The Compton Scattering ｜ Where Astronomy Meets Quantum Mechanics ｜ Derivation And Theory

Welcome to the channel The Secrets of the Universe. My name is Rishabh and this is the 8th video in the series of quantum mechanics. In today's video, we are going to learn about another important quantum phenomena, the Compton effect which has many applications in astrophysics also. So, today's video is titled Compton effect where astronomy meets quantum mechanics. As we saw in the Albert Einstein said that light is made up of small packets of energy each having energy h nu where h is the Planck's constant and nu is the frequency. Using this he was able to explain the photoelectric effect but in that year in 1905, he not only solved a long-standing problem of physics but also answered a question no one was asking. Since his childhood, Albert Einstein imagined what would he see if he traveled on a beam of light. His imagination took mathematical form in the year 1905 when he came up with special theory of relativity. As we approach the speed of light, Newton's laws of motion become invalid. They give absurd results. Consider a particle of mass m. It is traveling with a velocity v. Now classical physics says that its momentum p must be the product of its mass and velocity. Right? But in the case of particles of light, in the case of photons, rest mass is zero. So, according to classical physics, the momentum must be zero and hence the energy of the photons or the particles of light must also be zero. But that's not the case. We know that a particle of light travels in space and time. So, it must be carrying some momentum. Also, when you step out, you feel the warmth of the sun. This means those photons or the particles of light, they must be carrying some energy. So, how can we explain this mathematically? Einstein found a way out. Einstein, using special theory of relativity, devised a new formula that is obeyed by each and every particle in the universe. That formula is e square is equal to p square c square plus m square p4. Here, e is the energy of the particle, p is the momentum of the particle, c is the speed of light, m is the rest mass of the particle. Now, if you apply this formula to the photons. So, for the photons or the particles of light, m is equal to zero. So, you get e square is equal to p square c square which becomes e is equal to p into c. So, this is the formula of the energy of a photon. From this formula, you see that momentum is equal to e over c. Now, when you substitute the Planck's law over here, e is equal to h nu, you get h nu over c. I have substituted the value of energy which Planck gave in the year 1900 in this formula. So, I have the final relation of the momentum of the photon, p is equal to h nu by c. This is the final momentum of the photon. So, using Planck's law and the theory of relativity, Einstein not only solved the problem of momentum of a photon but also laid the foundation of antimatter because this relation was the one that was used by Paul Dirac in his derivation of the Dirac equation that ultimately led to the discovery of antimatter. We'll discuss antimatter and the Dirac equation in the coming videos of the series. So, having covered these basics, we are now ready to understand what is Compton effect and why is it so important in quantum mechanics. X-rays were first discovered in the year 1895. So, in the first quarter of the 20th century, that is from 1900 to 1925, research on how do x-rays interact with matter was well underway. In 1923, Arthur Compton observed something that is now considered to be the most conclusive evidence of the particle nature of light. So, the setup of the Compton experiment is as follows. We have a source of x-rays that is emitting monochromatic x-rays at a wavelength lambda. These x-rays hit a piece of graphite and are scattered off at an angle theta. Then, we have a detector D that can detect the scattered x-rays at any angle. Now, where does the problem lie? The problem lies over here. The photons of the x-rays that are scattered have a different wavelength lambda dash than what is being emitted by the source lambda. This is a problem because according to the classical theory of physics, when electromagnetic waves are scattered off the charged particles, their frequency, energy and wavelength must remain the same. But Compton observed that the wavelength of the scattered x-ray photons is different from what is being emitted by the source. Remember, these are highly energetic x-ray photons. So, the scattering of the x-ray photons or the highly energetic photons by charged particles such as electrons is known as the Compton effect. Now, let us see how mathematically Compton could explain the scattering of x-rays and the change in the wavelength of x-rays using the particle nature of light. So, Compton explained this entire phenomena by considering it as a process of collision between two particles. Suppose, we have a photon, x-ray photon that is traveling towards a stationary electron. After scattering, the