Solving 4 simultaneous equations

In summary, the person is trying to solve for the transmission and reflection coefficients for a quantum potential barrier using equations and unknowns. They started by getting four equations with four unknowns, but are being a retard and can't figure it out. They say that after some more wasted time and errors, they arrived at two equations for the transmission and reflection coefficients in terms of the unknowns F. However, they say that if they just plug the equations of C in either of the last two equations, they will get a cyclical result, without providing any new equations. So, the person needs to solve for F in terms of A, then take the resultings Fs from C and D and substitute back into the equations for A+
  • #1
joex444
44
0
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)
 
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  • #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
 
  • #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.
 
  • #4
joex444 said:
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.

dont do that

that would make a bigger mess

just work with what you have now
 
  • #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.
 
  • #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]
 
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  • #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 into get B(A,F) and then substitute to get B(A)??
 
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Related to Solving 4 simultaneous equations

What is the definition of simultaneous equations?

Simultaneous equations are a set of equations that are solved together to find the values of multiple variables.

How many equations are needed to solve 4 simultaneous equations?

At least 4 equations are needed to solve 4 simultaneous equations. However, more equations may be required depending on the complexity of the equations.

What are the methods used to solve 4 simultaneous equations?

There are multiple methods that can be used to solve 4 simultaneous equations, including substitution, elimination, and graphing. These methods involve manipulating the equations and solving for one variable at a time.

What is the importance of solving 4 simultaneous equations in scientific research?

Solving 4 simultaneous equations is important in scientific research as it allows for the determination of multiple unknown variables in a system. This can help in understanding complex relationships and making predictions.

What are the common challenges in solving 4 simultaneous equations?

Some common challenges in solving 4 simultaneous equations include dealing with large numbers or fractions, identifying the most efficient method to solve the equations, and avoiding errors in calculations.

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