Inferences from the General Relativity Metric

[Anamitra Palit]

Metric:

[Signature;(+,-,-,-)]
\[c^2 d\tau^2=c^2 g_{00} dt^2-g_{11} dx^2-g_{00} dy^2-g_{00} dz^2\]  (1)

[In (1) g_00,g_11,g_22 and g_33 are the absolute values of the metric coefficients,the system of coordinates being  orthogonal]
By our choice let 
\[cdt=dx=dy=dz\]  (2)
Dividing both sides of  (1) by (cdt)^2
We have
\[ \left(\frac {d\tau}{dt}\right)^2=g_{00}-g_{11}-g_{22}-g_{33}\]  (3)
Since the left side of (3) is positive we have,
\[g_{00}\ge g_{11}+g_{22}+g_{33}\](4)
Taking Schwarzschild geometry 

\[1-\frac{2Gm}{c^2r}\ge \left[1-\frac{2Gm}{c^2r}\right]^{-1}+r^2+r^2{\sin}^2\theta\] (5)
The above is not true for large r

With \[c^2 d\tau^2=c^2 \left[1-\frac{2Gm}{c^2r}\right]dt^2-\left[1-\frac{2Gm}{c^2r}\right]^{-1} dr^2-r^2 d\theta^2-r^2 \sin^2\theta d\phi^2\]

\[ cdt=kdr=ld\theta=md\phi \]

Each term above has the dimension of lengthk is dimensionless while l and m have the dimensions of length.

\[ \left(\frac {d\tau}{dt}\right)^2=\left[1-\frac{2Gm}{c^2r}\right]dt^2-\frac{1}{k^2}\left[1-\frac{2Gm}{c^2r}\right]^{-1}dr^2-\frac{1]{l^2}r^2d\theta^2-\frac{1}{m^2}r^2 \sin^2\theta d\phi^2\]

 

\[\left[1-\frac{2Gm}{c^2r}\right]\ge \frac{1}{k^2}\left[1-\frac{2Gm}{c^2r}\right]^{-1}+\frac{1]{l^2}r^2-+\frac{1}{m^2}r^2 \sin^2\theta \] (5')

 

As a special case we may set k=1 unit,l=1 unit,m=1 unit

We now do have

\[\left[1-\frac{2Gm}{c^2r}\right]\ge \left[1-\frac{2Gm}{c^2r}\right]^{-1}+r^2+r^2 sin^2\theta \] (5'')

 

In the rest of the discussion we are considering the rectangular Cartesian system.
At a given point on the manifold we take,[by our choice],
\[cdt=kdx=ldy=mdz\]
We divide both sides of (1) by (cdt)^2
\[ \left(\frac {d\tau}{dt}\right)^2=g_{00}-\frac{1}{k^2}g_{11}-\frac{1}{l^2}g_{22}-\frac{1}{m^2}g_{33}\](6)

\[\left(\frac {d\tau}{dt}\right)^2=g_{00}-Kg_{11}-Lg_{22}-Mg_{33}\](7)
For arbitrary K,L and M, we observe,
(1)
   \[ g_{00}-Kg_{11}-Lg_{22}-Mg_{33}\ge 0\](2)
   \[ g_{00}-Kg_{11}-Lg_{22}-Mg_{33}=g_{00}-g_{11}-g_{22}-g_{33}\]
   
These observations are not valid ones. The manifold is becoming threadbare.  

Two of the statements with out \[  \], have been given out in plaintext since the codes were not getting parsed

 

 

Edited January 19, 2021 by Anamitra Palit

[Anamitra Palit]

[One may have to refresh the page for viewing the formulas and the equations]

1.  In our discussion we have considered time like separations at a point so that d tau^2>0

2. Towards  the end of the last post..For arbitrary K,l and M we observe, point (2) has to be ignored

3. We had \[cdt=kdr=ld\theta=md \phi\].All three components in the last line have the dimensions of length. k is dimensionless while l and m have the dimensions of length.

Dividing the Schwarzschild metric by (cdt)^2 we obtain

\[\left(\frac {d\tau}{dt}\right)^2=\left(1-\frac{2Gm}{c^2 r}\right)-\frac{1}{k^2}\left(1-\frac{2Gm}{c^2 r}\right)^{-1}-\frac{1}{l^2}  r^2-\frac{1}{m^2}r^2 \sin^2 \theta\]

In the last equation we may set k=1 ,l=m=1 unit of length.

