Einstein originally gave the formula as

$$M = \mu + \frac{E_0}{c^2}.\tag{17}$$

In which $\mu$ was the mass of the system. Today, we more commonly get taught that the energy is in relation to the square root (written with modern notation as):

$$E = \sqrt{m_0^2c^4 + p^2c^2}.$$

The first equation does not generally follow the second, since $a + b \ne \sqrt{a^2 + b^2}$?

Of course, it is true that $a + b = \sqrt{a^2 + b^2}$ can be true if for the rest system if $b=0$ - is this the kind of idea that was being used?

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    $\begingroup$ Which of his papers has the first formula? $\endgroup$
    – G. Smith
    Feb 25, 2019 at 5:19
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    $\begingroup$ Unfortunately, I cannot make sense of $M=\mu+E_0/c^2$ because as far as I can tell from the paper, he is using $M$ to mean relativistic mass, $\mu$ to mean rest mass, and $E_0$ to mean rest energy. But with that those meanings this is wrong. He also says “we can arbitrarily assign the zero-point of $E_0$”, which is something else I don’t understand. In general, I have found that reading old papers is often extremely confusing compared with the relative clarity of modern expositions. $\endgroup$
    – G. Smith
    Feb 25, 2019 at 6:47
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    $\begingroup$ The only way the equation makes sense to me is if $E_0$ is the kinetic energy, but I find nothing in the paper that indicates this is so. $\endgroup$
    – G. Smith
    Feb 25, 2019 at 6:56
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    $\begingroup$ BTW, in my opinion you should think of the second equation in the form $m^2=E^2-\mathbf{p}^2$ (in units with $c=1$), which says “mass is the Lorentz-invariant length of the energy-momentum four-vector $(E, \mathbf{p})$”. This stresses the Minkowski geometry of spacetime, the relativity of energy and momentum, and the absoluteness of mass, all in one beautifully simple geometric equation. $\endgroup$
    – G. Smith
    Feb 25, 2019 at 7:01
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    $\begingroup$ Yes is just as posing v=0 to get E=mc2. In that original formula rest mass is the fraction, mu is the gain of inertia and m the relativistic mass. But I can't see in the link how mu was introduced. What I do here is based on what we know :) $\endgroup$
    – Alchimista
    Feb 25, 2019 at 8:37

1 Answer 1


In § 9 of his paper, Einstein defined the kinetic energy (eq. 14) as:

$\frac{\mu c^{2}}{\sqrt{1-\frac{q^{2}}{c^{2}}}}+const$

where $\mu$ is the invariant rest mass. In § 11 he went further and defined the total energy of a uniformly moving system, consisting of the above kinetic energy as well as the additional internal energy $E_0$ (work, thermodynamic energy, etc) as measured in its rest system, obtaining the equation (16a):


from which he concludes that the (invariant) mass of the system is given by (17):


In modern terminology, $E_{0}/c^{2}$ is the invariant mass of the additional internal energy, $\mu$ the invariant mass without additional internal energy, and $m$ is the invariant total mass or total rest energy (there is no "relativistic mass" involved in Einstein's example). Using $m$ in (16a), the relativistic energy equation obtains its modern form


Note, that Einstein's treatment is based on Planck's previous treatment in 1907 (to whom Einstein refers in his paper as well), who defined invariant mass $M$ as:


with $pV_{0}$ as the additional internal energy in terms of pressure and volume.

Now let's talk about the history of the energy-momentum relation: We simply take the above equation for relativistic energy, as well as the equation for relativistic momentum (given by Planck in 1906 and in 1907 as well as Einstein in 1907), which have the form in modern notation

$E=\frac{mc^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ and $p=\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Then we solve $p$ for $v$, inserting the result into $E$, which gives after some algebra and simplification:


  • $\begingroup$ Very clear and concise. $\endgroup$ Dec 16, 2019 at 13:22

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