Finding mass with dirac delta function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 3K views
dingo_d
Messages
199
Reaction score
0

Homework Statement



Distribution of matter is given in cylindrical coordinates:

[tex]\rho(\vec{r})=\frac{1}{\rho}\delta(\rho^2-10\rho+9)\delta\left(\frac{z^2-a^2}{z^2+a^2}\right)\delta(\cot(\phi))[/tex]

where [tex]a>0[/tex] is a constant. Find the complete mass of the object.


Homework Equations



The mass of the object is given as:

[tex]M=\int_{\partial V}\rho(\vec{r})dV[/tex]

So in order to calculate the integral I need to expand the deltas with this formula:

[tex]\delta(f(x))=\sum_{i=1}^n\frac{\delta(x-x_i)}{|f'(x_i)|}[/tex], where [tex]|f'(x_i)|[/tex] is the derivative of the function evaluated at the zeroes of the function [tex]x_i[/tex].

Integral in cylindrical coordinate system is:

[tex]\int_0^\infty\rho d\rho\int_0^{2\pi} d\phi\int_{-\infty}^\infty dz[/tex]

The Attempt at a Solution



I have expanded the first delta:

[tex]\delta(\rho^2-10\rho+9)=\frac{\delta(\rho-1)}{8}+\frac{\delta(\rho-9)}{8}[/tex].

I'm having problem with the second one. Do I 'brake' it with partial fractions?

Third one (which I'm not sure I got it right):

[tex]\delta(\cot(\phi))=\sum_{m=-\infty}^\infty\delta\left(\phi-\left(m+\frac{1}{2}\right)\pi\right)[/tex]

So my problem is: how do I solve the second delta and I'm not quite sure I got the third one right. How should I integrate it?
 
Physics news on Phys.org
I see no problem in the second delta. Just find the zeros and calculate the derivatives.
Your answer for the third delta seems to be right.
 
So the second one is:
[tex]a\left[\delta(z-a)+\delta(z+a)\right][/tex]?

But how do I calculate the integral? With [tex]\phi[/tex] part? When I put it into Mathematica I get different answers:
For putting sum in front of the integral:
[tex]\frac{1}{2}\theta(3 - 2 m, \pi + 2 m \pi)[/tex]
And for putting integral in front of the sum I have [tex]\frac{7}{\pi}[/tex], which is weird because shouldn't I get the same results no matter which goes first?

But that's on the side note, on my exam I won't have Mathematica :D, so how to calculate that?
 
The integral of delta function is:

[tex] \int_a^b f(x) \delta(x-c)\,dx = \begin{cases}<br /> f(c), & a < c < b; \\<br /> f(c)/2, & c = a \;\text{or}\; c = b; \\<br /> 0, & \text{otherwise}.<br /> \end{cases}[/tex]

I think it's not a problem to calculate it without Mathematica.

In cylindrical coordinates

[tex] 0 \leq \phi < 2\pi.[/tex]

You should keep only two terms of the infinite sum.
 
Maxim Zh said:
The integral of delta function is:

[tex] \int_a^b f(x) \delta(x-c)\,dx = \begin{cases}<br /> f(c), & a < c < b; \\<br /> f(c)/2, & c = a \;\text{or}\; c = b; \\<br /> 0, & \text{otherwise}.<br /> \end{cases}[/tex]

I think it's not a problem to calculate it without Mathematica.

In cylindrical coordinates

[tex] 0 \leq \phi < 2\pi.[/tex]

You should keep only two terms of the infinite sum.

Those terms are random or? I'm confused because I have only solved simple integrals with delta function in my class, never with these sums :(
 
No, they are not random.
When you integrate the sum

[tex] \sum_m \delta(\phi - \phi_m)[/tex]

only two terms fit the condition

[tex]\phi_m \in (0, 2\pi).[/tex]

The other terms are not integrated and give zero.
 
I see, that's because I'm only need two zeros in that interval, so I'm only using those two! Thanks!
 
Last edited: