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12

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 ...


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

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 ...


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.


4

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 = ...


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 ...


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

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

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

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

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 ...


2

Annihilation can happen when all the quantum numbers of two colliding particles add up to zero. It might be electron on positron, proton on antiproton, neutron on antineutron , quark on antiquark etc.The force responsible depends on the possible interactions of the annihilating particles. In the case of electron positron annihilation it is primarily the ...


2

Electron and positron can annihilate in free space, but a single photon cannot be turned into electron-positron pair because conservation of energy-momentum cannot be satisfied. The nucleus in pair production absorbs some of the momentum.


2

If you look at a photon in a vacuum, there is no preferred inertial frame of reference. And the energy of the photon is dependent upon which frame is chosen. So some frames have more than the needed energy, others less. It would be really awkward to have pair production in reference frames that lack the needed energy.


2

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 ...


2

The Heisenberg uncertainty principle applies only to operators satisfying canonical commutation rules. This is the case for corresponding components of position and momentum operators, but not for the energy operator (Hamiltonian), which has no associated conjugate partner. (Conjugate pairs of selfadjoint operators necessarily have unbounded spectrum, while ...


2

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 ...


2

Presumably what you want to do is think in the center-of-momentum frame. In this frame, there is no net momentum, just energy. So presumably you want to convert to a system of particles at rest, two of which are protons and the rest of which are unspecified. How much energy is left over to go into the unspecified other stuff? Although I think it's most ...


2

It's important to note that electron-positron production is well-established for photons or charged particles above the threshold of ~1 MeV total energy. For high-energy charged particles which lose energy by generating "showers" of less-energetic particles, it's the usually the primary energy loss mechanism. Your link is suggesting a photon-photon collider ...


2

You have a photon that comes in and splits into $e^-e^+$. One of the electrons escapes, the other has a short segment, where it sheds another photon. The second electron now escapes and the photon is absorbed by the nucleus. The center electron segment and the second photon can be off the mass shell.


2

Assuming that $\tau^-\tau^+$ can form a bound state similarly to positronium($e^-e^+$), all we need is the form of the ground state of positronium, specifically that it is proportional to the reduced mass of the pair: $\mu = \frac{m_1m_2}{m_1+m_2}$. Knowing that positronium's ground state is $(-13.6/2)= -6.8$ eV and that the new reduced mass for the Tau ...


2

It depends what you mean by "electrostatic energy". When we are talking of pair production we are talking of physics at the quantum mechanics framework. FEYNMAN DIAGRAMS for pair production by a gamma ray (left) or an electron (right). These represent the processes in the preceding sketch. Lets take the simplest diagram on the left: a photon ...


1

Quick answer: No. Not so quick answer: momentum conservation. You emit photons with momentum $\vec p$, your spaceship gets propelled with momentum $-\vec p$. When the photons get converted into pairs, they have the same momentum, your spaceship goes the same. Now, I take you want to use the pairs as an ionic motors. You take the electron positron pairs and ...


1

The atom takes so little kinetic energy becuse it is far more massive $ ( E_{0~Fe}=53 \frac{GeV}{c^{2}})$ than the electron $(E_{0}^{e-}=0.511 \frac{MeV}{c^{2}} )$.


1

As I said in the comments it really depends on what happens. Here I will assume the protons are colliding heads-on and after collision, all the particles will move with the same speed(so the square of their total momentum will be minimum). $$E_1=\sqrt{m_p^2c^4+p_1^2 c^2} \\ E_2=\sqrt{m_p^2c^4+p_2^2 c^2} \\ \Rightarrow E=E_1+E_2$$ After collision, as I ...


1

Have a look at this article My comments: For fig A: Photon photon collisions are not the primary creators of particle pairs, but granted it is a possibility in the very dense matter of the early universe. For fig. B: Once created they move on a geodesic without acceleration with the momentum given from the balance in the local area of the interaction. ...



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