The Schwarzschild Metric - A Simple Case

[pervect]

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

<br /> I = \sqrt {\int g_{ij} \dot{x}^i \dot{x}^j d\lambda}<br />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.

 

[Passionflower]

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.

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,
<br /> \begin{align*}<br /> t_2(R_1) &amp;= \int^{R_1}_{R_{max}-1} \left( \frac{dt}{dr} \right)_2 dr<br /> \end{align*}<br />

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

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

See also a few postings before where this formula is discussed.

So you assume a fall from a stationary position right?

If we take the integral of your formula we get:

<br /> \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) <br />

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?

 

[yuiop]

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

<br /> \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:

<br /> 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) <br />

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.

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

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

to begin with.

If we have an equation:

<br /> \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.

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:

<br /> <br /> \int _{{r0}}^{{r1}}\!{\frac {1}{\sqrt {1-{\frac {{\it rs}}{r}}<br /> }}}\,{dr}<br /> <br />

I interpret that as the length measured by a series of stationary infinitesimal rulers laid end to end from r0 to r1.

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.

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.

 

[starthaus]

If we have an equation:

<br /> \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.

 

[yuiop]

..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.

 

[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:

<br /> \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:

<br /> 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) <br />

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:

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

where:

<br /> Q = \sqrt{R/2m -1 } <br />

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?

<br /> \frac{dr}{ds}=\sqrt{r_s/r}<br />

and

dL&#039; = 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?

<br /> \frac{dr}{ds}=\sqrt{r_s/r}<br />

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

and

dL&#039; = dL\sqrt{(1-r_s/r)(1-(dr/ds)^2)}

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