# Simple problem driving me nuts :-)

1. Feb 24, 2008

### Fredrik

Staff Emeritus
I'm embarrassed to ask, but I think this will save me some time...

I'm trying to use the condition $\Lambda^T\eta\Lambda=\eta$ to show that $\Lambda_{0i}=-\Lambda_{i0}$, where i=1,2,3. This is the algebraic version of the physically obvious condition that if the velocity associated with a homogeneous Lorentz transformation is $\vec{v}$, then the velocity associated with its inverse is $-\vec{v}$. This should be easy, but I don't see it.

($X_{0i}$ is row 0, column i, of the matrix X. I'm putting all the indices downstairs because I feel that's less confusing when calculations include transposes of matrices).

Last edited: Feb 24, 2008
2. Feb 25, 2008

### Fredrik

Staff Emeritus
I gave it another shot, and it turned out that the "obvious" condition isn't true. If my math is correct this time, the general form of $\Lambda$ is

$$\Lambda=\gamma\begin{pmatrix}1 & v^T R \\ -v & -R\end{pmatrix}$$

where v is a 3x1 matrix that we can think of as "their" velocity in "our" frame, and R is a 3x3 matrix that isn't orthogonal in general, but is orthogonal in the limit v $\rightarrow$ 0 (so its components must depend on v). The velocity that corresponds to $\Lambda^{-1}$ the same way that v corresponds to $\Lambda$ is -RTv. So it's only =-v when R=0, i.e. when we have a pure boost.

3. Feb 25, 2008

### DrGreg

If we ignore the possibility of a spatial rotation and consider boosts only, a boost in an arbitrary direction ought to be

$${\Lambda^a}_b=\begin{pmatrix}\gamma & -\gamma \textbf{v}^T \\ -\gamma \textbf{v} & \textbf{I} + (\gamma - 1)\frac{\textbf{v} \textbf{v}^T}{\textbf{v}^T \textbf{v}}\end{pmatrix}$$

(assuming c = 1 or equivalently x0 = ct).

However, when you say you are "putting all the indices downstairs" I'm not sure whether you mean you are ignoring the index position or whether you really are calculating $$\Lambda_{ab}$$ which differs from both $$\Lambda^{ab}$$ and $${\Lambda^a}_b$$.

4. Feb 25, 2008

### Fredrik

Staff Emeritus
Thank you. That makes my R equal to

$$-\frac{1}{\gamma}(I+(\gamma-1)\frac{vv^T}{v^Tv})$$

I should get $-\gamma v^T$ when I calculate $\gamma v^T R$, and it turns out I do:

$$\gamma v^T R=-v^T(I+(\gamma-1)\frac{vv^T}{v^Tv})=-v^T-(\gamma-1)v^T=-\gamma v^T$$

And, yes, when I wrote $\Lambda_{i0}$ in the OP I was referring to components of the tensor which in the abstract index notation would appear as $\Lambda^a{}_b$. I find that notation slightly easier to use in the type of calculations I had to do to get this result.

Hm, do you know if the general form of $\Lambda$ (i.e. not a pure boost) is the same except that the identity matrix in your expression is replaced with an orthogonal matrix that's independent of v? Maybe it's more complicated than that.

Last edited: Feb 25, 2008
5. Feb 26, 2008

### DrGreg

I'm no expert in this, but my geometrical intuition says it ought to be either $$Q \Lambda$$ or $$\Lambda Q$$, where $$\Lambda$$ is as above and Q is a space-only rotation, i.e. in the form

$$Q = \begin{pmatrix}1 & \textbf{0}^T \\ \textbf{0} & \textbf{Q}_0 \end{pmatrix}$$

where Q0 is a 3 x 3 orthogonal matrix ($$\textbf{Q}_0^T \textbf{Q}_0 = \textbf{I}$$).

Perhaps someone with more experience in this area could confirm that?