1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relative energy and momentum

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of rest mass [itex]m_0[/itex] is moving the S-frame at a velocity of [itex]u=(c/2, c/2, c/2) [/itex]. Calculate:

    a) its energy E and its momentum p in the S-frame

    b) E' and p' in the S'-frame moving at [itex]v=(c/2, 0,0)[/itex] relative to S



    2. Relevant equations

    [itex]v_x = \frac{v_x-u}{\sqrt{1-(v_xu)/c^2}}[/itex]

    [itex]E= \gamma m_0 c^2[/itex]

    [itex]p = \gamma m_0 u[/itex]


    3. The attempt at a solution
    I was given the solutions, I just have difficulties understanding. The solution to part a) is fairly simple, simply calculating [itex]\gamma = \frac{1}{\sqrt{1-(u/c)^2}}[/itex] where [itex](u/c)^2 = 3/4[/itex]. Then [itex]E= \gamma m_0 c^2 = 2m_0c^2[/itex] and [itex]p = \gamma m_0 u = 2m_0(c/2, c/2, c/2) = (m_0c, m_0c, m_0c)[/itex]


    It's part b) that's giving me problems. The solution given is as follows:

    LT for E and p. [itex]\gamma(v) = \frac{1}{(1-1/4)^{1/2}} = 2/ \sqrt{3}[/itex]. [itex]E' = \gamma (v) (E-vp_x) = (2/ \sqrt{3})(2-1/2)m_0c^2 = \sqrt{3}m_0c^2[/itex]

    [itex]p_x' = \gamma(v)(p_x - vE/c^2) = 2/ \sqrt{3} (m_0c - c/2*2m_0) = 0[/itex]. So then since [itex]p_y'=p_y=m_0c[/itex] and [itex]p_z'=p_z=m_0c[/itex], we get [itex]p' = (0, m_0c, m_0c)[/itex].


    The following is how I thought the question should be solved.

    Note that [itex]v_x' = 0, v_y' = c/2, v_z' = c/2[/itex] so that [itex]u = (0, c/2, c/2)[/itex]. Then we would find [itex]\gamma = \frac{1}{\sqrt{1-(u^2/c^2)}}[/itex] where [itex]u^2 = (c/2)^2+(c/2)^2 = 1/2c^2[/itex] so that [itex]\gamma = \sqrt{2}[/itex].

    Then using we find that [itex]E= \gamma m_0 c^2 = \sqrt{2} m_0 c^2[/itex].

    We also get [itex]p = \gamma m_0 u = \sqrt{2} m_0 (0, c/2, c/2) = (0, m_0c/\sqrt{2}, m_0c/ \sqrt{2})[/itex].


    I understand the logic of the given solution for momentum, but I don't understand where I made the mistake in my solution. Also, does anyone know how the two equations for relative energy and momentum were derived? I hadn't seen these two equations before I looked at the solutions.

    I'm completely at a loss for the given solution for energy ... I'm confused why exactly we used [itex]\gamma (v)[/itex], when shouldn't we generally use the relative velocity of the particle with respect to that particular frame (so take [itex]v=(0,c/2,c/2)[/itex] with respect to the S'-frame)??

    Any help would be appreciated. Thanks!
     
    Last edited: Feb 12, 2013
  2. jcsd
  3. Feb 12, 2013 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can't just add velocities to transform between reference frames. Velocities don't add that way in special relativity. So it's NOT true that u' = u - v.

    In more detail: it looks like you tried to used the Galilean rule for the addition of velocities (which is wrong and only holds approximately true at low speeds), when you should use the special relativistic rule. Check it out:

    http://en.wikipedia.org/wiki/Velocity-addition_formula

    EDIT: It appears that you are familiar with the correct special relativistic formula for the addition of velocities, since you posted it in your Relevant Equations! So I find it puzzling that you would make this mistake.
     
  4. Feb 12, 2013 #3
    Hmm, I'm not quite sure where you believe I added velocities. I used Lorentz transformations to find that [itex]v_x'=0, v_y=c/2, v_z=c/2[/itex]. Where [itex]v_x' = \frac{v_x-u}{1-\frac{uv_x}{c^2}} = \frac{c/2-c/2}{1-\frac{(c/2)(c/2)}{c^2}} = 0[/itex] and [itex]v_y' = v_z = \frac{v_y-u}{1-\frac{uv_x}{c^2}} = \frac{c/2-0}{1-\frac{(0)(c/2)}{c^2}} = c/2[/itex]. It just occurred to me now that instead of using [itex]u=(u_x,u_y,u_z)[/itex], should I instead use [itex]u=||u||[/itex]?

    For the [itex]u^2=(c/2)^2+(c/2)^2[/itex], I was attempting to find the norm of u = [itex]\sqrt{u_x^2+u_y^2+u_z^2}[/itex] where each u_x, u_y, and u_z were all in the same frame (S'-frame).
     
    Last edited: Feb 12, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relative energy and momentum
Loading...