[MUSIC] 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. So this is the situation, and we want to see what happens with this wave packet. So, this wave packet goes according to 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 and, if one stays stationary of this radius, the relation between proper time and coordinate time is as follows. 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). So you know from this relation what we have is that omega at radius r is the initial frequency was omega0. Omegas the radius r is equal to the following. Relation. But g00 is just a component of the this component. So as a result, this is approximately equal to omega 0 times episolon/rg. From here though, from this relation that this is equal to this, we can find that epsilon. Well, from this relation one can see that epsilon/r is approximately equal to the minus exponent t-r/ rg. 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, 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. That a solution of Maxwell equation for a wave packet in this gravitational potential is like this. 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 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. 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? Geometric optic solution of this equation. This is the behavior and the conditions that we are discussing, that the wave packet started from here and climbed up the potential. Now we want to discuss the spectrum corresponding to this guy. 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 0, square root of (r*rg)* -t-r /(2rg). And this integral is not hard to take and it is proportional to 2omega0 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. 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. So what remains to be said is the following. We want to discuss another theme. We want to complete our lecture with a discussion of another theme. 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