“Work, Weber’s Critical Radius, Energy”

Einstein’s $E=mc^2$ is the most famous formula in history, but what does it mean? Here we emphasize that Einstein derived the formula from mathematics in 1905, but Wilhelm Weber derived the formula from his fundamental electodynamic force law. Weber’s derivation leads to the amazing prediction of 2 equal to 1 plus 1 as a stable electronic system.

Consider the problem of work and moving particles “upstream” of the potential. Recall the Coulomb-type expression that “opposite charges attract” and “like charges repel”. If we arranged some charged particles on a plate, then in a short time these particles would either be repelled outwards to the boundary of the plate or the opposite charges would cancel on the interior. Again this is with respect to the Coulomb model $V(r)=1/r$.

But Weber’s model has the following amazing prediction: “opposite charges attract except at a small critical distance $r=r_c$ where the charges acquires a negative inertial mass and the Weber force becomes repulsive”. Furthermore, “like charges repel except at a small critical distance where the Weber force becomes attractive”. This latter prediction leads to Weber’s *planetary model of the atom”.

In the (cgs) units the critical radius $r_c$ is computed as $r_c=\frac{1}{m c^2}$ where $m$ is the relative inertial mass of one of the particles. The proof of this formula is an application of Newton’s second law, that $m a = \bf{F}$ where $\bf{F}$ is Weber’s force.

Now the question arises: if we have two identical but opposite electrically charged particles, say $(-1)e$ and $(+1)e$, where $e$ is a small mass, then how much work is required to drive $(-1)e$ and $(+1)e$ to within Weber’s critical radius $r_c$?

Weber comments in his final memoir [ref] that an infinite amount of work would be required, however no technical details are provided. In fact this author thinks its evident that a large finite amount of work is required, and the computation of this work required is the subject of this post.

How to compute Work: We imagine a particle moving through a potential $U$ along a trajectory $\gamma.$ Our goal is to determine how much work is required to breach (“pass through”) the critical radius $r_c=\frac{1}{mc^2}$ in units where $4 \pi\epsilon_0 = 1$.

We begin assuming the particle’s trajectory is rectilinear. This means the tangent $\gamma'$ and $\gamma$ are parallel for all values of the parameter. Moreover we can use the Riemannian idea of parameterizing the curve by arclength. Then we ask how much work is required to move the particle through $U$ at a constant rate of speed, namely $||\gamma'||=1$.

Weber’s force is conservative, so there is some path independance, however the terminal conditions are not fixed a priori. It might happen that breaching the Weber radius is easier if the particles velocity is nearly parallel to the “virtual” surface of the Weber’s critical radius. That is to say, the curvature term $r r''/c^2$ in Weber’s force law might sometimes reduce the work needed, depending on the sign.

In terms of the relational variables recall the useful formula/definition \[ r' = \frac{\bf{r} \cdot \bf{r'}}{r}.\] In our setting we find \[r'=\frac{\gamma \cdot \gamma'}{||\gamma||}.\] Applying the Cauchy-Schwartz inequality, we find \[r'=1.\] I.e. equality is obtained in Cauchy-Schwartz because $\gamma, \gamma'$ are parallel by hypothesis and $||\gamma'||=1$.

Moreover the rectilinear motion implies $r''=0$, i.e. the trajectory has zero curvature, since the direction of the particle does not change. Therefore Weber’s force leads to work being computed by the integral \[W=(1-\frac{1}{2c^2})\int_{+\infty}^{r_c} \frac{1}{r^2} dr = (1-\frac{1}{2c^2})mc^2=mc^2-m/2.\] So what is the total energy of the above system? The total energy is the work required $mc^2 - m/2$ plus the initial kinetic energy $T_i=m|\gamma'|^2/2=m/2$. Therefore the total energy required to breach the Weber critical radius is $E=W+T_i=mc^2$.

N.B. All the while our particles are “travelling upstream”. So the above computation indicates that if a particle is given sufficient energy (i.e. sufficient kinetic energy, then the particle could breach the Weber critical radius, and arriving at the Weber radius with almost zero kinetic energy).

N.B. The integral representing $W$ is parameterization independant. Therefore the work required by the unit-parameterized path is not overly specialized, but represents the general computation.

The above discussion was restricted to rectilinear trajectories. But the possibility remains that curvilinear (“spiralling”) trajectories require less work to breach the Weber radius. Thus while it appears that breaching the Weber radius via rectilinear paths requires large energy $\approx mc^2$, perhaps the spiralling paths – where the curvature term maintains a definite sign – are the more interesting.

Problem: Determine the minimum energy required for an isolated two-body system to breach the Weber critical radius.

Answer: The minimum energy required to breach the critical radius is $E=mc^2$.

This is consequence of the fact that $\bf{F}$ is a conservative force, therefore dependant only on the initial and terminal states, and not the path taken. Moreover the work done by a particle traversing a path $\gamma$ depends only on the difference in potential energies. This implies that the above evaluation in the rectilinear case is essentially the same for all paths from some initial point to to within the critical radius.

Einstein’s NonPhysical Derivation

k we apparently all know that $E=mc^2$ is one of the great achievements of Albert Einstein. He first published the idea in 1905 in a paper called “Does the inertia of a body depend upon its energy content?”

But what does the formula mean?

Supposedly it’s one of the greatest formulas in history, yet what is the content of this supposed formula?

We give the reader a few minutes to consider what they understand about this formula. For example, what can you say about it?

