Manifolds: local & global coordinate charts

  • #26
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It is a straight-forward generalisation of the corresponding SR expression. It is going to locally correspond to the same thing - the product of the wave vector with the observer 4-velocity.
I'm ashamed to say that I've never seen the derivation of the expression in SR either. Would you be able to show me, and/or suggest a book to read up on it?!

Construct the parallel transport around a loop as the contributions from several infinitesimal loops. This can be done since Minkowski space is simply connected. The change in the parallel transported vector around an infinitesimal loop is proportional to the curvature tensor, which for a flat space is zero everywhere.
Is it essentially then, that ##\delta v^{\mu}\propto R^{\mu}_{\;\nu}v^{\nu}## (where ##\delta v^{\mu}## denotes the change in the components of a vector under parallel transport around an infinitesimal loop and ##R^{\mu}_{\;\nu}## is the Ricci curvature tensor) and for a flat space ##R^{\mu}_{\;\nu}=0##, hence ##\delta v^{\mu}=0## and the components of a vector are invariant under parallel transport?!
 
  • #27
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I'm ashamed to say that I've never seen the derivation of the expression in SR either. Would you be able to show me, and/or suggest a book to read up on it?!
If you are considering a photon, it is not purely SR. You then also need to add QFT on top. If you want to do it hand-wavingly you can refer to the de Broglie frequency of the photon and study the corresponding 4-frequency, whose zeroth component is frequency that according to de Broglie theory is proportional to energy.

But if you want to fo it properly for a quantum particle, you need to study QFT. Referring to photons in SR without using QFT is often quite misleading. You may be better off computing the energy momentum tensor from classical electrodynamics to find the energy density and energy currents related to a particular wave.

Is it essentially then, that δvμ∝Rμνvνδvμ∝Rνμvν\delta v^{\mu}\propto R^{\mu}_{\;\nu}v^{\nu} (where δvμδvμ\delta v^{\mu} denotes the change in the components of a vector under parallel transport around an infinitesimal loop and RμνRνμR^{\mu}_{\;\nu} is the Ricci curvature tensor) and for a flat space Rμν=0Rνμ=0R^{\mu}_{\;\nu}=0, hence δvμ=0δvμ=0\delta v^{\mu}=0 and the components of a vector are invariant under parallel transport?!
No, this is true only in a two-dimensional manifold where the curvature tensor only has one independent component. In general, you will need the (rank 4) Riemann curvature tensor along with the directions that define the infinitesimal loops. But the concept is the same.
 
  • #28
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If you are considering a photon, it is not purely SR. You then also need to add QFT on top. If you want to do it hand-wavingly you can refer to the de Broglie frequency of the photon and study the corresponding 4-frequency, whose zeroth component is frequency that according to de Broglie theory is proportional to energy.

But if you want to fo it properly for a quantum particle, you need to study QFT. Referring to photons in SR without using QFT is often quite misleading. You may be better off computing the energy momentum tensor from classical electrodynamics to find the energy density and energy currents related to a particular wave
Ah ok, I'll have to look into that then. Thanks for all the info.

No, this is true only in a two-dimensional manifold where the curvature tensor only has one independent component. In general, you will need the (rank 4) Riemann curvature tensor along with the directions that define the infinitesimal loops. But the concept is the same.
Would it be something like
$$\delta v^{\mu}= \delta x^{\alpha}\delta x^{\beta}R^{\mu}_{\;\alpha\nu\beta}v^{\beta}$$ where ##\delta x^{\alpha}## and ##\delta x^{\beta}## quantify infinitesimal displacements along the two different directions around an infinitesimal closed loop.
 
