The Schwarzschild Metric - A Simple Case

[yuiop]

No, your formula is wrong because:

1. you are using an inappropriate formula to begin with (the worst mistake)

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

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

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 have stated at least 3 times that the starting integrand used by myself, Passionflower and DrGreg, specifically:

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

is incorrect, but now finally you have contradicted yourself and admitted it was correct all along, in your last post on the subject:

...which is just as wrong since, in this case, the correct integrand is:

$$ds=\frac{dr}{\sqrt{1-r_s/r}}=(1+r_s/(2r))dr$$
(see Rindler 11.5)

Since you used the wrong integrand, you also obtained the incorrect integral. If you did it right, you would have obtained a much nicer formula:

$$\Delta S=r_2-r_1+r_s/2(ln(r_2)-ln(r_1))$$

The expression on the end of the Rindler equation in chapter 11.5 is an approximation and what Rindler actually said was:

$$dl = \left(1-\frac{2m}{r}\right)^{-1/2}\, dr \, \approx \, \left(1+\frac{m}{r}\right)\, dr$$

which in terms of the forms and variables you are using can be written as:

$$ds=\frac{dr}{\sqrt{1-r_s/r}}\, \approx \, (1+r_s/(2r)) \, dr$$
DO you see the wriggly equals sign? That means "approximately equal" and is not an exact expression. These approximations get increasingly inaccurate near the event horizon (the region the OP is interested in). Pervect has demonstrated this with a numerical example:

I don't have Rindler, but the left side of this equation is obviously not equal to the right. What gives?

i.e if r_s=1 and r=2, the left side is dr/sqrt(.5) , the right side is 1.25dr;

As I said before, you almost never check equations numerically so you often fail to spot when when different looking expressions are in fact the same and you fail to spot when an equality is in fact an approximation. So take a tip from Pervect and carry out a numerical check in future, before jumping in (or look for the wriggly "approximately equal" symbol).

So if we discard the approximate expression and your derived integrations from it, because they useless near the event horizon and because the OP has several times expressed an interest in exact solutions, we end up the correct initial integrand as given by Rindler and DrGreg is:

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

Now I know from past experience you never admit your are wrong, even when it is proven you are wrong and you never apologise to others when you falsely accuse them of being wrong, so I won't hold my breath waiting.

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

Let me remind of you part of the OP with a direct quote:

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

 

[yuiop]

This is my formula:

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

I'm not quite sure where the mistake is, but Maple says that your formula is definitely not the integral 1/sqrt(1-1/r).Maple gives
$$\sqrt {r \left( r-1 \right) }+1/2\,\ln \left( -1+2\,r+2\,\sqrt {r \left( r-1 \right) } \right)$$

If we take Passionflower's result for the integral of 1/sqrt(1-1/r):

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

and use the fact that ln(x) = (1/n)*ln(x^n) then we get:

$$\sqrt {r \left( r-1 \right) }+1/2 \ln \left(\left[\sqrt {r}+\sqrt {r-1} \, \right]^2 \right)$$

$$\sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( r +(r-1) +2\sqrt{r}\sqrt{r-1} \right)$$

$$\sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( -1 + 2r +2\sqrt{r(r-1)} \right)$$

Your result and Passionflower's results are in fact the same.

I demonstrated earlier in post #100 https://www.physicsforums.com/showthread.php?p=2915154#post2915154 that DrGreg's result and my result are also the same.

In fact your result, my result and DrGreg's result are all equivalent, but just have different forms.

 

[yuiop]

I'm not quite sure where the mistake is, but Maple says that your formula is definitely not the integral 1/sqrt(1-1/r).


Maple gives
$$\sqrt {r \left( r-1 \right) }+1/2\,\ln \left( -1+2\,r+2\,\sqrt {r \left( r-1 \right) } \right)$$

or equivalently (different by a constant factor0

$$\sqrt {r \left( r-1 \right) }+1/2\,\ln \left( r+\sqrt {r \left( r-1 \right) }-1/2 \right)$$

which does give the right answer when differentiated. There may be alternate forms...

Your second form is effectively:

$$\sqrt {r \left( r-1 \right) }+1/2\,\ln \left( -1+2\,r+2\,\sqrt {r \left( r-1 \right) } \right) - \ln (2)$$

where -ln(2) is simply a constant of integration. This is not a problem when taking the definite integral as the constant of integration cancels out.

