The Schwarzschild Metric - A Simple Case

[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:

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

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:

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

...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

$$r_2-r_1<\Delta S$$

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:

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

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, $$r_s<<r_1$$, 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, $$r_s<<r_1$$, 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 $$r_s<<r$$ 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 $$r_s<<r$$ 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:

$$\frac{1}{2} a \tau^2$$

Now taking rs=1 and l=0.1

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

$$a = {1 \over r^3} l$$

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

$$2/3\,{{\it ri}}^{3/2}-2/3\,{{\it ro}}^{3/2}$$

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

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...

 

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

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...

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 $$r_s<<r$$. 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 $$r_s<<r$$

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

Edited to add:

The RHS is the Taylor expansion for $$r_s<<r$$. 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:

$$\frac{1}{2} a \tau^2$$

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.