The Schwarzschild Metric - A Simple Case

[starthaus]

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?

"proper distance to the Schwarzschild radius"? You sure about the question?

 

[Passionflower]

"proper distance to the Schwarzschild radius"? You sure about the question?

Yes very sure. What seems to be the problem?

The nice thing about this formula is that we do not care if the particle fell from apogee or from infinity since it is expressed in coordinate velocity and r coordinate which should be sufficient to determine the proper distance to the Schwarzschild radius.

 

[yuiop]

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?

The coordinate distance between unconnected particles does indeed increase*. In the previous post I was referring to the coordinate length being shorter than the proper length of a rigid falling rod. Please note the subtle difference between a rigid rod and a series of unconnected particles. You really are not paying attention but just jumping at every seeming opportunity to attack me and getting it wrong most of the time.

Even in the case of the coordinate distance between unconnected particles increasing as they fall, the proper distance between the particles is greater than the coordinate distance between the particles.

* The coordinate distance between the particles increases in the particular case of the end point being at 2 Rs. If the end point was lower nearer 1 Rs, then the effective slow down in coordinate velocity as the event horizon is approached, means the trailing particles start to catch up with the leading particles in coordinate terms and the coordinate distance starts to decrease. This can be seen in the diagram I uploaded in a recent post. In all cases the coordinate separation of the particles is always shorter than the proper separation and the coordinate length of a rigid falling rod is always shorter than the proper length of the rigid rod.

Got it now?

 

[starthaus]

Yes very sure. What seems to be the problem?

The problem is that it makes no sense.

 

[Passionflower]

The problem is that it makes no sense.

I think it makes perfect sense. Could you explain why you think that it makes no sense?

For instance if the coordinate velocity is zero it is simple, when we assume R_S=1 we get:

$$\sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

 

[starthaus]

The coordinate distance between unconnected particles does indeed increase*. In the previous post I was referring to the coordinate length being shorter than the proper length of a rigid falling rod. Please note the subtle difference between a rigid rod and a series of unconnected particles. You really are not paying attention but just jumping at every seeming opportunity to attack me and getting it wrong most of the time.

Don't take it personally, I am not attacking you, I am just pointing out the unsupported hacks that contradict each other from post to post.

Even in the case of the coordinate distance between unconnected particles increasing as they fall, the proper distance between the particles is greater than the coordinate distance between the particles.

But you failed to establish this relationship. Hacking the GR equation for terminal velocity back into the SR length contraction formula does not count as a valid derivation.

 

[starthaus]

I think it makes perfect sense. Could you explain why you think that it makes no sense?

You can calculate proper distances only locally, in a small vicinity of the observer co-moving with the rod. What you are asking for is , in effect, $$r-r_s$$, which is not a proper distance.

 

[yuiop]

Don't take it personally, I am not attacking you, I am just pointing out the unsupported hacks that contradict each other from post to post.

But you failed to establish this relationship. Hacking GR equations back into SR length contraction formula does not count as a valid derivation.

It is valid, because locally over very short distances (infinitesimal) space is flat (Minkowskian) even in the curved space near a massive body and the equations of SR apply. If you do not know that, then you better learn it fast. It is fundamental. Pervect has explained that in greater detail and more formally in several posts in this thread and the related thread and you would do well to read his posts carefully and learn.

 

[starthaus]

It is valid, because locally over very short distances (infinitesimal) space is flat (Minkowskian) even in the curved space near a massive body and the equations of SR apply.

Problem is, the SR length contraction has been derived for inertial frames.You can't write it blindly for accelerated frames.

If you do not know that, then you better learn it fast. It is fundamental. Pervect has explained that in greater detail and more formally in several posts in this thread and the related thread and you would do well to read his posts carefully and learn.

Doesn't make your derivation valid. Quite the opposite.
To his credit, pervect did not advocate the kind of ugly hack that you are attempting. Quite the opposite.

 

[Passionflower]

You can calculate proper distances only locally, in a small vicinity of the observer comoving with the rod. What you are asking for is , in effect $$r-r_s$$, which is not a proper distance.

