# Special Relativity: Dynamics

### Contents

#### Problems on Forces and Acceleration

Problem : In the lab frame a particle has velocity (vx, vy) = (0.6c, 0) acceleration (ax, ay) = (2, 3). What force is observed to be acting on the particle in the instantaneous inertial frame of the particle?

First let us calculate the force in the lab frame. This is given by: (Fx, Fy) = m(γ3ax, γay), where γ = = 5/4. Thus (Fy, Fx) = ((5/4)3×2,(5/4)*3) = (125/32, 15/4). For a frame such as the particle's, moving along with speed 0.6c with respect to the lab we have Fx' = Fx and Fy' = γFy. Thus the force in the instantaneous inertial rest frame of the particle is (Fx', Fy') = (125/32, 75/16) = 25/16(5/2, 3).

Problem : If p = γ2mvx, what would Fx be?

We need the quantity F = = . First let us calculate :

 = = = 2γ4va

Then:

 F = = m(v + γ2) = m(2γ4v2a + γ2a) = maγ(2v2γ2 -1) = γ3ma(1 + v2/c2)

Problem : Use the definition of force found in Section 3 and what you know about the relationship between force and energy to find an expression for the relativistic energy.

We have Fdx = dE so integrating both sides:

 Fdx = (γ3ma)dx = γ3mvdx = γmc2|v2v1

Which is ΔE for a particle changing from v1 to v2 . Of course, the energy change is offset by an equivalent change in potential energy such that total energy is conserved.

Problem : Define a quantity bi = . Let (bx, by) = γ3, γ similar to the relation for force. How does this quantity transform between a stationary frame and a frame moving with speed v in the x-direction?

We know that (bx, by) = (γ3, γ). In the moving frame γ = 1 so: (bx', by') = ,. But because of time dilation and length contraction this becomes:

 (bx', by') = γ3, γ3 = γ4, γ3

Thus, bx = and by = .

Problem : In some frame F' which is moving at speed v in the x-direction with respect to frame F, there is a mass attached the end of a spring. The spring is constraining to vibrate along a straight line which lies at an angle θ' to the x'-direction in frame F'. In frame F, what is the direction of the spring oscillation, the direction of the acceleration of the mass, and the direction of force on the mass?

In the stationary frame the frame F appears to be moving with speed v in the opposite direction. Hence, the projection of the spring oscillation on the x-axis must be reduced by a factor γ due to length contraction. Thus tanθ = γtanθ'. Since the mass is being pulled in the direction of the spring oscillation (this fact must be the same in any inertial frame), the acceleration must be in this same direction (tanθ = γtanθ'). From what we learned about force transformations between frames, the x-component of the force remains the same, but the y component is decreased by a factor γ; thus, the angle between the horizontal and the force is μ, where tanμ = .