# The Schwarzschild Metric - A Simple Case

• Passionflower
In summary: Rs \over {2r}}}} = \sqrt{1-{Rs \over {2r}}} = \sqrt{1-{1 \over {2r}}}So, the Schwarzschild coordinate length of a stationary ruler of proper length L = 1 (using units Rs=1) that starts at Schwarzschild radial coordinate r = 1 and extends to Schwarzschild radial coordinate r = 2 is:L*sqrt{1-{1 \over {2r}}} = L*sqrt{1-{1 \over {2*2}}} = L*sqrt
Passionflower
The Schwarzschild Metric - A Simple Case of How to Calculate!

There is thread open at https://www.physicsforums.com/showthread.php?t=431407 about tidal effects but there may be too many question or the chunk asked is simply to large to handle. At any rate, perhaps it is better to have a very simple question answered first.

Assume we have the following case:

Mass: 0.5
Two test clocks FRONT and BACK (FRONT always has a lower R coordinate value than BACK)

Suppose the tests clocks start to free fall from infinity with a ruler distance of 1.

Let's assume that the clocks, by having little rockets or a super rigid cable (I know this can't be the case but we have to start somewhere if we want to make any calculations), at all times maintain a ruler distance of 1.

If anyone wants to chance these initial conditions fine, please then come with an alternative, use coffee ground, penguins, whatever you like, the objective is that we can calculate something not what tidal forces do in general terms.

Now let's consider the case when the FRONT clock reaches the Schwarzschild coordinate: R = 2.

Then we can calculate the Schwarzschild coordinate of the BACK clock by solving:

$$\sqrt {x \left( x-1 \right) }-\sqrt {2}+\ln \left( {\frac {\sqrt {x}+ \sqrt {x-1}}{1+\sqrt {2}}} \right) =1$$

This results in x = 2.757600642

It is perfectly understandable we do not get 3 because the ruler distances between R and R+1 increase for smaller values of R (See this graph https://www.physicsforums.com/attachment.php?attachmentid=28480&d=1285288006 ).

Now we can compute the (inertial) accelerations for both R values using:

$${m \over r^2} { 1 \over \sqrt{1- {r_0 \over r}}$$

Which gives:

FRONT Clock: 0.1767766952
BACK Clock: 0.08235933775 (as opposed to 0.06804138176 at R= 3!)

Now apparently tidal forces can be expressed in terms of accelerations, so now how do we go from here? If we assume there is no cable or rockets the clocks should be father apart due to tidal accelerations, but how far exactly?

Let's stay with the example and the given initial numbers so we can have a numerical example.

Who can fill me in, or perhaps correct me where I am going wrong.

Edited: fixed a mistake in the Latex formula

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Hi Passionflower. I started a second thread here https://www.physicsforums.com/showthread.php?t=431881 that I thought was a cut down version of your question, but with hindsight, I realize it is almost identical. Sorry about that. It is too late to delete the second thread.
Passionflower said:
Now let's consider the case when the FRONT clock reaches the Schwarzschild coordinate: R = 2.

Then we can calculate the Schwarzschild coordinate of the BACK clock by solving:

$$\sqrt {x \left( x-1 \right) }-\sqrt {2}+\ln \left( {\frac {\sqrt {x}+ \sqrt {\sqrt {x}-1}}{\sqrt {2}+\sqrt {\sqrt {2}-1}}} \right) =1$$

This results in x = 2.781138972

Could you explain exactly where you get this equation from?

It looks a lot like, (but not identical to) the equation for the stationary ruler length (RL) between x = r2 = r and x1 = r1 = Rs = 1 where Rs is the Schwarzschild radius:

$$RL = \sqrt{x(x-1)} + ln\left(\sqrt{x} + \sqrt{(x-1)}\right)$$

See https://www.physicsforums.com/showpost.php?p=1828783&postcount=22
and https://www.physicsforums.com/showpost.php?p=1830747&postcount=33

(The r2 and r1 is the notation I was using in the old thread.)

Are you sure your equation is for the Schwarzschild coordinate length of a free falling (rigid) ruler? I am not saying it is not correct. I just haven't seen it before.

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Oops, I see the problem, the formula is to get the ruler length is:

$$\sqrt {{\it ro}\, \left( {\it ro}-{\it rs} \right) }-\sqrt {{\it ri}\, \left( {\it ri}-{\it rs} \right) }+{\it rs}\,\ln \left( {\frac { \sqrt {{\it ro}}+\sqrt {{\it ro}-{\it rs}}}{\sqrt {{\it ri}}+\sqrt {{ \it ri}-{\it rs}}}} \right)$$

Obviously if we use rs = ri = 1 then the formula can be greatly simplified.

I corrected the original posting accordingly.

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I'm not quite sure what the point of this is - the objective is to "calculate something"?? What is it that your'e trying to calculate?

If we go onto the nitty-gritty details , there's also a question:

Suppose the tests clocks start to free fall from infinity with a ruler distance of 1.

Let's assume that the clocks, by having little rockets or a super rigid cable (I know this can't be the case but we have to start somewhere if we want to make any calculations), at all times maintain a ruler distance of 1.

Which is it? Both clocks can free-fall, or one clock can have a rocket, or both clocks could have rockets. I suspect that what you are doing is assuming both clocks free-fall. Then you shouldn't be surprised that the tidal forces as they both free-fall stretch them apart.

pervect said:
I'm not quite sure what the point of this is - the objective is to "calculate something"?? What is it that your'e trying to calculate?
...
I suspect that what you are doing is assuming both clocks free-fall. Then you shouldn't be surprised that the tidal forces as they both free-fall stretch them apart.
You asked me what to calculate:

How much will they be stretched apart when they are both in free fall (so no cable) when the front clock reaches R=2?

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Passionflower said:
Now let's consider the case when the FRONT clock reaches the Schwarzschild coordinate: R = 2.

Then we can calculate the Schwarzschild coordinate of the BACK clock by solving:

$$\sqrt{x*(x-1)}-\sqrt{2}+ln \left(\frac{sqrt{x}+\sqrt{x-1}}{1+sqrt{2}}\right) = 1$$

This results in x = 2.757600642

OK, x-2 = 0.7576.. is the Schwarzschild coordinate length of stationary ruler of proper length L = 1 with its lowest point at Schwarzschild radial coordinate 2 (using units Rs=1).

If the stationary ruler is progressively reduced from 1 to a very small size, then the Schwarzschild coordinate length contraction factor of the ruler converges to $\sqrt{2} = 0.707106$ which is proportional to a gravitational time dilation factor of $\sqrt{(1-Rs/2)} = \sqrt{(1-1/2)}$ and is also equivalent to the velocity time dilation factor of SR when we use the escape velocity v which is equal to the local velocity an object obtains when it falls from infinity so that the length contraction of the stationary ruler is equal to $\sqrt{(1-v^2)}$ where the terminal velocity of the hypothetical falling object works out to be $v=\sqrt{2}$.

