The Schwarzschild Metric - A Simple Case

[Passionflower]

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

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

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

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

<br /> {1 \over r^3} dr<br />

 

[starthaus]

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 />

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

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

Try solving the problem starting from the correct integrand:

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

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

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

 

[starthaus]

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

For our example (r_s = 1):

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

<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!

...which is totally incorrect.

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

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

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

 

[Passionflower]

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

Which formula do you think is not correct?

 

[starthaus]

Which formula do you think is not correct?

I asked you where the integrand

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

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

 

[Passionflower]

I asked you where the integrand

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

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

This is very disruptive Starthaus you need to read better before you claim something is wrong. I am talking about a particle not an extended object, the integral is perfectly correct.

 

[starthaus]

This is very disruptive Starthaus you need to read better before you claim something is wrong. I am talking about a particle not an extended object, the integral is perfectly correct.

No, it isn't correct for a particle either. I asked you several times where you got it from and you never answered. Where did you get it from?

 

[Passionflower]

No, it isn't correct for a particle either. I asked you several times where you got it from and you never answered. Where did you get it from?

Looking at the formula that you think is right it appears you simply do not understand the difference between taking the integral from r2 to r1 for a particle free falling from infinity and from a particle free falling from an apogee at r2. Both yuiop and I are talking about a situation free falling from infinity.

Please in the future be more careful in calling other people wrong, first read carefully before you jump into conclusions.

 

[starthaus]

Looking at the formula that you think is right it appears you simply do not understand the difference between taking the integral from r2 to r1 for a particle free falling from infinity and from a particle free falling from an apogee at r2. Both yuiop and I are talking about a situation free falling from infinity.

The formula that you are trying to use is also wrong if you consider a particle free-falling from infinity, I also gave you the correct formula for a particle free-falling from infinity (posts 54, 98). I am asking you again, where did you get your formula from? You claimed it is from a textbook, what textbook?

Please in the future be more careful in calling other people wrong, first read carefully before you jump into conclusions.

I am being very careful.

 

[Passionflower]

The formula that you are trying to use is also wrong if you consider a particle free-falling from infinity, I also gave you the correct formula for a particle free-falling from infinity (posts 54, 98). I am asking you again, where did you get your formula from? You claimed it is from a textbook, what textbook?

Since I remember you have Rindler (2nd edition) take a look in chapter 11.5 page 236 at the top you see the integral for ruler distance. If you make 2m=1 you get exactly the same integrand as I use.

I attached an image of the page in question.

 

[starthaus]

Since I remember you have Rindler (2nd edition) take a look in chapter 11.5 page 236 at the top you see the integral for ruler distance. If you make 2m=1 you get exactly the same integrand as I use.

Thank you, I had a look. You are misunderstanding chapter 11.5, it has nothing to do with free-falling particles. If you want to learn about free-falling particles (falling from a fixed distance or from infinity), this comes up later, in chapter 11.8.

 

[Passionflower]

You are misunderstanding chapter 11.5, it has nothing to do with free-falling particles. If you want to learn about free-falling particles (falling from a fixed distance or from infinity), this comes up later, in chapter 11.8.

As I wrote before you should really read carefully:

Check my posting https://www.physicsforums.com/showpost.php?p=2915076&postcount=99 again.

I quote myself from that posting:

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 />

As I said read carefully before you call wrong.

 

[starthaus]

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

For our example (r_s = 1):

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

<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 />

...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 above is a transcendental equation in r_2-r_1, so, it cannot be solved symbolically. The only thing one can say with certitude is that

r_2-r_1&lt;\Delta S

i.e. the coordinate length is less than the proper length.

With r_s=1 you should get:


r_2-r_1+1/2*(ln(r_2)-ln(r_1))=r_2-r_1+ln(\sqrt{r_2/r_1})

Anyway, if you want a comparison between coordinate, radar and proper distances, then Rindler 11.5 gives the answer to your question.
You need to pay attention, the above derivation applies only for a static (unmoving) rod, so all your attempts at using this kind of math, do NOT answer the case of a falling rod, as described in your OP. If you want to find out the length of a ruler falling radially in a gravitational field, then 11.5 is not your answer, you need to proceed to 11.8. A much more difficult problem as I tried to point out to you several times. If you want to know how to answer the problem of a moving rod, then you need to start with the integrand I gave you a few times already.