Therefore,

\[\left(1-\frac{2Gm}{c^2 r}\right)\ge \frac{1}{k^2}\left(1-\frac{2Gm}{c^2 r}\right)^{-1}+\frac{1}{l^2}r^2+\frac{1}{m^2}r^2 \sin\theta\]

In the last equation has to hold for any  k,l and m for which dtau^2 is positive.[as a particular instance we may set k=1 ,l=m=1 unit of length] and for arbitrary r  beyond the event horizon:dt,dx,dy and dz are not present but their rlative sizes are present. 

For large values of 'r' the metric automatically becomes space like unless we make dtheta and dphi extraordinarily small with respect to dr and dt

 

 

Plain text versions of the two stated formulas:

[dtau/dt]^2=[1-2Gm/c^2r]-(1/k^2)[1-2Gm/c^2r]^(-)-(1/l^2)r^2-(1/m^2)r^2sin^2 theta

Since time like separations are being considered on (+,-,-,-)

[1-2Gm/c^2r]>(1/k^2)[1-2Gm/c^2r]^(-)+(1/l^2)r^2+(1/m^2)r^2sin^2 theta

the above inequation will not hold for large r.

For a general type of a metric given the space time location we have to be careful with the relative sizes of the coordinate intervals in order to have a time like or a space like interval.

Edited January 19, 2021 by Anamitra Palit

[Markus Hanke]

I’m sorry, but I can’t make heads nor tails of this at all. 

What kind of coordinates (t,x,y,z) are you using here? Are these Cartesian coordinates, or Schwarzschild coordinates, or something else entirely?
And what is your reasoning behind the metric (1)? This form of metric is incompatible with both Schwarzschild coordinates and Cartesian coordinates in a Schwarzschild spacetime, so please explain how you arrived at this ansatz.

[Anamitra Palit]

8 hours ago, Markus Hanke said:

I’m sorry, but I can’t make heads nor tails of this at all. 

What kind of coordinates (t,x,y,z) are you using here? Are these Cartesian coordinates, or Schwarzschild coordinates, or something else entirely?
And what is your reasoning behind the metric (1)? This form of metric is incompatible with both Schwarzschild coordinates and Cartesian coordinates in a Schwarzschild spacetime, so please explain how you arrived at this ansatz.

(t,x,y,z) could represent  coordinates of four dimensional space in a general manner; they could be Cartesian,Spherical or any other system.  

Equation (1) of the initial post represents the most general type of a General Relativity metric in the orthogonal system:

\[c^2d\tau^2= c^2g_{00} d t^2-g_{11} dx^2-g_{22} dy^2-g_{33}  dz^2 \] 

[Inadvertently,with the string post of the discussion, after copy pasting I forgot to change the suffix with g_ii to appropriate values.]

Edited January 19, 2021 by Anamitra Palit

[Markus Hanke]

15 hours ago, Anamitra Palit said:

(t,x,y,z) could represent  coordinates of four dimensional space in a general manner; they could be Cartesian,Spherical or any other system.  

Ok. As a little tip - conventionally, writing (t,x,y,z) will imply Cartesian coordinates to most readers; if you want to indicate a general coordinate basis, it is better to use the notation \({x^0,x^1,x^2,x^3}\), as it is less ambiguous.

15 hours ago, Anamitra Palit said:

[Inadvertently,with the string post of the discussion, after copy pasting I forgot to change the suffix with g_ii to appropriate values.]

Ok, that’s an important difference.

On 1/19/2021 at 7:30 AM, Anamitra Palit said:

For a general type of a metric given the space time location we have to be careful with the relative sizes of the coordinate intervals in order to have a time like or a space like interval.

I’m not sure what you mean by “relative sizes”?

[studiot]

1 hour ago, Markus Hanke said:

16 hours ago, Anamitra Palit said:

(t,x,y,z) could represent  coordinates of four dimensional space in a general manner; they could be Cartesian,Spherical or any other system.  

