1. Classical escape velocity is given as: Ve = sqr(2GM/r). Which is based on Ke = mV^2/2 where: mV^2/2 = GMm/r. Relativistic escape velocity: Were: mc^2(1/sqr(1 - V^2/c^2) - 1) = GMm/r We get: Ve = sqr(2GM/r + G^2M^2/(r^2c^2)/(1 + GM/(rc^2). Were the escape velocity is never greater than the speed of light. Where light can always escape, and there are no singularities nor black holes. Check the math here.

2.

3. Originally Posted by 37818
Classical escape velocity is given as:
Ve = sqr(2GM/r).

Which is based on
Ke = mV^2/2

where:
mV^2/2 = GMm/r.

Relativistic escape velocity:

Where:
mc^2(1/sqr(1 - V^2/c^2) - 1) = GMm/r

We get:
Ve = sqr(2GM/r + G^2M^2/(r^2c^2)/(1 + GM/(rc^2).

Were the escape velocity is never greater than the speed of light.
Where light can always escape, and there are no singularities nor black holes.
Check the math here.
the approach appears solid.
you've calculated the energy a falling object will have then calculated the relativistic velocity.

4. I guess that just demonstrates that you can't use special relativity in a situation where general relativity is required.

5. Originally Posted by 37818
Classical escape velocity is given as: Ve = sqr(2GM/r). Which is based on Ke = mV^2/2 where: mV^2/2 = GMm/r. Relativistic escape velocity: Were: mc^2(1/sqr(1 - V^2/c^2) - 1) = GMm/r We get: Ve = sqr(2GM/r + G^2M^2/(r^2c^2)/(1 + GM/(rc^2). Were the escape velocity is never greater than the speed of light. Where light can always escape, and there are no singularities nor black holes. Check the math here.
As Strange said, inside a gravitational field you cannot use special relativity - the full general relativistic treatment is required. The maths above aren't wrong per se, they are just not applicable to this scenario.

6. Originally Posted by Markus Hanke
Originally Posted by 37818
Classical escape velocity is given as: Ve = sqr(2GM/r). Which is based on Ke = mV^2/2 where: mV^2/2 = GMm/r. Relativistic escape velocity: Were: mc^2(1/sqr(1 - V^2/c^2) - 1) = GMm/r We get: Ve = sqr(2GM/r + G^2M^2/(r^2c^2)/(1 + GM/(rc^2). Were the escape velocity is never greater than the speed of light. Where light can always escape, and there are no singularities nor black holes. Check the math here.
As Strange said, inside a gravitational field you cannot use special relativity - the full general relativistic treatment is required. The maths above aren't wrong per se, they are just not applicable to this scenario.
Then given that. Why? And what would be the correct math then?

The math shows:
sqr(1 - (v/c)^2) = 1 /(1 + GM/(rc^2))

That since objects, including photons are not singularities, but have dimensions which give different distances of r. Which results in an acceleration effect - gravity do to relativity. Maybe wrong. But the math works. If it is wrong, there should be calculations to show why it is approximately correct. I think it is at least a good approximation.

Now that being said.

look at:
sqr(1 - 2GM/(rc^2)) = sqr(1 - (v/c)^2).

wjere: v = sqr(2GM/r)

7. Originally Posted by 37818
That since objects, including photons are not singularities, but have dimensions which give different distances of r. Which results in an acceleration effect - gravity do to relativity. Maybe wrong. But the math works. If it is wrong, there should be calculations to show why it is approximately correct.
I really don't know what that means.
It is a good approximation only for weak gravitational fields. If you are talking about strong fields like the one of black holes, you need to use the geodesic equations of GR :

Geodesic - Wikipedia, the free encyclopedia

The solutions to these ( which in turn depend on the metric tensor obtained from the Einstein field equations ) yield geodesics, i.e. trajectories of moving particles. Inside the event horizon of a black hole, all geodesics run inwards towards the singularity, thus no escape is possible.