 

[yuiop]

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

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

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

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,

Here is a quote from page 35 of http://books.google.com/books?id=oP...th" hypersurface&pg=PA36#v=onepage&q&f=false":

$$v_{loc} = \left(\frac{2m}{r}\right)^{-1/2} \qquad \, \qquad \, \qquad \, (2.33)$$

.. Thus the locally measured velocity of infall increases steadily with decreasing r and in the limit as $r \rightarrow 2m$ we get $v_{loc} \rightarrow 1$ i.e. in physical units the locally measured speed tends to c).

In the preamble to that statement the author makes it clear that by $v_{loc}$ he is talking about the velocity measured by an observer located on a static shell at coordinate r. Thus my statement is supported by a book and you are wrong.

Almost every statement you have made in this thread has been shown to be wrong. I make mistakes (and you have had much fun pointing out where I have admitted I have made a mistake in old threads) but I have an excuse, I am simply an armchair amateur enthusiast. You on the other hand claim to be a mathematician and experimental physicist by profession. What is your excuse? I find it hard to believe you are as inept as your postings make out and can only assume you are deliberately posting stuff that you know is wrong just to be provocative and disruptive. Pretty much the definition of trolling. Or are you really just that inept?

 

[starthaus]

You have stated at least 3 times that the starting integrand used by myself, Passionflower and DrGreg, specifically:

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

is incorrect, but now finally you have contradicted yourself and admitted it was correct all along, in your last post on the subject:

I have repeatedly told both you and Passionflower that you are using the wrong approach since you are trying to force the stationary solution for the case of the moving rod.

The expression on the end of the Rindler equation in chapter 11.5 is an approximation and what Rindler actually said was:

$$dl = \left(1-\frac{2m}{r}\right)^{-1/2}\, dr \, \approx \, \left(1+\frac{m}{r}\right)\, dr$$

which in terms of the forms and variables you are using can be written as:

You must have missed the three posts where I explained that the above is a Taylor expansion valid only for $$r>>r_s$$. This is basic calculus.

As I said before, you almost never check equations numerically so you often fail to spot when when different looking expressions are in fact the same and you fail to spot when an equality is in fact an approximation.

There was nothing to check numerically since the Rindler formula was derived via Taylor expansion for $$r>>r_s$$. This is is quite clear from the Rindler book and from my explanation.

Anyway, this is just a red herring since both of you are attempting to use the wrong formalism for the problem at hand. I am going to ask you the same question I asked Passionflower: how is the Rindler solution derived? Where does $$ds=\frac{dr}{\sqrt{1-r_s/r}}$$ come from? If you can manage to show the math, you will learn by yourself why your approach is incorrect.

 

[starthaus]

If we take Passionflower's result for the integral of 1/sqrt(1-1/r):

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

and use the fact that ln(x) = (1/n)*ln(x^n) then we get:

$$\sqrt {r \left( r-1 \right) }+1/2 \ln \left(\left[\sqrt {r}+\sqrt {r-1} \, \right]^2 \right)$$

$$\sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( r +(r-1) +2\sqrt{r}\sqrt{r-1} \right)$$

$$\sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( -1 + 2r +2\sqrt{r(r-1)} \right)$$

Your result and Passionflower's results are in fact the same.

I demonstrated earlier in post #100 https://www.physicsforums.com/showthread.php?p=2915154#post2915154 that DrGreg's result and my result are also the same.

In fact your result, my result and DrGreg's result are all equivalent, but just have different forms.

True, yet your attempt to force the stationary solution for a non-stationary problem renders the above useless. The difference is DrGreg wasn't trying to force the above as a solution for the OP problem, you and Passion are doing that because you don't understand the conditions under which the integrand was derived.

 

[Mentz114]

Stretching in G-P coords

I've finally come up with a proposal for the stretching factor in Gullstrand-Painleve coords.

A test body falls from r₁ -> r₂ ( being at r₁ when t=0) in time T₁, so
$$T_1 &= \int_{r_1}^{r_2}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( r_2^{3/2}- r_1^{3/2} \right)$$
Another test body is on the same geodesic, but is at r₁+L at time t=0. It will fall to point x in time T₂
$$T_2 &= \int_{r_1+L}^{x}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( x^{3/2}- (L+r_1)^{3/2} \right)$$
We can set T₁-T₂=0 and solve for x, which gives
$$x={\left( {\left( L+r_1\right) }^{\frac{3}{2}}+{r_2}^{\frac{3}{2}}-{r_1}^{\frac{3}{2}}\right) }^{\frac{2}{3}}$$
and the new separation is given by r₂-x.

The algebra and calculus was done with Maxima. This is the only sensible looking result I've managed so far so it would be gratifying if it was right.

[edit] changed second integral

 

[starthaus]

I've finally come up with a proposal for the stretching factor in Gullstrand-Painleve coords.

