I Proper Acceleration: Why d^2x^α/dτ^2 is Not Ideal

[kent davidge]

Why $d^2 x^\alpha / d\tau^2$ is not considered a good expression for the proper acceleration (of a massive particle whose proper time is $\tau$ and coordinates are $x^\alpha$)?

 

[SiennaTheGr8]

It's just a matter of terminology.

The quantity you've given is the four-acceleration.

The term proper acceleration refers to the three-acceleration as measured in the traveler's instantaneous rest frame.

They're related, though:

• In the traveler's instantaneous rest frame, the time component of the four-acceleration vanishes and the three-vector spatial component is just the proper acceleration.
• It follows that the magnitude of the four-acceleration is equal to the magnitude of the proper acceleration.

 

[PeterDonis]

Why $d^2 x^\alpha / d\tau^2$ is not considered a good expression for the proper acceleration

Who says it isn't?

 

[PeterDonis]

The term proper acceleration refers to the three-acceleration as measured in the traveler's instantaneous rest frame.

According to whom?

 

[stevendaryl]

Why $d^2 x^\alpha / d\tau^2$ is not considered a good expression for the proper acceleration (of a massive particle whose proper time is $\tau$ and coordinates are $x^\alpha$)?

Well, in two dimensions, consider an object that is traveling in a circle of radius $R$ at constant angular velocity. In polar coordinates:

$x^r = R$
$x^\theta = \omega \tau$

So $\frac{d^2 x^\alpha}{d\tau^2} = 0$. But obviously, it's undergoing centripetal acceleration.

The error is that velocity is a vector, which can be written in the form $V = V^\alpha e_\alpha$. Velocity can change either by having the components $V^\alpha$ change, or by having the basis vectors $e_\alpha$ change. So a correct approach is not to use components, but to use vectors:

$V = V^\alpha e_\alpha = \frac{dx^\alpha}{d\tau} e_\alpha$
$A = \frac{dV}{d\tau} = \frac{d^2 x^\alpha}{d\tau^2} e_\alpha + \frac{dx^\alpha}{d\tau} \frac{d e_\alpha}{d\tau}$

To compute $\frac{d e_\alpha}{d\tau}$, you use the chain rule: $\frac{d e_\alpha}{d\tau} = (\nabla_\beta\ e_\alpha) \frac{d x^\beta}{d\tau}$

By definition, $\nabla_\beta e_\alpha = \Gamma^\gamma_{\beta \alpha} e_\gamma$. So

$\frac{dx^\alpha}{d\tau} \frac{d e_\alpha}{d\tau} = \frac{dx^\alpha}{d\tau} \Gamma^\gamma_{\beta \alpha} e_\gamma = \frac{dx^\gamma}{d\tau} \Gamma^\alpha_{\beta \gamma} e_\alpha$ (renaming the dummy indices $\alpha$ and $\gamma$ to $\gamma$ and $\alpha$). So

$A = (\frac{d^2 x^\alpha}{d\tau^2} + \Gamma^\alpha_{\beta \gamma} \frac{dx^\gamma}{d\tau} \frac{dx^\beta}{d\tau}) e_\alpha$

So $A^\alpha = \frac{d^2 x^\alpha}{d\tau^2} + \Gamma^\alpha_{\beta \gamma} \frac{dx^\gamma}{d\tau} \frac{dx^\beta}{d\tau}$

The $\Gamma^\alpha_{\beta \gamma}$ takes into account the variation of the basis vectors from point to point.

 

[kent davidge]

Thank you @stevendaryl , very helpful answer.

 

[SiennaTheGr8]

Yes, I should have qualified my comment by noting that usage varies.

There are at least 3 competing conventions for using the term proper:

1. To describe frame-dependent quantities as measured in the relevant (instantaneous) rest frame. Almost everybody does this for proper time, proper distance, and proper length. Many (most?) people do this for proper acceleration and proper force, too, though generally not for "proper velocity" (which would always be the zero three-vector).
2. To describe the proper-time derivatives of frame-dependent quantities (e.g., to call $d \vec p / d \tau$ "proper force," or to call $d \vec r / d \tau$ "proper velocity" instead of "celerity").
3. As another word for four-vectors (so "proper velocity" would be the four-velocity, and "proper acceleration" would be the four-acceleration).

I'm partial to the first of these conventions. It strikes me as the most consistent and the most useful.

 

[kent davidge]

@SiennaTheGr8 , looking on @stevendaryl 's post, I think maybe you mean that the proper acceleration equals my expression in the OP in a locally inertial frame? Instead of

In the traveler's instantaneous rest frame, the time component of the four-acceleration vanishes and the three-vector spatial component is just the proper acceleration

 

[Orodruin]

The quantity you've given is the four-acceleration.

I think that it is worth explicitly pointing out that, as shown in #5, this is only true in standard coordinates on Minkowski space. The generalised expression for the 4-acceleration is $A=\nabla_V V$, with $V$ being the 4-velocity.

 

[SiennaTheGr8]

@SiennaTheGr8 , looking on @stevendaryl 's post, I think maybe you mean that the proper acceleration equals my expression in the OP in a locally inertial frame?

No, the "proper acceleration" I was talking about in that post is a three-vector, not a four-vector.

 

[pervect]

Why $d^2 x^\alpha / d\tau^2$ is not considered a good expression for the proper acceleration (of a massive particle whose proper time is $\tau$ and coordinates are $x^\alpha$)?

That expression is a 4-vector, and is usually called the "four-acceleration".

So we'd write $a^i = d^2 x^i / d\tau^2$, and call $a^i$ the four-acceleration.

The magnitiude of $a^i$, namely $\sqrt{|g_{ij}\, a^i \,a^j|} = \sqrt{|a_j \,a^j|}$ is usually called the proper acceleration. The absolute value signs may or may not be needed, depending on sign conventions, but it's safer to include them.

It's basically just a matter of semantics, and style. I wouldn't guarantee that nobody has ever called $d^2 x^\alpha / d\tau^2$ the proper acceleration, but I'd call it the 4-acceleration myself.

 

[SiennaTheGr8]

Here's what I meant, @kent davidge.

In Minkowski/Lorentz coordinates in flat spacetime, with $c=1$, the four-acceleration is:

$\vec A = \dfrac{d \vec V}{d \tau} = \dfrac{d}{d \tau} \left( \cosh \phi, \, \hat v \sinh \phi \right) = \left( \sinh \phi \, \dfrac{d \phi}{d \tau}, \, \hat v \cosh \phi \, \dfrac{d \phi}{d \tau} + \dfrac{d \hat v}{d \tau} \, \sinh \phi \right)$,

where $\vec V$ is the four-velocity, $\tau$ is the proper time, and $\phi$ is the rapidity $\tanh^{-1} v$. In the traveler's instantaneous rest frame ($\phi = 0$), that's just:

$\vec A_{rest} = \left( 0, \, \hat v \, \dfrac{d \phi}{d \tau} \right)$.

But in the rest frame it's also the case that $d \phi / d \tau = dv / dt = a$ and $\hat v = \hat a$, so:

$\vec A_{rest} = \left( 0, \, \vec a \right)$.

 

[SiennaTheGr8]

But in the rest frame it's also the case that ... $\hat v = \hat a$

Come to think of it, this is kind of a strange thing to say.

Perhaps somebody could set me straight—is it "okay" to resolve the acceleration vector into components parallel and perpendicular to the velocity in the rest frame like this?