# Question about applying the Lorentz Transformation to velocity 4-vectors

1. Mar 20, 2009

### ygolo

My question deals not with the Lorentz Tranformation itself, but the matrix representation of it:

I see readily how the space-time 4-vector: $$x^{\mu}=\left( c \ast t, x, y, z\right)$$ transforms approptiately so that $$x^{\acute{\mu}}=\Lambda_{v}^{\acute{\mu}} \ast x^{\mu}=\left( \gamma \ast \left( c \ast t - \beta \ast x\right), \gamma \ast \left( x- \beta \ast c \ast t \right), y, z\right)$$.

I also see how a properly aligned Energy-Momentum 4-vector $$p^{\mu}=\left( \frac{E}{c}, \left|\vec{p}\right|, 0, 0\right)$$ transforms appropriately so that $$p^{\acute{\mu}}=\Lambda_{v}^{\acute{\mu}} \ast p^{\mu}=\left( \gamma \ast \left(\frac{E}{c} - \beta \ast \left|\vec{p}\right| \right), \gamma \ast \left( \left|\vec{p}\right| - \beta \ast \frac{E}{c} \right), 0, 0 \right)$$.

It is rather simple to use the transformed position four-vector to get the inverse of the velocity addition formula: $$\acute{v}=\frac{v-v_s}{1-\left(\frac{v_s \ast v}{c^2}\right)}$$

...and its not too much harder to use $$v = \frac{\left|\vec{p}\right| \ast c^2}{E}$$, and the result of the Energy-momentum 4-vector, to arrive at the same formula.

However, if I use an appropriately aligned the velocity 4-vector, $$\zeta^{\mu}=\left(\gamma \ast c, \gamma \ast \left|\vec{v}\right|, 0, 0 \right)$$, I am not able to get the correct formulas. I am not even able to get the result that the speed of light is constant.

I don't know if there is some algabraic trickery needed, or if there is something more fundamental I am missing (I suspect that I am missing something fundamental).

So I ask if someone can do one of the following:
1. Derive the velocity addition formula (or its inverse) by using the velocity 4-vector and the matrix reprecentation of the Lorentz Transformation
2. Explain why it is misguided to attempt this.

2. Mar 20, 2009

### Fredrik

Staff Emeritus
I'd say that that matrix is the Lorentz transformation, and that the system of equations that corresponds to it is a trick that someone invented to make it less clear what we're really doing. (Only half kidding).

See e.g. #9 here. Note that I'm using units such that c=1.

You can also do it just by multiplying two Lorentz transformation matrices together and see what you get in the position(s) that's supposed to represent the velocity. I suggest you try that too. You don't have to let a Lorentz transformation act on anything to derive the velocity addition law.

Not sure what formulas you're not able to get. To see that the speed of light is invariant, find the eigenvectors of $\Lambda$. Start by finding the eigenvalues from the equation

$$0=\det(\Lambda-\lambda I)$$

Last edited: Mar 20, 2009
3. Mar 21, 2009

### ygolo

I guess the conceptual component I was missing was the "re-normalization" (for lack of a better word) of the resulting vector so that the first component was $$\gamma \ast c$$ (or $$c$$) depending on which version of velocity you are using.

I'm not sure what the mathematical justification for it is though. Neither the position 4-vector, nor the Energy-Momentum 4-vector needed "renormalization" (again I'm not sure what to call it).

I suppose all of this, "knowing the units of things implicitly" comes naturally to those who have worked with the theory for a while, but it is a bit annoying for new-commers.