photon and the electron travel in different directions. So, we have the photon that is traveling at an angle theta with respect to the original direction and the electron also travels in another direction. The momentum before and after the collision have changed. So, in order to solve this mathematically, the problem of Compton effect, all you have to do is to consider two important laws of conservation. Conservation of linear momentum and conservation of energy before and after the collision. Let's start with the conservation of momentum. So, before the collision, we have the momentum of photon p plus momentum of electron. Since electron is stationary, its momentum before the collision is zero. So, p plus zero, this is before, is equal to the momentum of the particles after the scattering process. So, we have momentum of photon after collision is p dash. So, it has changed, p dash plus. Now, since electron has also started moving, it has gained momentum, let's say p e. So, from this, we get p e is equal to p minus p dash. You take the square of both these sides, the magnitude of these vectors. So, we get p e whole square is equal to p minus p dash whole square. You expand the square, p square plus p dash square minus 2 p p dash cos of theta because these are vectors. So, we have to take the angle between them, that is theta, cosine of the angle theta. So, this is p e dash. Now, you have to substitute the value of momentum that we discussed in the first part of the video. So, the momentum of the photon before collision was h nu by c. So, you take the square plus. Now, since after the collision, its frequency has changed to nu dash, we get h nu dash by c whole square. So, this is the difference. Before the collision, the momentum is h nu by c and after the collision or scattering, the momentum is h nu dash by c minus 2. So, now you substitute the values over here. You get h square over c square nu nu dash cos of theta. So, this you take out h square over c square common. h square over c square comes out common. We get nu square plus nu dash square minus 2 nu nu dash cos of theta. This is p e square. Now, next you have to apply the conservation of energy. So, for conservation of energy, energy before collision that is, now photon is travelling with energy e plus the kinetic energy of the electron before the collision is 0 because it is stationary. But according to relativity, it does have what is known as the rest mass energy e is equal to m c square. So, in the energy equation, electron contributes its rest mass energy m e c square where m e is the mass of the electron. So, this is before the scattering and after scattering, energy of the photon becomes e dash plus energy, total energy. Now, since it also contains the kinetic energy part, let us say it becomes e e dash. We know the value of e and e dash but we do not know the value of this parameter, the total energy of the electron. So, we make the use of this formula which we discussed, Einstein's formula for total energy of the particle. So, which is e square is equal to p square c square plus m square c 4 and you substitute this value over here by taking its under root. So, let me go over here e. So, let me substitute all the values now. Energy of the photon before scattering is h nu plus m e c square remains the same plus is equal to e dash is h nu dash energy of the photon after scattering plus. Now, you have to substitute the value of this over here under root of p e momentum of the electron square c square plus m e square c 4. So, what I have done is we did not know the value of the total energy of the electron over here because it contained the kinetic energy and the rest mass energy. So, we made use of Einstein's equation of energy which is obeyed by each and every particle. So, it is also obeyed by electron and we took its square root and substituted it over here. Now, we have the energy in terms of momentum. This value we have calculated just now from the conservation of momentum over here. So, now what you have to do is you have to substitute this entire value. We have the value of p e square into this equation and see what we get. So, h nu plus m e c square is equal to h nu dash plus under root of. Now, see we have a c square term with p e square and we also have a c square term over here in the denominator. So, this c square and this c square gets cancelled. So, we are only left with the numerator. So, we are left with h square into nu square. I am writing this value over here plus nu dash square minus 2 nu nu dash cos of theta plus this term remains as it is m e square c, right. Now, what you have to do is you have to take out h common from here h square. You take h square out of the square root. So, this remains the same left hand side h nu plus m e c square is equal to h nu dash plus you take out h square. So, after you take out h square from the under root, you get only h under root nu square plus nu dash square minus 2 nu nu dash cos of theta. Now, since there was no h square term in this m e