**Summary:**

*Starting with the vector version of the Lorentz transform for a frame's position, we work backwards and arrive at the expressions and assumptions needed to derive it. The derivation is not in the notes. The transform is simply given along with a statement that it can be easily checked.*

The notes derive the generic vector form of the Lorentz transforms in three dimensions. The position transform is simply stated as:

$\vec{r}^{\prime} = \vec{r} + \dfrac{\gamma - 1}{v^2}\left(\vec{v}\cdot\vec{r}\right)\vec{v}-\gamma\vec{v}t$

A little massaging will make it more clear what's going on. First, the mystery of the $v^2$

$\vec{r}^{\prime} = \vec{r} + \left(\gamma - 1\right)\left(\dfrac{\vec{v}}{v}\cdot\vec{r}\right)\dfrac{\vec{v}}{v}-\gamma\vec{v}t$

OK, so, we're taking the component of the $\vec{r}$ displacement vector that lies along $\vec{v}$ and then laying that in the $\vec{v}$ direction. The $v^2$ was just to get us back to unit vectors in all cases. The $\left(\gamma - 1\right)$ expression is interesting because it will come up in the Thomas precession. Actually, I think we're looking at the exact same thing here. the $\left(\gamma - 1\right)$ terms winds up being the contribution to the Thomas precession for each boost along a circular path.

Let's write down the one dimensional transform that we're aping just to look for any similarities.

$x^{\prime} = \gamma\left(x-vt\right)$

Expanding the $\left(\gamma - 1\right)$ term out gives us,

$\dfrac{1}{\sqrt{1-v^2}} - 1$

$ = \dfrac{1 - \sqrt{1-v^2}}{\sqrt{1-v^2}}$

meh

Let's go down another path. Starting again at

$\vec{r}^{\prime} = \vec{r} + \left(\gamma - 1\right)\left(\dfrac{\vec{v}}{v}\cdot\vec{r}\right)\dfrac{\vec{v}}{v}-\gamma\vec{v}t$

Let's take the middle term and write it as

$\left(\gamma - 1\right)\left(\dfrac{\vec{v}}{v}\cdot\vec{r}\right)\dfrac{\vec{v}}{v}$

$ = \gamma |r_v|\hat{v} - |r_v|\hat{v}$

Now, if you look at what's going on, we're pulling out the portion of $\vec{r}$ along $\vec{v}$ and replacing it with a $\gamma$ scaled version of itself, exactly as we used the $\gamma$ scaled version of $x$ in the one dimensional equation. Length contraction only happens along the direction parallel to the velocity. To account for this, the expression first pulls the uncontracted component of the position vector in this direction out and then replaces it with the length contracted version.

**Question**: This shows once again that the only direction that matters in special relativity is the direction parallel to the velocity. Is there a proof of this somewhere?

**Question**: If we were asked to derive this, is the above reasoning in reverse acceptable? If not, what subtleties does it miss,and what do they teach?

## Tuesday, August 5, 2014

### Deriving the Vector Lorentz Transform, Backwards: EM II notes 2014_08_05

Posted by Hamilton Carter at 6:40 PM

Labels: physics, special relativity

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