A test body falls from r₁ -> r₂ ( being at r₁ when t=0) in time T₁, so
$$T_1 &= \int_{r_1}^{r_2}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( r_2^{3/2}- r_1^{3/2} \right)$$
Another test body is on the same geodesic, but is at r₁+L at time t=0. It will fall to point x in time T₂
$$T_2 &= \int_{r_1+L}^{x}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( x^{3/2}- (L+r_1)^{3/2} \right)$$
We can set T₁-T₂=0 and solve for x, which gives
$$x={\left( {\left( L+r_1\right) }^{\frac{3}{2}}+{r_2}^{\frac{3}{2}}-{r_1}^{\frac{3}{2}}\right) }^{\frac{2}{3}}$$
and the new separation is given by r₂-x.

The algebra and calculus was done with Maxima. This is the only sensible looking result I've managed so far so it would be gratifying if it was right.

[edit] changed second integral

This is a step in the right direction, the formalism is very nicely set and the math is very clear. I can see a major problem with the approach, $$T$$ is proper time, so you do not want $$T_1=T_2$$. What you want is $$t_1=t_2$$, that is you want to mark the ends of the rod at the same time in the distant observer's frame, not in the rod frame. You can do that using Schwarzschild coordinates as well. When you do that, you get a very nasty transcendental equation, not due to the use of coordinates but due to the condition $$t_1=t_2$$. I solved this for the more general case of the rod falling from a finite height and I got a much nastier equation.

It is good to see that you are using the correct integrand:

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

as opposed to the incorrect ones attempted by yuoip and Passionflower.

 

[Mentz114]

I can see a major problem with the approach, T is proper time, so you do not want T₁=T₂ .

Is this because the G-P coordinate t is actually proper time ?

 

[starthaus]

Is this because the G-P coordinate t is actually proper time ?

I don't think it is. Anyways, the error is easy to fix:

-start with coordinate speed for rod falling from $$r_0$$:

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

-make $$r_0->\infty$$, therefore:

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

So, you need to replace your integrand with the modified one:

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

For $$r>>r_s$$ you can try the Taylor expansion:

$$\sqrt{r_s/r}+\sqrt{r/r_s}$$

The rest of the algorithm works exactly the same. If you feel particularly adventurous, you may want to try the case when $$r_0$$ is finite. Beware, you will get a much uglier formula.

 

[Mentz114]

I was assuming the given velocity was wrt coordinate time, but you say
$$\sqrt{2m/r} = \frac{dr}{d\tau}$$
and I have no reason right now not to believe you.

So, you need to replace your integrand with the modified one...

I did that calculation some time ago and it does get very messy.

 

[starthaus]

I was assuming the given velocity was wrt coordinate time, but you say
$$\sqrt{2m/r} = \frac{dr}{d\tau}$$
and I have no reason right now not to believe you.

Yes, more generally:

$$\sqrt{2m/r-2m/r_0} = \frac{dr}{d\tau}$$I don't expect you to take anything on faith, I wrote the complete derivation https://www.physicsforums.com/blog.php?b=1957 . Please have a look at the file "RadialMotion1" and please check the results, all the necessary equations are derived from base principles, this is the proper and scientific thing to do and I am glad that we are interacting the way two scientists should interact .

I did that calculation some time ago and it does get very messy.

Such is life, we don't always get "nice" formulas. :-)

 

[yuiop]

I've finally come up with a proposal for the stretching factor in Gullstrand-Painleve coords.

A test body falls from r₁ -> r₂ ( being at r₁ when t=0) in time T₁, so
$$T_1 &= \int_{r_1}^{r_2}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( r_2^{3/2}- r_1^{3/2} \right)$$
Another test body is on the same geodesic, but is at r₁+L at time t=0. It will fall to point x in time T₂
$$T_2 &= \int_{r_1+L}^{x}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( x^{3/2}- (L+r_1)^{3/2} \right)$$
We can set T₁-T₂=0 and solve for x, which gives
$$x={\left( {\left( L+r_1\right) }^{\frac{3}{2}}+{r_2}^{\frac{3}{2}}-{r_1}^{\frac{3}{2}}\right) }^{\frac{2}{3}}$$
and the new separation is given by r₂-x.

This is essentially the equation given by passionflower in the closely related parallel thread "https://www.physicsforums.com/showpost.php?p=2904752&postcount=21"".

The free fall from infinity does not pose a problem in terms of discovering proper time if we switch things around and consider proper time in terms of 'proper time till ultimate doom', e.g. the arrival at the singularity. See the attached graph for a comparison between proper and coordinate time. As you can see coordinate time never gets beyond R_s while proper time gets all the way to the singularity.

