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Special Relativity: Dynamics

Force and Acceleration

Problems on Four-vectors

Problems on Forces and Acceleration

Force in one dimension

For the sake of simplicity in this section we will switch to units in which c = 1 . This seems like a strange and confusing thing to do, but in fact simplifies things immensely. In doing this we just ignore all factors of c and if we need them back at the end (of working a problem, say) we can just check where units of m/s are missing. In so-called relativistic units, p = γmv , as before, and E = γm . It is good to get used to c = 1 because many advanced treatments of Special Relativity use it extensively.

Unfortunately the old Newtonian law is not much good to us in Special Relativity because our concept of velocity has undergone a radical change. Instead we must define the force on an object as the rate of change of momentum:

F =    

Clearly when p = mv , this reduces to the Newton's Second Law. But we saw in the section on relativistic momentum that p = γmv . Of course this is now complicated by the fact that for a changing velocity, γ is also changing with time. So:

= = = γ 3 va    

Since a = . Therefore we have:

F = = m( v + γ ) = ma(γ 3 v 2 + γ) = γ 3 ma    

We can also relate this to the derivative of the relativistic energy with respect to space:

= = m = γ 3 mv    

But v = = = a , so:

= γ 3 ma = F =    

This last statement is by far the most important: we have found that for p = γmv and E = γm , the rate of change of momentum over time equals the rate of change of energy over space.

Force in 2-dimensions

In Special Relativity, force in two dimensions can become a strange, unintuitive concept. Most strangely, it is not always true that force points in the same direction as the acceleration of an object! Even though we are working in two, and not three, dimensions we can use the vector equation:

   

Consider a particle moving in the x -direction, with a force acting on it . The momentum is given by:

   

Note that we are still in units where c = 1 . We can take the derivative of this with respect to time and use the fact that v y = 0 initially:


     
= m + ,( + |vy=0      
m (,      
= m(γ 3 a x, γa y)      

Thus the force is not proportional to the acceleration. The first component of the force vector agrees with what we derived in one dimension, but the y -component only has a single γ factor. This occurs because, assuming v y = 0 initially γ changes when v x changes but not when v y changes. Our conclusion is that it is easier to accelerate something in the direction transverse to its motion.

Say we have a force acting on a particle in its instantaneous inertial rest frame (it can only be instantaneous since the particle is accelerating due to the force on it) F' . Say F' is moving with speed v along the x -direction relative to another frame F . How can we relate the components of the force in the two frames? In F we have from above:

(F x, F y) = m γ 3 , γ    

In the instantaneous inertial frame γ = 1 so:

(F x', F y') = m ,    

By computing the appropriate length and time transformations from the Lorentz formulas we find that:

(F x', F y') = m γ 3 , γ 2    

Two factors of γ come from the time dilation ( t 2 ) and the additional factor on the x -component comes from a length contraction in that direction only. Thus the components of force transform as F x = F x' and F y = . The transverse force is a factor of γ larger in the particle's frame.

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