# I Proper Acceleration in GR

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1. Oct 3, 2016

### novice_hack

I know that in Special Relativity, proper acceleration is understood as: a*ga^3, where ga is the lorentz term and 'a' is coordinate acceleration. Is there a corresponding expression for proper acceleration within the various geodesics that result from solutions to Einstein's Field Equations? If so, I would like to know what the expression is for proper acceleration in the Schwarzschild Geodesic. If not, could someone explain to me why not?

2. Oct 3, 2016

### Orodruin

Staff Emeritus
This is not really correct and it depends on the direction of acceleration. The proper acceleration is the acceleration of an object in its own instantaneous rest frame and is also the acceleration an object will feel subjected to.

The proper acceleration along any geodesic is zero by definition.

3. Oct 3, 2016

### Ibix

Proper acceleration is the more fundamental quantity - it's what you yourself measure with your accelerometers when you accelerate. So probably better to say that coordinate acceleration is $a=a_0/\gamma^3$. As Orodruin says, however, this is only true if the acceleration and velocity are parallel. It's $a=a_0/\gamma^2$ (Edit: not $a=a_0/\gamma$ as I originally wrote) if they are perpendicular, and somewhere between for other cases.

Pretty much the definition of a geodesic is that the proper acceleration of anything following it is zero. It's the generalisation of a straight line to curved spacetime. That's why you are weightless in free fall. If you do undergo proper acceleration you do not followa geodesic. Unfortunately there is no unique answer for the relationship between coordinate and proper acceleration in general because there is no standard choice for coordinates.

If you specify a coordinate system and a spacetime then the question can be answered. Be prepared for it to be position and coordinate velocity dependent. Also for it to be fairly arbitrary and meaningless.

Last edited: Oct 4, 2016
4. Oct 3, 2016

### SiennaTheGr8

Isn't the perpendicular case $a_0 = \gamma^2 a$?

I believe the general formula for proper acceleration is: $a_0 = \gamma^3 a / \gamma_\bot$, where $\gamma_\bot = (1 - v^2_\bot / c^2)^{-1/2}$, and $v_\bot$ is the component of $\vec v$ that's perpendicular to $\vec a$.

So if $\vec v \, \bot \, \vec a$, then $v_\bot = v$ and $\gamma_\bot = \gamma$. Thus, $a_0 = \gamma^2 a$.

(Unless I'm wrong!)

5. Oct 4, 2016

### Ibix

Yes. Thanks - corrected above.