Can Gravitational Binding Energy Cause Negative Effective Mass?

[Neutrinos02]

Hello,

in nuclear physics we have a mass defect by the binding energy of the nuclides.
A similar effect appears in the theory of gravitation induced by the gravitational binding energy, which reduces the mass.
But for example at the ISCO of an Kerr black hole we have binding energys about ~0.4 mc² so if a particle would go to r→0 the binding energy E_B→-∞ and so the effective mass will be negative (and the mass defect will also be infinite)?

What is the problem about this thoughts?

Thanks for help.

 

[PeterDonis]

A similar effect appears in the theory of gravitation induced by the gravitational binding energy, which reduces the mass.

To be precise, it reduces the mass of a gravitationally bound object, as measured from a distance, compared to the sum of the masses of a bunch of very small objects, all with negligible gravity, that can be bound together to form the one single bound object.

for example at the ISCO of an Kerr black hole we have binding energys about ~0.4 mc²

In other words, if we took an object at infinity with a mass $m$, and put it into orbit in the ISCO of a Kerr black hole with mass $M$, the final mass of the composite system, measured from a distance, would be $M + (1 - 0.4) m = M + 0.6 m$.

so if a particle would go to r→0

I assume you mean it would fall through the horizon? There are no stable orbits at $r = 0$, or at any $r$ inside the radius of the ISCO, which is still outside the horizon.

the binding energy EB→-∞

No; the concept of "binding energy" doesn't make sense for an object that isn't in a stable bound state. An object that falls through the horizon just adds its mass to the hole. In other words, if we take an object at infinity with a mass $m$, and let it fall into a black hole of mass $M$, the final mass of the system (which will just be a slightly larger black hole) will be $M + m$. No "binding" has occurred.

This brings up a key point which is often overlooked: in order to take an object at infinity with mass $m$, and put it into a stable orbit around a black hole of mass $M$, such that the final mass is something less than $M + m$ (as in the ISCO example above), we have to extract energy from the system. For example, suppose we take the object at infinity and slowly lower it to the radius of the ISCO, allowing it to also gradually gain angular velocity as we lower it, so that by the time it reaches the radius of the ISCO, it also has just the right angular velocity to maintain orbit there. Then we will find that, in that process, we had to extract energy $0.4m$ and let that energy escape to infinity (or capture it very far away, essentially at infinity). If we just let the object free-fall into the hole, without controlling its descent, we aren't extracting any energy, so it all gets added to the mass of the hole.

 

[Neutrinos02]

No; the concept of "binding energy" doesn't make sense for an object that isn't in a stable bound state. An object that falls through the horizon just adds its mass to the hole. In other words, if we take an object at infinity with a mass mm, and let it fall into a black hole of mass MM, the final mass of the system (which will just be a slightly larger black hole) will be M+mM + m. No "binding" has occurred.

But the particle gets a potential energy of -∞ doesn't this influence the mass?

This brings up a key point which is often overlooked: in order to take an object at infinity with mass mm, and put it into a stable orbit around a black hole of mass MM, such that the final mass is something less than M+mM + m (as in the ISCO example above), we have to extract energy from the system. For example, suppose we take the object at infinity and slowly lower it to the radius of the ISCO, allowing it to also gradually gain angular velocity as we lower it, so that by the time it reaches the radius of the ISCO, it also has just the right angular velocity to maintain orbit there. Then we will find that, in that process, we had to extract energy 0.4m0.4m and let that energy escape to infinity (or capture it very far away, essentially at infinity). If we just let the object free-fall into the hole, without controlling its descent, we aren't extracting any energy, so it all gets added to the mass of the hole.

From what would the energy extract and how?

 

[PAllen]

Let's take the simple Newtonian collapse. If a bunch of small bodies coalesce without radiating, you end up with a body hotter than the infalling bodies. This collapsed body would have mass measured from a distance exactly the same as the sum of masses of the original bodies. After it radiates down to the same temperature as the original bodies, the mass will be less by the binding energy. Normally, of course, the radiation occurs continuously during the collapse, but you still typically end up with a hotter body that doesn't directly display the nominal mass defect until it cools.

