How to calculate the Riemann curvature at r=2GM?

[Bishal Banjara]

Hello lbix,
Just take it as my innocence and reply me, does the Kruskal-Szekeres coordinate transformation passes all the test of Schwarzschild Solution like light bent, mercury precision,...?

 

[Orodruin]

Hello lbix,
Just take it as my innocence and reply me, does the Kruskal-Szekeres coordinate transformation passes all the test of Schwarzschild Solution like light bent, mercury precision,...?

Predictions are not coordinate dependent and the KS coordinates are just a different set of coordinates on the Schwarzschild spacetime. Perhaps a result of how it is often being taught with emphasis on coordinates and coordinate transformations, students often make this mental mistake. Bottom line: The spacetime does not change just because you change coordinates.

 

[Bishal Banjara]

Predictions are not coordinate dependent and the KS coordinates are just a different set of coordinates on the Schwarzschild spacetime. Perhaps a result of how it is often being taught with emphasis on coordinates and coordinate transformations, students often make this mental mistake. Bottom line: The spacetime does not change just because you change coordinates.

Hello Orodruin,
I argue that if we change the coordinate there will be the change in Riemann curvature tensor (like at the case of the event horizon), doesn't this effect on the line element itself. Do we have some sortie reference in that regard? Thank you for your response.

 

[Orodruin]

I argue that if we change the coordinate there will be the change in Riemann curvature tensor (like at the case of the event horizon), doesn't this effect on the line element itself.

No, this is wrong. Vectors and tensors do not change when you change coordinates - their components change but that is only in response to a change in the basis used to describe them. In the case of the Schwarzschild spacetime, the Schwarzschild coordinates are singular at $r = 2MG$ and therefore not a good basis to use, meaning that you cannot use it properly at that coordinate.

 

[rrogers]

Hello Orodruin,
I argue that if we change the coordinate there will be the change in Riemann curvature tensor (like at the case of the event horizon), doesn't this effect on the line element itself. Do we have some sortie reference in that regard? Thank you for your response.

Remember that these equations are a description of "real" things. It turns out this description is a little complicated, but very accurately describes what we observe in intermediate range physics: colliding stars, black holes...

 

[Bishal Banjara]

If so, then let us assume that the $$g_{tt}$$ and $$g_{rr}$$ being like in the form $$(1+2GM/r)$$ and $$(1+2GM/r)^{-1}$$ respectively. Do such tests remain independent again where we have no singularity at r=2GM?

 

[Orodruin]

If so, then let us assume that the $$g_{tt}$$ and $$g_{rr}$$ being like in the form $$(1+2GM/r)$$ and $$(1+2GM/r)^{-1}$$ respectively. Do such tests remain independent again where we have no singularity at r=2GM?

I am sorry, but it is completely unclear what "tests" you are referring to. What you have quoted here are the components of the metric in Schwarzschild coordinates. In a different coordinate system, those components will be different, but the metric itself is still the same metric.

 

[Bishal Banjara]

I am sorry, but it is completely unclear what "tests" you are referring to. What you have quoted here are the components of the metric in Schwarzschild coordinates. In a different coordinate system, those components will be different, but the metric itself is still the same metric.

I mean the precision of Mercury, light bent,..and so on. Does this form of metrics still don't effect on these tests?

 

[kent davidge]

I mean the precision of Mercury, light bent,..and so on. Does this form of metrics still don't effect on these tests?

The form of the metric does not matter, because the theory is about the manifold (space-time). The tests are meant to check that gravity is the change in the underlying mesh of space-time. The form of tensors, like the metric, you need for doing calculations. The same way we represent a vector as an arrow and we do not need to know about any coordinate system nor what its components are. But if you are to solve a problem requesting you to give, say, a speed, then you need to go into numbers. Then you need to know the components.

So for actually computing the tests of General Relativity, you need to know the components, which is done through a coordinate system. But they do not affect the tests in the sense you seem to think.