So let's take the simplest case (we assume R_s=1 to make it even simpler) of an observer with zero coordinate velocity the proper distance to the Schwarzschild radius is not generally:

$$\sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

It works, as you claim, only in a small region?

All I am asking for is a simple formula that works also in case the coordinate velocity is not zero.

 

[starthaus]

So let's take the simplest case (we assume R_s=1 to make it even simpler) of an observer with zero coordinate velocity the proper distance to the Schwarzschild radius is not:

$$\sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

It works, as you claim, only in a small region?

How did you arrive to the above?

 

[yuiop]

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?

The radar distance as measured from r2, to the event horizon is:

$$(rS*LN\left( \frac{r2-rS}{r1-rS}\right) +r2 - r1)*\sqrt{1-\frac{rS}{r2}}$$

When r1=rS the radar distance is infinite, which we sort of expect.

The integrated or ruler distance is given by:

$$\sqrt{r2*(r2-rS)} - \sqrt{r1*(r1-rS)} + rS*\left(LN\left(\sqrt{r2} + \sqrt{(r2-rS)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-rS)}\right)\right)$$

and the really weird aspect of the above equation is that the answer is real and finite even when r1=rS. That always seemed strange to me. Since it presumably physically impossible to have one end of a stationary ruler located exactly at the event horizon, you might think the equation would spit out a complex number or something in that situation.

Unfortunately, neither of the above equations tell us about the proper distance to the event horizon according to a moving observer and I am not sure if that can be defined. The last equation is the closest thing to a notion of proper distance to the event horizon but it applies to a stationary ruler.

[EDIT] I see you have already worked out the stationary case for the above formula, in the post you made while I was posting this:

So let's take the simplest case (we assume R_s=1 to make it even simpler) of an observer with zero coordinate velocity the proper distance to the Schwarzschild radius is not generally:

$$\sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

It works, as you claim, only in a small region?

All I am asking for is a simple formula that works also in case the coordinate velocity is not zero.

P.S. That formula works over extended distances. It is the distance measured a physical ruler (or lots of very short ideal measuring rods laid end to end) extending from r2 to r1. When the rod is stationary with respect to r1 and r2 then the rest frame of the rod is well defined. If the r1 end of the ruler is located at the event horizon (stationary) and the other end of the rod is attached to a moving observer at r2, then which frame is the rod at rest in? At rest in the frame of the moving observer or at rest with respect to Shwarzschild coordinates?

In order to define proper distances over extended regions for a falling observer, you need (as Pervect mentioned) a notion of simultaneity over extended distances and a suitable coordinate system. One such coordinate system which may provide a way forward in resolving the problems posed in this tread is Kruskal-Szekeres coordinates. Observers that are stationary in KS coordinates are not exactly free-falling, but falling in such a way that they maintain constant separation. They would therefore experience some proper acceleration as they fall and naturally free falling particles tend to fall faster.

In the above KS diagram, the curved line from F to F' represents the path of a freely falling particle with apogee at F. Observers that are stationary in KS coordinates have paths that are vertical and remain a constant KS coordinate distance apart. Light paths are at 45 degrees in KS coordinates, so it easy to see that the radar distance between the "stationary observers" is constant in terms of KS coordinate time. Since the free falling geodesic is curved, the "falling" KS observers moving on vertical lines are not true free falling inertial observers. Wikipedia gives the equations for KS coordinates and how to transform them to Schwarzschild coordinates so it might be interesting to see how a pair of KS falling observers that maintain a constant KS coordinate separation look like in Schwarzschild coordinates.

 

[Passionflower]

I am not sure if that can be defined.

That would be really odd.

If it is that hard we should perhaps take it in steps, the next step should be we assume the given coordinate velocity is constant which will work all the way up to but not including the EH.

Another approach is perhaps that we work the other way around, e.g. by formulating the proper time it takes to reach the EH for a given coordinate velocity.

P.S. That formula works over extended distances.

That is what I think as well, I was simply asking Starthaus the question, let's wait what Starthaus has to say about it.

How did you arrive to the above?

You integrate (1-1/r)^-1/2 dr between 1 and r.

 

[starthaus]

You integrate (1-1/r)^-1/2 dr between 1 and r.

