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So now we already have finished our discussion of

the black hole type solutions in General Cereal for.

In the remaining part of this lecture we want to discuss two complicated phenomena.

The first one is, we want to discuss behavior

of the electromagnetic waves in the vicinity of the black hole.

So in particular, we want to consider the following situation

that there is a black hole already created.

So this is r=rg.

In the very vicinity of this black hole

the radius r0 which is rg+epsilon.

There is a creation of a wave packet.

There is a wave packet created at this position.

And climbs out from the gravitational potential and goes to some radius r,

which is much greater than r0, which is approximately rg.

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the Full Length Law, because it goes along the radius.

So, its equation of motion is given by this relation.

O=dS squared=(1-rg/r) dt

squared- dr squared /(1- rg/r).

From here one can find that time, which takes for

this wave packet to climb from this from this position to this position,

is equal to the integral from

r0 to r of dr/1-(rg/r)).

Well that follows from here.

This is just equivalent to r-r0+rg.log (r-rg/r0-rg).

And now, we use the fact

that r is much

greater than r0,

hence much greater than rg.

This is approximately equal to r+rg*log

(r/epsilon).

So this establishes a relation between Position radius

at which, how long does it take?

How much time does it take for

the wave packet to climb from radius r0 to a fixed radius r?

As this guy, this wave packet climbs up, the gravitational potential.

It performs a work against

gravitational force, hence its energy is decreasing.

Because its energy is decreasing,

it is natural to expect that its frequency also is decreasing.

How its increasing can be seen from the fact that at radius,r,

the relation between proper time

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But this time defines for us the frequency, because,

omega is proportional to 1/delta t.

So the frequency is just, clicking rate.

So if this is shorter this is bigger, frequency is bigger.

As a result we have the following relation that this guy,

omega times

g00(r) omega 1 at

g00(r)=l to omega 2 at g00(r2).

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As a result, as wave packet climbs up and

time goes by, its frequency behaves according to this law,

that omega is approximately

equal to omega 0*square root of (r/rg)*

the exponent of -t-r/(2rg).

And as a result the phase of the electric magnetic wave,

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which is equal to the integral from 0 to t dt'

omega(t') behaving like this.

It is approximately equal to -t2 omega 0 *square

root of (r*rg) * exponent

of -t -r/(2rg+ con).

Irrelevant for further discussion.

So this is a behavior of the face, what did we obtain?

Let me clarify, what did we obtain.

We basically obtained the following fact.

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Which letter should I use?

I probably should use k*k squared.

So, this is a solution of this equation.

As k goes to 0, k goes to 0, this

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wave equation following for Maxwell equation describing some component,

either of electromagnetic vector potential or

component of electric field or magnetic field.

So in this limit as K goes to zero, we obtain so-called Eikonall equation.

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So basically, what we did, we found the solution of

this equation in the conditions that we have established,

that it started from here and climbed up.

So we have obtained this guy, this is what the meaning of all these manipulations.

So now one can see what one obtains, where we are?

We have obtained the following situation, that

basically we implicitly have found the solution

of this wave equation in the eikonal approximation.

Eikonal, or quasiclassical, or geometric optic approximation.

So this is the equation appearing in the geometric optic, where instead of

waves like solution of this equation, we obtain straight line.

How you say it?

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So, to this wave packet that we are discussing.

We want to make it Fourier expansion,

which is the integral from 0 to infinity over dt times

exponent over here* the exponent

of this guy -2i omega

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square root of

(r*rg) to the power of (2i

omega rg)* exponent of (-pi omega rg)*,

eta gamma function (-2i omega g).

And in this equation we dropped off all the factors that do not depend on omega,

we kept only the factors depending on omega.

This is the spectrum that we obtained for

the wave packet climbing up

from the potential, the geometry.

And the spectrum then is just square root of the modulus of this function.

It is proportional to the -2 pi

omega rg*the gamma function of

(-2i omega rg) squared.

And this is approximately, well,

this is actually equal to pi/omega

g (1/ minus exponent

of( 4pi omega rg -1).

But geometric optic approximation is valid in the limit when omega,

when the wavelengths of the packet is much less than rg.

Which means that this is true.

So only this approximation or this manipulation is correct.

This is approximately equal to (pi*omega

rg) *exponent (-4pi omega rg).

So this all have been obtained in the dramatic optic approximation.

We in no way used quantum mechanics anything,

so h bar was not present anywhere.

But if we restore it there and define that energy as just h bar times omega.

Then this guy is proportional

to the Boltzmann distribution,

where T=h bar/4pi rg.

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So, this is so-called Hawking temperature.

Hawking temperature describing Hawking radiation.

Of course, Hawking radiation is much more complicated process,

it's quantum mechanical.

Demands a much deeper study than what we have done.

We just observe that if there is a source of radiation in the vicinity

of the horizon, due to the geometry, due to the specific Schwarzchild geometry.

This actually exponential factor follows from the specific

of the Schwarzchild geometry.

And due to the specifics of the Schwarzchild geometry,

this wave packet as it climbs up from the black hole to infinity, thermalizes.

So it behaves as if it is described by the thermal spectrum.

Hawking radiation, of course, demands a different derivation.

It is due to the change of the ground state of the quantum field theory.

So, of course, this is not a proof of the Hawking radiation, but

this is a hint why the Hawking radiation should be thermal.

If there is something created in the vicinity of the horizon,

it will become thermal in the infinity.