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Math Help - Dimensional Regularization

  1. #1
    Forum Admin topsquark's Avatar
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    Dimensional Regularization

    This may be a bit out of place here...

    Does anyone know anything about "dimensional regularization" of integrals? It is one method to renormalize integrals in Quantum Field Theory. I'm not sure where it crops up in Math, but it's probably somewhere. (Odd are no Physicist could come up with something this clever! I'm a Physicist, so I can say that. )

    Anyway, the basic method is this: We start with the divergent integral:
    \int d^4x F(x^{\mu}). The form of the function won't come into play here, so don't worry about it. The idea is to isolate the divergent part of the integral by letting the dimension of the integral be variable, then at the end of the calculation take \lim_{d \rightarrow 4} of the resulting expression. It sounds cheesy, but it does work, at least for certain forms of F.

    Now I was thinking about something I saw in another thread...a trick question, but the logic seems to apply here. Consider the equation:
    x^2=x+x+...+x+x where x appears on the RHS x times. Now take the derivative: 2x=1+1+...+1+1=x. Apparently, then, 2=1. The reason this fails is that we are taking the derivative of x that is variable on the LHS, but a constant on the RHS.

    Aren't we doing something similar in the integral above? I have yet to see a case in QFT where the dimension is anything but an integer, so taking the limit as d approaches 4 seems a bit ridiculous as a dimension of d+ \epsilon for small \epsilon doesn't exist.

    I admit that the method works, but how can it work??

    -Dan
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    This may be a bit out of place here...

    Does anyone know anything about "dimensional regularization" of integrals? It is one method to renormalize integrals in Quantum Field Theory. I'm not sure where it crops up in Math, but it's probably somewhere. (Odd are no Physicist could come up with something this clever! I'm a Physicist, so I can say that. )

    Anyway, the basic method is this: We start with the divergent integral:
    \int d^4x F(x^{\mu}). The form of the function won't come into play here, so don't worry about it. The idea is to isolate the divergent part of the integral by letting the dimension of the integral be variable, then at the end of the calculation take \lim_{d \rightarrow 4} of the resulting expression. It sounds cheesy, but it does work, at least for certain forms of F.

    Now I was thinking about something I saw in another thread...a trick question, but the logic seems to apply here. Consider the equation:
    x^2=x+x+...+x+x where x appears on the RHS x times. Now take the derivative: 2x=1+1+...+1+1=x. Apparently, then, 2=1. The reason this fails is that we are taking the derivative of x that is variable on the LHS, but a constant on the RHS.

    Aren't we doing something similar in the integral above? I have yet to see a case in QFT where the dimension is anything but an integer, so taking the limit as d approaches 4 seems a bit ridiculous as a dimension of d+ \epsilon for small \epsilon doesn't exist.

    I admit that the method works, but how can it work??

    -Dan
    Maybe the process is telling you something about the structure of space
    time. maybe this is indicative of the Hausdorff dimension of Space Time
    being close to but not actually 4?

    (Google for Hausdoff dimension for an explanation)

    RonL
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