 

[Passionflower]

...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)

I admit I made a typing mistake in entering the integrand, but that does not invalidate the rest.

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:

r_2-r_1+r_s/2(ln(r_2)-ln(r_1))

No my formula is ok for r_s=1

With r_1=1 you should get:

r_2-1+r_s/2*ln(r_2)

No just take a numerical example: if we take rs=1, r1=1 and r2=2 I get: 2.295587149 while your formula gives: 1.346573590 which is wrong.

So yes, I admit there was one mistake typing in the integrand but that has no further influence on the rest.

 

[starthaus]

I admit I made a typing mistake in entering the integrand, but that does not invalidate the rest.

You got the wrong integrand, meaning that you got the wrong integral.

No my formula is ok for r_s=1

I don't think so, I adjusted the answer in order to deal with making r_s rather than r_1 equal to 1 and your formula is just as wrong.

 

[Passionflower]

You got the wrong integrand, meaning that you got the wrong integral.

No, I told you I typed it in wrong the rest is fine.

I don't think so, I adjusted the answer in order to deal with making r_s rather than r_1 equal to 1 and your formula is just as wrong.

So are you saying that 2.295587149, the answer I get is wrong?

This is my formula:

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

for r=2 I get 2.29558714, which is correct.

 

[starthaus]

No, I told you I typed it in wrong the rest is fine.So are you saying that 2.295587149, the answer I get is wrong?

Read the results in post 113. Carefully.

 

[starthaus]

I admit I made a typing mistake in entering the integrand, but that does not invalidate the rest.No my formula is ok for r_s=1No just take a numerical example: if we take rs=1, r1=1 and r2=2 I get: 2.295587149 while your formula gives: 1.346573590 which is wrong.

Umm, r_s&lt;&lt;r_1, remember? So you can't make r_1=r_s.
Rather than pasting numbers at random, I suggest that you read chapter 11.5 from end to end.

 

[Passionflower]

Read the results in post 113. Carefully.

Well yes or no? If we have rs=1, r1=1 and r2 is 2 is the answer 2.295587149 or is it 1.346573590?

Umm, r_s&lt;&lt;r_1, remember? So you can't make r_1=r_s.
Rather than pasting numbers at random, I suggest that you read chapter 11.5 from end to end.

What are you talking about the distance all the way up to rs is finite.

Please answer the question, what is the right answer 2.295587149 or is it 1.346573590?

 

[starthaus]

Well yes or no? If we have rs=1, r1=1 and r2 is 2 is the answer 2.295587149 or is it 1.346573590?

The formula assumes r_s&lt;&lt;r so your attempt to make r_1=r_s makes no sense. Like I said, instead of wasting your time pasting numbers into formulas, why don't you read the one page chapter 11.5? I promise you, you would be learning a lot more.

 

[Passionflower]

The formula assumes r_s&lt;&lt;r so your attempt to make r_1=r_s makes no sense. Like I said, instead of wasting your time pasting numbers into formulas, why don't you read the one page chapter 11.5? I promise you, you would be learning a lot more.

You mean your formula assumes that, mine is just fine. You are simply wrong and instead pasted an approximation while my formula is exact. Why you call my formula wrong? Do you perhaps think an approximation is better?

Again the proper distance between r2 and r1 is finite even when r1=rs.Please answer the question, what is the right answer 2.295587149 or is it 1.346573590?

 

[starthaus]

You are simply wrong and instead pasted an approximation while my formula is exact. Why you call my formula wrong? Do you perhaps think an approximation is better?

You need to think a little as to how the formula was derived. I asked you before, do you know how to derive the integrand? This question is key, because if you knew where the integrand was derived, you would have known that it is applicable only for a static rod extending between the Schwarzschild coordinates r_1 and r_2. Your OP is trying to deal with a rod falling radially, remember? The integrand (and the integral) you are using are not applicable for a moving rod. Do you know why?

Again the proper distance between r2 and r1 is finite even when r1=rs.

For example, for Earth, r_s=9mm. So, your rod will never reach r_s. This is why.

Please answer the question, what is the right answer 2.295587149 or is it 1.346573590?

There are no answers to meaningless questions.

 

[Passionflower]

For example, for Earth, r_s=9mm. So, your rod will never reach r_s. This is why.

