# Energy left of body falling down to Schwarzschild radius

1. May 16, 2013

### Agerhell

Let us consider the case of a body falling down radially towards a Schwarzschild black hole. The velocity of the body is as high as it would be if it had been falling from standstill at infinity.

When the body finally fuses with the black hole (get infinitely close to the Schwarzshild radius), how much of the energy,mass it had at infinity is added to the black hole?

Is the answer the same whether we use the usual anisotropic Schwarzschild coordinates or isotropic coordinates?

When the falling object comes so close to the black hole that it starts to slow down, there will be radiation losses, I presume, so not all the energy the falling object contained at infinity will be added to the black hole.

2. May 16, 2013

### WannabeNewton

3. May 16, 2013

### Naty1

Do you expect something exceptional outside the horizon of a BH??

Just use any gravitating body..like earth....

4. May 16, 2013

### pervect

Staff Emeritus
All of it (ignoring any radiation loss, that is).

Yes, for the same physical system, the energy at infinity will be the same. Note that coordinate values for $r_i$, the radial coordinate in the isotropic chart, and $r$, the radial coordinate in the standard Schwarzschild chart, will be different for the same physical system because they are defined differently.

For instance, the event horizon will be located at $r_i = GM/2c^2$ , while it will be located at $r = 2GM/c^2$

What you call "slowing down" is purely a coordinate effect. It won't cause any radiation. I wouldn't even say it "slows down", a static observer would only see the velocity increase.

I don't recal seeing any expressions for radiation emitted due to radial infalls, the textbook cases that I can recall analyzed spiral infalls using various approximation schemes.

5. May 16, 2013

### Bill_K

Hm, yes, but in lowest order the radiation is |d3Q/dt3|2 where t is coordinate time. Looks to me like the "slowing down" will cause the radiation to decrease as the object nears the horizon.

6. May 16, 2013

### PAllen

http://arxiv.org/abs/1012.2028

7. May 16, 2013

### Agerhell

The bottom left field of Table 1 says $E_{total}=1.04\times10^{-2}m^2/M$ this maybe is to be interpreted as if the infalling object is a million times lighter than the black hole, one part in one hundred million of its original energy will be lost as radiation?
----

If we have for instance, a proton and an antiproton stationary positioned infinitely close to the Schwarzschild radius of a black hole, smash them together (so slow that we can ignore kinetic energy) and measure the frequency of the resulting radiation high up from the gravitational field it will be infinitely redshifted, correct? The interpretation I would make (a correct interpretation?) is that objects that are stationary close to the Schwarzschild radius of black hole contain virtually no energy.

No if we drop a proton and an antiproton down on a black hole they will ultimately end up infinitely close to the Schwarzschild radius and moving infinitely slow. If they then smash into each other and we measure the frequency of the resulting radiation high up from the black hole it will again be infinitely redshifted and the interpretation would be that the same elementary particles contain less energy if they are deep in a gravitational well.

These two situations seems identical to me and I would assume that this is because most of the energy of the elementary particle is lost as radiation when it closes in on the Scharzschild radius and slows down, but this is perhaps wrong?

I do not now if a black hole will be able to catch only electromagnetic radiation or also gravitational radiation. If so the black hole could be able to catch (some,a lot?) of the original energy of the elementary particle falling down as trapped radiation even though the particles themselves will loose all their energy...

8. May 16, 2013

### WannabeNewton

Well, the conserved energy of the particle's orbit is given by $E = -mu^{a}\xi_{a}$ where $\xi^{a}$ is the time-like killing vector field. If the object is stationary then it follows an orbit of $\xi^{a}$ i.e. $u^{a} = \frac{\xi^{a}}{(-\xi^{c}\xi_{c})^{1/2}}$. If we evaluate this in the coordinate basis of Schwarzschild coordinates we get $E = m(-\xi^{\mu}\xi_{\mu})^{1/2} = m(1 - \frac{2GM}{r})^{1/2}$ so for $r$ very close to the Schwarzschild radius, $E$ is very small.

Klein-Gordon waves, electromagnetic waves, and gravitational waves can all be absorbed by black holes. There is a phenomena that occurs with waves incident on rotating black holes called superradiance. In superradiance, a wave incident on the black hole has a part absorbed by the black hole which carries negative energy into the black hole and the reflected part will then have greater amplitude and energy than the incident wave so this allows for a form of energy extraction from rotating black holes. Klein-Gordon waves, electromagnetic waves, and gravitational waves all allow superradiance within a given range involving their frequency and the angular velocity of the horizon.

9. May 17, 2013

### pervect

Staff Emeritus
Thank you. While I don't follow all the details, a cursory reading indicates that the total radiated gravitatonal wave energy for an object of mass m falling into a black hole of mass M, given m<<M, is approximately:

.01 (m/M) m c^2

10. May 17, 2013

### pervect

Staff Emeritus
I read it the same way.

So far I'd agree with your interpretation. While I would suggest replacing "infinitely close to the black hole" with "in the limit as we approach the event horizon", I essentially agree.