How did you arrive to this? Give me some physical context, please.

 

[yuiop]

Another approach is perhaps that we work the other way around, e.g. by formulating the proper time it takes to reach the EH for a given coordinate velocity.

That is at least doable, and it should also be fairly easy to determine how far apart two freefalling clocks/observers are at any given equal proper time. I am not sure what the physical significance of that would be, but it may throw out something interesting.


Another approach that might simulate a falling rigid rod it to assume all parts of the rod have the same velocity as the front or middle of the rod. At least with a clearly defined velocity for all parts of the rod it should be easy to obtain its integrated length.

 

[starthaus]

Another approach that might simulate a falling rigid rod it to assume all parts of the rod have the same velocity as the front or middle of the rod. At least with a clearly defined velocity for all parts of the rod it should be easy to obtain its integrated length.

You can't do that since it is contradicted by the physics of the problem.
Indeed, the coordinate speed of a particle is :

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

It is easy to show that the above speed increases as r decreases up to the point $$r=3r_s$$

Indeed:

$$\frac{dv}{dr}=1/2(3r_s/r-1)\sqrt{r_s/r^3}$$

This means that the front of the rod passes has a higher speed than the tail at any given moment.
It also means that the distance between the front and the tail increases as r decreases, i.e. the rod is getting "spagettified" (extruded). Indeed:

$$v(r)-v(r+\Delta r)=\Delta r*\frac{dv}{dr}$$

For $$\Delta r$$ fixed, the above difference between two points on the rod increases as r decreases by virtue of the fact that $$\frac{dv}{dr}$$ is also increasing as it can be easily established.

To make matters even more complicated, the internal (electromagnetic) forces are fighting the "spagettifaction". What is happening to the rod is not treatable as a simple kinematics problem and, even less, as a hack on the SR length contraction.

 

[Passionflower]

Well let's start with a point instead of a rod, as so far nobody has come and showed the proper distance to r_s with respect to r and v_c(r).

Starthaus, could you tell me what the point is of your question, do you disagree with the formula? In that case I suggest to open a separate topic on this as I think this is standard textbook stuff. Or do you want to derive it from the Gaussian curvature? If that is the case I would also suggest you open a separate topic on this as it is not relevant to this topic. Or perhaps you want to illustrate that I am wrong by assuming we can have a proper distance when that distance is changing over coordinate time? In that case please come to the point, why would you think that a proper distance when that distance changes over coordinate time cannot be mathematically defined? I think an approach could be to start by keeping the second derivative = 0 so that the velocity remains constant, you can do this all the up to, but not including, the EH. Then we can address the general case where the second derivative is not 0.

 

[starthaus]

Well let's start with a point instead of a rod, as so far nobody has come and showed the proper distance to r_s with respect to r and v_c(r).

Starthaus, could you tell me what the point is of your question, do you disagree with the formula?

I suspect that the formula is most likely incorrect. Without a complete derivation , I cannot be sure. Can you show me how you arrived to the integrand?

In that case I suggest to open a separate topic on this as I think this is standard textbook stuff.

Which textbook did you get it from?

 

[yuiop]

You can't do that since it is contradicted by the physics of the problem.
Indeed, the coordinate speed of a particle is :

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

It is easy to show that the above speed increases as r decreases up to the point $$r=3r_s$$

Indeed:

$$\frac{dv}{dr}=1/2(3r_s/r-1)\sqrt{r_s/r^3}$$

This means that the front of the rod passes has a higher speed than the tail at any given moment.
It also means that the distance between the front and the tail increases as r decreases, i.e. the rod is getting "spagettified" (extruded). Indeed:

$$v(r)-v(r+\Delta r)=\Delta r*\frac{dv}{dr}$$

For $$\Delta r$$ fixed, the above difference between two points on the rod increases as r decreases by virtue of the fact that $$\frac{dv}{dr}$$ is also increasing as it can be easily established.

To make matters even more complicated, the internal (electromagnetic) forces are fighting the "spagettifaction". What is happening to the rod is not treatable as a simple kinematics problem and, even less, as a hack on the SR length contraction.

What you call my "hack on the SR length contraction" was an equation that applied to an infinitesimal particle where "spagettification" does not occur. Infinitesimal is very small, even smaller than an atom and is in fact as small is required to make the local spacetime flat (so negligable tidal forces). For a non infinitesimal rod some form of integrated length is required and I am aware as you are, as to how non trivial that is.

Now all the equations you give are for free falling particles. For a rigid rod parts of the rod are not free falling. By definition the rigidity of the rod prevents that. You said "front of the rod passes has a higher speed than the tail at any given moment", but Passionf has specified in the OP that we can use rockets to accelerate parts of the rod to ensure that its proper length remains constant by whatever means necessary. So while natural free fall might mean that the rear clock is moving slower than the front clock at any given moment (and here you should define "moment" according to what reference frame) we can attach a rocket to the rear clock to make sure it goes at the same speed (insert according to what observer here) as the front clock if that is convenient to us.

It is not unlike Born rigid acceleration, where all parts of a system are individually accelerated by the exact required amount to ensure constant proper length is obtained in the accelerating system. We are just trying to make the GR analogue of that.

 

[yuiop]

I suspect that the formula is most likely incorrect. Without a complete derivation , I cannot be sure. Can you show me how you arrived to the integrand?

Which textbook did you get it from?

DrGreg derives that equation in this thread. https://www.physicsforums.com/showthread.php?t=248015 There is no need to spend any more time on this diversion.

 

[starthaus]

What you call my "hack on the SR length contraction" was an equation that applied to an infinitesimal particle where "spagettification" does not occur.

A hack is always a hack no matter what extenuating circumstances you are trying to invoke.

Infinitesimal is very small, even smaller than an atom and is in fact as small is required to make the local spacetime flat (so negligable tidal forces). For a non infinitesimal rod some form of integrated length is required and I am aware as you are, as to how non trivial that is.

This simply means that your attempt at hacking length contraction in GR is unusable.

Now all the equations you give are for free falling particles. For a rigid rod parts of the rod are not free falling. By definition the rigidity of the rod prevents that. You said "front of the rod passes has a higher speed than the tail at any given moment", but Passionf has specified in the OP that we can use rockets to accelerate parts of the rod

The "rocket" bit is over the top, especially since the OP was about a free-falling rod.

to ensure that its proper length remains constant by whatever means necessary.

This only means that the problem cannot be treated kinematically since you are now dealing not only with the tidal and electromagnetic (internal) forces but also with the forces exerted by the two rockets.

So while natural free fall might mean that the rear clock is moving slower than the front clock at any given moment (and here you should define "moment" according to what reference frame) we can attach a rocket to the rear clock to make sure it goes at the same speed (insert according to what observer here) as the front clock if that is convenient to us.

Why entitles you to apply different rules to the spatial coordinate than to the time coordinate?

It is not unlike Born rigid acceleration, where all parts of a system are individually accelerated by the exact required amount to ensure constant proper length is obtained in the accelerating system. We are just trying to make the GR analogue of that.

I see, you plan to put a little rocket at infinitesimal distances from each other.
The point is that the OP problem is much more complicated so it cannot be reduced to simple kinematics.

 

[starthaus]

DrGreg derives that equation in this thread. https://www.physicsforums.com/showthread.php?t=248015 There is no need to spend any more time on this diversion.

All I can see is the post where he tells you that you got things wrong.

 

[yuiop]

You can't do that since it is contradicted by the physics of the problem.
Indeed, the coordinate speed of a particle is :

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

It is easy to show that the above speed increases as r decreases up to the point $$r=3r_s$$

Indeed:

$$\frac{dv}{dr}=1/2(3r_s/r-1)\sqrt{r_s/r^3}$$

This means that the front of the rod passes has a higher speed than the tail at any given moment.
It also means that the distance between the front and the tail increases as r decreases, i.e. the rod is getting "spagettified" (extruded).