Just for reference, the terminal velocity of an object falling from infinity as measured by a local observer at x (or escape velocity) is calculated from:

$$v = \sqrt{\frac{Rs}{x}}$$

Anyway, none of this directly confirms what the coordinate length or the radar length of the falling ruler is according to any observer. That is what we need to find out.

yuiop said:
OK, x-2 = 0.7576.. is the Schwarzschild coordinate length of stationary ruler of proper length L = 1 with its lowest point at Schwarzschild radial coordinate 2 (using units Rs=1).
Indeed. But here we assume there are no tidal effects, which is not correct but it is a start.

But since in reality there are tidal effects, thus the distance between the two clocks must have been increased. The question is how much? Is the tidal effect perhaps exactly offsetted by the increasing distance between the R coordinates? If so then the answer is: 1.300118429 However that would leave the question how we calculate the tidal effect in the other direction, e.g. now the two clocks are perpendicular to the field.

yuiop said:
If the stationary ruler is progressively reduced from 1 to a very small size, then the Schwarzschild coordinate length contraction factor of the ruler converges to $\sqrt{2} = 0.707106$ which is proportional to a gravitational time dilation factor of $\sqrt{(1-Rs/2)} = \sqrt{(1-1/2)}$ and is also equivalent to the velocity time dilation factor of SR when we use the escape velocity v which is equal to the local velocity an object obtains when it falls from infinity so that the length contraction of the stationary ruler is equal to $\sqrt{(1-v^2)}$ where the terminal velocity of the hypothetical falling object works out to be $v=\sqrt{2}$.
What do you mean by reduced in size? Are you talking about a ruler's proper length? Or the length with respect to another observer?

Looking at the geometry of the Schwarzschild solution it appears we can put more rulers of proper length 1 between R and R+1 the smaller the value of R gets all the way towards the Schwarzschild radius.

yuiop said:
Just for reference, the terminal velocity of an object falling from infinity as measured by a local observer at x (or escape velocity) is calculated from:

$$v = \sqrt{\frac{Rs}{x}}$$

Anyway, none of this directly confirms what the coordinate length or the radar length of the falling ruler is according to any observer. That is what we need to find out.
Agreed, but the first thing I think we need to be sure of is the proper length.

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Passionflower said:
Indeed. But here we assume there are no tidal effects, which is not correct but it is a start.

But since in reality there are tidal effects, thus the distance between the two clocks must have been increased.

I don't think tidal effects affect the calculation of the length of the stationary ruler at x but you are right that they do affect the distance between two free falling clocks. You have already asked the key question:
Passionflower said:
How much will they be stretched apart when they are both in free fall (so no cable) when the front clock reaches R=2?

and hopefully someone can provide a definitive answer/ equation to that question.

yuiop said:
I don't think tidal effects affect the calculation of the length of the stationary ruler at x but you are right that they do affect the distance between two free falling clocks. You have already asked the key question:

and hopefully someone can provide a definitive answer/ equation to that question.
And perhaps the easy answer is the distance between R2 and R3 :)

Working out the details gets very messy. Basically, though, if you assume the rod is very short, the rate of change of it's length should be just be

$$\frac{1}{2} a \tau^2$$

where a is -GM/r^3 * L, r being the schwarzschild coordinate of the center of the rod. And I've also assumed that the acceleration is moderate enough that we don't need to use realtivity to calculate the distance.

Basically, the equations of motion - and also the equations for the tidal forces - are the same as the Newtonian equations, except that we replace t by tau

Passionflower said:
The question is how much? Is the tidal effect perhaps exactly offsetted by the increasing distance between the R coordinates? If so then the answer is: 1.300118429
I have a hunch that when we have finally figured it all out there will be some weird coincidental offset/self-cancelling effect, but I not sure exactly what it will be yet.

However that would leave the question how we calculate the tidal effect in the other direction, e.g. now the two clocks are perpendicular to the field.
Horizontal distances measured by a local observer and a Schwarzschild observer are the same and tidal effects are not observed here. However, horizontal lengths are slightly longer when a free falling observer uses radar to measure them because an observer in a free-falling elevator "sees" light paths as being curved over extended distances, rather than traveling in a straight line. When two particles are free falling such that they are always the same height as each other above the source they progressively move towards each other because they each follow their own radial geodesic. It would be nice if the radial radar distance between the two particles appears to be constant to a free falling observer, but I am not sure if that quite works out.

What do you mean by reduced in size? Are you talking about a ruler's proper length? Or the length with respect to another observer?
Sorry, I should have been clearer here. I did not mean progressively changing over time. I meant if we chose a ruler with an initially shorter proper length, then the ratio of proper length to coordinate length converges to a ratio of 0.707106 for a ruler with one end at Schwarzschild radial coordinate 4GM/c^2. I.e this is the gravitational length contraction factor of an infinitesimal ruler at that location.

Looking at the geometry of the Schwarzschild solution it appears we can put more rulers of proper length 1 between R and R+1 the smaller the value of R gets all the way towards the Schwarzschild radius.
Agree, with the additional observation that even when R is exactly 2GM/c^2 the proper ruler length is always finite unlike the radar distance.

Let me expand on my previous remark just bit

The relativistic differential equation for the falling object r(tau) is(dr/tau)^2 = 2GM(1/r - 1/R_max)

If we consider a Newtonian object falling radially from some height R_max, conservation of energy gives us the same equation

.5*m*v^2 = GmM(1/r - 1/R_max)

but v = dr/dt and the m of the falling object cancels out so

dr/dt = 2GM(1/r - 1/R_max)So - except for replacing the Newtonian time t with the proper time tau, using Schwarzschild coordinates makes the equation for the falling object and the expression for the tidal forces similar to the Newtonian case. Unfortunately, it's still rather messy...

yuiop said:
I don't think tidal effects affect the calculation of the length of the stationary ruler at x but you are right that they do affect the distance between two free falling clocks. You have already asked the key question:
Well that is another interesting question, as the ruler tries to resist stretching. So perhaps we can use some formula of opposing forces (I am careful here since his is GR and perhaps we cannot use this approach). Can we quantify the rigidity of the ruler and see how it would behave under decreasing values of R? But again I suggest, only after we are absolutely sure the first step is indeed correct.

yuiop said:
Horizontal distances measured by a local observer and a Schwarzschild observer are the same and tidal effects are not observed here.

When two particles are free falling such that they are always the same height as each other above the source they progressively move towards each other because they each follow their own radial geodesic.
Well isn't it supposed to be volume preserving? If something is stretched from one direction something must be pushed as well from another direction. Am I wrong here? You say they come together so the same question applies, what is the distance between them with respect to the R coordinate. Since the coordinates are polar it is a little bit more complicated.

I am for the moment ignoring radar distances, not because they are not interesting but for clarity's sake, once the proper distance is ironed out, I like to talk about the radar distance.

yuiop said:
when R is exactly 2GM/c^2 the proper ruler length is always finite unlike the radar distance.
Indeed, and so is the area and the volume.

pervect said:
So - except for replacing the Newtonian time t with the proper time tau, using Schwarzschild coordinates makes the equation for the falling object and the expression for the tidal forces similar to the Newtonian case. Unfortunately, it's still rather messy...
I am trying to understand why the increase in proper distance of the two clocks due to the tidal accelerations would depend on tau? It cannot be expressed in terms of R, R+1 and the Schwarzschild radius?

