2
$\begingroup$

If the metric is a (0,2)-tensor then it transforms twice covariently under a basis transformation, but in special relativity the metric seems to be defined globally as:

$$\eta_{\mu\nu}=\begin{pmatrix}-1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 &&1&&0\\0&&0&&0&&1\end{pmatrix}$$

in most texts I've seen introducing tensors when we change between basis we also have to change the metric, so why does the Minkowski metric not have to be transformed in the same way when we change our basis in Lorentz transformations:

$$\eta_{\mu\nu}'=\Lambda_{\mu}^{\nu}\Lambda_{\mu}^{\nu}\eta_{\mu\nu}$$

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
10
$\begingroup$

It does transform, but into itself. A Lorentz transformation is defined to leave the Minkowski metric unchanged after its transformation. That's what makes the speed of light invariant.

Improve this answer
$\endgroup$
5
  • $\begingroup$ That makes sense now thank you, however I've now written it out explicitly and the only way I can get the metric to be unchanged under the transformation is with the order $\eta'_{\mu\nu}=\Lambda_{\mu}^{\nu}\eta_{\mu\nu}\Lambda_{\mu}^{\nu}$, but in general with a (0,2)-tensor I see the transformation written as $\eta'_{\mu\nu}=\Lambda_{\mu}^{\nu}\Lambda_{\mu}^{\nu}\eta_{\mu\nu}$ when changing basis, why are the transformations on either side of the metric? From what I've seen this is what you would do for a (1,1)-tensor. $\endgroup$
    – Charlie
    Commented Dec 7, 2019 at 18:48
  • 1
    $\begingroup$ All of your tensor equations are invalid. You can have only two indices the same, not three. But I understand what you meant. In index notation the order in which tensors are written makes no difference. The order matters when you use matrix notation. $\endgroup$
    – G. Smith
    Commented Dec 7, 2019 at 18:55
  • $\begingroup$ Ah you're right my mistake, I've just been copy-pasting to save myself time, are you saying that those two equations should return the metric either way? I have done the multiplication by hand and used a matrix multiplier online and gotten an answer that doesn't return a diagonal matrix for the second order of multiplication in my comment above. $\endgroup$
    – Charlie
    Commented Dec 7, 2019 at 18:58
  • 2
    $\begingroup$ A discussion about how the tensor-index notation relates to the matrix notation would be too lengthy for these comments. In general, I recommend that you not get too attached to the matrix notation because it doesn’t work for tensors with three or more indices. Just think of the tensor index notation as a sum of terms, summing over each contracted index from 0 to 3 so you have 16 terms. When you have a simple sum of terms, the order of products in each term doesn’t matter. $\endgroup$
    – G. Smith
    Commented Dec 7, 2019 at 19:04
  • $\begingroup$ Ok thank you for your help! $\endgroup$
    – Charlie
    Commented Dec 7, 2019 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.