As you say, the after r = 3Rs the coordinate speed of the falling particle decreases, and the following particles start to catch up and in terms of coordinate distance de-spagettification happens as the event horizon approaches. This however is coordinate distance and no doubt in the proper distance between particles is still increasing even below r = 3Rs and all the way to the event horizon, but we do not have an equation to prove that. that is one of thing Passionflower and myself are trying to establish. What is the proper distance in the rest frame of the falling particles? Does a ball of coffee granules maintain constant volume (as Weyl curvature suggests it should) as measured in the rest frame of the falling granules? it is easy to work out from what we have already established that the volume does not remain constant from a Schwarzschild observers point of view.

 

[yuiop]

All I can see is the post where he tells you that you got things wrong.

That is because you see only what you want to see and stopped reading. In the very next line DrGreg says

"However the ruler distance to the event horizon is the integral ..."

and gives a formula. This formula for the integral is established and confirmed by other posters later in the thread. That thread is a model of cooperation and constructive contributions by members of this forum in the good old days. You could learn a lot from it. Back then I foolishly believed that it was not possible to have a ruler distance to the event horizon. As soon as DrGreg demonstrated how it was calculated I realized he was right. Do you also now realize that his formula is right?

 

[yuiop]

All I can see is the post where he tells you that you got things wrong.

There are 3 pages and 39 posts in that thread and all you can see is one line of one single post where someone said I got something wrong. Obviously you did not see post 33 where I went to the trouble to post the full correct equation for everyone's future reference. Now why is it that all you see is that one line where DrGreg says I got something wrong and why do you feel the need to post that in this thread? Ah yes, it is called an "ad hominem" attack. It is the fallacious logic that if you establish that someone got something wrong in the past then EVERYTHING that person says must be wrong. I consider it a personal attack. Consider yourself reported.

 

[pervect]

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 ?

What the tidal tensor (or the appropriate component of the Riemann ) measures is the relative acceleration of geodesics. So, if you have two particles, initially at relative rest in their "local lorentz frame" (defined because they are both close together and moving at the same velocity), both following geodesics, the tidal tensor is basically the second derivative of the distance between geodesics with respect to the proper time of a particle following the geodesic.

Take a look at most any textbook derivation of the geodesic deviation equation to see the actual mathematical definition.

One generally considers that one has a one-parameter group of geodesics, i.e. one has some selector parameter n that defines a unique geodesic curve for every value of n. You select which geodesic by some selector parameter n, and you select "how far" along the geodesic by some affine parameter s.

The tangent vector of the geodesic, the partial derivitive with respect to the affine parameter s, $\partial / \partial s$ at any point can be intuitively thought of as the "local time vector" of the particle following the geodesic. One needs to exploit some "gauge degree of freedom" to make the separation vector between geodesics independent of the specific affine parameterization of s for each geodesic. When this is done, the separation vector between geodesics, the $\partial / \partial n$, becomes perpendicular to the tangent vector $\partial / \partial s$. (The issue with the affine paramterization is sometimes called stretch-out).

In less formal, more "feel good" but less precise language (some might find the less precise language puzzling, for which I apologize, I hope most consider it useful) one makes the local distance between geodesics perpendicular to the "local time" along the geodesic.

The stress measured on a born-rigid rod that's falling is the flip size of this equation. The amount of proper acceleration needed to keep the two ends of the rod the same distance apart is the same as the relative acceleration of the two geodesics passing through their endpoints at zero velocity (the zero velocity relative to the instantaneous local Lorentz frame of the falling rod).

 

[starthaus]

As you say, the after r = 3Rs the coordinate speed of the falling particle decreases, and the following particles start to catch up and in terms of coordinate distance de-spagettification happens as the event horizon approaches. This however is coordinate distance and no doubt in the proper distance between particles is still increasing even below r = 3Rs and all the way to the event horizon, but we do not have an equation to prove that. that is one of thing Passionflower and myself are trying to establish.

Given the fact that $$r_s$$ is much smaller than the radius of the celestial bodies (for example, for Earth, $$r_s=9mm$$. this is not relevant to our discussion.

What is the proper distance in the rest frame of the falling particles?

I don't think that this problem is tractable for reasons explained above.

 

[starthaus]

That is because you see only what you want to see and stopped reading. In the very next line DrGreg says

"However the ruler distance to the event horizon is the integral ..."

