## Christoffel Symbol

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3. The attempt at a solution

Does the Christoffel symbol $$\Gamma$$ have a dimension in physics? And if it does, what is its dimension?

Thank you!
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 The Christoffel symbols are arrays of real numbers. They are dimensionless.

 Quote by physicus The Christoffel symbols are arrays of real numbers. They are dimensionless.
Why do you say this? Since the metric tensor is dimensionless, and the Christoffel symbols are the derivatives of the metric tensor with respect to the coordinates, shouldn't the Christoffel symbols have dimensions of 1/length?

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## Christoffel Symbol

To OP: Can you think of any equation which contains the Christoffel symbols? That should help you answer the question.

 Quote by TSny To OP: Can you think of any equation which contains the Christoffel symbols? That should help you answer the question.
Yes, I can think of this equation:

$$\frac{\partial e_i}{\partial U^j} = \Gamma_{ij}^k e_k$$

But not really sure is that makes it easier for me. I can certainly think of the derivatives case for the christoffel symbols. I had decided before anyone posted here, that the symbol would have 1/length because of the metric derivatives. But the responses here seem to be... a bit mixed.
 I am reading that the Rindler Christoffel Symbols are dimensionless.
 The curvature tensor is made up completely of christoffel symbols, you'd expect that to have a dimension of length.

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 Quote by help1please Yes, I can think of this equation: $$\frac{\partial e_i}{\partial U^j} = \Gamma_{ij}^k e_k$$ But not really sure is that makes it easier for me. I can certainly think of the derivatives case for the christoffel symbols. I had decided before anyone posted here, that the symbol would have 1/length because of the metric derivatives. But the responses here seem to be... a bit mixed.
I'm not familiar with the Uj notation, but your reasoning based on the derivatives of the metric is correct. The Christoffel symbols have dimension of 1/length.

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 Quote by help1please The curvature tensor is made up completely of christoffel symbols, you'd expect that to have a dimension of length.
If I look at how the curvature tensor is constructed from Christoffel symbols and their derivatives, I don't get that the curvature tensor has dimension of length. What is the dimension of the product of two Christoffel symbols? What is the dimension of the derivative of a Christoffel symbols with respect to a coordinate xj?
 $$(\partial_{\nu} \Gamma_\mu - \Gamma_\mu \partial_\nu) - (\partial_{\mu} \Gamma_\nu - \Gamma_\nu \partial_\mu) + \Gamma_\nu \Gamma_\mu - \Gamma_\mu \Gamma_\nu$$ Bracketed expressions are the derivatives of connection $$\frac{\partial \Gamma_\mu}{\partial x^{\nu}} - \frac{\partial \Gamma_\nu}{\partial x^{\mu}} + \Gamma_\nu \Gamma_\mu - \Gamma_\mu \Gamma_\nu$$ I know the first lot of terms from the curvature tensor are zero] $$\partial_{\mu}\partial_{\nu} - \partial_{\nu}\partial_{\mu}$$
 But,you seem to be telling me that the dimensions of the Christoffel symbols are indeed 1/length?

 Quote by TSny I'm not familiar with the Uj notation, but your reasoning based on the derivatives of the metric is correct. The Christoffel symbols have dimension of 1/length.
U^j comes from U^iU^j U^k for curvilinear coordinates.

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 Quote by help1please But,you seem to be telling me that the dimensions of the Christoffel symbols are indeed 1/length?
They have dimension of 1/length if all the coordinates xj have the dimension of length. Without thinking, that's what I was assuming .

However, generally the coordinates do not need to have dimension of length (for example, in spherical coordinates where some of the coordinates are angles.) For a given coordinate system it is possible for different Christoffel symbols to have different dimesions.
 Also, since the Christoffel symbols have dimension 1/length, the curvature tensor has dimensions 1/(length^2). This is exactly what you'd expect. Consider the Gaussian curvature of a 2D surface - it is the product of the two principle curvatures. Since each principle curvature is equal to 1/(radius of curvature), the Gaussian curvature has dimensions 1/(length^2).

 Quote by phyzguy Also, since the Christoffel symbols have dimension 1/length, the curvature tensor has dimensions 1/(length^2). This is exactly what you'd expect. Consider the Gaussian curvature of a 2D surface - it is the product of the two principle curvatures. Since each principle curvature is equal to 1/(radius of curvature), the Gaussian curvature has dimensions 1/(length^2).
Right, I understand that. Thank you.
 Recognitions: Gold Member Homework Help Look here to see some explicit examples of Christoffel symbols and curvature tensor components: http://onlinelibrary.wiley.com/doi/1...2061.app12/pdf For example, if you use spherical coordinates in flat spacetime you can see that some of the Christoffel symbols have dimension 1/length while others are dimensionless (see equations L.14). All the curvature tensor components are zero in this case because the spacetime is flat. For the Schwarzchild metric, again you find some of the Christoffel symbols have dimesion 1/length while others are dimensionless (see equations L.21). Some components of the curvature tensor have dimension 1/length^2, while others are dimensionless (see equations L.22). [The symbol m in these expressions has the dimension of length.]
 ty very much

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