Who is talking about the Earth, how many more excuses are you going to find to circumvent admitting you were wrong in calling my formula wrong. I admit I made a typing mistake in writing down the integrand but the formula is correct.

One can't answer meaningless questions.

I take that as you realize you are wrong.

I welcome anyone to calculate my and Starthaus' answer and see which one is actually correct.

I don't know how many posting wasted because of all this nonsense. If someone makes a typo just mention what you think is the problem instead of saying it is wrong and going on and on about vageries. This is not the first time. This is not very conductive to a good discussion.

I have wasted enough time on this so this is my last posting about this particular thing. Starthaus if you want to contribute great, if you find mistakes fine, please mention it, but make it useful and not disruptive.

 

[starthaus]

Who is talking about the Earth, how many more excuses are you going to find to circumvent admitting you were wrong in calling my formula wrong. I admit I made a typing mistake in writing down the integrand but the formula is correct.

It was just an example to show you what happens if you think that physics is about plugging in numbers mindlessly.

I welcome anyone to calculate my and Starthaus' answer and see which one is actually correct.

This is a totally pointless exercise since your "solution" does NOT apply to your problem. You have been trying for about 100 posts to force the solution for static (unmoving) rods to the case of a radially falling rod. I have been telling you several different ways that what you are doing is plain wrong.

 

[Passionflower]

As I wrote before since the integrand for proper distance for a stationary observer multiplied by the length contraction factor from a free falling particle from infinity exactly cancels out, the proper length in this free falling frame is simply dr.

Now let's add the formula member Pervect gave before:

He wrote that, assuming the rod is very short, the rate of change of it's length should be just:

<br /> \frac{1}{2} a \tau^2<br />

Now taking rs=1 and l=0.1

We know the tidal acceleration of such a rod l would be

<br /> a = {1 \over r^3} l<br />

Now for a free falling particle falling from infinity the total proper time over a range ro to ri is:

<br /> 2/3\,{{\it ri}}^{3/2}-2/3\,{{\it ro}}^{3/2}<br />

So far so good but if we plug in Pervect's formula the rate of change appears to depend on the initial r value, which seems to indicate that this formula is not applicable to a free falling from infinity scenario. So which is the one in case we have a free fall from infinity?

The dependence on the initial r value can be demonstrated graphically, I included three graphs. The first one shows the proper time from R=20 until we reach the singularity, the next one shows a starting value of ro=10 and the other one shows a starting value of ro=20. For readability we sample up to r=5 to keep the graph from exploding upwards. As you can see the rate of change depends on the initial r value.

Notice that we are discussing a free fall from infinity case, the starting value is simply where we start to sample the information.

 

[pervect]

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

or equivalently (different by a constant factor0

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

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

 

[Passionflower]

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

or equivalently (different by a constant factor0

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

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

I do not know who you are referring to Starthaus or me but I verified in Maple that my formula is exact. I checked the above formula you gave here and that one is also not correct, are you sure you use the exact integral and not the approximation that Rindler gave?

 

[starthaus]

As I wrote before since the integrand for proper distance for a stationary observer multiplied by the length contraction factor from a free falling particle from infinity exactly cancels out, the proper length in this free falling frame is simply dr.

This is pure nonsense, you are are still trying to force the solution for the stationary rod to answer your original OP about a free-falling rod. Physics is not a collection of nonsensical hacks.

 

[Passionflower]

This is pure nonsense, you are are still trying to force the solution for the stationary rod to answer your original OP about a free-falling rod. Physics is not a collection of nonsensical hacks.

For a reference in the literature see:

"Black holes: An Introduction" By Derek J. Raine, Edwin George Thomas, page 36

http://books.google.com/books?id=oP...gth" hypersurface&pg=PA36#v=onepage&q&f=false

It is becoming clear to me that your prime motive here in this forum is to disrupt conversations.

 

[pervect]

...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)

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;

 

[Passionflower]

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;

I attached a scan of that page a few postings ago. Rindler clearly indicates it is an approximation. Starthaus likely misread that and assumed that my formula was wrong based on the approximation, but in fact my formula is exact.

Here is the link to the posting: https://www.physicsforums.com/showpost.php?p=2915507&postcount=110

 

[starthaus]

For a reference in the literature see:

"Black holes: An Introduction" By Derek J. Raine, Edwin George Thomas, page 36

http://books.google.com/books?id=oP...gth" hypersurface&pg=PA36#v=onepage&q&f=false

It is becoming clear to me that your prime motive here in this forum is to disrupt conversations.