  • #29
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Would it be something like
δvμ=δxαδxβRμανβvβδvμ=δxαδxβRανβμvβ​
\delta v^{\mu}= \delta x^{\alpha}\delta x^{\beta}R^{\mu}_{\;\alpha\nu\beta}v^{\beta} where δxαδxα\delta x^{\alpha} and δxβδxβ\delta x^{\beta} quantify infinitesimal displacements along the two different directions around an infinitesimal closed loop.
Something like that, but not exactly. Be careful with which indices you contract with what. With the definition ##R(\partial_a,\partial_b)\partial_c = R_{cab}^d \partial_d##, the contraction should be ##\delta v^a = \delta x^b \delta x^d R_{cbd}^a v^c##.
 
  • #30
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Something like that, but not exactly. Be careful with which indices you contract with what. With the definition ##R(\partial_a,\partial_b)\partial_c = R_{cab}^d \partial_d##, the contraction should be ##\delta v^a = \delta x^b \delta x^d R_{cbd}^a v^c##.
Ah ok, what would it be exactly? (I know that parallel transport along a curve would be ##\delta v^{\mu}=-\Gamma^{\mu}_{\;\nu\alpha}v^{\alpha}\delta^{\nu}##). Is the point essentially that, on a flat manifold, ## R_{cbd}^a=0##, and so ##\delta v^{\mu}=0##, i.e. the components of the vector remain unchanged under parallel transport in flat space?!
Yes, apologies about the indices, I was trying to think of it off the top of my head and couldn't exactly remember which indices contract with which :frown:
 
  • #31
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Ah ok, what would it be exactly?
What would what be? In order to derive the relation with the Riemann tensor, you need to do several parallel transports (4 to be exact) and remember to take into account that the Christoffel symbols will change (hence the appearance of the derivatives of the Christoffel symbols in the Riemann tensor).

Is the point essentially that, on a flat manifold, Racbd=0Rcbda=0 R_{cbd}^a=0, and so δvμ=0δvμ=0\delta v^{\mu}=0, i.e. the components of the vector remain unchanged under parallel transport in flat space?!
No, again we are talking about a parallel transport around a closed loop. If the components change during an arbitrary parallel transport depends on the coordinate system and whether the Christoffel symbols are zero in that particular system or not (remember that the Christoffel symbols being zero is not equivalent to zero curvature!). For example, the components of a vector in Euclidean space do change under parallel transport if you use spherical coordinates. The point is that when you parallel transport around a loop you end up with a vector in the same tangent space as you started in! (In other words, a closed loop defines a type (1,1) tensor at the start/end-point through parallel transport along it and so finding ##\delta v = 0## is a coordinate independent statement.)
 
  • #32
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What would what be? In order to derive the relation with the Riemann tensor, you need to do several parallel transports (4 to be exact) and remember to take into account that the Christoffel symbols will change (hence the appearance of the derivatives of the Christoffel symbols in the Riemann tensor).
Ah ok, I'll have to work through it then.

No, again we are talking about a parallel transport around a closed loop. If the components change during an arbitrary parallel transport depends on the coordinate system and whether the Christoffel symbols are zero in that particular system or not (remember that the Christoffel symbols being zero is not equivalent to zero curvature!). For example, the components of a vector in Euclidean space do change under parallel transport if you use spherical coordinates. The point is that when you parallel transport around a loop you end up with a vector in the same tangent space as you started in! (In other words, a closed loop defines a type (1,1) tensor at the start/end-point through parallel transport along it and so finding δv=0δv=0\delta v = 0 is a coordinate independent statement.)
Good point, I hadn't fully thought that through.

Thanks for all of your help, I really appreciate it!
 
  • #33
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Ah ok, I'll have to work through it then.
One thing to note is that it is usually much (much!) easier to work it through if you start the parallel transport at the middle of the loop and perform the parallel transport in the two different directions. This should also be done in any introductory text on the calculus on manifolds.
 
  • #34
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One thing to note is that it is usually much (much!) easier to work it through if you start the parallel transport at the middle of the loop and perform the parallel transport in the two different directions. This should also be done in any introductory text on the calculus on manifolds.
Thanks for the tips, are there any books that you would recommend in particular?