Ok. As a little tip - conventionally, writing (t,x,y,z) will imply Cartesian coordinates to most readers; if you want to indicate a general coordinate basis, it is better to use the notation x0,x1,x2,x3 , as it is less ambiguous.

@Anamitra Palit

I would like to add to this my comment to the OP.

"(t,x,y,z) could represent  coordinates of four dimensional space in a general manner"

No they couldn't.

t has the wrong physical dimensions to make this a physical space of four dimensions.

You can only make an abstract 'space' (in the mathematical sense) with mixed dimensions on the coordinate axes.

It is important to always keep this distinction in mind.

 

[joigus]

There is no such thing as "absolute values of the metric coefficients" with an invariant meaning. Also, I don't think there's an invariant meaning to the concept of "relative sizes of the coordinate intervals".

Another methodological comment on my part. If you're serious about relativity, try not to build up your arguments from coordinate patches. You're going to make mistakes. The literature is full of similar arguments, in the years before intrinsic formalisms were developed, which were proved to be wrong. The technique in GR today is: You use intrinsic formalism --vectors, forms, and tensors formed from them-- to establish the general results, and then go to a coordinate patch for a particular calculation with a particular distribution of energy-momentum.

Allowed coordinate transformations in GR are diffeomorphisms, which means they're infinitely differenciable and never change the invariants of the metric --because their inverses are also differentiable--, so it's possible that you're going to a mathematical no-man's land. Nobody would bother to check due to your always working in coordinates.

[studiot]

47 minutes ago, joigus said:

Allowed coordinate transformations in GR are diffeomorphisms, which means they're infinitely differenciable and never change the invariants of the metric --because their inverses are also differentiable--, so it's possible that you're going to a mathematical no-man's land. Nobody would bother to check due to your always working in coordinates.

To add to this.

Mathematical continuity is defined in terms of the inverse functions, not the functions themselves.

This has implications for any calculus you choose to adopt.

[joigus]

10 hours ago, studiot said:

Mathematical continuity is defined in terms of the inverse functions, not the functions themselves.

 

Is it? I wasn't aware of this.

[studiot]

12 hours ago, joigus said:

23 hours ago, studiot said:

Mathematical continuity is defined in terms of the inverse functions, not the functions themselves.

 

Is it? I wasn't aware of this.

A detailed explanation of what I mean would be off topic in this thread so I will start another one for that purpose.

I will let you know when I have posted it.

[joigus]

On 1/19/2021 at 5:47 AM, Anamitra Palit said:

These observations are not valid ones. The manifold is becoming threadbare.  

 

I don't understand this. How can a manifold "become" anything? And of all things, "threadbare"?

On 1/19/2021 at 8:30 AM, Anamitra Palit said:

For large values of 'r' the metric automatically becomes space like unless [...]

Can you make this argument a little clearer? Metrics don't become space-like. Intervals do. I deeply mistrust Schwarzschild coordinates. I think they're good only to solve Einstein's field equations.

8 hours ago, studiot said:

A detailed explanation of what I mean would be off topic in this thread so I will start another one for that purpose.

I will let you know when I have posted it.

OK. I'm looking forward to it. Thank you, @studiot.

[Markus Hanke]

11 hours ago, joigus said:

I deeply mistrust Schwarzschild coordinates. I think they're good only to solve Einstein's field equations.

Well, they are useful to represent the perspective of a stationary observer at infinity. The crucial point to realise is that that is the only situation where measurements of space and time made in Schwarzschild coordinates actually coincide with what physically happens, since such measurements are always purely local in curved spacetimes. Anywhere else other than for a stationary observer at infinity, Schwarzschild coordinates are only a bookkeeping device, but they do not necessarily reflect what actually happens there, locally speaking. They also don’t cover the entirety of the spacetime. Many students of GR either do not understand this, or refuse to acknowledge it, since it goes against Newtonian intuition. The unfortunate result is all manner of misconceptions and misunderstandings. 

So you have good reason to mistrust Schwarzschild coordinates - they can be useful in certain circumstances, but they are also dangerous if not understood correctly. I think the only reason why pretty much every GR text uses them is because they are algebraically simple.

[joigus]

13 minutes ago, Markus Hanke said:

Well, they are useful to represent the perspective of a stationary observer at infinity.

Observers at infinity are misinformed.

Great summary, Hanke.