Using (MTW 25.38):

$$-2/3\,{\it rs}\, \left( {\frac {r}{{\it rs}}} \right) ^{3/2}$$

This is the proper time remaining until the clock falling from infinity before it arrives at the singularity. For the elapsed proper time as the clock falls from r1 to r2 if free-falling from initially at rest at infinity the equation is:

$$\Delta tau = \frac{2}{3}r_s \left( \left(\frac{r_1}{r_s}\right)^{3/2} - \left(\frac{r_2}{r_s}\right)^{3/2} \right)$$

It is clear from the above that your T1 and T2 are elapsed proper times and unfortunately I have to agree with Starthaus that setting T1-T2=0 does not give the proper distance and it is the coordinate time difference that has to be set to zero.

 

[starthaus]

This is essentially the equation given by passionflower in the closely related parallel thread "https://www.physicsforums.com/showpost.php?p=2904752&postcount=21"".



This is the proper time remaining until the clock falling from infinity before it arrives at the singularity. For the elapsed proper time as the clock falls from r1 to r2 if free-falling from initially at rest at infinity the equation is:

$$\Delta tau = \frac{2}{3}r_s \left( \left(\frac{r_1}{r_s}\right)^{3/2} - \left(\frac{r_2}{r_s}\right)^{3/2} \right)$$

It is clear from the above that your T1 and T2 are elapsed proper times and unfortunately I have to agree with Starthaus that setting T1-T2=0 does not give the proper distance and it is the coordinate time difference that has to be set to zero.

Thank you, in the meanwhile Mentz and I have solved the problem. Correctly.

 

[yuiop]

I was assuming the given velocity was wrt coordinate time, but you say
$$\sqrt{2m/r} = \frac{dr}{d\tau}$$
and I have no reason right now not to believe you.I did that calculation some time ago and it does get very messy.

There are various ways to express the velocity of a particle initially at rest rest at R that falls to r.

The "proper velocity" (Coordinate distance divided by proper time of the falling object is:

$$\frac{dr}{dtau} = \sqrt{2m/r - 2m/R}$$

The local velocity according to a shell observer using his local rulers and clocks as the object passes is:

$$\frac{dr'}{dt'} = \frac{\sqrt{2m/r - 2m/R}}{\sqrt{1-2m/R}}$$

The Schwarzschild coordinate velocity in either case is:

$$\frac{dr}{dt} = (1-2m/r) \frac{\sqrt{2m/r - 2m/R}}{\sqrt{1-2m/R}}$$

All these have been given at some point in this thread, but I thought it might be useful to have them all in the same place and clearly defined.

There are obvious simplifications if R = infinity.

 

[yuiop]

Thank you, in the meanwhile Mentz and I have solved the problem. Correctly.

To me, "a step in the right direction" and "A major problem with the approach" is a little short of solving the problem.

This is a step in the right direction, the formalism is very nicely set and the math is very clear. I can see a major problem with the approach, $$T$$ is proper time, so you do not want $$T_1=T_2$$. What you want is $$t_1=t_2$$, that is you want to mark the ends of the rod at the same time in the distant observer's frame, not in the rod frame. You can do that using Schwarzschild coordinates as well. When you do that, you get a very nasty transcendental equation, not due to the use of coordinates but due to the condition $$t_1=t_2$$. I solved this for the more general case of the rod falling from a finite height and I got a much nastier equation.

You have also come nowhere near solving the corresponding Schwarzschild coordinates of a rigid (constant proper length) falling object.

 

[starthaus]

To me, "a step in the right direction" and "A major problem with the approach" is a little short of solving the problem.

You must have missed the fact that the "problem with the approach" has been fixed. All you are left to do is to introduce the correct integrand (as opposed to the incorrect one that you've been pushing throughout the thread) and evaluate the integral. Turn the crank kind of stuff, use your favorite canned software.

 

[starthaus]

There are various ways to express the velocity of a particle initially at rest rest at R that falls to r.

The "proper velocity" (Coordinate distance divided by proper time of the falling object is:

$$\frac{dr}{dtau} = \sqrt{2m/r - 2m/R}$$

The local velocity according to a shell observer using his local rulers and clocks as the object passes is:

$$\frac{dr'}{dt'} = \frac{\sqrt{2m/r - 2m/R}}{\sqrt{1-2m/R}}$$

The Schwarzschild coordinate velocity in either case is:

$$\frac{dr}{dt} = (1-2m/r) \frac{\sqrt{2m/r - 2m/R}}{\sqrt{1-2m/R}}$$

All these have been given at some point in this thread, but I thought it might be useful to have them all in the same place and clearly defined.

