Are There Unresolver Foundational Issues With GR
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Bill Hobba - 01 Apr 2007 03:26 GMT In another post I mentioned there are no unresolved foundational issues with GR. And to the best of my knowledge there isn't. But an actual knowledgeable poster thought I was joking. Does anyone know what of any actual unresolved issues? No cranks please - although I doubt it will stop anyone.
Thanks Bill
Eric Gisse - 01 Apr 2007 04:33 GMT > In another post I mentioned there are no unresolved foundational issues with > GR. And to the best of my knowledge there isn't. But an actual > knowledgeable poster thought I was joking. Does anyone know what of any > actual unresolved issues? No cranks please - although I doubt it will stop > anyone. None that I'm aware of, outside the usual noise about singularities and not-being-quantum.
> Thanks > Bill Bill Hobba - 01 Apr 2007 05:41 GMT >> In another post I mentioned there are no unresolved foundational issues >> with [quoted text clipped - 6 lines] > None that I'm aware of, outside the usual noise about singularities > and not-being-quantum. Same here - maybe someone like Steve Carlip will give an answer.
Thanks Bill
>> Thanks >> Bill Koobee Wublee - 01 Apr 2007 06:27 GMT > In another post I mentioned there are no unresolved foundational issues with > GR. And to the best of my knowledge there isn't. But an actual > knowledgeable poster thought I was joking. Does anyone know what of any > actual unresolved issues? No cranks please - although I doubt it will stop > anyone. There are no unresolved foundational issues with GR because the mathematical foundation that GR is based is wrong for at least a hundred years. Since a hypothesis in physics has to be backed up by solid mathematics in which physics is still not mathematics, GR's foundation is actually very faulty. Here are the basics.
1. There are many ways to write the geodesic equation where each way would yield a different set of connection coefficients. Christoffel's way is not unique, and the Christoffel symbols are not unique.
2. The covariant derivative was an operator suggested by Ricci based on the non-unique Christoffel symbols of the second kind. For each set of connection coefficients to the geodesic equation, there is one such operator where each such operator is very different from the other.
3. The Riemann curvature tensor was derived through the covariant derivative. It was designed by Ricci. There are many but mathematically different ways to present the Riemann curvature tensor. Ricci only found one.
4. The Ricci curvature tensor was designed by Ricci's faithful student Levi-Civita from the Riemann curvature tensor.
5. The Lagrangian to the Einstein-Hilbert action was designed or cooked up as an alchemist would have done in the medieval times by Hilbert himself. The main ingredient was none other than the Ricci curvature tensor.
6. The Einstein field equations are derived by dividing that Lagrangian Hilbert pulled out of his *ss with each of the elements to the metric. Since there are 16 elements, there are indeed 16 such equations.
7. The Schwarzschild metric is one of the infinite numbers of solutions to the ordinary differential equations (the field equations) for static spacetime. The creation of a black hole is strictly how one would embrace one non-unique solution to another. It allows anyone to play God through a pen and a piece of paper.
The above represent the mathematical foundation of GR. It shows how the curvature in spacetime is shaped by the mathematical tool invented by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert, Schwarzschild, and Einstein are His prophets.
However, the general concept of the metric being invariant is totally wrong. The curvature in spacetime must be invariant. This should be a no-brainer to accept. This invariant geometry is described by the dot product of the metric itself and the choice of coordinate. If the choice of coordinate is varying from one observer to another, the metric must vary with the choice of coordinate system to define an invariant geometry. This is a concept taught in elementary school where these kids do understand, but the GRists seem to have great difficulties grasping. <sigh>
There are more. Need I to continue? These should be enough for one night.
Thanks for asking.
Eric Gisse - 01 Apr 2007 07:14 GMT > > In another post I mentioned there are no unresolved foundational issues with > > GR. And to the best of my knowledge there isn't. But an actual [quoted text clipped - 7 lines] > solid mathematics in which physics is still not mathematics, GR's > foundation is actually very faulty. Here are the basics. Here comes another Koobee Wublee "sh.t and run" post.
> 1. There are many ways to write the geodesic equation where each way > would yield a different set of connection coefficients. Christoffel's > way is not unique, and the Christoffel symbols are not unique. What is interesting is that you continue to assert this despite having no evidence that it is true.
I asked you to prove this statement once before, and you just ran off. It surprises me not in the least to see it again. Isn't it interesting to see how often you criticize people for not 'showing the math' when you have never, not even once, shown yours?
> 2. The covariant derivative was an operator suggested by Ricci based > on the non-unique Christoffel symbols of the second kind. For each > set of connection coefficients to the geodesic equation, there is one > such operator where each such operator is very different from the > other. An even more idiotic conclusion which follows from a dumb premise.
> 3. The Riemann curvature tensor was derived through the covariant > derivative. It was designed by Ricci. There are many but > mathematically different ways to present the Riemann curvature > tensor. Ricci only found one. Show us another, then show that they are different.
> 4. The Ricci curvature tensor was designed by Ricci's faithful > student Levi-Civita from the Riemann curvature tensor. [quoted text clipped - 3 lines] > Hilbert himself. The main ingredient was none other than the Ricci > curvature tensor. Have you already forgotten what Bill Hobba was asking about?
> 6. The Einstein field equations are derived by dividing that > Lagrangian Hilbert pulled out of his *ss with each of the elements to > the metric. Since there are 16 elements, there are indeed 16 such > equations. Duh. 4x4 = 16.
At any rate, why is that a problem? There is more than one path to the field equations.
> 7. The Schwarzschild metric is one of the infinite numbers of > solutions to the ordinary differential equations (the field equations) > for static spacetime. The creation of a black hole is strictly how > one would embrace one non-unique solution to another. It allows > anyone to play God through a pen and a piece of paper. BZZZT, TRY AGAIN.
The field equations are _partial_ differential equations but the symmetry of the situation allows them to be reduced to ordinary differential equations, whose solutions are unique if they satisfy the ODEs as well as the boundary conditions.
> The above represent the mathematical foundation of GR. It shows how > the curvature in spacetime is shaped by the mathematical tool invented > by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert, > Schwarzschild, and Einstein are His prophets. You sure do seem to whine a lot about who did what.
> However, the general concept of the metric being invariant is totally > wrong. The curvature in spacetime must be invariant. This should be [quoted text clipped - 5 lines] > where these kids do understand, but the GRists seem to have great > difficulties grasping. <sigh> The only one who is having "difficulties grasping" here is YOU. YOU have been told _countless_ times that while the coordinate representation of a tensor is obviously coordinate dependent, the tensor itself is not. Shortly after being told that, it has also been asked of you to prove the assertion wrong with a counter-example or some kind of actual mathematical proof.
To date, all you have been able to do is do a coordinate transformation and say "LOOK! THEY ARE DIFFERENT!" without actually proving that they are different. Cmon KW, follow your own maxim: SHOW THE MATH.
For reference, here are some of your previous dumbass claims that you have never proven.
* The Minkowski metric does not admit a Lagrangian. I explicitly derived the Lagrangian for you....and you never posted to that thread again. * A coordinate transformation can introduce curvature. You kept asserting it, but you never proven it. * A tensor is not invariant under a coordinate transformation. You continue asserting it, but you have never proven it. * The surface area for space of constant radius for the Schwarzschild metric is 4pi*r^2. Despite the careful proof to the contrary by JanPB, you continue to assert that it is.
> There are more. Need I to continue? These should be enough for one > night. Yes, how about some actual mathematics to prove your dumbass assertions?
haha like that will ever happen. You don't even have a firm grasp of multivariable calculus.
> Thanks for asking. Koobee Wublee - 05 Apr 2007 07:23 GMT <yawn> The supposed-to-be educated don't have anything worthwhile to say. I guess I have some time left to address their apprentices.
> Here comes another Koobee Wublee "sh.t and run" post. Not really. I reserve the right not to answer any personal insults. <shrug>
> > 1. There are many ways to write the geodesic equation where each way > > would yield a different set of connection coefficients. Christoffel's > > way is not unique, and the Christoffel symbols are not unique. > > What is interesting is that you continue to assert this despite having > no evidence that it is true. Don't blame me on your failure in mathematics. <shrug>
> I asked you to prove this statement once before, and you just ran off. > It surprises me not in the least to see it again. Isn't it interesting > to see how often you criticize people for not 'showing the math' when > you have never, not even once, shown yours? I have addressed them already in my past posts.
