The Schwarzschild Metric - A Simple Case

[yuiop]

. 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.

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

$$r_s=\frac{2GM}{c^2}$$

Local velocity at r in Schwarzschild coordinates is:

$$v = \frac{dr/dt}{1-r_s/r}$$

This means that your equation:

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

is equivalent to my equation:

$$d\tau=dt *\sqrt{1-r_s/r}*\sqrt{1-(v/c)^2}$$

where v is the local velocity.

Please don't go spoiling yet another thread with petty red herrings again. Why not try and be constructive for a change and either state where you think the errors are (if any) or what you think the correct solutions are.

 

[starthaus]

Local velocity at r in Schwarzschild coordinates is:

$$v = \frac{dr/dt}{1-r_s/r}$$

This means that your equation:

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

is equivalent to my equation:

$$d\tau=dt *\sqrt{1-r_s/r}*\sqrt{1-(v/c)^2}$$

where v is the local velocity.

Sure. Which means that your formula for length contraction is wrong. Precisely what I've been telling you.

 

[starthaus]

..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.

No, it is basic math.

 

[yuiop]

Sure. Which means that your formula for length contraction is wrong. Precisely what I've been telling you.

My formula is for the special case of an object that is initially stationary and released from infinity. I made it clear that I was talking about that special case in the post where I derived it and it simplifies things a bit. You could calculate a more general formula for an object with arbitrary initial velocity at infinity or for an object with dropped from an arbitrary height. If you want to derive the more general case, then feel free to "fill your boots"

 

[starthaus]

My formula is for the special case of an object that is initially stationary and released from infinity.

Point is, your "formula" is wrong.

I made it clear that I was talking about that special case in the post where I derived it and it simplifies things a bit.

You produced three contradictory formulas with no derivation whatsoever. This is the problem.

 

[pervect]

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?

The formula for what? Your terminology is so nonstandard , it's hard to tell what you're asking. And you seem to keep changing your mind, too :-(.

Once upon a time you did ask

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.

Well, right from the start there's the obvious question:

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.

To which your reply was:

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

So I guess you're looking for the paramaterized equations of two worldlines, falling into a black hole, from different heights, as a first step, and looking to figure out how they separate as the next.

And I gather you're not so much interested in something that's easier to calculate that may actually illustrate the physics better - like the tidal force on an infalling, reasonably rigid rod. At least not from your responses to date.

So, I'll point you at MTW on pg 663, the formula for radial infall for an object that is intitally stationary at r=R. It's given in parametric form in 25.28

r=(R/2)*1 + cos eta
tau = (R/2)*(R/2M)^.5 (eta + sin eta)

the expression for t as a function of eta is given on pg 666 in 25.37 and is so messy I don't care to type it in.

I have the impresion you had MTW already? If that's actually what you're looking for, then you should have your formula. If you don't have MTW, AND if that's what you're really looking for, I'd be willing to type it in, if I must. At one point, you were using some OTHER equation from MTW, which represented a fall from infinity - but I think you've realized the error there? I couldn't find the post, maybe it was in some other thread.

So, if we take two curves with nearby values of R, we now have to measure the distance between the two curves, the proper distance, needs some discussion. My goal isn't to do the work for you, it's to get you to illustrate some understanding of the problem. So far, I don't feel very successful, I must say.

The separation between the two worldlines is going to be increasing with time. What you really need to do to define the separation or distance is to unambiguously describe the path that you consider to be spacelike along which you are going to integrate ds. That's the point of the SR example - in that case, there is only one clear answer.

You've got a number of choices: you can select one of the two bodies, and a particular time point on it, and do a simple linear projection from it to the second body, using the notion that the bodies are close enough together that you are in a sufficiently flat space-time. This is the easiest approach.

Slightly more sophisticated, one might draw a spacelike geodesic that's perpendicular to the timelike geodesic of the selected falling body, and measure the length along the geodesic. But I don't think you have the math skills to do that. And I don't see the point in doing it for you.

The point is, when you define some notion of simultaneity, you define a space-time split, and the distance becomes defined. But, until you define that notion of simultaneity (which takes some discussion, and you don't seem to understand the problem), you can't interpret the invariant interval in terms of time and space.

 

[Passionflower]

So, I'll point you at MTW on pg 663, the formula for radial infall for an object that is intitally stationary at r=R. It's given in parametric form in 25.28

r=(R/2)*1 + cos eta
tau = (R/2)*(R/2M)^.5 (eta + sin eta)

the expression for t as a function of eta is given on pg 666 in 25.37 and is so messy I don't care to type it in.

