Radial Motion in Schwarzschild's Geometry

[DrGreg]

I simply meant that the time component and the radial component of the acceleration vector are calculated to be zero by the ground observer. Just use the formula in posting 23.

OK, so you are saying the 4-acceleration (tensor) is zero for a free-falling object, which is correct. And in fact the magnitude of the 4-acceleration is the proper acceleration. This all agrees with what I said in post #32.

I'm still not clear what you mean by "physical acceleration".

 

[Anamitra]

The acceleration four vector has zero components for radial geodesic motion in Schwarzschild geometry.
Now these components are covariant derivatives. A covariant derivative has an ordinary derivative and an affine connection containing the Christoffel symbols.
The two parts are cancelling out here.
But in curved space time the ordinary derivative is not a rate measurer because it does not take into account the parallel transport.The entire thing[including the affine connection] is the rate measurer. So the radial component of the acceleration measures the rate of change of velocity component in the radial direction.

Let us consider the Earth itself in this context.Should an observer sitting on the Earth's surface record zero radial acceleration for a falling body?

 

[DrGreg]

The acceleration four vector has zero components for radial geodesic motion in Schwarzschild geometry.
Now these components are covariant derivatives. A covariant derivative has an ordinary derivative and an affine connection containing the Christoffel symbols.
The two parts are cancelling out here.
But in curved space time the ordinary derivative is not a rate measurer because it does not take into account the parallel transport.The entire thing[including the affine connection] is the rate measurer.

OK so far.

So the radial component of the acceleration measures the rate of change of velocity component in the radial direction.

That is true only in coordinates where the Christoffel symbols vanish at the event of interest, viz. a locally-inertial frame.

Let us consider the Earth itself in this context. Should an observer sitting on the Earth's surface record zero radial acceleration for a falling body?

Obviously not, so your previous sentence must be wrong. The observer is observing local coordinate acceleration in local coordinates. He would have to do a calculation with Christoffel symbols to work out the 4-acceleration.

Roughly speaking, the 4-acceleration measures acceleration relative to a free-falling observer.

 

[Anamitra]

If I am standing on the surface of a planet and an object is falling radially towards me. If it is at a distance from me what formula should I use to calculate the acceleration[in curved spacetime]?
[What formula should we use to define acceleration in such a situation?]

 

[Passionflower]

If I am standing on the surface of a planet and an object is falling radially towards me. If it is at a distance from me what formula should I use to calculate the acceleration[in curved spacetime]?
[What formula should we use to define acceleration in such a situation?]

Since the proper acceleration of the object is zero there is only a coordinate acceleration. At a distance and in curved spacetime I do not believe there is one unique physical answer to this question.

 

[Anamitra]

Any observer sitting on a planet meets your first to conditions: coordinate acceleration of zero and non-zero physical acceleration.

This was said in relation to a falling body [which is at a distance from the observer on the planet]
By coordinate acceleration Pallen seems to indicate the components of the acceleration four vector.

I would request Pallen to define [by providing the formula] physical acceleration which he says is non zero

 

[PAllen]

This was said in relation to a falling body [which is at a distance from the observer on the planet]
By coordinate acceleration Pallen seems to indicate the components of the acceleration four vector.

I would request Pallen to define [by providing the formula] physical acceleration which he says is non zero

Coordinate acceleration is second derivative of spacelike coordinates by coordinate time. For example, second derivative of r by t in Schwarzschild coordinates for a radial world line (free fall, static, or otherwise).

By physical acceleration, I meant acceleration you can feel and measure 'inside a black box'. That is, to me, physical acceleration = proper acceleration.

Thus, a planet surface observer has zero coordinate acceleration but nonzero physical acceleretion. A free fall observer has non-zero coordinate acceleration but zero physical acceleration.

However, let's forget physical acceleration as a term, since it is non-standard, and stick to proper acceleration since it is standard. Also, coordinate acceleration is completely well defined in any given coordinate system.

