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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 ...

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 ...

6

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 $E = ... 6 Basics You need to be able to generate a pair-creation event and be able to image it well enough to know what it was. Getting a pair conversion event Pair creation calls for the highest energy gamma you can get and as much mass in the chamber as you can arrange. The odds of getting a pair-conversion event are graphed in figure 31.17 of the 2013 ... 6 Other way to see why this is impossible is to look at inverse process: why annihilating positron and electron can't give up only one photon? Imagine these two particles at rest near each other (or look at center-of-mass system). They will annihilate giving 1MeV of energy, but single photon can't pick this energy up by itself because it would also have E/c of ... 5 Pair production of electron/positron happens in the electric field of the atoms to satisfy conservation laws and the same will be true for off mass shell Z0 going into a neutrino antineutrino pair, interacting with the weak field of the atoms. The equivalent to the photon weak interaction mediator is the Z0 and neutrino antineutrino pairs can be formed ... 5 The answer by @gns-ank covers the kinematics of why. Below I tackle the Why does a photon "split" into an electron and positron, and not just bounce off the nucleus in your comment to his answer. In general physics can answer "why" in a nested way, like russian dolls. In the end, the kernel answer is "because it does". In this case though we are in ... 5 You can't simultaneously conserve energy and linear momentum. Let the photon have energy$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 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 ... 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 ...

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

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

look at energy-momentum conservation: $$p_\gamma = p_1+p_2$$ the photon has invariant mass 0 wheras the electron and positron have mass $m_e$ $$p_\gamma^2 = (p_1+p_2)^2 = p_1^2+p_2^2+2p_1\cdot p_2$$ $$0 = 2m_e^2 + 2p_1\cdot p_2$$ $$-m_e^2= p_1\cdot p_2 = E_1E_2-|\vec{p_1}||\vec{p_2}|cos\theta > E_1E_2-|\vec{p_1}||\vec{p_2}| = E_1E_2(1-\beta_1\beta_2) ... 4 Check if momentum can be conserved. That ought to do the trick. 3 Question 3: One time and two space dimensions for simplicity (t, x, y). Photon travelling in +x direction. Photon four momentum is (\frac{E}{c}, p_x, 0). It's null so$$ \frac{E^2}{c^2}-p_x^2=0$$So$$p_x = \frac{E}{c} $$E=h\nu, so photon four momentum is (\frac{h\nu}{c}, \frac{h\nu}{c}, 0)  To keep it simple, assume the electron/positron are ... 3 The reply by Emilio Pisanty to your other question also pertains here. But to prove the impossibility of pair production of a photon in vacuum it is not necessary to go into the mathematics of the Lorenz transformations further than you have already done. When one uses valid algebra and from two paths reaches a different answer one has already proven that ... 3 Pair production is not the same as decay of a particle. A particle can decay into two components according to its decay probability without needing an extra interaction. A lamda in its rest frame will decay into a proton and a pion, for example, within a predictable decay time . There is no rest frame for the photon since its mass is 0 and it is always ... 3 My previous answer is beside the point now that the question has been edited. There is a simpler example of a question of this type which has been analyzed in great detail in the literature, and that involves an electric charge which is stationary in a gravitational field. Since the power radiated is nonzero if a charge is accelerated one might, by the ... 3 The pair production is only possible due to relativistic quantum physics and one needs to describe all these processes by the so-called "quantum field theory" or its generalization (well, string theory is the only example). In quantum field theory, electromagnetic fields are, just like all other fields, quantized. All of the configurations of the ... 3 There is no tree level vertex for a neutrino scattering off of a photon, so you have two choices: Weak Drell-Yan if the progenitors are leptons. That is$$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

"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 ...

3

in a similar manner in which a fission reaction is moderated I think you are confusing ideas here, and it is leading you to ask a somewhat nonsensical question. The individual nuclear fission events don't change at all in a moderated nuclear pile. What does change is the odds of a neutron from one causing another and how long it take for that ...

2

The occurrence of pair production within the detector depends on gamma energy and detector material. The occurrence of escape peaks in addition to a given full-energy peak depends on detector geometry and sample geometry. If the gamma energy is large enough to make pair production relevant, the photon may disappear and be replaced by an electron and a ...

2

I talked to my friend about this, and he said something along the lines of "the pair production particles aren't 'real' particles, and they can't really interact with anything", though I'm not too sure I understand what that means. Your friend is correct. A virtual particle is a mathematical construct, it is a creation of the mathematical model used ...

2

"Pair production is when a particle-antiparticle pair is produced from a single gamma photon. The gamma photon must have enough energy to produce that much mass. Pair-production usually happens near a nuclear, which helps to conserve momentum." It is a hand waving way of describing pair production. Gammas are photons of high energy by definition. ...

2

Gamma decay is the emission of photons. You are thinking of $\beta$ decay. When the particle is decaying, if it emits a $W^{-}$boson, it will subsequently decay and create an electron ($e^-$), and an electron antineutrino ($\overline{v}_e$), the antimatter particle to an electron neutrino. It will also flip a neutron into a proton. This is known as ...

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