To review my special relativity I tried to work out the inverse lorentz transformation explicitly. This led to a lot of confusion; I would like to ask what the issue was with the assumptions I made in the last steps & what the actual expression should be, in index notation. In other SE posts it seems index notation is avoided, maybe precisely because of this confusion.

The derivation follows. We start with

$$\Lambda^\alpha_{\,\,\mu} \,g_{\alpha \beta} \Lambda^\beta _{\,\,\nu} = g_{\mu \nu}$$

Contract with/multiply by $g^{\nu \sigma}$

$$\Lambda ^\alpha_{\,\,\mu} \, g_{\alpha \beta} \Lambda^\beta _{\,\,\nu} g^{\nu \sigma}= g_{\mu \nu}g^{\nu \sigma}=\delta_\mu ^{\,\,\sigma}$$

Write $g_{\alpha \beta} \Lambda^\beta _{\,\,\nu} g^{\nu \sigma}=\Lambda_\alpha^{\,\,\sigma}$. Then we have

$$\Lambda^\alpha _{\,\,\mu}\Lambda_\alpha^{\,\,\,\sigma}=\delta_\mu ^{\,\,\sigma} \tag 1$$

We would like to identify this as matrix multiplication, and thus identify the rightmost $\Lambda$ in this equation as the inverse. To do this, the inner indices need to be the same. Therefore we first take the transpose: $$\Lambda_\alpha^{\,\,\sigma}=(\Lambda^T)_\sigma^{\,\,\alpha} \tag 2$$

Note 1: The answer to this question is that $(2)$ was incorrect and we actually have $\Lambda_\alpha^{\,\,\sigma}=(\Lambda^T)^\sigma_{\,\,\alpha}$. The correct derivation continues with Equation $(3)$.

Plugging into $(1)$ gives

$$\Lambda^\alpha _{\,\,\mu}(\Lambda^T)_\sigma^{\,\,\alpha}=(\Lambda^T)_\sigma^{\,\,\alpha}\Lambda^\alpha _{\,\,\mu}=\delta_\mu ^{\,\,\sigma}$$

...which should be the answer. However, ever since taking the transpose, this equation is just plain weird: the indices do not match on the RHS and LHS so it is no longer lorentz covariant.

This suggests that taking the transpose actually implicitly inverts the transformation properties of each index, changing upper <-> lower in addition to just permuting them:


In this case, we have

$$\Lambda^\alpha _{\,\,\mu}(\Lambda^T)^\sigma_{\,\,\alpha}=(\Lambda^T)^\sigma_{\,\,\alpha}\Lambda^\alpha _{\,\,\mu}=\delta_\mu ^{\,\,\sigma} \tag 3$$

However, now there is a new problem: It doesn't work! As noted in this question, if you write the matrix representation of a Lorentz boost, the transpose is clearly not the inverse.

So, how can I write the components of the inverse? And what went wrong here? Also greatly appreciated would be a source where all of this is very explicitly done. I did not find MTW or Griffiths especially clear.

Note 2: I submitted an answer which explains that Equation $(3)$ is in fact consistent.


3 Answers 3


I'm sorry to post my own answer, but to be very honest I don't think that either of the answers really got to the heart of the question. However, they did inspire more thought, and after also reading more I think I have understood what is going on. I will post it here and leave it open to criticism.

Question #1: Which is the correct form of the transpose? $$\Lambda_\alpha^{\,\,\sigma}=(\Lambda^T)_\sigma^{\,\,\alpha}? \,\,\,\,\,\,\,\,\,\,\text{(just swap indices)}$$ $$\Lambda_\alpha^{\,\,\sigma}=(\Lambda^T)^\sigma_{\,\,\alpha}? \,\,\,\,\,\,\,\,\,\text{(keep the co/contravariance)} $$

The second form of the transpose is in fact correct: otherwise indices are inconsistent, and QMechanic also states it in his answer here. Thus the transpose reads:


Previously, I had thought of this as permuting the indices and then swapping each of them from co- to contra-variant afterwards. I think the better interpretation is that when swapping the order of indices, upper indices are still upper and lower indices are still lower (simply swapping them in place does not give this).

