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$3$-acceleration can not be constant in a relativistic system. Because $\vec a^2$ is not Lorentz invariant. Does it mean that Lorentz invariance works only for $4$-vectors? How this should be understood properly? Does it mean to define the constant acceleration which Lorentz invariant the condition $\vec a^2$ is not enough?

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    $\begingroup$ Yes, Lorentz invariance pertains to $4$-vectors. I'm having a hard time parsing your last question. To calculate the $4$-vector acceleration, differentiate the $4$-vector velocity with respect to proper time. $\endgroup$ – Cinaed Simson Jan 26 at 19:34
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Four-vectors are not invariant. They are co-variant, i.e. they, by which I mean the components of the vector, change in a very specific fashion when one changes coordinates.

Scalars, on the other hand can be Lorentz-invariant. Though one has to be careful: for example, charge density is 1d, but it is not invariant (see length contraction).

The Lorentz-covariance is not that scary, you simply have to put it in terms that make sense for all observers equally well. For example I could define a 3-vector

$\mathbf{V}=V^x \mathbf{\hat{x}}+V^y \mathbf{\hat{y}}+V^z \mathbf{\hat{z}}$

in my reference frame, and then extend it to four-dimensional vector by specifying that, for me, this vector has no temporal component:

$V=V^x \partial_x + V^y \partial_y + V^z \partial_z$

Now this prescription is sufficient for any space-time observer.

Regarding the acceleration. The four-acceleration is related to the curvature of the world-line of the observer under acceleration. So if your world-line is $r^\mu=r^\mu\left(\tau\right)$ ($\tau$ is the proper time). Then

four-velocity: $\frac{dr^\mu}{d\tau}=u^\mu$

four-acceleration: $\frac{d u^\mu}{d\tau}=a^\mu$ (assuming flat space-time and trivial coordinates)

It is possible to show that in any reference frame where you are instantaneously at rest (you cannot have a full rest-frame if you are accelerating), $u^\mu=\left(c,\mathbf{0}\right)^\mu$, $a^\mu=\left(0,\mathbf{a}\right)$, where $\mathbf{a}$ is your usual acceleration.

Then $a^\mu a_\mu = -\mathbf{a}.\mathbf{a}=-a^2$

So the squared magnitude of your acceleration, i.e. the acceleration you would measure with accelerometer, is actually the magnitude of your four-acceleration, and is therefore invariant (as any other magnitude of a four-vector).


Misner-Thorne-Wheeler in II.6 have a good section about the acceleration in Special Relativity. Basically, non-intertial frame is associated with straight world-line, whereas the worldline of an accelerating observer is curved. However in regions of space small enough for cuvature of accelerating observer to be insignificant, one can talk about observer's rest frame, but with great care. This sound wooly, so let us do the full calculation.

So, you are accelerating observer. I am intertial. I see your world-line as $x^\mu=\left(ct,\,\mathbf{r}\right)$. Irrespective of whether you are accelerating or not, my time $t$ is related to your proper time $\tau$ through $\frac{dt}{d\tau}=\gamma=1/\sqrt{1-v^2/c^2}$, where $\mathbf{v}=d\mathbf{r}/dt$. Think of it as a tilt of your world-line relative to my worldline.

Your four-velcity is $u^\mu=\frac{dx^\mu}{d\tau}=\gamma\frac{dx^\mu}{dt}=\gamma\left(c,\,\mathbf{v}\right)^\mu$

As you move and accelerate your four-velocity will change, the tensorial quantity that quanitfies this change is the four-acceleration:

$a^\mu=\frac{Du^\mu}{D\tau}=u^\nu \nabla_\nu u^\mu = \frac{du^\mu}{d\tau}+\Gamma^\mu_{\nu\kappa}u^\nu u^\kappa=\gamma \frac{du^\mu}{dt}+\Gamma^\mu_{\nu\kappa}u^\nu u^\kappa$

The $\Gamma$-term is the Christoffel symbol that allows to work with any coordinate system. If my coordinates are plain Eucledian then $\Gamma=0$. Lets keep it this way. The derivative in the first term can be evaluated in a usual form. Using $\mathbf{a}=d\mathbf{v}/dt$, I find:

$a^\mu=\gamma^4 \cdot\frac{\mathbf{v}.\mathbf{a}}{c^2}\cdot\left(c,\, \mathbf{v}\right)^\mu + \gamma^2 \cdot \left(0,\,\mathbf{a}\right)^\mu$

The magnitude of this four-vector is:

$a^\mu a_\mu = -\gamma^4 a^2 - \gamma^6 \left(\mathbf{v}.\mathbf{a}\right)^2/c^2$

Now, if you are instantaneusly at rest relative to me, then: $a^\mu=\left(0,\,\mathbf{a}\right)^\mu$ and $a^\mu a_\mu = -a^2$, so the magnitude of the acceleration I would observe, would be related directly to the magnitude of your four-acceleration, but only at that instance in time. Later a different observer, in intertial motion relative to me, would observe the same effect.etc. This is what I meant by instantaneous rest frame.

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  • $\begingroup$ I am not saying 4-vectors are invariant. It is meant that the spacetime distance is invariant under Lorentz transformations. Thank you! $\endgroup$ – Constantin Jan 27 at 4:18
  • $\begingroup$ "you cannot have a full rest-frame if you are accelerating" Do you mean by this the presence of acceleration even in a comoving with the accelerating object frame? $\endgroup$ – Constantin Jan 27 at 4:23
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    $\begingroup$ Reference frames are not accelerating. But as you are accelerating your velocity will change. At any specific time I can imagine an intertial observer relative to whom your velocity is negiligible. See edit above for a more general treatment $\endgroup$ – Cryo Jan 28 at 0:10

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