Acceleration of a static test-particle

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1. Mar 2, 2012

Agerhell

What is the correct expression for the acceleration of a static test-particle in coordinate time according to the Schwarzshild solution? I am a bit confused. I would like it to be the same as classically, $d\bar{v}/dt=-\frac{GM}{r^2}\hat{r}$, but according to "reflections on relativity":

http://www.mathpages.com/rr/s6-04/6-04.htm

The expression for the acceleration of a static test-particle is only the same as classically if you replace coordinate time with proper time, (see equation 6). I am guessing that this means that according to the Schwarzschild solution the acceleration of a static test-particle in coordinate time is

$$\frac{d^2 r}{dt^2} = - \frac{GM / {r^2}} { (1 - {{2GM} {/} {rc^2}}) }$$

Is this correct? I would like to have it that relativity will only come in to play when the test-particle has started moving but according to the above it will also play a role in the static case...

2. Mar 2, 2012

Bill_K

Ok, we have to be careful what we mean. If you really truly mean the acceleration, as in aμ = δvμ/δτ, the acceleration of a particle falling freely is zero. If the particle is 'static', i.e. remains at constant r, the radial component of the acceleration is, as you say, GM/r2, but this is outward.

To change the two d/dτ's to d/dt's one must divide by two factors of γ = dt/dτ, making it (r - 2GM)GM/r3. (I think you have the correction factor upside down.)

3. Mar 2, 2012

Staff: Mentor

But you also have to correct for the radial coordinate r not being the same as physical distance in the radial direction; that correction cancels one of the factors of gamma. The correct expression is:

$$a = \frac{GM}{r^{2} \sqrt{1 - \frac{2 G M}{c^{2} r}}}$$

4. Mar 2, 2012

Staff: Mentor

"Relavity" includes general relativity as well as special relativity. The "acceleration" you are trying to compute is really an effect of GR, not SR, so the extra factor that comes into the formula is not a function of how fast the particle is moving (SR), but how strong gravity is (GR). You will note that the extra factor goes to unity as R gets large compared to GM/c^2; for a case like the Earth or the Sun, R >> GM/c^2, so the correction is very small and the "classical" Newtonian formula works fine.

5. Mar 2, 2012

yuiop

See post #1 of this old thread https://www.physicsforums.com/showthread.php?p=2710548#post2710548

(Note that the second equation given for the acceleration measured by a local observer is the same as the one given by Peter, just expressed in a different format.)

Just to give you an idea where the equation comes from, start with the equation given by mathpages for the local acceleration in terms of the proper time and coordinate distance:

$$\frac{d^2 r}{d\tau^2} = \frac{GM}{r^2}$$

and given that from the Schwarzschild metric the relation between proper time and coordinate time is:

$$\frac {d\tau}{dt} = \sqrt{1-2GM/rc^2}$$

then:

$$\frac{d^2 r}{d\tau^2}\frac{d\tau^2}{dt^2} = \frac{GM}{r^2}\frac{d\tau^2}{dt^2}$$

$$\frac{d^2 r}{dt^2} = \frac{GM(1-2GM/rc^2)}{r^2}$$

which is the acceleration in purely coordinate terms (as given by Bill_K).

Similarly if you convert coordinate distance (dr) to proper distance (dr') you obtain the proper acceleration as measured locally as:

$$\frac{d^2 r'}{dt^2} = \frac{GM}{r^2 \sqrt{1-2GM/rc^2}}$$

(as given by PeterDonnis).

Last edited: Mar 2, 2012
6. Mar 2, 2012

Bill_K

r of course is the Schwarzschild radius. The expression is correct as I gave it.

EDIT: r is the Schwarzchild radial coordinate

Last edited: Mar 2, 2012
7. Mar 2, 2012

Staff: Mentor

I'm confused; I thought "r" was supposed to be the radial coordinate, and "2M" (or 2GM/c^2 in conventional units) was the Schwarzschild radius.

After reading yuiop's post, I see that we were giving expressions for two different things. The expression I was trying to give, as yuiop noted, was for *proper* acceleration--what would actually be measured by an accelerometer that is "static" at a constant radial coordinate r.

8. Mar 2, 2012

9. Mar 2, 2012

pervect

Staff Emeritus
The expressions being written down are for the proper acceleration of the worldline of a static observer - coordinate time isn't used or needed.

If you want to somehow convert this to "coordinate time" rather than something that's specifiable without using coordinates, you'll have to do this bit of interpretation for yourself - I don't quite follow why you'd want to do that. I'm hoping that perhaps you didn't really want to.

The logical answer for the "coordinate accleration" is zero, because the coordinates of a static observer are not changing with respect to coordinate time by definition. So what people are posting is the proper acceleration that an accelerometer would measure.

In special relativity you would write

$$a = \frac{d u}{d \tau}$$

where a is the 4-accleration vector, and u is the four-velocity vector. Given the acceleration 4-vector, it's invariant length would give the magnitude of the proper acceleration of the worldline, which is what the formulas you're being given are all about.

In general relativity, you have to replace the differential with a covariant derivative. I'm not sure if you're familiar with them, and if you are what sort of notation you prefer, so I"m not going to comment further on how to take the covariant derivative, as it would probalby be more confusing than helpful without knowing more about your background. (Which I seem to recall you mentioned once, but I'm afraid I forgot). You also use the metric in GR to compute the length of the four-vector.

I'm hoping that you do know about four-vectors...

Last edited: Mar 2, 2012