I have the impresion you had MTW already? If that's actually what you're looking for, then you should have your formula. If you don't have MTW, AND if that's what you're really looking for, I'd be willing to type it in, if I must. At one point, you were using some OTHER equation from MTW, which represented a fall from infinity - but I think you've realized the error there? I couldn't find the post, maybe it was in some other thread.

What error? In my posting I clearly indicated that is what I was looking for.

All I know is that you skimp over my postings with apparent preconceived notions, then if I respond you never even bother to reply to my responses.

My goal isn't to do the work for you, it's to get you to illustrate some understanding of the problem. So far, I don't feel very successful, I must say.

Well if you do not want to bother giving the answers that is alright, let's just assume the answers are all crystal clear to you but you just want to keep them to yourself, perhaps you think it is not worthy to write down formulas and stick out your neck in giving solutions. In my experience sometimes we think we understand an issue but when we are forced to actually calculate it we find out we only thought we knew the problem, and then we face our egos. Surely that must not apply to you, you seem to know it all and have a goal to educate all those who do not know, not by trying or giving solutions, no other people do that, you only give hints or 'illustrations of understanding'. That is certainly your prerogative.

 

[Mentz114]

r=(R/2)*1 + cos eta
tau = (R/2)*(R/2M)^.5 (eta + sin eta)

This has been quoted from MTW, which I don't have. The mathpages 'Reflections on Relativity' has a very similar treatment and also gives the worldline parameterized by t. I thought this would be a way to compare the two frames but I'm not so sure now, having tried some calculations. It is difficult to compare separated events in GR.

But if an ideal rod falls radially, aligned along a radius, then the velocity differential must cause stresses and there must be a way to put a number on it. It is just the rr component of the tidal tensor, but what does that component ( +2m/r^3 ) signify, quantitatively ? Is it an acceleration, i.e. a velocity gradient of dr/dtau in the r direction ?

 

[yuiop]

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...

Small correction. That last equation should be (dr/dt)^2 = 2GM(1/r - 1/R_max).

Could you confirm that the tau mentioned in the first (relativistic) equation is the proper time a stationary clock at r, rather than the proper time of a clock attached to the falling object?

 

[yuiop]

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

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

then we are in a position to partly answer your question.

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.

I have noticed a problem with the proposed solution above. It assumes the velocity of each part of the rigid falling rod is simply the free fall velocity of a particle dropped from infinity. For a rigid rod this would not be exactly true for all parts of the rod because some parts of the rod not at the centre of gravity will be forced to fall at rate which is not the natural free fall rate. The equation actually represents the proper spatial separation of a string of unconnected particles with negligible mutual gravitational interaction (e.g coffee grounds). The even worse news, is that even though we know the proper length between the leading and trailing particles at the lower Schwarzschild coordinates, we do not know the initial proper length between the particle at infinity as the particles will have been stretched apart by tidal forces. More analysis needed. :-(

 

[Passionflower]

Yuiop, could you comment on Mentz's thoughts on the matter (as expressed in the other topic), are they identical with your ideas? I may have applied your formula in the wrong context while I probably applied it correctly with Mentz's approach. Take a look at the generated graphs in the other topic and if you are on the same line.

I have noticed a problem with the proposed solution above. It assumes the velocity of each part of the rigid falling rod is simply the free fall velocity of a particle dropped from infinity. For a rigid rod this would not be exactly true for all parts of the rod because some parts of the rod not at the centre of gravity will be forced to fall at rate which is not the natural free fall rate. The equation actually represents the proper spatial separation of a string of unconnected particles with negligible mutual gravitational interaction (e.g coffee grounds). The even worse news, is that even though we know the proper length between the leading and trailing particles at the lower Schwarzschild coordinates, we do not know the initial proper length between the particle at infinity as the particles will have been stretched apart by tidal forces. More analysis needed. :-(

Very true, the simplest approach would be to replace the rod with two clocks with rockets that constantly calculate the center between them (as if it were a rod) and accelerate the proper way (front clock and black clock accelerate towards this center. The good news is that we can calculate the tidal acceleration at r (but I suspect it will also depend on the proper velocity at that point).

Second point: Can't we just define the length to be one at infinity and as the curvature is zero there there is no tidal effect?

