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This is a spin off from another thread ( https://www.physicsforums.com/showthread.php?t=310455&page=2 ) to avoid hijacking the original thread.

DrGreg offered this expression for Potential Energy from the textbook "General Relativity" by Woodhouse:

[tex]PE = m_o c^2*\sqrt{1 - 2GM/(rc^2)}[/tex]

where [itex]m_o[/itex] is rest mass of the test particle, G is the gravitational constant, M is the mass of the large gravitational body, c is the speed of light and r is the radius of the test particle from the centre of M.

Setting units of c=G=1 (geometrised units) and [itex]\gamma = 1/ \sqrt{1-2GM/(rc^2)}[/itex] for the gravitational gamma factor, the equation simplifies to:

[tex]PE = m_o/\gamma}[/tex]

To find out how potential energy is converted into Kinetic Energy (KE) we need to find the velocity of the falling particle at a given height.

The equation for the local velocity of a free falling object is:

[tex]v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}[/tex]

where R is the release height or apogee of the free falling particle. See http://www.mathpages.com/rr/s6-07/6-07.htm for the source of this equation.

When R is set to infinity and G=c=1 the equation simplifies to:

[tex]v = \sqrt{ 1 - {1}/{\gamma^2}}[/tex]

The Total Energy (TE) of a particle is given by:

[tex]TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}[/tex]

where Rest mass Energy [itex] (RE) = m_o c^2[/itex], Momentum Energy (ME) = pc and momentum [itex] (p) = \gamma m_o v [/itex].

The Kinetic Energy (KE) of a falling particle is that part of the Total Energy that is not rest energy so KE = TE-RE.

It should be noted that the gravitational gamma factor is numerically equal to the velocity gamma factor of special relativity for the velocity of an object that has fallen from infinity at any given height.

We are now in a position to do a numerical example.

Initial quanties at infinity per unit mass of the test particle.

Velocity = 0.0

RE = 1.0

PE = 1,0

TE = 1.0

After the particle has fallen to a height (r) of 2.1*(Schwarzschild radius) and assuming a mass of 10,000 units for the black hole

The gravitational gamma factor is: [tex] \gamma = 4.582576 [/tex]

The terminal velocity is [tex] v = 0.975900 [/tex] (97% of the speed of light)

RE = 1.0

PE = 0.218218

ME = 4.472136

KE = 3.582576

TE = 4.582576

A problem arises here. By assumed Rest mass Energy is invariant it turns out that energy is not conserved as promised by Baez. (See http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html "The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls.")

The Total Energy of the particle after falling is greater than the combined total of the initial Rest Energy of the particle and the Potential Energy of the particle at infinity.

What if Potential Energy and Rest mass Energy are the same thing?

Here is the numerical example reworked with the new assumption PE=RE :

Initial quanties at infinity per unit mass of the test particle.

Velocity = 0.0

RE = 1.0

PE = 1,0

TE = 1.0

After the particle has fallen to a height (r) of 2.1*(Schwarzschild radius) and assuming a mass of 10,000 units for the black hole

The gravitational gamma factor is: [tex] \gamma = 4.582576 [/tex]

The terminal velocity is [tex] v = 0.975900 [/tex] (97% of the speed of light)

RE = 0.218218

PE = 0.218218

ME = 0.975900

KE = 0.781782

TE = 1.000000

Now, by assuming the Rest Mass Energy and gravitational Potential Energy of a falling particle are the same thing, energy is conserved, and the gain in Kinetic Energy is a result of loss of Potential Energy which is the same thing as the loss of Rest mass.

The implication, if correct, is that the Rest mass Energy of a particle falling from infinity is completely converted to Kinetic Energy by the time it gets to the event horizon. In effect the falling massive particle becomes a photon at the event horizon which is consistent with the particle attaining the speed of light by the time it gets to the event horizon. If it is true that a particle has no rest mass as it reaches the event horizon then that would be something very difficult to transform away by use of some alternative metric. For example it would not be possible to have a physical observer (with rest mass) fall smoothly and continuously through the event horizon. If it is not true, the equation for PE in the Woodhouse textbook is wrong OR Baez's claim that energy is conserved in Schwarzschild coordinates is wrong, OR the equation for the terminal velocity of a falling particle from the mathpages website is wrong OR my calculations are wrong.

It should be noted that the Potential Energy of a particle goes to zero at the event horizon and the final velocity goes to c at the event horizon for ANY initial height that the particle is dropped from, not just infinity.

Also, as mentioned in previous thread, a photon has no Rest mass Energy and as such it would seem that it is not possible for a photon to have gravitational Potential Energy, so if energy is conserved in Schwarzschild coordinates the Total Energy of a photon does not change as it rises or falls.