In Peter's case, you would have to do work to slow down a body falling from afar, or it would have to somehow get rid of KE or it would be too fast for the ISCO.

 

[PeterDonis]

the particle gets a potential energy of -∞

No, it doesn't; the concept of "potential energy" doesn't make sense at or inside the horizon, and if you're analyzing free-fall motion into the hole, "potential energy" isn't telling you anything useful anyway, even outside the horizon.

From what would the energy extract and how?

From the object. Otherwise, as PAllen says, it won't be moving at the right speed for a stable orbit at the ISCO.

 

[Neutrinos02]

I read in a textbook, that the mass of a star is less then the sum of the pieces from which the star is made of? So there should be a mass defect?

From the object. Otherwise, as PAllen says, it won't be moving at the right speed for a stable orbit at the ISCO.

I'm not sure if I understand how a object can loose mass in a stable orbit?

After it radiates down to the same temperature as the original bodies, the mass will be less by the binding energy. Normally, of course, the radiation occurs continuously during the collapse, but you still typically end up with a hotter body that doesn't directly display the nominal mass defect until it cools.

Ok, but in this case the obejct will also loose mass if it falls into a star or a black hole? Or would the mass only reduced in the case of the star? Because the radiation can't escape from the black hole?

 

[PeterDonis]

I read in a textbook, that the mass of a star is less then the sum of the pieces from which the star is made of?

There is a sense in which this is true. Suppose we take $10^{57}$ or so atoms, mostly hydrogen but with some others thrown in, that are in a very big cloud, so big that each atom is far enough from all the others to have negligible gravitational influence on any other one. All the atoms are at rest relative to each other. Then the total mass of the cloud will just be the sum of the rest masses of all the atoms (where "rest masses" here means "the rest masses that we measure in the laboratory").

Now we take the cloud and collapse it into a star like the Sun (the number of atoms I gave is roughly the number of atoms in the Sun, though of course in the Sun they are a plasma with separate electrons and nuclei, not bound into atoms). The mass of the star will be less than the original total mass of the cloud. And in order to form the star, we will have had to extract energy from the system equal to the difference in the masses. (The simplest way to extract the energy would be to let the atoms collide as the cloud collapses, and extract the radiation from the collisions.) Extracting that energy is key; if we don't--if, for example, we confine the cloud in a box that prevents any radiation or anything else from escaping during the collapse process--then the final mass of the system will be the same as the initial mass, since there's no way for it to lose any energy or radiate anything away and therefore no way for it to change its mass.

I'm not sure if I understand how a object can loose mass in a stable orbit?

You're missing the point. The object starts out at rest at infinity. If we just let it free-fall to the radius of the ISCO, it won't be in a stable orbit; it will be moving too fast. So if we want to put it into a stable orbit in the ISCO, we have to slow it down (and we also have to change its direction so it is moving tangentially instead of radially). The process of slowing it down will extract energy from the system, and that energy extraction reduces the mass of the system, as observed from infinity. (Note that local measurements of the object's mass will show it to be unchanged; it's only measurements of the system as a whole from a large distance that will show a reduction in mass.)

in this case the obejct will also loose mass if it falls into a star or a black hole?

Only if we control the process of falling in order to extract energy from it. If we just let the object free-fall into the star or the hole, the total mass of the final system (star or hole + object that fell in) will be the same as the sum of the masses of the pieces beforehand (star or hole + object).

Or would the mass only reduced in the case of the star? Because the radiation can't escape from the black hole?

It would be possible in principle for some energy to be radiated away when the object hits the surface of the star; but probably not much, because a star doesn't have a solid surface. Nothing at all would be radiated away if the object free-falls into a black hole, even in principle.