> > 2. The covariant derivative was an operator suggested by Ricci based > > on the non-unique Christoffel symbols of the second kind. For each [quoted text clipped - 3 lines] > > An even more idiotic conclusion which follows from a dumb premise. You don't understand what covariant derivative is all about. <shrug>
> > 3. The Riemann curvature tensor was derived through the covariant > > derivative. It was designed by Ricci. There are many but > > mathematically different ways to present the Riemann curvature > > tensor. Ricci only found one. > > Show us another, then show that they are different. If you have the math skills, you can see it yourself. If not, you are on your own. <shrug>
> > 4. The Ricci curvature tensor was designed by Ricci's faithful > > student Levi-Civita from the Riemann curvature tensor. No protest?
> > 5. The Lagrangian to the Einstein-Hilbert action was designed or > > cooked up as an alchemist would have done in the medieval times by > > Hilbert himself. The main ingredient was none other than the Ricci > > curvature tensor. > > Have you already forgotten what Bill Hobba was asking about? No. Have you?
> > 6. The Einstein field equations are derived by dividing that > > Lagrangian Hilbert pulled out of his *ss with each of the elements to > > the metric. Since there are 16 elements, there are indeed 16 such > > equations. > > Duh. 4x4 = 16. Yes. <shrug>
> At any rate, why is that a problem? There is more than one path to the > field equations. No, there is only one solid way to derive the field equations.
> > 7. The Schwarzschild metric is one of the infinite numbers of > > solutions to the ordinary differential equations (the field equations) [quoted text clipped - 3 lines] > > BZZZT, TRY AGAIN. 7. The Schwarzschild metric is one of the infinite numbers of solutions to the ordinary differential equations (the field equations) for static spacetime. The creation of a black hole is strictly how one would embrace one non-unique solution to another. It allows anyone to play God through a pen and a piece of paper.
> The field equations are _partial_ differential equations but the > symmetry of the situation allows them to be reduced to ordinary > differential equations, whose solutions are unique if they satisfy the > ODEs as well as the boundary conditions. 'The field equations are _partial_ differential equations' PERIOD. <shrug>
> > The above represent the mathematical foundation of GR. It shows how > > the curvature in spacetime is shaped by the mathematical tool invented > > by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert, > > Schwarzschild, and Einstein are His prophets. > > You sure do seem to whine a lot about who did what. This is what history is all about. <shrug>
> > However, the general concept of the metric being invariant is totally > > wrong. The curvature in spacetime must be invariant. This should be [quoted text clipped - 7 lines] > > The only one who is having "difficulties grasping" here is YOU. Wrong. GR concepts and math is so simple to me. It will be so simple to you too if you understand what I am saying. <shrug>
> YOU > have been told _countless_ times that while the coordinate > representation of a tensor is obviously coordinate dependent, the > tensor itself is not. Shortly after being told that, it has also been > asked of you to prove the assertion wrong with a counter-example or > some kind of actual mathematical proof. You have been told COUNTLESS TIMES by me that holding the metric to be invariant is absurd and physically invalid.
> To date, all you have been able to do is do a coordinate > transformation and say "LOOK! THEY ARE DIFFERENT!" without actually > proving that they are different. Cmon KW, follow your own maxim: SHOW > THE MATH. I did. You must be sleeping. <shrug>
> For reference, here are some of your previous dumbass claims that you > have never proven. > > * The Minkowski metric does not admit a Lagrangian. I explicitly > derived the Lagrangian for you....and you never posted to that thread > again. Your derivation is stupid and nonsense. I reserve the right not to answer any posts so unintelligently put together. You need to get over with that. <shrug>
> * A coordinate transformation can introduce curvature. You kept > asserting it, but you never proven it. I never said that.
> * A tensor is not invariant under a coordinate transformation. You > continue asserting it, but you have never proven it. Whatever a tensor is. However, the metric is dependent on the choice of the coordinate system. <shrug>
> * The surface area for space of constant radius for the Schwarzschild > metric is 4pi*r^2. Despite the careful proof to the contrary by JanPB, > you continue to assert that it is. Yes, the surface area of a sphere to any observer is always 4pi R^2 regardless how much space is curved where R is the observed radius. <shrug>
> > There are more. Need I to continue? These should be enough for one > > night. > > Yes, how about some actual mathematics to prove your dumbass > assertions? Like what?
> haha like that will ever happen. You don't even have a firm grasp of > multivariable calculus. I can always surprise you that I promise. <yawn>
Eric Gisse - 05 Apr 2007 10:41 GMT > <yawn> The supposed-to-be educated don't have anything worthwhile to > say. I guess I have some time left to address their apprentices. Yes, retirement....all those hours, so many things to fill them with.
> > Here comes another Koobee Wublee "sh.t and run" post. > [quoted text clipped - 9 lines] > > Don't blame me on your failure in mathematics. <shrug> Interesting - your failure to disclose the relevant mathematics is somehow *my* failure.
Ignorance is strength, war is peace, freedom is slavery...
> > I asked you to prove this statement once before, and you just ran off. > > It surprises me not in the least to see it again. Isn't it interesting > > to see how often you criticize people for not 'showing the math' when > > you have never, not even once, shown yours? > > I have addressed them already in my past posts. Isn't it interesting how the proofs always end up in 'past posts'?
Perhaps you could show me the 'past posts' that contain the proofs or you claims - or would you like to shift the blame to me again?
> > > 2. The covariant derivative was an operator suggested by Ricci based > > > on the non-unique Christoffel symbols of the second kind. For each [quoted text clipped - 15 lines] > If you have the math skills, you can see it yourself. If not, you are > on your own. <shrug> Your failures are not mine. Either you are able to, or you are not. Since you steadfastly refuse to document your claims, it appears that you are not.
> > > 4. The Ricci curvature tensor was designed by Ricci's faithful > > > student Levi-Civita from the Riemann curvature tensor. > > No protest? The claim isn't worth arguing about since who developed what has no influence on the mathematics.
> > > 5. The Lagrangian to the Einstein-Hilbert action was designed or > > > cooked up as an alchemist would have done in the medieval times by [quoted text clipped - 13 lines] > > Yes. <shrug> Just making sure you agree since you seem to have trouble with the idea that there are 16 equations.
> > At any rate, why is that a problem? There is more than one path to the > > field equations. > > No, there is only one solid way to derive the field equations. "Solid" is a fungible concept, especially when the answer does not change.
> > > 7. The Schwarzschild metric is one of the infinite numbers of > > > solutions to the ordinary differential equations (the field equations) [quoted text clipped - 9 lines] > one would embrace one non-unique solution to another. It allows > anyone to play God through a pen and a piece of paper. ODEs have ONE *unique* solution. This has been explained to you before - crack open any worthwhile ODE text [Boyce and DiPrima is adequately deep, as an example] and read it.
The Schwarzschild solution is the *unique* vacuum solution to the field equations. This has been explained to you before. Open up a textbook and read about the proof of Birkhoff's theorem.
> > The field equations are _partial_ differential equations but the > > symmetry of the situation allows them to be reduced to ordinary [quoted text clipped - 3 lines] > 'The field equations are _partial_ differential equations' PERIOD. > <shrug> True, but all those components of the Ricci tensor being equal to zero via it being a vacuum solution, as well as the spherical symmetry via constraining the form of the metric anzatz happens to make the relevant quantities the solutions of a few ODEs rather than PDEs.
You can see an adequate demonstration of this in any proof of Birkhoff's theorem. Of course, that would require you to go to the library and _read_ a book on GR which by all accounts you have, and will, never do.
FYI, you seem to be confused - are the field equations PDEs or ODEs? Do you know the difference?
> > > The above represent the mathematical foundation of GR. It shows how > > > the curvature in spacetime is shaped by the mathematical tool invented [quoted text clipped - 19 lines] > Wrong. GR concepts and math is so simple to me. It will be so > simple to you too if you understand what I am saying. <shrug> Perhaps you could spend some time and write down this simple math rather than argue for months?
> > YOU > > have been told _countless_ times that while the coordinate [quoted text clipped - 5 lines] > You have been told COUNTLESS TIMES by me that holding the metric to be > invariant is absurd and physically invalid. Yes, but my statements have the backing of mathematical proofs. Your statements have the backing of...what, exactly?
I am yet to see you cite one single reference that agrees with your contrary viewpoint, much less an actual proof of any of them.
> > To date, all you have been able to do is do a coordinate > > transformation and say "LOOK! THEY ARE DIFFERENT!" without actually > > proving that they are different. Cmon KW, follow your own maxim: SHOW > > THE MATH. > > I did. You must be sleeping. <shrug> Just because I am typing this in bed does not mean I am sleeping.
I saw your "proof", not only did you transform the line element incorrectly, you failed to realize it even after it was carefully explained to you by both myself and JanPB. Your entire proof consisted of "hey, I did a transformation and the metric looks different! It must be different!", and of course you couldn't even do the transformation correctly.
