4
$\begingroup$

I'm studying special relativity.

A general Lorentz transformation is defined by $\Lambda^T\eta\Lambda=\eta$.

Now, \begin{align} \eta'^{\mu\nu} &= \Lambda^\mu_{\;\;\alpha}\Lambda^\nu_{\;\;\beta}\eta^{\alpha\beta}\\ &= (\Lambda\eta\Lambda^T)^{\mu\nu} \end{align}

How does this equal $\eta^{\mu\nu}$? All we know is that $\Lambda^T\eta\Lambda=\eta$.

$\endgroup$
2
  • $\begingroup$ What difficulty exactly do you have here? If you look at your definition, then for all $\Lambda$, $\Lambda^T$ is also a Lorentz transformation... $\endgroup$
    – ACuriousMind
    Jul 22, 2020 at 15:10
  • $\begingroup$ @ACuriousMind Can you help me see how that follows? $\endgroup$
    – Atom
    Jul 23, 2020 at 4:19

3 Answers 3

4
$\begingroup$

As you pointed out, the matrix form for the transformation equation of $\eta$ is:

$$\eta' = \Lambda \eta \Lambda^T$$

It's possible to show that $\eta' = \eta$, given that $\eta = \Lambda^T \eta \Lambda$, recalling that $\eta = \eta^{-1}$:

$$\begin{align} \eta &= \Lambda^T \eta \Lambda \\ \Lambda \eta \eta &= (\Lambda \eta \Lambda^T) \eta \Lambda \\ \Lambda &=(\Lambda \eta \Lambda^T) \eta \Lambda \end{align}$$

This means that $\Lambda \eta \Lambda^T = \eta$ as well, so $\eta = \eta'$.

Alternatively, we can use the fact that $\eta = \Lambda^T \eta \Lambda$ must hold for all Lorentz transformations $\Lambda$ and if $\Lambda$ is a Lorentz transformation, then so is $\Lambda^{-1}$. Taking the inverse of both sides:

$$\begin{align} \eta = \Lambda^{-1} \eta {(\Lambda^{-1})}^T \end{align}$$

However, what actually is $\Lambda^{-1}$ is semantics since this equation holds for all $\Lambda$ anyway, so we could just as well write $\eta = \Lambda \eta \Lambda^T$.

One final way to show that $\eta' = \eta$ is through the transformation of $\eta_{\mu \nu}$ instead of $\eta^{\mu \nu}$:

$$\begin{align} \eta'_{\mu \nu} &= \Lambda^{\alpha}{}_{\mu} \Lambda^{\beta}{}_{\nu} \eta_{\alpha \beta} \\ &= (\Lambda^T)_{\mu}{}^{\alpha} \eta_{\alpha \beta} \Lambda^{\beta}{}_{\nu} \end{align}$$

which has the direct matrix interpretation of $\eta' = \Lambda^T \eta \Lambda = \eta$. This can then also be used to show why $\Lambda^T \eta \Lambda = \Lambda \eta \Lambda^T$.

$\endgroup$
3
  • $\begingroup$ I don't see how you get the second equation in \begin{align} \eta &= \Lambda^T \eta \Lambda \\ \Lambda \eta \eta &= (\Lambda \eta \Lambda^T) \eta \Lambda \\ \Lambda &=(\Lambda \eta \Lambda^T) \eta \Lambda. \end{align} $\endgroup$
    – Atom
    Jul 26, 2020 at 14:01
  • $\begingroup$ @Atom I multiplied both sides of the first equation by $\Lambda \eta$ and then grouped terms since matrix multiplication is associative. $\endgroup$
    – Shrey
    Jul 26, 2020 at 14:10
  • $\begingroup$ Oh, my bad! I see it now! +1 $\endgroup$
    – Atom
    Jul 26, 2020 at 14:18
3
$\begingroup$

Here's a no-nonsene approach to the question. The metric $\eta$ is an object that acts on pairs of vectors $v, w$. In matrix notation, we can write that as $$\eta(v, w) = v^{T} \eta w$$ Now what happens if we apply the transformation $\Lambda$ to these vectors? We get: $$\eta(\Lambda v, \Lambda w) = (\Lambda v)^{T} \eta (\Lambda w) = v^{T}(\Lambda^{T} \eta \Lambda)w = v^T \eta w = \eta(v, w)$$ Hence the transformation $\Lambda$ does preserve the metric. In fact it's not hard to convince yourself that $\Lambda$ preserves the metric (for all possible pairs of vectors) if and only if the identity $\eta = \Lambda^t \eta \Lambda$ holds.

Now, with regards to your derivation, I suspect the problem is somewhere down the line your index notation is switching what should be lower indices with upper indices. So it's not surprising you end up transposed.

$\endgroup$
4
  • $\begingroup$ $\eta=\Lambda^T\eta\Lambda$ is actually equivalent to $\eta=\Lambda\eta\Lambda^T$ because $\Lambda$ is unitary. $\endgroup$ Jul 26, 2020 at 0:16
  • $\begingroup$ What are you talking about? If $\Lambda$ were 'unitary' (you probably mean orthogonal) it would preserve the Euclidean metric, not the Minkowski metric. $\endgroup$
    – Pedro
    Jul 26, 2020 at 0:26
  • $\begingroup$ Right, orthogonal. I'm too used to complex matrices! I mean that ${(\Lambda^{-1})^\mu}_\nu={\Lambda_\nu}^\mu$. See, for example, my answer, or equation 36.15 in this reference: bohr.physics.berkeley.edu/classes/209/f02/lorentz.pdf. $\endgroup$ Jul 26, 2020 at 3:11
  • $\begingroup$ Nevermind. Shrey cleared this up for me. You are right that the issue is with switching upper and lower indices. +1 $\endgroup$ Jul 26, 2020 at 20:04
2
$\begingroup$

The Lorentz transformations are in particular coordinate transformations and the metric tensor is well, a tensor therefore it transforms as a tensor (left hand side of the following equation) and we demand it leaves the metric unchanged (right hand side of the equation): $$\eta_{\alpha\beta} = \Lambda^{\mu}_{\;\;\alpha} \Lambda^{\nu}_{\;\;\beta}\eta_{\mu\nu}\tag{1}\label{eq:lorentz}$$ or the contra-variant version of the metric if you will $$\eta^{\alpha\beta} = \Lambda^{\alpha}_{\;\;\mu} \Lambda^{\beta}_{\;\;\nu}\eta^{\mu\nu},$$ thus arriving to the defining property of the Lorentz transformations. The problems begin when you try to attach a matrix to the objects involved, so the transpose is to be taken with care. My suggestion is to stick to the tensor notation in realize a bit more geometrically what it means. We are looking for transformations that leave the space-time intervals unchanged: $$\Delta s^2 = t^2 - \vec{x}\cdot\vec{x} = x^\mu\, \eta_{\mu\nu}\, x^\nu$$ Let us then explore what happens when we transform the coordinates according to a Lorentz transformation $$x'^\mu = \Lambda^\mu_{\;\;\nu}\,x^\nu,\tag{2}\label{eq:coords}$$ by computing the space-time interval in these new coordinates: $$\Delta s'^2 = x'^\mu x'_\mu = x'^\mu \eta_{\mu\nu} x'^\nu = \Lambda^\mu_{\;\;\beta}\, x^\beta\, \eta_{\mu\nu}\,\Lambda^\nu_{\;\;\alpha}\, x^\alpha = x^\beta (\Lambda^\mu_{\;\;\beta}\,\eta_{\mu\nu}\,\Lambda^\nu_{\;\;\alpha})x^\alpha$$ and now using the definition \eqref{eq:lorentz} of the Lorentz transformations we get $$\Delta s'^2 = x^\beta \eta_{\beta\alpha}x^\alpha = \Delta s^2$$ so it matches the interval in the original coordinates as we wanted.

For the sake of completeness if we have the convention that contra-variant (index up) means a column, then you can obtain/define the transpose of the coordinate transformation \eqref{eq:coords}, by contracting with the metric and arrive to: $$ x'_\mu = (\eta_{\mu\alpha}\Lambda^\alpha_{\;\;\beta}\eta^{\beta\nu})\,x_\nu\equiv x_\nu (\Lambda^T)_{\;\;\mu}^{\nu}$$ with this you can see that your property coincides with my equation \eqref{eq:lorentz}.


Not recommended for kids at home

To see how this is immediately your equation, remember that if you want to map these tensors into matrices, the first index is to be interpreted as pointing to the rows and the second index to the columns (up or down do not mean anything for this matter, just the order of appearance of the indices). So contracting the first indices of two objects is equivalent to contracting the second index of the first object transposed with the first index of the second object (same thing for sums where both indices are the second index). $$(A B)^{\;\,k}_{j} = A^{j}_{\;i} B^{i k} = (A^T)^{i}_{\;j} B^{i k} $$ notice that as matrices the above equation is ok, however the tensor structure is lost. Your equation is using this "matrix" map of the tensors, so when you write the metric tensor, the indices have a fixed meaning,

$$\eta^{\alpha\beta} = \Lambda^{\alpha}_{\;\;\mu} \Lambda^{\beta}_{\;\;\nu}\eta^{\mu\nu} = (\Lambda^T)_{\mu}^{\;\;\alpha}\Lambda^{\beta}_{\;\;\nu} \eta^{\mu\nu}$$ so it means you have a matrix of rows indexed by $\alpha$ and columns index by $\beta$ the last expression just tells you how to compute it. That is we are adding a row against a row. For this reasons I don't recommend considering all tensors as matrices, a mixed (1,1) tensor, does have such natural interpretation but it is otherwise to be used with care. I really hope you see the issue is just a matter of trying to write down what is multiplied with what.

$\endgroup$
13
  • 1
    $\begingroup$ This is not what I ask. I ask to show the Lorentz invariance of $\eta$ not of spacetime intervals. But thanks for clearing up my notation misconceptions. :) $\endgroup$
    – Atom
    Jul 22, 2020 at 13:19
  • 1
    $\begingroup$ It is the same thing. Read the last part, perhaps that helps more. $\endgroup$
    – ohneVal
    Jul 22, 2020 at 13:21
  • $\begingroup$ Can you please show how? I just want to show that $\eta'=\eta$. $\endgroup$
    – Atom
    Jul 22, 2020 at 13:23
  • 1
    $\begingroup$ Excellent answer! +1. I liked the caution about interpreting tensors as matrices. I have been doing this for years and never heard it put so succinctly and clearly $\endgroup$
    – Dale
    Jul 22, 2020 at 15:52
  • $\begingroup$ Can the down voter explain the motivation? $\endgroup$
    – ohneVal
    Jul 23, 2020 at 6:49

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.