Riemann Tensor knowing Christoffel symbols (check my result)

  • #1
I need to find all the non-zero components of the Riemann Tensor in a two-dimensional geometry knowing that the only two non-zero components of the Christoffel symbols are:

[tex]\Gamma^x_{xx}=\frac{1}{x}[/tex] and [tex]\Gamma^y_{yy}=\frac{2}{y}[/tex]

knowing that: [tex]R^\alpha_{\beta\gamma\delta}=\partial_\gamma \Gamma^\alpha_{\delta\beta}-\partial_\delta \Gamma^\alpha_{\gamma\beta}+\Gamma^\epsilon_{\delta\beta}\Gamma^\alpha_{\gamma\epsilon}-\Gamma^\epsilon_{\gamma\beta}\Gamma^\alpha_{\delta\epsilon}[/tex]

The result I have obtained is that all the components of the Riemann curvature tensor are zero. Is this correct? If it is, what does it mean that all the components are zero?

MY PROCEDURE HAS BEEN:

the only plausible non-zero components of the Riemann curvature tensor are:
[tex]R^\alpha_{\beta\gamma\delta}=\partial_\gamma \Gamma^\alpha_{\delta\beta}-\partial_\delta \Gamma^\alpha_{\gamma\beta}+\Gamma^x_{\delta\beta}\Gamma^\alpha_{\gamma x}+\Gamma^y_{\delta\beta}\Gamma^\alpha_{\delta y}-\Gamma^x_{\gamma\beta}\Gamma^\alpha_{\delta x}-\Gamma^y_{\gamma\beta}\Gamma^\alpha_{\delta y}[/tex]

[tex]\alpha=x\quad:\quad R^x_{\beta\gamma\delta}=\partial_\gamma \Gamma^x_{\delta\beta}-\partial_\delta \Gamma^x_{\gamma\beta}+\Gamma^x_{\delta\beta}\Gamma^x_{\gamma x}+\Gamma^y_{\delta\beta}\Gamma^x_{\delta y}-\Gamma^x_{\gamma\beta}\Gamma^x_{\delta x}-\Gamma^y_{\gamma\beta}\Gamma^x_{\delta y}[/tex]

[tex]\text{ }\quad\quad \Longrightarrow \quad R^x_{xxx}=\partial_x \Gamma^x_{xx}-\partial_x \Gamma^x_{xx}+\Gamma^x_{xx}\Gamma^x_{xx}+\Gamma^y_{xx}\Gamma^x_{xy}-\Gamma^x_{xx}\Gamma^x_{xx}-\Gamma^y_{xx}\Gamma^x_{xy}=0[/tex]

[tex]\alpha=y\quad:\quad R^y_{\beta\gamma\delta}=\partial_\gamma \Gamma^y_{\delta\beta}-\partial_\delta \Gamma^y_{\gamma\beta}+\Gamma^x_{\delta\beta}\Gamma^y_{\gamma x}+\Gamma^y_{\delta\beta}\Gamma^y_{\delta y}-\Gamma^x_{\gamma\beta}\Gamma^y_{\delta x}-\Gamma^y_{\gamma\beta}\Gamma^y_{\delta y}[/tex]

[tex]\text{ }\quad\quad \Longrightarrow \quad R^y_{yyy}=\partial_y \Gamma^y_{yy}-\partial_y \Gamma^y_{yy}+\Gamma^y_{yy}\Gamma^y_{yx}+\Gamma^y_{yy}\Gamma^y_{yy}-\Gamma^y_{yy}\Gamma^y_{yx}-\Gamma^y_{yy}\Gamma^x_{yy}=0[/tex]

Therefore, all the components of the Riemann Tensor are zero.


Thanks!!!
 
Last edited:

Answers and Replies

  • #2
Orodruin
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Same comment as on your other thread.
 
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  • #3
Orodruin
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Your conclusion is correct and your argument holds. However, your argument would be made more concise if, instead of checking the cases for ##\alpha##, you used the antisymmetry of the curvature in the two last indices.


If it is, what does it mean that all the components are zero?
Are you familiar with the geometrical interpretation of the curvature tensor?
 
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  • #4
Are you familiar with the geometrical interpretation of the curvature tensor?
No, not really. Could you illustrate it for me? Thank you!!
 
  • #5
Orodruin
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How was the curvature tensor introduced to you?
 
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  • #6
How was the curvature tensor introduced to you?
Purely mathematical. Two years ago I took a course on differential geometry. It wasn't until this year I started studying General Relativity, but I'm learning it on my own.
 
  • #7
Orodruin
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Purely mathematical.
This does not help. Geometry (also differential geometry) is a mathematics subfield so I would think even a ”pure maths” intro would mention the geometrical interpretation. The question I am asking is how it was presented to you.
 