 

[yuiop]

Yuiop, could you comment on Mentz's thoughts on the matter (as expressed in the other topic), are they identical with your ideas? I may have applied your formula in the wrong context while I probably applied it correctly with Mentz's approach. Take a look at the generated graphs in the other topic and if you are on the same line.

I am busy reformulating my ideas so I cannot give a definitive answer at this point. One possible problem with Mentz's approach is that he assumed that when t=0 at R(front) and R(back) that tau will also be synchronised front and back. Since the two clocks are different heights (and at different velocities when falling they are running at different rates (the lower clock signals will appear redshifted from the top) and we cannot assume the falling clocks remain synchronised without modification. In short, there is a difference in what is simultaneous for the falling clocks and what is simultaneous according the Schwarzschild observer at infinity.

... the simplest approach would be to replace the rod with two clocks with rockets that constantly calculate the center between them (as if it were a rod) and accelerate the proper way (front clock and black clock accelerate towards this center. The good news is that we can calculate the tidal acceleration at r (but I suspect it will also depend on the proper velocity at that point).

This might work, but I suspect the mathematical solution is not trivial.

Second point: Can't we just define the length to be one at infinity and as the curvature is zero there there is no tidal effect?

We sure can and that was my intention with the first attempted solution. The trouble is we are back at square one. What is R2=R(trailing) when R1=R(leading) is at 2Rs? R1 and R2 are the limits of integration in the equation I gave and they unknown if we assume the initial proper length at infinity.

What we need is some guiding principle like the volume remains constant (according to some observer) as the object falls. That means we need someone who really knows the physical meaning of Weyl and Ricci curvature in Schwarzschild coordinates and how to calculate it.

 

[yuiop]

So, I'll point you at MTW on pg 663, the formula for radial infall for an object that is intitally stationary at r=R. It's given in parametric form in 25.28

r=(R/2)*1 + cos eta
tau = (R/2)*(R/2M)^.5 (eta + sin eta)

the expression for t as a function of eta is given on pg 666 in 25.37 and is so messy I don't care to type it in.

I think that should be r = (R/2)*(1+cos(eta))

which of course defines the parameter eta as: acos(2r/R - 1).

I already posted the expression for t as a function of eta in an old thread, so it not too much trouble for me change the symbols and copy and paste it here:

$$t = (R/2 + 2M)*eta*Q + 2M*\ln\left(\frac{Q+ \tan(eta/2)}{Q- \tan(eta/2)}\right) + QR/2 \, \sin(eta)$$

where:

$$Q = \sqrt{R/2m -1 }$$

See https://www.physicsforums.com/showpost.php?p=1834328&postcount=21

 

[yuiop]

If you relax the requirement to start at infinity and start from a finite height such as R1=100m and R2=101m, then I think it will be relatively straightforward to calculate where the trailing edge is, when the leading edge is at 4m, for a system of unconnected free-falling particles. That would at least answer one of your original questions and this automatically takes tidal stretching into account. The difficult part is then calculating what happens when the tidal stretching is resisted by a rigid cable or rockets.

 

[starthaus]

I have noticed a problem with the proposed solution above. It assumes the velocity of each part of the rigid falling rod is simply the free fall velocity of a particle dropped from infinity.

No, it isn't.If that is what you want , then the correct equation is something different from anything you posted.

For a rigid rod this would not be exactly true for all parts of the rod because some parts of the rod not at the centre of gravity will be forced to fall at rate which is not the natural free fall rate.

This is only part of the problem with your solution.

The equation actually represents the proper spatial separation of a string of unconnected particles with negligible mutual gravitational interaction (e.g coffee grounds).

It is much worse than that. If you wanted to solve that problem, then the starting formalism is different from anything you posted so far.

The even worse news, is that even though we know the proper length between the leading and trailing particles at the lower Schwarzschild coordinates, we do not know the initial proper length between the particle at infinity as the particles will have been stretched apart by tidal forces. More analysis needed. :-(

Yes, a LOT more. Since the above is a very difficult problem, I suggest that you start with a perfectly rigid rod , one that exhibits constant proper length, like the ones used by SR.

 

[starthaus]

Small correction. That last equation should be (dr/dt)^2 = 2GM(1/r - 1/R_max).

