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How can one prove that $\mathbf{\Lambda^T \eta \Lambda} = \mathbf{\eta}$ in special relativity, where $\mathbf{\Lambda}$ is the Lorentz transformation and $\eta$ is the Lorentz metric? Also, how does this correspond to the intuitive definition of the invariance of the Lorentz metric that is $s(\mathbf{a}, \mathbf{b}) = s(\mathbf{\Lambda a}, \mathbf{\Lambda b})$?

If we could stay clear of anything explicitly related to tensors or Einstein notation here, that would be great, since they haven't been introduced at this stage in the course that I'm taking.

Qmechanic
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Noldorin
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  • possible duplicate? https://physics.stackexchange.com/q/230495/84967 – AccidentalFourierTransform Mar 02 '21 at 23:39
  • @AccidentalFourierTransform Kind of, but not so much, because it uses the advanced tensor notation. I was hoping for a fairly succinct answer. – Noldorin Mar 02 '21 at 23:44
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    I wrote the top answer; it does not use advanced tensor notation, and honestly I don't think you should expect an answer more elementary and low-level than that one. But I am not voting to close this post. Perhaps someone will come up with an even simpler answer! – AccidentalFourierTransform Mar 02 '21 at 23:50
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    See $\boldsymbol \S,$ A in my answer here : Show that any proper homogeneous Lorentz transformation may be expressed as the product of a boost times a rotation. – Frobenius Mar 02 '21 at 23:54
  • See Andrew Steane’s answer from yesterday: https://physics.stackexchange.com/q/617752/ – G. Smith Mar 02 '21 at 23:55
  • I'm afraid they're both over my head. – Noldorin Mar 03 '21 at 00:05
  • Have you tried writing out a boost and the Minkowski metric as matrices and doing the calculation explicitly? It's not exactly a general proof but you can at least see that the relationship holds for that example. If you haven't covered enough material in your course yet to understand the linked answers perhaps it would be best to wait until you have. – Charlie Mar 03 '21 at 00:13
  • @AccidentalFourierTransform I'll have a read of your answer there in any case, as it does look generally elucidative, even if Thomas Fritsch's answer below is perhaps the most direct answer to my question. Cheers. – Noldorin Mar 03 '21 at 01:28
  • @Charlie Yeah, it works, but it's terribly unenlightening. – Noldorin Mar 03 '21 at 01:41
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    A general remark for all iterations of this question: you have to say what you are taking as "special relativity". Weinberg, for example, starts with $\Lambda^T\eta\Lambda = \eta$ as the basic postulate of special relativity, so it is just axiomatically true -- you can then only talk about how you derive results from it and whether they match with our other conceptual/empirical considerations. I am not saying that thus, this question is moot -- but it is just hard for one to answer when the question doesn't mention what they are starting with. –  Mar 03 '21 at 03:43
  • @DvijD.C. That's fair enough. I stated it in another question on the SE that led to this, but indeed I should have made it clear here that I was assuming Einstein's original postulates, with the derivation of the Lorentnz transformation (via linear algebra) taken as a given. – Noldorin Mar 03 '21 at 03:48

2 Answers2

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Let $\mathbf{a}$ and $\mathbf{b}$ be some arbitrary 4-vectors. Begin with the invariance of the product $s(\mathbf{a}, \mathbf{b})$ with respect to Lorentz transformation $\mathbf{\Lambda}$: $$s(\mathbf{a}, \mathbf{b}) = s(\mathbf{\Lambda a}, \mathbf{\Lambda b})$$

Using the definition $s(\mathbf{a}, \mathbf{b})=\mathbf{a}^T \eta \mathbf{b}$ both for the left and the right side we get: $$\begin{align} \mathbf{a}^T \eta \mathbf{b} &= (\mathbf{\Lambda}\mathbf{a})^T \eta \mathbf{\Lambda}\mathbf{b} \\ &= \mathbf{a}^T \mathbf{\Lambda}^T \eta \mathbf{\Lambda}\mathbf{b} \end{align}$$

Since the above is true for every $\mathbf{a}$ and $\mathbf{b}$ we can immediately conclude: $$\eta = \mathbf{\Lambda}^T \eta \mathbf{\Lambda}$$