8. Originally Posted by Markus Hanke
Originally Posted by 37818
That since objects, including photons are not singularities, but have dimensions which give different distances of r. Which results in an acceleration effect - gravity do to relativity. Maybe wrong. But the math works. If it is wrong, there should be calculations to show why it is approximately correct.
I really don't know what that means. It is a good approximation only for weak gravitational fields. If you are talking about strong fields like the one of black holes, you need to use the geodesic equations of GR : Geodesic - Wikipedia, the free encyclopedia The solutions to these ( which in turn depend on the metric tensor obtained from the Einstein field equations ) yield geodesics, i.e. trajectories of moving particles. Inside the event horizon of a black hole, all geodesics run inwards towards the singularity, thus no escape is possible.
Which gives us Newton's equation for velocity where: sqr(1 - 2GM/(rc^2)) = sqr(1 - (v/c)^2) so v = sqr(2GM/r) in relativity. http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf http://physics.ucsd.edu/students/cou...61.26jan11.pdf

9. Originally Posted by 37818
Which gives us Newton's equation for velocity where: sqr(1 - 2GM/(rc^2)) = sqr(1 - (v/c)^2) so v = sqr(2GM/r) in relativity. http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf http://physics.ucsd.edu/students/cou...61.26jan11.pdf
I lost you now. What is your point or question ? I am not certain what you are trying to show / ask.
Are your trying to demonstrate that black holes can't exist ?

10. Originally Posted by Markus Hanke
Originally Posted by 37818
Which gives us Newton's equation for velocity where: sqr(1 - 2GM/(rc^2)) = sqr(1 - (v/c)^2) so v = sqr(2GM/r) in relativity. http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf http://physics.ucsd.edu/students/cou...61.26jan11.pdf
I lost you now. What is your point or question ? I am not certain what you are trying to show / ask.
Are your trying to demonstrate that black holes can't exist ?

The fact is the black hole theory is based on the Lorentz transformation sqr(1 - (v/c)^2) where v = sqr(2GM/r). That does not work correctly with the orbit of mercury for example. Yet the tenser field theory calculations give us 2GM/RC^2 for the velocity squared as used in the Lorentz transformation.

If we use the relativistic kinetic engery equation to calculate escape velocity we get v = sqr(2GM/r + (GM/(rc))^2)/(1 + GM/(rc^2)). Where light can always escape the gravitional well. That is all I'm pointing out. I find this interesting, and yes this suggests that black holes do not really exist. Yet there are astronomical observations of centers of galaxies which are interpreted by scientists to be evidence of predicted black holes.

What I'm asking if the latter is indeed the case, how do we algebraically come up with a better solution?

11. Originally Posted by 37818
The fact is the black hole theory is based on the Lorentz transformation
No, black hole theory is actually based on the Interior Schwarzschild Metric, a solution to the Einstein Field Equations of General Relativity in the interior of a mass distribution :

wherein

This solution describes the curvature of space-time in the interior of a massive, non-rotating, uncharged body. Upon further investigation one finds that a gravitational collapse occurs when the gravitational pressure R diverges beyond . In essence, gravity exceeds all other known forces and all matter collapses in on itself. The critical initial mass is in the region of 30 solar masses; any star with a mass exceeding this limit can collapse into a black hole.

12. Originally Posted by 37818
If we use the relativistic kinetic engery equation to calculate escape velocity we get v = sqr(2GM/r + (GM/(rc))^2)/(1 + GM/(rc^2)).
Ok, let's look at this step by step.
This is obviously something which occurs in space outside the black hole, so we need to use the vacuum solution to the field equations. If we assume that the black hole is static, non-rotating and uncharged, we get the Schwarzschild Metric as simplest solution :

Now let's see what happens if a photon is emitted at the event horizon ( or below ) of a black hole, and travels outwards towards an external observer at some distance . How long does it take the photon to get to the outside observer ? Here's the answer :

This means that a photon emitted at the event horizon of a black hole takes infinitely long to travel back to an outside observer, or alternatively that the escape velocity at the event horizon or below exceeds the speed of light.
Once could also go the other route and calculate the length of the geodesic of a photon between the event horizon and some outside observer, and one would find that it is infinitely long. Either way, the photon can not reach the outside observer.

Thus black holes are in perfect agreement with GR.

13. Originally Posted by Markus Hanke
Originally Posted by 37818
The fact is the black hole theory is based on the Lorentz transformation
No, black hole theory is actually based on the Interior Schwarzschild Metric, a solution to the Einstein Field Equations of General Relativity in the interior of a mass distribution :

wherein

This solution describes the curvature of space-time in the interior of a massive, non-rotating, uncharged body. Upon further investigation one finds that a gravitational collapse occurs when the gravitational pressure R diverges beyond . In essence, gravity exceeds all other known forces and all matter collapses in on itself. The critical initial mass is in the region of 30 solar masses; any star with a mass exceeding this limit can collapse into a black hole.

