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 socalled
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:
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:
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 2dimensions
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
xdirection, 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
ycomponent 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 xdirection 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}) = mγ^{3}, γ 

In the instantaneous inertial frame
γ = 1 so:
By computing the appropriate length and time transformations from the
Lorentz formulas we find that:
(F_{x}', F_{y}') = mγ^{3}, γ^{2} 

Two factors of
γ come from the time
dilation (
t^{2}) and the
additional factor on the
xcomponent 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.