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

[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.