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## Lorentz transformation of the metric tensors.

Posted by peeterjoot on January 16, 2011

Following up on the previous thought, it is not hard to come up with an example of a symmetric tensor a whole lot simpler than the electrodynamic stress tensor. The metric tensor is probably the simplest symmetric tensor, and we get that by considering the dot product of two vectors. Taking the dot product of vectors $a$ and $b$ for example we have

\begin{aligned}a \cdot b = a^\mu b^\nu \gamma_\mu \cdot \gamma_\nu\end{aligned} \hspace{\stretch{1}}(4.17)

From this, the metric tensors are defined as

\begin{aligned}\eta_{\mu\nu} &= \gamma_\mu \cdot \gamma_\nu \\ \eta^{\mu\nu} &= \gamma^\mu \cdot \gamma^\nu\end{aligned} \hspace{\stretch{1}}(4.18)

These are both symmetric and diagonal, and in fact equal (regardless of whether one picks a $+,-,-,-$ or $-,+,+,+$ signature for the space).

Let’s look at the transformation of the dot product, utilizing the transformation of the four vectors being dotted to do so. By definition, when both vectors are equal, we have the (squared) spacetime interval, which based on the speed of light being constant, has been found to be an invariant under transformation.

\begin{aligned}a' \cdot b'= a^\mu b^\nu L(\gamma_\mu) \cdot L(\gamma_\nu)\end{aligned} \hspace{\stretch{1}}(4.20)

We note that, like any other vector, the image $L(\gamma_\mu)$ of the Lorentz transform of the vector $\gamma_\mu$ can be written as

\begin{aligned}L(\gamma_\mu) = \left( L(\gamma_\mu) \cdot \gamma^\nu \right) \gamma_\nu\end{aligned} \hspace{\stretch{1}}(4.21)

Similarily we can write any vector in terms of the reciprocal frame

\begin{aligned}\gamma_\nu = (\gamma_\nu \cdot \gamma_\mu) \gamma^\mu.\end{aligned} \hspace{\stretch{1}}(4.22)

The dot product factor is a component of the metric tensor

\begin{aligned}\eta_{\nu \mu} = \gamma_\nu \cdot \gamma_\mu,\end{aligned} \hspace{\stretch{1}}(4.23)

so we see that the dot product transforms as

\begin{aligned}a' \cdot b' = a^\mu b^\nu ( L(\gamma_\mu) \cdot \gamma^\alpha ) ( L(\gamma_\nu) \cdot \gamma^\beta ) \gamma_\alpha\cdot\gamma_\beta= a^\mu b^\nu {L_\mu}^\alpha{L_\nu}^\beta\eta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.24)

In particular, for $a = b$ where we have the invariant interval defined by the condition $a^2 = {a'}^2$, we must have

\begin{aligned}a^\mu a^\nu \eta_{\mu \nu}= a^\mu a^\nu {L_\mu}^\alpha{L_\nu}^\beta\eta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.25)

This implies that the symmetric metric tensor transforms as

\begin{aligned}\eta_{\mu\nu}={L_\mu}^\alpha{L_\nu}^\beta\eta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.26)

Recall from 3.16 that the coordinates representation of a bivector, an antisymmetric quantity transformed as

\begin{aligned}T^{\mu \nu} \rightarrow T^{\sigma \pi} {L_\sigma}^\mu {L_\pi}^\nu.\end{aligned} \hspace{\stretch{1}}(4.27)

This is a very similar transformation, but differs from the bivector case where our free indexes were upper indexes. Suppose that we define an alternate set of coordinates for the Lorentz transformation. Let

\begin{aligned}{L^\mu}_\nu = L(\gamma^\mu) \cdot \gamma_\nu.\end{aligned} \hspace{\stretch{1}}(4.28)

This can be related to the previous coordinate matrix by

\begin{aligned}{L^\mu}_\nu = \eta^{\mu \alpha } \eta_{\nu \beta } {L_\alpha}^\beta. \end{aligned} \hspace{\stretch{1}}(4.29)