Then please explain how is derived from this in the Schwarzschild solution,

14. Originally Posted by 37818
Then please explain how is derived from this in the Schwarzschild solution,
I am not certain where you get the above formula from, or what it even means. The leading minus is confusing.
I can only assume that this is meant to be an escape velocity; under GR you get this by inserting the Schwarzschild Metric into the geodesic equations, which yields the energy condition

One can set the constant to zero by choosing appropriate units of measurement, and then obtains

which is in agreement with usual formula for escape velocity for a given distance r from the center of gravity :

Escape velocity - Wikipedia, the free encyclopedia

EDIT : Oh, I get now what you are trying to do ( sorry, not used to this way of doing things ). If you insert the above velocity formula into the Lorentz factor, you get the formula you had quoted earlier, without the leading minus. I assume that leading minus is a mistake.

15. then an infalling object should release and infinite amount of energy.

16. Originally Posted by granpa
then an infalling object should release and infinite amount of energy.
No it doesn't. Why would you think so ?

17. Originally Posted by Markus Hanke
Originally Posted by 37818
Then please explain how is derived from this in the Schwarzschild solution,
I am not certain where you get the above formula from, or what it even means. The leading minus is confusing.
I can only assume that this is meant to be an escape velocity; under GR you get this by inserting the Schwarzschild Metric into the geodesic equations, which yields the energy condition

One can set the constant to zero by choosing appropriate units of measurement, and then obtains

which is in agreement with usual formula for escape velocity for a given distance r from the center of gravity :

Escape velocity - Wikipedia, the free encyclopedia

EDIT : Oh, I get now what you are trying to do ( sorry, not used to this way of doing things ). If you insert the above velocity formula into the Lorentz factor, you get the formula you had quoted earlier, without the leading minus. I assume that leading minus is a mistake.
As it is found in this equation:

ds^2 = - (1 - 2GM/(rc^2)) dt^2 + dr^2/{1 - 2GM/(rc^2))+ r^2 d theta^2 + r^2 sin^2 theta d phi^2

How is the arrived at?

18. Originally Posted by 37818

As it is found in this equation:

ds^2 = - (1 - 2GM/(rc^2)) dt^2 + dr^2/{1 - 2GM/(rc^2))+ r^2 d theta^2 + r^2 sin^2 theta d phi^2

How is the arrived at?
The above is the Schwarzschild Metric, which is a vacuum solution to the Einstein Field Equations, and describes space-time in the surroundings of a static, non-rotating, uncharged spherical mass. I am not certain where you got the above form of the metric from, normally the convention is to give the metric tensor a signature of (+,-,-,-), so that the line element looks like this :

This is the conventional form as normally found in most GR textbooks. In order to derive this, one starts with a set of assumptions like so :

1. The mass is spherically symmetric
2. The mass is static
3. The mass does not have angular momentum
4. The mass does not carry electrical charge

Given these one can then make an ansatz to the ( as yet unknown ) metric, which is really only a generalization of the normal line element in standard spherical coordinates :

reducing the problem to two unknown metric coefficients A(r) and B(r); this is possible because of the spherical symmetry of the body of mass, and it being static. Now, in vacuum ( remember we are looking at the space surrounding the body of mass, not its interior ) one has

and thus the Einstein field equations reduce to

Inserting our ansatz into this tensor equation yields a system of partial differential equations for the two unknown functions A(r) and B(r), which can be solved exactly and analytically. Doing the maths one finds

The form you have given ( where did you get that from, anyway ? ) then results from inserting the A(r) and B(r) into the relativistic Lorentz factor, though I haven't seen the Schwarzschild Metric written in that form before.

