0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Which of his papers has the first formula? $\endgroup$ – G. Smith Feb 25 at 5:19
  • $\begingroup$ In fear of the paper not being in english, you can follow the same relationship in the nomenclature section of en.wikipedia.org/wiki/Energy-mass_equivalence $\endgroup$ – Gareth Meredith Feb 25 at 5:40
  • $\begingroup$ It is equation (17) in this (translated) paper: einsteinpapers.press.princeton.edu/vol2-trans/300 $\endgroup$ – G. Smith Feb 25 at 5:58
  • $\begingroup$ thank you. Other than that, any idea's how history went from the first equation., to taking the square root? $\endgroup$ – Gareth Meredith Feb 25 at 6:32
  • $\begingroup$ 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 at 6:47
3
$\begingroup$

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):

$E=\left(\mu+\frac{E_{0}}{c^{2}}\right)\frac{c^{2}}{\sqrt{1-\frac{q^{2}}{c^{2}}}}$

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

$m=\mu+\frac{E_{0}}{c^{2}}$

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

$E=\frac{mc^{2}}{\sqrt{1-\frac{q^{2}}{c^{2}}}}$

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:

$M=\frac{E_{0}+pV_{0}}{c^{2}}$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.