There are obvious simplifications if R = infinity.

They've been in https://www.physicsforums.com/blog.php?u=241315 (including their derivations from base principles) for months. You only needed to figure out what was the correct integrand to use (definitely not he one you've been pushing).

 

[yuiop]

I have repeatedly told both you and Passionflower that you are using the wrong approach since you are trying to force the stationary solution for the case of the moving rod.

I made it clear in post 82 of this thread and earlier in the related thread that that formula is for a stationary ruler. I was simply defending DrGreg when you said his equation for a static ruler distance in Schwarzschild coordinates is wrong and we have had to waste dozens of posts proving DrGreg equation was correct in the context it was given. Now that you have forced us to waste time proving that DrGreg's equation was correct (and both passionflower and myself expressly stated we did not want to waste time on it, you say oh yeah, that formula that you said was incorrect is actually correct but it not relevant.) Here is post 82:

The integrated or ruler distance is given by:

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

...

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.

I was aware and always have been that the above equation is for a stationary ruler in Schwarzschild coordinates.

There was nothing to check numerically since the Rindler formula was derived via Taylor expansion for $$r>>r_s$$. This is is quite clear from the Rindler book and from my explanation.

If you knew that the Rindler formula was an approxiamtion, then why on Earth did you change Rindler's "approximately equals" symbol to an "exactly equal" symbol and then proceed to integrate the approximate equation and declare you integration of the approximate equation to be correct and and Passionflower's integration of the exact equation incorrect? Your responses are barely those of a sane person sometimes.

 

[starthaus]

I was aware and always have been that the above equation is for a stationary ruler in Schwarzschild coordinates.

Yet, you kept pushing it as a solution for the problem. Why?
If you knew your solution was wrong all along why did you keep pushing it?

 

[Mentz114]

Thanks everyone, I understand the free falling frame much better now. Good job listing the various velocities, yuiop.

The Wiki page on G-P is rather good. It shows the essentials concisely, including the coordinate transformation which I couldn't find before and had to work out myself.

I'll leave you guys to argue about who said what, when and how often.

 

[yuiop]

I don't think it is. Anyways, the error is easy to fix:

-start with coordinate speed for rod falling from $$r_0$$:

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

-make $$r_0->\infty$$, therefore:

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

So, you need to replace your integrand with the modified one:

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

After you do this, all you end up with is where the clocks are after they fallen for an equal coordinate time and this has already been done in this thread and is not the proper distance of the falling clocks (as far as I can tell). Anybody is welcome to try and convince me otherwise.

 

[starthaus]

After you do this, all you end up with is where the clocks are after they fallen for an equal coordinate time and this has already been done in this thread

...incorrectly , by you. Several times.

and is not the proper distance of the falling clocks

What ever gives you the idea that the solution gives the proper distance? Do you even understand the basic formulas?

(as far as I can tell). Anybody is welcome to try and convince me otherwise.

Given that you've been pushing the wrong solution for 100+ posts, your opinion doesn't mean much.

 

[Mentz114]

After you do this, all you end up with is where the clocks are after they fallen for an equal coordinate time and this has already been done in this thread and is not the proper distance of the falling clocks (as far as I can tell). Anybody is welcome to try and convince me otherwise.

I was dealing with the question - given an initial coordinate separation, what will the separation be after a certain coordinate time has passed ?

And I think that's what I got, eventually.

[edit]yuiop - I think your opinions have merit.

 

[yuiop]

Thanks everyone, I understand the free falling frame much better now. Good job listing the various velocities, yuiop.

The Wiki page on G-P is rather good. It shows the essentials concisely, including the coordinate transformation which I couldn't find before and had to work out myself.

Just had a quick look at GP coordinates on Wiki as you suggested. In radial motion terms only the GP metric is:

$$d tau^2 = (1-2m/r) dt_r^2 - 2\sqrt{(2m/r)} dt_r dr - dr^2$$

Now for $dt_r = 0$ the above metric appears to give $d tau^2 = dr^2$ (with c=1) so that in the free falling frame the proper distance derivative is the same as the Schwarzschild coordinate distance derivative. Maybe Passionflower was onto something The integrated distance between freefalling objects in GP coordinates is the same as the integrated distance between two stationary objects in Schwarzschild coordinates. Not quite sure what the significance of that is yet.

 

[yuiop]

After you do this, all you end up with is where the clocks are after they fallen for an equal coordinate time and this has already been done in this thread and is not the proper distance of the falling clocks (as far as I can tell). Anybody is welcome to try and convince me otherwise.

