What is the Invariant Magnitude of 4-Acceleration?

[copernicus1]

I know that if you go an accelerated observer's frame, the 4-acceleration is $$a^\mu=(0,{\bf a})$$ which means the magnitude is just $$-{\bf a}^2$$ which should be invariant. But I'm having a hard time showing this from the general expression for the 4-acceleration. I get to $$a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$ But I don't know where to go from here. Any ideas or am I just wrong somehow?

Thanks!

 

[PeterDonis]

I'm having a hard time showing this from the general expression for the 4-acceleration.

What general expression are you starting from?

 

[copernicus1]

I'm starting from the expression on this page (which I derived for myself so I think it's correct):

http://en.m.wikipedia.org/wiki/Four-acceleration

 

[PeterDonis]

I'm starting from the expression on this page

Yes, that will work. How are you getting from that to the second expression you gave in your OP?

 

[Chestermiller]

I don't think the three vector a is the same in the third equation as in the first and second equations. If you want to be working with the same 4 acceleration, you have to apply the Lorentz Transformation to its components.

Chet

 

[Bill_K]

I don't think the three vector a is the same in the third equation as in the first and second equations. If you want to be working with the same 4 acceleration, you have to apply the Lorentz Transformation to its components.

Chet

He's right, you know! In the first and second equation, a is the acceleration 3-vector in the instantaneous rest frame. In the third equation, a is the acceleration 3-vector in an arbitrary rest frame. They agree if you set v = 0 and γ = 1, but not otherwise.

 

[copernicus1]

I was under the impression {\bf a} was an arbitrary acceleration 3-vector either way? There's nothing in the Wikipedia article that puts any constraints on {\bf a}.

This is the way I went about it:

The 4-acceleration is: $$a^\mu=\left(\gamma^4\frac{{\bf v}\cdot{\bf a}}c,\gamma^2{\bf a}+\gamma^4\frac{\bf v\cdot a}{c^2}{\bf v}\right).$$

The magnitude should be given by $$a_\mu a^\mu=\gamma^8\frac{\left|{\bf v\cdot a}\right|^2}{c^2}-\gamma^4|{\bf a}|^2-\gamma^8\frac{|{\bf v\cdot a}|^2}{c^4}|{\bf v}|^2-2\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$

The first and third terms can be combined to give $$\left(1-\frac{|{\bf v}|^2}{c^2}\right)\gamma^8\frac{\left|{\bf v\cdot a}\right|^2}{c^2}=\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$ So we now have $$a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$

 

[copernicus1]

Oh I think I see what you're saying. So then, is my answer for a_\mu a^\mu the general expression for the norm? This is what I've been trying to find. If so, it's no wonder professors don't spend much time talking about it---it's hideous! :-)

 

[dauto]

You can still improve it as
a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.
a_\mu a^\mu=-\gamma^6[\gamma^{-2}|{\bf a}|^2-\frac{|{\bf v\cdot a}|^2}{c^2}].
a_\mu a^\mu=-\gamma^6[(1-\frac{|{\bf v}|^2}{c^2})|{\bf a}|^2-\frac{|{\bf v}|^2|{\bf a}|^2(cos \theta)^2}{c^2}].
a_\mu a^\mu=-\gamma^6 |{\bf a}|^2[1-\frac{|{\bf v}|^2}{c^2}(1+ (cos \theta)^2)].

That is, if that is an improvement...