Radial Motion in Schwarzschild's Geometry

[Anamitra]

We consider the geodesic equation:

(1):
[tex]
\frac{ d^2 x^\alpha}{ d \tau^2} =
- {{\Gamma}^{\alpha}}_{\beta \gamma} \frac{d x^{\beta}}{d{\tau}} \frac{d x^{\gamma}}{d {\tau}}
[/tex]
For radial motion in Schwarzschild geometry

(2):
[tex]\frac{d^2 r}{d \tau^2} =
- {M / {r^2}} { (1 - {{2M} {/} {r}}) }
{( \frac{dt}{d \tau} )}^2
+ {M} {/}{r^2}
{( 1- {2M}{/}{r} )}^{-1}
{( \frac{dr}{d \tau} )}^2[/tex]

Again for radial motion we have

(3):
[tex]{d}\tau^{2} =
{(}{1}{-}{2M}{/}{r}{)}{dt}^{2}
{-}{{(}{1}{-}{{2M}{/}{r}}{)}}^{-1}{dr}^{2}[/tex]

Dividing both sides of equation (3) by [tex]{{d}{\tau}}^{2}[/tex] we have,
[tex]{1}{=}{(}{1}{-}{{2M}{/}{r}}{)}{{(}{\frac{dt}{{d}{\tau}}}{)}^{2}{-}{{(}{1}{-}{{2M}{/}{r}}{)}}^{-1}{(}{\frac{dr}{{d}{\tau}}{)}}^{2}}[/tex] ------------------------------ (4)
Using relation (4) in equation (2)
[tex]{\frac{{d}^{2}{r}}{{d}{\tau}^{2}}}{=}{-}{\frac {M}{{r}^{2}}}[/tex]
The inverse square law is valid accurately if proper time is used.Here 'r' represents coordinate distance along the radius

 

[Mentz114]

The 'r' in those equations is only equivalent to the Newtonian r in a very weak field.

 

[Anamitra]

Observations:
1. For non relativistic motion[in regions of weak space-time curvature] proper time may be approximated by coordinate time.[tex]{{[}{(}{d}{\tau}{/}{c}{)}}{\approx}{dt}{]}[/tex] and c=1 for the natural system of units.
2. If space time curvature is weak coordinate separation and physical separation are approximately equal

We have Newton's Universal law of Gravitation in its exact form. It is to be noted that in the natural system of units c=1. If we do not choose this system we have to consider "c" right from the beginning.I mean right from Schwarzschild's metric[t has to be replaced by "ct":This achieves the consistency of dimensions.]
We arrive at Newton's Universal Law of Gravitation provided the first two conditions are satisfied.

 

[Anamitra]

In the last equation of posting 1 we may use the following substitutions:

1.[tex]{\tau}{\approx}{ct}[/tex]
2. [tex]{M}{--}{>}{GM}{/}{c}^{2}[/tex]

We get the Universal Law of Gravitation remembering that we are in a region of weak spacetime curvature and the v<<c.
We must also note that in our formalism [tex]{d}{\tau}[/tex] should have the dimension of length.

 

[Mentz114]

That is what I meant. In the weak field Newton's laws are recovered and there are several ways to show this.

 

[Anamitra]

The last equation in post 1 is of a general nature inside the system---general in the sense that we may consider strong curvature regions and high speed particles.It is also of a simple form, reminding one of the universal law of gravitation.One can perform simple calculations in relation to objects falling in regions of strong spacetime curvature. The mathematics would be quite similar to the classical ones.

 

[Mentz114]

An observer who remains at a constant radius r in the Schwarzschild vacuum experiences a proper acceleration

[tex]
\frac{m/r^2}{\sqrt{1-2m/r}}
[/tex]

( in Schwarzschild coordinates) which reflects the fact that there is a coordinate singularity at r=2m. For large r the factor [itex]1/\sqrt(1-2m/r)[/itex] tends to 1 and the Newtonian value is recovered.

