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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?

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up vote 8 down vote accepted

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.

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sorry, I should have made it clear that the frame is moving along the negative x axis with velocity v, which I've now edited in. – Larry Harson May 25 '12 at 1:56
@LarryHarson: If you want to get zero, you need a frame moving with the particle, along the positive x-axis with velocity v. I know what you did, it was clear from the formulas. – Ron Maimon May 25 '12 at 2:31
the lhs is zero regardless of whether the proper frame is moving in the postive or negative x direction. – Larry Harson May 25 '12 at 14:15
@LarryHarson: Absolutely not. That's nonsense. Please think about it. The above is a complete answer, you are confused about boost direction. See comments below – Ron Maimon May 25 '12 at 18:16
the lhs is zero because it's the proper frame of the particle, independent of the other frame!! – Larry Harson May 26 '12 at 18:15

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.

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-1: Sorry. Are we back to before the time of negative numbers? "v" is a signed number in the Lorentz transformation, and the signs are consistent. The OP is confused because "boosting in the positive x-direction" can mean "boosting so that the frame are moving in the positive x direction" or "boosting so that the objects are moving in the positive x direction", and the two give opposite sign. The way to fix the sign is by intuition, because people half the people use one sense of the term and half use the other. – Ron Maimon May 25 '12 at 18:20
@RonMaimon in the LTs boosted along the x-axis in any direction as used by Einsten, v is a positive number which is why one set of transformations are different to the other through a sign change. The op is very clear about the particle and its frame moving along the negative x direction of the other frame – John McVirgo May 26 '12 at 1:07
I removed the downvote, but it isn't reasonable to give such a detailed explanation of a sign error. The OP has to sort this out himself, there is no efficient way to explain signs. – Ron Maimon May 26 '12 at 2:28
this looks like what I've done cos i've used the second transformation and so thanks +1 – Larry Harson May 26 '12 at 21:45
@LarryHarson: The two transformations are the same up to a change of sign, the second transformation with a positive v will not boost you to the rest frame, it is boosting in the wrong direction. This answer is not very good--- you should learn the signed transformation, with signed numbers. – Ron Maimon May 27 '12 at 3:19

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