Also, one thing I'm slightly confused over now is, if one wishes to compare two vectors at different points in a flat space, then one can uniquely parallel transport one of the vectors to the other and compare them at the same point in a well defined manner. However, this is not around a closed loop and so the components of the parallel transported vector will change, in general (unless one uses Cartesian coordinates), so how can one meaningfully compare the two vectors (for example, suppose it is the same vector, but at two different points, with the same components at both points)? (Apologies, this may be a stupid question - it's a bit late at night and my brain has gone a bit to mush)
 
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  • #35
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However, this is not around a closed loop and so the components of the parallel transported vector will change, in general (unless one uses Cartesian coordinates), so how can one meaningfully compare the two vectors (for example, suppose it is the same vector, but at two different points, with the same components at both points)?
If the vectors are in the same vector space, you can compare them using the normal addition/subtraction in that vector space. It does not matter what the components are in some arbitrary coordinate system. You simply do the parallel transport and compare the result with the other vector at the point.
 
  • #36
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Also, one thing I'm slightly confused over now is, if one wishes to compare two vectors at different points in a flat space, then one can uniquely parallel transport one of the vectors to the other and compare them at the same point in a well defined manner. However, this is not around a closed loop and so the components of the parallel transported vector will change, in general (unless one uses Cartesian coordinates), so how can one meaningfully compare the two vectors (for example, suppose it is the same vector, but at two different points, with the same components at both points)? (Apologies, this may be a stupid question - it's a bit late at night and my brain has gone a bit to mush)
Am I just being stupid here, since it is natural that, in a non Cartesian coordinate basis, the basis vectors will vary from point to point so one would expect the components of a vector (with respect to this basis) to vary as one parallel transports the vector from one point to another, in order to keep it parallel to itself. The important point of why one can compare two vectors residing in different tangent spaces (at different points) in flat space is that the path connecting the two tangent spaces, along which one parallel transports one vector to the other to compare them, is unique, and so the comparison of vectors residing in different tangent spaces (in flat space) is a well defined concept?!
(Of course, in a curved space, it is not meaningful to compare vectors in two different tangent spaces (at least from a physical perspective) since the path one parallel transports one vector to the other along is not unique and so it is not well defined, since parallel transport along different paths will yield different results).
 
  • #37
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If the vectors are in the same vector space, you can compare them using the normal addition/subtraction in that vector space. It does not matter what the components are in some arbitrary coordinate system. You simply do the parallel transport and compare the result with the other vector at the point.
Ah ok. Would what I put in my post above (post #36) be correct at all then?
 
  • #38
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The important point of why one can compare two vectors residing in different tangent spaces (at different points) in flat space is that the path connecting the two tangent spaces, along which one parallel transports one vector to the other to compare them, is unique, and so the comparison of vectors residing in different tangent spaces (in flat space) is a well defined concept?!
No, the important point is that the parallel transport is independent of the path (again, as long as the manifold is simply connected). Therefore it does not matter which path you select, the result will be the same.
 
  • #39
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No, the important point is that the parallel transport is independent of the path (again, as long as the manifold is simply connected). Therefore it does not matter which path you select, the result will be the same.
So, just to check, in the case of a more general manifold (with a "curved" geometry), parallel transport is not independent of the path taken between two tangent spaces and so comparison of vectors at two different points is not well defined, in the sense that the result depends on the path taken?! Is this why in GR, the velocity of a distant galaxy, for example, is not well defined, since the result is dependent on the path we use to parallel transport its velocity vector to the tangent space to our location?
 
  • #40
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So, just to check, in the case of a more general manifold (with a "curved" geometry), parallel transport is not independent of the path taken between two tangent spaces and so comparison of vectors at two different points is not well defined, in the sense that the result depends on the path taken?!
Right.

Is this why in GR, the velocity of a distant galaxy, for example, is not well defined, since the result is dependent on the path we use to parallel transport its velocity vector to the tangent space to our location?
Yes.
 
  • #41
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Right.


Yes.
Ok, great. I think things are starting to become a little clearer now. Thanks!
 

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