# Is a black hole's mass zero according to a distant observer?

1. Dec 16, 2012

### johne1618

According to the Schwarzchild metric an interval of proper time $d\tau$ at a fixed distance $r$ from an object of mass $M$ is related to an interval of co-ordinate time $dt$ measured by a distant observer:
$$d\tau = dt \sqrt{1 - \frac{2GM}{rc^2}}$$
This is just the gravitational time dilation formula. Rearraging this expression we get
$$\frac{1}{dt} = \frac{1}{d\tau} \sqrt{1 - \frac{2GM}{rc^2}}$$
Now according to quantum mechanics the total energy of any system is proportional to its quantum frequency.

Thus the rest mass of a quantum system, $m_0$, is inversely proportional to the proper period of its quantum oscillation $d\tau$
$$m_0 \propto \frac{1}{d\tau}$$
The mass of the system from the point of view of the distant observer who uses co-ordinate time $t$ is
$$m \propto \frac{1}{dt}$$
Substituting into the rearranged time dilation formula we get
$$m = m_0 \sqrt{1 - \frac{2GM}{rc^2}}$$
Thus as a mass $m_0$ is lowered towards a gravitating mass $M$, the mass as seen by a distant observer at $r=\infty$, $m$, is reduced.

If the gravitating body is a black hole then as the mass approaches the Schwarzchild radius it disappears from the point of view of the distant observer.

If we imagine the whole mass of the black hole as having been lowered into its own metric then does this imply that the mass/energy of the black hole is zero from the point of view of a distant observer?

2. Dec 16, 2012

### Staff: Mentor

For the appropriate definition of "mass", yes. By that definition, though, "mass" is equivalent to "total energy". You don't even need quantum mechanics to show this; the same equation holds classically for the energy of an object at a finite radius r, compared to its energy "at infinity". But "total energy" is not the same as, for example, "number of atoms in the object"; that must be constant for the derivation you give to be valid. So none of the "substance" in the object changes.

(You could test this, for example, by measuring the object's mass locally, for example by pushing it with a known force and measuring its acceleration. Any such local measurement would find the object's mass to be m, *not* m0. This is one illustration of the fact that energy is frame-dependent in relativity.)

You also left out a key qualification in your equation: it applies to an object that is *static*, i.e., that remains at a constant radius r. And static objects can only exist outside the horizon. So there can never be an object with a total energy of zero according to your equation, because such an object would have to be static at the horizon, and that's not possible.

Finally, you have not considered *how* an object would get from "infinity" to a finite radius r; adding that to the picture makes it clearer what your energy calculation really means. Suppose that I attach an object with energy at infinity m to a rope; one end of the rope is held static "at infinity", and the other end, with the object attached, is slowly lowered, so that work can be extracted by the lowering process (for example, say the rope is attached to a system of gears that spins up a flywheel, storing energy in it). When the object is lowered to some radius r, its total energy will now be m0 instead of m; and the difference, m - m0, is just the energy that we stored in the flywheel.

So what you are really calculating is the maximum amount of work that can be extracted by a process that starts with the object static "at infinity", and ends with the object static at some finite radius r. (It's "maximum" because all that energy can only be extracted as work if the system is perfectly efficient: for example, in my setup above, there would have to be zero friction in the gears and the flywheel.) But again, an object can't be static at the horizon, so you can never extract *all* of the energy an object has at infinity by such a process.

No. The object that collapses to form the hole is not static, so your calculation doesn't apply to it.

3. Dec 16, 2012

### zonde

Can you explain what do you compare? Is it two observers and one gravitating object? Or is it one observer and collapsing gravitating object at different stages of collapse?

Because if we speak about two observers and the same gravitating object we would attribute different viewpoint to the difference between observers themselves. Then it does not make much sense to say that one observer sees the mass as non zero while the other sees it as zero.

If you ask about collapsing mass at different stages of collapse then there we can speak about binding energy released by collapsing body. And your question might be understood as: Can the gravitating body release all it's mass as binding energy?

4. Dec 17, 2012

### pervect

Staff Emeritus
It's not true that the mass of a black hole is zero in any sense I've ever sen or read about.

To oversimplify slightly:

If you have a 1kg mass at infinity and you drop it into a black hole on the free fall trajectory, the energy at infinity of the free-falling mass stays constant as it falls. When it finally enters the black hole, the black hole gains 1kg of mass.

If you have a 1kg mass at infinity , and lower it down with an idealized rope, the situation is different. You could, for instance, wind the rope around a shaft and use the falling mass to turn the shaft and generate power. The black hole gains less than 1kg of mass in this case, but the generator does useful work.

Lowering stuff into a black hole with a rope to generate energy isn't really practical, but infalling dust can and does heat up and radiate a lot of energy away as it spirals into a black hole, energy that escapes towards infinity and doesn't add to the mass of the black hole.

I've seen some figures for the amount of energy that typically gets radiated away - unfortunately I dont recall the references or the values, but I do recall it can be a substantial fraction of the rest energy (rest mass) of the infalling object.

Warning: while this makes it sound like GR always has a well defined notion of energy, this is misleading. However, in this particular case (static observers) there is a reasonable definition of energy one can use.