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[Linear Algebra]- Simultaneous equation. Deriving the general solution.

  1. Jul 19, 2011 #1

    I need help with solving the following simultaneous equation. I tried using matrix solution using Gaussian elimination but it became too difficult early on.

    Known constants are a,A,b, and B.
    Known variables are x and y.
    Unknown variables are m and n.

    (a[itex]\times[/itex]x)[itex]/[/itex](A - x) + (b[itex]\times[/itex]y)[itex]/[/itex](B - y) =
    (a[itex]\times[/itex]m)[itex]/[/itex](A - m) + (b[itex]\times[/itex]n)[itex]/[/itex](B - n)

    I tried the matrix solution below. I set the known constants to 1 for the second row.

    [(a[itex]\times[/itex]x)[itex]/[/itex](A - x) (b[itex]\times[/itex]y)[itex]/[/itex](B - y) = c1 ]
    [(x)[itex]/[/itex](1 - x) (y)[itex]/[/itex](1 - y) = c2]
  2. jcsd
  3. Jul 19, 2011 #2


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    Hey randombill and welcome to the forums.

    I'm sorry but I can't figure out what linear system this is representing.

    Also I notice you have a few division signs in your post. Linear systems have only variables with powers of 1, so if those any of the denominators involve variables, you won't be able to use linear algebra to find a solution (if one exists).
  4. Jul 19, 2011 #3
    Ok, thanks for your help. I guess that solves the problem.

    So I guess numerically it may be possible.
  5. Jul 19, 2011 #4


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    This NOT a "system" of equations, it is a single equation in two unknowns and so has no unique solution.

    This is NOT a matrix (matrices do not have "=" inside them) nor does it correspond to the equation you had before. Assuming the "= c1" and "= c2" were supposed to be outside the matrix, this corresponds to the equations
    [tex]\frac{ax}{A- x}+ \frac{by}{B-y}= c1[/tex]
    [tex]\frac{x}{1- x}+ \frac{y}{1-y}= c2[/tex]

    Not the single equation you had before. I have no idea what problem you really meant!
  6. Jul 19, 2011 #5

    So I suppose this equation would have the same situation and the square roots make it even more difficult to work with.

    [itex] \frac{mv1}{\sqrt{1-v1/c^{2}}}[/itex] + [itex] \frac{mv2}{\sqrt{1-v2/c^{2}}}[/itex] = [itex] \frac{mv3}{\sqrt{1-v3/c^{2}}}[/itex] + [itex] \frac{mv4}{\sqrt{1-v4/c^{2}}}[/itex]

    where v1 and v2 are the initial velocities and v3 and v4 are the final velocities and v1,v2 are known.
  7. Jul 19, 2011 #6
    Ah! Finally it becomes clear what you're working on! You're trying to figure out what are the possible outcomes of a collision between two particles, right?

    In one dimension, if you know v_1 and v_2, then the outcome (v_3 and v_4) is almost completely determined (assuming an elastic collision). But you need another equation: conservation of energy. One way to find the solution is to boost into the center of mass frame. In that frame, where v_1 = -v_2, there are only two possible results: (1) v'_3 = v'_1, v'_4 = v'_2, or (2) v'_3 = -v'_1, v'_4 = -v'_2. (v' = velocity in the CM frame.) So then the only problem is to find the boost that gets you in the CM frame.

    Hope this helps.
  8. Jul 19, 2011 #7
    Can you give an example, please. I'm not sure what equations to use.
  9. Jul 19, 2011 #8
    Actually, in this case, where the two particles are equal in mass, the result is very simple. You either have v3 = v1, v4 = v2, or v3 = v2, v4 = v1.

    You've written the conservation of momentum equation. Let's call momentum p. In relativity, the energy of a particle is related to the momentum by [itex]e^2-p^2c^2=m^2c^4[/itex], so [itex]e=\sqrt{m^2c^4+p^2c^2}[/itex]. Your energy equation is e1 + e2 = e3 + e4, which works out to
    \sqrt{\frac{c^6 m^2}{c^2-\text{v1}^2}}+\sqrt{\frac{c^6 m^2}{c^2-\text{v2}^2}}=\sqrt{\frac{c^6 m^2}{c^2-\text{v3}^2}}+\sqrt{\frac{c^6 m^2}{c^2-\text{v4}^2}}

    It should be obvious, looking at that equation and the conservation of momentum equation that you've already posted, that the two answers I gave are solutions. It takes a little more work to show that they're the only solutions, but actually, no messy solving of equations is necessary. Just notice that momentum always has the same sign as velocity, and always increases as velocity increases. This tells you that, if you have a solution of the momentum equation, any other solution will require compensating changes in v3 and v4 -- if one goes up, the other has to come down. Energy, on the other hand, is always positive, and increases as the absolute value of velocity. If you think hard about this, you'll see that it means there are only two ways to solve both equations.
  10. Jul 19, 2011 #9
    It is obvious when the masses are the same but does the above equation work for unequal masses and if not, how would I solve for that.

  11. Jul 19, 2011 #10
    The equations work for unequal masses, except of course that, instead of having just "m" everywhere, you'll have m_1 and m_2 in the parts of the expressions referring to the momenta and energies of particles 1 and 2. Solving the equations is a lot messier when the masses are unequal (except for some simple special cases). I would have to think a bit about how to do that. Maybe someone else who is familiar with this problem will jump in.
  12. Jul 19, 2011 #11
    I know its not easy, I was expecting you to surprise me but thanks anyways. Hopefully someone will chime in.
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