Wave Mechanics help-for optics course

[nissanztt90]

Wave Mechanics help--for optics course

Homework Statement

Consider a stretch string of length L along the x-axis in a stationary vibration mode of the form

z(x, t) = z₀sin((2*pi*n/L)x) cos(omega*t)

where n is greater than or equal to 2 and is a given integer number and y₀ and $$\omega$$ are constant (i think they meant z₀). The left endpoint of the string is located at the origin x = y = z = 0 and the right endpoint is located at x = L, y = z= 0

An electrically charged particle with charge q>0 is glued to the string at a distance a from the left endpoint.

Write down the radiation fields E_RAD(r, t) and B_RAD(r, t) of the electric charge using spherical coordinates (r, $$\theta$$,$$\phi$$) centered at x = a, y = z = 0, assuming that the vibration amplitude z₀ is much smaller than |r|. Express your answer as a vector field written in terms of the unit vectors e_r, e_$$\theta$$, e_$$\phi$$.

There is more to it, but this is all i need help with for now.

Homework Equations

$$\vec{E}$$_RAD($$\vec{r}$$, t) = (-q/(4*pi*epsilon₀c²))(1/r)($$\vec{a}$$(t-r/c) X $$\vec{e}$$_r) X $$\vec{e}$$_r

By X i mean vector cross product, not multiplication, just so there's no confusion.

$$\vec{B}$$_RAD($$\vec{r}$$, t) = (1/c)$$\vec{e}$$_r X $$\vec{E}$$_RAD($$\vec{r}$$, t)

The Attempt at a Solution

So my first guess was to convert the form of the vibration mode to spherical coordinates and i got...

z(x, t) = rcos$$\theta$$((2*pi*n/L)(rsin$$\theta$$cos$$\phi$$))cos(omega*t)

So now, i sub my z(x, t) equation in spherical coordinates in for e_r in my E_RAD equation, correct? I believe the E_RAD equation will only be expressed in the unit vector e_r since this is only along the x-axis where $$\theta$$ and $$\phi$$ = 0?

 

[gabbagabbahey]

You've expressed your $z(x,t)$ in spherical coordinates centered at the origin. To express it in SC's centered at (a,0,0), use the fact that $x-a=rsin(\theta)cos(\theta)$. Also, $z_0$ is just a constant so leave it as is. Once you have $z(r,\theta)$ use the fact that $x(r,\theta)=y(r,\theta)=0$ and $r^2=x^2+y^2+z^2$ to find an expression for $\vec{r}(t)$. Then differentiate it twice to find $\vec{a}(t)$. What does that make $\vec{a}(t-r/c)$?

 

[nissanztt90]

Ok so z(r, $$\theta$$) = z₀sin(((2*pi*n)/L)*(rsin$$\theta$$cos$$\phi$$ + a))cos(omega*t)...

Since x(r, $$\theta$$) and y(r, $$\theta$$) = 0, r² = z², $$\vec{r}$$(t) = z(r, $$\theta$$)? And then differentiate twice with respect to t?

So $$\vec{r}$$'(t) = z₀sin(((2*pi*n)/L)*(rsin$$\theta$$cos$$\phi$$ + a)* -omega*sin(omega * t)...

So $$\vec{r}$$''(t) = $$\vec{a}$$(t) = z₀sin(((2*pi*n)/L)*(rsin$$\theta$$cos$$\phi$$ + a)* -omega²cos(omega * t)?

Which makes $$\vec{a}$$(t)(t - r/c) = $$\vec{a}$$(t) = (z₀sin(((2*pi*n)/L)*(rsin$$\theta$$cos$$\phi$$ + a)* -omega²cos(omega * t))*t - (z₀sin(((2*pi*n)/L)*(rsin$$\theta$$cos$$\phi$$ + a)* -omega²cos(omega * t)* t) / c?

 

[gabbagabbahey]

Actually, $r(t)=z(r,\theta)$, so $\vec{r}(t)=z(r,\theta)\hat{e}_r$. Remeber to use the product rule when taking the derivative, because $\hat{e}_r$ is not constant.

 

[nissanztt90]

I thought $$\vec{e}$$_r was just a unit vector? Isnt it the same as i as in i j k for cartesian coords? I am not sure how i would take the derivative of it it since there is no given value for it in the problem...the second derivative i worked out though was...

-omega²cos(omega*t) - 2d$$\vec{e}$$_r/dt (omega*sin(omega*t)) + d²$$\vec{e}$$_r/dt (cos(omega*t))

I left a large chunk of the equation out to make it easier to type/read. I am pretty sure it was a constant so it wouldn't effect the differentiation, correct?