I think what you're really looking for here is how one observer measures another's acceleration using some reasonable measuring system. That is more complex, and only well defined in respect to some measuring system. I think a good way to get a handle on that for your scenario is to define Fermi-normal coordinates for a specified static (planetary surface) observer, and compute coordinate acceleration of a nearby free-falling world line in those coordinates. I do not know of a link for where to find such a calculation. I know that the general method for doing this is covered in MTW, among other places.

 

[Passionflower]

I think a good way to get a handle on that for your scenario is to define Fermi-normal coordinates for a specified static (planetary surface) observer, and compute coordinate acceleration of a nearby free-falling world line in those coordinates.

Yes I agree with Paul here. When you want to measure the acceleration of the free falling object exactly at the location of he stationary observer it is a lot simpler, but due to the curvature of spacetime the meaning of velocity at a distance is not unambiguous.

 

[George Jones]

I think a good way to get a handle on that for your scenario is to define Fermi-normal coordinates for a specified static (planetary surface) observer, and compute coordinate acceleration of a nearby free-falling world line in those coordinates. I do not know of a link for where to find such a calculation. I know that the general method for doing this is covered in MTW, among other places.

I have done the calculation (I don't know of any explicit references), and the result is very surprising. It is instructive to first consider acceleration in special relativity. See equation (16) and Figure 1 from

http://arxiv.org/abs/gr-qc/0406118.

 

[Mentz114]

I have done the calculation (I don't know of any explicit references), and the result is very surprising. It is instructive to first consider acceleration in special relativity. See equation (16) and Figure 1 from

http://arxiv.org/abs/gr-qc/0406118.

Yes, it is a surprise to see that happening in SR.

In the Schwarzschild geometry the EOM for radial motion is

$$\ddot{r}= \Gamma^r_{rr} \dot{r}^2+\Gamma^r_{tt} \dot{t}^2$$

the overdot indicates a derivative wrt s, an affine parameter, as always. I hope to show that the acceleration changes sign. After substituting the Christoffel symbols and some algebra I get

$$\ddot{r}=\frac{M}{r^2} \left( \frac{\dot{r}^2 - \dot{t}^2(1-2M/r)^2}{1-2M/r} \right)$$The numerator of the term in brackets has a zero and changes sign when
$$\dot{r}^2 = \dot{t}^2(1-2M/r)^2$$

taking the square root and dividing by $\dot{t}$ this becomes

$$\frac{dr}{dt}=(1-2M/r)$$

so when $dr/dt$ exceeds the RHS the acceleration becomes positive - repulsion ?
This is obviously a coordinate dependent effect and will only be 'seen' by the observer at infinity.

If I've made a mistake please tell me.

 

[Anamitra]

Some simple calculations for radial geodesic motion [in Schwarzschild Geometry]

$${a}^{\alpha}{=}{{u}^{\beta}}{{\nabla}_{\beta}}{u}^{\alpha}$$
"a" represents four acceleration and "u" four velocity
Now coming to the time and the radial components of acceleration we have,

(1):
$${a}^{t}{=}{u}^{t}{{\nabla}_{t}}{u}^{t}{+}{u}^{r}{{\nabla}_{r}}{u}^{t}$$

(2):
$${a}^{r}{=}{u}^{t}{{\nabla}_{t}}{u}^{r}{+}{u}^{r}{{\nabla}_{r}}{u}^{r}$$

For radial geodesic motion each of the above relations reduce to zero value, individually.

But it would be interesting to consider the sum of the second terms in the equations (1) and (2). That would represent the sum of the contributions in the radial direction that one gets from the differentiation of the time component and the radial component of the velocity four vector. Since each of these terms results from covariant differentiation they represent rate measurement correctly in curved spacetime,taking into account parallel transport.

We could use the relations:

(3):
$${g}_{tt}{{(}{u}^{t}{)}}^{2} {-} {g}_{rr} {({u}_{r})}^{2} {=}{1}$$

(4):
$${(}{1}{-} \frac{2M}{r}{)} {{u}^{t}}{=}{e}$$
where "e" is a constant

We get equation (4) since the metric is independent of time and (1,0,0,0) is a Killing vector.

The sum[of the second terms in eqn (1) and (2)] works out to a non-zero value[in general] where $\frac{{\partial}{u}^{r}}{{\partial}{t}}$ is a dominating term.This becomes more convincing for r>>2M.

 

[Anamitra]

In posting 46, I have said --"But it would be interesting to consider the sum of the second terms in the equations (1) and (2)".
Here I have meant the sum of the second terms on the right hand side of equations(1) and (2)

 

[Anamitra]

We consider the sum:

(1):
$${u}^{r}{\nabla}_{r}{u}^{r}{+}{u}^{r}{\nabla}_{r}{u}^{t}$$
$${=} {u}^{r}{(}{\frac{{\partial}{u}^{r}}{{\partial}{r}}}{+}{{\Gamma}^{r}}_{rr}{u}^{r}{)}{+}{u}^{r}{(} {\frac{{\partial}{u}^{t}}{{\partial}{r}}} {+} {{\Gamma}^{t}}_{rt}{u}^{t}{)}$$
$${=} {u}^{r}{\frac{{\partial}{u}^{r}}{{\partial}{r}}}{+}{u}^{r}{\frac{{\partial}{u}^{t}}{{\partial}{r}}} {+} {{\Gamma}^{r}}_{rr}{{u}^{r}}^{2} {+} {{\Gamma}^{t}}_{rt}{u}^{t}{u}^{r}$$

Evaluation of the first two terms:
$${ \frac{{d}{u}^{r}}{{ d}{\tau}}}{=}{\frac{{\partial}{u}^{r}}{{\partial}{r}}}{u}^{r}{+}{\frac{{\partial}{u}^{r}}{{\partial}{t}}}{u}^{t}$$

But
$$\frac{{\partial}{u}^{r}} {{\partial}{t}}{=}{0}$$
[From the last two equations of the last posting]

Therefore,

(2):
$${ \frac{{d}{u}^{r}}{{ d}{\tau}}}{=}{\frac{{\partial}{u}^{r}}{{\partial}{r}}}{u}^{r}$$
Again,
$${ \frac{{d}{u}^{t}}{{ d}{\tau}}}{=}{\frac{{\partial}{u}^{t}}{{\partial}{r}}}{u}^{r}{+}{\frac{{\partial}{u}^{t}}{{\partial}{t}}}{u}^{t}$$
But,
$${\frac{{\partial}{u}^{t}}{{\partial}{t}}}{=}{0}$$
[From the second last equation of the last posting]

Therefore,

(3):
$${ \frac{{d}{u}^{t}}{{ d}{\tau}}}{=}{\frac{{\partial}{u}^{t}}{{\partial}{r}}}{u}^{r}$$

So the sum of the first two terms of equation (1) work out to:

(4):
$${ \frac{{d}{u}^{r}}{{ d}{\tau}}}{+}{ \frac{{d}{u}^{t}}{{ d}{\tau}}}$$
$${=}{-}{{ \Gamma}^{r}}_{tt}{{u}^{t}}^{2}{-}{{ \Gamma}^{r}}_{rr}{{u}^{r}}^{2}{-}{{ \Gamma}^{t}}_{tr}{u}^{t}{u}^{r}$$

[The last term of equation (4)includes two terms due to interchange of r and t. But one should not interchange r and t in the last term of equation (1)]
One may evaluate expression (1) by using equations (2) (3) and (4) in (1). The last two equations of posting 46 have to be used along with the relevant values of the Christoffel symbols.
The expression obtained has an inverse square term[wrt “r”] which is a dominant one especially for sufficiently large values of r.