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Algebra, how to separate a term in the sum of two roots

  1. Apr 16, 2014 #1
    My mind has gone blank and ive suddenly forgotten basic algebra, please could someone give me direction on how to make P the subject of this equation?

    E = (P^2 C^2 + M^2 C^4)^1/2 + (P^2 C^2)^1/2

    thanks for any help
     

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  2. jcsd
  3. Apr 16, 2014 #2
    oh i can just square each term, time for a break!

    mods please feel free to delete, sorry
     
  4. Apr 16, 2014 #3

    Mentallic

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    That image is terribly low quality, but what I could make of it is that it's

    [tex]E=\sqrt{P^2C^2+M^2C^2}+\sqrt{P^2C^2}[/tex]

    This equation seems similar to the energy-mass equivalence, but anyway, are there any restrictions on these variables? If [itex]P,C\geq 0[/itex] for example, then [itex]\sqrt{P^2C^2}=PC[/itex] and then you could move that to the other side and then square both sides to get rid of the square root.
     
  5. Apr 16, 2014 #4

    Mentallic

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    Be careful!

    [tex]\left(\sqrt{a}+\sqrt{b}\right)^2\neq a+b[/tex]
     
  6. Apr 16, 2014 #5
    hmm, yes just realised this, no restrictions, just a basic algebra problem.

    but if i take PC to the other side and square both sides i'll have:

    (E-PC)^2 on the left hand side? which still leaves me questioning how to get P out?

    if i multiply out (E-PC)^2 and simply i get

    E^2 - 2EPC = 2P^2 C^2 + M^2 C^2

    im not sure it should be this difficult
     
    Last edited: Apr 16, 2014
  7. Apr 16, 2014 #6

    SammyS

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    That's a quadratic equation in P.

    It might work out a bit more neatly if you divide through by C2 first.
     
  8. Apr 16, 2014 #7
    yes thought that may be an option, still would complicate the question further. im going to have to re-evaluate how to do the problem.

    thanks for everyones input!

    for those interested or if you are bored and have time here is the full solution that i am working through:

    http://inside.mines.edu/~lwiencke/PH300/F13/homework/hw4_solutions.pdf

    the bit im struggling with is on page 4, where it says "2pn=pe=...." in the equations above it, it solves the 4th equation down for pe.

    thanks again!
     
  9. Apr 16, 2014 #8

    SammyS

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    What does ##\displaystyle E_e = \left(p_e^2 c^2 + m_e^2 c^4 \right)^{1/2} ## have to do with the OP ?
     
  10. Apr 17, 2014 #9
    its the equation below it that he has rearranged, its very much complicated by the sudden change in notation (why oh why?) , so that how come i was trying to figure out the rearrangement of the eqn in the op.

    ive circled it below, and indicated what it goes to.
     

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  11. Apr 18, 2014 #10

    vela

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    The solution you're trying to work through is doing things the hard way. I imagine that whoever wrote it went around in circles with the algebra for a while and simply wrote down the end result. One of the tricks to doing relativity is figuring out how to avoid this unnecessary algebra in the first place. Try this approach instead:

    Let ##p_e## be the magnitude of the electron's momentum. Conservation of energy gives you
    $$E_\mu = E_e + E_\nu + E_\bar{\nu}.$$ Rewrite that as
    $$E_\mu - E_e = E_\nu + E_\bar{\nu}.$$ Using conservation of momentum, you should be able to show the righthand side is equal to ##p_e c##. Now square both sides and use the fact that ##(m_ec^2)^2 = E_e^2 - (p_e c)^2##. You'll end up with an equation you can easily solve for ##E_e## in terms of the masses.

    If you want to an even simpler method, learn how to use four-vectors.
     
  12. Apr 19, 2014 #11
    Excellent, will try this thank you!!!

    And I agree, I think the person writing the solution had similar difficulties and gave it to someone else who simply wrote down the end result, hence the change in notation lol

    But you are right, it is only a 4 mark question and shouldnt be this complicated. thanks for everyones help!
     
  13. Apr 20, 2014 #12
    hm spoke too soon, i get the first bit but then if i square both sides i.e. (E(u) - E(e))^2 i end up with E(u)^2 + E(e)^2 - 2E(u)E(e), see photo below.
     

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  14. Apr 20, 2014 #13

    vela

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    That's right. Keep going. Eliminating ##(p_e c)^2## from both sides, you have
    $$(m_\mu c^2)^2 + (m_e c^2)^2 - 2(m_\mu c^2)E_e = 0,$$ which you can solve for ##E_e## to get
    $$E_e = \frac{(m_\mu c^2)^2 + (m_e c^2)^2}{2 m_\mu c^2}.$$ Now solve for ##(p_e c)^2## using ##(m_e c^2)^2 = E_e^2 - (p_e c)^2## and plugging in the expression for ##E_e##.
     
  15. Apr 21, 2014 #14
    ahh i substituted for E(e) for some reason which complicated the problem, not sure why i did that since E(e) is what i'm trying to find :-/

    but anyhow THANKS for all your help!!!!
     
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