# Magnitude of 4-acceleration

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
Mentor
2020 Award
I'm having a hard time showing this from the general expression for the 4-acceleration.

What general expression are you starting from?

PeterDonis
Mentor
2020 Award

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

Chestermiller
Mentor
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.

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 $$a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$

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! :-)

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...

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