Work-Energy
Theorem
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Back to Special Relativity
The
work done on a particle under the force F in moving it from r1
to r2
is defined as
Let
us now replace the v in terms of p to help the integration and
make it easier. The momentum is given by p = mv = gm0v
Þ v = p/gm0v.
Substituting this expression into Eq. (1) gives
where
. Eq. (2) can now be expressed as
Integrate by parts using the template
Let u
= g-1
dv = dp2
du
= - g-2dg
v = p2
From the
definition of g we obtain
Substituting
into Eq. (5) we obtain
The
last integral on the right side can be shown to be
Differentiating
both sides verifies this fact. Eq. (7) can now be expressed as
Substituting
this back into Eq. (4) gives, omitting limits of integration for the moment
Solving
for W gives
Choose
the constant of integration such that the kinetic energy K = 0 when
W = 0. This gives a constant of E0
= m0c2,
i.e. objects rest energy. The kinetic energy is therefore given by
where E
= mc2.
Therefore the Eq. (1) becomes
This
can be expressed as
If
the force can be expressed as
where
the second term on the right side of Eq. (15) is defined such that
Substituting
Eq. (13) into Eq. (1) gives
Equating
with Eq. (13) we obtain
which
can be written as
If we
now add on the rest energy E0
= m0c2,
which is a
constant during the motion of the particle, to each side we get
the total energy which we’ll label T. Note that E = K + E0
= inertial energy so that we can express Eq. (19) as