> > For reference, here are some of your previous dumbass claims that you > > have never proven. [quoted text clipped - 6 lines] > answer any posts so unintelligently put together. You need to get > over with that. <shrug> Why is it "stupid and nonsense"? Did you not understand it? Would you like me to go through it again, slowly?
I derived it straight from the metric without referring to the more general formula in what was about a third of a page of math on paper. However, it was derived from the more general formula in about a line here:
http://groups.google.com/group/sci.physics.relativity/msg/872133b163bba475?dmode =source
Isn't it interesting that there are now two disproofs or your asinine statement? How many do you require before you admit you are wrong and drop it?
> > * A coordinate transformation can introduce curvature. You kept > > asserting it, but you never proven it. > > I never said that. Yes, you did.
http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4?dmode =source
Firefox already had the relevant search phrase in its' history from the last time you did this song and dance.
> > * A tensor is not invariant under a coordinate transformation. You > > continue asserting it, but you have never proven it. > > Whatever a tensor is. However, the metric is dependent on the choice > of the coordinate system. <shrug> No, the metric _components_ are dependent. The metric itself is not.
Learn to tell the difference.
> > * The surface area for space of constant radius for the Schwarzschild > > metric is 4pi*r^2. Despite the careful proof to the contrary by JanPB, [quoted text clipped - 3 lines] > regardless how much space is curved where R is the observed radius. > <shrug> Yet the mathematical proof of a contrary claim hasn't been refuted by you. Why is that?
Why is it that everyone but you substantiates their arguments with mathematics? Why is it you are the only one who will see the proof and ignore the result because it proves you wrong?
> > > There are more. Need I to continue? These should be enough for one > > > night. [quoted text clipped - 8 lines] > > I can always surprise you that I promise. <yawn> Oh good. Then prove the statement that the Schwarzschild metric has a surface area of 4pi*r^2 for constant t and r.
Barry - 05 Apr 2007 18:58 GMT > ODEs have ONE *unique* solution. This has been explained to you before > - crack open any worthwhile ODE text [Boyce and DiPrima is adequately > deep, as an example] and read it. As everybody knows, I don't know any maths or physics, never having been to school.
In my ognorance, I thought that ODE's were somewhat analogous to quadratic equations.
More specifically, I used to think that an "n"th-order ODE would have "n" linearly independent solutions and that any linear combination of those solutions would also be a solution.
I really must look at one of those text books one of these days.
Barry
Eric Gisse - 05 Apr 2007 20:22 GMT > > ODEs have ONE *unique* solution. This has been explained to you before > > - crack open any worthwhile ODE text [Boyce and DiPrima is adequately [quoted text clipped - 5 lines] > In my ognorance, I thought that ODE's were somewhat analogous to > quadratic equations. Not really.
The characteristic polynomial for a linear, constant coefficient ODE is a n'th order polynomial. But that is about it.
> More specifically, I used to think that an "n"th-order ODE would have > "n" linearly independent solutions and that any linear combination of > those solutions would also be a solution. That is true only for linear ODEs.
> I really must look at one of those text books one of these days. > > Barry Koobee Wublee - 05 Apr 2007 19:19 GMT > > <yawn> The supposed-to-be educated don't have anything worthwhile to > > say. I guess I have some time left to address their apprentices. > > Yes, retirement....all those hours, so many things to fill them with. I look forward towards my own retirement.
> > Don't blame me on your failure in mathematics. <shrug> > > Interesting - your failure to disclose the relevant mathematics is > somehow *my* failure. No, your failure to understand the relevant mathematics is not my fault. <shrug>
> Ignorance is strength, war is peace, freedom is slavery... This can only be embraced by someone as crooked as Gisse and incompetence too.
> > I have addressed them already in my past posts. > > Isn't it interesting how the proofs always end up in 'past posts'? Yes, it is called history. <shrug>
> Perhaps you could show me the 'past posts' that contain the proofs or > you claims - or would you like to shift the blame to me again? You can find them.
> > If you have the math skills, you can see it yourself. If not, you are > > on your own. <shrug> > > Your failures are not mine. Either you are able to, or you are not. > Since you steadfastly refuse to document your claims, it appears that > you are not. We are talking about you and your mathematical skills. Or rather lack of it. <shrug>
> > 4. The Ricci curvature tensor was designed by Ricci's faithful > > student Levi-Civita from the Riemann curvature tensor. [quoted text clipped - 3 lines] > The claim isn't worth arguing about since who developed what has no > influence on the mathematics. But this is about history. <shrug>
> Duh. 4x4 = 16. > > > Yes. <shrug> > > Just making sure you agree since you seem to have trouble with the > idea that there are 16 equations. Wrong.
> > No, there is only one solid way to derive the field equations. > > "Solid" is a fungible concept, especially when the answer does not > change. Your loss, not mine.
> ODEs have ONE *unique* solution. This has been explained to you before > - crack open any worthwhile ODE text [Boyce and DiPrima is adequately > deep, as an example] and read it. Not if they have some dependencies on each other.
> The Schwarzschild solution is the *unique* vacuum solution to the > field equations. This has been explained to you before. Open up a > textbook and read about the proof of Birkhoff's theorem. This has been explained to you numerous times by me on why the Schwarzschild solution is not unique. <shrug>
> > 'The field equations are _partial_ differential equations' PERIOD. > > <shrug> [quoted text clipped - 8 lines] > library and _read_ a book on GR which by all accounts you have, and > will, never do. Birkhoff's theorem states that any spherically symmetric solution to the field equations in vacuum must be static and asymptotically flat. Spherically symmetric means the geometry in spacetime must be independent of any angular displacements. Static means the metric is independent of time. Therefore, it really does not say the Schwarzschild metric is the only solution. However, there are other solutions to the field without evoking the Cosmological constant that would fit an expanding universe. Therefore, Birkhoff's theorem is wrong on that account. <shrug>
> FYI, you seem to be confused - are the field equations PDEs or ODEs? Oh, trying to trick me to do your homework for you.
> Do you know the difference? Yes.
> > Wrong. GR concepts and math is so simple to me. It will be so > > simple to you too if you understand what I am saying. <shrug> > > Perhaps you could spend some time and write down this simple math > rather than argue for months? I did. <shrug>
> > You have been told COUNTLESS TIMES by me that holding the metric to be > > invariant is absurd and physically invalid. > > Yes, but my statements have the backing of mathematical proofs. Your > statements have the backing of...what, exactly? No, your statements are BS. <shrug>
> I am yet to see you cite one single reference that agrees with your > contrary viewpoint, much less an actual proof of any of them. That is because I am right, and I am the first to point out the errors made by the forefathers of GR. <shrug> For example, the metric must be observer dependent.
> > I did. You must be sleeping. <shrug> > > Just because I am typing this in bed does not mean I am sleeping. Close enough. <shrug>
> I saw your "proof", not only did you transform the line element > incorrectly, you failed to realize it even after it was carefully > explained to you by both myself and JanPB. Your entire proof consisted > of "hey, I did a transformation and the metric looks different! It > must be different!", and of course you couldn't even do the > transformation correctly. You have failed to understand what I did. <shrug>
> > Your derivation is stupid and nonsense. I reserve the right not to > > answer any posts so unintelligently put together. You need to get [quoted text clipped - 5 lines] > I derived it straight from the metric without referring to the more > general formula in what was about a third of a page of math on paper.
> However, it was derived from the more general formula in about a line > here: > > http://groups.google.com/group/sci.physics.relativity/msg/872133b163bba475 What is this? Two buffoon trying to out-design his own version of the Lagrangian?
> Isn't it interesting that there are now two disproofs or your asinine > statement? How many do you require before you admit you are wrong and > drop it? Not once.
> * A coordinate transformation can introduce curvature. You kept > asserting it, but you never proven it. [quoted text clipped - 4 lines] > > http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4 Where is the statement saying a coordinate transformation can introduce curvature?
> Firefox already had the relevant search phrase in its' history from > the last time you did this song and dance. Good. Keep it for eternity.
> > Whatever a tensor is. However, the metric is dependent on the choice > > of the coordinate system. <shrug> > > No, the metric _components_ are dependent. The metric itself is not. That is total BS. If you have a different lens to a pair of glasses, the prescription is different.
> Learn to tell the difference. Learn to try not to BS.
> > Yes, the surface area of a sphere to any observer is always 4pi R^2 > > regardless how much space is curved where R is the observed radius. > > <shrug> > > Yet the mathematical proof of a contrary claim hasn't been refuted by > you. Why is that? Yes, it did.
> Why is it that everyone but you substantiates their arguments with > mathematics? Why is it you are the only one who will see the proof and > ignore the result because it proves you wrong? You are entitle to hallucinate in your bed.
> > I can always surprise you that I promise. <yawn> > > Oh good. Then prove the statement that the Schwarzschild metric has a > surface area of 4pi*r^2 for constant t and r. Not now. It is time for you to go to sleep. Make sure you dream of Koobee Wublee coming to get you.
