Delta Dirac: $\phi=-\pi+\epsilon$

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Homework Help Overview

The discussion revolves around the properties and evaluation of the Dirac delta function, specifically in the context of angular coordinates defined between (-π, π). The original poster questions the value of the delta function when the angle φ is set to -π + ε.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the delta function's argument being -2π + ε and question whether the original poster intended to use φ = π + ε instead. There are discussions about the evaluation of the delta function under the specified angular constraints and the implications for an equation involving T and k.

Discussion Status

Some participants have provided insights into the behavior of the Dirac delta function, noting that it yields zero unless its argument is zero. There is an ongoing exploration of the integration limits and whether φ can equal π, with differing interpretations being considered.

Contextual Notes

Participants are working within the constraint that φ is defined between (-π, π), which raises questions about the behavior of the delta function at the boundaries of this interval.

alejandrito29
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if [tex]\phi[/tex] is a angular coordinate , between ([tex]-\pi,\pi[/tex])

¿how much is [tex]\delta(\phi-\pi)[/tex] with [tex]\phi=-\pi+\epsilon[/tex]?
 
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The Dirac delta function yield zero unless it's argument is zero, in which case it yields 1 (this is an oversimplification, but it should do for the present discussion). In your case, the argument of the delta function is [tex]-2*\pi + \epsilon[/tex], so it should be zero. Did you mean to type, [tex]\phi=\pi + \epsilon[/tex]? In that case, the argument of the delta function would be just [tex]\epsilon[/tex], and then you need to get a bit more specific about how you are defining the delta function. Have you looked at this thread? https://www.physicsforums.com/showthread.php?t=73447
 
SpectraCat said:
The Dirac delta function yield zero unless it's argument is zero, in which case it yields 1 (this is an oversimplification, but it should do for the present discussion). In your case, the argument of the delta function is [tex]-2*\pi + \epsilon[/tex], so it should be zero. Did you mean to type, [tex]\phi=\pi + \epsilon[/tex]? In that case, the argument of the delta function would be just [tex]\epsilon[/tex], and then you need to get a bit more specific about how you are defining the delta function. Have you looked at this thread? https://www.physicsforums.com/showthread.php?t=73447

then the dirac delta evaluated in (-2pi+epsilon) is 0 or infinite?
 
specifically, i need to find T, in the follows equation:

[tex]\delta(\phi-\pi)+k=T\delta(\phi-\pi)[/tex]
where [tex]\phi[/tex] is between ([tex]-\pi,\pi[/tex])
 
With that specific range the delta distribution is zero; phi can't become equal to pi. So you get the equation 0 + k = T*0.
 
haushofer said:
With that specific range the delta distribution is zero; phi can't become equal to pi. So you get the equation 0 + k = T*0.

but, if i do:

[tex]\int^{\pi-\epsilon}_{-\pi+\epsilon}\delta(\phi-\pi)d\phi+\int^{\pi-\epsilon}_{-\pi+\epsilon} k=T\int^{\pi-\epsilon}_{-\pi+\epsilon}\delta(\phi-\pi)[/tex]

is correct??

pd: [tex]\phi[/tex] can be equal to [tex]\pi[/tex]...[tex]\phi[/tex] is between [tex](-\pi,\pi)[/tex]
 
Last edited:
You want to integrate from pi-epsilon to pi+epsilon. That will include the relevant range of the dirac delta.
 

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