Those are my observations on the subject. I look forward to the members of this forum telling me where I have gone wrong.

DrGreg offered this expression for Potential Energy from the textbook "General Relativity" by Woodhouse:

[tex]PE = m_o c^2*\sqrt{1 - 2GM/(rc^2)}[/tex]

where [itex]m_o[/itex] is rest mass of the test particle, G is the gravitational constant, M is the mass of the large gravitational body, c is the speed of light and r is the radius of the test particle from the centre of M.

Setting units of c=G=1 (geometrised units) and [itex]\gamma = 1/ \sqrt{1-2GM/(rc^2)}[/itex] for the gravitational gamma factor, the equation simplifies to:

[tex]PE = m_o/\gamma}[/tex]

To find out how potential energy is converted into Kinetic Energy (KE) we need to find the velocity of the falling particle at a given height.

The equation for the local velocity of a free falling object is:

[tex]v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}[/tex]

where R is the release height or apogee of the free falling particle. See http://www.mathpages.com/rr/s6-07/6-07.htm for the source of this equation.

When R is set to infinity and G=c=1 the equation simplifies to:

[tex]v = \sqrt{ 1 - {1}/{\gamma^2}}[/tex]

The Total Energy (TE) of a particle is given by:

[tex]TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}[/tex]

where Rest mass Energy [itex] (RE) = m_o c^2[/itex], Momentum Energy (ME) = pc and momentum [itex] (p) = \gamma m_o v [/itex].

The Kinetic Energy (KE) of a falling particle is that part of the Total Energy that is not rest energy so KE = TE-RE.

It should be noted that the gravitational gamma factor is numerically equal to the velocity gamma factor of special relativity for the velocity of an object that has fallen from infinity at any given height.

We are now in a position to do a numerical example.

Initial quanties at infinity per unit mass of the test particle.

Velocity = 0.0

RE = 1.0

PE = 1,0

TE = 1.0

After the particle has fallen to a height (r) of 2.1*(Schwarzschild radius) and assuming a mass of 10,000 units for the black hole

The gravitational gamma factor is: [tex] \gamma = 4.582576 [/tex]

The terminal velocity is [tex] v = 0.975900 [/tex] (97% of the speed of light)

RE = 1.0

PE = 0.218218

ME = 4.472136

KE = 3.582576

TE = 4.582576

A problem arises here. By assumed Rest mass Energy is invariant it turns out that energy is not conserved as promised by Baez. (See http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html "The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls.")

The Total Energy of the particle after falling is greater than the combined total of the initial Rest Energy of the particle and the Potential Energy of the particle at infinity.

What if Potential Energy and Rest mass Energy are the same thing?

Here is the numerical example reworked with the new assumption PE=RE :

Initial quanties at infinity per unit mass of the test particle.

Velocity = 0.0

RE = 1.0

PE = 1,0

TE = 1.0

After the particle has fallen to a height (r) of 2.1*(Schwarzschild radius) and assuming a mass of 10,000 units for the black hole

The gravitational gamma factor is: [tex] \gamma = 4.582576 [/tex]

The terminal velocity is [tex] v = 0.975900 [/tex] (97% of the speed of light)

RE = 0.218218

PE = 0.218218

ME = 0.975900

KE = 0.781782

TE = 1.000000

Now, by assuming the Rest Mass Energy and gravitational Potential Energy of a falling particle are the same thing, energy is conserved, and the gain in Kinetic Energy is a result of loss of Potential Energy which is the same thing as the loss of Rest mass.

The implication, if correct, is that the Rest mass Energy of a particle falling from infinity is completely converted to Kinetic Energy by the time it gets to the event horizon. In effect the falling massive particle becomes a photon at the event horizon which is consistent with the particle attaining the speed of light by the time it gets to the event horizon. If it is true that a particle has no rest mass as it reaches the event horizon then that would be something very difficult to transform away by use of some alternative metric. For example it would not be possible to have a physical observer (with rest mass) fall smoothly and continuously through the event horizon. If it is not true, the equation for PE in the Woodhouse textbook is wrong OR Baez's claim that energy is conserved in Schwarzschild coordinates is wrong, OR the equation for the terminal velocity of a falling particle from the mathpages website is wrong OR my calculations are wrong.

It should be noted that the Potential Energy of a particle goes to zero at the event horizon and the final velocity goes to c at the event horizon for ANY initial height that the particle is dropped from, not just infinity.

Also, as mentioned in previous thread, a photon has no Rest mass Energy and as such it would seem that it is not possible for a photon to have gravitational Potential Energy, so if energy is conserved in Schwarzschild coordinates the Total Energy of a photon does not change as it rises or falls.

Those are my observations on the subject. I look forward to the members of this forum telling me where I have gone wrong.

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