special relativity - What's the purpose of the arbitary additive constants in Einstein's Inertia of Energy Paper?

In Einstein's paper: Does the Inertia of a Body Depend upon its Energy content? he introduces arbitary additive constants whose purpose I'm confused about.

The paper has a frame $(x,y,z)$ where a body at rest emits plane waves of light in opposite directions, each with an energy $\frac 1 2 L$ at an angle $+\theta$, $-\theta$ to the $x$ axis where the energy of the body before and after transmission is $E_0$ and $E_1$. This process is measured in a frame $(\xi,\eta,\zeta)$ moving along $x$ with velocity $v$, where the energies of the body before and after are $H_0$ and $H_1$. Subtracting the total energy in frame $(\xi,\eta, \zeta)$ from $(x,y,z)$

$$H_0 − E_0 − (H_1 − E_1) = L\left(\frac {1} {\sqrt{1 - \frac {v^2}{c^2}}} -1\right)$$

He writes with my emphasis

The two differences of the form H − E occurring in this expression have simple physical significations. H and E are energy values of the same body referred to two systems of co-ordinates which are in motion relatively to each other, the body being at rest in one of the two systems (system (x, y, z)). Thus it is clear that the difference H − E can differ from the kinetic energy K of the body, with respect to the other system $(\xi,\eta,\zeta)$, only by an additive constant C, which depends on the choice of the arbitrary additive constants of the energies H and E

Obviously the same energy scale is used in both frames when measuring $E$ and $H$, so what's the purpose of these arbitarary additive constants?

Answer

Suppose I have a body that (on some measurement scale) has an energy of $E$ in the rest frame. It will have energy $H=\gamma E$ in the moving frame, so the difference is $H-E = E(\gamma - 1)$. Now suppose I want to change to a different energy scale with the same units but a different additive constant, so that I now consider the rest energy to be $E'=E+E_0$, with $E_0$ being an arbitrary constant. Now the energy in the moving frame will be $H'= \gamma E' = \gamma(E+E_0) = H + \gamma E_0$, and the difference will be $H'-E' = (E+E_0)(\gamma - 1) = (H-E) + E_0(\gamma - 1)$. I think that Einstein's $C$ is the $E_0(\gamma-1)$ term here.