# Successive boosts

1. Nov 2, 2006

### actionintegral

Hi Friends,

From time to time I have seen transformations replaced by a succession of infinitesimal transformations. The end result ends up being an exponent.

My knowledge of this is vague and I would like to look into it more seriously. Particularly I am interested in describing the lorentz transformation as a sequence of infinitesimal boosts.

Can someone point me to the first thing I need read to understand this?

Thanks

2. Nov 2, 2006

### actionintegral

Hi, ActionIntegral,

I'll be more than happy to take a stab at it! Just expand the lorentz transform in a taylor series in v about v=0 keeping only the first order.

You can raise this operator to the nth power. Somehow this becomes an exponent but I haven't figured that part out yet.

3. Nov 2, 2006

### actionintegral

That's ok - I wouldn't want you to do all the work for me!

4. Nov 2, 2006

### pervect

Staff Emeritus
What I think you're trying to do is find the "generator" of the Lorentz group.

http://en.wikipedia.org/wiki/Generating_set_of_a_group

But you want to do this for a continuous group. The Lorentz transformation is a "group" in the abstract algebra sense with the group operation being the successive application of transforms, because the result is associative f x (g x h) = (f x g) x h, and has an inverse. However, the Lorentz group is an infinite group. This is called a "Lie group".

So what you need to do is to read up on Lie groups (specifically the generators of Lie groups). Or see if you can find a mathemetician.

Hope this helps.

5. Nov 2, 2006

### robphy

Let \Theta=\left[ \begin {array}{cc} 0&1\\\noalign{\medskip}1&0\end {array} \right]\theta.
Formally, write $$\exp(\Theta)=I+\Theta+\Theta^2/2!+\Theta^3/3!+\ldots$$. Do you recognize $$\exp(\Theta)$$?

6. Nov 2, 2006

### actionintegral

Both good answers - i'll read up on it. thanks to all three of you!