Timeline for Why is $F_1(0) = 1$? (form factor in QED)
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| May 18 at 22:23 | history | edited | User3141 | CC BY-SA 4.0 |
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| May 18 at 22:21 | answer | added | 11zaq | timeline score: 1 | |
| May 18 at 21:53 | comment | added | User3141 | @Prahar Perhaps to clarify my point; I quote from you, "P&S defines e as the electric charge. It follows from that definition that $F_1(0)=1$". This is like saying that I defined the bare mass to be the physical mass and hence by definition, the self interaction term is zero (without any "counter terms"). Obviously, the last statement would be incorrect; the value of the self interaction term doesn't depend on what our definition of bare and physical mass happens to be. Likewise, the form factors are defined from the vertex correction diagrams, independent of our definition of "$e$". | |
| May 18 at 21:46 | comment | added | User3141 | @Prahar The precise value of $F_1(0)$ would, as the form factor is defined in P&S, depend on the contribution from the radiative corrections, whose form would depend on the Feynman rules for QED, not the precise value "$e$"; it can't magically happen to equate to 1 when "$e$" is put equal to (in SI units) -1.6$\cdot10^{-19} C$ and be different from 1 for other values. If you argue that you have to add "counterterms" in general (although, again, not familiar with renormalization), then how can you know that you won't have to do that here? | |
| May 18 at 21:35 | comment | added | Prahar | BTW, there's nothing wrong with defining the bare mass as the physical mass of the electron. As long as you add counterterms correctly to renormalize the UV divergences, there is nothing wrong with this. It's simply a choice. | |
| May 18 at 21:33 | comment | added | Prahar | Sure, you can define $e$ as a coupling constant and NOT the electric charge. Then, the value of $F_1(0)$ will depend on your precise definition of the coupling constant. P&S defines $e$ as the electric charge. It follows from that definition that $F_1(0)=1$. | |
| May 18 at 21:16 | comment | added | User3141 | @Prahar No, perhaps you didn't quite get my point; suppose, for the sake of argument, that the value of $F_1(0)$ found by summing up the contribution to all orders was 2. Then, we could only conclude that the coupling constant was one half the physical charge. $e$ is defined to be the QED coupling constant, not (in general) the physical charge. If we followed your advice and just "defined" the bare mass in classical EM to be the physical mass, then we would of course run into trouble down the line. How can we conclude that not a similar thing could happen here? | |
| May 18 at 21:11 | comment | added | Prahar | Given the form factors, we find that the electron charge is measured to be $e F_1(0)$. However, by definition, $e$ is the electron charge. Therefore, $F_1(0)=1$. | |
| May 18 at 21:10 | history | edited | User3141 | CC BY-SA 4.0 |
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| May 18 at 21:00 | history | asked | User3141 | CC BY-SA 4.0 |