What ever gives you the idea that the solution gives the proper distance? Do you even understand the basic formulas?

When I say "is not the proper distance" I clearly do not mean that I have the idea that it is the proper distance. However one of the major things we have been trying to solve in this thread is the proper distance of the falling clocks. In the OP passionflower wanted to know the coordinate separation of two clocks that maintain constant proper distance apart as they fall. In order to do that you need to know how to define proper constant distance for the clocks. Since you claim to solved the questions posed in the OP this implies that you solved how to define proper distance in the reference frame of the falling clocks. Clearly you have not done that and your proposed solution does not solve that.

Given that you've been pushing the wrong solution for 100+ posts, your opinion doesn't mean much.

I gave a proposed solution early on in the thread that I also withdrew early on when I realized the problem was much more complicated than I originally assumed. Since then I have not proposed any definitive solutions to the problems in the OP. Although you have said the problem is intractable, the fact that think you have a solution only tells me you have not realized the full complexity of the problems posed in the OP. If you have a real solution, you can post a numerical solution to the problem posed in the OP and you can do this without fear of giving away any valuable secret information about the formulas used, as I know you think this it is detrimental to the educational process of readers if you are not not vague and evasive and provide concise accurate information.

 

[yuiop]

Yet, you kept pushing it as a solution for the problem. Why?
If you knew your solution was wrong all along why did you keep pushing it?

Show me one post where I claimed the the equation for the static ruler distance between two Schwarzschild coordinates is the proper length of a falling ruler?

 

[pervect]

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

works just fine. And I think I see how to sketch out a solution that most of us will agree with.

(It appears that some other people wrote up similar ideas while I was writing this - and it appears that if yuiop agrees with me, strathaus will claim he's wrong, so I don't expect universal agreement ::smiley face::).

If we take r_s =1 or M = 1/2, and R = infinity, we can use the simpler equations MTW 25.38 to find the Schwarzschild t coordinate as a function of the r coordinate. It also gives us proper time, $\tau$, though we don't really need it.

$$\tau = -\frac{2}{3} \, r^{\frac{3}{2}}$$

$$t = -\frac{2}{3} \, r^{\frac{3}{2}} - 2 \, r^{\frac{1}{2}} + ln\frac{r^{\frac{1}{2}}+1}{r^{\frac{1}{2}}-1}$$

Now, all we need to do to make this a one-parameter group of curves is to use time-translation symmetry. We simply add some small number $\epsilon$ to the schwarzschild time coordinate to represent the second particle, which works because of the time-translation symmetry of the metric, in fancier language $\partial / \partial t$ is a "Killing vector". So we have

$$t\left(r, \epsilon\right) = -\frac{2}{3} \, r^{\frac{3}{2}} - 2 r^{\frac{1}{2}} + ln\frac{r^{\frac{1}{2}}+1}{r^{\frac{1}{2}}-1} + \epsilon$$

Assume that the leading edge of the rod, with the parameter $\epsilon =0$, has some r-coordinate $r_0$ at some t-coordinate $t_0$.

Then we need to find the radial coordinate $r_1$ of the trailing rod at the same time coordinate $t_0$ by solving

$$t\left(r_1, \epsilon) = t_0 = t\left(r_0, 0\right)$$

i.e.
$$-\frac{2}{3} \, r_{1}^{\frac{3}{2}} - 2 r_{1}^{\frac{1}{2}} + ln\frac{r_{1}^{\frac{1}{2}}+1}{r_{1}^{\frac{1}{2}}-1} + \epsilon = -\frac{2}{3} \, r_{0}^{\frac{3}{2}} - 2 r_{0}^{\frac{1}{2}} + ln\frac{r_{0}^{\frac{1}{2}}+1}{r_{0}^{\frac{1}{2}}-1}$$

Then we use
$$\sqrt {r_{1} \left( r_{1}-1 \right) }+\ln \left( \sqrt {r_{1}}+\sqrt {r_{1}-1} \right) - \sqrt {r_{0} \left( r_{0}-1 \right) } - \ln \left( \sqrt {r_{0}}+\sqrt {r_{0}-1} \right)$$

to integrate between $r_0$ and $r_1$ to get the length as seen by a static observer for the falling rod. To get the proper length, we multiply by gamma, based on the radial velocity which we know is 1 / $\sqrt{r}$.