19. Originally Posted by Markus Hanke
Originally Posted by 37818

As it is found in this equation:

ds^2 = - (1 - 2GM/(rc^2)) dt^2 + dr^2/{1 - 2GM/(rc^2))+ r^2 d theta^2 + r^2 sin^2 theta d phi^2

How is the arrived at?
The above is the Schwarzschild Metric, which is a vacuum solution to the Einstein Field Equations, and describes space-time in the surroundings of a static, non-rotating, uncharged spherical mass. I am not certain where you got the above form of the metric from, normally the convention is to give the metric tensor a signature of (+,-,-,-), so that the line element looks like this :

This is the conventional form as normally found in most GR textbooks. In order to derive this, one starts with a set of assumptions like so :

1. The mass is spherically symmetric
2. The mass is static
3. The mass does not have angular momentum
4. The mass does not carry electrical charge

Given these one can then make an ansatz to the ( as yet unknown ) metric, which is really only a generalization of the normal line element in standard spherical coordinates :

reducing the problem to two unknown metric coefficients A(r) and B(r); this is possible because of the spherical symmetry of the body of mass, and it being static. Now, in vacuum ( remember we are looking at the space surrounding the body of mass, not its interior ) one has

and thus the Einstein field equations reduce to

Inserting our ansatz into this tensor equation yields a system of partial differential equations for the two unknown functions A(r) and B(r), which can be solved exactly and analytically. Doing the maths one finds

The form you have given ( where did you get that from, anyway ? ) then results from inserting the A(r) and B(r) into the relativistic Lorentz factor, though I haven't seen the Schwarzschild Metric written in that form before.

It is given as the Schwarzschild line solution
In this file: http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf

What I'm interested in is understanding the justification for using simply 2GM/(rc^2) in the equation. Instead of (2GM/(rc^2) + (GM/(rc^2)^2)/(1 + GM/(rc^2))^2

20. Originally Posted by 37818
It is given as the Schwarzschild line solution
In this file: http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf

What I'm interested in is understanding the justification for using simply 2GM/(rc^2) in the equation. Instead of (2GM/(rc^2) + (GM/(rc^2)^2)/(1 + GM/(rc^2))^2
I see, thank you. It is the same solution in a different set of coordinates; this is always allowed since the Einstein equation is invariant under choice of different coordinates.
There is no "justification" as such - it is simply the solution you obtain from the Einstein equations. It's purely maths.

21. Originally Posted by Markus Hanke
Originally Posted by 37818
It is given as the Schwarzschild line solution
In this file: http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf

What I'm interested in is understanding the justification for using simply 2GM/(rc^2) in the equation. Instead of (2GM/(rc^2) + (GM/(rc^2)^2)/(1 + GM/(rc^2))^2

I see, thank you. It is the same solution in a different set of coordinates; this is always allowed since the Einstein equation is invariant under choice of different coordinates.
There is no "justification" as such - it is simply the solution you obtain from the Einstein equations. It's purely maths.
mv^2/2 = GMm/r

which gives us v=sqr(2GM/r)

mc^2 (1 / sqr(1 - (v/c)^2) - 1) = GMm/r

where
sqr(1- (v/c)^2) = 1/(1 + GM/(rc^2))

which gives us
v = sqr( c^2 - c^2 (1 /(1 + GM/(rc^2)))^2)

which is
v = sqr(2GM/r + (GM/(rc))^2))/(1 + GM/(rc^2))

Where escape velocity is a matter of energy required to escape.

22. Originally Posted by 37818
Originally Posted by Markus Hanke
Originally Posted by 37818
It is given as the Schwarzschild line solution
In this file: http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf

What I'm interested in is understanding the justification for using simply 2GM/(rc^2) in the equation. Instead of (2GM/(rc^2) + (GM/(rc^2)^2)/(1 + GM/(rc^2))^2

I see, thank you. It is the same solution in a different set of coordinates; this is always allowed since the Einstein equation is invariant under choice of different coordinates.
There is no "justification" as such - it is simply the solution you obtain from the Einstein equations. It's purely maths.
mv^2/2 = GMm/r

which gives us v=sqr(2GM/r)

mc^2 (1 / sqr(1 - (v/c)^2) - 1) = GMm/r

where
sqr(1- (v/c)^2) = 1/(1 + GM/(rc^2))

which gives us
v = sqr( c^2 - c^2 (1 /(1 + GM/(rc^2)))^2)

which is
v = sqr(2GM/r + (GM/(rc))^2))/(1 + GM/(rc^2))

Where escape velocity is a matter of energy required to escape.
You need to use the GR field equations as demonstrated in posts 13 and 17, not the classical, non-relativistic energy relation. In relativistic terms the energy-momentum relations are different - refer here for an explanation for both the special relativistic and general relativistic cases ( sorry, I'm at work and in a rush, so I can just give the reference for now ) :