 

[Bishal Banjara]

The form of the metric does not matter, because the theory is about the manifold (space-time). The tests are meant to check that gravity affects space-time, affecting the "underlying mesh". The form of tensors, like the metric, you need for doing calculations. The same way we represent a vector as an arrow and we do not need to know about any coordinate system nor what its components are. But if you are to solve a problem requesting you to give, say, a speed, then you need to go into numbers. Then you need to know the components.

So for actually computing the tests of General Relativity, you need to know the components, which is done through a coordinate system. But they do not affect the tests in the sense you seem to think.

But if you are to solve a problem requesting you to give, say, a speed, then you need to go into numbers. Then you need to know the components. Don't these tests are the results of respective components and get affected due to such change in metric values?

 

[kent davidge]

Don't these tests are the results of respective components?

No, think of it this way: you are at the entrance of a tunnel, the result is at the exit and the components are the tunnel. They are a path linking you to the result, but the result is out there whether or not you have the components. Therefore they cannot affect the results.

 

[Bishal Banjara]

No, think of it this way: you are at the entrance of a tunnel, the result is at the exit and the components are the tunnel. They are a path linking you to the result, but the result is out there whether or not you have the components. Therefore they cannot affect the results.

This means that it is alike the coordinate transformation as like the KS. Am I right?

 

[Orodruin]

No, think of it this way: you are at the entrance of a tunnel, the result is at the exit and the components are the tunnel. They are a path linking you to the result, but the result is out there whether or not you have the components. Therefore they cannot affect the results.

Stated a little differently: The numbers you use in all your intermediate computations will, in general, depend on your choice of coordinates. However, regardless of the coordinate system, the numbers will in the end conspire to produce the same result.

 

[Bishal Banjara]

Stated a little differently: The numbers you use in all your intermediate computations will, in general, depend on your choice of coordinates. However, regardless of the coordinate system, the numbers will in the end conspire to produce the same result.

This means I could reach the top hill from all sides around (conceptually linking the least action principle).
My doubts are clear now. Thank you very much for all of you!

 

[Ibix]

A concrete example would be the electric field from a point charge at the origin of a Cartesian coordinate system. The field at (x,y,z) in the matching coordinate basis is $E^\mu=(E_0x/r^3,E_0y/r^3,E_0z/r^3)$. But you can switch to spherical polar coordinates and say that the field at $(r,\theta,\phi)$ is $E^{\mu'}=(E_0/r^2,0,0)$ in that coordinate basis. The electric field vector at any point is the same in either system - only the representation has changed. Similar remarks apply to one-forms and higher rank tensors.

When you make a measurement, it'll turn out that what you actually measure is a Lorentz scalar (necessarily, or your instrument readings would be coordinate dependent). For example, one thing you could do to measure the electric field is have a test charge attached to a spring and constrained to move in one dimension. What you actually measure is the inner product of the Lorentz force and the direction of (constrained) motion, which is a scalar. If you let that direction be parallel to a basis vector then this is equal to the matching component of the electric field vector expressed in that basis multiplied by the charge. But you are deriving the component from the scalar you measured, and if you make three linearly independent measurements at a point you can express the result in any basis you like.

 

[PeterDonis]

let us assume that the

$$
g_{tt}
$$

and

$$
g_{rr}
$$

being like in the form

$$
(1+2GM/r)
$$

and

$$
(1+2GM/r)^{-1}
$$

respectively.

Are you intending these to be the standard Schwarzschild coordinate forms? If so, you've written them down wrong, there should be minus signs, not plus signs.

If you intend these to be as you have written them, with the plus signs, then they must be the metric coefficients in some other coordinate chart, and you're going to need to specify what chart and how you are obtaining these coefficients. You can't just wave your hands and write down anything you like.

 

[PeterDonis]

What you have quoted here are the components of the metric in Schwarzschild coordinates.

