3
$\begingroup$

So, in the case of special relativity, we look for transformations relating inertial coordinates that leave the spacetime interval invariant and these transformations turn out to be generated by three Lorentz boosts and three spatial rotations.

But at the same time, I couldn't understand why does this count as a restriction as the spacetime interval is supposed to be a scalar (it has two covariant indices that come from the metric and two covariant indices).

Now for any coordinate transformation, a scalar shouldn't change. Therefore, the invariance of the spacetime interval should be an obvious deduction, given that it is a scalar and not a constraint imposed by Lorentz transformations.

I am sure that there is some mistake in this interpretation, could someone point it out?

Edit: So, we look for transformations that leave the spacetime interval invariant. If I write in index notation: $$ dS^2 = \eta_{\mu\nu}dx^{\mu}dx^{\nu} $$ Now suppose under a coordinate transform, $$ x^{\mu} \rightarrow x^{\mu'}=\Lambda^{\mu}_{\nu}x^{\nu}$$

$$dS^2 \rightarrow dS'^2 = \eta_{\mu'\nu'}dx^{\mu'}dx^{\nu'}$$ For the speed of light to be invariant as we go from one inertial frame to another, we demand that these transformations be such that, $$dS^2 = dS'^2$$ The condition we get from this, $$\Lambda^T\eta\Lambda = \eta$$ This characterises Lorentz transformations. But my confusion is the following: Saying, $dS^2 = dS'^2$, to me seems like something that should be always true since it is a scalar. How is it different from saying $\Phi(x^{\mu}) = \Phi'(x^{\mu'})$, where $\Phi$ is a scalar-valued function.

Edit2: Am I already restricting the coordinate transformations I can take if I demand, $\Phi(x^{\mu}) = \Phi'(x^{\mu'})$?

$\endgroup$
  • $\begingroup$ I'm not quite seeing what you are getting at. You might need to clarify. It sounds to me like you are saying "the invariance of the interval leads to the Lorentz transformations, but why does the Lorentz transformation leave the interval invariant?" I must be not be reading what you wrote in the way you intended. $\endgroup$ – garyp Jun 13 at 5:36
  • $\begingroup$ I will try to clarify, what I meant. $\endgroup$ – Yorozuya Jun 13 at 6:14
  • $\begingroup$ The infinitesimal line interval is a differential $2$-form. $\endgroup$ – Cinaed Simson Jun 13 at 6:26
  • $\begingroup$ @CinaedSimson, I thought the metric would be the differential 2-form, which takes two vectors and maps them to a scalar? I apologise if I'm wrong since I don't understand the language of differential geometry very well. $\endgroup$ – Yorozuya Jun 13 at 6:45
  • 1
    $\begingroup$ Try this in one dimension. $x$ is a real number. $x^2$ is also a real number, i.e. a scalar. Replace $x$ with $2x$. Is the value of $x^2$ unchanged? $\endgroup$ – WillO Jun 13 at 11:23
3
$\begingroup$

You are completely correct that the invariance of the spacetime interval under any arbitrary coordinate transformation follows from the fact that it is a scalar in a (pseudo) Riemannian manifold. As such the mere invariance of the spacetime interval places no restrictions on the allowable coordinate transformations and therefore cannot be used to derive the Lorentz transform.

Therefore, the invariance of the spacetime interval is not the property used to derive the Lorentz transform. The key property is not just the invariance of the interval, but the invariance of the form of the interval. In other words, we require not only $ds^2=ds’^2$ but instead we require $-dt^2+dx^2+dy^2+dz^2=-dt’^2+dx’^2+dy’^2+dz’^2$ which is a much stronger condition.

For example, a transformation of the spatial coordinates to polar coordinates satisfies the former property, as expected, but it violates the latter condition and therefore is not a Lorentz transform.

Edit: Now, in your edit you post some derivation that confuses you. There are two points which are leading to the confusion.

First, for a general coordinate transform you should write $ds’^2=g_{\mu’\nu’}dx^{\mu’}dx^{\nu’}$ because at this point you have not proven that the metric has the Minkowski form. And in fact for a general coordinate transform it does not (consider for example a spatial rotation).

Second, by rewriting the above it becomes immediately apparent that we do not get the condition $\Lambda^T \eta \Lambda = \eta$ from the invariance of the interval. This condition now must be added and doing so allows us to place restrictions on $\Lambda$ in order to obtain the Lorentz transform.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So, this is very much related to the root cause of my doubt. The problem is that, demanding $dS^2 = dS'^2$ turned out to be sufficient to show that the metric is invariant. $\endgroup$ – Yorozuya Jun 13 at 21:19
  • $\begingroup$ I think you made two mistakes in your derivation in the edit. I will comment on those in the answer. $\endgroup$ – Dale Jun 14 at 11:34
  • 2
    $\begingroup$ +1: This is the correct answer, the other answer is more or less incorrect in the context. As with most equivalent results in physics, you can either prove A given B or B given A. It's almost always a personal preference as to which direction you prefer. $\endgroup$ – Dvij D.C. Jun 14 at 12:34
  • 2
    $\begingroup$ To be explicitly clear, @Yorozuya, notice that what Dave is pointing out is that we require the metric components to be invariant under transformation, not just the interval. Just requiring the interval to be invariant, you'd get the trivial relation $g'=\Lambda^T g \Lambda$, i.e., just the usual transformation rule for a second rank tensor. But if you further require that $g=g'=\eta$, i.e., the metric is invariant and is Minkowskian in both the frames, then you get $\eta=\Lambda^T \eta\Lambda$. $\endgroup$ – Dvij D.C. Jun 14 at 12:34
1
$\begingroup$

It seems that you think the Lorentz transformations is a result of requiring the spacetime interval to be invariant.

It is actually the other way round. The spacetime interval being invariant is a result obtained from the Lorentz Transformations.

While the Lorentz Transformations is a result of the speed of light having the same value $c$ in all inertial frames.

Thus the chain of thought should be like this:

Speed of light same in all inertial frames $\rightarrow$ Lorentz Transformations $\rightarrow$ Spacetime interval invariant under Lorentz Transformations.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ The order for the chain of thought doesn't really relate to my question. I am asking as to why the invariance of the spacetime under a set of coordinate transformations that relate two inertial frames restrict the allowed transformations. $\endgroup$ – Yorozuya Jun 13 at 6:39
  • 2
    $\begingroup$ I may have misunderstood your question then. I just thought that it wasn't natural to try to start with invariance of the spacetime interval to derive the Lorentz transformations. $\endgroup$ – TaeNyFan Jun 13 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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