square c 4, you have to divide it by h square to balance the equation over h square. So, this is what we get finally. Now, you cut out h from here. So, if you cut h, you are left with nu plus m e c square over h because we have cancelled out h. There was no h over here. So, we have to divide it by h is equal to we have actually divided both the sides by h nu dash plus under root of this entire term nu square plus nu dash square minus 2 nu nu dash cos of theta plus m e square c 4 over h square. Now, what we have to do is you take this to the left hand side nu minus nu dash plus m e c square over h is equal to and this remains the same nu square plus nu dash square. This is the simple algebra nothing much algebra in trigonometry m e square c 4 over h square. Now, in order to solve this equation, there is a simple trick. You have to take the square of both these sides, but instead of applying a plus b plus c whole square on the left hand side, you consider this a and this as b and apply simple a plus b whole square is equal to a square plus 2 a b plus b square. When you solve this and simplify this, you get 1 by nu dash. I have told you the trick how to do it. Just take the square by considering it as one term. It will make the calculation easier. 1 over nu dash minus 1 over nu is equal to h over m e c square into 1 minus cos theta. Just a couple of steps more and you will get to this relationship. This is the relationship in terms of frequency. So, in order to convert it to wavelength, you use the simple result c is equal to the relationship between frequency and the wavelength. So, when you substitute this over here, you get lambda dash minus lambda is equal to h over m e into c 1 minus cos of theta. This is the final relationship of the Compton effect. So, there are two important things to note about this expression. The first is that the shift in the wavelength that is lambda dash minus lambda only depends on the angle of scattering. It does not depend on the radiation or the intensity of the radiation from source. Second is that when is the maximum shift occurring? That is what is the value of the maximum shift? The maximum of this expression occurs when this 1 minus cos theta is at maximum because h is blank constant, m e is the mass of electron and c is the speed of light. These three are fundamental constants. The only varying parameter is 1 minus cos theta. Its maximum occurs when the photons that are travelling are completely back scattered. In that case, cos theta is minus 1. So, lambda dash minus lambda maximum is equal to. So, cos theta gives minus 1. This becomes 2 h over m e into c. Now, when you substitute the values of all these constants, you get this relationship to be 0.05 angstroms or 5 into 10 raised to power minus 12 meters. This number is very small for the case of visible light because for visible light, the wavelength is in the range of 4000 to 8000 angstroms. So, this number 0.05 angstroms is very less as compared to the visible range that is from 4000 to 8000 angstroms. So, Compton effect is negligible for low energy photons. In that case, Thomson scattering or the Thomson effect is more relevant. Now, let's discuss the applications of Compton effect in astrophysics. It's one of the many places where astrophysics and quantum physics meet. On that board, I have drawn the spectrum of a galaxy named Mercurian 421. So, it's a graph between the flux and the log of the frequency of the galaxy. It's a blackbody spectrum. Remember, we discussed in the previous videos how stars can also behave like blackbodies. If you notice carefully, there are two parts of this spectrum. The first is the soft part which is also known as the low energy peak and the second is the hard part which is also known as the high energy peak. Now, this soft part of the spectrum, the soft low energy peak can be easily explained using realistic temperatures, millions of degrees Celsius. At these temperatures, molecules break into atoms and these atoms are ionized to create the fourth state of matter that is plasma. In this particular galaxy, matter is circling around the supermassive black hole and due to collisions of atoms, the accretion disk is heating up. But, the hard part of this spectrum is too energetic to be explained by colliding atoms. So, how do we explain it? The answer is inverse Compton effect. The hard part of the spectrum is just a set of photons from the soft spectrum which have been boosted in energy and frequency and collisions with the beam of energetic electrons. So, the photons of the soft part because of the inverse Compton effect, they get a boost in energy and they appear as the hard part of the spectrum which is highly energetic. So, the inverse Compton effect has many applications in high energy astrophysics i.e. in understanding the nature of the active galaxies. It is also used in understanding the cosmic microwave background radiation. So, this was all about the Compton effect and today's video. 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