Gravity Wells steal Energy & Mass ??

[Widdekind]

Please ponder the Gravitational Potential Well of a massive central body:

[math]U = - \frac{G \; M}{R}[/math]

Crucially, Gravitational Potential Energy is negative. Thus, other bodies placed into this Gravity Well will lose Energy -- and, hence, by Einstein's Mass-Energy Equivalence (E = m c²), they should lose Mass, equal to:

[math]\Delta M = \frac{U}{c^{2}} = - \frac{G \; M}{c^{2} \; R}[/math]

Plugging in appropriate numbers, this yields Fractional Mass Loss effects of order:

• At Earth's surface -- 10^-9
  
• At Earth's orbit -- 10^-8
  
• At Sun's surface -- 10^-5
  
• At Neutron Star's surface -- 10^-2
  
• At Black Hole Event Horizon -- 1
  

Thus, by these admittedly primitive classical arguments, objects passing through a Black Hole's Event Horizon have lost more Energy than their Rest Mass Energy Equivalent (!!), ignoring of course any gains of Kinetic Energy*.

*

It seems strange, to this author, that, from the perspective of a distant observer,

Time

slows for objects accelerated to large

positive

Kinetic Energies

, and for objects dropped down into large

negative

Potential Energies

.

[D H]

You are mixing and matching Newtonian and relativistic mechanics here. This is not a good idea in general and is a very bad idea in this case. Gravitational force is a fictional force in general relativity. There is no such thing as gravitational potential energy in general relativity.

 

The mass-energy an infalling object adds to a black hole is the object's energy-at-infinity. That the object follows a geodesic into the black hole doesn't change that.

[Martin]

Please ponder the Gravitational Potential Well of a massive central body...

 

Thus, by these admittedly primitive classical arguments...

 

Wait Widdy! Don't you mean 1/2 instead of 1? I'll make the change and see if you like it:

 

• At Earth's surface -- 10^-9
  
• At Earth's orbit -- 10^-8
  
• At Sun's surface -- 10^-5
  
• At Neutron Star's surface -- 10^-2
  
• At Black Hole Event Horizon -- 1/2
  

 

I hope you hear the wise note of caution in DH's post. However people do apply classical and semiclassical analysis to black holes, and it can be interesting to see what you get.

 

In this case if you had a pulley machine and could lower a mass, say like Bernard Madoff, or a broken television set, down to the black hole horizon and have that turn the pulley and generate electricity. Then you would get a large fraction of the energy that you would get by converting Bernard Madoff into pure energy. Say by reacting him with another Madoff made of antimatter.

 

But using the pulley machine I think you wouldn't get the whole energy equivalent, only 50%.

Nothing in this world is perfect.

 

This would be another interesting chapter to put in a textbook, or a discussion to have with a class, if you should ever teach a Fun Astrophysics course.

 

The students will be asking questions like "what happens to the mass???!!!"

If you measure the mass of a bh, and also a kilogram brick a long ways away. And you drop the brick in. Then the end mass of the bh will be the old mass plus the mass of the brick.

 

But if you lower it down gradually, extracting the energy and using it, say to heat many many cups of coffee. And then you cut the brick loose at the horizon. Then the mass of the bh will only be equal to the old mass plus 1/2 of the mass of the brick.

 

And the students will ask "what happened to the other half kilogram???!!!" and the answer is that all those cups of coffee are now more massive by a total of half a kilogram because of the heat that went into them. Cups of hot coffee weigh more.

 

Is that right? Or did I make a mistake? Notice that the Schw radius is 2GM/c^2, not GM/c^2. You forgot a factor of two in your calculation (something physics teachers habitually do.) If I made an error let me know please. In any case this would make a nice 15 minute class discussion and might lead some of the students to learn a whole lot more about gravity and black holes on their own. "Black holes as an energy source" etc etc.

Edited June 9, 2009 by Martin

[Widdekind]

You are mixing and matching Newtonian and relativistic mechanics here. This is not a good idea in general and is a very bad idea in this case. Gravitational force is a fictional force in general relativity. There is no such thing as gravitational potential energy in general relativity.

 

The mass-energy an infalling object adds to a black hole is the object's energy-at-infinity. That the object follows a geodesic into the black hole doesn't change that.

 

Wow. How does GR explain the infalling particle's increased energy, [math]\gamma m c^{2}[/math] ? Where does the KE come from ?

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Merged post follows:

Consecutive posts merged

...

Is that right? Or did I make a mistake? Notice that the Schw radius is 2GM/c^2, not GM/c^2. You forgot a factor of two in your calculation (something physics teachers habitually do.) If I made an error let me know please. In any case this would make a nice 15 minute class discussion and might lead some of the students to learn a whole lot more about gravity and black holes on their own. "Black holes as an energy source" etc etc.

 

That's true. I was only making an Order-of-Magnitude estimate, since I suspected that something was amiss, w.r.t. GR vs. Newtonian.

[D H]

Wow. How does GR explain the infalling particle's increased energy, [math]\gamma m c^{2}[/math] ? Where does the KE come from ?

Short, snide answer:

It comes from doing physics incorrectly.