Thomas Fritsch
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  • Uff, how very simple! I should have seen this myself. Thank you, in any case. – Noldorin Mar 03 '21 at 01:27
  • Actually, if I'm not mistaken, there's a little error above (in notation, essentially). We should instead write that $s(0, \mathbf{a}) = \mathbf{a}^T \mathbf{\eta} \mathbf{a}$, right? The same argument still works, however. – Noldorin Mar 03 '21 at 01:56
  • Or even $s(\mathbf{a}, \mathbf{b}) = (\mathbf{b - a})^T \mathbf{\eta} (\mathbf{b - a})$, if one prefers, right? Thought I think the former is nice. – Noldorin Mar 03 '21 at 01:58
  • $s(\mathbf{a}, \mathbf{b}) = \mathbf{a}^T\eta\mathbf{b}$ is correct. $s(\mathbf{0}, \mathbf{a}) = \mathbf{0}^T\eta\mathbf{a} = 0$. – Adam Zalcman Mar 03 '21 at 01:59
  • @AdamZalcman That's certainly not the in agreement with the definition of the Lorentz metric though. $s$ is meant to be the Lorentz metric (as a function of two events), per my original post. How are you interpreting it? – Noldorin Mar 03 '21 at 02:00
  • Oh, sorry, I just read the answer and was under the impression that $s$ was Minkowski inner product. In any case, metric and norm can be expressed in terms of $s$ in the usual way, so any transformation preserving one preserves the other (IIRC, the converse follows from a polarization identity). – Adam Zalcman Mar 03 '21 at 02:09
  • @AdamZalcman Ah, no worries. I probably should have been clearer. Thomas was likely under the same impression. I didn't know such a thing as the "Minkowski inner product" even existed until now — does it have any physical significance to take the Minkowski inner product between two different vectors, i.e., $\mathbf{x^T \eta y} $? – Noldorin Mar 03 '21 at 02:13
  • does it have any physical significance to take the Minkowski inner product between two different vectors Yes. In this answer, which you accepted, $\mathbf a$ and $\mathbf b$ are two different vectors. – G. Smith Mar 03 '21 at 03:52
  • In my opinion the first requirement is to keep the "indefinite norm" of a 4-vector $:\mathbf{a}:$ invariant : $s(\mathbf{a}, \mathbf{a}) = s(\mathbf{\Lambda a}, \mathbf{\Lambda a})$. That this norm is induced by the inner product $s(\mathbf{a}, \mathbf{b}) = \mathbf{a}^T\eta\mathbf{b}$ which it's proved to be also invariant, $s(\mathbf{a}, \mathbf{b}) = s(\mathbf{\Lambda a}, \mathbf{\Lambda b})$, is a consequence. – Frobenius Mar 03 '21 at 06:47
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    @Frobenius Yeah, although I did state that $s$ was the Lorentz metric in my original question, that also works. (The usual way to write the norm would be $|\mathbf{a}|$ or $||\mathbf{a}||$, as well.) Nonetheless, since the inner product, norm, and metric can all be defined in terms of each, it doesn't really matter which we use here. – Noldorin Mar 03 '21 at 16:05
  • @G.Smith That still hasn't highlighted any physical significance. This is just a side-point though; something I'm curious about but is not directly related to my question. The Minkowski metric has a physical significance because it it both an extension of Euclidean distance to 1+3d spacetime, in a sense, and it is the square of the proper time elapsed between two events. – Noldorin Mar 03 '21 at 16:06
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$x^u x_u = n_{u\beta} x^{\beta} x^u = n_{u \beta} \Lambda^u _\gamma x^\gamma \Lambda^\beta _\omega x^\omega = n_{u \beta} \Lambda^\beta _\omega \Lambda^u _\gamma x^\gamma x^\omega$

$[m] \equiv$ matrix of m.

By the definition of matrix multiplication,

$n_{u \beta} \Lambda^\beta _\omega = n \Lambda =[n \Lambda]^u _\omega$

$\to$

$\Lambda^u _\gamma [n \Lambda]^u _\omega=[\Lambda^T]^{\gamma} _u [n \Lambda]^u _\omega = [\Lambda^T n \Lambda]^\gamma _{\omega}$

and

$x^T [n]^u _b x = \text{a number} =\overline{x}^T [\Lambda^T n \Lambda]^\gamma _{\omega} \overline{x}$

$\to$

$ [n]^u _b = [\Lambda^T n \Lambda]^\gamma _{\omega}$

because these matrices define the same dot product in their respective bases.

$n = \Lambda^T n \Lambda$.