No, he hasn't, because he put in plus signs instead of minus signs.

 

[Bishal Banjara]

Are you intending these to be the standard Schwarzschild coordinate forms? If so, you've written them down wrong, there should be minus signs, not plus signs.

If you intend these to be as you have written them, with the plus signs, then they must be the metric coefficients in some other coordinate chart, and you're going to need to specify what chart and how you are obtaining these coefficients. You can't just wave your hands and write down anything you like.

It was just a different value of metric I wrote by assuming as a variable value whether it is still true to itself again. And there is nothing more.

 

[PeterDonis]

It was just a different value of metric I wrote by assuming as a variable value whether it is still true to itself again.

I have no idea what you mean by this. You can't just write down any "variable value" you like and expect it to describe the same metric (spacetime geometry).

 

[Bishal Banjara]

I have no idea what you mean by this. You can't just write down any "variable value" you like and expect it to describe the same metric (spacetime geometry).

If you had seen my previous comments, you would understand what I meant. I am simply focusing that the form of the variable metric values for Schwarzschild line element (like as I mentioned above) will keep revision for the Classical test like precession of mercury, light bent,...or not.

 

[PeterDonis]

If you had seen my previous comments, you would understand what I meant.

I have read all of your previous comments. I don't think the problem is my not understanding what you meant; I think the problem is that you don't understand what you are doing. See below.

I am simply focusing that the form of the variable metric values for Schwarzschild line element (like as I mentioned above)

You are still missing my point: how do you know that the metric values you wrote down in post #41 are for the Schwarzschild line element? You can't just assume that. You have to prove it.

I don't think the metric values you wrote down in post #41 have anything at all to do with the Schwarzschild line element; they don't describe that spacetime geometry. Which means those values are completely irrelevant to the question it seems like you are trying to ask.

The question it seems like you are trying to ask is: if we change coordinate charts but keep the spacetime geometry the same, will all of the invariants, such as curvature invariants at $r = 2M$, stay the same? The answer to that question is yes. But what you wrote in post #41 is irrelevant to that question, because the metric coefficients you wrote in post #41 do not represent a change of coordinate charts while keeping the spacetime geometry (Schwarzschild) the same.

 

[Bishal Banjara]

I have read all of your previous comments. I don't think the problem is my not understanding what you meant; I think the problem is that you don't understand what you are doing. See below.
You are still missing my point: how do you know that the metric values you wrote down in post #41 are for the Schwarzschild line element? You can't just assume that. You have to prove it.

I don't think the metric values you wrote down in post #41 have anything at all to do with the Schwarzschild line element; they don't describe that spacetime geometry. Which means those values are completely irrelevant to the question it seems like you are trying to ask.

The question it seems like you are trying to ask is: if we change coordinate charts but keep the spacetime geometry the same, will all of the invariants, such as curvature invariants at $r = 2M$, stay the same? The answer to that question is yes. But what you wrote in post #41 is irrelevant to that question, because the metric coefficients you wrote in post #41 do not represent a change of coordinate charts while keeping the spacetime geometry (Schwarzschild) the same.

I was in search of this kind of critical answer. Thank you! Ok, let's assume I have such kind of metric values as I wrote (I am intending not to modify Schwarzschild metric) by derivation itself. Then, it will not be Schwarzschild, right? And I think this change will bring the change in such predictive values for such Tests, am I right?

 

[Orodruin]

I was in search of this kind of critical answer. Thank you! Ok, let's assume I have such kind of metric values as I wrote (I am intending not to modify Schwarzschild metric) by derivation itself. Then, it will not be Schwarzschild, right? And I think this change will bring the change in such predictive values for such Tests, am I right?

Your metric is senseless in the framework of GR. First of all, it does not even have the correct signature. Second, you cannot just write down a metric and hope that it describes anything useful. Third, even if it did, it would have a Einstein tensor (and therefore energy-momentum tensor) that would be different from the one of the Schwarzschild spacetime and therefore not describe the spacetime outside a point mass. It would be like writing down a random gravitational potential in Newtonian gravity in the vain hope that it would describe something.