 

[yuiop]

If we take r_s =1 or M = 1/2, and R = infinity, we can use the simpler equations MTW 25.38 to find the Schwarzschild t coordinate as a function of the r coordinate. It also gives us proper time, $\tau$, though we don't really need it.

$$\tau = -\frac{2}{3} \, r^{\frac{3}{2}}$$

$$t = -\frac{2}{3} \, r^{\frac{3}{2}} - 2 \, r^{\frac{1}{2}} + ln\frac{r^{\frac{1}{2}}+1}{r^{\frac{1}{2}}-1}$$

I got a similar result for t by starting with the coordinate velocity dr/dt = (1-2m/r)*sqrt(2m/r) of a falling object that was initially stationary at infinity and then using 2m=1 and integrating dt = sqrt(r)/(1-1/r) dr, except my signs were reversed, but I won't worry about that too much right now. A small simplification of the equation for t can be brought about by noting that:

$$ln\frac{r^{\frac{1}{2}}+1}{r^{\frac{1}{2}}-1} = 2 \, arcoth(r)$$

valid for r>1 which is OK if we stay outside the event horizon.

...

Then we use
$$\sqrt {r_{1} \left( r_{1}-1 \right) }+\ln \left( \sqrt {r_{1}}+\sqrt {r_{1}-1} \right) - \sqrt {r_{0} \left( r_{0}-1 \right) } - \ln \left( \sqrt {r_{0}}+\sqrt {r_{0}-1} \right)$$

to integrate between $r_0$ and $r_1$ to get the length as seen by a static observer for the falling rod. To get the proper length, we multiply by gamma, based on the radial velocity which we know is 1 / $\sqrt{r}$.

So far I have only skimmed your solution, but broadly it seems a reasonable approach. One small observation is that the last part where you suggest multiplying by gamma based on radial velocity, there seems to be the minor problem that for an extended string of particles, the velocities will be different (I think) depending on where the particles are located. Are you proposing a sort of average velocity or having an arrangement where the particles fall at the same velocity or simply relying on the spatial separation being small enough that differences in individual falling velocities is insignificant?

[EDIT] After a quick re-read I now think you mean the last one because you have specified a small separation. [/EDIT]

Forgive me if I have missed anything significant. This is just a first pass on your proposed solution.

 

[pervect]

So far I have only skimmed your solution, but broadly it seems a reasonable approach. One small observation is that the last part where you suggest multiplying by gamma based on radial velocity, there seems to be the minor problem that for an extended string of particles, the velocities will be different (I think) depending on where the particles are located.

Correct. If you have a long enough rod, rather than integrate

$$\int_{r1}^{r2} \frac{dr}{\sqrt{1-1/r}}$$

and then multiply the result by gamma, which varies, you want to integrate instead

$$\int_{r1}^{r2} \gamma\left( r \right) dr = \int_{r1}^{r2} \frac{dr}{1-1/r}$$

since in this case $\gamma$ = 1/sqrt(1-v^2) = 1/sqrt(1-1/r)

You can view this as breaking up the long rod into a bunch of short ones, with the evenly spaced "tic marks" of $\epsilon$ marking the divisions, and summing up the distances between the segments of the rod

You might want to do this all the time anyway - it's an easier integral :-)

Note that this means there is no single"frame" for the falling rod. We break the long rod up into a series of short ones, small enough where we can use the flat-space approximation of a "frame" for each segment.

Are you proposing a sort of average velocity or having an arrangement where the particles fall at the same velocity or simply relying on the spatial separation being small enough that differences in individual falling velocities is insignificant?

My original post was for a short enough rod where the velocity change was small enough, the above revision is needed to handle a longer one.

 

[starthaus]

When I say "is not the proper distance" I clearly do not mean that I have the idea that it is the proper distance.

LOL, you clearly don't understand what the formulas mean.

However one of the major things we have been trying to solve in this thread is the proper distance of the falling clocks.

Again? Same mistake? And you still claim that you understand the problem?

Since then I have not proposed any definitive solutions to the problems in the OP.

Then why do you push the incorrect stationary solution in each one of your posts?

 

[starthaus]

Show me one post where I claimed the the equation for the static ruler distance between two Schwarzschild coordinates is the proper length of a falling ruler?

Each post you pushed the integrand $$\frac{dr}{\sqrt{1-r_s/r}}$$ or its integral $$\sqrt{r*(r-r_s)} + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)$$. Meaning all your posts where you made an attempt at solving the problem.

 

[starthaus]

I've finally come up with a proposal for the stretching factor in Gullstrand-Painleve coords.