Err, no, it is much more complicated than that:

$$\frac{dr}{dt}=(1-r_s/r)\sqrt{1-\frac{1-r_s/r}{1-r_s/r_0}}$$

where:

$$r_s$$=Schwarzschild radius
$$r_0$$=initial coordinate

For fall from infinity

$$r_0=\infty$$

so, you get:$$\frac{dr}{dt}=(1-r_s/r)\sqrt{r_s/r}$$

 

[Passionflower]

If you relax the requirement to start at infinity and start from a finite height such as R1=100m and R2=101m, then I think it will be relatively straightforward to calculate where the trailing edge is, when the leading edge is at 4m, for a system of unconnected free-falling particles. That would at least answer one of your original questions and this automatically takes tidal stretching into account. The difficult part is then calculating what happens when the tidal stretching is resisted by a rigid cable or rockets.

Actually as I explained with 'time to ultimate doom' the fall from infinity is not a problem at all. We can pick any starting coordinate we want and assume both the front and the back came from infinity. This approach I think has several advantages, not it the least that once we have the right formula we can then express it in terms of proper time left and take it down past the event horizon as well.

Well that is encouraging, so say we start from R=10 (but they come free falling from infinity - I would not like the inattentive reader to claim I make yet another mistake), and then take it down a coordinate time of 5 and then all we need to do is take the back down from R=11 with the same coordinate time. And that would be our falling two coffee grind particles including a tidal stretching freebie?

Is everybody absolutely sure it is really that simple? If so then if we take the proper distance pre and post we can simply subtract the pre proper length from the post proper length and voila we have a difference in length in terms of R coordinate distance and coordinate time (and then of course we can also calculate proper distance and proper time for each grind perspective as well). I certainly hope we can say that as that would get us in the direction of the solution. Assuming I understand correctly, and please correct me if I am wrong Mentz, this is also the approach that Mentz suggested.

Since the above is a very difficult problem

You are most welcome to join the club and come with formulas we can plug in the example (RS=1, FRONT=10, BACK=11 both came free falling from infinity). Then we can graph it and look at it.

I suggest that you start with a perfectly rigid rod , one that exhibits constant proper length, like the ones used by SR.

That is an option, but it seems to me that you need to operationally define how this is obtained (for instance my suggestion of two rockets at each instant adjusting their accelerations towards each other in order to maintain a free falling virtual center) since each scenario will give a different result. But perhaps I am wrong and are all of them equivalent.

 

[Mentz114]

Is everybody absolutely sure it is really that simple? If so then if we take the proper distance pre and post we can simply subtract the pre proper length from the post proper length and voila we have a difference in length in terms of R coordinate distance and coordinate time (and then of course we can also calculate proper distance and proper time for each grind perspective as well). I certainly hope we can say that as that would get us in the direction of the solution. Assuming I understand correctly, and please correct me if I am wrong Mentz, this is also the approach that Mentz suggested.

I confirm this is what I suggested for independently falling clocks. But I wonder if saying that "after proper time T seconds the clocks are separated by d meters according to the observer at infinity" is devoid of meaning.

In the case of the falling rod, if we are dealing with a short rod the problem can be handled by Newtonian tidal theory because nothing relativistic is happening in the local frame ( as someone has pointed out earlier). If we are thinking of objects thousands of kilometers long or bigger, it isn't realistic and probably pointless. But the principle that tidal forces are proportional to the velocity gradient helps.

 

[starthaus]

That is an option, but it seems to me that you need to operationally define how this is obtained (for instance my suggestion of two rockets at constantly accelerating towards each other to maintain a free falling virtual center) since each scenario will give a different result. But perhaps I am wrong and are all of them equivalent.

This simple case can be dealt with. The correct starting formula is:

$$\frac{dr}{ds}=\sqrt{r_s/r-r_s/r_0}$$
$$\frac{dr}{dt}=(1-r_s/r)\sqrt{1-\frac{1-r_s/r}{1-r_s/r_0}}$$

If you make

$$r_0=\infty$$

as you are suggesting, things get even simpler:

$$\frac{dr}{ds}=\sqrt{r_s/r}$$
$$\frac{dr}{dt}=(1-r_s/r)\sqrt{r_s/r}$$

where:

$$r_s$$=Schwarzschild radius
$$r_0$$=initial coordinate

 

[yuiop]

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...

Small correction. That last equation should be (dr/dt)^2 = 2GM(1/r - 1/R_max).