The link explains that the the proper length is tied to the coordinate length via the expression:

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

It then goes to explain the adjustment for a free-falling observer that moves at terminal velocity. This has nothing to do with the statement of your problem, it is a different problem altogether. Are you copying formulas from different books in order to see what sticks?

 

[starthaus]

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

The RHS is the Taylor expansion for r_s&lt;&lt;r. This enables Rindler to produce a much more useful formula than the one Passionflower and yuoip have been torturing throughout this thread.

 

[Passionflower]

It then goes to explain the adjustment for a free-falling observer that moves at terminal velocity. This has nothing to do with the statement of your problem, it is a different problem altogether. Are you copying formulas from different books in order to see what sticks?

It has everything to do with the problem as I many times stated the problem is a case of free falling from infinity.

Clearly you start to become close to being harrasing, please change your tone.

The RHS is the Taylor expansion for r_s&lt;&lt;r

It appears you start to realize your mistake in calling my formula wrong.

Edited to add:

The RHS is the Taylor expansion for r_s&lt;&lt;r. This enables Rindler to produce a much more useful formula than the one Passionflower and yuoip have been torturing throughout this thread.

Now it is absolutely clear you know you were wrong.

Perhaps you now can read "Black holes: An Introduction" by Derek J. Raine, Edwin George Thomas, page 35 and 36 and realize that you actually called that "pure nonsense".

 

[starthaus]

I attached a scan of that page a few postings ago. Rindler clearly indicates it is an approximation. Starthaus likely misread that and assumed that my formula was wrong based on the approximation, but in fact my formula is exact.

Here is the link to the posting: https://www.physicsforums.com/showpost.php?p=2915507&postcount=110

No, your formula is wrong because:

1. you are using an inappropriate formula to begin with (the worst mistake)
2. you wrote down the incorrect integrand (you claim it was a typo)

 

[starthaus]

It has everything to do with the problem as I many times stated the problem is a case of free falling from infinity.

Whether the rod is falling from a finite distance or from infinity is not an excuse to hack in formulas used for stationary rods. This is the mistake that you have been repeating throughout the thread. You picked up the formula for a stationary rod and you kept pasting it (with an assortment of smaller mistakes) throughout the thread. When this is pointed out to you , you simply ignore the corrections. I am going to ask you again, do you even know how the formula was derived? Do you know the physical restrictions? Do you know how to derive it yourself? Come one, it is one line of calculations, let's see if you can do it.

It appears you start to realize your mistake in calling my formula wrong.

Your formula is wrong because it is inappropriate for the problem you are trying to solve.Period.

 

[Passionflower]

Hopefully we can now continue this topic without further interruptions. If a moderator has some extra time, I and I am sure many others, would appreciate it if he or she could remove the bickering as the topic is very interesting.

I attached a pdf file taken from:

"Black holes: An Introduction" By Derek J. Raine, Edwin George Thomas, page 35 and 36

which I think is relevant, but not the complete solution, to our problem.

Remember the case is about a free falling from infinity not from the apogee!

As you can see in the attached document the integral to obtain the proper distance is multiplied by the length contraction factor and the conclusion is that the proper distance is simply r in this case. Provided the rod is short enough we now have the means to investigate how the proper length changes for decreasing values of r (or increasing values of tau).

So if you take a look at the provided graphs in some earlier posting you can indeed see a stretch factor but I believe the formula Pervect provided is applicable to a fall from apogee and not one falling from infinity (e.g. E=1). Since in both cases (r=10 and r=20) the rod falls from infinity the stretch factor should be identical for identical r values which as you can see is not the case.

This is the formula in question:

<br /> <br /> \frac{1}{2} a \tau^2<br /> <br />

After we verified we got the correct formula, we can finally consider the more difficult case where the rod is longer and we cannot consider the whole rod to be free falling.

 

[starthaus]

Hopefully we can now continue this topic without further interruptions. If a moderator has some extra time, I and I am sure many others, would appreciate it if he or she could remove the bickering as the topic is very interesting.

I attached a pdf file taken from:

"Black holes: An Introduction" By Derek J. Raine, Edwin George Thomas, page 35 and 36

which I think is relevant, but not the complete solution, to our problem.