 

 

Medium length answer:

It comes from doing physics incorrectly in a non-inertial frame. In Newtonian mechanics it can very convenient to use non-inertial frames to model some processes (e.g., the Earth's atmosphere). However, when you use a non-inertial frame you need to invoke fictitious forces and fictitious energies to make the physics come out right. Those fictitious forces and energies of course vanish when you do physics from the perspective of an inertial frame.

 

If you are doing physics in an inertial frame you need don't need to explain where the coriolis energy, [math]-m \boldsymbol v \times (\boldsymbol{\omega}\times \boldsymbol r)[/math], comes from. It doesn't exist in an inertial frame. The same goes for the relativistic kinetic energy [math]\gamma m c^{2}[/math]. It doesn't exist in an inertial frame.

 

 

Long-winded answer:

So what are inertial frames in general relativity? The equivalence principle is the key to answering this question. Suppose you are in an enclosed, windowless space capsule. You release a ball and it falls toward the bottom of the capsule. How can you distinguish between the capsule being at rest on the surface of a sufficiently large planet versus being out in space far from any massive object with rockets making the capsule accelerate? You can't. The gravity field looks on the planet is approximately uniform, and the deviation from uniformity is too small to measure within the confines of the tiny capsule. Newtonian mechanics says that a reference frame based on the accelerating capsule is a non-inertial frame while a reference frame based on the planet-bound capsule is an inertial frame.

 

Now imagine that the capsule is out in space and its thrusters are inactive. How can you distinguish between the capsule being in orbit about some planet versus being very, very from any massive object? Once again, there is no way to distinguish between the two situations. Yet Newtonian mechanics says that a capsule-based reference frame is non-inertial in the first situation and inertial in the latter. There appears to be something wrong with the Newtonian definition of an inertial frame if the equivalence principle is correct.

 

The Newtonian concept of an inertial frame is in a sense deeply flawed. If a capsule-based frame is not an inertial frame for the accelerating capsule in deep space it had better not be an inertial frame in the case when the capsule is at rest on the surface of a planet. If a capsule-based frame is an inertial frame in the case of the non-accelerating capsule in deep space it had better be an inertial frame in the case of a capsule in orbit. The solution: The origin of an inertial frame in general relativity is a free-falling point.

 

Another flaw in the Newtonian concept of an inertial frame is that inertial frames have infinite extent in Newtonian mechanics. Inertial frames only have local extent in general relativity. Suppose the orbiting space capsule is sufficiently large. If you place a ball at rest at the capsule's center of gravity the ball will stay in place. If you place a ball at rest far from the capsule's center of gravity the ball will begin to fall because the planet does not have a uniform gravitational field. Given a sufficiently large capsule, sufficiently sensitive equipment, and sufficiently long enough periods of time, you can distinguish between being in orbit versus being adrift in deep space.

 

Finally, to answer the question: That relativistic kinetic energy arises from doing physics incorrectly. You are assuming inertial frames have an infinite extent. They don't.

[Widdekind]

(Thanks DH for the fuller, Relativistic, explanation.)

 

 

Brashly combining Newton & Einstein again, would deny the existence of Black Holes. For, from our Escape Velocity Equation, Relativistically revised, we have that:

[math]KE = -U[/math]

[math]\left( \gamma - 1\right) \; m \; c^{2} = \frac{G \; M}{R}[/math]

And, the LHS of that equation is unbounded. Thus, for any positive & finite Mass & Radius, you can find some finite Relative Velocity ([math]0 \leq \beta \leq 1[/math]), such that your KE exceeds your GPE.

 

CONCLUSION (?!?): Sufficiently relativistically-fast particles can always plow up out of any Gravity Well. Only the sufficiently slower particles would be bound back to the Gravitating Body.

 

Likewise, Light could be claimed to Gravitate, by Mass-Energy Equivalence:

[math]F_{g} = - \frac{G \; M \; \left(\frac{h \; \nu}{c^{2}}\right) }{R^{2}}[/math]

 

However, for Light, there would still be some sort of Black Hole criterion, requiring:

[math]h \; \nu > \frac{G \; M \; \left(\frac{h \; \nu}{c^{2}}\right)}{R}[/math]

[math] 1 > \frac{G \; M}{c^{2} \; R}[/math]

which would seem to make the "Schwarzschild Radius" half of its Relativistic value (being the same radius, at which a massive body's GPE exceeds its RME-eq., as per previous posts).

Edited June 12, 2009 by Widdekind

[D H]

Brashly combining Newton & Einstein again, ...

Why are you brashly embarrassing yourself again? What you wrote was nonsense.

[Widdekind]

I only pursued the "semi-Relativistic" scenario some more, "for fun", "to see what happened".

 

 

Would you (DH) please explain how GR can account for Martin's observation, that the "GPE" gained by particles sinking into Gravity Wells can be radiated away ?? For, if "GPE" is fictitious, akin to Coriolis Energy, then, how could GR account for a particle being able to produce Light Energy as it nears other massive bodies ?? Where could that Radiated energy come from, if not from the fictitious GPE ??

 

(Thanks again in advance.)