Furthermore, I have read all your posts in the thread and I agree with @PeterDonis , it is very difficult to understand what you mean because it seems you do not control the basic terminology.

However, if you do have a spacetime that has the correct metric signature, is a solution to the EFEs for some reasonable energy-momentum tensor, then it will generally be different from the Schwarzschild spacetime.

 

[PeterDonis]

Ok, let's assume I have such kind of metric values as I wrote (I am intending not to modify Schwarzschild metric) by derivation itself. Then, it will not be Schwarzschild, right?

So you are intending not to modify the Schwarzschild metric, but you are asking if the metric will not be Schwarzschild? This makes no sense.

The values of observable quantities related to the spacetime geometry (which I think is what you mean by "Tests") will obviously be different for different spacetime geometries. As far as I can tell, that answers the actual question you are asking. But that answer has nothing whatever to do with what coordinates you choose.

 

[Bishal Banjara]

So you are intending not to modify the Schwarzschild metric, but you are asking if the metric will not be Schwarzschild? This makes no sense.

The values of observable quantities related to the spacetime geometry (which I think is what you mean by "Tests") will obviously be different for different spacetime geometries. As far as I can tell, that answers the actual question you are asking. But that answer has nothing whatever to do with what coordinates you choose.

Your metric is senseless in the framework of GR. First of all, it does not even have the correct signature. Second, you cannot just write down a metric and hope that it describes anything useful. Third, even if it did, it would have a Einstein tensor (and therefore energy-momentum tensor) that would be different from the one of the Schwarzschild spacetime and therefore not describe the spacetime outside a point mass. It would be like writing down a random gravitational potential in Newtonian gravity in the vain hope that it would describe something.

Furthermore, I have read all your posts in the thread and I agree with @PeterDonis , it is very difficult to understand what you mean because it seems you do not control the basic terminology.

However, if you do have a spacetime that has the correct metric signature, is a solution to the EFEs for some reasonable energy-momentum tensor, then it will generally be different from the Schwarzschild spacetime.

Sorry that I made mistake with signature. I was intended to write $$-(1+2GM/R)$$ for $$g_{tt}$$ metric. However, if you do have a spacetime that has the correct metric signature, is a solution to the EFEs for some reasonable energy-momentum tensor, then it will generally be different from the Schwarzschild spacetime. I think this point will be applied then.

 

[Orodruin]

Actually, your proposed metric (with the change of signature) will be a solution to the EFE's in vacuum. In the typical derivation of the Schwarzschild metric you end up with a differential equation of the form
$$
\partial_r ( r f(r)) = 1,
$$
where $f(r)$ is the coefficient of $- dt^2$ in the line element. The solution to this is obviously
$$
r f(r) = r + C,
$$
where $C$ is an integration constant. This tells you nothing about the sign of $C$ and generally you have
$$
f(r) = 1 + \frac{C}{r}.
$$
Asking for the correct behaviour in the weak field limit then identifies $C = - 2MG$, but there is nothing preventing you from putting $C$ positive apart from the fact that we have not observed anything that actually behaves like that (essentially the weak field limit becomes a limit where a point "mass" repels rather than attracts). Celestial objects would move on approximate hyperbolae rather than approximate ellipses, etc. The case $C > 0$ is therefore just not physically interesting.

 

[Bishal Banjara]

The case C>o is therefore just not physically interesting. It is not the matter of interest, its the matter of reality. Then such teste-parameters would get varied?

 

[PeterDonis]

Then such teste-parameters would get varied?

This question has already been answered:

The values of observable quantities related to the spacetime geometry (which I think is what you mean by "Tests") will obviously be different for different spacetime geometries.

There is no point in continuing to repeat the same answer. Thread closed.