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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.

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  • $\begingroup$ Hint: draw a 4-velocity vector and break it into components... in the resulting right-triangle [in Minkowski spacetime geometry] compare the hypotenuse with the adjacent side. Do the same for a 4-momentum vector. $\endgroup$
    – robphy
    Feb 1, 2018 at 2:35

3 Answers 3

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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: RRGP-4vectorQ

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 $v$).

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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.

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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.

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

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