Special Relativity: Dynamics

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.

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.

Solution for Problem 2 >>

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

But c ² γ _wy is given by:

c 2 γ wy	=
	=	(c 2 - u y v y)

So the overall result is: Mmγ _vy γ _uy (c ² - u _y v _y) , which is the same as above.

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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.

Solution for Problem 3 >>

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

Then using the definition of the 4-vector we have:

= 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'

Which is the same as the inner product in some other inertial frame moving past with speed v .

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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.

Solution for Problem 4 >>

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’ =

Solving for w yields:

w =

Which is the velocity addition formula.

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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.

Particle approaching each other at an angle.

Solution for Problem 5 >>

In the frame of an outside observer the 4-velocities of the two particles are (γ _v c, γ _v v cosθ, - γ _v v sinθ, 0) and ( γ _v c , γ _v v cosθ , γ _v v sinθ , 0 ). If we let w be the speed of one particle as viewed by the other and rotate so that the relative motion is along the x -axis, then the 4-velocities in the frame of one of the particles are (γ _w c, γ _w w, 0, 0) and (c, 0, 0, 0) . The invariance of the inner product says:

(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 2cos 2θ) = γ w

where we used cos ² θ - sin² θ = cos 2θ in the last equality. Expanding out for the γ s, squaring, and solving for w yields:

w =
âá’w =

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