So, how do we go from here? What formula do we have to apply here to get the proper distance of the BACK clock when the front clock reaches R=2?

The inertial coordinate acceleration at r of a free falling clock (free fall from infinity) is:

$$\frac{m}{r^2}$$

While we can convert that in terms of proper time by dividing by:

$$\sqrt{1-{r_0 \over r}}$$

to get the inertial acceleration.

So do we need difference between the coordinate accelerations of clock FRONT and BACK? But the question is, what is the R value for the BACK clock?

For instance:
Coordinate acceleration R=2: 0.125
Coordinate acceleration R=3: 0.0.05555555556

In proper time:
R=2: 0.1767766952
R=3: 0.06804138176

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Passionflower said:
Well that is another interesting question, as the ruler tries to resist stretching. So perhaps we can use some formula of opposing forces (I am careful here since his is GR and perhaps we cannot use this approach). Can we quantify the rigidity of the ruler and see how it would behave under decreasing values of R? But again I suggest, only after we are absolutely sure the first step is indeed correct.
Yes, we would have to somehow quantify the rigidity of the ruler and I suspect that is going to make things more complicated than they already. We should in principle be able to get an exact equation for two unconnected free falling particles and that is the place to start and a key question. Next, we might be able to obtain an equation for a hypothetical perfectly rigid rod and then a realistic rod should be somewhere between the two results. If by the first step, you mean the first equation we have obtained for the ruler distance between two Shwarzschild radial coordinates, then yes I am fairly confident that equation is correct.
Well isn't it supposed to be volume preserving? If something is stretched from one direction something must be pushed as well from another direction. Am I wrong here? You say they come together so the same question applies, what is the distance between them with respect to the R coordinate. Since the coordinates are polar it is a little bit more complicated.
I am not sure about the volume preserving property. Baez and Penrose mention this property but are rather vague about who measures it and when it applies. On this webpage,http://math.ucr.edu/home/baez/gr/ricci.weyl.html Baez starts by talking specifying curvature at a point in terms of 20 numbers and then concludes:
When we are in truly empty space, there's no Ricci curvature, so actually our ball of coffee grounds doesn't change volume. But there can be Weyl curvature due to gravitational waves, tidal forces, and the like. Gravitational waves and tidal forces tend to stretch things out in one direction while squashing them in the other. So these would correspond to our ball changing into an ellipsoid! Just as we hoped.

Similarly, when a ball of coffee grounds falls freely through outer space in the Earth's gravitational field, it feels no Ricci curvature, only Weyl curvature. So the "tidal forces" due to some coffee grounds being near to the Earth than others may stretch the ball into an ellipsoid, but not change its volume.
It appears that his final conclusion is essentially a Newtonian one in the Earth's gravitational field where the Ricci curvature is negligable and since Ricci curvature causes the change in volume, there is (almost) no change in volume in the Earth example and only the Weyl curvature (the shape changing property) is significant. However, this shape changing is predicted by simple Newtonian considerations.

Consider an example. A 2 sphere of penguins with radius w1 (and volume (4/3)Pi*w1^2*h1 where w1 = h1 because the height and width of the sphere are the same) is released from height r1 = 100. When the centre of this sphere falls to height r2 = 2 the volume of the spheroid is equal to (4/3)Pi*w2^2*h2. Now from purely geometrical considerations the ratio of the width w2 of the prolate spheroid to the original width of the sphere w1 is proportional to r1/r2 so we can equate (if the volumes are the same) (4/3)Pi*w1^3 with (4/3)Pi*r1^2/r2^2*h2 and solve to find that for the volume to be constant, h2 has to have the value 2,500 if the original h1 was 1. This is huge ratio and the spheriod is now longer than the height the sphere was dropped from so there is obviously a problem! It might still work out OK if the experiment is terminated when the first penguin arrives at r=2 and the centre of the spheroid is much higher up. Depending on whether you like penguins or not, we might come back to this later.

I am for the moment ignoring radar distances, not because they are not interesting but for clarity's sake, once the proper distance is ironed out, I like to talk about the radar distance.
Proper distance is hard to nail down. I like to think of it as the ruler distance when the ruler distance is measured by a series of very short rods that all calibrated by radar. Pervect explains this better here: https://www.physicsforums.com/showthread.php?t=431881

This ruler distance is a sort of average of the radar distance measured from one end and the radar distance measured from the other end. However, the proper distance defined as a physical ruler length may let us down when considering a free falling ruler. This has yet to be determined.

One thing I am fairly certain about from discussions in other threads is the proper distance between two points is not the distance measured by a CMIRF.

I am trying to understand why the increase in proper distance of the two clocks due to the tidal accelerations would depend on tau? It cannot be expressed in terms of R, R+1 and the Schwarzschild radius?
Offhand, I think the tau factor is built into the Schwarzschild metric. Consider two observers that jump from a very high tower. A jumps first holding a ruler upwards and B jumps when the top end of the ruler passes him. The height of the tower and short distance between A and B ensures that initially their clock are running at about the same rate. As they fall, B pulls away from A and A sees the nearest end of the ruler going away from him. As B falls, her clock is time dilated by the gravitational time dilation and a velocity dependent time dilation factor. Now if the ruler is not length contracting, this slowing down of B's clock will cause B to perceive the radar length of the ruler to getting shorter over time and while in coordinate terms A and b appear to be getting further apart it is "conceivable" that this time dilation of B's clock might cause B to consider the distance AB to be constant, by a cancelling out effect. (maybe). Now if radar distance is a measure of proper length, then B considers the proper length of the ruler to be getting shorter over time. There is another thing to consider. The coordinate speed of light is slower by a factor of the inverse gravitational time dilation factor squared lower down and this tends to increase radar distances measured by B. Now this almost exactly cancels out the shortening effect of B's clock slowing down but any real length contraction of the falling ruler due to gravity or relative velocity will be detectable by B's radar measurements. This I think will be the surprise result if we ever resolve your original question fully. The length of a free falling (and fairly rigid) elevator changes over time and is detectable by a free falling observer inside the elevator. In other words an observer at rest with the elevator may be able to detect its length contraction! The reason this is "allowed" is because while the observer and the centre of mass of the free-falling elevator have inertial motion, the top and bottom of the elevator are not inertial and not free falling in the proper sense, because there motion is constrained by being physically attached to the centre of the mass of the elevator. This fine distinction was mentioned by Pervect.

So, how do we go from here? What formula do we have to apply here to get the proper distance of the BACK clock when the front clock reaches R=2?

