The Schwarzschild Metric - A Simple Case

[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).

 

[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]

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

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

..and I just showed that in post 6.

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?

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&#039; = 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&#039; = L_o \sqrt{(1-v&#039;^2/c^2)}. The velocity of an object that was initially at rest and dropped from infinity according to the LSO is v&#039;/c =\ sqrt{(r_s/r)} so we can rewrite the last equation in this special case as L&#039; = L_o \sqrt{(1-r_s/r)}. Now the transformation from local length L' to Schwarzschild coordinate length L is L = L&#039;\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&#039;}{dt&#039;}

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&#039; = 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&#039; = L_o \sqrt{(1-v&#039;^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&#039;/c =\ sqrt{(r_s/r)} so we can rewrite the last equation in this special case as L&#039; = 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&#039;}{dt&#039;}

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&#039;}{dt&#039;}

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.

 

[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:

<br /> \sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)<br />

 

[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:

<br /> <br /> \sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)<br /> <br />

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:

<br /> <br /> \sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)<br /> <br />

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:

<br /> (rS*LN\left( \frac{r2-rS}{r1-rS}\right) +r2 - r1)*\sqrt{1-\frac{rS}{r2}} <br />

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

The integrated or ruler distance is given by:

<br /> \sqrt{r2*(r2-rS)} - \sqrt{r1*(r1-rS)}<br /> + rS*\left(LN\left(\sqrt{r2} + \sqrt{(r2-rS)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-rS)}\right)\right)<br />

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:

<br /> <br /> \sqrt {r}\sqrt {r-1}+\ln \left( \sqrt {r}+\sqrt {r-1} \right)<br /> <br />

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&lt;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:

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

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:

<br /> <br /> {1 \over \gamma} = \sqrt{1-v_{local}^2}<br /> <br />

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

<br /> <br /> \sqrt{1 \over r}<br /> <br />

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

<br /> <br /> \int _{1}^{r_1}\!{\sqrt {1-{r}^{-1}}\over \sqrt {1-{r}^{-1}}}{dr}<br /> <br />

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&lt;r_0

The formula I gave in post 33 is:

<br /> \sqrt{r2*(r2-r_s)} - \sqrt{r1*(r1-r_s)}<br /> + r_s*\left(LN\left(\sqrt{r2} + \sqrt{(r2-r_s)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-r_s)}\right)\right)<br />

This is the definite integral. The indefinite integral is:

<br /> \sqrt{r*(r-r_s)} + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)<br />

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)<br />

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

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

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

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

\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<br />

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.