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I (QED) The initial mass and the correction cancelling out

  1. Jul 22, 2017 #1
    The prefix is a bit irrelevant

    This is on renormalisation.

    How do they cancel out? Isn't it adding? So the mass experimental = m + (c2correction) so how do you cancel out the m and correction? I'm new to this area (just finished watching lectures by Richard Feynman, specifically a 4 lecture series on QED in 1979 in New Zealand).

    Is it that the experimental mass is already confirmed by experiment and because the initial mass isn't important because of a change of theory (pre-interaction to electron interacting with photons) we can just change the number so the mass experimental is a finite one? Can you give me an example of an experiment having already been done?
     
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  3. Jul 22, 2017 #2

    Orodruin

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    The bare mass is not an observable, nor is the radiative correction. On the other hand, the physical mass is an observable.
     
  4. Jul 22, 2017 #3
    Physical mass being c2 or the experimental mass? I'm just thinking about basic maths. You can't cancel out a c2correction by adding an m because wouldn't that be c2infinity + m? How do I make m a value to cancel out c2infinity or c2whateverlargenumberitis ?
     
  5. Jul 22, 2017 #4
    Also what's the importance of the bare mass and the radiative correction not being an observable?
     
  6. Jul 22, 2017 #5

    Orodruin

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    Both the bare mass and the correction are formally infinite. The observable is the physical mass, which is what you measure in experiments.
     
  7. Jul 22, 2017 #6
    Is the bare mass the initial mass pre interaction? If so how is it infinite? What is the physical mass? Lastly how in that equation would you cancel out the correction and pre interaction mass? My thanks for your response so far
     
  8. Jul 22, 2017 #7

    Vanadium 50

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    Let $$x=4 + 1/r - 1/r$$.

    If you are going to assign a value to x at r=0, what would that value be? Does it matter that 1/r is infinite?
     
  9. Jul 22, 2017 #8
    No but what about 4? Isn't 4 or the initial mass important in the renormalisation?
     
  10. Jul 22, 2017 #9

    Orodruin

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    No, in this parallel 4+1/r is.
     
  11. Jul 22, 2017 #10
    Why? Surely the answer would be 4 but in the equation we don't have a finite number for the initial mass. As in is the 4 the min?
     
  12. Jul 22, 2017 #11

    Orodruin

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    This is a likeness. You can never know what the bare mass is (please use the accepted terminology). It is not measurable nor computable on its own. What you can measure is the physical mass, which is a combination of the bare mass and the radiative correction.
     
  13. Jul 22, 2017 #12
    Thank you for your responses.
    Just to confirm the radiative correction is the amplitude of all possible positions of the photon being ejected and being reabsorbed in an area being infinity or a non finite number? My only problem with it is if it's m +1/r and r is infinity how do we make m negative infinity? Is that what we're doing?
     
  14. Jul 22, 2017 #13

    Vanadium 50

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    In the analogy - and it's only an analogy - the 4+1/r can represent the bare mass and the 1/r represents the corrections. Both are formally infinite, but both are unobservable.
     
  15. Jul 22, 2017 #14
    Thank you for your response. Where does c2 come into this?
     
  16. Jul 22, 2017 #15

    Nugatory

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    It's the conversion factor between mass and energy, needed because we've chosen to use different units for the two.

    Any time that ##c## (or any other dimensionful constant) shows up in a formula, it's just an artifact of your choice of units.
     
  17. Jul 22, 2017 #16

    radium

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    The bare mass is the coupling in the Lagrangian. When you compute something like the electron self energy, with dimensional regularization in an epsilon expansion (very commons), you will find terms that will diverge taking epsilon to zero. You will then choose some renormalization scheme, like minimal subtraction or the on shell subtraction scheme to choose counter terms giving you the renormalized mass. The counter terms and renormalized are different depending on the renormalization scheme.

    The onshell subtraction identifies the renormalized mass as the pole in the renormalized propagator. The pole mass is the physical mass. It is independent of the scheme used to identify finite parts of the counter terms. By contrast, the minimal subtraction scheme chooses counterterms which include only infinite parts. The renormalized mass (bare mass+counter terms) is not equal to the pole mass and hence is not a physical mass.
     
  18. Jul 22, 2017 #17
    Thank you for your response. So how do you cancel out the min and the correction to leave behind just the c2 or whatever percent we go to (mass experimental = m + c2correction)? Because say the correction is 10 billion and I make m = -10billion it's still -10billion + 1/137x10billion which doesn't cancel out
     
    Last edited: Jul 22, 2017
  19. Jul 22, 2017 #18
    With your 4+1/r representing the bare mass and 1/r representing the correction, can the bare mass have another value e.g. 4+1/w while the correction has the 1/r? that way because they're both infinites you can just cancel them both out of themselves independently?
     
  20. Jul 22, 2017 #19

    Vanadium 50

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    The analogy is not the theory. You can't stretch it.
     
  21. Jul 22, 2017 #20
    What do you mean and what's the result of not being able to stretch it?
     
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