JanPB - 05 Apr 2007 20:28 GMT [...]
> Birkhoff's theorem states that any spherically symmetric solution to > the field equations in vacuum must be static and asymptotically flat. No, it doesn't say that. You've fallen into the standard "static" trap. What it says is that the solution _must have coefficients independent of the "t" variable_. (I guess you never went through solving the EFE assuming spherical symmetry _only_ or else you'd see it clearly from one of the relevant ODEs).
Of course independence of the "t" variable (together with the diagonal form of the metric presumed for the derivation) does translate into "static" outside the horizon but it doesn't imply static inside (in fact, it's not static inside as everyone knows).
> Spherically symmetric means the geometry in spacetime must be > independent of any angular displacements. Static means the metric is > independent of time. Therefore, it really does not say the > Schwarzschild metric is the only solution. It does. You misunderstood the statement of the theorem. I just noticed Wikipedia also states this incorrectly - is it where you get your physics from?
> However, there are other > solutions to the field without evoking the Cosmological constant that > would fit an expanding universe. Therefore, Birkhoff's theorem is > wrong on that account. <shrug> No, it is you who is wrong. It's not a big deal, just study it some more and don't waste your time on phantoms.
-- Jan Bielawski
Androcles - 05 Apr 2007 23:10 GMT > [...] >> >> Birkhoff's theorem states that any spherically symmetric solution to >> the field equations in vacuum must be static and asymptotically flat. > > No, it doesn't say that. Hahaha!
> You've fallen into the standard "static" > trap. What it says is that the solution _must have coefficients > independent of the "t" variable_. Oh wow! HAHAHA!
> (I guess Guessing is always the Polish answer.
Koobee Wublee - 06 Apr 2007 01:28 GMT > > Birkhoff's theorem states that any spherically symmetric solution to > > the field equations in vacuum must be static and asymptotically flat. > > No, it doesn't say that. You've fallen into the standard "static" > trap. I don't get this 'static trap' of yours.
> What it says is that the solution _must have coefficients > independent of the "t" variable_. Is that not what 'static' means?
> (I guess you never went through > solving the EFE assuming spherical symmetry _only_ or else you'd see > it clearly from one of the relevant ODEs). Yes, I did.
> Of course independence of the "t" variable (together with the diagonal > form of the metric presumed for the derivation) does translate into > "static" outside the horizon but it doesn't imply static inside (in > fact, it's not static inside as everyone knows). Inside, outside, then inside? Done with dancing around? Make up your mind. Do you not know Birkhoff's theorem only applies to vacuum? Is that outside or inside? Outside or inside of what?
> > Spherically symmetric means the geometry in spacetime must be > > independent of any angular displacements. Static means the metric is > > independent of time. Therefore, it really does not say the > > Schwarzschild metric is the only solution. > > It does. You misunderstood the statement of the theorem. You need to understand the theorem first before making that judgment. <shrug>
> I just > noticed Wikipedia also states this incorrectly - is it where you get > your physics from? So, just how many schools of physics are there? I suppose you only endorse one school of interpretations. Does that sound familiar? My God is always more powerful than someone else's. <shrug>
You must have studied under Reverend Hammond.
> > However, there are other > > solutions to the field without evoking the Cosmological constant that [quoted text clipped - 3 lines] > No, it is you who is wrong. It's not a big deal, just study it some > more and don't waste your time on phantoms. Done that. You are still wrong.
JanPB - 06 Apr 2007 01:48 GMT > > > Birkhoff's theorem states that any spherically symmetric solution to > > > the field equations in vacuum must be static and asymptotically flat. [quoted text clipped - 8 lines] > > Is that not what 'static' means? Independence of a variable means "static" (ignoring the difference between "static" and "stationary" for the time being) if that variable is timelike. In Schwarzschild's chart the coordinate labelled by "t" is not timelike inside the horizon, so independence of "t" there means that the region is spatially homogeneous in that direction.
> > (I guess you never went through > > solving the EFE assuming spherical symmetry _only_ or else you'd see [quoted text clipped - 10 lines] > mind. Do you not know Birkhoff's theorem only applies to vacuum? Is > that outside or inside? Outside or inside of what? I'm just using the standard shortcut: "outside" means "r>2M", "inside" means "r<2M". Easier to type.
> > > Spherically symmetric means the geometry in spacetime must be > > > independent of any angular displacements. Static means the metric is [quoted text clipped - 13 lines] > endorse one school of interpretations. Does that sound familiar? My > God is always more powerful than someone else's. <shrug> This is just a question of mathematics, so interpretations don't enter into it. Either there are multiple solutions to EFE in the sph. symm. case or not. The answer is the solution is unique.
> You must have studied under Reverend Hammond. That would be the sight! :-)
> > > However, there are other > > > solutions to the field without evoking the Cosmological constant that [quoted text clipped - 5 lines] > > Done that. You are still wrong. I know that I cannot convince you here - it's impossible to teach a complex subject on an ASCII BB. Some readers might be interested in learning more of that stuff though.
-- Jan Bielawski
Koobee Wublee - 06 Apr 2007 08:20 GMT > Independence of a variable means "static" (ignoring the difference > between "static" and "stationary" for the time being) if that variable > is timelike. In our conversation, you have foliaged the word 'static' into 'stationary' and 'non-timelike'. The best BS'ers are the ones who can come up more words to describe one thing. Was it not one of the Confucius sayings?
> In Schwarzschild's chart the coordinate labelled by "t" > is not timelike inside the horizon, so independence of "t" there means > that the region is spatially homogeneous in that direction. You still allow yourself to be trapped in the mental black hole. <shrug> In the meantime, we are discussing about solutions in vacuum.
> > Inside, outside, then inside? Done with dancing around? Make up your > > mind. Do you not know Birkhoff's theorem only applies to vacuum? Is > > that outside or inside? Outside or inside of what? > > I'm just using the standard shortcut: "outside" means "r>2M", "inside" > means "r<2M". Easier to type. Inside does not apply with any mathematics we have. So, how the heck can you apply the Birkhoff's theorem inside the event horizon? You do not have any mathematics to back you back in this case.
> > So, just how many schools of physics are there? I suppose you only > > endorse one school of interpretations. Does that sound familiar? My > > God is always more powerful than someone else's. <shrug> > > This is just a question of mathematics, so interpretations don't enter > into it. May I remind you that you have not addressed any mathematics!
> Either there are multiple solutions to EFE in the sph. symm. > case or not. The answer is the solution is unique. Mechanically, indeed there are infinite number of independent and unique solutions. Philosophically, you have interpreted all solutions are identical due to your misconception that the metric is invariant. <shrug>
> > You must have studied under Reverend Hammond. > > That would be the sight! :-) Indeed. :-)
> > Done that. You are still wrong. > > I know that I cannot convince you here - it's impossible to teach a > complex subject on an ASCII BB. Some readers might be interested in > learning more of that stuff though. The outcome of a physical hypothesis can only be decided in the arena of mathematics.
Ps. I know you have backed away from how an observer would observe the surface area of a sphere in curved space. You and Mr. McCullough were wrong about that one. Please tell your student/admirer Gisse about this one. So, he can have a good-night sleep. Thank you.
JanPB - 06 Apr 2007 17:34 GMT > > Independence of a variable means "static" (ignoring the difference > > between "static" and "stationary" for the time being) if that variable [quoted text clipped - 4 lines] > come up more words to describe one thing. Was it not one of the > Confucius sayings? I just prefer to focus on one issue at a time. The issue at hand was that deriving Schwarzschild's solution yielded particular metric coefficients not depending on the "t" variable which you concluded meant the metric was necessarily static. All I said was that this implication follows only for locations outside the horizon (r>2M) where the letter "t" denotes a timelike coordinate.
> > In Schwarzschild's chart the coordinate labelled by "t" > > is not timelike inside the horizon, so independence of "t" there means > > that the region is spatially homogeneous in that direction. > > You still allow yourself to be trapped in the mental black hole. > <shrug> In the meantime, we are discussing about solutions in vacuum. This is pure rhetoric.
> > > Inside, outside, then inside? Done with dancing around? Make up your > > > mind. Do you not know Birkhoff's theorem only applies to vacuum? Is [quoted text clipped - 4 lines] > > Inside does not apply with any mathematics we have. Of course it does.
> So, how the heck > can you apply the Birkhoff's theorem inside the event horizon? You do > not have any mathematics to back you back in this case. I've just told you. The derivation of the solution works both inside and outside and in either case one of the Einstein equations says that the metric coefficients (in the presumed diagonal form of the metric) do not depend on the "t" variable. Examining the signs of the coefficients this implies that the metric is static outside and spatially homogeneous inside.