If we examine how the coordinates of $x^2$ transform in thier lower index representation we find

\begin{aligned}{x'}^2 = x_\mu x_\nu {L^\mu}_\alpha {L^\nu}_\beta \eta^{\alpha \beta} = x^2 = x_\mu x_\nu \eta^{\mu \nu},\end{aligned} \hspace{\stretch{1}}(4.30)

and therefore find that the (upper index) metric tensor transforms as

\begin{aligned}\eta^{\mu \nu} \rightarrow\eta^{\alpha \beta}{L^\mu}_\alpha {L^\nu}_\beta .\end{aligned} \hspace{\stretch{1}}(4.31)

Compared to $4.27$ we have almost the same structure of transformation. Are these the same? Does the notation I picked here introduce an apparent difference that does not actually exist? We really want to know if we have the identity

\begin{aligned}L(\gamma_\mu) \cdot \gamma^\nu\stackrel{?}{=}L(\gamma^\nu) \cdot \gamma_\mu,\end{aligned} \hspace{\stretch{1}}(4.32)

which given the notation selected would mean that ${L_\mu}^\nu = {L^\nu}_\mu$, and justify a notational simplification ${L_\mu}^\nu = {L^\nu}_\mu = L^\nu_\mu$.

# The inverse Lorentz transformation

To answer this question, let’s consider a specific example, an x-axis boost of rapidity $\alpha$. For that our Lorentz transformation takes the following form

\begin{aligned}L(x) = e^{-\sigma_1 \alpha/2} x e^{\sigma_1 \alpha/2},\end{aligned} \hspace{\stretch{1}}(5.33)

where $\sigma_k = \gamma_k \gamma_0$. Since $\sigma_1$ anticommutes with $\gamma_0$ and $\gamma_1$, but commutes with $\gamma_2$ and $\gamma_3$, we have

\begin{aligned}L(x) = (x^0 \gamma_0 + x^1 \gamma_1) e^{\sigma_1 \alpha} + x^2 \gamma_2 + x^3 \gamma_3,\end{aligned} \hspace{\stretch{1}}(5.34)

and after expansion this is

\begin{aligned}L(x) = \gamma_0 ( x^0 \cosh \alpha - x^1 \sinh \alpha ) +\gamma_1 ( x^1 \cosh \alpha - x^0 \sinh \alpha )+\gamma_2+\gamma_3\end{aligned} \hspace{\stretch{1}}(5.35)

In particular for the basis vectors themselves we have

\begin{aligned}\begin{bmatrix}L(\gamma_0) \\ L(\gamma_1) \\ L(\gamma_2) \\ L(\gamma_3)\end{bmatrix}=\begin{bmatrix}\gamma_0 \cosh \alpha - \gamma_1 \sinh \alpha \\ -\gamma_0 \sinh \alpha + \gamma_1 \cosh \alpha \\ \gamma_2 \\ \gamma_3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.36)

Forming a matrix with $\mu$ indexing over rows and $\nu$ indexing over columns we have

\begin{aligned}{L_\mu}^\nu =\begin{bmatrix}\cosh \alpha &- \sinh \alpha & 0 & 0 \\ -\sinh \alpha & \cosh \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.37)

Performing the same expansion for ${L^\nu}_\mu$, again with $\mu$ indexing over rows, we have

\begin{aligned}{L^\nu}_\mu =\begin{bmatrix}\cosh \alpha & \sinh \alpha & 0 & 0 \\ \sinh \alpha & \cosh \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(5.38)

This answers the question. We cannot assume that ${L_\mu}^\nu = {L^\nu}_\mu$. In fact, in this particular case, we have ${L^\nu}_\mu = ({L_\mu}^\nu)^{-1}$. Is that a general condition? Note that for the general case, we have to consider compounded transformations, where each can be a boost or rotation.

# References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.