References: W. Greiner & J. Reinhardt, *Field Quantization*, Springer-Verlag (1996), Chapter 2, Section 2.4.

Arthur Jaffe, *Lorentz transformations, rotations and boosts*, online notes available (at time of writing, Sep 2016) here.

Continuing our examination of general Lorentz transformations, we can now complete the demonstration that a general Lorentz transformation is the product of a pure boost (motion at a constant velocity) multiplied by a pure rotation. We’ll follow Corollary IV.2 in Jaffe’s article.

In the last post, we saw that we could write a general Lorentz transformation in the form

where is the 4-vector of a spacetime event, is the Lorentz transformation as a matrix, is a matrix with complex elements and a hat over a symbol means we’re looking at the complex matrix representing that object. We also saw in the last post that this representation restricts

Jaffe goes through a rather involved proof that the transformation defined by 1 is a member of the physically relevant group with and , but this involves a lot of somewhat obscure matrix theorems that I don’t want to get into here, and these techniques don’t seem to be required for the rest of the demonstration, so we’ll just accept this fact for now.

What we really want to do is find out how we can calculate given the matrix . We can do this by using the result we got earlier for the components of the 4-vector :

where the are four Hermitian matrices:

We can invert 2 to get

Reverting back to the matrix (no hat), we have

We used 1 in the fourth line and 2 in the fifth line. Comparing the first and last lines, we see that

where in the last line we used the fact that all the are Hermitian so that .

In order for to be a valid Lorentz transformation, clearly its elements must be real numbers. We can show this is true as follows. The complex conjugate is represented by drawing a bar over a quantity. We get

We can now use the fact that the trace of a product of matrices remains unchanged if we cyclically permute the order of multiplication. In particular . Also, since the trace of a matrix is equal to the trace of its transpose. In 17, we can set and and use the fact that the are all Hermitian so that :

where in the third line we cyclically permuted the matrices in the trace. Thus the elements of are real.

Now we consider two cases. First, suppose that , where is a unitary matrix, so that . From 16 we find that is, using :

The other elements in the first row and first column of are all zero, as we can see by using 16 again:

since for . A similar argument works for the first column of as well:

For the other elements, we have

That is

so that

where is a matrix, and the 0s represent 3 zero components in the top row and first column. In other words, when , is a pure rotation.

The other case we need to examine is when , where is a Hermitian matrix, so that . In that case, from 16

so is a symmetric matrix. (We used two cyclic permutations in the trace here.) Although we haven’t proved that a symmetric Lorentz transformation always represents a pure boost, this has been verified (see, for example, Wikipedia; I can’t be bothered going through it all here).

Now we are ready to get our final result. To do this, we need to use a theorem from matrix algebra which says that every matrix in the group (that is, a matrix with complex elements and determinant +1) has a unique polar decomposition into a strictly positive Hermitian matrix and a unitary matrix , so that we always have

To connect this with what we’ve done above, we can define

[The square root of a matrix is defined to be the matrix so that .] This definition is consistent with being Hermitian, since

Thus if we restrict to be the positive square root, we must have .

The definition is also consistent with being unitary, since

[We define to be the inverse of .]

Therefore, we can uniquely decompose any Lorentz transformation into

that is, the product of a pure rotation and a pure boost.