A test body falls from r₁ -> r₂ ( being at r₁ when t=0) in time T₁, so
$$T_1 &= \int_{r_1}^{r_2}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( r_2^{3/2}- r_1^{3/2} \right)$$
Another test body is on the same geodesic, but is at r₁+L at time t=0. It will fall to point x in time T₂
$$T_2 &= \int_{r_1+L}^{x}\sqrt(r/2m) dr &= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( x^{3/2}- (L+r_1)^{3/2} \right)$$
We can set T₁-T₂=0 and solve for x, which gives
$$x={\left( {\left( L+r_1\right) }^{\frac{3}{2}}+{r_2}^{\frac{3}{2}}-{r_1}^{\frac{3}{2}}\right) }^{\frac{2}{3}}$$
and the new separation is given by r₂-x.

The algebra and calculus was done with Maxima. This is the only sensible looking result I've managed so far so it would be gratifying if it was right.

[edit] changed second integral

Replace the incorrect integrand with the correct one:

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

and you get the correct solution, expressed in coordinate time, as it should be.

 

[pervect]

Now

While finding a closed for solution for $r_1$ as a function of $r_0, \epsilon$ from the following is probably impossible

$$-\frac{2}{3} \, r_{1}^{\frac{3}{2}} - 2 r_{1}^{\frac{1}{2}} + ln\frac{r_{1}^{\frac{1}{2}}+1}{r_{1}^{\frac{1}{2}}-1} + \epsilon = -\frac{2}{3} \, r_{0}^{\frac{3}{2}} - 2 r_{0}^{\frac{1}{2}} + ln\frac{r_{0}^{\frac{1}{2}}+1}{r_{0}^{\frac{1}{2}}-1}$$we can do the usual and consider a sufficiently short rod. Let $\Delta_r$ = $r_{1} - r_{0}$. Then for a short enough rod:

$$\Delta_{r} = -\frac{\epsilon}{dt/dr} = \frac {\epsilon\, \left( r-1 \right) }{{r}^{3/2}}$$

Now $\Delta_{r}$ is a coordinate difference and not really a length of any sort. The length relative to a static observer will be $\Delta_{r} / sqrt(1-1/r)$, and the proper length will be $\Delta_{r} / (1-1/r)$ - for reasons I mention in the last post, basically gamma = 1/sqrt(1-v^2) and v=1/sqrt(r), where v is the local velocity seen by a static observer.

Solving for the proper length, one gets $\epsilon / \sqrt{r}$.

 

[Passionflower]

Assume that the leading edge of the rod, with the parameter $\epsilon =0$, has some r-coordinate $r_0$ at some t-coordinate $t_0$.

Then we need to find the radial coordinate $r_1$ of the trailing rod at the same time coordinate $t_0$ by solving

$$t\left(r_1, \epsilon) = t_0 = t\left(r_0, 0\right)$$

i.e.
$$-\frac{2}{3} \, r_{1}^{\frac{3}{2}} - 2 r_{1}^{\frac{1}{2}} + ln\frac{r_{1}^{\frac{1}{2}}+1}{r_{1}^{\frac{1}{2}}-1} + \epsilon = -\frac{2}{3} \, r_{0}^{\frac{3}{2}} - 2 r_{0}^{\frac{1}{2}} + ln\frac{r_{0}^{\frac{1}{2}}+1}{r_{0}^{\frac{1}{2}}-1}$$

I am running some numbers to verify I understand it.

If we take the front clock at R=40 we have 180.9850160 time units left.
Now let's take an epsilon of 13 time units, this will give us a back coordinate value of R=41.98100155. Thus the coordinate distance between the front and the back is 1.98100155 which gives a proper distance (for a static observer) of 2.005621996. The two are not identical due to the curvature of spacetime.

I calculated r for t=-60-0 for the front and t=-73-13 for the back and plotted the results. I also calculated the coordinate and proper distance (for a static observer).

Now I do see a stretch both in coordinate and proper time. Is this the tidal stress? One problem with this approach is that we cannot run this past the event horizon, the major spaghettification happens closer to r=0, rather than changing coordinate charts is there any way we can use proper time for this?

$$\int_{r1}^{r2} \gamma\left( r \right) dr = \int_{r1}^{r2} \frac{dr}{1-1/r}$$

since in this case $\gamma$ = 1/sqrt(1-v^2) = 1/sqrt(1-1/r)

I thought the proper distance for a free falling observer is simply dr because the integral becomes the integral for a static observer times 1/gamma and these two factors exactly cancel in this case. (See the page from Raine's and Thomas' book on Black Holes I quoted: https://www.physicsforums.com/attachment.php?attachmentid=28805&d=1286253576)

$$\int_{r1}^{r2} {1 \over \gamma(r)} \frac{dr}{\sqrt{1-1/r}} = dr$$