Hello,

I am learning general relativity by applying the Schwarzschild Metric
(SM) using geometrical units (G=c=1).

I'd like to know if my method at solving the following problem is
correct because I am new to such things. I give an outline of my
thought process in hopes that if I am wrong, that someone will kindly
point out my error.

Consider two clocks. One on the surface of the Earth, and one at a some
height above the surface of the Earth. If the elevated clock moves
around the Earth once in the east direction, the clocks will be out of
sync if they initially read the same. I'd like to determine how much
they are out of sync in a simple way and without having to calculate
the portion that results from the one clock moving from 0 elevation to
its height. Now I know that there are various causes for a time
difference to result: the difference in the gravitational potential
field, and the relative velocity of the two clocks.

My thought process was that the spacetime is described by the SM with
the central mass being the Earth, and that there are two observers that
move about the central mass in circles of different radii. Hence there
is not a change in the 'r' coordinate for either observer, thus 'dr =
0', and I can take 'theta = PI/2' and 'dtheta = 0', since the observers
can be thought to move about in the same plane. Thus the metric is
greatly simplified. A distant observer can be thought to watch the
movement of both observers and it is that idea that allows for the
equating of two SM equations: one for each observer, since each can be
related to the time interval of the distant observer. The result is the
following equation:

dT_a = (1-2M/r_a)/(1-2M/r_b)*( (dT_b)^2 + (r_b)^2 + (dPhi_b)^2 ) - (r_a
* dPhi_a)^2

where dT is the proper time, M the mass of the Earth, r is the
r-coordinate as measured by the distant observer, dPhi is the change in
rotation angle, and the 'a' and 'b' subscripts refer to observer 'a'
and observer 'b'.

The equation gives the change in proper time of observer 'a' in terms
of its r-coordinate, and that of 'b' and the respective changes in
angle, and the change in proper time of 'b'. To determine the time
difference, to solve the problem stated, dT_a can be determined using
the info and then the difference would be dT_a - dT_b.

A potential problem that I see with the above equation is that the
r-coordinates will not be the distances from the center of the earth to
the radius of each observer, and so must be determined from the
measured values. That is, the measured radius = Integral from 0 to r of
1/SQRT(1-2M/p)dp, and so given the radius of the observer 'a', it is to
be determined the r-coordinate 'r'.

I'd appreciate any help in regards to this problem, especially in
regards to my reasoning process, as I seek to improve my understanding
and abilities.

Thanks,

Max wrote:
Hello,

I am learning general relativity by applying the Schwarzschild Metric
(SM) using geometrical units (G=c=1).

I'd like to know if my method at solving the following problem is
correct because I am new to such things. I give an outline of my
thought process in hopes that if I am wrong, that someone will kindly
point out my error.

Consider two clocks. One on the surface of the Earth, and one at a some
height above the surface of the Earth. If the elevated clock moves
around the Earth once in the east direction, the clocks will be out of
sync if they initially read the same. I'd like to determine how much
they are out of sync in a simple way and without having to calculate
the portion that results from the one clock moving from 0 elevation to
its height. Now I know that there are various causes for a time
difference to result: the difference in the gravitational potential
field, and the relative velocity of the two clocks.

My thought process was that the spacetime is described by the SM with
the central mass being the Earth, and that there are two observers that
move about the central mass in circles of different radii. Hence there
is not a change in the 'r' coordinate for either observer, thus 'dr =
0', and I can take 'theta = PI/2' and 'dtheta = 0', since the observers
can be thought to move about in the same plane. Thus the metric is
greatly simplified. A distant observer can be thought to watch the
movement of both observers and it is that idea that allows for the
equating of two SM equations: one for each observer, since each can be
related to the time interval of the distant observer. The result is the
following equation:

dT_a = (1-2M/r_a)/(1-2M/r_b)*( (dT_b)^2 + (r_b)^2 + (dPhi_b)^2 ) - (r_a
* dPhi_a)^2

where dT is the proper time, M the mass of the Earth, r is the
r-coordinate as measured by the distant observer, dPhi is the change in
rotation angle, and the 'a' and 'b' subscripts refer to observer 'a'
and observer 'b'.

The equation gives the change in proper time of observer 'a' in terms
of its r-coordinate, and that of 'b' and the respective changes in
angle, and the change in proper time of 'b'. To determine the time
difference, to solve the problem stated, dT_a can be determined using
the info and then the difference would be dT_a - dT_b.

A potential problem that I see with the above equation is that the
r-coordinates will not be the distances from the center of the earth to
the radius of each observer, and so must be determined from the
measured values. That is, the measured radius = Integral from 0 to r of
1/SQRT(1-2M/p)dp, and so given the radius of the observer 'a', it is to
be determined the r-coordinate 'r'.

