Light entering a blackhole

Lbeard · 13-02-2013 3:45pm #1

I don't know the maths of General Relativity

But I was thinking. When light tries to escape a black hole, it is red shifted. Is it red shifted to a point, the photons have an infinitely small energy?

Conversely, when light enters a black hole, it is blue shifted. Unless there is some constraint (like the wavelength not being shorter than the Planck length?) a single photon would reach immense energies very quickly. Similarly, any mass on entering a black hole would have immense kinetic energy on impact.

I know no one really knows what happens when matter enters a black hole

Smiles35 · 17-02-2013 12:58pm

Lbeard wrote: »

Similarly, any mass on entering a black hole would have immense kinetic energy on impact.

I'd say it would. Someday we might be able to get a reading from that.

Lbeard · 17-02-2013 3:51pm

Smell Badda wrote: »

I'd say it would. Someday we might be able to get a reading from that.

Well, that's another problem with black holes. You might be able to drop a detector in, getting it back out will be a problem - you'd need an infinitely strong cable.

I was roughly going to try and calculate what kind of energy something falling into a black hole would have.

I was trying some equations but I hit a snag.

Momentum is given by

$gif.latex?p%20=%20mv$

Integrate to get the kinetic energy.

$gif.latex?E%20=%20\frac{1}{2}%20mv^{2}$

But of course since the object that falls in will rapidly reach relativistic speeds - the Lorrentz factor is needed

$gif.latex?p%20=%20mv\gamma$

I'm not sure my integration is correct (in fact I think it's wrong) but the kinetic energy equation will now look like

$gif.latex?E%20=%20\frac{1}{2}\gamma%20mv^{2}$

Then this bit has me really confused.

Since $gif.latex?\gamma%20=%20\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Setting c = 1

$gif.latex?\gamma%20=%20\frac{1}{\sqrt{1-\frac{1^{2}}{1^{2}}}}%20=%20\frac{1}{\sqrt{0}}%20=%20\infty$

Which then mean $gif.latex?E%20=%20\frac{1}{2}\gamma%20mv^{2}%20=%20\frac{1}{2}\infty%20mv^{2}$

If I'm correct, the kinetic energy of any object falling into a black hole, is infinite.

Justin1982 · 17-02-2013 4:04pm

Lbeard wrote: »

I know no one really knows what happens when matter enters a black hole

Well we know theoretically that an observer placed outside a black hole can never see an object entering (ie. crossing the event horizon) a black hole. You can only know what happens by falling in yourself.

They most likely will need a theory of everything to be able to tell us what really happens inside a black hole as gravity on the small scale needs a quantum theory of gravity.

Lbeard · 17-02-2013 6:52pm

Justin1982 wrote: »

Well we know theoretically that an observer placed outside a black hole can never see an object entering (ie. crossing the event horizon) a black hole. You can only know what happens by falling in yourself.

What I have been thinking, but have no idea if it's correct. As an object falls into the black hole's gravity well, the light from the object will be red shifted. If gravity is equal to the speed of light, the Lorrentz factor will reach infinity which means the light can never escape. An event horizon nothing can escape from. Of the many interesting things that happen, light trying to escape reaches an infinitely low energy, light entering reaches an infinitely high energy.

But. If the accelerating of the universe (2011 Nobel Prize) applies everywhere even in black holes. If it is some feature of space and not necessarily anything to do with the existence of dark matter, then the Lorrentz factor never reaches infinity - there is negative gravity. Then I would assume there is no event horizon. And light may escape but it is shifted to such a low energy in its' escape from the gravity well, we cannot observe it.

They most likely will need a theory of everything to be able to tell us what really happens inside a black hole as gravity on the small scale needs a quantum theory of gravity.

I don't have a theory on that. At a guess I might say the Higgs field causes a distortion in space. To find that out you might need to build a stellar sized particle accelerator.

Smiles35 · 18-02-2013 4:31pm

Lbeard wrote: »

light may escape but it is shifted to such a low energy in its' escape from the gravity well, we cannot observe it.

You'd be in good company on this one, Professor Hawking proposes some thermal radiation does be emitted.

Capt'n Midnight · 18-02-2013 5:49pm

At the Schwartzchild radius the escape velocity is the speed of light.

NOTHING , not even light can escape from inside that. You get an event horizon where events beyond are undetectable.

Due to random fluctuations in energy in the universe you can get random particles and anti-particles formed. Happens everywhere , eventually. Low energy areas just take a lot longer for a statistical spike.

If the particle / anti particle are formed on opposite sides of the event horizon then there is a tiny chance that something on the outside will escape - Hawking radiation.

Eventually, and we are talking about long after supposed proton decay, this gradual leakage will eventually allow a black hole to slowly evaporate. And finally the Schwartzchild radius will shrink to a size that will allow radiation to escape and it'll explode. Or something like that

Light entering a blackhole

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