- #1
PirateFan308
- 94
- 0
Homework Statement
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
Homework 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]
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: