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18

Another way of solving such problems is to go to another reference frame, where you obviously don't have enough energy. For example you've got a $5 MeV$ photon, so you think that there is plenty of energy to make $e^-e^+$ pair. Now you make a boost along the direction of the photon momentum with $v=0.99\,c$ and you get a $0.35 MeV$ photon. That is not ...


11

Yes, there is a simple explanation! The pair production process $\gamma \to e^+ + e^-$ is forbidden by energy-momentum conservation, so it doesn't happen in either frame. One way to see this is that, as a massless object, the photon has "the most possible momentum for its energy", as it doesn't have the extra rest mass-energy. So if you try to make the ...


8

In the beginning, let's investigate the pair production. We know from relativity that mass can be equivalent to energy and if we set the most important physical constants to one $\hbar = c = \varepsilon = 1$, then we have the following relation (assuming that we produce electron-positron pair): $$ E = \omega = \gamma_1 m_{e^-} + \gamma_2 m_{e^+} $$ Where $\...


8

Pair production cannot just happen. It requires some other particle/object to balance out the momentum. Lets suppose this object is a nucleus. So the interaction looks to the fast moving observer like a blue shifted photon scattering off a stationary nucleus and producing an $e\bar{e}$ pair. To us it looks like a nucleus moving at relativistic speed ...


7

This process is the result of the cooperation of two theories of nature: (i) Special relativity: This is a huge topic to study but we shall only need a small part of it, and perhaps the most famous one, which tells us this $E=mc^2$. This equation shows us that matter and energy are equivalent and interchangeable. For example, if an amount of energy $E = ...


6

Other way to see why this is impossible is to look at inverse process: why annihilating positron and electron can't give up only one photon? Imagine these two particles at rest near each other (or look at center-of-mass system). They will annihilate giving 1MeV of energy, but single photon can't pick this energy up by itself because it would also have E/c of ...


6

Basics You need to be able to generate a pair-creation event and be able to image it well enough to know what it was. Getting a pair conversion event Pair creation calls for the highest energy gamma you can get and as much mass in the chamber as you can arrange. The odds of getting a pair-conversion event are graphed in figure 31.17 of the 2013 ...


6

There are three main attenuation processes for high energy photons. The photoelectric effect and Compton scattering are more important at lower energies and their cross-section decreases monotonically with increasing energy. On the other hand, the pair production cross-section takes over and increases towards higher energies. The sum of these is what is ...


5

Pair production of electron/positron happens in the electric field of the atoms to satisfy conservation laws and the same will be true for off mass shell Z0 going into a neutrino antineutrino pair, interacting with the weak field of the atoms. The equivalent to the photon weak interaction mediator is the Z0 and neutrino antineutrino pairs can be formed ...


5

The answer by @gns-ank covers the kinematics of why. Below I tackle the Why does a photon "split" into an electron and positron, and not just bounce off the nucleus in your comment to his answer. In general physics can answer "why" in a nested way, like russian dolls. In the end, the kernel answer is "because it does". In this case though we are in ...


5

look at energy-momentum conservation: $$p_\gamma = p_1+p_2$$ the photon has invariant mass 0 wheras the electron and positron have mass $m_e$ $$p_\gamma^2 = (p_1+p_2)^2 = p_1^2+p_2^2+2p_1\cdot p_2$$ $$0 = 2m_e^2 + 2p_1\cdot p_2$$ $$-m_e^2= p_1\cdot p_2 = E_1E_2-|\vec{p_1}||\vec{p_2}|cos\theta > E_1E_2-|\vec{p_1}||\vec{p_2}| = E_1E_2(1-\beta_1\beta_2) &...


5

You can't simultaneously conserve energy and linear momentum. Let the photon have energy $E_{\gamma} = p_{\gamma} c$ and the electron have energy $E_{-}^{2} = p_{e}^{2}c^2 + m_{e}^{2}c^4$ and an analogous expression for the positron. Suppose the electron and positron depart from the interaction site with an angle $2\theta$ between them. Conservation of ...


4

Check if momentum can be conserved. That ought to do the trick.


4

Pair production is not the same as decay of a particle. A particle can decay into two components according to its decay probability without needing an extra interaction. A lambda in its rest frame will decay into a proton and a pion, for example, within a predictable decay time . There is no rest frame for the photon since its mass is $0$ and it is always ...