> > > So, just how many schools of physics are there? I suppose you only > > > endorse one school of interpretations. Does that sound familiar? My [quoted text clipped - 4 lines] > > May I remind you that you have not addressed any mathematics! I'll do it in my next post then. I was only addressing the general flow of logic in these posts as you had this false implication at the forefront: [Birkhoff]=>[static].
> > Either there are multiple solutions to EFE in the sph. symm. > > case or not. The answer is the solution is unique. [quoted text clipped - 3 lines] > are identical due to your misconception that the metric is invariant. > <shrug> No, they are all identical in the sense of being physically indistinguishable by any means.
> > > You must have studied under Reverend Hammond. > [quoted text clipped - 13 lines] > Ps. I know you have backed away from how an observer would observe > the surface area of a sphere in curved space. That's total news to me. Surface areas are integrals of 2D volume forms, and these are given in terms of the metric - there is no wiggle room here, everything is set.
> You and Mr. McCullough > were wrong about that one. Please tell your student/admirer Gisse > about this one. So, he can have a good-night sleep. Thank you. More rhetoric. In general n-dim. volume = Integral(dvol), where "dvol" is the volume form (the Hodge dual of 1). No one can "back away" from it.
-- Jan Bielawski
JanPB - 06 Apr 2007 21:31 GMT [...]
> > May I remind you that you have not addressed any mathematics! > > I'll do it in my next post then. I was only addressing the general > flow of logic in these posts as you had this false implication at the > forefront: [Birkhoff]=>[static]. Here is the relevant mathematics. You'll see precisely how in the Schwarzschild case the independence of metric coefficients of t follows from the EFE, not staticity everywhere.
We seek a spherically symmetric metric of signature 2 satisfying the Einstein equation in vacuum:
R_ab = 0
The assumption of symmetry allows us to assume WLOG that the metric has the general form like so:
ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2,
where r is a coordinate labelling the spheres of symmetry and:
dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
It turns out that even this reasonably nice general form results in an unmanageable computational mess so one simplifies the general form further by a local change of coordinates which cleverly removes the "dr dt" term and also relabels the spheres of symmetry by their areas (divided by 4pi). (I can supply more details on this final coordinate change if you wish.) This results in the following two possible simplified, diagonal, forms consistent with the given signature constraint:
either:
ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dO^2 [1]
or:
ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
where A(r,t) and B(r,t) are some functions which are never zero (or else the metric would not be rank-4).
NB: this local coordinate change happens to rely on dividing by certain function, so any solutions for metric coefficients we may obtain by proceeding from these two general forms may not be defined over the spacetime events for which the function we divided by is zero.
[In fact (getting slightly ahead of myself here) this is precisely what happens: the function we divide by is zero for r=2M, hence the (in)famous exclusion of the locus r=2M from the domain of the metric coefficients in the Schwarzschild form. There is of course no reason to presume this is an actual physical exclusion as it potentially results from the man-made work-saving transformation. So whether it's a true singularity must be verified separately. It turns out - as is well known - that the metric is well-defined for r=2M, it is merely non-diagonalizable there hence the forms [1] and [2] break down there - it's just a case of being too greedy/lazy.]
OK, so we have those two forms potentially satisfying the EFE. The maximal domain possible (unless the EFE themselves somehow end up restricting it) is:
-infty < t < +infty, 0 < r < +infty, 0 < theta < 2pi, 0 < phi < pi.
I'm going to skip the exact details of deriving the equations below (we can go into it later if necessary). I'm using Cartan's method of moving frames, and after computing the connection and curvature forms the Einstein equation R_ab=0 reduces to the following four PDEs:
R_tt = 0 R_rr = 0 R_theta theta = 0 R_rt = 0
...with the remaining equations being either identities or equivalent to those four ones. (For example, R_phi phi = R_theta theta.)
The exact forms of the four left-hand sides are, few pages of calculations later, respectively (excuse the mess, it's the fourth, simplest, equation that's interesting to us):
(1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0 (1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0 -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0 (2/rAB^2)*dB/dt = 0
...where "-+" means use the "-" sign for the first general form of the diagonal metric [1], use "+" for the second [2]. Also "d" obviously denotes partial derivatives.
Note the fourth equation R_rt = 0 says then (cancelling the factor in front):
dB/dt = 0
...in other words, B(r,t) is independent of t, both in case [1] and [2]. This means that B is only a function of r which means that the first two PDEs simplify quite a bit:
(1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0 (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
This is actually a pair of _ODEs_ (r is the only variable in them) which remarkably simplify to the following (denoting the r-derivatives by primes now):
A'/A + B'/B = 0
i.e.,
(ln |A|)' + (ln |B|)' = 0
hence:
|AB| = C (a positive constant of integration)
i.e.,
AB = C (an arbitary nonzero constant of integration).
The important thing as far as Birkhoff is concerned is that this means that:
B = constant/A
...in other words B(r,t) is independent of t as well, in both [1] and [2].
Plugging A(r) and B(r)=1/A(r) into the remaining (third) PDE yields another _ordinary_ differential equation whose exact form and details of the solution are less interesting to us at this point. Suffice to say it yields the well-known Schwarzschild form of the metric components:
ds^2 = -(1 - 2M/r) dt^2 + 1/(1 - 2M/r) dr^2 + r^2 dO^2 (case [1])
...with the domain restricted to r > 2M, and:
ds^2 = (2M/r - 1) dt^2 - 1/(2M/r - 1) dr^2 + r^2 dO^2 (case [2])
...with the domain restricted to 0 < r < 2M.
[As I mentioned earlier, the case of r=2M must be verified separately, as logically at this point we don't know if the exclusion is real or merely due to our coordinate change in the beginning.]
A quick examination of these two formulas shows that they can be written as a single equation by a simple sign switch:
ds^2 = -(1 - 2M/r) dt^2 + 1/(1 - 2M/r) dr^2 + r^2 dO^2
...with the domain restricted to r>0 AND r=/=2M.
We note that this solution is static for r>2M and spatially homogeneous for r<2M. It is also unique over this domain as it's a solution to an ODE system.
Incidentally, since the coordinate denoted by "r" inside the horizon is timelike, it's instructive to switch around the letters "r" and "t" in case [2] because of our strong mental habit of thinking of "t" as denoting time:
ds^2 = -1/(2M/t - 1) dt^2 + (2M/t - 1) dr^2 + t^2 dO^2,
...with r arbitrary and t > 0 only (we "fall off" spacetime at t=0).
This is what the metric inside "really" is.
-- Jan Bielawski
Androcles - 06 Apr 2007 22:49 GMT > We seek a spherically symmetric metric of signature 2 satisfying the > Einstein equation Another week, another post full of idiocies about another book full of idiocies.
Eric Gisse - 07 Apr 2007 03:59 GMT On Apr 6, 1:49 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> wrote:
> > We seek a spherically symmetric metric of signature 2 satisfying the > > Einstein equation > > Another week, another post full of idiocies about another book full of > idiocies. Then why participate in this newsgroup?
JanPB - 07 Apr 2007 04:28 GMT On Apr 6, 2:49 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> wrote:
> > We seek a spherically symmetric metric of signature 2 satisfying the > > Einstein equation > > Another week, another post full of idiocies about another book full of > idiocies. Wait a moment, which book? I was just reading David Parlett's "History of Card Games" but I guess that's not what you meant?
-- Jan Bielawski
Androcles - 07 Apr 2007 13:35 GMT > On Apr 6, 2:49 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> > wrote: [quoted text clipped - 6 lines] > Wait a moment, which book? I was just reading David Parlett's "History > of Card Games" but I guess that's not what you meant? You tell me, they are your words, idiot. Have a pleasant divide-by-zero, Hawking is as good at it as Einstein was. He even wrote "A Brief History of Time", which is briefer than the history of card games but funnier.
Koobee Wublee - 07 Apr 2007 01:53 GMT > We seek a spherically symmetric metric of signature 2 satisfying the > Einstein equation in vacuum: > > R_ab = 0 Or (G_ab = 0)
> The assumption of symmetry allows us to assume WLOG that the metric > has the general form like so: WLOG?
> ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2, > [quoted text clipped - 18 lines] > > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2] If you are describing the same invariant spacetime, ds^2, then your dr, dt must be different due to your coordinate transformation.
ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
I don't agree with (r^2 dO^2).
> where A(r,t) and B(r,t) are some functions which are never zero (or > else the metric would not be rank-4). Rank-4?
> NB: this local coordinate change happens to rely on dividing by > certain function, so any solutions for metric coefficients we may > obtain by proceeding from these two general forms may not be defined > over the spacetime events for which the function we divided by is > zero. This transformation is not necessary a division transformation. Why do you think it is so?