The inertial coordinate acceleration at r of a free falling clock (free fall from infinity) is:

$$\frac{m}{r^2}$$

While we can convert that in terms of proper time by dividing by:

$$\sqrt{1-{r_0 \over r}}$$

to get the inertial acceleration.

$$\frac{m}{r^2}$$

is the acceleration measured by taking the coordinate distance fallen (over a short time interval) and dividing by the proper time of the falling object. It is not the coordinate acceleration measured by an observer at infinity and is not the proper acceleration and is not the acceleration measured by a local observer.

$$\sqrt{1-{r_0 \over r}}$$

is the proper acceleration measured by an accelerometer of a stationary or hovering object. For a free falling object the proper acceleration is zero.

For arbitrary velocities, see post #345 (yes 345!) of this old thread https://www.physicsforums.com/showpost.php?p=2747788&postcount=345

and this different older thread https://www.physicsforums.com/showthread.php?p=1807379#post1807379 that goes into more detail about the meaning of $m/{r^2}$ in GR (which I was very confused about at the time.) Fortunately that last thread was a short thread that was rapidly resolved.

Just verifying if we perhaps disagree, or more likely, I misunderstand something:

yuiop said:
$$\frac{m}{r^2}$$

is the acceleration measured by taking the coordinate distance fallen (over a short time interval) and dividing by the proper time of the falling object. It is not the coordinate acceleration measured by an observer at infinity and is not the proper acceleration and is not the acceleration measured by a local observer.
Right, exactly what I mentioned as well right? So we agree here?

yuiop said:
$$\sqrt{1-{r_0 \over r}}$$

is the proper acceleration measured by an accelerometer of a stationary or hovering object. For a free falling object the proper acceleration is zero.
Correct, that is why I wrote inertial acceleration.
Are you saying the inertial acceleration is not the same magnitude as the proper acceleration in case the clock would be stationary in the field?

For arbitrary velocities it indeed gets harder, if I am not mistaken a velocity over sqrt(2) (or sqrt(3) coordinate) will cause a decelleration, as does light, in the field. But that is another interesting topic and issue (at least for now).

As for radar distance, at this stage I honestly think it will confuse matters, just as throwing in other coordinate charts or tensor forms.

yuiop said:
It appears that his final conclusion is essentially a Newtonian one in the Earth's gravitational field where the Ricci curvature is negligable and since Ricci curvature causes the change in volume, there is (almost) no change in volume in the Earth example and only the Weyl curvature (the shape changing property) is significant. However, this shape changing is predicted by simple Newtonian considerations.
In the Schwarzschild case we do not have to worry about Ricci curvature. So it looks like it is the same as Newton with the difference that we have to integrate the distance because spacetime is not flat.

yuiop said:
Consider an example. A 2 sphere of penguins with radius w1 (and volume (4/3)Pi*w1^2*h1 where w1 = h1 because the height and width of the sphere are the same) is released from height r1 = 100. When the centre of this sphere falls to height r2 = 2 the volume of the spheroid is equal to (4/3)Pi*w2^2*h2. Now from purely geometrical considerations the ratio of the width w2 of the prolate spheroid to the original width of the sphere w1 is proportional to r1/r2 so we can equate (if the volumes are the same) (4/3)Pi*w1^3 with (4/3)Pi*r1^2/r2^2*h2 and solve to find that for the volume to be constant, h2 has to have the value 2,500 if the original h1 was 1. This is huge ratio and the spheriod is now longer than the height the sphere was dropped from so there is obviously a problem! It might still work out OK if the experiment is terminated when the first penguin arrives at r=2 and the centre of the spheroid is much higher up. Depending on whether you like penguins or not, we might come back to this later.
Yes clearly the sphere will not become an elipse, but more like a some stretched out oildrop.

yuiop said:
Proper distance is hard to nail down. I like to think of it as the ruler distance when the ruler distance is measured by a series of very short rods that all calibrated by radar. Pervect explains this better here: https://www.physicsforums.com/showthread.php?t=431881
Honestly I do not see why we have to do it like that, we would just drop a measuring tape down to see how far below the other clock is. Proper distance is a very good point to start with, radar distance obviously has the additional issue with that fact that the coordinate speed of light is variable.

yuiop said:
Offhand, I think the tau factor is built into the Schwarzschild metric.
I really do not see how tau could be of influence to the problem. All we want to know as a first step is the R value of the trailing clock given the R value of the leading clock. This value seems independent of anyone's clock.

pervect said:
Working out the details gets very messy. Basically, though, if you assume the rod is very short, the rate of change of it's length should be just be

$$\frac{1}{2} a \tau^2$$

where a is -GM/r^3 * L, r being the schwarzschild coordinate of the center of the rod. And I've also assumed that the acceleration is moderate enough that we don't need to use realtivity to calculate the distance.

Basically, the equations of motion - and also the equations for the tidal forces - are the same as the Newtonian equations, except that we replace t by tau
If you know the solution in Schwarzschild coordinates I would appreciate it you show it or at least help finding a solution. I suppose it I was still not clear enough that I do not want a weak field or short distance solution. The whole point, for me at least, is to learn how to calculate it. If you do not know it it is all right, if you do know, even partially, I and I suspect many others will appreciate your help.

The question is still open:

We have two clocks with an initial coordinate distance of 1 free falling. As soon as the front clock reaches R=2, what is the R value of the back clock?

Can you help answering this question?

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yuiop said:
In other words an observer at rest with the elevator may be able to detect its length contraction! The reason this is "allowed" is because while the observer and the centre of mass of the free-falling elevator have inertial motion, ...the top and bottom of the elevator are not inertial and not free falling in the proper sense, because there motion is constrained by being physically attached to the centre of the mass of the elevator. This fine distinction was mentioned by Pervect.

Hi yuiop Wouldn't this seem to imply that a free falling particle inside the elevator near the ceiling would have motin toward the ceiling and likewise for a particle towards the floor?
If this is not the case then in what sense would those parts of the system not be inertially freefalling?
In another context : if there is an elastic sphere and an equivalent sphere of disassociated particles
freefalling together then we expect the particle sphere to spread out radially over time and the elastic sphere to either : 1) inertially fall based on it's COM , retaining shape while coaccelerating with the particles initially at the center of the particle sphere. With the particles at the front now accelerating ahead and the particles at the rear falling behind.
Or 2) Deform comparably to the particles but perhaps to a lesser degree.
In either case, if we assume the elastic sphere to be hollow and large enough to contain the particle sphere it would seem to mean the particles moving in both directions relative to the leading and trailing hemispheres , no??

Passionflower said:
Right, exactly what I mentioned as well right? So we agree here?Correct, that is why I wrote inertial acceleration.
Are you saying the inertial acceleration is not the same magnitude as the proper acceleration in case the clock would be stationary in the field?
Inertial acceleration is not a term I normally use, so if you want define inertial acceleration as "is the acceleration measured by taking the coordinate distance fallen (over a short time interval) and dividing by the proper time (squared) of the falling object." then that is fine by me. I am short of time so I don't have time to carefully answer your question here.

In the Schwarzschild case we do not have to worry about Ricci curvature. So it looks like it is the same as Newton with the difference that we have to integrate the distance because spacetime is not flat.
I think in the end we might have to consider Ricci curvature in strongly curved spacetime.

Austin0 said:
Hi yuiop Wouldn't this seem to imply that a free falling particle inside the elevator near the ceiling would have motin toward the ceiling and likewise for a particle towards the floor?
Yes.

