Recast of an expression containing Diracs [tex]\delta[/tex]-function

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In summary, the conversation is about a problem with rearranging an expression involving spherical harmonics and a completeness relation. The question is about justifying a rearrangement of the right-hand side of the expression, which involves a delta function and cosine. After several attempts, the person is still unsure about how to justify the rearrangement and asks for help. The expert notes that the rearrangement is not the same unless the variable phi is restricted to a certain range, and explains that cosine is "one to one" in this range which is important for the delta function. The person still has questions about the rearrangement and the role of the 1/sin(theta) term.
  • #1
johnpatitucci
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Hey there,


I got a problem with one recast of an expression which pops up by considering the completeness relation of common spherical harmonics:

[tex]\sum_{l=0}^{\infty} \sum_{m=-l}^{+l} Y_{lm}(\theta,\phi)Y^{*}_{lm}(\theta^{\prime},\phi^{\prime}) = \frac{1}{sin(\theta)} \delta(\theta - \theta^{\prime}) \delta(\phi - \phi^{\prime})[/tex]

The question is about the rhs which is sometimes recast like

[tex]\frac{1}{sin(\theta)} \delta(\theta - \theta^{\prime}) \delta(\phi - \phi^{\prime}) = \delta(\cos(\theta) - \cos(\theta^{\prime})) \delta(\phi - \phi^{\prime})[/tex]

After several attempts I just can't explain how to justify that rearrangement. Could you help me please ?
 
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  • #2
It isn't quite the same thing, unless you restrict [itex]\phi[/itex] to be between [itex]-\pi/2[/itex] and [itex]\pi/2[/itex]. In that range, cosine is "one to one" so that [itex]cos(\phi)- cos(\phi)[/itex] if and only if [itex]\phi- \phi'[/itex] which is all that matters for the delta function.
 
  • #3
Thanks for your reply.

I still don't get it. Cos(x) is not injective between -[tex]\pi /2[/tex] and [tex]\pi /2[/tex] as it is symmetric within this range (take e.g. [itex] \pm \frac{\pi}{4}[/itex] and you obtain [itex] \cos(\pm\frac{\pi}{4}) = \frac{1}{\sqrt(2)} [/itex] ). It would be true if [tex] 0 \leq x \leq \pi [/tex]. But nevertheless, what happens to the [tex] 1 / \sin(\theta)[/tex] term?
 

What is a Dirac delta function?

The Dirac delta function, denoted by [tex]\delta[/tex], is a mathematical function that is zero everywhere except at the origin, where it is infinite. It is often used in physics and engineering to model point sources or impulses.

How can an expression containing a Dirac delta function be recast?

An expression containing a Dirac delta function can be recast by using certain properties and identities of the delta function, such as its sifting property and its relation to the Heaviside step function. This allows for simplification and manipulation of the expression.

What is the significance of a Dirac delta function in mathematical models?

The Dirac delta function is useful in mathematical models as it allows for the representation of point sources or impulses, which often occur in real-world systems. It also helps in solving differential equations and performing integrals involving discontinuous functions.

What are some common applications of the Dirac delta function?

The Dirac delta function has numerous applications in various fields, including physics, engineering, signal processing, and probability theory. It is used to model point masses in mechanics, electric charges in electromagnetism, and point sources in fluid mechanics. It is also used in Fourier analysis and in solving differential equations.

Are there any limitations to using the Dirac delta function in mathematical models?

While the Dirac delta function is a powerful tool in mathematical models, it has some limitations. It is an idealized function and does not exist in physical reality. Additionally, it can lead to certain mathematical inconsistencies, such as division by zero, if not used properly.

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