Can Singularities Be Modeled as a Conventional Field?

AI Thread Summary
The discussion centers on the feasibility of modeling singularities as a conventional field using discrete Dirac delta functions. The participants debate whether a field can represent various magnitudes of singularities, highlighting that a Dirac delta function lacks a value until integrated. They emphasize that the integral of a function with a Dirac delta function effectively selects a specific value from the function. The conversation also touches on the mathematical implications of this selection process in quantum mechanics, illustrating how Dirac delta functions can isolate components of vectors. Overall, the dialogue explores the relationship between singularities and conventional field theory through the lens of integration and normalization.
Loren Booda
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Is it possible to have a field f([pard](x)) - fractal or otherwise - where [pard](x) are discrete Dirac delta functions, and f interrelates the various magnitudes of those singularities as a conventional field would for points over a continuum?
 
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I think the phrase "various magnitudes of all the singularities" is a contradiction in terms. a δ doesn't have a value until it's integrated, as in ∫f(x)δ(x-a)dx = f(a).
 
selfAdjoint,

Couldn't local Dirac singularities in my proposed field represent also various magnitudes obeying their point-by-point (distribution) normalization through overall integration?
 
Loren, go back and look at the integral I posted, notice that f(x) in there. It could be anything. What the Dirac δ does is to pick out a particular value of any function that you integrate. Dirac modeled it on the finite case of a vector like (0,1,0). If you inner multiply that by any arbitrary vector (a,b,c) you get
(a,b,c)(0,1,0) = 0*a + 1*b + 0*c = b
(so it picks out the second component, and if you used a 1 in a different place you would pick out a different component. Now in QM math the integral of the product of two "functions" is an inner product in the algebra of those functions, so δ(x-a) in the integral picks out the "a-value component" of the function.
 
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