Timeline for Renormalized mass
Current License: CC BY-SA 4.0
33 events
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| S May 18 at 22:30 | history | suggested | Gabriel Ybarra Marcaida | CC BY-SA 4.0 |
Corrected \not{p} to \not p to properly display the slashes.
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| May 18 at 15:24 | review | Suggested edits | |||
| S May 18 at 22:30 | |||||
| Oct 10, 2022 at 19:56 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 10, 2022 at 19:22 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 10, 2022 at 19:15 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 10, 2022 at 18:54 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 10, 2022 at 18:48 | history | edited | MadMax | CC BY-SA 4.0 |
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| Jan 13, 2021 at 8:27 | comment | added | TheQuantumMan | Oh, wait. I think I misunderstood something. In your last equation, the physical mass (pole) is also dependent on the momentum. Is the pole of the propagator -the physical mass- also the running mass? I might be so confused that I distinguished the two. | |
| Jan 13, 2021 at 8:14 | comment | added | TheQuantumMan | I see. So, what is "the" mass of the electron? It certainly doesn't seem to make sense to to be renormalized one (depends on definition). It would make sense to define the physical mass as either the pole of the propagator or the running mass (in analogy with the running coupling being physical; the electron's running charge produces observable corrections to the Coulomb potential, leading to things like the Lamb shift). So what is the physical mass? | |
| Jan 12, 2021 at 21:19 | history | edited | MadMax | CC BY-SA 4.0 |
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| Jan 12, 2021 at 21:11 | comment | added | MadMax | The renormalized mass is just the $p$ independent term (constant term) in the Tailor expansion of the renormalized inverse propagator $G^{-1} \sim \not{p}- (m_r + c_1p^2 + c_2p^4 + ...)$. You can redefine renormalized mass to whatever value you want. For example, $m_r' = m_r + c_1p_0^2 + c_2p_0^4 + ...$. Now if you let $p_0^2 = m_p^2$, you will have $m_r' = m_p$. Note that $m_r'$ is NOT the constant term of the expansion around $p = 0$ any more. | |
| Jan 12, 2021 at 17:20 | comment | added | TheQuantumMan | If I may ask, since the physical mass seems to be the only mass (compared to the renormalized and the effective, with the latter running with momentum) that has some absolute sense (we don't need a reference energy/momentum to fix it), is it considered as being "the" mass? If so, why don't we just fix the renomalized mass to be equal to the physical one, since we can freely choose the reference energy at which we define the renormalized mass? | |
| Oct 19, 2018 at 14:08 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 20:45 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 20:37 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 17:07 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 16:56 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 16:48 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 16:37 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 16:31 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 16:17 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 16:04 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 15:49 | comment | added | MadMax | Right. please see update. | |
| Oct 18, 2018 at 15:47 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 15:47 | comment | added | AccidentalFourierTransform | To be clear, what you call $e(p^2)$ is what is usually denoted by $\Gamma^\mu(p_1,p_2)$, right? The 1PI vertex function associated to $\langle\psi(p_1)A^\mu(p_2)\bar\psi(p_3)\rangle\delta(p_1+p_2+p_3)$? | |
| Oct 18, 2018 at 15:47 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 15:41 | comment | added | MadMax | Thank you @AccidentalFourierTransform, see updates. | |
| Oct 18, 2018 at 15:40 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 15:32 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 15:19 | comment | added | AccidentalFourierTransform | The renomalised charge is not momentum-dependent. In $\mathrm{OS}$ it is a constant ($\approx0.3$), and in $\mathrm{MS}$ it depends on the mass $\mu$. More generally, it may depend on the cutoff scale or any other mass scale introduced by the renormalisation scheme; but it is not, in any case, momentum-dependent. | |
| Oct 18, 2018 at 15:16 | history | edited | MadMax | CC BY-SA 4.0 |
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| Oct 18, 2018 at 15:11 | history | answered | MadMax | CC BY-SA 4.0 |