# 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?

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.

• I think he's saying that some energies measured in a frame have an absolute zero such as kinetic or electromagnetic, whereas others such as gravitational potential energy have an arbitary point chosen as zero and will therefore differ from kinetic by an arbitary constant in some frame. Looking on the web, there's still a lot of confusion over interpreting this paper. – John McVirgo Jun 16 '12 at 20:10
• That's how I interpreted it too, from what you quoted in your question - that the rest energy is defined with respect to an arbitrary zero, but that kinetic energy should be zero when the object is at rest. In retrospect I could have expressed that more clearly in my answer. – Nathaniel Jun 16 '12 at 20:57

I agree with Nathaniel's answer, but I'd like to offer a different way of looking at this which I would claim is both more historically natural and easier to understand because it's more concrete.

You, as Einstein's reader, are presumed to have read the first 1905 paper and to have understood its material about kinematics, electromagnetism, and the kinetic energy of the electron (not of a macroscopic body). But you don't know anything else about relativistic mass, energy, or momentum, and you don't know anything about how these quantities transform except in the nonrelativistic limit. Let's consider two examples of types of energy.

You know about potential energy, and you know that it depends on the distance between objects. Einstein describes a body, such as a lantern, that emits rays of light in opposite directions. If you imagine that the lantern contains particles separated by various distances and interacting through various forces, then it contains some amount of potential energy $U$, which depends on the distances between the particles. Nonrelativistically, $U$ is frame-independent, but you know from the first 1905 paper that relativistically the distances between the particles will contract when we transform from the $(x,y,z)$ rest frame to the new frame $(\xi,\eta,\zeta)$. If you express $U$ relative to its value when the lantern is disassembled into atoms, then presumably $|U|$ will increase because of the decreased distance. You have no way of calculating this increase, but you expect it to exist.

From reading section 10 of the earlier paper, you also know the relativistic expression for the kinetic energy of an electron, and you know that it differs from the nonrelativistic expression $(1/2)mv^2$. An object such as a lantern contains some amount of thermal energy $E_T$. In nonrelativistic mechanics, you know that there's a theorem stating that if a system of particles conserves energy in one frame, then you can infer that it would conserve energy in another frame if and only if momentum is also conserved. This theorem vastly simplifies classical mechanics by, e.g., allowing us to ignore the thermal energy of objects in elastic collisions. But the theorem assumes the exact $v^2$ dependence of kinetic energy on velocity. The relativistic dependence isn't $v^2$. The lantern is made out of some materials that have some microscopic structure, containing some amount of thermal energy $E_T$. This $E_T$ is going to transform in some complicated, nonlinear way because of the relativistic transformation of the microscopic kinetic energies. You have no idea how to calculate the transformation exactly, but you expect that because the relativistic expression for kinetic energy grows faster than $v^2$ at large velocities, the lantern in the $(\xi,\eta,\zeta)$ will probably have some excess of thermal energy relative to what you would have expected nonrelativisticvally.

When Einstein constructs a difference like $H_0-E_0$, you therefore expect a bunch of different terms to contribute to it physically. At a minimum, you expect it to contain:

1. the macroscopic kinetic energy $K_0$ of the lantern

2. an anomalous relativistic effect on the potential energy $U$

3. an anomalous relativistic effect on the thermal energy $E_T$

The constant $C$ is meant to absorb effects like 2 and 3.

With modern hindsight, we know that mass-energy is the timelike part of the energy-momentum four-vector, and therefore all of these effects can be calculated in a uniform and simple way. As Nathaniel says, the transformation from the rest frame $(x,y,z)$ to the $(\xi,\eta,\zeta)$ frame simply multiplies the lantern's mass-energy by a factor of $\gamma$. But Einstein hasn't established that yet, and that's why he can't assume anything about $C$ other than that it's a constant that "doesn't change during the emission of the light."

• The assumption that $C$ doesn't change during the emission of the light is equivalent to assuming that the potential and thermal energy of the lantern in one frame will be the same before and after the emission of light. How such an assumption is justified? it's not that I find it wrong, however Einstein just stated that $C$ won't change without any further elaboration or explanation, and for me it's not quite obvious why should this be the case. – Omar Nagib Sep 1 '15 at 7:16

Einstein is simply referring to the fact that energy, length,... are always measured with respect to an arbitrary origin, and what is physical is not the absolute value of the measure, but the difference of the measures before and after. In other words, what you measure is always a difference between two states. In fact, the equation can be written initially as $$(E_1 - E_0) - (H_1 -H_0) = H_0 − E_0 − (H_1 − E_1) = L\left(\frac {1} {\sqrt{1 - \frac {v^2}{c^2}}} -1\right)$$ and the constant $C$ appears if the observer in the frame A, measuring the energies E, uses a different origin than the observer in frame B, measuring the energies H, but it has also to be combined with the relativistic effects as shown by Nathaniel (where both observers use the same origin $E_0$, but they are in a rectilinear and uniform movement)

• -1 he didn't bother with different frames using different standards of measurements in his 1905 paper, so it's safe to assume he didn't do the same here, otherwise he would have said so. – Physiks lover Apr 5 '13 at 21:01
• But this is exactly what he says in ".... by an additive constant C, which depends on the choice of the arbitrary additive constants of the energies H and E". In general, let's choose a metric system such that in the frame A, we measure $E = E_0 + a_0$ and in the frame B, we measure $H =H_0 + b_0$, whatever $a_0$ and $b_0$ which are here to determine what benchmark you use (you can choose 0, I can choose 1 billion), then in this case, $C = b_0 - a_0$. Therefore, you see that the difference you measure $H-E$ relies on your choice of benchmark ! – Tom SymplMech Apr 5 '13 at 22:05
• Moreover, let's say that you use the same ruler in both frame, but due to the relativistic effect (as the contraction of the length), as shown by @Nathaniel, you will have now as reference $b_0 = \gamma a_0$. Anyway, we don't measure absolute values, only differences between 2 states which give you the physical energy spend during the movement :for two observers with the same conditions,$(E_1 - E_0)_a = (E_1 - E_0)_b$ independently of the observers. – Tom SymplMech Apr 5 '13 at 22:18
• Ok then, so why didn't Einstein also account for different energy scales, such as temperature, having different multiplicative factors? – Physiks lover Apr 5 '13 at 22:35
• could you clarify what you mean ? – Tom SymplMech Apr 5 '13 at 23:02