If this is not the case then in what sense would those parts of the system not be inertially freefalling?
It is the case.

In another context : if there is an elastic sphere and an equivalent sphere of disassociated particles
freefalling together then we expect the particle sphere to spread out radially over time and the elastic sphere to either : 1) inertially fall based on it's COM , retaining shape while coaccelerating with the particles initially at the center of the particle sphere. With the particles at the front now accelerating ahead and the particles at the rear falling behind.
Or 2) Deform comparably to the particles but perhaps to a lesser degree.
In either case, if we assume the elastic sphere to be hollow and large enough to contain the particle sphere it would seem to mean the particles moving in both directions relative to the leading and trailing hemispheres , no??
I think case 2 is the most realistic, with case 1 being the case for a hypothetical but unrealistic infinitely rigid sphere. Also bear in mind when we say "changing shape" or "changing volume" we have to be careful to define according to which observer.

Passionflower said:
There is thread open at https://www.physicsforums.com/showthread.php?t=431407 about tidal effects but there may be too many question or the chunk asked is simply to large to handle. At any rate, perhaps it is better to have a very simple question answered first.

Assume we have the following case:

Mass: 0.5
Two test clocks FRONT and BACK (FRONT always has a lower R coordinate value than BACK)

Suppose the tests clocks start to free fall from infinity with a ruler distance of 1.

Let's assume that the clocks, by having little rockets or a super rigid cable (I know this can't be the case but we have to start somewhere if we want to make any calculations), at all times maintain a ruler distance of 1.

If anyone wants to chance these initial conditions fine, please then come with an alternative, use coffee ground, penguins, whatever you like, the objective is that we can calculate something not what tidal forces do in general terms.

Now let's consider the case when the FRONT clock reaches the Schwarzschild coordinate: R = 2.

If we can attach any validity to the equation I derived in the other thread:

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

The above equation can be solved for r2 when we know r1 as:

$$r2 = r_s + r_s W \left(\left(\frac{r1}{r_s}-1\right) \, \exp\left(\frac{r1}{r_s}+\frac{\Delta S}{r_s} -1 \right)$$

We know all the variables. r1=2, Rs = 1 and the proper length of the rigid ruler at infinity (or anywhere else for that matter as we are assuming the proper length does not change) is $\Delta S =1$. The bad news is that you need mathematical software that can compute the Lambert W function shown as simply W() in the above equation, in order to obtain a numerical solution to your question.

Just for completeness, you can compute r1 from:

$$r1 = r_s + r_s W \left(\left(\frac{r2}{r_s}-1\right) \, \exp\left(\frac{r2}{r_s}-\frac{\Delta S}{r_s} -1 \right)$$

yuiop said:
If we can attach any validity to the equation I derived in the other thread:

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

The above equation can be solved for r2 when we know r1 as:

$$r2 = r_s + r_s W \left(\left(\frac{r1}{r_s}-1\right) \, \exp\left(\frac{r1}{r_s}+\frac{\Delta S}{r_s} -1 \right)$$
1. Expressing $$r_2$$ as a function of $$r_1$$ is totally meaningless since they are arbitrary limits of integration, so there cannot be any correlation between them.

2. There is no justification for the

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad$$

to begin with.

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yuiop said:
If we can attach any validity to the equation I derived in the other thread:

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$
So basically this is light travel coordinate time right? We also use this formula (multiplied by 2) to calculate the radar coordinate distance between two R values.

So to refresh the initial question:

RS=1
Free fall from infinity
As rigid as possible
FRONT lower R as BACK
Proper distance: 1

When FRONT reaches R=2 what R value is BACK?

So applying your formula we would get: R=2.557145599 as opposed to the 2.757600642 I calculated based on the proper distance formula.

My proper distance remains 1 while your changes, at least in the way I calculate it: freeze the 'frame' and take the proper distance, to 0.7461229122

So the question here is: if we want to take the proper distance between two R values does it matter if it is from a stationary of moving object?

Let me ask you this, what do think is the physical interpretation for this integral:

$$\int _{{r0}}^{{r1}}\!{\frac {1}{\sqrt {1-{\frac {{\it rs}}{r}} }}}\,{dr}$$

starthaus said:
1. Expressing $$r_2$$ as a function of $$r_1$$ is totally meaningless since they are arbitrary limits of integration, so there cannot be any correlation between them.
Of course there is assuming we know Delta S, which we do in this example: Delta S = 1

If we make this a function and consider various values of R we see that it is almost linear except for r values very close to rs almost identical to the proper distance formula.

Now, if we plot out the differences between the two approaches in a R range from 1 to 10 we hardly see any difference, but if we 'zoom them out' we are seeing a kind of 'band', looking at this more detailed gives us a difference that maximizes (around 0.332) close to the event horizon.

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Passionflower said:
Of course there is assuming we know Delta S, which we do in this example: Delta S = 1

1. It is the other way around, $$\Delta S$$ is a function of the limits of integration

2. The integral shown is incorrect to begin with.

starthaus said:
1. It is the other way around, $$\Delta S$$ is a function of the limits of integration
Well it is calculated and plotted right in front of your eyes.

Passionflower said:
Well it is calculated and plotted right in front of your eyes.

It is calculated as a function of $$r_1$$ and $$r_2$$. This makes it pretty much useless for calculating $$r_2$$.

Here is an easy way to get the expressions for length contraction and time dilation in GR. Start with the Schwarzschild metric:

$$-ds^2=\left(1+\frac{2\Phi}{c^2 }\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(1+\frac{2\Phi}{c^2}\right)dt^2$$

If you want to calculate time dilation, make $$dr=d\theta=d\phi=0$$ . The reason is that you are trying to determine the relationship between two co-located events.

$$\frac{ds}{cdt}=\sqrt{1+\frac{2\Phi}{c^2}}$$Length contraction for the case of a non-rotating object is obtained by making $$d\theta=d\phi=0$$ (absence of rotation) and $$dt=0$$ (object endpoints are marked simultaneously bu the distant observer).

$$\frac{ds}{dr}=\frac{1}{\sqrt{1+\frac{2\Phi}{c^2}}}$$

Particularization:

For the radial gravitational field :

$$\Phi=-\frac{GM}{r}$$

so:

$$cdt=\frac{cd\tau}{\sqrt{1-\frac{2GM}{rc^2}}}$$
i.e. coordinate time appears dilated wrt proper time

$$dr=ds\sqrt{1-\frac{2GM}{rc^2}}$$

i.e. coordinate length appears contracted wrt proper length.

If you want proper length as a function of coordinate length, then:

$$ds=\frac{dr}{\sqrt{1-\frac{2GM}{rc^2}}}$$

Integrate the above and you get the correct expression for proper length as a function of coordinate length.

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Passionflower said:
If you know the solution in Schwarzschild coordinates I would appreciate it you show it or at least help finding a solution. I suppose it I was still not clear enough that I do not want a weak field or short distance solution. The whole point, for me at least, is to learn how to calculate it. If you do not know it it is all right, if you do know, even partially, I and I suspect many others will appreciate your help.