The standard explanations are not so useful. They say that a body of mass possesses some intrinsic energy which is independant of the observer frame. And this intrinsic energy is computed – according to Einstein’s suggestion – as $m_0c^2$, where $m=m_0$ is the inertial rest mass of the object.

Now here enters another idea, that the inertial mass of an object increases with its velocity, i.e. we have $E=mc^2 = \beta m_0 c^2$ where $\beta$ is one of the Lorentz-type beta factor arising so frequently in special relativity, that is $\beta = 1 /\sqrt{1-v^2/c^2}$.

Now look, for most students in university, or even adults, this is the end of the story. You can take physics in undergraduate college or university, and I think this is the standard treatment.

You can ask “Why is the formula true?”, and the teacher might say “You won’t need to prove it on the exam, don’t worry about it!”, or “Well, I don’t know, nobody ever asked me, but I guess you can read Einstein’s work”, or “I don’t know, but look at wikipedia.”

And reading Einstein’s original paper is a good idea, although it won’t likely be satisfying. Another good reference is Levi-Civita’s “Absolute Differential Calculus” textbook which has very detailed mathematical review of Einstein’s “physical principles”.

In Levi-Civita’s textbook, the formula is derived via an argument using Hamilton’s principle. It’s something like this: we start with the usual Lagrangian $L$, and find the equations of motion are $\delta \int L = 0$, where $\delta \int$ is the variational derivative of the functional $\alpha \mapsto \int_\alpha L$. (Here $\alpha$ denotes a path, not the earlier Lorentz gamma factor.)

Actually in Einstein and Levi-Civita’s approach, which is based on Hamilton principle. The starting point is a Lagrangian $L$ defined on the statespace, and then with Hamiltonian $H=L^*$. The Hamiltonian principle says \[\delta \int L=0.\] But Einstein introduces the variational equation \[\delta \int c^2 = 0.\] And he says this equation is trivial like \[\delta \int dt =0 .\]

So Einstein introduces the Lagrangian $c^2 - L$ and the Hamilton principle becomes \[\delta \int (c^2 - L) = 0.\]

This apparently trivial modification of the Lagrangian is what introduces (imports) the constant $c^2$ energy term in the Hamiltonian.

We omit the details from Levi-Civita, but the main formula obtained is \[H^*=c^2 - L = c^2 \sqrt{ 1-\beta -2U/c } = c^2 + v^2/2 - U,\] since $L = v^2/2 + U$. Einstein remarks that if $v=0$ and $U=0$, i.e. if the potential energy and kinetic energy vanish, then there still remains an intrinsic energy represented by $c^2$, i.e. the Hamiltonian $H^*$ does not vanish.

In the above argument, we remark that it’s not always clear that $U=0$ is meaningful or nonarbitrary. As well known, it’s the gradient of the potential $\nabla U$ which enters into the dynamical equations, and not the scalar values of $U$. So on this point we are not entirely persuaded that $v=0, U=0$ represents an intrinsic energy.

So in summary, what’s the story behind this amazing formula $E=mc^2$ ? In the above argument, there is no story except trivial mathematics. There is no physics!

Levi-Civita makes this same remark in his textbook, but points a posteriori to radioactive substances.

Weber’s Physical Interpretation of E=mu.c^2 ?

Wilhelm Weber (1804 – 1891) was a great physicist mathematician who succeeded C.F. Gauss as directory of the Gottingen Observatory. Weber was the first physicist to define $c$, with Kirchoff using Weber’s force equations, as the velocity of electrical impulses in a material wire.

In his works on particle electrodynamics, he discovered in 1860s an amazing physical model of $E=\mu c^2$ where $\mu$ is the reduced mass of the system of electric particles. We have written on this amazing fact in a previous post “Weber’s Critical Radius, Work, and E=mc2 Formulas”

We repeat the fundamental physical idea (and this is what is totally absent in Einstein’s derivation).

Consider two equal and identical unit electric charges $q_1$ and $q_2$ which are separated by some relative distance $r>0$. The Couloumbian motto is that “like charges repel, and opposite charges attract”. So a force is required to push the charges $q_1$ and $q_2$ closer together. In otherwords, there is work that needs be done to push the identical charges $q_1, q_2$ together.

What Weber discovererd in 1860s (maybe earlier) was that there is a critical distance, where if a sufficient amount of work is performed, and the charges $q_1, q_2$ become sufficiently close $r< r_c $ (passing through Weber’s critical radius $r_c$) then there is a sign in Weber’s force law which predicts that the equal charged particles become attractive within the critical radius $r< r_c$. In otherwords it’s possible for two-body system of net charge $-2$ $(=-1 -1)$ to be a stable system with large potential energy! There is a large amount of potential energy within this system because alot of work was performed to push the particles through the critical radius.

More specifically, Weber’s force predicts that the amount of work required to push the particles through the critical radius is $E= \mu c^2$, where $\mu$ is the reduced mass of the system.

This is amazing fact! This is a physical explanation of what the formula means. The formula represents an amount of work which has been invested into the system. Moreover, Weber’s formula predicts that this potential energy is stored in surprisingly stable many body systems, e.g. two-body systems of the form $-2=-1-1$.

I don’t know if the reader can appreciate how EPIC this idea of Weber’s is. It’s the beginning of something incredible, basically returning physics to classical paradigm, and appreciating Wilhelm Weber’s tremendous contributions. Here I have been much influenced by AKT Assis’ recent translation into English (and Spanish, Portugese) of Wilhelm Weber’s collected works (in four volumes).