 

[gabbagabbahey]

$\hat{e}_r$ is a unit vector; but unlike $\hat{i},\hat{j},\hat{k}$ it is not necessarily constant in time. But in either case, I misled you a bit; the real $\vec{r}(t)$ for the source charge is just $\vec{r}(t)=z(t)\hat{k}$ and so $\vec{a}(t)=\ddot{z}(t)\hat{k}$ (which you correctly calculated the magnitude of). To find $\vec{a}(t-r/c)$, just substitute $(t-r/c)$ for $t$. What is $(\hat{k} \times \hat{e}_r) \times \hat{e}_r$? What does that make [itex]\vec{E}_{rad}(\vec{r},t)[/tex]?

 

[nissanztt90]

Wait...what is k? Did you mean ($$\vec{a}$$(t -r/c) X $$\vec{e}$$_r) X $$\vec{e}$$_r? I am not sure where that k came from.

 

[gabbagabbahey]

$\hat{k}$ is the unit vector in the z-direction. Since $\vec{a}$ points in the z-direction,
$$(\vec{a}(t-v/c) \times \hat{e}_r) \times \hat{e}_r=a(t-v/c)[(\hat{k} \times \hat{e}_r) \times \hat{e}_r]$$

 

[nissanztt90]

Alright I've been trying to find out exactly what $$\vec{e}$$_r is...it is the unit vector of the r component, correct? And isn't the unit vector just a scalar? Does it have a value in this case--same with k...what is the value of these vectors? I know how to do a cross product of two vectors, but i don't understand what I am supposed to "plug in". I apoligize for all these questions that i should probably know the answers to, its been awhile since I've taken Calculus, I've got 3 books open in front of me and i feel like i keep going in circles.

 

[gabbagabbahey]

$\hat{e}_r$ and $\hat{k}$ are both unit vectors not scalars. The fact that they are unit vectors means that they each have a magnitude of 1.
First use the BAC-CAB triple product rule on $(\hat{k} \times \hat{e}_r) \times \hat{e}_r$
What do you get when you do that?

 

[nissanztt90]

Ok, so the unit vector k would be <0, 0, 1> correct? Since its in the z direction? And the unit vector $$\vec{e}$$_r = xsin$$\theta$$cos$$\phi$$ + ysin$$\theta$$sin$$\phi$$ + zcos$$\theta$$, and since again since since $$\theta$$ and $$\phi$$ are both 0 since its along the z axis means $$\vec{e}$$_r is also <0, 0, cos$$\theta$$> which would be <0,0,1>?
So with the triple product rule all I am getting is <0, 0, 1> - <0, 0, 1> = 0, but that doesn't seem right since it would make all of E_RAD = 0.

 

[gabbagabbahey]

That's not the triple product rule. This is:
$$(\hat{k} \times \hat{e}_r) \times \hat{e}_r=-\hat{e}_r \times (\hat{k} \times \hat{e}_r)=\hat{k}(-\hat{e}_r \cdot \hat{e}_r)-\hat{e}_r(-\hat{e}_r \cdot \hat{k})=\hat{e}_r(\hat{e}_r \cdot \hat{k}) -\hat{k}$$

Now, since the answer is supposed to be in terms of spherical coordinates; you should express $\hat{k}$ in terms of the SC unit vectors instead of the other way around.

$$\hat{k}=cos(\theta)\hat{e}_r-sin(\theta)\hat{e}_{\theta}$$

What do you get when you use this to simplify the above triple product?

Also, theta and phi are not necessarily zero; E_rad is a function of the field Point at which it is measured $\vec{r}$ While the source point is along thee z-axis, the field point need not be. The theta's and phi's and r's in the expression for E_rad represent the coordinates of the field point, not the source charge.

 

[nissanztt90]

Was i correct about the definition of the unit vector $$\vec{e}$$_r?

Then i just substitute them both into $$\hat{e}$$_r($$\hat{e}$$_r $$\cdot$$ $$\hat{k}$$)-$$\hat{k}$$?

Wouldnt i have to express one in terms of the other? Also, how did you derive k?

It seems like there's a significant piece of information I am missing here, like some sort of basic knowledge of this material that i don't have.

 

[gabbagabbahey]

I'm assuming you meant:

$$\hat{e}_r=sin(\theta)cos(\phi)\hat{i}+sin(\theta)sin(\phi)\hat{j}+cos(\theta)\hat{k}$$

There is no need to use this relation though, since your answer is supposed to be expressed in terms of $\hat{e}_r,\hat{e}_{\theta},\& \hat{e}_{\phi}$.