He never derived anything close to what you are claiming. In fact, in post 35, he's still telling you that you still have things wrong in here, after your post 33 where you claim that you obtained the correct formula. The reason is simple, the correct formula is:

$$ds=\frac{dr}{\sqrt{r_s/r-r_s/r_0}}$$, $$r<r_0$$

I showed it earlier in this thread (post 54). For $$r_0=\infty$$ the formula reduces to:


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

This formula for the integral is established and confirmed by other posters later in the thread.

I don't see anyone else "establishing" and/or "confirming" your formula. This is irrelevant anyway since the formula you are attempting to use is incorrect to begin with.

That thread is a model of cooperation and constructive contributions by members of this forum in the good old days. You could learn a lot from it. Back then I foolishly believed that it was not possible to have a ruler distance to the event horizon.

You believe a lot of things. Unfortunately, "belief" is not a good approach to physics, proper derivation from base principles is.

 

[Passionflower]

It looks like yuiop was absolutely correct when he last week mentioned that things can magically can cancel out (at least for a free falling particle from infinity):

For our example (r_s = 1):

Since the proper distance from r₁ to r_s for a stationary observer (v_local=0) is:

$$\int _{1}^{r_1}\!\sqrt {1-{r}^{-1}}{dr} = \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

we can calculate the proper distance for an observer who has a non zero local velocity.

We need to length contract this distance by the following factor:

$${1 \over \gamma} = \sqrt{1-v_{local}^2}$$

Thus for instance for a free falling particle from infinity (E=1) we have a local velocity of:

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

Now the 'magic' when we multiply the two before we integrate we get:

$$\int _{1}^{r_1}\!{\sqrt {1-{r}^{-1}}\over \sqrt {1-{r}^{-1}}}{dr}$$

Thus the proper distance for a free falling particle from infinity (E=1) is simply dr!

One thing I did not realize, like many things, is that one could calculate the proper time till ultimate doom based on the tidal acceleration assuming the distance is small irrespective of the mass of the Schwarzschild metric. So when we are in a rocket radially falling into a black hole (from infinity) and we measure the tidal acceleration between the top and the bottom and we know the length of the rocket we can calculate how much longer we are going to be alive.

So this is the first step the proper distance for a particle falling from infinity is simply r. So how do we go from here?

 

[yuiop]

He never derived anything close to what you are claiming. In fact, in post 35, he's still telling you that you still have things wrong in here, after your post 33 where you claim that you obtained the correct formula. The reason is simple, the correct formula is:

$$ds=\frac{dr}{\sqrt{r_s/r-r_s/r_0}}$$, $$r<r_0$$

The formula I gave in post 33 is:

$$\sqrt{r2*(r2-r_s)} - \sqrt{r1*(r1-r_s)} + r_s*\left(LN\left(\sqrt{r2} + \sqrt{(r2-r_s)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-r_s)}\right)\right)$$

This is the definite integral. The indefinite integral is:

$$\sqrt{r*(r-r_s)} + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)$$

Now some simple algebraic manipulation of the above expression:

$$\Rightarrow \sqrt{(1-r_s/r)}*r + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)$$

$$\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(\left(\sqrt{r} + \sqrt{(r-r_s)}\right)^2}\right)$$

$$\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(r + 2\sqrt{r(r-r_s)}+(r-r_s)\right)$$

$$\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(2r + 2\sqrt{r(r-r_s)}-r_s\right)$$

$$\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(2r + 2r\sqrt{(1-r_s/r)}-r_s\right)$$

$$\Rightarrow \sqrt{(1-r_s/r)}*r + \left(r_s*LN\left(\, -r_s + 2*\left(1 + \sqrt{(1-r_s/r)}\right)*r \right)\right)/2$$

This is the same as the equation posted by DrGreg in post 18 https://www.physicsforums.com/showpost.php?p=1827990&postcount=18 of that thread where he works out the integral as:

If you replace r by x and r_s by a, the site gives formula that works perfectly well at x=a.

