Ratio of relativistic energy to rest mass energy

Question

Consider a relativistic particle, of mass $m$, moving in Minkowski space. Show that $\frac{dt}{d\tau}$ is the ratio of the energy to the rest-mass energy of the particle.

I'm having trouble with this problem. I know that the rest mass energy is $E_{res}^2=(mc^2)^2$ while the (relativistic) energy is $E_{rel}^2=(mc^2)^2 + (pc)^2$

How do I relate $E_{rel}$ and $E_{res}$ to t and $\tau$?

I also know that $t = \gamma\tau$ where $\tau$ is the proper time and $\gamma$ is the Lorentz factor, but I don't know how this will help.

Answer 1

Along the lines of my comment,
the answer to "How do I relate $E_{rel}$ and $E_{res}$ to t and τ?" is "similar triangles".

For a 4-vector $\tilde Q$, the components are:

A proper-time displacement 4-vector $\tilde \tau$ (with magnitude $\tau$) and the energy-momentum 4-vector $\tilde p$ (with magnitude $p=mc$) are both proportional to the 4-velocity (whose slope is $v$).

Answer 2

It can be shown that the total relativistic energy of a particle of mass $m$ travelling with Lorentz factor $\gamma$ is given by: $$ E = \gamma m c^2 $$ Recognizing $mc^2$ as the rest mass of the particle, you can see that $\gamma$ is the ratio of the particle's energy to its rest mass. You can use this with the formula you provided involving proper time to demonstrate the relationship you are looking for.

Answer 3

We can use the definition of the spacetime interval to work out the relationship between $dt$ and $d\tau$: $$ds^2=c^2d\tau^2=c^2dt^2-dx^2$$ Dividing by $c^2dt^2$, we get: $$\left(\frac{d\tau}{dt}\right)^2=1-\frac{v^2}{c^2}$$ $$d\tau=dt\sqrt{1-\frac{v^2}{c^2}}$$ $$\frac{dt}{d\tau}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma$$

In general, the total energy of a particle is: $$E=\gamma mc^2$$ And the energy in its proper frame (rest frame) is: $$E_{0}=mc^2$$

Dividing $\tfrac{E}{E_{0}}$, we get the Lorentz factor.

$$\frac{E}{E_{0}}=\gamma$$