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Special relativity. energy

  1. Jun 22, 2008 #1

    jk4

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    1. The problem statement, all variables and given/known data
    In its own frame of reference, a proton takes 5 min to cross the Milky Way galaxy, which is about [tex]10^{5}[/tex] light-years in diameter.
    (a) What is the approximate energy of the proton in electronvolts?
    (b) About how long would the proton take to cross the galaxy as measured by and observer in the galaxy's reference frame?


    2. Relevant equations
    [tex]T = \frac{T_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

    [tex]L = L_{0} \sqrt{1-\frac{v^{2}}{c^{2}}}[/tex]

    mass of proton = [tex]1.6726x10^{-27} kg[/tex]


    3. The attempt at a solution
    I assume the first thing I need to find is the velocity of the proton. Is that right?

    first I calculate the distance in meters:
    so [tex]10^{5}[/tex] light-years = [tex]9.45x10^{20} m[/tex]

    Then I set up the equation with time in seconds (5min = 300s) (going to leave out units):
    and [tex]T_{0}[/tex] is found by (distance/velocity) so (the distance above in meters)/v
    [tex]300 = \frac{9.45x10^{20}}{v \sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

    Anyways, when I run into problems when I try and solve this.

    Also the book shows the answer, which I have copied below exactly how it appears in book:
    ~[tex]10^{19} eV;[/tex] ~[tex]10^{5} y[/tex]

    So I'm assuming they used some kind of approximations, because If I use those values and work backwards I get a velocity of 1c I'm assuming the real velocity is like .99999c ect.. But my calculator just rounds off after so many digits.
     
  2. jcsd
  3. Jun 22, 2008 #2

    Hootenanny

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    Good :approve:. All the remains now is some algebraic manipulation, which can be a little awkward but is solvable non the less. I suggest you start by bringing the denominator to the LHS, dividing through by 300 and then squaring both sides.

    P.S. Thanks for taking the time to lay out your problem properly. It's a nice to see someone who take pride in their work and puts some effort into presentation.
     
  4. Jun 22, 2008 #3

    jk4

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    well I tried doing it like you said, and I got as far as:
    [tex]v^{2}-\frac{v^{4}}{c^{2}} = 3.15 x 10^{18}[/tex]

    So from there, I tried doing the quadratic equation. by substituting u in for [tex]v^{2}[/tex] so I had:
    [tex]-\frac{1}{c^{2}}u^{2} + u - 3.15 x 10^{18} = 0[/tex]

    but then because 4ac is bigger than [tex]b^{2}[/tex] I get an imaginary number.

    Also, to make sure I'm not doing it wrong I plugged the entire equation into my calculator that has a "solve" feature, and it gave me the answer "false"....
     
  5. Jun 22, 2008 #4

    Hootenanny

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    There must be an error somewhere, but I can't seem to spot it. In any case, a much more straight-forward approach would be to use four-vectors. Have you come across four-vectors before?
     
  6. Jun 22, 2008 #5

    jk4

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    only briefly, as in I've read about them and I know spacetime x,y,z,ict is one, but I have never seen them used in any calculations, and wouldn't know how to begin.
     
  7. Jun 22, 2008 #6

    jk4

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    alright, I think I may be on to something. I decided to start from scratch, and think of it from the protons point of view as if the galaxy was moving, so the size of the galaxy then becomes (in the protons frame of reference):
    [tex]L = 9.45x10^{20}\sqrt{1-\frac{v^{2}}{c^{2}}}[/tex]

    then the time it takes (5min or 300seconds) is equal to the distance over the velocity so:
    [tex]\frac{9.45x10^{20}\sqrt{1-\frac{v^{2}}{c^{2}}}}{v} = 300[/tex]

    and it is much easier to solve for v. If I find exact values though I run into the same probelm I did when working backwards, I get a speed of 1c. However, if I do the same calculation with rounding I get 0.9996c for the velocity but, then when I finish finding the energy with velocity 0.9996c I get [tex]3.3x10^{10} eV[/tex] whereas the book says the answer is ~[tex]10^{19} eV[/tex], so clearly something is still wrong.

    I'm still thinking that it has to do with the fact I had to round to get something other than 1c. I bet the answer just has so many 9's after the decimal that my calculator treats it as exactly 1, and this would explain why the book put a "~" in front of the answer, they must have had to use some approximation methods unknown to me.
    unless of course I'm still just getting it wrong lol.
     
  8. Jun 22, 2008 #7

    Dick

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    The approximation they likely used is that the proton's velocity is so close to c, that you can approximate that the time needed to cross the galaxy from a 'rest' observers viewpoint is really close to 10^5 years. So the gamma factor is 10^5*years/(5*min). Then just multiply the rest energy of the proton by that gamma factor. To solve it using your exact approach, you forgot to square the constant part 3.15*10^(18). But even then, you'd better be prepared to keep a lot of significant figures around.
     
    Last edited: Jun 22, 2008
  9. Jun 22, 2008 #8

    Dick

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    Now I think that's right. In your first attempt I think you had a gamma factor upside down. But still v is still so close to c you are going to have a hard time solving for it. Just solve for sqrt(1-v^2/c^2). Take v/c~1.
     
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