> [In fact (getting slightly ahead of myself here) this is precisely > what happens: the function we divide by is zero for r=2M, hence the [quoted text clipped - 6 lines] > non-diagonalizable there hence the forms [1] and [2] break down there > - it's just a case of being too greedy/lazy.] This makes no sense. It has the signature of Archimedes Plutonium. Are you the real person behind this handle?
> OK, so we have those two forms potentially satisfying the EFE. The > maximal domain possible (unless the EFE themselves somehow end up [quoted text clipped - 14 lines] > R_theta theta = 0 > R_rt = 0 No, with (3), you only get
** Rt't' = 0 ** Rr'r' = 0 ** RO'O' = 0
Recall
ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
You have reduced the spacetime from 4 variables to 3, remember?
> ...with the remaining equations being either identities or equivalent > to those four ones. (For example, R_phi phi = R_theta theta.) [quoted text clipped - 7 lines] > -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0 > (2/rAB^2)*dB/dt = 0 There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).
Actually, with time derivative, these equations are a lot more complicated than you have shown above.
> ...where "-+" means use the "-" sign for the first general form of the > diagonal metric [1], use "+" for the second [2]. Also "d" obviously [quoted text clipped - 4 lines] > > dB/dt = 0 Therefore showing (dB/dt = 0) is just BS.
> ...in other words, B(r,t) is independent of t, both in case [1] and > [2]. This means that B is only a function of r which means that the [quoted text clipped - 8 lines] > > A'/A + B'/B = 0 Bullsh*t. Your differential equations involve a second derivative of A! Where is it?
> i.e., > [quoted text clipped - 7 lines] > > AB = C (an arbitary nonzero constant of integration). Don't you feel like a prophet if you know the answer before hand?
> The important thing as far as Birkhoff is concerned is that this means > that: [quoted text clipped - 3 lines] > ...in other words B(r,t) is independent of t as well, in both [1] and > [2]. Only with your mathematical errors, that is so.
[The rest snipped due to the obvious mathematical error.]
Also, you have conveniently assumed (C^2 = r^2). See (3).
That is how the Birkhoff's theorem as the author of Wikipedia defined it.
Assuming (C^2 = r^2) which translates to spherically symmetric. For every C^2, there is a new set of metric that satisfies the field equations. Have you finally understood Mr. Crothers?
Assuming (static) which translates to all elements of the metric independent of time.
Then, you end up with much, much, much, and much simpler equations described below.
** (r / B^2) (dB/dr) - (1 / B) + 1 = 0 ** (r / A / B) (dA/dr) + (1 / B) - 1 = 0
These are actually not Ricci tensor equations but rather Einstein tensor equations. With all these simplifications, they are all the same.
For the spacetime described below,
ds^2 = A dt^2 - B dr^2 - C dO^2
Where (C = r^2)
You will see that you can easily solve for B to be the following.
** B = 1 / (1 + K / r)
Where (K = integration constant)
And A follows.
** A = K' (1 + K / r)
Where (K' = another integration constant)
For (C = (r + K)^2),
You get
** A = K' / (1 + K / r) ** B = 1 + K / r
In this case, you don't get a black hole.
For (C = ((r^3 + K^3)^(1/3) - K)^2),
You end up with Schwarzschild's original solution which does not manifest any black holes either.
There are infinite numbers of solutions more.
Eric Gisse - 07 Apr 2007 03:55 GMT > > We seek a spherically symmetric metric of signature 2 satisfying the > > Einstein equation in vacuum: [quoted text clipped - 7 lines] > > WLOG? "Without loss of generality"
It is a phrase you should be used to.
> > ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2, > [quoted text clipped - 21 lines] > If you are describing the same invariant spacetime, ds^2, then your > dr, dt must be different due to your coordinate transformation. They aren't coordinate transformations - they are representations of the same manifold just with different signatures - remember the diag(-1,1,1,1) and the diag(1,-1,-1.-1) representations of Minkowski space?
> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3] > > I don't agree with (r^2 dO^2). Why? ds^2 is a squared distance - it needs to have units of squared distance.
> > where A(r,t) and B(r,t) are some functions which are never zero (or > > else the metric would not be rank-4). [quoted text clipped - 23 lines] > This makes no sense. It has the signature of Archimedes Plutonium. > Are you the real person behind this handle? Of course it doesn't make sense - you are confused by C^n notation for being n-times differentiable and you don't know what WLOG means.
What he is saying is that you can't represent the manifold at r=2GM with a metric that has no cross terms.
> > OK, so we have those two forms potentially satisfying the EFE. The > > maximal domain possible (unless the EFE themselves somehow end up [quoted text clipped - 26 lines] > > You have reduced the spacetime from 4 variables to 3, remember? It isn't 3 variables - the variables are _still_ (t,r,theta,phi). Remember the definition of dO he wrote down for you?
> > ...with the remaining equations being either identities or equivalent > > to those four ones. (For example, R_phi phi = R_theta theta.) [quoted text clipped - 9 lines] > > There is no way you can get ((2/rAB^2)*dB/dt=0) from (3). Why not?
You have the metric anzatz - go derive the R_tr equation. You will find that he is, in fact, right if you do it correctly.
> Actually, with time derivative, these equations are a lot more > complicated than you have shown above. No, they are not.
> > ...where "-+" means use the "-" sign for the first general form of the > > diagonal metric [1], use "+" for the second [2]. Also "d" obviously [quoted text clipped - 6 lines] > > Therefore showing (dB/dt = 0) is just BS. *sigh*
It is the only way for the R_tr equation to hold for all t,r that exclude trivial A,B. I am amazed that you have trouble with something as basic as *this*.
> > ...in other words, B(r,t) is independent of t, both in case [1] and > > [2]. This means that B is only a function of r which means that the [quoted text clipped - 11 lines] > Bullsh*t. Your differential equations involve a second derivative of > A! Where is it? It went away. He integrated both equations with respect to r.
How can someone claim, with a straight face, that all of differential geometry is wrong when that someone struggles with regular one dimensional integration?
> > i.e., > [quoted text clipped - 9 lines] > > Don't you feel like a prophet if you know the answer before hand? Don't you feel like a jackass for insulting someone who obviously spent a lot of time writing down something you should already know, which you will not understand anyway?
Integrate the equations - you get ln(AB) = C. Which is equivalent to AB = C, where C in both cases are _completely arbitrary_ constants. Is that enough detail, or would you like it done in even more baby steps?
I guess JanPB was entirely unreasonable in expecting that the guru of Riemann be capable of understanding the concept of "integration".
> > The important thing as far as Birkhoff is concerned is that this means > > that: [quoted text clipped - 5 lines] > > Only with your mathematical errors, that is so. *laughs*
> [The rest snipped due to the obvious mathematical error.] > > Also, you have conveniently assumed (C^2 = r^2). See (3). YOU ARE THE ONE WHO WROTE (3). THE FUCKUP IS YOURS.
> That is how the Birkhoff's theorem as the author of Wikipedia defined > it. Who cares? You don't have the education to differentiate between good and bad yet - if ever.
> Assuming (C^2 = r^2) which translates to spherically symmetric. For > every C^2, there is a new set of metric that satisfies the field > equations. Have you finally understood Mr. Crothers? Hellooo he never wrote C^2 - you were the one who introduced it. Is your head that far up your a.s that you have to invent completely nonsensical arguments?
[snip remaining junk which is based upon a false premise]
Koobee Wublee - 07 Apr 2007 06:08 GMT > > WLOG? > > "Without loss of generality" Boy, you need know how to read Mr. Bielawski's mind.
> It is a phrase you should be used to. No, I won't.
It is about an hour and a half since Mr. Bielawski posted his last post. It is certain that he is trying to find another mathemagic trick for a show. In the meantime, you are here to keep the rest entertained. That is OK.
> > > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2] > [quoted text clipped - 5 lines] > diag(-1,1,1,1) and the diag(1,-1,-1.-1) representations of Minkowski > space? Yes, I know. <shrug>
> > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3] > > > I don't agree with (r^2 dO^2). > > Why? ds^2 is a squared distance - it needs to have units of squared > distance. Right. I still don't agree with (r^2 dO^2). It should be more general such as (C^2 dO^2) where C is a function of r.
> > > [In fact (getting slightly ahead of myself here) this is precisely > > > what happens: the function we divide by is zero for r=2M, hence the [quoted text clipped - 12 lines] > Of course it doesn't make sense - you are confused by C^n notation for > being n-times differentiable and you don't know what WLOG means. I am not laughing. Can you find something better to say while the magician is trying to come up with another mathemagic trick?
> What he is saying is that you can't represent the manifold at r=2GM > with a metric that has no cross terms. Oh, that's it! What does that mean? Notice I am still not laughing.
