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 = |
= = = γ ^{3} va |
F = = m( v + γ ) = ma(γ ^{3} v ^{2} + γ) = γ ^{3} ma |
= = m = γ ^{3} mv |
= γ ^{3} ma = F = |
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:
= m + ,( + |_{vy=0} | |||
m (, | |||
= m(γ ^{3} a _{x}, γa _{y}) |
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 |
(F _{x}', F _{y}') = m , |
(F _{x}', F _{y}') = m γ ^{3} , γ ^{2} |
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