I'd appreciate any help in regards to this problem, especially in
regards to my reasoning process, as I seek to improve my understanding
and abilities.

Thanks,

xxein: So you would like to know how to predict satellite clocks also?
I can give you a very simple solution.

Find the escape velocity from the center of attraction (gravity -
Earthcenter) for each component. This is given by GR as 2*M*c^2/r)^.5
with M in meters of light. Next, determine the path velocity for each
component. Use the sidereal consideration to compute the time for each
revolution.

Add the squares and take the sqrt result of each component. This gives
a velocity that determines their individual clockrates. 1-(v/c)^2)^.5
is the standard formula. Compare. Set one to 1 if you like and
examine the ratio.

The above is NOT what GR teaches but it gives the exact same result of
comparison. If you wanted to compute more, you'd better get a
supercomputer to program in all the variables. This is why we send
space missions out on a monitored and realtime communication link that
uses only Newton and Keppler as a base.

Do you know how to solve the problem using the SM, or if my equation
seems correct? I'm trying to learn to solve such problems using the
equations of general relativity itself.

Thanks,

"Max" wrote in message
oups.com...
| Do you know how to solve the problem using the SM, or if my equation
| seems correct? I'm trying to learn to solve such problems using the
| equations of general relativity itself.
|
| Thanks,


I gave you the answer, are you a ****ing simpleton?
Androcles.

Max wrote:
My thought process was that the spacetime is described by the SM with
the central mass being the Earth, and that there are two observers that
move about the central mass in circles of different radii. Hence there
is not a change in the 'r' coordinate for either observer, thus 'dr =
0', and I can take 'theta = PI/2' and 'dtheta = 0', since the observers
can be thought to move about in the same plane. Thus the metric is
greatly simplified.

This is fine.

A distant observer can be thought to watch the
movement of both observers and it is that idea that allows for the
equating of two SM equations: one for each observer, since each can be
related to the time interval of the distant observer.

This is probably fine too, depending on what exactly you want to work out.

The result is the following equation:

dT_a = (1-2M/r_a)/(1-2M/r_b)*( (dT_b)^2 + (r_b)^2 + (dPhi_b)^2 ) - (r_a
* dPhi_a)^2

That's what I get by equating dt_a and dt_b and solving for dT_a, except
that "(r_b)^2 + (dPhi_b)^2" should be "(r_b)^2 * (dPhi_b)^2" and "dT_a"
should be "(dT_a)^2".

The equation gives the change in proper time of observer 'a' in terms
of its r-coordinate, and that of 'b' and the respective changes in
angle, and the change in proper time of 'b'. To determine the time
difference, to solve the problem stated, dT_a can be determined using
the info and then the difference would be dT_a - dT_b.

Well, that's a differential quantity, not a time interval of the sort you'd
measure. You derived a correct mathematical relationship amongst some
quantities, but I'm still confused about what physical setup you're
interested in. You might try dividing everything by dt^2, so you get

(dT_a/dt)^2 = (1-2M/r_a)/(1-2M/r_b)*((dT_b/dt)^2 + r_b^2 (dPhi_b/dt)^2)
- r_a^2 (dPhi_a/dt)^2

These quantities all have straightforward physical interpretations. Assuming
'a' and 'b' are in circular equatorial orbits around the central mass,
(dT_a/dt) is the rate of a's clock as seen (via light signals) by the
distant observer, (dPhi_a/dt) is a's orbital rate as seen by the distant
observer, and similarly for b.

A potential problem that I see with the above equation is that the
r-coordinates will not be the distances from the center of the earth to
the radius of each observer, and so must be determined from the
measured values. That is, the measured radius = Integral from 0 to r of
1/SQRT(1-2M/p)dp, and so given the radius of the observer 'a', it is to
be determined the r-coordinate 'r'.

The r coordinate is actually the more physically meaningful of the two. It's
1/(2pi) times the circumference of a circle around the center of mass. You
can measure this even if the central mass is a black hole, but you can't
directly measure a in this case and it's not even clear what it means. If
the central mass isn't a black hole then you can measure a, but the
relationship to r is more complicated than the one given by your integral
because the Schwarzschild solution is only valid in vacuum. Best to stick
with r.

-- Ben

You really are a simpletom xxein. You claim to have studied this stuff
for 40 years and yet you can't even figure out how to get the ration
for

dT_a/dT_b = ........