Err, no, it is much more complicated than that:

$$\frac{dr}{dt}=(1-r_s/r)\sqrt{1-\frac{1-r_s/r}{1-r_s/r_0}}$$

where:

$$r_s$$=Schwarzschild radius
$$r_0$$=initial coordinate

If you had read Pervect's post carefully (the one I was responding to), you would have noticed that he was talking about the Newtonian analogue of the correct relativistic differential equation:

(dr/tau)^2 = 2GM(1/r - 1/R_max)

 

[yuiop]

I suggest that you start with a perfectly rigid rod , one that exhibits constant proper length, like the ones used by SR.

That is an option, but it seems to me that you need to operationally define how this is obtained (for instance my suggestion of two rockets at each instant adjusting their accelerations towards each other in order to maintain a free falling virtual center) since each scenario will give a different result.

This simple case can be dealt with. The correct starting formula is:

$$\frac{dr}{ds}=\sqrt{r_s/r-r_s/r_0}$$
$$\frac{dr}{dt}=(1-r_s/r)\sqrt{1-\frac{1-r_s/r}{1-r_s/r_0}}$$

If you make

$$r_0=\infty$$

as you are suggesting, things get even simpler:

$$\frac{dr}{ds}=\sqrt{r_s/r}$$
$$\frac{dr}{dt}=(1-r_s/r)\sqrt{r_s/r}$$

where:

$$r_s$$=Schwarzschild radius
$$r_0$$=initial coordinate

You have completely missed the point of passionflower's question. You said use a rigid rod of constant proper length, and Passion asked the astute question of how you were going to define the proper length of rigid rod in the case of a rod falling in curved space. This is much more complicated than you perhaps imagine. All you have done is given the velocity of a falling particle which is not what Passion asked for. The equations for the velocity of a falling particle have already been given in various forms in posts #6, #12 and #33 of this thread and in posts #19, #20 and #21 in the closely related thread https://www.physicsforums.com/showthread.php?t=431881&page=2 and in this reference document given by Pervect http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111073v1.pdf so you have added nothing of substance.

 

[yuiop]

A question for Pervect:

Do you agree that for the special case of a very short (infinitesimal) rod of length dL that is initially at rest and dropped from infinity, that the following is true?

$$\frac{dr}{ds}=\sqrt{r_s/r}$$

and

$$dL' = dL\sqrt{(1-r_s/r)(1-(dr/ds)^2)} = dL \sqrt{(1-r_s/r)(1-r_s/r)} = dL(1-r_s/r)$$

where dL' is the length according to the Schwarzschild observer when the object is falling past Schwarzschild radial coordinate r?

If you do not agree, do you have any suggestions what the equation should be?

 

[starthaus]

If you had read Pervect's post carefully (the one I was responding to), you would have noticed that he was talking about the Newtonian analogue of the correct relativistic differential equation:

(dr/tau)^2 = 2GM(1/r - 1/R_max)

You may have wanted to write $$\frac{dr}{d\tau}$$ but you put down $$\frac{dr}{dt}$$. so your attempt at correcting his post is incorrect. Like all the other three formulas that you guessed in this thread.

 

[starthaus]

You have completely missed the point of passionflower's question. You said use a rigid rod of constant proper length, and Passion asked the astute question of how you were going to define the proper length of rigid rod in the case of a rod falling in curved space. This is much more complicated than you perhaps imagine. All you have done is given the velocity of a falling particle which is not what Passion asked for. The equations for the velocity of a falling particle have already been given in various forms in posts #6, #12 and #33 of this thread and in posts #19, #20 and #21 in the closely related thread https://www.physicsforums.com/showthread.php?t=431881&page=2 and in this reference document given by Pervect http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111073v1.pdf so you have added nothing of substance.

I simply corrected your incorrect "guesses", that's all.

 

[starthaus]

A question for Pervect:

Do you agree that for the special case of a very short (infinitesimal) rod of length dL that is initially at rest and dropped from infinity, that the following is true?

$$\frac{dr}{ds}=\sqrt{r_s/r}$$

This is correct (I just showed that in post 54).

and

$$dL' = dL\sqrt{(1-r_s/r)(1-(dr/ds)^2)}$$

Are you guessing the above? Because there is no derivation (again).

 

[yuiop]

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.

Small correction. That last equation should be (dr/dt)^2 = 2GM(1/r - 1/R_max).