No, it isn't relevant to the problem. The material that IS relevant to the problem can be found under "RadialMotion1" https://www.physicsforums.com/blog.php?b=1957 , I will not do it again, you are on your own.

As you can see in the attached document the integral to obtain the proper distance is multiplied by the length contraction factor and the conclusion is that the proper distance is simply r

This is pure nonsense, you do not even understand the books you are citing. If you want to solve this problem, you need to stop cutting and pasting irrelevant formulas and you need to start learning how to derive the solutions on your own.

 

[Passionflower]

This is pure nonsense, you do not even understand the books you are citing.

To get the tidal force we need the proper distance corresponding to a coordinate distance dr (because our height that we measure in metres is our proper height in free fall, not, in principle, a coordinate displacement in Schwarzschild coordinates, although the two will turn out to be numerically the same).

In a stationary frame the proper distance corresponding to a displacement dr is dr(1-2m/r)^-1/2. For the freely falling observer moving with the speed v_loc = (2m/r)^1/2, this length is contracted by a factor:

1/y = (1-v²_loc)^1/2 = (1-2m/r)^1/2.​

So the proper length in the free falling frame is:

y^-1dr(1-2m/r)^-1/2 = dr​

 

[starthaus]

To get the tidal force we need the proper distance corresponding to a coordinate distance dr

You need the connection between dr and ds. THIS is the problem you set up to solve.

(because our height that we measure in metres is our proper height in free fall, not, in principle, a coordinate displacement in Schwarzschild coordinates, although the two will turn out to be numerically the same).

You are NOT in free-fall, the rod IS.

In a stationary frame the proper distance corresponding to a displacement dr is dr(1-2m/r)^-1/2. For the freely falling observer moving with the speed v_loc = (2m/r)^1/2, this length is contracted by a factor:

1/y = (1-v²_loc)^1/2 = (1-2m/r)^1/2.​

The free-falling observer has nothing to do with your problem.

So the proper length in the free falling frame is:

y^-1dr(1-2m/r)^-1/2 = dr​

Feel free to continue with your nonsense, I gave you the tools to solve the problem, from here on you are on your own.

 

[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:

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

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

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

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

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

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

<br /> \sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( r +(r-1) +2\sqrt{r}\sqrt{r-1} \right) <br />

<br /> \sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( -1 + 2r +2\sqrt{r(r-1)} \right) <br />

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

or equivalently (different by a constant factor0

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

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

Your second form is effectively:

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

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&#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. (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&gt;&gt;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&gt;&gt;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):

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

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

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

<br /> \sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( r +(r-1) +2\sqrt{r}\sqrt{r-1} \right) <br />

<br /> \sqrt {r \left( r-1 \right) }+ 1/2 \ln \left( -1 + 2r +2\sqrt{r(r-1)} \right) <br />

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
<br /> T_1 &amp;= \int_{r_1}^{r_2}\sqrt(r/2m) dr<br /> &amp;= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( r_2^{3/2}- r_1^{3/2} \right)<br />
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
<br /> T_2 &amp;= \int_{r_1+L}^{x}\sqrt(r/2m) dr<br /> &amp;= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( x^{3/2}- (L+r_1)^{3/2} \right)<br />
We can set T₁-T₂=0 and solve for x, which gives
<br /> x={\left( {\left( L+r_1\right) }^{\frac{3}{2}}+{r_2}^{\frac{3}{2}}-{r_1}^{\frac{3}{2}}\right) }^{\frac{2}{3}}<br />
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
<br /> T_1 &amp;= \int_{r_1}^{r_2}\sqrt(r/2m) dr<br /> &amp;= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( r_2^{3/2}- r_1^{3/2} \right)<br />
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
<br /> T_2 &amp;= \int_{r_1+L}^{x}\sqrt(r/2m) dr<br /> &amp;= \frac{1}{3}\sqrt{ \frac{2}{m} } \left( x^{3/2}- (L+r_1)^{3/2} \right)<br />
We can set T₁-T₂=0 and solve for x, which gives
<br /> x={\left( {\left( L+r_1\right) }^{\frac{3}{2}}+{r_2}^{\frac{3}{2}}-{r_1}^{\frac{3}{2}}\right) }^{\frac{2}{3}}<br />
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-&gt;\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&gt;&gt;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.