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Wave Mechanics help-for optics course

  1. Oct 3, 2008 #1
    Wave Mechanics help--for optics course

    1. The problem statement, all variables and given/known data

    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 [tex]\omega[/tex] 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 ERAD(r, t) and BRAD(r, t) of the electric charge using spherical coordinates (r, [tex]\theta[/tex],[tex]\phi[/tex]) 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 er, e[tex]\theta[/tex], e[tex]\phi[/tex].

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

    [tex]\vec{E}[/tex]RAD([tex]\vec{r}[/tex], t) = (-q/(4*pi*epsilon0c2))(1/r)([tex]\vec{a}[/tex](t-r/c) X [tex]\vec{e}[/tex]r) X [tex]\vec{e}[/tex]r

    By X i mean vector cross product, not multiplication, just so theres no confusion.

    [tex]\vec{B}[/tex]RAD([tex]\vec{r}[/tex], t) = (1/c)[tex]\vec{e}[/tex]r X [tex]\vec{E}[/tex]RAD([tex]\vec{r}[/tex], t)


    3. 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[tex]\theta[/tex]((2*pi*n/L)(rsin[tex]\theta[/tex]cos[tex]\phi[/tex]))cos(omega*t)

    So now, i sub my z(x, t) equation in spherical coordinates in for er in my ERAD equation, correct? I believe the ERAD equation will only be expressed in the unit vector er since this is only along the x axis where [tex]\theta[/tex] and [tex]\phi[/tex] = 0?
     
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  3. Oct 3, 2008 #2

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    You've expressed your [itex]z(x,t)[/itex] in spherical coordinates centered at the origin. To express it in SC's centered at (a,0,0), use the fact that [itex]x-a=rsin(\theta)cos(\theta)[/itex]. Also, [itex]z_0[/itex] is just a constant so leave it as is. Once you have [itex]z(r,\theta)[/itex] use the fact that [itex]x(r,\theta)=y(r,\theta)=0[/itex] and [itex]r^2=x^2+y^2+z^2[/itex] to find an expression for [itex]\vec{r}(t)[/itex]. Then differentiate it twice to find [itex]\vec{a}(t)[/itex]. What does that make [itex]\vec{a}(t-r/c)[/itex]?
     
  4. Oct 3, 2008 #3
    Re: Wave Mechanics help--for optics course

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

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

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

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

    Which makes [tex]\vec{a}[/tex](t)(t - r/c) = [tex]\vec{a}[/tex](t) = (z0sin(((2*pi*n)/L)*(rsin[tex]\theta[/tex]cos[tex]\phi[/tex] + a)* -omega2cos(omega * t))*t - (z0sin(((2*pi*n)/L)*(rsin[tex]\theta[/tex]cos[tex]\phi[/tex] + a)* -omega2cos(omega * t)* t) / c?
     
  5. Oct 3, 2008 #4

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    Actually, [itex]r(t)=z(r,\theta)[/itex], so [itex]\vec{r}(t)=z(r,\theta)\hat{e}_r[/itex]. Remeber to use the product rule when taking the derivative, because [itex]\hat{e}_r[/itex] is not constant.
     
  6. Oct 3, 2008 #5
    Re: Wave Mechanics help--for optics course

    I thought [tex]\vec{e}[/tex]r was just a unit vector? Isnt it the same as i as in i j k for cartesian coords? Im 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[tex]\vec{e}[/tex]r/dt (omega*sin(omega*t)) + d2[tex]\vec{e}[/tex]r/dt (cos(omega*t))

    I left a large chunk of the equation out to make it easier to type/read. Im pretty sure it was a constant so it wouldnt effect the differentiation, correct?
     
  7. Oct 3, 2008 #6

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    [itex]\hat{e}_r[/itex] is a unit vector; but unlike [itex]\hat{i},\hat{j},\hat{k}[/itex] it is not necessarily constant in time. But in either case, I misled you a bit; the real [itex]\vec{r}(t)[/itex] for the source charge is just [itex]\vec{r}(t)=z(t)\hat{k}[/itex] and so [itex]\vec{a}(t)=\ddot{z}(t)\hat{k}[/itex] (which you correctly calculated the magnitude of). To find [itex]\vec{a}(t-r/c)[/itex], just substitute [itex](t-r/c)[/itex] for [itex]t[/itex]. What is [itex](\hat{k} \times \hat{e}_r) \times \hat{e}_r[/itex]? What does that make [itex]\vec{E}_{rad}(\vec{r},t)[/tex]?
     
  8. Oct 3, 2008 #7
    Re: Wave Mechanics help--for optics course

    Wait...what is k? Did you mean ([tex]\vec{a}[/tex](t -r/c) X [tex]\vec{e}[/tex]r) X [tex]\vec{e}[/tex]r? Im not sure where that k came from.
     
  9. Oct 3, 2008 #8

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    [itex]\hat{k}[/itex] is the unit vector in the z-direction. Since [itex]\vec{a}[/itex] points in the z-direction,
    [tex](\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][/tex]
     
  10. Oct 3, 2008 #9
    Re: Wave Mechanics help--for optics course

    Alright ive been trying to find out exactly what [tex]\vec{e}[/tex]r is...it is the unit vector of the r component, correct? And isnt 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 dont understand what im supposed to "plug in". I apoligize for all these questions that i should probably know the answers to, its been awhile since ive taken Calculus, ive got 3 books open in front of me and i feel like i keep going in circles.
     