Kinetic energy - Wikipedia, the free encyclopedia

23. Originally Posted by Markus Hanke
Originally Posted by 37818
Originally Posted by Markus Hanke
Originally Posted by 37818
It is given as the Schwarzschild line solution
In this file: http://www.physics.usu.edu/Wheeler/G...desicsPost.pdf

What I'm interested in is understanding the justification for using simply 2GM/(rc^2) in the equation. Instead of (2GM/(rc^2) + (GM/(rc^2)^2)/(1 + GM/(rc^2))^2

I see, thank you. It is the same solution in a different set of coordinates; this is always allowed since the Einstein equation is invariant under choice of different coordinates.
There is no "justification" as such - it is simply the solution you obtain from the Einstein equations. It's purely maths.
mv^2/2 = GMm/r

which gives us v=sqr(2GM/r)

mc^2 (1 / sqr(1 - (v/c)^2) - 1) = GMm/r

where
sqr(1- (v/c)^2) = 1/(1 + GM/(rc^2))

which gives us
v = sqr( c^2 - c^2 (1 /(1 + GM/(rc^2)))^2)

which is
v = sqr(2GM/r + (GM/(rc))^2))/(1 + GM/(rc^2))

Where escape velocity is a matter of energy required to escape.
You need to use the GR field equations as demonstrated in posts 13 and 17, not the classical, non-relativistic energy relation. In relativistic terms the energy-momentum relations are different - refer here for an explanation for both the special relativistic and general relativistic cases ( sorry, I'm at work and in a rush, so I can just give the reference for now ) :

Kinetic energy - Wikipedia, the free encyclopedia

Ke = mc^2(1 / sqr(1 - (v/c)^2) - 1).

So instead of mV^2/2 = GMm/r

Using instead, mc^2(1 / sqr(1 - (v/c)^2) - 1) = GMm/r, how is that classical?

In the field equation how is (1 - rs/r) derived?

24. Ok, what is actually your point ? You asked in the OP about how black holes can exist if the escape velocity stays below light speed, and I have answered that in posts 10 and 11. You then asked how the Schwarzschild Metric is obtained, and I answered that in posts 13 and 17.
So now, what exactly is your question ? If you ask clearly, then it will be much easier for me to respond.

25. Originally Posted by Markus Hanke
Ok, what is actually your point ? You asked in the OP about how black holes can exist if the escape velocity stays below light speed, and I have answered that in posts 10 and 11. You then asked how the Schwarzschild Metric is obtained, and I answered that in posts 13 and 17.
So now, what exactly is your question ? If you ask clearly, then it will be much easier for me to respond.
For some reason, which I have yet to understand, it is supposed that for gravity, sqr(1 - (v/c)^2)) = sqr(1 - 2GM/(rc^2)) based on the Schwarzschild radius of 2GM/c^2. Which gives us v = sqr(2GM/r) the same as mv^2/2 = GMm/r.

Algebraically that makes no sense. Where it should be mc^2(1 / sqr(1 - (v/c)^2) -1) = GMm/r, according to relativity.

How the Schwarzschild radius of 2GM/c^2 is derived using tenser field calculus is, I think, at issue. Something is not correct. What? And why?

The algebra? mc^2(1 / sqr(1 - (v/c)^2) -1) = GMm/r or sqr(1 - (v/c)^2)) = sqr(1 - 2GM/(rc^2))?

26. Originally Posted by 37818
How the Schwarzschild radius of 2GM/c^2 is derived using tenser field calculus is, I think, at issue.
I have shown you that already in post 17 - the factor in A(r) and B(r) is the Schwarzschild radius, it follows directly from solving the field equations ( sorry, perhaps the confusion comes from me not writing out the term in full, but abbreviating it with ).
More specifically, when solving the differential equations for A(r), one of the boundary conditions is that the final solution reduces to Newton's law in the case of weak gravitational fields - this condition determines the inevitable integration constant.
Does it make sense now ?

Algebraically that makes no sense. Where it should be mc^2(1 / sqr(1 - (v/c)^2) -1) = GMm/r, according to relativity.
In the presence of gravitational fields you are no longer in an inertial frame, which is why the above expressions do not make sense algebraically. If you use the field equations of General Relativity, as I have demonstrated in posts 13 and 17, you automatically arrive at the correct expression for the Schwarzschild radius.