4

There's a simple argument that massive bodies that are not black holes cannot emit Hawking radiation. Consider a single proton in its ground state. It does not emit radiation because if it were, it would have to decay to some lower-energy state. This eventually may happen when the proton decays, but the radiation is not black-body radiation, so it can't be ...


4

The general diffeomorphism symmetry in the target space is not a symmetry of the world line theory or, analogously, the world sheet theory! A general spacetime diffeomorphism changes the metric tensor $g_{\mu\nu}(X^\alpha)$ which plays the role of the "coupling constants" (coefficients defining the action, e.g. your exponent) in the world line or world sheet ...


4

It is all hidden in the QED Lagrangian: One can answer this question in a simple way in terms of the QED Lagrangian, at the electron-field interaction part: $L=\bar\psi(\partial_\mu\gamma^\mu-m_e+eA_\mu\gamma^\mu)\psi$ This tells us that interactions of the form: $e^++e^-\rightarrow\gamma$..................(1) $\gamma\rightarrow e^++e^-$..................


4

Let us say that the photon has created pair of massive particles. There must exist a reference frame ("center of mass") in which the total momentum of the two particles is zero. So by momentum conservation, in the same frame the photon had to have momentum zero. But photon cannot have momentum zero, because then it would have zero energy. So momentum is not ...


3

There is no tree level vertex for a neutrino scattering off of a photon, so you have two choices: Weak Drell-Yan if the progenitors are leptons. That is $$l + \bar{l} \to Z^0 \to \nu + \bar{\nu} \,.$$ A multi-step process involving a weak radiative correction. Say running a $W^\pm$ across the out-going lepton lines in a normal Drell-Yan diagram. Unlike the ...


3

Question 3: One time and two space dimensions for simplicity (t, x, y). Photon travelling in +x direction. Photon four momentum is $(\frac{E}{c}, p_x, 0)$. It's null so $$ \frac{E^2}{c^2}-p_x^2=0$$ So $$p_x = \frac{E}{c} $$ $E=h\nu$, so photon four momentum is $(\frac{h\nu}{c}, \frac{h\nu}{c}, 0) $ To keep it simple, assume the electron/positron are ...


3

Conservation of energy and conservation of momentum cannot really be separated, since energy and momentum are just different components of a relativistic 4-vector; different inertial observers will "split" this 4-momentum into energy and momentum in different ways, much like they will "split" spacetime into space and time in different ways. The real reason ...


3

The reply by Emilio Pisanty to your other question also pertains here. But to prove the impossibility of pair production of a photon in vacuum it is not necessary to go into the mathematics of the Lorenz transformations further than you have already done. When one uses valid algebra and from two paths reaches a different answer one has already proven that ...


3

My previous answer is beside the point now that the question has been edited. There is a simpler example of a question of this type which has been analyzed in great detail in the literature, and that involves an electric charge which is stationary in a gravitational field. Since the power radiated is nonzero if a charge is accelerated one might, by the ...


3

The pair production is only possible due to relativistic quantum physics and one needs to describe all these processes by the so-called "quantum field theory" or its generalization (well, string theory is the only example). In quantum field theory, electromagnetic fields are, just like all other fields, quantized. All of the configurations of the ...


3

"Rest mass" is probably a bit more abstract in modern science than what you seem to be thinking. It is simply this: if you can put yourself into an inertial frame such that a body is at rest relative to you, then that body's rest mass is defined as that body's total energy measured in this particular inertial frame - multiplied by $c^2$, if you want to ...


3

What is mass of elementary particles? It is the "length" of the four vector (p_x,p_y,-_z,E), for complex systems it is called rest mass. As the length of three dimensional vectors is not additive ( think adding two opposite momenta), rest masses are not additive, vector algebra has to be used. The invariant mass of your two gammas must be larger than the ...


3

Within current quantum field theory, it does not make sense to ask "how long" a particular process takes to occur. There is a certain probability that a particle and its antiparticle annihilate. But there is no concept of a "process of annihilation". There's the in-state (particle and anti-particle) and the out-state (products of the annihilation, usually ...



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