Integration of Dirac delta w/ different dimensionalities

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Discussion Overview

The discussion revolves around the integration of the Dirac delta function in different dimensionalities, specifically focusing on the implications of integrating a four-dimensional delta function over a three-dimensional momentum space. The conversation touches on theoretical aspects of the Dirac delta function in the context of physics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states the integral of the Dirac delta function in one dimension, \(\int dx f(x) \delta(x-a) = f(a)\), and questions how this extends to \(\int d^3x f(x) \delta^{(4)}(x-a)\) where \(x\) and \(a\) are four-momenta.
  • Another participant explains that the notation \(\delta^{(4)}(x - a)\) can be expressed as a product of delta functions in each dimension, leading to the conclusion that \(\int d^3x f(x) \delta^{(4)}(x-a) = f(\vec a) \delta(x^0 - a^0)\).
  • A further comment notes that the interpretation of the remaining delta function requires additional context, suggesting it is only meaningful under certain conditions, specifically when \(x^0 = a^0\).
  • A participant queries whether an integral of the form \(\int d^3p f(p) \delta(p^0-a^0)\) should be interpreted as \(\int d^3p f(a)\) given that the argument in the delta function is zero due to energy conservation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation and implications of integrating the Dirac delta function across different dimensions. There is no consensus on the best approach or interpretation of the remaining delta function after integration.

Contextual Notes

Limitations include the dependence on the definitions of the delta function in multiple dimensions and the conditions under which the remaining delta function is considered well-defined.

FredMadison
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Hi!

The Dirac delta satisfies

[tex]\int dx f(x) \delta(x-a) = f(a)[/tex]

But how about

[tex]\int d^3x f(x) \delta^{(4)}(x-a)[/tex]

Here, x and a are four-momenta, and the integral is over the regular 3-dim momentum.

How does the delta behave here?
 
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The notation you used usually means
[tex]\delta^{(4)}(x - a) = \delta(x^0 - a^0) \delta(x^1 - a^1) \delta(x^2 - a^2) \delta(x^3 - a^3)[/tex]

So in that case
[tex] \int d^3x f(x) \delta^{(4)}(x-a) = f(\vec a) \delta(x^0 - a^0)[/tex]
where [itex]a = (a^0, \vec a)[/itex].
 
Ok, thanks!
 
By the way, note that this is what you might call "physicist's notation".
Of course, technically the remaining delta makes no sense whatsoever, unless you put it under an integral sign again.

However, we usually interpret it as 'this quantity has the value f(a), but it is only well-defined* if x0 = a0' or something along those lines.

* well-defined being: having a physical meaning
 
So, if I end up with something like

[tex]\int d^3p f(p) \delta(p^0-a^0)[/tex]

where the argument in the delta is zero from energy conservation, you're saying that I should regard this as

[tex]\int d^3p f(a)[/tex]

?
 

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