Is there a subtlety to the Lorentz transformations one needs to be aware of?

Question

Playing around with the four acceleration $$\left(\gamma^4_u \frac{\vec a\cdot\vec u}{c^2}\vec u +\gamma^2_u\vec a,\, \gamma^4_u\frac{\vec a\cdot\vec u}{c}\right)$$ I found the time part for frame' moving with velocity $\vec v$ along the negative $x$ axis of the other frame, using the standard Lorentz transformation $$\gamma'^4_{u'}\frac{\vec a'\cdot\vec u'}{c^2} = \gamma_V\left(\gamma^4_u\frac{\vec a\cdot\vec u}{c^2} + \frac {V}{c^2}\left(\gamma^4_u \frac{\vec a\cdot\vec u}{c^2}\vec u+\gamma^2_u\vec a\right)\right)$$ Making the primed variables from the proper frame and noting $\vec u'= 0,\,\vec V = \vec u$ this reduces to $$\begin{align*}0 &= \gamma_V\left(\gamma^4_V\frac{ a_xV}{c^2} + \frac {V}{c^2}\left(\gamma^4_V \frac{a_xV^2}{c^2} +\gamma^2_Va_x\right)\right)\\ &= \gamma_V\left(\gamma^4_V\frac{ a_xV}{c^2} + \gamma^4_V\frac{ a_xV}{c^2}\right)\end{align*}$$

Which isn't correct, so what subtlety about the use of the Lorentz transform in this case have I missed?

Answer 1

Quick answer

Your only screw up was in boosting the wrong way--- you boosted so as to "double" the velocity (by velocity addition), not to zero it out. If you flip the sign of v,

$$ \gamma^4(a\cdot u) - v \cdot ( \gamma^4 (a\cdot u) u + \gamma^2 a ) $$

gives zero when v=u, since it becomes

$$ \gamma^2(a\cdot u) ( \gamma^2 - v^2 \gamma^2 - 1) $$

and $\gamma^2 - v^2\gamma^2 =1$ (the basic trignometric identity of cosh's and sinh's, or if you like, by explicitly substituting for $\gamma$)

How to not screw up

Your actual problem is much more serious, you are following the Wikipedia presentation! It is ridiculous and wrong to do relativity in three-vector form, and if you do, always start with a component calculation to know what you are doing, and derive all the formulas first, to get intuition and to check the formulas. The three vector forms are just a form of anti-pedagogy, designed to obfuscate very simple geometric expressions.

For example, in any three-vector form, the Lorentz transformation is ridiculously complicated looking:

$$ t' = \gamma_v (t - v\cdot x) $$ $$ x' = \gamma_v (x-vt) + (1-\gamma_v){(x\times v)\times v \over v^2}$$

(where I have set c=1, always always do this). The only point of the double-cross-product divided by the magnitude squared of v is to produce the component of x perpendicular to v, so as to cancel out the Lorentz contraction in the perpendicular direction. The reason is that the directions perpendicular behave differently than the directions along the motion, and the vector notation doesn't allow you to say this clearly.

This nonsense isn't doing anyone any favors, let alone yourself. The Lorentz transformation is as simple as a rotation, and 3-vector notation makes you and everyone else miserable and makes your calculations impossible to follow and impossible to check.

The four acceleration in the rest frame of the particle is, by definition, equal to the three acceleration in the rest frame.

$$a = (0, \vec{a}^0)$$

From this, you can see that the four acceleration is perpendicular to the velocity, and you can boost the four acceleration to velocity v in the x-direction, to see that it is equal to

$$(\gamma v a^0_x,\gamma a^0_x, a^0_y,a^0_z) $$

Why is this not the formula for the 4-acceleration? Because this is the acceleration in the rest-frame, the expression is in terms of the acceleration in the frame where the particle is moving with velocity v.

To find this expression, consider a particle whose trajectory has a velocity in the x-direction and a general acceleration:

$$ x(t) = (t,vt + {1\over 2} a_x t^2, {1\over 2} a_y t^2, {1\over 2} a_z t^2) $$

Now boost every point of the trajectory by -v in the x-direction to make the particle be at rest at t=0.

$$ x(t) = (\gamma_v(t-v^2t - {1\over 2} v a_x t^2), {1\over 2} \gamma a_x t^2, {1\over 2} a_y t^2, {1\over 2} a_z t^2)$$

In the rest frame, the time is $\gamma (1-v^2) t$ up to irrelevant higher orders, so it's $t'={1\over\gamma} t$, so that the trajectory is:

$$ x(t') = (t', {1\over 2} \gamma^3 a_x t'^3, {1\over 2} \gamma^2 a_y t'^2, {1\over 2} \gamma^2 a_z t'^2) $$

So that you can read off the 3-acceleration in the rest frame:

$$ a^0 = (0,\gamma^3 a_x, \gamma^2 a_y, \gamma^2 a_z) $$

and now boost this to find the four-acceleration

$$ a = (\gamma^4 v a_x, \gamma^4 a_x, \gamma^2 a_y, \gamma^2 a_z)$$

The Wikipedia answer reproduces this expression when v is in the x-direction, so it is correct. But it is hopelessly obfuscating the much-simpler component relations you can see above. This is always true--- never use 3d-vectors in relativity, it is the wrong formalism.

Answer 2

You haven't taken into acount the need to add the minus sign in $u_x = -v$ when boosting along the $-x$ direction, and this is a subtlety to Einstein's form of the Lorentz transformations one needs to be aware of.

There's one set of Lorentz transformations for boosting along the $+x$ axis, another set along the $-x$ axis, first derived by Einsten $$ x'=\gamma(x - vt)\quad t'=\gamma(t- \frac{vx}{c^2})$$ $$ x =\gamma(x' + vt')\quad t =\gamma(t'+ \frac{vx'}{c^2})$$

$v$ is a positive number in both, and the sign change of the boosted velocity is taken care of by the sign change in the transformations.

You've boosted along the $-x$ axis and used the correct second transformation and now just need to put the negative sign in: $u_x = -v$

$$\begin{align*}0 &= \gamma_V\left(-\gamma^4_V\frac{ a_xV}{c^2} + \frac {V}{c^2}\left(\gamma^4_V \frac{a_xV^2}{c^2} +\gamma^2_Va_x\right)\right)\\ &= \gamma_V\left(-\gamma^4_V\frac{ a_xV}{c^2} + \gamma^4_V\frac{ a_xV}{c^2}\right)=0\end{align*}$$

giving a consistent result.