Bill_K said:
Under a Lorentz transformation t = γ t0, where t is the time in an arbitrary rest frame and t0 is the proper time. So a reasonable guess to make is that E behaves the same way that time behaves, namely E = γ E0 where E0 is the energy in the rest frame.
It's better not to guess but to calculate:
The change of energy is
dE = F \cdot ds
Newton says
F = \frac{{dp}}{{dt}}
and
p = m \cdot v
This leads to
dE = m \cdot v \cdot dv + v^2 \cdot dm
Now we need an expression for the inertial mass m. If there is a dependence from velocity it should be the same for all bodies. Therefore I start with
m\left( v \right): = m_0 \cdot f\left( v \right)
where f(v) is a function of velocity independent from the body and from the frame of reference and m0 is the mass of the body in rest. So we already know
f\left( 0 \right) = 1
Due to isotropy f must also be symmetric
f\left( { - v} \right) = f\left( v \right)
Now let’s assume a body A with the velocity v and a body B at rest. Both bodies should have the same rest mass m0. The product C of a perfectly inelastic collision shall have the rest mass M0 and the velocity u. The conservation of momentum leads to
p = m_0 \cdot f\left( v \right) \cdot v = M_0 \cdot f\left( u \right) \cdot u
At this point I have to make a reasonable assumption (I will check it later): If energy is conserved the mass of C shall be the sum of the masses of A and B:
M_0 \cdot f\left( u \right) = m_0 \cdot f\left( v \right) + m_0 \cdot f\left( 0 \right)
This results in
f\left( v \right) = \frac{u}{{v - u}}
To get the velocity u I change to a frame of reference moving with the velocity v. Now body B moves with -v and body A is at rest. As the situation is symmetric the velocity of C is
u' = - u
The next steps depends on the transformation. Galilei transformation
u' = - u = u - v
leads to
f\left( v \right) = 1
Therefore in classical mechanics inertial mass is independent from the frame of reference and the change of Energy is reduced to
dE = m \cdot v \cdot dv
The integration leads to
E = E_0 + {\textstyle{1 \over 2}}m_0 \cdot v^2
In SRT Galilei transformation is replaced by Lorentz transformation
u' = - u = \frac{{u - v}}{{1 - \frac{{u \cdot v}}{{c^2 }}}}
Everything else remains unchanged. This leads to
m\left( v \right) = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}
Although this is often called "relativistic mass" it is still the good old inertial mass as used in Newton’s definition of momentum. With
dm = \frac{{m_0 \cdot v \cdot dv}}{{c^2 \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3 }} = \frac{{m \cdot v \cdot dv}}{{c^2 - v^2 }}
The change of Energy is
dE = dm \cdot c^2
Now Integration leads to
E = E_0 + \left( {m - m_0 } \right) \cdot c^2
With Einstein’s equation for rest mass and rest energy we get an expression for inertial mass and total energy:
E = m \cdot c^2
Due to the additivity of energy
E = \sum\limits_i {E_i } = \sum\limits_i {\left( {m_i \cdot c^2 } \right) = \left( {\sum\limits_i {m_i } } \right)} \cdot c^2 = m \cdot c^2
the inertial mass is additive:
m = \sum\limits_i {m_i }
Therefore my assumption above has been correct.
