Penrose's Zig-Zag Model and Conservation of Momentum

Question

I was reading through Penrose's Road to Reality when I saw his interesting description of the Dirac electron (Chapter 25, Section 2). He points out that in the two-spinor formalism, Dirac's one equation for a massive particle can be rewritten as two equations for two interacting massless particles, where the coupling constant of the interaction is the mass of the electron. In the Dirac formalism, we can write the electron field as $\psi = \psi_L + \psi_R$ where $\psi_L = \frac{1}{2}(1-\gamma_5)\psi$ and $\psi_R = \frac{1}{2}(1+\gamma_5)\psi$. Then the Lagrangian is:

$$ \mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi - m\bar{\psi}\psi $$ $$ \mathcal{L}=i\bar{\psi}_L\gamma^\mu\partial_\mu\psi_L+i\bar{\psi}_R\gamma^\mu\partial_\mu\psi_R - m(\bar{\psi}_L\psi_R + \bar{\psi}_R\psi_L) $$

This is the Lagrangian of two massless fields, one left-handed and one right-handed, which interact with coupling constant $m$.

He then pictorially explains his interpretation by drawing this interaction Feynman-diagram-style. The initial particle (L or R) travels at speed $c$ (with luminal momentum) until it "interacts" and transforms into the other particle (R or L) which also travels at speed $c$ but in the opposite direction. This "zig-zagging" of the particles causes the two-particle system to travel at a net velocity which is less than $c$, thus giving the electron a subluminal momentum, thereby granting it a mass of $m$. He later states that this interaction is mediated by the Higgs boson, which we get if we replace the coupling constant $m$ with the Higgs field.

I've tried to put this argument into a mathematical setting, deriving the massive propagator from the two massless propagators via perturbation methods, but what I can't seem to get around is the conservation of momentum. When a "zig" particle changes into a "zag", the direction changes and thus the new particle gains momentum out of "nowhere." If we involve Higgs, then the Higgs boson could carry away/grant the necessary momentum, but I want to know if the model can work without involving the Higgs. Is this possible?

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user FrancisFlute

Comments

The momentum is conserved. Feynman diagram is not a space-time diagram. It is a "momentum-energy" diagram. It could be seen as a superposition of a infinite sum (over space) of space-time diagrams wich, in Penrose idea, could be some specific space-time zig-zag diagrams. You may see the correspondence fig $25.2$ page $631$ in the book.

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Trimok

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But for each of the diagrams in the superposition, wouldn't momentum be conserved at each vertex?

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user FrancisFlute

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No. Without pronouncing about the validity of the zig-zag analogy, , "momentum/energy" Feynman Diagrams (the usual Feynman diagrams) $A(p_1,p_2)$ can be seen as Fourier transform of "space-time" Feynman Diagrams $A(x_1,x_2)$ (Fourier Transform is the superposition). When starting from space-time amplitudes, The sum on intermediary $x$ positions, make terms like $\int dp f(p) ~e^{ip(x_2 - x_1)}$ appear in $A(x_1,x_2)$. And this corresponds to a factor $\delta(p_1-p_2)$ for the momentum amplitude $A(p_1,p_2)$.

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Trimok

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More exactly, this does not make sense to speak about momentum, when looking at space-time diagrams.

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Trimok

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Oh, I see it now. I was taking the diagram too literally. If you just take the momentum representation of the propagators it should come out as a geometric series in $m^2$ giving the massive propagator. Thanks!

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user FrancisFlute

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Is it related to Feynman "checkerboard" zig-zag?

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user arivero

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@FrancisFlute You write well. I have no knowledge of this subject matter, but your writing is so clear I feel a textbook authored by you would be valuable to many.

This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Inquisitive