- #1

- 241

- 10

g

^{cd}g

_{ac}(dv

^{a}/ds) becomes (dv

^{d}/ds)

I seems to me that that only works if the metric matrix is diagonal.

(1) Is that correct?

(2) If so, that doesn't seem to be a legitimate limitation on the property of geodesics??

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- #1

- 241

- 10

g

I seems to me that that only works if the metric matrix is diagonal.

(1) Is that correct?

(2) If so, that doesn't seem to be a legitimate limitation on the property of geodesics??

- #2

Science Advisor

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(1) Is that correct?

No. ##g^{cd}g_{ac} = \delta^{d}{}{}_{a}## by definition of the inverse of any metric tensor, hence the desired result.

- #3

- 241

- 10

PS. Must the metric matrix always be symmetric? For example, a skew-symmetric matrix can produce the same ds^2 interval as a diagonal. Just curious.

- #4

Mentor

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- 8,690

Must the metric matrix always be symmetric? For example, a skew-symmetric matrix can produce the same ds^2 interval as a diagonal. Just curious.

It gives you the same ##ds^2## as some diagonal matrix, but it won't work for calculating the inner product of two different vectors.

- #5

Science Advisor

Gold Member

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PS. Must the metric matrix always be symmetric? For example, a skew-symmetric matrix can produce the same ds^2 interval as a diagonal. Just curious.

A metric is, by definition, symmetric. :)

- #6

Mentor

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it won't work for calculating the inner product of two different vectors.

To expand on this just a bit, the metric must be symmetric because the inner product is commutative; ##g(a, b) = g(b, a)## for any two vectors ##a## and ##b##. If you write this out in components, you get that ##g## must be a symmetric matrix.

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