> > Recall > [quoted text clipped - 3 lines] > > It isn't 3 variables - the variables are _still_ (t,r,theta,phi). Not anymore.
> Remember the definition of dO he wrote down for you? Yes.
> > > The exact forms of the four left-hand sides are, few pages of > > > calculations later, respectively (excuse the mess, it's the fourth, [quoted text clipped - 8 lines] > > Why not? Why yes?
> You have the metric anzatz - go derive the R_tr equation. You will > find that he is, in fact, right if you do it correctly. Through a mathemagic trik?
> > Actually, with time derivative, these equations are a lot more > > complicated than you have shown above. > > No, they are not. Says someone who has not gone through them. <shrug>
> > Therefore showing (dB/dt = 0) is just BS. > [quoted text clipped - 3 lines] > exclude trivial A,B. I am amazed that you have trouble with something > as basic as *this*. <sigh> You are drowning in BS, and you don't know it.
> > > (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0 > > > (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0 [quoted text clipped - 9 lines] > > It went away. He integrated both equations with respect to r. Come back after you have understood integration.
> How can someone claim, with a straight face, that all of differential > geometry is wrong when that someone struggles with regular one > dimensional integration? Easily if it is wrong which it is. <shrug>
> > Don't you feel like a prophet if you know the answer before hand? > > Don't you feel like a jackass for insulting someone who obviously > spent a lot of time writing down something you should already know, > which you will not understand anyway? No.
> Integrate the equations - you get ln(AB) = C. Which is equivalent to > AB = C, where C in both cases are _completely arbitrary_ constants. Is > that enough detail, or would you like it done in even more baby steps? Yes, but the differential equations don't indicate that result. <shrug>
> I guess JanPB was entirely unreasonable in expecting that the guru of > Riemann be capable of understanding the concept of "integration". Correct me if I am wrong. It is up to him to decide on that one.
> > > The important thing as far as Birkhoff is concerned is that this means > > > that: [quoted text clipped - 7 lines] > > *laughs* I am still not laughing.
> > Also, you have conveniently assumed (C^2 = r^2). See (3). > > YOU ARE THE ONE WHO WROTE (3). THE FUCKUP IS YOURS. Yes, I wrote that one as a correction. <shrug>
> > That is how the Birkhoff's theorem as the author of Wikipedia defined > > it. > > Who cares? You don't have the education to differentiate between good > and bad yet - if ever. I do, and I did.
> > Assuming (C^2 = r^2) which translates to spherically symmetric. For > > every C^2, there is a new set of metric that satisfies the field > > equations. Have you finally understood Mr. Crothers? > > Hellooo he never wrote C^2 I know.
> - you were the one who introduced it. Yes, as a correction.
> Is > your head that far up your a.s that you have to invent completely > nonsensical arguments? No.
Any other questions?
Eric Gisse - 07 Apr 2007 09:32 GMT > > > WLOG? > [quoted text clipped - 5 lines] > > No, I won't.
> It is about an hour and a half since Mr. Bielawski posted his last > post. It is certain that he is trying to find another mathemagic [quoted text clipped - 12 lines] > > Yes, I know. <shrug> If you actually knew you would not have not made that comment.
> > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3] > [quoted text clipped - 5 lines] > Right. I still don't agree with (r^2 dO^2). It should be more > general such as (C^2 dO^2) where C is a function of r. Then use C(r) in place of r^2. The argument that r^2*dO^2 appears in the metric is an argument from spherical symmetry - since you have such issue with it, start off with the most general metric and work from there.
> > > > [In fact (getting slightly ahead of myself here) this is precisely > > > > what happens: the function we divide by is zero for r=2M, hence the [quoted text clipped - 20 lines] > > Oh, that's it! What does that mean? Notice I am still not laughing. Defining "cross terms" is left as an exercise for those who have enough math to understand.
> > > Recall > [quoted text clipped - 5 lines] > > Not anymore. In what fantasy world?
The metric is manifestly a function of t,r,theta,phi - is your reading disability so strong that you cannot read what is in front of your face?
> > Remember the definition of dO he wrote down for you? > [quoted text clipped - 19 lines] > > Through a mathemagic trik? It is only magic to those who do not understand the process.
If you know how to obtain the Ricci tensor from a given metric - work out the R_tr equation. It will be equal to 0 via being a vacuum solution. If you *don't* know how to....well that's your problem and you shouldn't be making the claims you are making.
> > > Actually, with time derivative, these equations are a lot more > > > complicated than you have shown above. > > > No, they are not. > > Says someone who has not gone through them. <shrug> Actually I have.
> > > Therefore showing (dB/dt = 0) is just BS. > [quoted text clipped - 5 lines] > > <sigh> You are drowning in BS, and you don't know it. These tools have served me well thus far. Not my problem you are incapable of understanding their usage.
> > > > (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0 > > > > (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0 [quoted text clipped - 11 lines] > > Come back after you have understood integration. Well then what do _you_ get when you integrate both equations with respect to r? Did you even try before crying "BS" ?
> > How can someone claim, with a straight face, that all of differential > > geometry is wrong when that someone struggles with regular one > > dimensional integration? > > Easily if it is wrong which it is. <shrug> Did you even check?
This can be as interactive as you want to be - why not start off with the metric anzatz _he_ uses and calculate the relevant equations yourself. Or is that too much work to ask of you?
[...]
Koobee Wublee - 08 Apr 2007 05:19 GMT > They aren't coordinate transformations - they are representations of > the same manifold just with different signatures - remember the [quoted text clipped - 4 lines] > > If you actually knew you would not have not made that comment. No, I comment was totally irrelevant to that! <shrug>
> > Right. I still don't agree with (r^2 dO^2). It should be more > > general such as (C^2 dO^2) where C is a function of r. [quoted text clipped - 3 lines] > such issue with it, start off with the most general metric and work > from there. This was Mr. Bielawski's convention. I personally would use C(r). As long as it is C(r) in the general case and not r^2, I have no issues. <shrug>
I want to correct myself that 'spherically symmetric' means A, B, and C are independent of any angular displacement.
ds^2 = A dt^2 - B dr^2 - C dO^2
> Defining "cross terms" is left as an exercise for those who have > enough math to understand. Whatever! It is more like a call to another mathemagic trick.
> > > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3] > [quoted text clipped - 5 lines] > > In what fantasy world? Yours, of course.
> The metric is manifestly a function of t,r,theta,phi - is your reading > disability so strong that you cannot read what is in front of your > face? No. Try to find the problem in front of a mirror.
> > Through a mathemagic trik? > > It is only magic to those who do not understand the process. The trick is exposed. <shrug>
> If you know how to obtain the Ricci tensor from a given metric - work > out the R_tr equation. It will be equal to 0 via being a vacuum > solution. If you *don't* know how to....well that's your problem and > you shouldn't be making the claims you are making. To do so, it involves more assumptions. <shrug>
> > Says someone who has not gone through them. <shrug> > > Actually I have. I don't mean in your dreams, please.
> > <sigh> You are drowning in BS, and you don't know it. > > These tools have served me well thus far. I can see that. It makes you look awful.
> Not my problem you are > incapable of understanding their usage. I do. However if you don't which is the case, you are just drowning in your own BS and won't know it. <shrug>
> > Come back after you have understood integration. > > Well then what do _you_ get when you integrate both equations with > respect to r? Did you even try before crying "BS" ? Try to look up the answer in a textbook. <shrug>
> > Easily if it is wrong which it is. <shrug> > > Did you even check? Yes. <shrug>
> This can be as interactive as you want to be - why not start off with > the metric anzatz _he_ uses and calculate the relevant equations > yourself. Or is that too much work to ask of you? I have done that. Good night!
Eric Gisse - 08 Apr 2007 05:38 GMT [...]
> > This can be as interactive as you want to be - why not start off with > > the metric anzatz _he_ uses and calculate the relevant equations > > yourself. Or is that too much work to ask of you? > > I have done that. Good night! Good.
Show your work.
JanPB - 08 Apr 2007 19:45 GMT [...]
Something went wrong with my response, it looks like it never showed up. Fortunately, I'm in the habit of writing posts in my text editor as I lost a few posts before. So I'm reposting it now (and apologies if this thing does end up posted twice).
> > We seek a spherically symmetric metric of signature 2 satisfying the > > Einstein equation in vacuum: [quoted text clipped - 7 lines] > > WLOG? "Without loss of generality". This abbreviation is rarely seen in books but is otherwise very common.
> > ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2, > [quoted text clipped - 21 lines] > If you are describing the same invariant spacetime, ds^2, then your > dr, dt must be different due to your coordinate transformation. Yes, I have recycled the letters "r" and "t" from the first formula (the one with coefficients f, g, h, k) into the formulas [1] and [2]. BTW, did you notice the nonzero cross term "dr dt"? _This_ is the source of the computational complexity, not k(r,t) which is trivially replaced by r^2 (here I'm recycling "r" as usual).
> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3] > > I don't agree with (r^2 dO^2). It's simply switching from r to r', where:
r' = sqrt(k(r,t))
> > where A(r,t) and B(r,t) are some functions which are never zero (or > > else the metric would not be rank-4). > > Rank-4? Brush up on your linear algebra. Here it's synonymous with "nondegenerate".
> > NB: this local coordinate change happens to rely on dividing by > > certain function, so any solutions for metric coefficients we may [quoted text clipped - 4 lines] > This transformation is not necessary a division transformation. Why > do you think it is so? It's not _just_ a division - I said it only relied on it. The entire procedure involves cooking up an integrating factor. Here is a quickie summary. We have:
ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dt dr + k(r,t) dO^2 (*)
...and we want to get rid of the cross-term involving dt dr. We consider the 1-form:
f dt + (h/2) dr
We know there exists a function F(r,t) (an "integrating factor") which makes this form into a _closed_ form when multiplied by F:
d( F(f dt + (h/2) dr) ) = 0
...which means that locally there exists another function G(r,t) such that:
F(f dt + (h/2) dr) = dG (**)
Turns out if we use this G in the following change of variables:
t' = G(r,t) r = r theta = theta phi = phi
...then the crossterm dt dr drops out. Specifically, from (**) we have:
dt'^2 = F^2 ( f dt + (h/2) dr )^2 = = F^2 ( f^2 dt^2 + fh dt dr + (h^2/4) dr^2)
...which means (collecting terms):
f^2 dt^2 + fh dt dr = (1/F^2) dt'^2 - (h^2/4) dr^2
...and so DIVIDING BY f (this is the step I was referring to in my previous post):
f dt^2 + h dt dr = U dt'^2 + V dr^2
...where I substituted U(r,t) and V(r,t) for the actual algebraic expressions.
Now the reason it works is that when one plugs the LHS above into the expression (*) for ds^2, we cancel out the unwanted cross-term like so:
ds^2 = f dt^2 + g dr^2 + h dt dr + k dO^2 =
= U dt'^2 + (V + g) dr^2 + k dO^2
...and we have a diagonal form.(2/rAB^2)*dB/dt
> > [In fact (getting slightly ahead of myself here) this is precisely > > what happens: the function we divide by is zero for r=2M, hence the [quoted text clipped - 9 lines] > This makes no sense. It has the signature of Archimedes Plutonium. > Are you the real person behind this handle? I am quite real, yes. Do you know Archimedes' original name? I think he had like 5 name changes? The paragraph you quoted above simply describes what happens for those r and t for which the function f(r,t) equals zero (we've just divided by f, remember? so there is a possibility we threw out a portion of the domain just out of convenience).
> > OK, so we have those two forms potentially satisfying the EFE. The > > maximal domain possible (unless the EFE themselves somehow end up [quoted text clipped - 20 lines] > ** Rr'r' = 0 > ** RO'O' = 0 You get three equations if you assume A and B are functions of r only (i.e. independent of t), IOW if you assume a static metric from the beginning. Here we are interested in Birkhoff's theorem so we only assume spherical symmetry. Because A and B are thus functions of r and t, the rt-component of Ricci curvature is NOT identically zero, and is in fact equal to (2/rAB^2)*dB/dt. It's just a calculation - I typed up the details behind this calculation in TeX and posted them here: http://www.mastersofcinema.org/jan/sch.pdf
TeX's deafult typeface is rather ugly but it beats ASCII anytime :-)
> Recall > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3] > > You have reduced the spacetime from 4 variables to 3, remember? I haven't reduced anything in this sense. There are still 4 variables t,r,theta,phi. The Ricci rt-component just happens to involve dB/dt, so it's obviously identically zero if you presume B is a function of r only. But R_rt is not identically zero in general. Just compute (or read that PDF doc) and see for yourself.
> > ...with the remaining equations being either identities or equivalent > > to those four ones. (For example, R_phi phi = R_theta theta.) [quoted text clipped - 9 lines] > > There is no way you can get ((2/rAB^2)*dB/dt=0) from (3). That's right. You don't get the fourth equation out of the remaining three. It's just a fourth independent equation (there are 10 of them in general, 6 of which are either equivalent to the 4 above or are identities 0=0).
> Actually, with time derivative, these equations are a lot more > complicated than you have shown above. That's because you use Christoffel symbols for your calculations. I use Cartan's connection forms which are based on orthonormal moving frames instead of coordinate frames. This technology is well-worth learning as it offers significant simplification of the calculations. At the "bottom" both Christoffel and Cartan-based quantities are the same of course, but Cartan's approach makes clever use of certain hidden SKEW-symmetries (rather than Christoffel symmetries in the lower indices) which means you have less quantities to deal with: for example, there are only 6 independent entries in a 4x4 skew-symmetric matrix vs. 10 entries in a 4x4 symmetric matrix. Etc. MTW has a chapter on Cartan's calculus and books like Wald use this technique extensively.
> > ...where "-+" means use the "-" sign for the first general form of the > > diagonal metric [1], use "+" for the second [2]. Also "d" obviously [quoted text clipped - 6 lines] > > Therefore showing (dB/dt = 0) is just BS. You have to work harder than that :-) The fourth equation R_rt = 0 says that dB/dt=0. It just says what it says.
> > ...in other words, B(r,t) is independent of t, both in case [1] and > > [2]. This means that B is only a function of r which means that the [quoted text clipped - 11 lines] > Bullsh*t. Your differential equations involve a second derivative of > A! Where is it? Just multiply through 1/AB in the first equation and then add both equations together. The first term cancels out. Then note that d/dr(1/ B) = -B'/B^2 and you obtain the simple equation in A and B.
> > i.e., > [quoted text clipped - 9 lines] > > Don't you feel like a prophet if you know the answer before hand? It's just following the math - no room wiggle. When I was writing this I didn't know exactly where it lead (besides the general expectation of seeing Schwarzschild's solution in the end somewhere).
> > The important thing as far as Birkhoff is concerned is that this means > > that: [quoted text clipped - 7 lines] > > [The rest snipped due to the obvious mathematical error.] Conveniently pulling a rabbit out of the hat? At least tell us what mathematical errors lead to B=constant/A?
> Also, you have conveniently assumed (C^2 = r^2). See (3). This is a trivial coordinate change. I'm beginning to suspect where you (and Crothers) get this idea of metric being coordinate-dependent. It's the problem of physical units, isn't it? You both think that changing units of measurement is the same as changing coordinates, right? So when I change my unit of length - so you both think - the numerical values of, say, vectors lengths change also, hence the values of dot products change, and these values ARE the metric, so the metric changes too! Am I getting close?
> That is how the Birkhoff's theorem as the author of Wikipedia defined > it. > > Assuming (C^2 = r^2) which translates to spherically symmetric. For > every C^2, there is a new set of metric that satisfies the field > equations. Have you finally understood Mr. Crothers? C^2 makes no difference. Mr. Crothers is a confused soul.
> Assuming (static) which translates to all elements of the metric > independent of time. This whole exchange started with you knocking Birkhoff's theorem and how it implied the metric had to be static. So don't assume now that the metric is static.
> Then, you end up with much, much, much, and much simpler equations > described below. > > ** (r / B^2) (dB/dr) - (1 / B) + 1 = 0 > ** (r / A / B) (dA/dr) + (1 / B) - 1 = 0 Well, wonderful, but again - we are discussing how Birkhoff says spherical symmetry in Schwarzschild coordinates implies independence of t (which you mistook for static). We cannot assume that which we want to prove!
> These are actually not Ricci tensor equations but rather Einstein > tensor equations. With all these simplifications, they are all the [quoted text clipped - 26 lines] > > In this case, you don't get a black hole. You do - you simply neglected to use the correct domain (you changed coordinates but kept the domain the same - this is like saying "0 Celsius equals 0 Kelvin, therefore absolute zero doesn't exist"). You've changed the domain of r from r>0 to r>-K.
> For (C = ((r^3 + K^3)^(1/3) - K)^2), > > You end up with Schwarzschild's original solution which does not > manifest any black holes either. > > There are infinite numbers of solutions more. You've just altered the domain to exclude the black hole. This is not a different solution, it's just a subset of the good old Schwarzschild with a restriction of the radial coordinate. See: http://groups.google.com/group/sci.physics.relativity/msg/068e4e33d3fecf0d?hl=en&
-- Jan Bielawski
Eric Gisse - 08 Apr 2007 22:24 GMT [...]
I don't think he will ever reverse his position, to be honest. He has spent too many years on USENET claiming st |
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