Problem : Find the inner product of the 4-momenta for the following two particles: a particle of mass m moving with speed in the lab frame and a particle of mass M moving with velocity also in the lab frame.
Because the inner product is invariant it does not matter which frame we work in, therefore let us choose the simplest. This would be the frame of lab. The m particle has 4-momentum (γ _{vy } mc, 0γ _{vy } mv _{y}, 0) . The particle in motion has 4-momentum (γ _{uy } Mc, 0, γ _{uy } Mu _{y}, 0) . Thus the inner product is:
Mmγ _{vy } γ _{uy }(c ^{2} - v _{y} u _{y}) |
Problem : Calculate the same inner product as in the previous question, but now in a frame moving with one of the particles (or, if you already did it in such a frame, calculate it in the lab frame). Check that the result is the same.
In this frame the 4-momenta of the particle with speed v _{y} is (mc, 0, 0, 0) . The particle at rest in this frame sees the other particle moving in the y -direction with speed w _{y} = . Thus the 4-momentum is (γ _{wy } Mc, 0, γ _{wy } Mw _{y}, 0) . Thus the inner product is:
Mmc ^{2} γ _{wy } |
c ^{2} γ _{wy } | = | ||
= | (c ^{2} - u _{y} v _{y}) |
Problem : Prove that if A and B are 4-vectors, A.B their inner product is independent of the frame in which it is calculated.
The inner product in an arbitrary frame is given by:
A.BâÉáA _{0} B _{0} - A _{1} B _{1} - A _{2} B _{2} - A _{3} B _{3} |
= γ(A _{0}' + vA _{1}') γ(B _{0}' + vB _{1}') - γ(A _{1}' + vA _{0}') γ(B _{1}' + vB _{0}') - A _{2}'B _{2}' - A _{3}'B _{3}' | |||
= γ ^{2} A _{0}'B _{0}' + v(A _{0}'B _{0}' + A _{1}'B _{0}') + v ^{2} A _{1}'B _{1}' - γ ^{2} A _{1}'B _{1}' + v(A _{1}'B _{0}' + A _{0}'B _{1}') + v ^{2} A _{0}'B _{0}' - A _{2}'B _{2}' - A _{3}'B _{3}' | |||
= A _{0}'B _{0}'(γ ^{2} - γ ^{2} v ^{2}) - A _{1}'B _{1}'(γ ^{2} - γ ^{2} v ^{2}) - A _{2}'B _{2}' - A _{3}'B _{3}' | |||
= A _{0}'B _{0}' - A _{1}'B _{1}' - A _{2}'B _{2}' - A _{3}'B _{3}' | |||
âÉáA'.B' |
Problem : Derive the velocity addition formula using the invariance of the 4-velocity inner product. In other words, if in frame A, B moves to the right with speed v , and C moves to the left with speed u , find w , the speed of B with respect to C.
First we must compute the 4-velocities of B and C in the frame of A. They are (γ _{v} c, γ _{v} v, 0, 0) for B and (γ _{u} c, - γ _{u} u) for C. Next, what are the 4-velocities of B and C in C's frame: for B (γ _{w} c, γ _{w} w, 0, 0) and for C (c, 0, 0, 0) . We can take the inner product of each of these pairs and set them equal since the inner product is frame invariant:
(γ _{v} c, γ _{v} v, 0, 0).(γ _{u} c, - γ _{u} u, 0, 0) = (γ _{w} c, γ _{w}, 0, 0).(c, 0, 0, 0) | |||
âá’γ _{u} γ _{v}(c ^{2} + uv) = γ _{w} c ^{2} | |||
âá’ = |
w = |
Problem : Again using the invariance of the inner product, determine the speed of one particle as observed by the other as two particles approach each other with speed v along trajectories separated by an angle 2θ , as shown in the figure below.
(γ _{v} c, γ _{v} v cosθ, - γ _{v} v sinθ, 0).(γ _{v} c, γ _{v} v cosθ, γ _{v} v sinθ, 0) = (γ _{w} c, γ _{w} w, 0, 0).(c, 0, 0, 0)âá’γ _{v} ^{2}(1 - v ^{2}cos 2θ) = γ _{w} |
w = | |||
âá’w = |
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