Using

dTau/dt = (1-3M/r)^1/2

wrote:
Ben Rudiak-Gould wrote:
Max wrote:
My thought process was that the spacetime is described by the SM with
the central mass being the Earth, and that there are two observers that
move about the central mass in circles of different radii. Hence there
is not a change in the 'r' coordinate for either observer, thus 'dr =
0', and I can take 'theta = PI/2' and 'dtheta = 0', since the observers
can be thought to move about in the same plane. Thus the metric is
greatly simplified.

This is fine.

A distant observer can be thought to watch the
movement of both observers and it is that idea that allows for the
equating of two SM equations: one for each observer, since each can be
related to the time interval of the distant observer.

This is probably fine too, depending on what exactly you want to work out.

The result is the following equation:

dT_a = (1-2M/r_a)/(1-2M/r_b)*( (dT_b)^2 + (r_b)^2 + (dPhi_b)^2 ) - (r_a
* dPhi_a)^2

That's what I get by equating dt_a and dt_b and solving for dT_a, except
that "(r_b)^2 + (dPhi_b)^2" should be "(r_b)^2 * (dPhi_b)^2" and "dT_a"
should be "(dT_a)^2".

The equation gives the change in proper time of observer 'a' in terms
of its r-coordinate, and that of 'b' and the respective changes in
angle, and the change in proper time of 'b'. To determine the time
difference, to solve the problem stated, dT_a can be determined using
the info and then the difference would be dT_a - dT_b.

Well, that's a differential quantity, not a time interval of the sort you'd
measure. You derived a correct mathematical relationship amongst some
quantities, but I'm still confused about what physical setup you're
interested in. You might try dividing everything by dt^2, so you get

(dT_a/dt)^2 = (1-2M/r_a)/(1-2M/r_b)*((dT_b/dt)^2 + r_b^2 (dPhi_b/dt)^2)
- r_a^2 (dPhi_a/dt)^2

These quantities all have straightforward physical interpretations. Assuming
'a' and 'b' are in circular equatorial orbits around the central mass,
(dT_a/dt) is the rate of a's clock as seen (via light signals) by the
distant observer, (dPhi_a/dt) is a's orbital rate as seen by the distant
observer, and similarly for b.

A potential problem that I see with the above equation is that the
r-coordinates will not be the distances from the center of the earth to
the radius of each observer, and so must be determined from the
measured values. That is, the measured radius = Integral from 0 to r of
1/SQRT(1-2M/p)dp, and so given the radius of the observer 'a', it is to
be determined the r-coordinate 'r'.

The r coordinate is actually the more physically meaningful of the two. It's
1/(2pi) times the circumference of a circle around the center of mass. You
can measure this even if the central mass is a black hole, but you can't
directly measure a in this case and it's not even clear what it means. If
the central mass isn't a black hole then you can measure a, but the
relationship to r is more complicated than the one given by your integral
because the Schwarzschild solution is only valid in vacuum. Best to stick
with r.

-- Ben

Since Max is interested in doing this type of an analysis I'll add this
to your comments about the r_coordinate.

For the Max example the r_coordinate is a close approximation for
r_Earth and the Schwarzschild metric is a useful tool for deriving a
formula for analysing his example.

The following formula can be derived from the Schwarzschild metric.

dT/dt = ( 1 - 3M/r )^1/2

from which the ratio [dT_a / dT_b] wrt dt can be determined.

Deriving the formula from a 2 dimensional form of the Schwarzschild
metric setting dr=0:

dT^2 = (1 - 2M/r)dt^2 - 0 - r^2 dphi^2

Determining a useful expression for dphi:

L/m = r^2 (dphi/dT)

dphi = [(L/m)/r^2]dT

Substituting dphi^2 the metric becomes:

dT^2 = ( 1 - 2M/r)dt^2 - [(L/m)^2/r^2]dT^2

Dividing through by dt^2 will give the ratio (dT/dt)^2:

(dT/dt)^2 = (1 - 2M/r) - [(L/m)/r]^2(dT/dt)^2

Simplified:

(dT/dt)^2 = (1 - 2M/r)/ [1 + (L/m)^2/r^2]

Using the equation of motion:

(dr/dT)^2 = (E/m)^2 - (1 - 2M/r)[1 + (L/m)^2/r^2]

and taking the derivative of the effective potential wrt r, dividing
through by r^4 to express dV/dr as a quadratic, solving for 0, then r^2
can be expressed:

"dividing through by r^4 to express dV/dr as a quadratic, solving for
0....."

Correction to the above line: It should be, "multiplying through by r^4
................".



r^2 = [(L/m)^2]r - 3(L/m)^2

Substituting [(L/m)^2]r - 3(L/m)^2 for r^2 into (dT/dt)^2 = [..]
results in:

(dT/dt)^2=(1 - 2M/r)/[1 + (L/m)^2)/[(L/m)^2)r - 3(L/m)^2]

Simplified:

dTau/dt = (1 - 3M/r)^1/2


Bruce

Hello again!,

I thank you all for your help. It's been most useful and figured out
some misunderstanding that I had and am all the better for it. Thanks.

The following is my argument to solve the problem. It's much as was
said before, but slightly different.

I begin with the SM, and set the differentials of theta and 'r' to
zero, which results in:
(1) (dT)^2 = (1-2M/r)(dt)^2 - r^2(dPhi)^2

Since the information I have in term sof angular velocity is in terms
of local proper time, divide (1) by (dT)^2 and then solve for (dt/dT)^2
to arrive at

(2) (dt/dT)^2 = (1-2M/r)^(-1)(1+r^2(dPhi/sT)^2.

Equation 2 will be valid for both an observer on the Earth, given
subscript e, and an observer on a satellite, given subscript s.
Therefore, if it is solved for each observer, using their respective
information, then (2) and it's inverse can be multiplied together to
arrive at a ratio of the respective proper times.

(3) (dT_s/dT_e)^2 =
(1-2M/r_s)(1-2M/r_e)^(-1)(1+(r_e*w_e)^2)(1+(r_s*w_s)^2)^(-1),
where w is the respective angular velocity.

If dT_e is set to 1, then dT_s is found by (3), and the difference in
proper time between the Earth observer and the satellite observer will
be 1 - dT_s.

Concerning the 'r' coordinate. Since the mass of the Earth, represented
as M, is so small, curvature is very small as well and hence the 'r'
coordinates will be approximately equal to the given measured radius to
thee surface of the Earth and to the satellite. Hence, the problem has
been solved using the given information and approximation.

To my knowledge, the above is correct. I'd appreciate constructive
comments as to the solution given and other helpful comments too.

Thank you very much,

Max wrote:
Hello again!,

I thank you all for your help. It's been most useful and figured out
some misunderstanding that I had and am all the better for it. Thanks.

The following is my argument to solve the problem. It's much as was
said before, but slightly different.

I begin with the SM, and set the differentials of theta and 'r' to
zero, which results in:
(1) (dT)^2 = (1-2M/r)(dt)^2 - r^2(dPhi)^2

Since the information I have in term sof angular velocity is in terms
of local proper time, divide (1) by (dT)^2 and then solve for (dt/dT)^2
to arrive at

(2) (dt/dT)^2 = (1-2M/r)^(-1)(1+r^2(dPhi/sT)^2.

Equation 2 will be valid for both an observer on the Earth, given
subscript e, and an observer on a satellite, given subscript s.
Therefore, if it is solved for each observer, using their respective
information, then (2) and it's inverse can be multiplied together to
arrive at a ratio of the respective proper times.

(3) (dT_s/dT_e)^2 =
(1-2M/r_s)(1-2M/r_e)^(-1)(1+(r_e*w_e)^2)(1+(r_s*w_s)^2)^(-1),
where w is the respective angular velocity.

If dT_e is set to 1, then dT_s is found by (3), and the difference in
proper time between the Earth observer and the satellite observer will
be 1 - dT_s.

Concerning the 'r' coordinate. Since the mass of the Earth, represented
as M, is so small, curvature is very small as well and hence the 'r'
coordinates will be approximately equal to the given measured radius to
thee surface of the Earth and to the satellite. Hence, the problem has
been solved using the given information and approximation.

To my knowledge, the above is correct. I'd appreciate constructive
comments as to the solution given and other helpful comments too.

Thank you very much,

Max
That is probably right but very cumbersome for doing a real calculation
[ my hand held only gives answers to 9 decimal places].

Check out:

http://www.eftaylor.com/pub/projecta.pdf

Taylor developes a useful formula for answering your example. Taylor
uses the GPS.
Equation [2] approximates v^2 = r^2(dphi/dt)^2. This isn't
r^2(dphi/dT)^2 but for your example the ratio is ~ 1. I compared your
[3] to [3] in this project and the ratio for a solution to your [3],
(dT/dt)^2, and a solution for Taylor's [3] is 1. This isn't a ratio for
your [3] toTaylor's [3]. The project is in geometric units and for the
GPS:

v_sat = .000012896

v_Earth = .000001551

r_sat = 26.655x10^6 m

r_Earth = 6.371x10^6 m

M_Earth = .00444 m

BTW the book Exploring Black Holes might be something you would find
interesting. You can find sample chapters and projects at Edwin
Taylor's homepage

Thanks for the information. When I was looking for information on RAIN
frames, the book you wrote of, Exploring Black Holes, was cited. My
campus library doesn't carry it though, so I will see what I can gleam
from the extracts. I'll go and read the document you linked to and
appreciate your help.

Thanks again,