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  • #8
This does not help. Geometry (also differential geometry) is a mathematics subfield so I would think even a ”pure maths” intro would mention the geometrical interpretation. The question I am asking is how it was presented to you.
It was presented to me by its deffinition with Christoffel symbols. I was never explained the physical meaning behind it (geometrical meaning) to help me imagine it.
 
  • #9
Orodruin
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Now that is just bad teaching. The curvature tensor is related to the change of a vector that is parallel transported around a loop. This should be discussed in any introductory text on differential geometry.
 
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  • #10
haushofer
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See e.g. chapter 3 of Carroll's notes,

https://arxiv.org/abs/gr-qc/9712019

page 74 (eqn. 3.63) and 75 (eqn.3.65). In short, the Riemann tensor tells you how the orientation of a vector parallelly transported around an infinitesimal loop is changed. This is expressed as the commutator of the two corresponding covariant derivatives. To go with Orodruin, it's hard for me to believe you've never seen this definition; it's like teaching calculus and integration without mentioning the Riemann sum or definition of a derivative.
 
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  • #11
Orodruin
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This is expressed as the commutator of the two corresponding covariant derivatives
This is only true for vector fields that commute (such as the holonomic basis vectors). If the fields do not commute the loop does not close and you need to add a contribution from going along the curve that closes the loop, proportional to the commutator of the fields. The full expression is ##R(X,Y)Z = (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z##.
 
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  • #12
haushofer
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Yes, I was a bit too implicit.
 
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  • #13
stevendaryl
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This is only true for vector fields that commute (such as the holonomic basis vectors). If the fields do not commute the loop does not close and you need to add a contribution from going along the curve that closes the loop, proportional to the commutator of the fields. The full expression is ##R(X,Y)Z = (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z##.
@haushofer was talking about parallel-transporting vectors, not vector fields. Can we equally well take parallel transport of vectors to be fundamental, and define covariant derivatives of vector fields in terms of parallel transport, or take covariant derivatives of vector fields as fundamental and define parallel transport of vectors in terms of that?
 
  • #14
Orodruin
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@haushofer was talking about parallel-transporting vectors, not vector fields
This is a bit of a funny point. In order for ##\nabla_X Z## and ##[X,Y]## to be defined, ##X, Y, Z## need to be vector fields. However, the properties of the objects actually result in the curvature depending only on the vectors at the point in question (if not, it would not be a tensor). Either way, the parallel transport is of a vector ##Z## around a loop spanned by ##X## and ##Y##.

Can we equally well take parallel transport of vectors to be fundamental, and define covariant derivatives of vector fields in terms of parallel transport, or take covariant derivatives of vector fields as fundamental and define parallel transport of vectors in terms of that?
Are you asking if the parallel transport equations have a one-to-one correspondence with the connection? The answer to that question is yes. If you know how vectors behave under (all possible) parallel transports, then you know what the connection is, since you will know (in particular) ##\nabla_{\dot\gamma}X## for all curves ##\gamma## and all fields ##X## along ##\gamma##.
 
  • #15
stevendaryl
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This is a bit of a funny point. In order for ##\nabla_X Z## and ##[X,Y]## to be defined, ##X, Y, Z## need to be vector fields.
It seems to me that ##\nabla_X Z## only requires that ##Z## be a vector field, not ##X##.
 
  • #16
Orodruin
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It seems to me that ##\nabla_X Z## only requires that ##Z## be a vector field, not ##X##.
Yes, unfortunate formulation, but in the definition of the curvature, all fields have derivatives acting on them and therefore need to be fields. The point is it does not matter what the fields are as long as they agree with the corresponding vectors at the relevant point.
 
  • #17
stevendaryl
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Yes, unfortunate formulation, but in the definition of the curvature, all fields have derivatives acting on them and therefore need to be fields. The point is it does not matter what the fields are as long as they agree with the corresponding vectors at the relevant point.
I would say that the definition of curvature in terms of covariant derivatives requires all three arguments to be vector fields. But if you define curvature in terms of parallel transport, they don't need to be vector fields, it seems to me.
 
  • #18
Orodruin
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I would say that the definition of curvature in terms of covariant derivatives requires all three arguments to be vector fields. But if you define curvature in terms of parallel transport, they don't need to be vector fields, it seems to me.
Well, the point is that the curvature tensor is a tensor. It does not depend on the vectors it take being fields. You can extend the vectors ##XYZ## to arbitrary vector fields at will and ##R(X,Y)Z## will still be the same. As such, it is clear that the definition of the curvature tensor only requires vectors in ##T_p M##, not sections of ##TM##. However, expressing the tensor in terms of the commutator of the covariant derivatives, as done in #10, does require the extension of ##XYZ## to sections (even if the actual extensions are irrelevant). Either way, the parallel transport - connection link is there.
 

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