The complete relations between unit vectors can be derived in many ways and should be found in any good textbook dealing with vector calculus. You can easily generate them using the following relationships:

$$\begin{bmatrix} \hat{e}_r \\ \hat{e}_{\theta} \\ \hat{e}_{\phi} \end{bmatrix}} =\begin{bmatrix} sin(\theta)cos(\phi) & sin(\theta)sin(\phi) & cos(\theta) \\ cos(\theta)cos(\phi) & cos(\theta)sin(\phi) & -sin(\theta) \\ -sin(\phi) & cos(\phi) & 0 \end{bmatrix} \begin{bmatrix} \hat{i} \\ \hat{j} \\ \hat{k} \end{bmatrix}$$

And,

$$\begin{bmatrix} \hat{i} \\ \hat{j} \\ \hat{k} \end{bmatrix} =\begin{bmatrix} sin(\theta)cos(\phi) & cos(\theta)cos(\phi) & -sin(\phi) \\ sin(\theta)sin(\phi) & cos(\theta)sin(\phi) & cos(\phi) \\ cos(\theta) & -sin(\theta) & 0 \end{bmatrix} \begin{bmatrix} \hat{e}_r \\ \hat{e}_{\theta} \\ \hat{e}_{\phi} \end{bmatrix}$$

Since you have $\hat{k}$ in terms of $\hat{e}_r,\& \hat{e}_{\theta}$ all you need to do is substitute it into the triple product and use the fact that the unit vectors $\hat{e}_r,\hat{e}_{\theta},\& \hat{e}_{\phi}$ are orthonormal.

 

[nissanztt90]

Where I am getting confused here is i thought the dot product is between two vectors...what is the e_r vector that i need to perform the dot product calculation with k?

Or is it...

e_r ² $$\cdot$$ cos($$\theta$$)e_r² - sin($$\theta$$)e_$$\theta$$e_r - cos($$\theta$$)e_r - sin($$\theta$$)e_$$\theta$$

And for some reason this Latex isn't really working for me...its not displaying the dot product symbol or theta in the correct spot.

 

[gabbagabbahey]

You have to type the whole expression in between the [_tex] and [_/tex] delimiters (without the _'s of course), not just part of it. If you click on any of the LaTeX images from my posts, you will see the code that generated it.

The $\hat{e}_r$'s are vectors.

For instance,

$$\hat{e}_r \cdot \hat{k}=\hat{e}_r \cdot(cos(\theta)\hat{e}_r-sin(\theta)\hat{e}_{\theta})=cos(\theta)(\hat{e}_r \cdot \hat{e}_r )-sin(\theta) (\hat{e}_r \cdot \hat{e}_{\theta})=cos(\theta)(1)-sin(\theta) (0)=cos(\theta)$$

Do you follow that?

 

[nissanztt90]

I follow it, except i don't know how you got the values 1 and 0 for the dot products.

If you could quickly explain that and also how $$\hat{k}= cos(\theta)\hat{e}_r - sin(\theta)\hat{e}_r$$ i would appreciate it.

 

[gabbagabbahey]

The unit vectors $\hat{e}_r,\hat{e}_{\theta},\& \hat{e}_{\phi}$ are othonormal. Each has magnitude of 1 and they are perpendicular to each other. Since they are perpendicular,

$$\hat{e}_r \cdot \hat{e}_{\theta}=\hat{e}_r \cdot \hat{e}_{\phi}=\hat{e}_{\phi} \cdot \hat{e}_{\theta}=0$$

And since they each have magnitude 1,

$$\hat{e}_r \cdot \hat{e}_r=\hat{e}_{\theta} \cdot \hat{e}_{\theta}=\hat{e}_{\phi} \cdot \hat{e}_{\phi}=1$$

As for how I got the expression for $\hat{k}$; I looked it up in my trusty Griffith's "Introduction to Electrodynamics" text. I recommend you crack open a linear algebra textbook and do a quick review of unit vectors and different coordinate systems. Keep in mind, some authors call $\hat{e}_r,\hat{e}_{\theta}, \hat{e}_{\phi}$ by $\hat{r},\hat{\theta}, \hat{\phi}$ and $\hat{i},\hat{j},\hat{k}$ by $\hat{x},\hat{y},\hat{z}$ or $\hat{e}_x,\hat{e}_y,\hat{e}_z$

 

[nissanztt90]

Ok, so $$(\hat{k} \times \hat{e}) \times \hat{e} = -sin(\theta)\hat{e}_r$$

$$\hat{e}_r\cdot\hat{k} = cos(\theta)$$

$$\hat{e}_r(cos(\theta)) - \hat{k} = \hat{e}_r(cos(\theta)) - (\hat{e}_r(cos(\theta)) - sin(\theta)\hat{e}_\theta) = -sin(\theta)\hat{e}_\theta$$

So...

$$a(t-v/c)[(\hat{k} \times \hat{e}_r) \times \hat{e}_r] = z_0 sin((2\pi n)/L*(rsin(\theta)cos(\phi)))\omega^2 cos(\omega(t-r/c)) * -sin(\theta)\hat{e}_\theta$$

Which makes E_RAD =

$$-q/(4\pi\epsilon_0 c^2)(1/r)(z_0 sin((2\pi n)/L*(rsin(\theta)cos(\phi)))\omega^2 cos(\omega(t-r/c)) * -sin(\theta)\hat{e}_\theta)$$

 

[gabbagabbahey]

Looks good to me (although you should probably cancel the negative signs in front of q and sin(theta) with each other).

Now , How about B_rad?

 

[nissanztt90]

So $$\vec{B}_{RAD}(\vec{r}, t) = (1/c) *\vec{e}_r \times \vec{E}_{RAD}(\vec{r}, t)$$

Im lost again with what values i should be using for these vectors to do a cross product calculation, so I am guessing that's not the correct way to do it...is there some identiy again that i should know for the next step?

 

[gabbagabbahey]

Well, you just showed that $\vec{E}_{rad}$ points in the $\hat{e}_{\theta}$ direction, so $\hat{e}_r \times \vec{E}_{RAD}(\vec{r}, t)=E_{RAD}(\vec{r}, t) (\hat{e}_r \times \hat{e}_{\theta})$.

Using the right hand rule; $\hat{e}_r \times \hat{e}_{\theta}=\hat{e}_{\phi}$

 

[nissanztt90]

So then $$\vec{B}_{RAD}(\vec{r}, t) = (1/c)(q/(4\pi\epsilon_0 c^2)(1/r)(z_0 sin((2\pi n)/L*(rsin(\theta)cos(\phi)))\omega^2 cos(\omega(t-r/c))*sin(\theta)\hat{e}_\theta ^2))$$

 

[gabbagabbahey]

Not quite, when I write $E_{RAD} (t-r/c)$ I'm referring to the magnitude of $\vec{E}_{RAD} (t-r/c)$; so

$$\vec{B}_{RAD}(t-r/c)=\frac{E_{rad}(t-r/c)}{c} \hat{e}_{\phi}=\frac{q \omega ^2 z_0 sin \left( \frac{2 \pi n}{L} rsin(\theta)cos(\phi) \right)cos(\omega (t-r/c))sin(\theta)}{4 \pi \epsilon_0 r c^3} \hat{e}_{\phi}$$

 

[nissanztt90]

Thats actually what i meant but for some reason i though $$\hat{e}_r \times \hat{e}_\theta = \hat{e}_\theta$$...dont know why...i know the cross product is a vector orthogonal to both...and all $$\hat{e}_\phi$$ is othogonal to both...anyway...also...i forgot about the a value this whole time...shouldnt it be $$rsin\theta cos\phi - a$$?

The next part is "What are the values of the radiation fields $$\vec{E}_{RAD}(\vec{r}, t)$$ and $$\vec{B}_{RAD}(\vec{r}, t)$$ if $$a = \frac{kL}{n}$$, for some integer numer k, 0 < k < n?"

Im guessing i just plug it into a in the equation to get $$\frac{q \omega ^2 z_0 sin \left( \frac{2 \pi n}{L} rsin(\theta)cos(\phi) - \frac{kL}{n}\right)cos(\omega (t-r/c))sin(\theta)}{4 \pi \epsilon_0 r c^3} \hat{e}_{\phi}$$? Or do i rearrange the equation and substitute it in for n or L?

Lastly...is there any energy input required in order to maintain the string in the above vibration mode with constant amplitude $$z_0$$? If this is the case, what is the average energy required per unit time? We may use any formulas derived in class without proof...

The equation for power (energy/time) we derived in class is...

Power = $$\frac{q^2}{6\pi\epsilon_0 c^3} * a(t - \frac{r}{c})^2$$

I just plug $$a(t- \frac{r}{c})$$ in and simplify, correct?

 

[gabbagabbahey]

I think it should be $rsin\theta cos\phi +a$. Other than that, it looks good to me.