  11. Oct 3, 2008 #10

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    [itex]\hat{e}_r[/itex] and [itex]\hat{k}[/itex] 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 [itex](\hat{k} \times \hat{e}_r) \times \hat{e}_r[/itex]
    What do you get when you do that?
     
  12. Oct 4, 2008 #11
    Re: Wave Mechanics help--for optics course

    Ok, so the unit vector k would be <0, 0, 1> correct? Since its in the z direction? And the unit vector [tex]\vec{e}[/tex]r = xsin[tex]\theta[/tex]cos[tex]\phi[/tex] + ysin[tex]\theta[/tex]sin[tex]\phi[/tex] + zcos[tex]\theta[/tex], and since again since since [tex]\theta[/tex] and [tex]\phi[/tex] are both 0 since its along the z axis means [tex]\vec{e}[/tex]r is also <0, 0, cos[tex]\theta[/tex]> which would be <0,0,1>?
    So with the triple product rule all im getting is <0, 0, 1> - <0, 0, 1> = 0, but that doesnt seem right since it would make all of ERAD = 0.
     
  13. Oct 4, 2008 #12

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    That's not the triple product rule. This is:
    [tex](\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}[/tex]

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

    [tex]\hat{k}=cos(\theta)\hat{e}_r-sin(\theta)\hat{e}_{\theta}[/tex]

    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 [itex]\vec{r}[/itex] 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.
     
    Last edited: Oct 4, 2008
  14. Oct 4, 2008 #13
    Re: Wave Mechanics help--for optics course

    Was i correct about the definition of the unit vector [tex]\vec{e}[/tex]r?

    Then i just substitute them both into [tex]\hat{e}[/tex]r([tex]\hat{e}[/tex]r [tex]\cdot[/tex] [tex]\hat{k}[/tex])-[tex]\hat{k}[/tex]?

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

    It seems like theres a significant piece of information im missing here, like some sort of basic knowledge of this material that i dont have.
     
    Last edited: Oct 4, 2008
  15. Oct 4, 2008 #14

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    I'm assuming you meant:

    [tex]\hat{e}_r=sin(\theta)cos(\phi)\hat{i}+sin(\theta)sin(\phi)\hat{j}+cos(\theta)\hat{k}[/tex]

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

    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:

    [tex]\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} [/tex]

    And,

    [tex]\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} [/tex]

    Since you have [itex]\hat{k}[/itex] in terms of [itex]\hat{e}_r,\& \hat{e}_{\theta}[/itex] all you need to do is substitute it into the triple product and use the fact that the unit vectors [itex]\hat{e}_r,\hat{e}_{\theta},\& \hat{e}_{\phi}[/itex] are orthonormal.
     
    Last edited: Oct 4, 2008
  16. Oct 4, 2008 #15
    Re: Wave Mechanics help--for optics course

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

    Or is it...

    er 2 [tex]\cdot[/tex] cos([tex]\theta[/tex])er2 - sin([tex]\theta[/tex])e[tex]\theta[/tex]er - cos([tex]\theta[/tex])er - sin([tex]\theta[/tex])e[tex]\theta[/tex]

    And for some reason this Latex isnt really working for me...its not displaying the dot product symbol or theta in the correct spot.
     
  17. Oct 4, 2008 #16

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    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 [itex]\hat{e}_r[/itex]'s are vectors.

    For instance,

    [tex]\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)[/tex]

    Do you follow that?
     
  18. Oct 4, 2008 #17
    Re: Wave Mechanics help--for optics course

    I follow it, except i dont know how you got the values 1 and 0 for the dot products.

    If you could quickly explain that and also how [tex]\hat{k}= cos(\theta)\hat{e}_r - sin(\theta)\hat{e}_r[/tex] i would appreciate it.
     
  19. Oct 4, 2008 #18

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

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

    [tex]\hat{e}_r \cdot \hat{e}_{\theta}=\hat{e}_r \cdot \hat{e}_{\phi}=\hat{e}_{\phi} \cdot \hat{e}_{\theta}=0[/tex]

    And since they each have magnitude 1,

    [tex]\hat{e}_r \cdot \hat{e}_r=\hat{e}_{\theta} \cdot \hat{e}_{\theta}=\hat{e}_{\phi} \cdot \hat{e}_{\phi}=1[/tex]

    As for how I got the expression for [itex]\hat{k}[/itex]; 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 [itex]\hat{e}_r,\hat{e}_{\theta}, \hat{e}_{\phi}[/itex] by [itex]\hat{r},\hat{\theta}, \hat{\phi}[/itex] and [itex]\hat{i},\hat{j},\hat{k}[/itex] by [itex]\hat{x},\hat{y},\hat{z}[/itex] or [itex]\hat{e}_x,\hat{e}_y,\hat{e}_z[/itex]
     
  20. Oct 4, 2008 #19
    Re: Wave Mechanics help--for optics course

    Ok, so [tex](\hat{k} \times \hat{e}) \times \hat{e} = -sin(\theta)\hat{e}_r[/tex]

    [tex]\hat{e}_r\cdot\hat{k} = cos(\theta)[/tex]

    [tex]\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[/tex]

    So...

    [tex]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[/tex]

    Which makes ERAD =

    [tex]-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)[/tex]
     
  21. Oct 4, 2008 #20

    gabbagabbahey

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    Re: Wave Mechanics help--for optics course

    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?
     
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