The question is still open:

We have two clocks with an initial coordinate distance of 1 free falling. As soon as the front clock reaches R=2, what is the R value of the back clock?

Can you help answering this question?

Passionflower: I've been reading these threads and thinking about this problem, which is an interesting one, but I have a question. Is this question really well posed? In Schwarzschild coordinates, the front clock never reaches R=2, so the question has no answer. What am I missing?

starthaus said:
If you want proper length as a function of coordinate length, then:

$$ds=\frac{dr}{\sqrt{1-\frac{2GM}{rc^2}}}$$

Integrate the above and you get the correct expression for proper length as a function of coordinate length.
What you say is correct, but are you actually following the whole discussion?

What the charts show is:

1. The solution of this integral with the proper length and one R coordinate known, so the other R value is obtained.

2. The solution of the light travel time integral with the length and one R coordinate known, so the other R value is obtained.

Assuming I understand yuiop's position, if you follow the discussion then yuiop is of the opinion that we only can use the formula you quote above, which is also the one I use, if the object in question is stationary. I think that is does not matter as long as we take the same coordinate time. yuiop proposes to use the other formula in case the object is not stationary. Yuiop's formula is actually the integral representing light travel distance between two coordinate values.

phyzguy said:
Passionflower: I've been reading these threads and thinking about this problem, which is an interesting one, but I have a question. Is this question really well posed? In Schwarzschild coordinates, the front clock never reaches R=2, so the question has no answer. What am I missing?
Why do you think the front clock never reaches R=2?

Notice that RS, which is the Schwarzschild radius, is 1 in the example.

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Time to resolve a SR question here, first, before any progress is going to be made. We've got a moving ruler in the flat space-time of SR. If we can't agree on how to measure it's length in the lab frame, and its proper length, we aren't going to make any progress with the relativistic case.

$$$\begin{picture}(62,90)(0,0) \linethickness{0.3mm} \multiput(0,0)(0.12,0.54){167}{\line(0,1){0.54}} \put(20,90){\vector(1,4){0.12}} \linethickness{0.3mm} \multiput(40,0)(0.12,0.54){167}{\line(0,1){0.54}} \put(60,90){\vector(1,4){0.12}} \linethickness{0.3mm} \qbezier(5,20)(5,20)(5,20) \qbezier(5,20)(5,20)(5,20) \put(5,20){\vector(0,0){0.12}} \linethickness{0.3mm} \multiput(5,20)(0.12,0.12){417}{\line(1,0){0.12}} \put(55,70){\vector(1,1){0.12}} \linethickness{0.3mm} \put(0,0){\line(1,0){40}} \put(40,0){\vector(1,0){0.12}} \put(0,0){\makebox(0,0)[cc]{}} \put(5,5){\makebox(0,0)[cc]{}} \put(0,0){\makebox(0,0)[cc]{}} \put(-3,-3){\makebox(0,0)[cc]{A}} \put(42,-3){\makebox(0,0)[cc]{B}} \put(-1,20){\makebox(0,0)[cc]{C}} \put(60,70){\makebox(0,0)[cc]{D}} \put(22,93){\makebox(0,0)[cc]{E}} \put(62,93){\makebox(0,0)[cc]{F}} \end{picture}$$$

The diagram is a space-time diagram, with time running up the page, and space running across the page.

On this space-time diagram, AE and BF are the worldlines representing the two endpoints of a moving ruler.

Can we resolve that AB represents the length of the ruler in the lab frame, which is also a coordinate difference in the lab frame, because events A and B occur "at the same time" in the lab frame, but that CD, and NOT AB, represents the proper length of the ruler, because events C and D occur at the same time in the frame of the ruler?

The diagram is not to scale - the intent is that lline CD will appear to be orthogonal to line AE and AF if one transforms to the ruler's frame, or if you express them as vectors in any frame, the Lorentz invariant dot product AE with CD will be zero.

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PassionFlower said:
We have two clocks with an initial coordinate distance of 1 free falling. As soon as the front clock reaches R=2, what is the R value of the back clock?

Can you help answering this question?

Is this what's required ?

Two clocks are dropped from $r=R_{max}$ and $r=R_{max}-1$ at $t=0$, in such a way that they are synchronised at $t=0$, showing times $\tau_1=\tau_2=0$. So the releases appear simultaneous in the clock frames.
When the leading clock (subscript 2) has fallen to $r=R_1$,
\begin{align*} t_2(R_1) &= \int^{R_1}_{R_{max}-1} \left( \frac{dt}{dr} \right)_2 dr \end{align*}

at $t=t_2$ the position of the first clock is found by solving for the unknown radius $X$ in this
equation,
\begin{align*} t_2= \int_{R_{max}}^X \left( \frac{dt}{dr} \right)_1 dr \end{align*}

Given the worldline of a freely falling body parameterized by coordinate time, the required radius can be found ( but probably not in a closed form).

Timelike radial geodesics parameterized by t are given in the link below but the equations are laborious to set up so I have not done any calculations with this yet.

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

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pervect said:
Time to resolve a SR question here, first, before any progress is going to be made. We've got a moving ruler in the flat space-time of SR. If we can't agree on how to measure it's length in the lab frame, and its proper length, we aren't going to make any progress with the relativistic case.

$$$\begin{picture}(62,90)(0,0) \linethickness{0.3mm} \multiput(0,0)(0.12,0.54){167}{\line(0,1){0.54}} \put(20,90){\vector(1,4){0.12}} \linethickness{0.3mm} \multiput(40,0)(0.12,0.54){167}{\line(0,1){0.54}} \put(60,90){\vector(1,4){0.12}} \linethickness{0.3mm} \qbezier(5,20)(5,20)(5,20) \qbezier(5,20)(5,20)(5,20) \put(5,20){\vector(0,0){0.12}} \linethickness{0.3mm} \multiput(5,20)(0.12,0.12){417}{\line(1,0){0.12}} \put(55,70){\vector(1,1){0.12}} \linethickness{0.3mm} \put(0,0){\line(1,0){40}} \put(40,0){\vector(1,0){0.12}} \put(0,0){\makebox(0,0)[cc]{}} \put(5,5){\makebox(0,0)[cc]{}} \put(0,0){\makebox(0,0)[cc]{}} \put(-3,-3){\makebox(0,0)[cc]{A}} \put(42,-3){\makebox(0,0)[cc]{B}} \put(-1,20){\makebox(0,0)[cc]{C}} \put(60,70){\makebox(0,0)[cc]{D}} \put(22,93){\makebox(0,0)[cc]{E}} \put(62,93){\makebox(0,0)[cc]{F}} \end{picture}$$$

The diagram is a space-time diagram, with time running up the page, and space running across the page.

On this space-time diagram, AE and BF are the worldlines representing the two endpoints of a moving ruler.

Can we resolve that AB represents the length of the ruler in the lab frame, which is also a coordinate difference in the lab frame, because events A and B occur "at the same time" in the lab frame, but that CD, and NOT AB, represents the proper length of the ruler, because events C and D occur at the same time in the frame of the ruler?

