot{x}^\mu = (e^{-1} e) \dot{x}^\mu[/itex] as required. However, this has come from the E-L equation not [itex]\frac{\delta I}{\delta x^\mu}=0[/itex] as required. How do I sort this?
 
  • #19
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[itex]\frac{\partial I}{\partial x^\mu}=0[/itex]

is wrong, you should have it

[itex]\frac{\delta I}{\delta x^\mu}=0[/itex]

which is the variational derivative of the Lagrangian. Which is what you have computed with my help.
Aaah! Of course! Thanks.
 
  • #20
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[itex]\frac{\partial I}{\partial x^\mu}=0[/itex]

is wrong, you should have it

[itex]\frac{\delta I}{\delta x^\mu}=0[/itex]

which is the variational derivative of the Lagrangian. Which is what you have computed with my help.
Ok. So a couple of other things:

(i) Is the choice of e independent of the choice of lambda or does fixing one, fix the other through eqn (2.5)?

(ii) Looking at equation (2.8), is there a mistke on the LHS? Surely t should be contracted with V? How does this define a geodesic? Can we reduce it to the standard geodesic equation?
 
  • #21
dextercioby
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(i) equation 2.5 just shows that the einbein is not an independent variable. It can be computed once we know the x^mu (lambda) function.

(ii) there's indeed a mistake in equation 2.8. The V should vanish and t should have a free index.
 
  • #22
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(i) equation 2.5 just shows that the einbein is not an independent variable. It can be computed once we know the x^mu (lambda) function.

(ii) there's indeed a mistake in equation 2.8. The V should vanish and t should have a free index.
Yes but (2.8) can be expanded to

[itex]D_{(\lambda)} t^\mu = \dot{t}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu t^\rho[/itex]

But [itex]t^\mu = \frac{d x^\mu}{dt}[/itex]

So [itex]D_{(\lambda)} t^\mu = \ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho[/itex]

So surely, in order to describe a geodesic, we want to set this equal to 0 in order to reproduce our familiar "geodesic equation". However, here he has set it equal to [itex]f( x(\lambda)) t^\mu[/itex]. Why is that?

And presumably, (2.12) is supposed to read [itex]D_{(\lambda)} V^\mu = f( \lambda) V^\mu[/itex]?

And I don't see how (2.13) implies the Einbein is constant? When I expand it out, I just get the original geodesic equation?
 
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  • #23
dextercioby
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I don't see your issue with 2.8 it's just a recast of the original 2.6 with the einbein part denoted by f(x).

latentcorpse said:
And presumably, (2.12) is supposed to read ... ?
he's sloppy again. the V^mu is put in 2.8 i/o 2.12. Forgive him.

2.13 clarifies your problem. For a timelike geodesic, the lambda parameter is no longer arbitrary by proportional with the proper time. This means, either by 2.5 or 2.6 that the einbein is a mere constant wrt tau/lambda.
 
  • #24
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I don't see your issue with 2.8 it's just a recast of the original 2.6 with the einbein part denoted by f(x).
Ok. My problem is that we have the well known geodesic equation

[itex]\ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho = 0[/itex] (*)

Now look at (2.8).

If we expand the LHS, we get

[itex]D_{(\lambda)} t^\mu = \ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho[/itex]

Now by the geodesic equation (*), the RHS should be zero. Instead, he has that it is somehow equal to [itex]f(x) t^\mu[/itex]?

he's sloppy again. the V^mu is put in 2.8 i/o 2.12. Forgive him.

2.13 clarifies your problem. For a timelike geodesic, the lambda parameter is no longer arbitrary by proportional with the proper time. This means, either by 2.5 or 2.6 that the einbein is a mere constant wrt tau/lambda.
Ok. Thanks. That's fine.

Then I have a couple of things about section 2.2:

If [itex]x^\mu \rightarrow x^\mu - \alpha k^\mu[/itex]. How does this correspond to [itex]e \rightarrow e[/itex]? Looking at (2.5), and assuming [itex]k^\mu[/itex] is a constant vector then it the [itex]\dot{x}[/itex] terms are unaffected but won't the metric change as well?

Then, I had a bash at proving (2.18):

If we make the transformation, we find that [itex]I[x,e] \rightarrow I[x,e] + \int d \lambda e^{-1} \left( - \alpha \dot{k}^\mu \dot{x}^\nu g_{\mu \nu} - \alpha \dot{k}^\nu \dot{x}^\mu g_{\mu \nu} - \dot{x}^\mu \dot{x}^\nu g_{\mu \nu}( x^\sigma - \alpha k^\sigma) \right) + O ( \alpha^2 )[/itex]

Clearly, that third term is wrong but what is the first order term going to look like - I'm not sure?
Basically, can you tell me what [itex]g_{\mu \nu} ( x^\sigma - \alpha k^\sigma ) = \dots[/itex] when we expand? I thought it was going to be [itex]g_{\mu \nu} ( x^\sigma) + \alpha k^\sigma g_{\mu \nu , \sigma} + \dots[/itex] but the indices in those first two terms don't match up !

Thanks!
 
  • #25
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I don't see your issue with 2.8 it's just a recast of the original 2.6 with the einbein part denoted by f(x).



he's sloppy again. the V^mu is put in 2.8 i/o 2.12. Forgive him.

2.13 clarifies your problem. For a timelike geodesic, the lambda parameter is no longer arbitrary by proportional with the proper time. This means, either by 2.5 or 2.6 that the einbein is a mere constant wrt tau/lambda.
And, on a lighter note, can you put me out of my misery and explain the picture on the front cover of Wald to me?
 

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