Electrodynamics of a moving particle

[fuchini]

Homework Statement

There is a charged particle (charge=q) moving on the x-axis such that $x(t)=A\,sin(\omega t)$. Prove that:
$$\int <\rho>\,dV=q$$

Homework Equations

We have the following equations:
$$<\rho>=\frac{1}{T} \int_0^T \rho\, dt$$

Where $T=\frac{2 \pi}{\omega}$

The Attempt at a Solution

So for this particle we have the following charge density $\rho=q\,\delta(z)\delta(y)\delta(x-A\,sin(\omega t))$ therefore:

$$<\rho>=\frac{\omega}{2\pi}q\,\delta(z)\delta(y) \int_0^T \delta(x-A\,sin(\omega t))\,dt$$

However, we know that $\delta(f(t))=\sum \frac{\delta(t-t_i)}{|f^\prime (t_i)|}$ for every root of $f(t)$. So we have $\delta(x-A\,sin(\omega t))=\delta(x/A-sin(\omega t))/|A|$ where $f(t)=x/A-sin(\omega t)$.

Considering the integral is over one period I'll only take one root into account:

$$\delta(x-A\,sin(\omega t))=\frac{1}{|A|}\frac{\delta(t-sin^{-1}(x/A)/\omega)}{\omega cos(sin^{-1}(x/A))}=\frac{\sqrt{A^2-x^2}}{\omega}\delta(t-sin^{-1}(x/A)/\omega)$$

Then, the average value is:

$$<\rho>=\frac{\sqrt{A^2-x^2}}{\omega}\frac{\omega}{2\pi}q\,\delta(z)\delta(y) \int_0^T \delta(t-sin^{-1}(x/A)/\omega)\,dt=\frac{\sqrt{A^2-x^2}}{2\pi}q\,\delta(z)\delta(y)$$

Finally we have:

$$<\rho>=\int \frac{\sqrt{A^2-x^2}}{2\pi}q\,\delta(z)\delta(y)\, dV =\frac{q}{2\pi}\int \sqrt{A^2-x^2}\,dx$$

This is as far as I go, please help me if there is any mistake and how can I get the final result. Thanks a lot.

 

[nrqed]

Homework Statement

There is a charged particle (charge=q) moving on the x-axis such that $x(t)=A\,sin(\omega t)$. Prove that:
$$\int <\rho>\,dV=q$$

Homework Equations

We have the following equations:
$$<\rho>=\frac{1}{T} \int_0^T \rho\, dt$$

Where $T=\frac{2 \pi}{\omega}$

The Attempt at a Solution

So for this particle we have the following charge density $\rho=q\,\delta(z)\delta(y)\delta(x-A\,sin(\omega t))$ therefore:

$$<\rho>=\frac{\omega}{2\pi}q\,\delta(z)\delta(y) \int_0^T \delta(x-A\,sin(\omega t))\,dt$$

However, we know that $\delta(f(t))=\sum \frac{\delta(t-t_i)}{|f^\prime (t_i)|}$ for every root of $f(t)$. So we have $\delta(x-A\,sin(\omega t))=\delta(x/A-sin(\omega t))/|A|$ where $f(t)=x/A-sin(\omega t)$.

Considering the integral is over one period I'll only take one root into account:

$$\delta(x-A\,sin(\omega t))=\frac{1}{|A|}\frac{\delta(t-sin^{-1}(x/A)/\omega)}{\omega cos(sin^{-1}(x/A))}=\frac{\sqrt{A^2-x^2}}{\omega}\delta(t-sin^{-1}(x/A)/\omega)$$
[/itex]

But $x$ is itself a function of time. In the following steps you proceed as if x was a time independent parameter.

The easy (and only way, I think) to do it is to do the volume integral before doing the time integral. Then the calculation is trivial.

 

[fuchini]

Thanks a lot! It's far easier that way. I have one final question, does this hold:

$$\int f(x) \delta(x^\prime-x(t)) dx^\prime=f(x(t))$$

Thanks again.

 

[nrqed]

Thanks a lot! It's far easier that way. I have one final question, does this hold:

$$\int f(x) \delta(x^\prime-x(t)) dx^\prime=f(x(t))$$

Thanks again.

Yes, that's correct. To see this, just imagine picking a certain value of t. Then it is obviously true at that time. But it must be true for any value of t so it is a valid relation even given that x(t) is a function of time.

You are welcome