Here is where I start to disagree strongly. Unless it radiates an energy away, the energy at infinity of any object dropped into a black hole is a constant of motion, as has been mentioned.

So, in your specific example, if you drop a proton and an antiproton down a black hole from at rest at infinity, they will have a constant energy at infinity, the same energy they had at infinity.

From the previous argument, we know that if the particles were moving so slow that their kinetic energy could be neglected, that they would have very little energy at infinity. But we know this can't be the case, because the energy is a constant of motion.

Thus, the conclusion one can draw from this is that the particles, dropped from at rest at infinity, are NOT going slow in the sense that they are "so slow that we can ignore kinetic energy".

You are being misled by the coordinate velocities being low into thinking that the kinetic energy is low. Unfortunately, as this example illustrates, the coordinate velocities have little to no physical significance and are actively misleading you when you interpret them as if they had the usual physical significance.

If you consider what happens in the frame of a static observer close to the black hole, it's much less confusing. The particles are moving nearly at the speed of light, so most of their energy is kinetic energy. But while locally they've gained energy, their energy-at-infinity is constant. The kinetic energy they gained, plus their rest energy, when redshifted to infinity, remains constant.

The coordinate velocities gives you incorrect ideas about the magnitude of the kinetic energy, one of several reasons why I *really* don't recommend using coordinate velocities. They generally mislead one about the physics, unless the metric coefficients are close to the Minkowskii ones - which is definitely not true with large redshifts approaching infinity.

Because the velocities measured by a static observer are, by definition, the velocities measured in a nearly Minkowskiian metric, their physical interpretation is relatively straightforwards. One still needs to convert local energy into energy at infinity, of course.

11. May 18, 2013

### pervect

Staff Emeritus
To expand on the correct expression for E:

Suppose we write the metric in the isotropic form as follows:

$-c^2 \, d\tau^2 = -c^2 \, g_{00} dt^2 + g_{11} \left( dx^2 + dy^2 + dz^2 \right)$

where $g_{00} = \left( \frac{1-GM/2rc^2}{1+GM/2rc^2} \right)^2 \quad g_{11} = \left( 1 + GM/2rc^2 \right)^4$

Then $E = g_{00} \left( \frac{dt}{d \tau} \right)$ is the the general relativistic equation for the energy, which is a constant of motion for any infalling particle, written out in component form.

We can write:
$\left( \frac{d \tau}{dt} \right)^2 = g_{00} - \frac{g_{11}}{c^2} \left[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 \right]$

let $\dot{s}^2 = \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2$

where $\dot{s}$ is the not-very-physically signfificant "coordinate velocity".

Then we can write
$\frac{dt}{d\tau} = \frac{1} { {\sqrt{g_{00} -\frac{ g_{11}}{c^2} \dot{s}^2}} } = \frac{1} { \sqrt{g_{00}} \sqrt{1 - \frac {\dot{s}^2 } {c_{loc}^\,2} } }$

where
$c_{loc}\,^2 = \frac {g_{00} c^2}{g_{11}}$

Then
$E = E = g_{00} \left( \frac{dt}{d \tau} \right) = \frac {\sqrt{g_{00}}} { \sqrt{1 - \left( \frac {\dot{s} } {c_{loc}} \right)^2 } }$

If we interpret $c_{loc}$ as the local speed of light, then we can see that this has a semi-intuitive form. The energy gets redshifted by the gravitational time-dilation factor $\sqrt{g_{00}}$, and it gets multipled by a gamma factor $\frac{1}{\sqrt{1-\beta^2}}$ where $\beta$ is the ratio of the coordinate velocity of the infalling particle to the coordinate velocity of light at the same spot.

So objects moving near the local speed of light have a high gamma factor.

Note that we need to use isotropic coordinates in order to make the coordinate velocity of light independent of direction.

Note that due to our choice of coordinates, the energy at the event horizon is formally undefined, as it equal to the ratio of zero over zero. Using L'hopital's rule is one way around this, though the best way is to choose better coordinates that don't have the coordinate singularity at the horizon.

12. May 19, 2013

### Naty1

These particle radiation discussions drive me nuts!! I need some help...

pervect posts, #10:

'close to the black hole event horizon' is the context here, right?

Is this description different from a gravitational source such as a planet? If so how?

Agerhell posts:
Same questions.

Thanks

13. May 19, 2013

### pervect

Staff Emeritus
The context here is that you have a particle free-falling from at rest at infinity. And also (as you remark) that the particle has fallen so that it is close to the event horizon.

I'm basically trying to point out that said particle must have a large amount of kinetic energy.

Angerhell's argument is , as I understand it, basically that it can't have a large amount of kinetic energy, because it comes to a stop (in isotropic Schwarzschild coordinates he favors) as it approaches the event horizon. My point is that the coordinate expression (which are already known to be singular at the horizon) are misleading.

I also go through a rather detailed calculation to come up with an expression for the energy of an infalling particle in the chosen coordinates as a subsequent post, i'm not sure if you've seen it.