Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Lorentz transformation of an antisymmetric tensor

Posted by peeterjoot on January 14, 2011

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I have an ancient copy of [1] from the library right now (mine is on order still) for my PHY450H1S course (relativistic electrodynamics). Given the transformation rule for a first rank tensor

\begin{aligned}A_{i} = \alpha_{im} A'_{m},\end{aligned} \hspace{\stretch{1}}(1.1)

they list the transformation rule for a second rank tensor as

\begin{aligned}A_{ik} = \alpha_{im} \alpha_{kl} A'_{ml}.\end{aligned} \hspace{\stretch{1}}(1.2)

This isn’t motivated in any way. Let’s compare to transformation of a bivector expressed in the Dirac basis, transformed by outermorphism. That’s specifically a transformation of a antisymmetric tensor (once expressed in components anyways), but should provide some intuition.

It is also worthwhile to note that there are some old fashioned notational quirks in this text (at least the old version that I have currently borrowed). Specifically, they uses Latin indexes four vectors with Greek indexes for three vectors, completely opposite to what appears to be the current conventions. They also don’t use upper and lower indexes to keep track of bookkeeping. I’ll use the conventions I’m used to for now.


I’ll use conventions from [2] using the Dirac basis, with a preference for index upper coordinates, and express a vector as

\begin{aligned}x = x^\alpha \gamma_\alpha = x_\alpha \gamma^\alpha,\end{aligned} \hspace{\stretch{1}}(2.3)

Here the basis pairs \{\gamma_\mu\} and \{\gamma^\mu\} are reciprocal frames with \gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu. I’ll have no need for any specific metric convention here.

The dot and wedge products used will be defined in terms of their Clifford Algebra formulation

\begin{aligned}a \cdot b &= \frac{1}{{2}} (a b + b a) \\ a \wedge b &= \frac{1}{{2}} (a b - b a).\end{aligned} \hspace{\stretch{1}}(2.4)

Transformation of the coordinates.

Let’s assume our transformation is linear, and we will denote its action on vectors as follows

\begin{aligned}x' = L(x) = x^\alpha L( \gamma_\alpha).\end{aligned} \hspace{\stretch{1}}(3.6)

Extracting coordinates for the transformed coordinates (assuming a non-moving frame where the unit vectors on both sides are the same), we have after dotting with \gamma^\mu

\begin{aligned}{x'}^\mu = \left( {x'}^\alpha \gamma_\alpha \right) \cdot \gamma^\mu= x^\alpha \left( L( \gamma_\alpha) \cdot \gamma^\mu \right) \end{aligned} \hspace{\stretch{1}}(3.7)

Now introduce a coordinate representation for the transformation L

\begin{aligned}L( \gamma_\alpha) \cdot \gamma^\mu  = {L_\alpha}^\mu,\end{aligned} \hspace{\stretch{1}}(3.8)

so our transformation rule for the four vector coordinates becomes

\begin{aligned}{x'}^\mu = x^\alpha {L_\alpha}^\mu.\end{aligned} \hspace{\stretch{1}}(3.9)

We are now ready to look at the transformation of a bivector (a quantity having a rank two antisymmetric tensor representation in coordinates), and see how the coordinates transform.

Let’s transform by outermorphism of the transformed vector factors the bivector

\begin{aligned}c = a \wedge b \rightarrow a' \wedge b'.\end{aligned} \hspace{\stretch{1}}(3.10)

First we’ll need the coordinate representation of the bivector before transformation. We dot with \gamma^\nu \wedge \gamma^\mu to pick up the desired term

\begin{aligned}(a \wedge b) \cdot (\gamma_\nu \wedge \gamma_\mu)&= a^\alpha b^\beta (\gamma_\alpha \wedge \gamma_\beta) \cdot (\gamma^\nu \wedge \gamma^\mu) \\ &= a^\alpha b^\beta ( \gamma_\alpha {\delta_\beta}^\nu -\gamma_\beta {\delta_\alpha}^\nu ) \cdot \gamma^\mu \\ &= a^\alpha b^\beta ( {\delta_\alpha}^\mu {\delta_\beta}^\nu -{\delta_\beta}^\mu {\delta_\alpha}^\nu ) \\ &= a^\mu b^\nu -a^\nu b^\mu \\ \end{aligned}

If we introduce a rank two tensor now, say

\begin{aligned}T^{\mu\nu} = a^\mu b^\nu -a^\nu b^\mu,\end{aligned} \hspace{\stretch{1}}(3.11)

we recover our bivector with

\begin{aligned}a \wedge b = \frac{1}{{2}} T^{\alpha \beta} \gamma_\alpha \wedge \gamma_\beta.\end{aligned} \hspace{\stretch{1}}(3.12)

Now let’s look at the coordinate representation of the transformed bivector. It will also be helpful to make use of the identity that can be observed above from the initial coordinate extraction

\begin{aligned}(\gamma_\alpha \wedge \gamma_\beta) \cdot (\gamma^\nu \wedge \gamma^\mu) = {\delta_\alpha}^\mu {\delta_\beta}^\nu -{\delta_\beta}^\mu {\delta_\alpha}^\nu \end{aligned} \hspace{\stretch{1}}(3.13)

In coordinates our transformed bivector is

\begin{aligned}a' \wedge b' = a^\sigma {L_\sigma}^\alpha b^\pi {L_\pi}^\beta \gamma_\alpha \wedge \gamma_\beta,\end{aligned} \hspace{\stretch{1}}(3.14)

and we can proceed with the coordinate extraction by taking dot products with \gamma^\nu \wedge \gamma^\mu as before. This gives us

\begin{aligned}( a' \wedge b' ) \cdot (\gamma^\nu \wedge \gamma^\mu) &= a^\sigma {L_\sigma}^\alpha b^\pi {L_\pi}^\beta \gamma_\alpha \wedge \gamma_\beta \\ &= a^\sigma {L_\sigma}^\alpha b^\pi {L_\pi}^\beta ( {\delta_\alpha}^\mu {\delta_\beta}^\nu -{\delta_\beta}^\mu {\delta_\alpha}^\nu  ) \\ &= a^\sigma {L_\sigma}^\mu b^\pi {L_\pi}^\nu - a^\sigma {L_\sigma}^\nu b^\pi {L_\pi}^\mu \\ &= a^\sigma {L_\sigma}^\mu b^\pi {L_\pi}^\nu-a^\pi {L_\pi}^\nu b^\sigma {L_\sigma}^\mu \\ &= (a^\sigma b^\pi - a^\pi b^\sigma) {L_\sigma}^\mu {L_\pi}^\nu\\ &= T^{\sigma \pi} {L_\sigma}^\mu {L_\pi}^\nu\\ \end{aligned}

We are able to conclude that the bivector coordinates transform as

\begin{aligned}T^{\mu \nu} \rightarrow T^{\sigma \pi} {L_\sigma}^\mu {L_\pi}^\nu.\end{aligned} \hspace{\stretch{1}}(3.15)

Except for the lowering index differences this verifies the rule 1.2 from the text.

It would be reasonable seeming to impose such a tensor transformation rule on any antisymmetric rank 2 tensor, and in the text this is also imposed as the rule for transformation of symmetric rank 2 tensors. Do we have a simple example of a rank 2 symmetric tensor that can be expressed geometrically? The only one that comes to mind off the top of my head is the electrodynamic stress tensor, which isn’t exactly simple to work with.


[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.


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