Here Pervect effectively said:

0.5*m*(dr/dt)^2 = GmM(1/r - 1/R_max) $\Rightarrow$ dr/dt = 2GM(1/r - 1/R_max)

which by simple algebra should have been:

0.5*m*(dr/dt)^2 = GmM(1/r - 1/R_max) $\Rightarrow$ (dr/dt)^2 = 2GM(1/r - 1/R_max)

O.K. It was just a small typo on his part and I was fixing it to avoid confusion to future readers. To which your response was:

Err, no, it is much more complicated than that:

$$\frac{dr}{dt}=(1-r_s/r)\sqrt{1-\frac{1-r_s/r}{1-r_s/r_0}}$$

You gave the relativistic equation, when Pervect was talking about the Newtonian version. I had already pointed this out to you when I said:

If you had read Pervect's post carefully (the one I was responding to), you would have noticed that he was talking about the Newtonian analogue of the correct relativistic differential equation:

(dr/tau)^2 = 2GM(1/r - 1/R_max)

Obviously you have still not read Pervect's post carefully because you next response was:

You may have wanted to write $$\frac{dr}{d\tau}$$ but you put down $$\frac{dr}{dt}$$. so your attempt at correcting his post is incorrect.

I put dr/dt instead of dr/dtau because I was talking about Pervect's Newtonian equation which was in terms of dr/dt. There is no distinction between coordinate time dt and proper time dtau in Newtonian equations, only universal time. Surely you know this? Pervect and I were talking about NEWTONIAN equations, not relativistic equations, so no, I did not mean to write and nor should I have written dr/dtau.

So for the second time, READ PERVECT'S POST CAREFULLY.

Your responses are bordering on illogical.

I suggest that you start with a perfectly rigid rod , one that exhibits constant proper length, like the ones used by SR.

That is an option, but it seems to me that you need to operationally define how this is obtained...

This simple case can be dealt with. The correct starting formula is:

$$\frac{dr}{ds}=\sqrt{r_s/r-r_s/r_0}$$
$$\frac{dr}{dt}=(1-r_s/r)\sqrt{1-\frac{1-r_s/r}{1-r_s/r_0}}$$

You have completely missed the point of passionflower's question.

I simply corrected your incorrect "guesses", that's all.

Again, an illogical response, because the posts you were responding to were nothing to do with any equations (correct or otherwise) I have posted.

Passionflower asked you how to operationally define the proper length of a falling rod to which your response was to quote some formulas for the velocity of a falling particle.
Are you now going to try and answer the question that Passionflower actually asked?

 

[yuiop]

$$\frac{dr}{ds}=\sqrt{r_s/r}$$

This is correct (I just showed that in post 54).

..and I just showed that in post 6.

$$dL' = dL\sqrt{(1-r_s/r)(1-(dr/ds)^2)} = dL \sqrt{(1-r_s/r)(1-r_s/r)} = dL(1-r_s/r)$$

where dL' is the length according to the Schwarzschild observer when the object is falling past Schwarzschild radial coordinate r?

Are you guessing the above? Because there is no derivation (again).

I did not show a derivation, because it is (almost) self evident.

In SR the coordinate length L' of moving object with proper length L and relative velocity v is $L' = L_o \sqrt{(1-v^2/c^2)}$. Locally, even in GR, the equations of SR apply and so the coordinate length of an object moving with local velocity v' = dr'/dt' according to a local static observer (LSO) is also simply $L' = L_o \sqrt{(1-v'^2/c^2)}$. The velocity of an object that was initially at rest and dropped from infinity according to the LSO is $v'/c =\ sqrt{(r_s/r)}$ so we can rewrite the last equation in this special case as $L' = L_o \sqrt{(1-r_s/r)}$. Now the transformation from local length L' to Schwarzschild coordinate length L is $L = L'\sqrt{(1-r_s/r)}$ so we can now write $L = L_o (1-r_s/r)$ and this is valid for very short (infinitesimal) lengths.

Q.E.D.

Just incase you are wondering, for the special case of a initially stationary object dropped from infinity:

$$\frac{dr}{dtau}=\sqrt{\frac{r_s}{r}} = \frac{dr'}{dt'}$$

where dtau is the proper time rate of a co-falling clock and dr' and dt' are measurements made by the LSO.

 

[starthaus]

I did not show a derivation, because it is (almost) self evident.

In SR the coordinate length L' of moving object with proper length L and relative velocity v is $L' = L_o \sqrt{(1-v^2/c^2)}$. Locally, even in GR, the equations of SR apply and so the coordinate length of an object moving with local velocity v' = dr'/dt' according to a local static observer (LSO) is also simply $L' = L_o \sqrt{(1-v'^2/c^2)}$.

You are guessing again.