The diagram is not to scale - the intent is that lline CD will appear to be orthogonal to line AE and AF if one transforms to the ruler's frame, or if you express them as vectors in any frame, the Lorentz invariant dot product AE with CD will be zero.

Maybe this is the time to fess up to a confusion I've never resolved in both SR and GR. I wonder how is it possible to define an invariant length at all? I raise this from a few points:

1) Rigid bodies are impossble in both SR and GR because they violate causality and lightspeed limit. A rigid ruler is truly and fundamentally impossible, which is why it is much better to measure lengths using light signals.

2) Ok (1) is not confusion, but well accepted. Now to work around (1), you try to say you can keep (with whatever rocket thrust needed) e.g. two rockets at constant invariant distance from each other. But what does this mean?

a) For events that can be connected by a timelike trajectory, you can define a unique invariant interval in SR, and (I think) a finite number of physically meaningful invariant invervals between them in GR (representing the different locally extremal paths between them for different regions of a curved geometiry).

b) For events that cannot be connected by a timelike path (this is what we presumably want for a ruler), the concept of an extremal path is much more problematic (at least to me). First, you cannot be minimizing pathlength within a time slice (a la Euclidean geometry) because there are no unique time slices, even in SR. On the other hand, if you allow arbitrary paths through spacetime (you can't restrict to timelike, since there are no timelike paths), there exists neither a maximum or a minimum extremal path. Obviously, circuitous routes can raise the the path length without limit. However, if you allow paths that go forward and backward through coordinate time, you can reduce the pathlength without limit. So no extremal. You might arbitrarily preclude time traveling paths, but is *this* possible to define in an invariant way (e.g. especially in GR where even timelike curves can be circular in theory, though many doubt in practice).

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One possible answer to my conundrum on invariant length is to use the parallel transport definition of geodesic (a curve that transports its tangent such that it remains tangent), rather than the variational definition. This would seem to fully resolve it in SR, while GR might need (in general) to accept several invariant distances between two causally disconnected events (spacelike).

Of course, then there is the philosophical issue of what it means to have an object (e.g. ruler) consisting of causally disconnected events?

PAllen said:
Maybe this is the time to fess up to a confusion I've never resolved in both SR and GR. I wonder how is it possible to define an invariant length at all? I raise this from a few points:

1) Rigid bodies are impossble in both SR and GR because they violate causality and lightspeed limit. A rigid ruler is truly and fundamentally impossible, which is why it is much better to measure lengths using light signals.

2) Ok (1) is not confusion, but well accepted. Now to work around (1), you try to say you can keep (with whatever rocket thrust needed) e.g. two rockets at constant invariant distance from each other. But what does this mean?

There's a fairly widely accepted definition of rigid motion, called Born Rigidity. While it can't be defined for rotating object, it works fine for non-rotating ones such as most of our thought experiments (elevators and such). But that doesn't seem to be the focus of your question, though it's one approach.

a) For events that can be connected by a timelike trajectory, you can define a unique invariant interval in SR, and (I think) a finite number of physically meaningful invariant invervals between them in GR (representing the different locally extremal paths between them for different regions of a curved geometiry).

b) For events that cannot be connected by a timelike path (this is what we presumably want for a ruler), the concept of an extremal path is much more problematic (at least to me). First, you cannot be minimizing pathlength within a time slice (a la Euclidean geometry) because there are no unique time slices, even in SR. On the other hand, if you allow arbitrary paths through spacetime (you can't restrict to timelike, since there are no timelike paths), there exists neither a maximum or a minimum extremal path. Obviously, circuitous routes can raise the the path length without limit. However, if you allow paths that go forward and backward through coordinate time, you can reduce the pathlength without limit. So no extremal. You might arbitrarily preclude time traveling paths, but is *this* possible to define in an invariant way (e.g. especially in GR where even timelike curves can be circular in theory, though many doubt in practice).

Well, I haven't seen this discussed specifically in the textbooks, but the argument that a space-like slice gives you a local minimum, and a timelike slice gives you a local maximum seems valid to me. Specifically, given some parameterized curve $x^i(\lambda)$, and it's partial derivitaves $\dot{x}^i = \partial x^i / \partial \lambda$ the quantity I is extermized by a geodesic

$$I = \sqrt {\int g_{ij} \dot{x}^i \dot{x}^j d\lambda}$$This suggests that the extreme point is a saddle point, rather than a maximum or a minimum.

Saddle points (rather than a true maximum or minimum) arising from the "principle of least action" are not unique to relativity. See for instance http://www.eftaylor.com/pub/Gray&TaylorAJP.pdf

Parallel transport also works fine, as you point out - a geodesic is often defined as a curve that parallel transports its own tangent vector.

pervect said:
Time to resolve a SR question here, first, before any progress is going to be made.
If someone comes up with the right formula progress is made.

You seem to give me the impression that you know this all. So why not simply write down the formula? Do you actually know the formula?

So all we need is a simple formula describing proper distance in terms of rs, r1, r1 and the velocities at r1 and r2. Five parameters, the Schwarzschild radius, two r coordinates and two coordinate velocities.

Mentz114 said:
Is this what's required ?

Two clocks are dropped from $r=R_{max}$ and $r=R_{max}-1$ at $t=0$, in such a way that they are synchronised at $t=0$, showing times $\tau_1=\tau_2=0$. So the releases appear simultaneous in the clock frames.
When the leading clock (subscript 2) has fallen to $r=R_1$,
\begin{align*} t_2(R_1) &= \int^{R_1}_{R_{max}-1} \left( \frac{dt}{dr} \right)_2 dr \end{align*}

at $t=t_2$ the position of the first clock is found by solving for the unknown radius $X$ in this
equation,
\begin{align*} t_2= \int_{R_{max}}^X \left( \frac{dt}{dr} \right)_1 dr \end{align*}

Given the worldline of a freely falling body parameterized by coordinate time, the required radius can be found ( but probably not in a closed form).

Timelike radial geodesics parameterized by t are given in the link below but the equations are laborious to set up so I have not done any calculations with this yet.

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

So you assume a fall from a stationary position right?

If we take the integral of your formula we get:

$$\Delta t = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

Same formula as a few postings ago but for a different purpose now?

Ok, let's solve the movement of the front clock

For your case let's assume FRONT=9 and BACK=10, they are both stationary at this point, note that this is different from an earlier scenario where we have the clock travel from infinity.

Let's 'run' the front clock from 9 to 8, we get:
Delta t=1.133531393

Feeding that back to get the R value for BACK from 10 to x we get:
BACK=8.986002823

It looks like coordinate distances between the two clocks are getting smaller for decreasing values of R.

Now let's see how the coordinate values and distance behave wrt falling R values using your function:

The graph shows the front clock (orange) in a straight line down to R=1 and the back clock (green) using your formula trailing. The red graph shows a clock R+5 removed from the front clock at the start of the experiment. The dotted blue line shows the coordinate distance between the front and clock R+5 removed to visualize the change in coordinate distance between them.

So coordinate distance drops with lower R values, the drop accelerates rapidly close to the event horizon to become zero at the event horizon.

Assuming this is the right step, how do we tie this in with tidal effect? Conceivably by calculating the proper distance and concluding the proper distance increases with lower values of R?

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yuiop said:
If we can attach any validity to the equation I derived in the other thread:

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

The above equation can be solved for r2 when we know r1 as:

$$r2 = r_s + r_s W \left(\left(\frac{r1}{r_s}-1\right) \, \exp\left(\frac{r1}{r_s}+\frac{\Delta S}{r_s} -1 \right)$$

We know all the variables. r1=2, Rs = 1 and the proper length of the rigid ruler at infinity (or anywhere else for that matter as we are assuming the proper length does not change) is $\Delta S =1$. The bad news is that you need mathematical software that can compute the Lambert W function shown as simply W() in the above equation, in order to obtain a numerical solution to your question.
starthaus said:
1. Expressing $$r_2$$ as a function of $$r_1$$ is totally meaningless since they are arbitrary limits of integration, so there cannot be any correlation between them.

2. There is no justification for the

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad$$

to begin with.

If we have an equation:

$$\Delta S = (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

..then it perfectly mathematically and physically valid to solve for r2 when we know the values of the all the other variables as we do in this case as Passionflower specified them all.

The justification of the equation is this. To a local observer, local measurements are just Lorentzian. The falling object is length contracted by a factor of gamma(v) according to the local observer, where v is the local velocity. To the Schwarzschild observer any measurement made by the local observer is subject to a transformation factor of gamma(g) which is the gravitational gamma factor. The coordinate length of the moving object according to the observer at infinity is L*gamma(v)*gamma(g) where L is the proper length of the falling object and falling object is infinitesimal. For an extended object the ingrated lengths have to be used. The important take home message is that the coordinate length of a moving rod is not the same as the coordinate length of the same rod when it stationary, in both SR and GR.
Passionflower said:
So the question here is: if we want to take the proper distance between two R values does it matter if it is from a stationary of moving object?
The proper distance between two R values is equivalent to the measurement made by a ruler that is at rest so that the ends of the given ruler remain at the two R values over time. This proper distance is not the same as the length of a moving ruler that momentarily spans those two coordinates. The difference is important if want to know the proper length of the falling ruler or elastic ball or whatever. We cannot assume the coordinate length of a object moving relative to the observer is the same as the coordinate length of a stationary object in either SR or GR.
Let me ask you this, what do think is the physical interpretation for this integral:

$$\int _{{r0}}^{{r1}}\!{\frac {1}{\sqrt {1-{\frac {{\it rs}}{r}} }}}\,{dr}$$
I interpret that as the length measured by a series of stationary infinitesimal rulers laid end to end from r0 to r1.

starthaus said:
It is calculated as a function of $$r_1$$ and $$r_2$$. This makes it pretty much useless for calculating $$r_2$$
See above.

Here is an easy way to get the expressions for length contraction and time dilation in GR. Start with the Schwarzschild metric:

$$-ds^2=\left(1+\frac{2\Phi}{c^2 }\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(1+\frac{2\Phi}{c^2}\right)dt^2$$

If you want to calculate time dilation, make $$dr=d\theta=d\phi=0$$ . The reason is that you are trying to determine the relationship between two co-located events.

$$\frac{ds}{cdt}=\sqrt{1+\frac{2\Phi}{c^2}}$$Length contraction for the case of a non-rotating object is obtained by making $$d\theta=d\phi=0$$ (absence of rotation) and $$dt=0$$ (object endpoints are marked simultaneously bu the distant observer).

$$\frac{ds}{dr}=\frac{1}{\sqrt{1+\frac{2\Phi}{c^2}}}$$

Particularization:

For the radial gravitational field :

$$\Phi=-\frac{GM}{r}$$

so:

$$cdt=\frac{cd\tau}{\sqrt{1-\frac{2GM}{rc^2}}}$$
i.e. coordinate time appears dilated wrt proper time

We know from many discussions in this forum that that is the relationship between the proper time and coordinate time of a clock that is stationary in the coordinates. For a moving clock the equation is:

$$cdt=\frac{cd\tau}{\sqrt{1-\frac{2GM}{rc^2}}\sqrt{1-\frac{v^2}{c^2}}}$$

where v is the local velocity.
$$dr=ds\sqrt{1-\frac{2GM}{rc^2}}$$

i.e. coordinate length appears contracted wrt proper length.

If you want proper length as a function of coordinate length, then:

$$ds=\frac{dr}{\sqrt{1-\frac{2GM}{rc^2}}}$$

Integrate the above and you get the correct expression for proper length as a function of coordinate length.

Again this the coordinate length of short stationary object. For a short moving object the equation is:

$$ds=\frac{dr}{\sqrt{1-\frac{2GM}{rc^2}}\sqrt{1-\frac{v^2}{c^2}}}$$

where v is the local velocity.

starthaus said:
1. It is the other way around, $$\Delta S$$ is a function of the limits of integration

2. The integral shown is incorrect to begin with.
Given all the above arguments, I do not think your above statements are valid.

yuiop said:
If we have an equation:

$$\Delta S = (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

..then it perfectly mathematically and physically valid to solve for r2 when we know the values of the all the other variables as we do in this case as Passionflower specified them all.

..not if $$\Delta S$$ is a function of $$r_1$$ and $$r_2$$ . You are simply using circular logic.
We know from many discussions in this forum that that is the relationship between the proper time and coordinate time of a clock that is stationary in the coordinates. For a moving clock the equation is:

$$cdt=\frac{cd\tau}{\sqrt{1-\frac{2GM}{rc^2}}\sqrt{1-\frac{v^2}{c^2}}}$$

where v is the local velocity.

The example I showed is for a clock at rest . The derivation shows clearly $$dr=0$$. It is the standard expression for gravitational time dilation.
If you want the correct formula for time dilation in radial motion, then you must not make $$dr=0$$ in the derivation. If you do this correctly, you will be getting the correct formula:

$$d\tau=dt *\sqrt{1-r_s/r}*\sqrt{1-(\frac{dr/cdt}{1-r_s/r})^2}$$

where

$$r_s=\frac{2GM}{c^2}$$
Again this the coordinate length of short stationary object. For a short moving object the equation is:

$$ds=\frac{dr}{\sqrt{1-\frac{2GM}{rc^2}}\sqrt{1-\frac{v^2}{c^2}}}$$

where v is the local velocity.

You would be hard-pressed to derive the above. Care to give it a try?
Given all the above arguments, I do not think your above statements are valid.

...which only means that your formula for $$\Delta S$$ is incorrect because your integrand is wrong. Exactly my point.

Last edited:

starthaus said:
..not if $$\Delta S$$ is a function of $$r_1$$ and $$r_2$$ . You are simply using circular logic.
This is just silly. If we have an equation such as t = d/v, then we can solve for v to get v = d/t and even though v is a function of d and t, there is nothing circular about it. The equation v = d/t is simply stating how v is a function of d and t.

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