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Solving 4 simultaneous equations

  1. Nov 14, 2006 #1
    Got 4 equations with 4 unknowns and I'm being a retard so I can't figure this thing out.

    k, k', alpha and beta are known constants.

    [tex] A + B = C + D [/tex]
    [tex] k(A-B) = k'(C-D) [/tex]
    [tex] Ce^{\alpha} + De^{-\alpha} = Fe^{\beta} [/tex]
    [tex] k'Ce^{\alpha} - k'De^{-\alpha} = kFe^{\beta} [/tex]

    alpha is ik'L, beta is ikL, k is [tex] \frac{\sqrt{2mE}}{\hbar} [/tex] and k' is [tex] \frac{\sqrt{2m(U_{0}+E)}}{\hbar} [/tex]

    Now, I can eliminate F by multiply the 3rd equation by k and setting the two equations equal.

    I'm trying to solve for B.

    Anybody got an idea where to start? (If you're interested, these are the smoothness conditions of a quantum potential barrier)
     
  2. jcsd
  3. Nov 14, 2006 #2
    there are actually five unknowns

    i think you should try and try and get a relation between C and D by eliminating F... tahts the first step that was shown to me

    after that find a relation between A and C by eliminating B

    are you trying to find the tranmission and reflection coefficients??

    if you find relations betwen each other they will cancel out once you try and find T and R
     
  4. Nov 14, 2006 #3
    Yes, I'm proving an equation for R. R is B*B / A*A, so I need to solve for B by eliminating C, D, and F

    I did get the relation thru eliminating F for C and D.

    Would converting [tex] e^{ik'L} = cos(k'L) + isin(k'L) [/tex] help? I'm trying it, but it looks worse.
     
  5. Nov 14, 2006 #4
    dont do that

    that would make a bigger mess

    just work with what you have now
     
  6. Nov 14, 2006 #5
    The relation for C and D you mean is
    [tex]k(Ce^{\alpha}+De^{-\alpha}) = k'(Ce^{\alpha}-De^{-\alpha})[/tex]

    I have absolutely no clue what to do after this. It sucks. I've wasted 13 legal sized pages of paper trying ideas. None work. I can get B in terms of A and C or A and D, but I can't eliminate the C or D terms instead of A.
     
  7. Nov 15, 2006 #6
    Now, I get:
    [tex] C = -\frac{k+k'}{k-k'}e^{-2\alpha}D [/tex]
    and I can obviously solve that backwards for D.

    So now I have an equation for C in terms of D. I'm solving for B. I'm having so much trouble with the fact that the powers of e are opposite. I know they don't drop out, because the equation I'm trying to prove is:

    [tex] R = \frac{B*B}{A*A} = \frac{sin^2(k'L)}{sin^2(k'L) + 4\frac{k'^2k^2}{(k^2-k'^2)^2}} [/tex]

    So I obviously need B in terms of A only. I can see how you would get a sine function based on [tex]e^{ix} = cos(x) + isin(x)[/tex]
     
    Last edited: Nov 15, 2006
  8. Nov 18, 2006 #7
    OK, ok, so after more wasted time and errors made I have arrived at the following two equations, for C and D in terms of F.
    [tex]C = \frac{(k+k')Fe^{\beta}}{2k'e^{\alpha}}[/tex]
    and
    [tex]D = \frac{(k'-k)Fe^{\beta}e^{\alpha}}{2k'}[/tex]

    Now, I'm thinking that if I just plug it into one of the first two equations I will obtain B in terms of A and F. The Fs will not cancel, so that won't help. Substituting these equations of C in either of the last two should just produce a cyclical result, giving me no new equations.

    So, what I'm thinking I have to do next is solve for F in terms of A. Then when I plug in C and D in A+B=C+D I can then take the resultings Fs from C and D and substitute my equation for F in there. Then, I should hopefully have B in terms of A.

    Is this a good way of doing it? Taking the 3rd and 4th to solve for C(F) and D(F) then taking the 1st and 2nd and solve for F(A) then take the 1st in its B(A,C,D) form and substitute back in to get B(A,F) and then substitute to get B(A)??
     
    Last edited: Nov 18, 2006
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