Integrate[1/Sqrt[1 - a/x], x] == Sqrt[1 - a/x]*x + (a*Log[-a + 2*(1 + Sqrt[1 - a/x])*x])/2

This is only one of many times that you have claimed that two equations that look different are not equivalent, when they are. If you are having trouble with the algebraic manipulations then here is a tip for you. Try sample variables in both equations and if you get the same numerical result every time, then the two equations are probably equivalent. This is not proof that they are equivalent, but it is a strong clue that you should look closer before declaring the equations are not equivalent. You claim to be a mathematician. I should not have to keep showing how to do these algebraic manipulations. In your haste to try and prove me wrong at every opportunity, you have shot yourself in the foot again.

 

[Passionflower]

Right and in addition to what yuiop writes if we take r_s=1 and want the proper distance to from r to r_s we get an even simpler formula:

$$\sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

In continuation we should probably address the issue of translation the acceleration differential (e.g. the tidal acceleration) in terms of some stretching factor in terms of the r coordinate, the local velocity at r and the height of the object in question, e.g. the spaghetti factor.

The tidal acceleration at r is (as always r_s=1):

$${1 \over r^3} dr$$

 

[starthaus]

The formula I gave in post 33 is:

$$\sqrt{r2*(r2-r_s)} - \sqrt{r1*(r1-r_s)} + r_s*\left(LN\left(\sqrt{r2} + \sqrt{(r2-r_s)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-r_s)}\right)\right)$$

This is the definite integral. The indefinite integral is:

$$\sqrt{r*(r-r_s)} + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)$$

...which are both equally incorrect since they have a common incorrect starting point, the integrand:

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

Try solving the problem starting from the correct integrand:

$$ds=\frac{dr}{\sqrt{r_s/r-r_s/r_0}}$$

This is only one of many times that you have claimed that two equations that look different are not equivalent, when they are. If you are having trouble with the algebraic manipulations then here is a tip for you. Try sample variables in both equations and if you get the same numerical result every time, then the two equations are probably equivalent. This is not proof that they are equivalent, but it is a strong clue that you should look closer before declaring the equations are not equivalent. You claim to be a mathematician. I should not have to keep showing how to do these algebraic manipulations. In your haste to try and prove me wrong at every opportunity, you have shot yourself in the foot again.

Physics is a lot more than applying software packages in blindly integrating meaningless expressions. If you start with the wrong integrand, don't be surprised when you get a useless expression, no matter how much you turn the crank on your integrating software package.

 

[starthaus]

It looks like yuiop was absolutely correct when he last week mentioned that things can magically can cancel out (at least for a free falling particle from infinity):

For our example (r_s = 1):

Since the proper distance from r₁ to r_s for a stationary observer (v_local=0) is:

$$\int _{1}^{r_1}\!\sqrt {1-{r}^{-1}}{dr} = \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1} \right)$$

we can calculate the proper distance for an observer who has a non zero local velocity.

We need to length contract this distance by the following factor:

$${1 \over \gamma} = \sqrt{1-v_{local}^2}$$

Thus for instance for a free falling particle from infinity (E=1) we have a local velocity of:

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

Now the 'magic' when we multiply the two before we integrate we get:

$$\int _{1}^{r_1}\!{\sqrt {1-{r}^{-1}}\over \sqrt {1-{r}^{-1}}}{dr}$$

Thus the proper distance for a free falling particle from infinity (E=1) is simply dr!

...which is totally incorrect.

One thing I did not realize, like many things, is that one could calculate the proper time till ultimate doom based on the tidal acceleration assuming the distance is small irrespective of the mass of the Schwarzschild metric. So when we are in a rocket radially falling into a black hole (from infinity) and we measure the tidal acceleration between the top and the bottom and we know the length of the rocket we can calculate how much longer we are going to be alive.

So this is the first step the proper distance for a particle falling from infinity is simply r. So how do we go from here?

You don't go anywhere since the formula you keep using is not correct.

 

[Passionflower]

You don't go anywhere since the formula you keep using is not correct.

Which formula do you think is not correct?

 

[starthaus]

Which formula do you think is not correct?

I asked you where the integrand

$$\frac{dr}{\sqrt{1-r_s/r}}$$

was coming from. I told you that it was most likely incorrect (turns out that it is, thus making the integral and all your and yuoip's calculations useless). You never answered. I gave you the correct integrand several times already. Try using it.