Say now I have an arbitrary field strength tensor $F$, and I want to boost it according to a Lorentz transformation matrix $(\Lambda)$

The transformation is given by $$ F^{'\mu \nu} = \Lambda^{\mu}_{\;\alpha} \Lambda^{\nu}_{\;\beta} F^{\alpha\beta} $$

The question is, how do I actually calculate this with actual values?

  • $\begingroup$ First you have to find an explicit form for the matrix $\Lambda$. I think the easiest way to do this is exponentiating a linear combination of the generators of Lorentz's Lie Algebra. Then you just sum over the repeated indices, and you do that for each component of $F^{\mu\nu}$ that you would like to find. $\endgroup$ – Federico Carta Jun 9 '14 at 17:28
  • $\begingroup$ If I have explicit matrix forms for both $\Lambda$ and $F^{\mu\nu}$ is there a nice rule of thumb or so to calculate $F^{'\mu\nu}$, $F^{'00} = \Lambda^0_{\; 0}\Lambda^0_{\; 0}F^{00} + .... $ is not only tedious but prone to silly mistakes $\endgroup$ – William Grunow Jun 9 '14 at 17:38
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    $\begingroup$ Can I ask why you need to exponentiate the generators of the Lie Algebra? I thought that was how spinors transform under Lorentz transformation. If you have an arbitrary tensor can't you just compose standard lorentz boost matrices? $\endgroup$ – user82235 Jun 9 '14 at 17:39
  • $\begingroup$ WilliamGrunow, no. I don't think any shortcut exist, fora a general form of $\Lambda$. @user82235, any element of the connected component to the identity of a Lie Group can be expressed as the exponential of a combination of generators of the algebra. Of course, if you find it simpler, you can use the "standard lorentz boost matrices". I personally prefer the other way, cause it allows me to remember by heart less things, and work out the same results. But either way is fine. $\endgroup$ – Federico Carta Jun 9 '14 at 17:43
  • $\begingroup$ @Grunow: I agree with Carta that it is simply a tedious task, but it may be easier for you to think of it as matrix multiplication. Since: $AB = \sum_{j} (A)_{ij}(B)_{jk}$ (where $(A)_{ij}$ is the elementh in the $i$th row and $j$th column of $A$) In your example, $F' = \Lambda F \Lambda^T$ $\endgroup$ – user82235 Jun 9 '14 at 17:46

Important Note.

As written, the transformation formula in the question is suppressing an important subtlety. The field strength tensor is a tensor field; it is a function on spacetime. As such, it's not just its indices that must transform under a Lorentz transformation, but also its argument. The full transformation is as follows: \begin{align} F'^{\mu\nu}(x') = \Lambda^\mu_{\phantom\mu\alpha}\Lambda^\nu_{\phantom\nu\beta}F^{\alpha\beta}(x), \qquad x' = \Lambda x \end{align} Or, put another way \begin{align} F'^{\mu\nu}(x) = \Lambda^\mu_{\phantom\mu\alpha}\Lambda^\nu_{\phantom\nu\beta}F^{\alpha\beta}(\Lambda^{-1}x) \end{align} Authors often suppress this fact in the notation, but I found it confusing when I first learned the subject, so I think it's dangerous to not be explicit.

As a terminological aside, the transformation of the tensor indices is often referred to as a target space transformation while the transformation of its argument is sometimes referred to as a spacetime transformation.

Target space transformation - efficient computation

An efficient way to compute an arbitrary target space transformation in this context is to note that when dealing with strings of contracted two-tensors, one can convert the expression into a sequence of matrix multiplications.

If we define matrices $\Lambda = (\Lambda^\mu_{\phantom\mu\nu})$ and $F = (F^{\mu\nu})$ where the left-most index is the "row" label and the right-most index is the "column" label, then we find that \begin{align} \Lambda^\mu_{\phantom\mu\alpha}\Lambda^\nu_{\phantom\nu\beta}F^{\alpha\beta} = \Lambda^\mu_{\phantom\mu\alpha}F^{\alpha\beta}\Lambda^\nu_{\phantom\nu\beta} = (\Lambda F\Lambda^T)^{\mu\nu} \end{align} where the expression in parenthesis denotes successive matrix multiplication.

