How does the Lorentz Factor appear in $E = c^2$?

Question

I recently found the derivation of the Lorentz factor here. It shows the ratio of time in different inertial frames to be equal to the Lorentz factor. I'm now trying to understand how this ratio relates to the equation
$_{rel}≡$.

Answer 1

We should start by saying that few physicists today would say the mass of a particle changes with the velocity. This was the view taken early in the development of special relativity, but these days we take the view that the only mass that matters is the rest mass and that this is a constant. If you are interested this topic is discussed in considerable detail in the answers to the question Why is there a controversy on whether mass increases with speed?

Where this all comes from is that the momentum of a relativistic particle is given by:

$$ \mathbf p = \gamma m \mathbf v \tag{1} $$

The notion of a relativistic mass originated by writing this as:

$$ \mathbf p = m_{rel} \mathbf v \tag{2} $$

i.e. keeping the Newtonian definition of momentum $p = mv$ and just replacing $\gamma m$ in equation (1) by the relativistic mass $m_{rel}$ to get equation (2). However the view these days is that we should restrict ourselves to using the rest mass and accept that the momentum is given by equation (1) instead.

So starting from this perspective you questions becomes why and how the factor of $\gamma$ appears in the equation (1) for the momentum. There is a rigorous way to derive this by using the principle of least action to derive the momentum from the action, but this is somewhat involved and luckily there is a simpler if less rigorous approach.

What we will do is assume the four-momentum $P$ is given by $P = mU$, where $U$ is the four-velocity. The four-vectors $P$ and $U$ are the relativistic versions of the Newtonian momentum and velocity. We start by writing the position of the particle as the four-vector:

$$ \mathbf X = (t, x, y, z) $$

then differentiating with respect to proper time $\tau$ to get the four-velocity:

$$ \mathbf U = \frac{d\mathbf X}{d\tau} = \left(\frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau} \right) $$

and finally multiplying by $m$ to get the four-momentum:

$$ \mathbf P = m\mathbf U = \left(m\frac{dt}{d\tau}, m\frac{dx}{d\tau}, m\frac{dy}{d\tau}, m\frac{dz}{d\tau} \right) \tag{3} $$

Now in equation (3) we are differentiating with respect to proper time $\tau$ not coordinate time $t$ but we can convert to a derivative wrt coordinate time using the chain rule. For example:

$$ \frac{dx}{d\tau} = \frac{dx}{dt}\frac{dt}{d\tau} $$

And this is where the factor of $\gamma$ creeps in because $dt/d\tau$ is just the time dilation factor i.e. the Lorentz factor $\gamma$. So $dx/d\tau$ is just equal to $\gamma dx/dt$ i.e. just $\gamma v_x$, where $v_x$ is the $x$ component of the non-relativistic three velocity. And making this substitution the expression for the four-momentum becomes:

$$ \mathbf P = \left(\gamma m, \gamma mv_x, \gamma mv_y, \gamma mv_z \right) = (\gamma m, \mathbf p) \tag{4} $$

where the $\mathbf p$ in equation (4) is just the non-relativistic three momentum:

$$ \mathbf p = \left(\gamma mv_x, \gamma mv_y, \gamma mv_z \right) = \gamma m \mathbf v $$

And that's the equation (1) that we started out with!