The velocity of an object that was initially at rest and dropped from infinity according to the LSO is $v'/c =\ sqrt{(r_s/r)}$ so we can rewrite the last equation in this special case as $L' = L_o \sqrt{(1-r_s/r)}$.

Err, no. It doesn't work this way. You can get the equations of SR as a limiting case from the equations of GR (for the case of null-curvature). You can't do yet another hack "derivation" of the equations of GR by plugging in GR-derived (aka $$dr/ds=\sqrt{r_s/r}$$) equations into the equations of SR. There is no concept of $$r_s$$ in SR to begin with.
The same way you can't derive new equations of SR by plugging in SR equations into Newtonian mechanics. You need to learn that physics is not a collection of hacks and "guesses".

 

[yuiop]

Actually as I explained with 'time to ultimate doom' the fall from infinity is not a problem at all. We can pick any starting coordinate we want and assume both the front and the back came from infinity. This approach I think has several advantages, not it the least that once we have the right formula we can then express it in terms of proper time left and take it down past the event horizon as well.

The problem is that the equations you gave for "time till doom" do not give the coordinate falling time.

Well that is encouraging, so say we start from R=10 (but they come free falling from infinity - I would not like the inattentive reader to claim I make yet another mistake), and then take it down a coordinate time of 5 and then all we need to do is take the back down from R=11 with the same coordinate time. And that would be our falling two coffee grind particles including a tidal stretching freebie?

As I said before, it is probably easier to relax the fall from infinity requirement (although we might try that later) and assume a drop from initially at rest from a finite height.

After a coordinate time of 5 the leading clock will have fallen to 9.99437 and in the same coordinate time the trailing clock will have fallen to 10.953 and the coordinate gap will have increased from 1 to 1.009 if both particles were initially at rest and released at the same coordinate time from heights 10 Rs and 11 Rs.

It is a bit more interesting if you let the particles fall further.

If we start with the same release heights and let the leading particle fall to height 2 Rs then the coordinate fall time is 2.64 and trailing clock is at 4.64 Rs.

I am using the equations given in post 48 to obtain these results, but I have to numerically solve them as the mathematical software I use is unable to find a symbolic solution for r when t is know.

Here is a list of results for various initial and final heights for the trailing clock, all with the leading clock starting one coordinate unit lower and end point determined by the leading clock arriving at 2 Rs.

11 : 4.64
101 : 9.81
1001 : 21.86
10001 : 48.42
100001 : 105.987
1000001 : 230.22
10000001 : 498.02
100000001 : 1075.06
1000000001 : 2318.31

Very approximately, each tenfold increase in the initial drop height results in a two fold increase in the coordinate separation at the end point.

Is everybody absolutely sure it is really that simple? If so then if we take the proper distance pre and post we can simply subtract the pre proper length from the post proper length and voila we have a difference in length in terms of R coordinate distance and coordinate time (and then of course we can also calculate proper distance and proper time for each grind perspective as well).

No, it is not that simple. All we have done is determined the coordinate separation of the unconnected free falling particles. We have still not determined the proper separation of the freely falling particles in the rest frame of the particles and I not even sure at this point if it is possible to define what that is. It is as you probably already aware, very difficult to operationally define the proper length of the falling particles for non infinitesimal separations. It does however give an indication of the tidal stretching in coordinate terms.

The attached diagram is a graph in Schwarzschild coordinates of the objects falling from 10 Rs and 11 Rs.

 

[starthaus]

Just incase you are wondering, for the special case of a initially stationary object dropped from infinity:

$$\sqrt{\frac{r_s}{r}} = \frac{dr'}{dt'}$$

dr' and dt' are measurements made by the LSO.

Err, this is incorrect. You might want to check your facts. Instead of making up formulas that are wrong, it would be good if you tried consulting a good book or doing your own derivations,

 

[yuiop]

You are guessing again.

Err, no. It doesn't work this way. You can get the equations of SR as a limiting case from the equations of GR (for the case of null-curvature). You can't do yet another hack "derivation" of the equations of GR by plugging in GR-derived (aka $$dr/ds=\sqrt{r_s/r}$$) equations into the equations of SR. There is no concept of $$r_s$$ in SR to begin with.
The same way you can't derive new equations of SR by plugging in SR equations into Newtonian mechanics. You need to learn that physics is not a collection of hacks and "guesses".

You really don't "get it".

You better have a word with Pervect too about these "crimes" against physics:

If you make the subdivisions small, you can use radar, or the notion of distance in the local Lorentz frame, to determine the distance between the subdivision. I think it's already been noted in previous threads that the radar distance back-front and front-back varies for a long accelerating ruler, hence the importance of "dividing it up" in this manner.

While the space-time of the Schwarzschild metric is curved, if you have a small enough ruler, you can ignore the curvature. Then you simply have the problem of a ruler with a known proper length, falling past you at some velocity. You already know how to solve this problem from SR, the problem is not essentially different

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...

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

You've got a number of choices: you can select one of the two bodies, and a particular time point on it, and do a simple linear projection from it to the second body, using the notion that the bodies are close enough together that you are in a sufficiently flat space-time. This is the easiest approach.

Ah yeah, I forgot. The only person you continually attack on this forum is me. Silly me.

 

[starthaus]

The problem is that the equations you gave for "time till doom" do not give the coordinate falling time.

As I said before, it is probably easier to relax the fall from infinity requirement (although we might try that later) and assume a drop from initially at rest from a finite height.

After a coordinate time of 5 the leading clock will have fallen to 9.99437 and in the same coordinate time the trailing clock will have fallen to 10.953 and the coordinate gap will have increased from 1 to 1.009 if both particles were initially at rest and released at the same coordinate time from heights 10 Rs and 11 Rs.

It is a bit more interesting if you let the particles fall further.

If we start with the same release heights and let the leading particle fall to height 2 Rs then the coordinate fall time is 2.64 and trailing clock is at 4.64 Rs.

I am using the equations given in post 48 to obtain these results, but I have to numerically solve them as the mathematical software I use is unable to find a symbolic solution for r when t is know.

Here is a list of results for various initial and final heights for the trailing clock, all with the leading clock starting one coordinate unit lower and end point determined by the leading clock arriving at 2 Rs.

11 : 4.64
101 : 9.81
1001 : 21.86
10001 : 48.42
100001 : 105.987
1000001 : 230.22
10000001 : 498.02
100000001 : 1075.06
1000000001 : 2318.31

Very approximately, each tenfold increase in the initial drop height results in a two fold increase in the coordinate separation at the end point.

So, according to the above, the distance between particles increases. In the previous post (#62), you were arguing that the distance decreases. Which one is correct?

 

[yuiop]

Just incase you are wondering, for the special case of a initially stationary object dropped from infinity:

$$\frac{dr}{dtau}=\sqrt{\frac{r_s}{r}} = \frac{dr'}{dt'}$$

where dtau is the proper time rate of a co-falling clock and dr' and dt' are measurements made by the LSO.

Err, this is incorrect. You might want to check your facts. Instead of making up formulas that are wrong, it would be good if you tried consulting a good book or doing your own derivations,

You might want to check this paper http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111073v1.pdf referenced by Pervect.

It defines the velocity in Eq2 as:

Following along the same lines, Frolov and Novikov recently [3, pp.19,20] add that
“The physical velocity v measured by an observer who is at rest in the Schwarzschild
reference frame situated in the neighborhood of the freely moving body is:

Eq 2 resolves to $$v = \left(\frac{2m}{r}\right)^{1/2}$$ when the particle is initally at rest and released from infinity.

It also defines the velocity in Eq 27 as:

$$v = \left(\frac{2m}{r}\right)^{1/2}$$

where v was earlier defined (near Eq 17) as:

where (v) is, accordingly to Eq.(8), the velocity of the particle with respect to a static observer (r = constant); i.e. while the particle travels a proper distance α^−1/2dr the observer measure a proper time given by α^1/2dt.

So there are some fact to stuff in your pipe and smoke.

 

[Passionflower]

Come on folks let's focus on the problem and the solution!

On point though we are looking for a solution in the Schwazschild metric. So approaching the problem from Newton is no problem but we do have to get it eventually expressed correctly using the Schwarzschild solution, that is the whole point of this topic.

What I think will help for the next step is the answer to this, seemingly simple, question

What is the formula for getting the proper distance to the Schwarzschild radius for a given R coordinate and coordinate velocity?

Anyone know this formula?

 

[starthaus]

You might want to check this paper http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111073v1.pdf referenced by Pervect.

It defines the velocity in Eq2 as:




Eq 2 resolves to $$v = \left(\frac{2m}{r}\right)$$ when the particle is initally at rest and released from infinity.

It also defines the velocity in Eq 27 as:

$$v = \left(\frac{2m}{r}\right)$$

You are compounding your mistakes.