Special Cases

The matrix multiplication above is made significantly easier provided the Lorentz transformation one is performing is special. In particular, suppose for instance that the Lorentz transformation is a boost along the $x$-direction. Then the matrix $\Lambda$ will have the following block-matrix form \begin{align} \Lambda = \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix} \end{align} where \begin{align} \lambda = \begin{pmatrix} \gamma & -\gamma\beta \\ -\gamma \beta & \gamma \end{pmatrix}, \qquad I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align} Therefore, if we write $F$ is 2-by-2 block matrix form; \begin{align} F = \begin{pmatrix} f_{11} & f_{12} \\ f_{12} & f_{22} \\ \end{pmatrix} \end{align} then we have \begin{align} F' &= \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix} \begin{pmatrix} f_{11} & f_{12} \\ f_{21} & f_{22} \\ \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix} \\ &= \begin{pmatrix} \lambda f_{11} & \lambda f_{12} \\ f_{21} & f_{22} \\ \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix}\\ &= \begin{pmatrix} \lambda f_{11}\lambda & \lambda f_{12} \\ f_{21}\lambda & f_{22} \\ \end{pmatrix} \end{align} so now one just needs do perform four 2-by-2 matrix multiplications, which is relatively easy. Boosts in the $y$- and $z$- directions are similarly straightforward.

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    $\begingroup$ I think it's worth pointing out that it's meaningful to transform the EM field without transforming the argument, when the boost is not meant as a coordinate transformation. $\endgroup$ – Muphrid Jun 10 '14 at 2:22
  • $\begingroup$ @Muphrid Would you mind being more explicit about what context/example you have in mind? $\endgroup$ – joshphysics Jun 10 '14 at 4:07
  • $\begingroup$ The consideration of local Lorentz transformations as one would do when using tetrad formalism. $\endgroup$ – Muphrid Jun 10 '14 at 4:20
  • $\begingroup$ Hi Josh, there's a mistake here! I think you're forgetting $f_{11}$ must be a matrix. The correct result is $\{\{\lambda f_{11}\lambda,\lambda f_{12}\},\{ f_{21}\lambda,f_{22} \} \}$ $\endgroup$ – user12029 Jun 3 '15 at 20:49
  • $\begingroup$ @NeuroFuzzy Yes indeed! Thanks for catching my sloppiness! Edited. $\endgroup$ – joshphysics Jun 4 '15 at 1:38

For special relativity, you can use clifford algebra rotors to save some work over computing the matrix representation of a Lorentz boost.

Rotors in a clifford algebra are the 3+1 analogues of quaternions, and they're nearly as easy to use. The big piece that you need to use them is the idea of clifford multiplication (also called the geometric product).

Let $e_t, e_x, e_y, e_z$ be basis vectors. First, there's the metrical part of the geometric product.

$$e_t e_t = -1, \quad e_x e_x = +1, \quad e_y e_y = +1, \quad e_z e_z = +1$$

The signs, of course, are part and parcel to choosing a metric signature.

We've chosen an orthogonal basis here, in which the other products are very simple: they all anticommute. Here, $a, b$ are particular indices (but it doesn't matter which indices they are).

$$e_a e_b = -e_b e_a$$

So $e_t e_x = -e_x e_t$, and so on.

Finally, the geometric product is associative, so an expression like $e_t e_x e_t$ can be manipulated as so:

$$e_t e_x e_t = e_t (e_x e_t) = -e_t (e_t e_x) = -(e_t e_t) e_x = + e_x$$

With the geometric product in place, we can begin to create rotors, our 3+1 analogues of quaternions. Observe the following curious fact about the "bivector" $e_t e_x$:

$$(e_t e_x)^2 = e_t e_x e_t e_x = (+e_x) e_x = +1$$

So we have a nonscalar object that squares to +1. Interesting. What happens if we put that into an exponential using a power series? Let $\phi$ be a scalar, and see that

$$\begin{align*}\exp(\phi e_t e_x) &= 1 + \phi e_t e_x + \frac{\phi^2}{2} (e_t e_x)^2 + \frac{\phi^3}{3!} (e_t e_x)^3 + \ldots \\ &= 1 + \frac{\phi^2}{2} + \ldots + e_t e_x (\phi + \frac{\phi^3}{3!} + \ldots)\\ &= \cosh \phi + e_t e_x \sinh \phi\end{align*}$$

This is a derivation of the Lorentz transformations, in terms of the rapidity $\phi = \tanh^{-1}(v/c)$. An exercise for those at home: try using the bivector $e_x e_y$ instead. Show that $(e_x e_y)^2 = -1$, and thus the power series of the exponential $\exp(\theta e_x e_y) = \cos \theta + e_x e_y \sin \theta$. This exponential approach emphasizes the similarities between boosts and rotations.

Now then, given a unit bivector $\hat B$ that describes the plane of rotation, and given a rapidity $\phi$, we can calculate a rotor $q$:

$$q = \exp(-\hat B \phi/2)$$

And we can use that rotor to rotate a vector $a$:

$$a' = q a q^{-1} = (\cosh \frac{\phi}{2} - \hat B \sinh \frac{\phi}{2}) a (\cosh \frac{\phi}{2} + \hat B \sinh \frac{\phi}{2})$$

Now, here's a fun trick: the formula works for a bivector like the Faraday tensor $F$, also! So you don't even have to rotate it twice the way you would have to do two matrix multiplies in index notation.


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