Remeber,
$$sin \left( \frac{2 \pi n}{L} (rsin(\theta)cos(\phi) + \frac{kL}{n}) \right)=sin \left( \frac{2 \pi n}{L} rsin(\theta)cos(\phi)+2\pi k\right)$$

What is the sine of any angle plus some integer multiple of 2pi?

 

[nissanztt90]

Correct about the +...not sure how kL/n = 2pi*k, but the sine of any angle + an integer multiple of 2pi is just the sine of the angle...2pi is just a shift in the sine graph...right?

 

[nissanztt90]

And for average energy/t i get...

$$\frac{q^2}{6\pi\epsilon_0 c^3} *<z^2_0 sin^2(\frac{2\pi n}{L} rsin\theta cos\phi + a)\omega^4 cos^2(\omega t - \frac{\omega r}{c}>_T$$

Can that be simplified any further?

 

[gabbagabbahey]

Correct about the +...not sure how kL/n = 2pi*k, but the sine of any angle + an integer multiple of 2pi is just the sine of the angle...2pi is just a shift in the sine graph...right?

the 2pik came from the fact that the 2pin/L multiplies the whole expression for x, not just the rsincos part. (2pin/L*kL/n=2pik) And yes, it is just a shift in the sine graph and is equal to the sine of the angle.

 

[gabbagabbahey]

And for average energy/t i get...

$$\frac{q^2}{6\pi\epsilon_0 c^3} *<z^2_0 sin^2(\frac{2\pi n}{L} rsin\theta cos\phi + a)\omega^4 cos^2(\omega t - \frac{\omega r}{c})>_T$$

Can that be simplified any further?

Yes, to compute the time average <...>_T, integrate the expression over one period $t:0 \rightarrow T=\frac{2\pi}{\omega}$ and then divide by the period, $T$. The only part of the expression that contains a t in it is the $cos^2(\omega (t-r/c))$ part. What is the integral over cos^2 over a full period?

 

[nissanztt90]

Im using the integral table laws of sins/cosines and for the integral of cos^2 I am using the integral of cos^2 = 1/2(x + sinxcosx)...after simplifying everything to what i believe is correct i end up with...

$$\frac{1}{2} (2\pi + sin(2\pi - \frac{\omega r}{c})cos(2\pi - \frac{\omega r}{c}) - sin(\frac{\omega r}{c})cos(\frac{\omega r}{c}))$$

 

[gabbagabbahey]

Huh?! I've never seen that trig ID before (because it is incorrect). I've seen $cos^2(x)=\frac{1}{2}(1-cos(2x)$ before though. There is an easier way though; the integral of cos^2 over a full period represents the area under cos^2 from 0 to T. Due to symmetry, the area under sin^2 will be the same so

$$\int_0^T cos^2(\omega(t-r/c))dt=\frac{1}{2} \left(\int_0^T cos^2(\omega(t-r/c))dt +\int_0^T sin^2(\omega(t-r/c))dt \right)$$

$$=\frac{1}{2} \int_0^T (cos^2(\omega(t-r/c)) + sin^2(\omega(t-r/c)))dt=\frac{1}{2} \int_0^T (1)dt=\frac{T}{2}$$

$$\Rightarrow \left< cos^2(\omega(t-r/c)) \right> _T=\frac{1}{2}$$

I think you are probably supposed to take a spatial average of P_rad as well, in that case, just integrate the r,theta and phi dependent terms over the volume of a sphere and divide by the volume.

 

[nissanztt90]

Thats because its not a trig identity, it was the integral, although i did it incorrectly anyway.

So the power simplified is..

$$\frac{q^2 \omega^4 z^2_0}{12\pi\epsilon_0 c^3} * sin^2(\frac{2\pi n}{L} rsin\theta cos\phi + a)$$

There is no mention of a spatial average in the hw, nor in my notes.

 

[gabbagabbahey]

Well, the question asks for the average energy per unit time. To me, that means spatial and temporal average. It makes no sense that the average power would depend on your position in the radiation field.

 

[nissanztt90]

I will ask the professor tomorrow what exactly he is looking for. The only thing i have in my notes for average power (that is the same as average energy per unit time right?) is:

power = $$\frac{q^2}{6\pi\epsilon c^3} *a(t-\frac{r}{c})$$

Looking over my notes I think we did do the spatial average during the derivation of the above formula...part the derivation consists of what you described on the previous page.

Then he gave us the average power forumla

power = $$\frac{q^2}{6\pi\epsilon c^3} *<a(t-\frac{r}{c})>_T$$