 

[Anamitra]

In the relation:
[tex]{\frac{{d}^{2}{r}}{{{d}{\tau}}^{2}}}{=}{-}{\frac{M}{{r}^{2}}}[/tex]

the left side is not a covariant derivative and it does not represent acceleration except in the inertial frames. But it may be used to solve problems in the r,theta,phi system which is most common for examples having spherical symmetry.

But for non-relativistic motion in regions of small spacetime curvature it may be taken to represent acceleration itself.

The formula is accurate for regions of strong curvature and also relativistic motion in the r,theta,phi system. But we must remember that the LHS does not represent acceleration in the general case.

 

[Anamitra]

The equation in the above posting is not a tensor equation.But when we are working within a particular system we can always choose a simple type of a formula that works well in it.In this case it works for relativistic motion and also for regions of strong curvature[spacetime].

 

[PAllen]

In the relation:
[tex]{\frac{{d}^{2}{r}}{{{d}{\tau}}^{2}}}{=}{-}{\frac{M}{{r}^{2}}}[/tex]

the left side is not a covariant derivative and it does not represent acceleration except in the inertial frames. But it may be used to solve problems in the r,theta,phi system which is most common for examples having spherical symmetry.

But for non-relativistic motion in regions of small spacetime curvature it may be taken to represent acceleration itself.

The formula is accurate for regions of strong curvature and also relativistic motion in the r,theta,phi system. But we must remember that the LHS does not represent acceleration in the general case.

An inertial frame is a free falling frame and shows no acceleration. A Newtonian analog would be the motion of radial geodesic relative to a static observer. Mentz114 gave you the formula for acceleration of a static observer relative to an instantaneously comoving inertial observer (I believe proper acceleration is equivalent to this). By symmetry, I would think this also gives coordinate acceleration of the free falling observer relative to Fermi-Normal coordinates of the static observer - which is what they would locally measure. I think no real meaning can be attached to second derivative of r by tau.

 

[Anamitra]

It not essential to have laws that span across all systems, preserving their form.We can always look for laws that hold correctly in a particular system.

The first postulate of relativity simply classifies laws into two compartments
1. Those that hold with unchanging form, spanning across all systems
2. Relations that hold correctly in the individual systems---that can help us in our calculations.

 

[PAllen]

It not essential to have laws that span across all systems, preserving their form.We can always look for laws that hold correctly in a particular system.

The first postulate of relativity simply classifies laws into two compartments
1. Those that hold with unchanging form, spanning across all systems
2. Relations that hold correctly in the individual systems---that can help us in our calculations.

Calculations of what? What we want to calculate are observables. I claim r does not represent locally measured distance for either a static observer or a free falling observer. For one thing, it is well known that these coordinates are not isotropic; any observer's local measurements will establish isotropic coordinates.

 

[Anamitra]

Let us consider a simple problem:

A person in a freely falling lift [ radial motion is being considered] notes his time of fall as the lift passes two coordinate labels r1 ar2[theta and phi same for both the points]What time difference should he observe?

Diff Equation:
[tex]\frac{{d}^{2}{r}}{{{d}{\tau}}^{2}} = {-} {\frac{M}{r^2}}[/tex]
We substitute:

[tex]\frac{dr}{d\tau} = v[/tex]

and then integrate,using suitable boundary conditions

The lift is falling through a region of strong curvature[spacetime curvature]

 

[PAllen]

Let us consider a simple problem:

A person in a freely falling lift [ radial motion is being considered] notes his time of fall as the lift passes two coordinate labels r1 ar2[theta and phi same for both the points]What time difference should he observe?

Diff Equation:
[tex]\frac{{d}^{2}{r}}{{{d}{\tau}}^{2}} = {-} {\frac{M}{r^2}}[/tex]

We substitute:

[tex]\frac{dr}{d\tau} = v[/tex]

and then integrate,using suitable boundary conditions

The lift is falling through a region of strong curvature[spacetime curvature]

You agree tau is the proper measure of time. The proper measure of distance for this observer is ds integrated along a spacelike path from r1 to r2 that is 4-orthogonal to the free fall world line. That is the only thing that would correspond to distance as measured by said observer. If you do this properly, I believe you would end up with Mentz114's formula. The concept of proper acceleration is a much faster route the same result.

 

[Anamitra]

You agree tau is the proper measure of time.

Tau simply represents proper time.In this situation it is the proper measure of time for the observer in the lift. For an observer on the planet the time elapsed between the lift passing between the two coordinate labels is not going to be this "proper time"

 

[PAllen]

Tau simply represents proper time.In this situation it is the proper measure of time for the observer in the lift. For an observer on the planet the time elapsed between the lift passing between the two coordinate labels is not going to be this "proper time"

It will just be proper time computed along an (r,theta,phi)=constant worldline.

The upshot is the clearly obvious requirement that you compute observables: proper time, proper distance (for rigid ruler distance) or null geodesic paths (for light based distance definition). Otherwise what you compute simply has no relation to what anyone would measure.

Obviously, many coordinate systems are designed to reflect measurements for some observer, and in SR they can do so globally. In GR, in general, they can reflect measurements only locally for one observer and it really is well known the standard Schwarzschild coordinates you are using don't directly reflect local measurements for either the static observer or the free falling observer deep in the gravity well.

 

[Anamitra]

If I am standing on the surface of the planet the clock would tick at a certain rate depending on the value of the metric coefficient at that point. As the other person moves along some trajectory his clock will tick at a different rate depending on the value of the metric coefficients at the points he is passing through.

The two events are:
1. The person starts from point A
2. The journey finishes at point B

I am standing at point C

 

[PAllen]

If I am standing on the surface of the planet the clock would tick at a certain rate depending on the value of the metric coefficient at that point. As the other person moves along some trajectory his clock will tick at a different rate depending on the value of the metric coefficients at the points he is passing through.

Correct, so you must compute proper time along each world line as the time component of anything expected to be a measurement. Similarly, to calculate anything about locally measured ruler distance, you compute proper length 4-orthogonal (based on the local metric) to a given world line. For neither observer does r difference represent a locally measured distance. So derivatives of r by anything are locally meaningless.

 

[Anamitra]

[tex]{\int ds}{\neq}{g}_{00}{(}t2-t1{)}[/tex]

in the general case.Here g00 is the metric coefficient at C[where I am standing]

t2-t1 denotes the coordinate time interval between the events as recorded by me at the point C

 

[PAllen]

[tex]{\int ds}{\neq}{g}_{00}{(}t2-t1{)}[/tex]

in the general case.Here g00 is the metric coefficient at C[where I am standing]

t2-t1 denotes the coordinate time interval between the events as recorded by me at the point C

Who said it does? It seems like you are not reading what I write.

 

[Anamitra]

The proper measure of distance for this observer is ds integrated along a spacelike path from r1 to r2 that is 4-orthogonal to the free fall world line. That is the only thing that would correspond to distance as measured by said observer. If you do this properly, I believe you would end up with Mentz114's formula. The concept of proper acceleration is a much faster route the same result.

If I know the labels r1 and r2 I can perform as simple integration to calculate tau2-tau1 by my formula.


What Mentz114 has suggested:
u.u=1
[tex]{u}^{\alpha}{=}{[}{{(}{1}{-}{2M}{/}{r}{)}}^{{-}{1}{/}{2}}{,}{0}{,}{0}{,}{0}{]}[/tex]
[u has only the time component since the moving observer is viewing the situation from his own frame.]

[tex]{a}^{\alpha}{=} {u}^{t}{(}{ \frac{{\partial}{u}^{\alpha}} {{\partial}{t}}} {+}{{{\Gamma}^{\alpha}}_{tt}}{u}^{t}{)}[/tex]

Or,
[tex]{a}^{\alpha}{=}{{{\Gamma}^{\alpha}}_{tt}}{{(}{u}^{t}{)}}^{2}[/tex]

[tex]{\Gamma}^{r}_{tt}{=}
{(}{1}{-}{{2M}{/}{r}}{)}{{M}{/}{r}^{2}}[/tex]

The other Christoffel symbols work out to zero value
Therefore,
[tex]{a}^{\alpha}{=}{[}{0}{,}{M}{/}{{r}^{2}}{,}{0}{,}{0}{]}[/tex]

Norm of Acceleration four vector=The formula given by Mentz114

How does this help us in solving the problem stated in posting 13?

 

[PAllen]

If I know the labels r1 and r2 I can perform as simple integration to calculate tau2-tau1 by my formula.

You have to integrate along a *spacelike* path 4-orthogonal to the free fall geodesic to get this observer's locally measured distance. A simple case would be a free fall observers starting at rest relative to a static observer. Then their initial 4-velocity will be the same as the static observer's, and the proper length to calculate would just be integrating -(d tau )*c for r from r1 to r2, t constant. (this will be 4-orthogonal to said instant 4-velocity). This would only be valid for the free fall observer for a very small distance. Or, viewed differentially, ds at the beginning of free fall will be given by:

ds = dr/sqrt (1-R/r) // R event horizon radius

d^2s/dtau^2 using dtau as you've computed it for the free fall world line, should give you the initial acceleration of nearby static objects as observed by the free fall observer.

 

[Mentz114]

This is probably surplus to rquirements in this company but, here's a simple way to get the general radial
geodesic for dropped test particles in the Schwarzschild spacetime.

Starting with a velocity [itex]V^\mu=(B,A,0,0)[/itex] where A,B are functions of r only, we normalise this to
get

[tex]
\begin{align*}
V^0 &=\frac{\sqrt{-2\,r\,M+{r}^{2}\,{A}^{2}+{r}^{2}}}{2\,M-r},\ \ \ V^1=A \\
U_0 &=\frac{\sqrt{-2\,r\,M+{r}^{2}\,{A}^{2}+{r}^{2}}}{r},\ \ \ U_1=-\frac{r\,A}{2\,M-r}
\end{align*}
[/tex]

Now we get the acceleration vector [itex]\nabla_\nu U_\mu V^\nu(=U_{\mu;\nu}V^\nu)[/itex] which

has two non-zero components
[tex]
\begin{align*}
\dot{U}_0 &= \frac{A\,M+{r}^{2}\,{A}^{2}\,\left( \frac{d}{d\,r}\,A\right) }{r\,\sqrt{-2\,r\,M+{r}^{2}\,{A}^{2}+{r}^{2}}} \\
\dot{U}_1 &= -\frac{M+{r}^{2}\,A\,\left( \frac{d}{d\,r}\,A\right) }{2\,r\,M-{r}^{2}}
\end{align*}
[/tex]
both of these will be zero if

[tex]
M+{r}^{2}\,A\,\left( \frac{d}{d\,r}\,A\right)=0
[/tex]

The solution of the trivial differential equation is

[tex]
A=\frac{dr}{d\tau}=\pm \sqrt{2M}\,\sqrt{\frac{1}{r}-\frac{1}{R_0}}\
[/tex]

where [itex]R_0[/itex] is a constant. As [itex]R_0 \rightarrow\ \infty[/itex] we recover the velocity of the test particle released from infinity, so it is reasonable to identify [itex]R_0[/itex] as the drop radius of the faller ( which follows from the fact that the velocity is zero when [itex]r=R_0[/itex]).

Anyway, I think it is correct and thought it might be of interest.

 

[Anamitra]

[tex]
M+{r}^{2}\,A\,\left( \frac{d}{d\,r}\,A\right)=0
[/tex]

Well, the above equation is nothing different from:

[tex]{\frac{{d}^{2}{r}}{{d}{\tau}^{2}}} {=} {-} \frac {M}{r^2}[/tex]

Which I got in post 1.

The above equation is always valid for geodesic motion---as one finds from post 1.
Therefore the time component and radial component of the acceleration vector must be zero for geodesic motion.

 

[Passionflower]

This is probably surplus to rquirements in this company but, here's a simple way to get the general radial
geodesic for dropped test particles in the Schwarzschild spacetime.

Starting with a velocity [itex]V^\mu=(B,A,0,0)[/itex] where A,B are functions of r only, we normalise this to
get

[tex]
\begin{align*}
V^0 &=\frac{\sqrt{-2\,r\,M+{r}^{2}\,{A}^{2}+{r}^{2}}}{2\,M-r},\ \ \ V^1=A \\
U_0 &=\frac{\sqrt{-2\,r\,M+{r}^{2}\,{A}^{2}+{r}^{2}}}{r},\ \ \ U_1=-\frac{r\,A}{2\,M-r}
\end{align*}
[/tex]

Now we get the acceleration vector [itex]\nabla_\nu U_\mu V^\nu(=U_{\mu;\nu}V^\nu)[/itex] which

has two non-zero components
[tex]
\begin{align*}
\dot{U}_0 &= \frac{A\,M+{r}^{2}\,{A}^{2}\,\left( \frac{d}{d\,r}\,A\right) }{r\,\sqrt{-2\,r\,M+{r}^{2}\,{A}^{2}+{r}^{2}}} \\
\dot{U}_1 &= -\frac{M+{r}^{2}\,A\,\left( \frac{d}{d\,r}\,A\right) }{2\,r\,M-{r}^{2}}
\end{align*}
[/tex]
both of these will be zero if

[tex]
M+{r}^{2}\,A\,\left( \frac{d}{d\,r}\,A\right)=0
[/tex]

The solution of the trivial differential equation is

[tex]
A=\frac{dr}{d\tau}=\pm \sqrt{2M}\,\sqrt{\frac{1}{r}-\frac{1}{R_0}}\
[/tex]

where [itex]R_0[/itex] is a constant. As [itex]R_0 \rightarrow\ \infty[/itex] we recover the velocity of the test particle released from infinity, so it is reasonable to identify [itex]R_0[/itex] as the drop radius of the faller ( which follows from the fact that the velocity is zero when [itex]r=R_0[/itex]).

Anyway, I think it is correct and thought it might be of interest.

Interesting in this case is that the relationship between the coordinate difference r between the test particle and the EH and the physical distance to the EH only depends on the local velocity wrt a stationary observer at that locality and is in fact the Lorentz contraction.

 

[PAllen]

Well, the above equation is nothing different from:

[tex]{\frac{{d}^{2}{r}}{{d}{\tau}^{2}}} {=} {-} \frac {M}{r^2}[/tex]

Which I got in post 1.

The above equation is always valid for geodesic motion---as one finds from post 1.
Therefore the time component and radial component of the acceleration vector must be zero for geodesic motion.

No one questioned this. The point being made is that r is not locally measured distance. So to get locally observed acceleration, you can't use r, you must use proper distance in the radial direction.

 

[Passionflower]

No one questioned this. The point being made is that r is not locally measured distance. So to get locally observed acceleration, you can't use r, you must use proper distance in the radial direction.

For a radially free falling observer falling from infinity the coordinate difference between two r-values is also the distance. However when the observer free falls from a given r-value we need to apply the Lorentz factor locally. If we then integrate this result we can get the actual distance between two given r-values.

 

[Anamitra]

Let us consider a particle released from rest [in Schwarzschild's Geometry]

[tex]{dr}_{p}{=}{g}_{rr}{dr}[/tex]

Where [tex]{dr}_{p}[/tex] is physical separation while dr is the coordinate separation in the radial direction

It would be interesting to observe a case where the following two conditions hold simultaneously:

Coordinate acceleration=0
Physical acceleration not equal to zeroNow the particle is released from rest from some coordinate label. It would never reach the next label since the coordinate acceleration is zero.But physical acceleration demands that it should.

This is definitely a contradiction.

 

[PAllen]

Let us consider a particle released from rest [in Schwarzschild's Geometry]

[tex]{dr}_{p}{=}{g}_{rr}{dr}[/tex]

Where [tex]{dr}_{p}[/tex] is physical separation while dr is the coordinate separation in the radial direction

It would be interesting to observe a case where the following two conditions hold simultaneously:

Coordinate acceleration=0
Physical acceleration not equal to zero


Now the particle is released from rest from some coordinate label. It would never reach the next label since the coordinate acceleration is zero.But physical acceleration demands that it should.

This is definitely a contradiction.

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

Your second statement is incorrect. As soon as you release an object, the force holding it in place (causing acceleration relative to inertial motion) ceases. Now it will follow inertial path, which will involve coordinate acceleration but no physical acceleration (inverse of static observer, which has physical acceleration but no coordinate acceleration).

 

[Anamitra]

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

I am sitting on a planet and I observe an apple getting detached from its stalk. It is not supposed to reach the next coordinate label[below it]--coordinate acceleration is zero.

But it should reach the ground--physical acceleration is not zero.

How does this happen?

 

[PAllen]

I am sitting on a planet and I observe an apple getting detached from its stalk. It is not supposed to reach the next coordinate label[below it]--coordinate acceleration is zero.

But it should reach the ground--physical acceleration is not zero.

How does this happen?

It's coordinate acceleration is zero while being held by the stalk. As soon as it is not held by the stalk (applying force to accelerate it away from inertial motion), it no longer has coordinate acceleration of zero.

Your second statement is false. When released, it ceases to have physical acceleration, and follows a free fall geodesic path until it hits the earth, when the ground pushes on it, preventing it from continuing its inertial path.

 

[DrGreg]

I am sitting on a planet and I observe an apple getting detached from its stalk. It is not supposed to reach the next coordinate label[below it]--coordinate acceleration is zero.

But it should reach the ground--physical acceleration is not zero.

How does this happen?

Just joining this thread, I'm not sure you've defined what you mean by "physical acceleration". Do you mean proper acceleration? If you do, then you've got it the wrong way round. A falling apple has non-zero coordinate acceleration (in coordinates in which the planet surface is at rest e.g. Schwarzschild coordinates) but zero proper acceleration.

 

[Anamitra]

Actually I did not mean proper acceleration by physical acceleration.

Now let me put it this way:

We assume that the cable of a lift breaks at the initial stage when it is motionless wrt the Earth [at some height] and the observer is falling with the lift under gravity.There is a glass window in the lift through which he can observe the outside during his free fall.Now there is a tall pole going up from the ground which has the coordinate labels[closely packed] marked on it.As the lift goes past each mark the observer notes the time by his clock[proper time].

He can always calculate the quantities:
[tex]{\frac{dr}{{d}{\tau}}}[/tex] and

[tex]\frac{d^2 r}{d \tau^2}[/tex]

He does not have to calculate the Christoffel Symbols since he is an inertial .He calculates the above quantities to be of non-zero value.

He gets,
[tex]\frac{d^2 r}{d \tau^2} {=} - \frac {M}{{r}^{2}}[/tex]

Therefore the ground observer calculates zero acceleration for both the time component and the radial component.

 

[DrGreg]

Therefore the ground observer calculates zero acceleration for both the time component and the radial component.

Sorry, I don't see how this follows from what preceded it. Actually, I don't even understand what it means. "Acceleration for the time component"?? And you still don't seem to have defined "physical acceleration".

 

[Anamitra]

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.