Question #2: With this understanding of the transpose, the result was

$$(\Lambda^T)^\sigma_{\,\,\alpha}\Lambda^\alpha _{\,\,\mu}=\delta_\mu ^{\,\,\sigma}$$

but the transpose of a boost matrix does not give its inverse (check by writing explicitly). Isn't this wrong?

It is not wrong. The subtlety here is in the indices: If the leftmost $\Lambda$ above were just the transpose of the rightmost $\Lambda$, the indices would be written $\Lambda_\sigma ^{\,\,\alpha}$ because the transpose does not leave them intact. Therefore the left $\Lambda$ is not the transpose of the right $\Lambda$, rather it is the transpose of the matrix $g\Lambda g^{-1}$ with components $g_{\alpha \beta}\Lambda ^\beta _{\,\,\gamma}g^{\gamma \sigma}$. Essentially it was just a confusion of notation. To make this explicit, one could write

$$((g\Lambda g^{-1})^T)^\sigma_{\,\,\alpha}\Lambda^\alpha _{\,\,\mu}=\delta_\mu ^{\,\,\sigma}$$

or just understand that in the equation appearing in Question #2, since the indices don't match the transposed indices, the multiplication of $g$ on both sides is implicit.

As for using the inverse, the most useful thing seems to be just to remember the earlier condition, copied below.

$$\Lambda^\alpha _{\,\,\mu}\Lambda_\alpha^{\,\,\,\sigma}=\delta_\mu ^{\,\,\sigma}$$

..even though the matrix multiplication is weird. This is exactly the expression you'd get if trying to transform a Lorentz-covariant expression like $a_\alpha b^\alpha$


In addition to the suggestion of walber97, I'd also propose that the transpose should not lower or raise indices. It just interchanges the order of the indices. So your 4th equation should perhaps be $$\Lambda^\alpha_{\,\,\mu}=(\Lambda^T)_\mu^{\,\,\alpha} . $$ The matrix multiplication then remains as a contraction between an upper and a lower pair of indices $$(\Lambda^T)_\mu^{\,\,\alpha} \Lambda_\alpha^{\,\,\,\sigma} = \delta_\mu^{\,\,\sigma}$$ Then the Lorentz transformation properties remain in tact.

  • $\begingroup$ I had actually considered this possibility in my original question :) I had just run into other problems when going that route which i wanted to ask about. But yes, this is the correct way. $\endgroup$ Commented Jan 26, 2019 at 23:46
  • $\begingroup$ The Lorentz transform is $not$ a tensor - it is a linear transformation where the origin doesn't change. The correct method is per @walber97 answer. The link by Qmechanic was for $(1,1)$ tensor. The answer by doublefelix is incorrect in the sense it implies the Lorentz transformation is a $(1,1)$ tensor. $\endgroup$ Commented Dec 9, 2019 at 20:32
  • $\begingroup$ Opps - I meant the answer by flippiefanus was incorrect in the sense it treats the Lorentz transform as a $(1,1)$ tensor - you're doublefelix. $\endgroup$ Commented Dec 9, 2019 at 20:40

I think you are wrong in the third equation.Instead of writing,
$$g_{\alpha \beta}\Lambda^{\beta} _{\nu}g^{\nu\sigma}=\Lambda_{\alpha}^{\sigma}$$
You should have to write $$g_{\alpha \beta}\Lambda^{\beta} _{\nu}g^{\nu\sigma}=(\Lambda^{-1}) _{\alpha}^{\sigma}.$$
(As per the question link you have shared)

  • $\begingroup$ I think it's a bit confusing to use the notation $\Lambda^{-1}$ before getting the result, that is what I wanted to derive. Also, the notation I used is common (see wikipedia for lorentz trafo for example, in the "Covariant Vectors" section). After your comment I used the notation $L$ instead, which was helpful to me though, so thank you. $\endgroup$ Commented Jan 26, 2019 at 22:36

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.