How do you calculate kinetic energy in a convention where the one-way speed of light is anisotropic?

Question

The question about measuring the one-way speed of light has been debated in several posts in this forum, and are treated in detail on this wiki page.

In relation to this debate, I wonder how you would calculate the kinetic energy of an object with mass $m$, if the one way speed of light is different from c?

We can reason that if the one-way speed of light, $c’$, is smaller than c in some direction, then this speed limit would apply to particles as well, $v’<c’$. If it was not so we would be able to detect superluminal particles from supernovas (The light of the supernova always arrives before the particle radiation, and since the order of local events are preserved, this is true in any given frame of reference)

A reasonable assumption would be that a given kinetic energy of mass $m$ would correspond to a constant fraction of $c’$. The equation describing the kinetic energy would then look like this.

$E_K=E_0 \left(\frac{1}{\sqrt{1-\frac{v'^2}{c'^2}}} -1\right)$

with $E_0$ being the rest energy of the object. (Here I made the assumption that $E_0=mc^2$ depends on the two-way speed of light, to keep the rest mass constant. Alternatively you would have to define a one-way mass so that $m'c'^2=mc^2$)

Effectively this would mean that the energy it requires to accelerate the mass $m$ to a given velocity in a given direction, would depend on the one-way speed of light in that direction. It seems that the inertia of the object would have a directional dependency. My best guess is that this effect is canceled out from the perspective of the accelerating observer due to the transport of his clock, and that you would, therefore, not experience this difference in inertia in reality.

Can anyone clarify this?

Answer 1

The fact that the two-way speed of light is constant, makes it theoretically impossible to measure the one-way speed of light. It turns the one-way speed of light into a matter of definition. And what value to define for the one-way speed is linked to the definition of simultaneity.

We usually use a definition of simultaneity that is natural to us: two events are deemed to be simultaneous, if light from the two events arrives at the mid-point at the same time. That is, they are defined to be simultaneous if we see them simultaneously from the mid-point. That definition equates to defining light speed to be isotropic. The light is defined to take the same time from both directions to the midpoint.

In order to measure the speed of some object, you need to set up two clocks some known distance apart, and measure how much time it takes for the object to travel that distance. For that to be a fair measurement, those clocks need to be properly synchronised before the measurement, consistent with the definition of simultaneity. That is, how they are synchronised depends on the real one-way speed of light. That automatically means that any speed measurement can always be seen as a fraction of the real speed of light in that direction. That is, using your terms, measuring a speed $v$ really means measuring $v/c = v'/c'$, with $c$ the defined one-way speed, and $c'$ the real one.

That was a long-winded introduction to get to your question. We assume the real one-way speed differs from $c$ in some direction, and there is some supernatural being who is actually able to detect that real speed of light. That supernatural view would indeed have different equations for pretty much all physics, including the kinetic energy. In that supernatural view, almost all physics equations would have to depend on the real one-way speed, in order to properly explain the measurements we do in the real world. Momentum and energy would also not be conserved in such a view.

The same would be required if we would define simultaneity, and with it the one-way speed of light, in a different way.

Answer 2

Indeed, as you suggest, the form of basic expressions for energy and momentum are different under the assumption that the one-way speed of light (OWSOL) is anisotropic. I will use Anderson's notation where the anisotropy of the OWSOL is defined by a parameter, $\kappa$, defined such that the OWSOL in the positive $x$ direction is $c/(1-\kappa)$ and the OWSOL in the negative $x$ direction is $c/(1+\kappa)$. Anderson's notation is related to Reichenbach's by $\kappa = 2 \epsilon-1$ with the standard isotropic convention for $\kappa=0$. I will also use units where $c=1$ for convenience.

If we start from a standard inertial frame using Einstein synchronization $(t,x,y,z)$ then we can transform to the Anderson anisotropic frame with $$t=T+ \kappa X$$$$(x,y,z)=(X,Y,Z)$$ Note that positions are unchanged in this frame, and all that is changed is the synchronization convention. This leads to the metric $$ds^2=-(dT+\kappa \ dX)^2+dX^2+dY^2+dZ^2$$ Note that this is the standard metric plus additional terms that depend on the OWSOL via $\kappa$. In particular, if you expand this expression there is a cross term $-2\kappa \ dT dX$ leads to changes in many of the standard physics equations. However, importantly, because this is merely a coordinate change, there is no change in the underlying physics and no experimental measurement is changed with this convention.

To derive the formula for KE in this metric, we first write the Lagrangian of a free particle in this metric: $$\mathcal L=m \sqrt{\left(\dot T +\kappa \ \dot X\right)^2-\dot X^2 - \dot Y^2 - \dot Z^2}$$

From this Lagrangian we can find the conserved quantities through the usual Lagrangian-based approach. This gives the conserved energy and the conserved momentum: $$E=m\frac{1+\kappa \ V_X}{\sqrt{\left(1+\kappa \ V_X\right)^2-V^2}}$$ $$ \vec p = \frac{m}{\sqrt{\left(1+\kappa \ V_X\right)^2-V^2}}\left( - \kappa - \kappa^2 \ V_X+V_X, V_Y, V_Z\right) $$ where $V$ is the speed in the Anderson frame and $V_X$ is the component of the velocity in the $X$ direction.

For the energy, if we set $(V_X,V_Y,V_Z)=(V,0,0)$ and do a Taylor series expansion around $V=0$ we get $$E\approx m+\frac{1}{2}m V^2 -\kappa m V^3$$ where we see the first term is the usual mass-energy, the second term is the usual kinetic-energy, and the third term is an additional kinetic-energy term that depends on both the velocity and the OWSOL.

Now, according to this formula if the OWSOL is anisotropic ($\kappa \ne 0$) then energy will also be anisotropic in the following sense. If a given amount of potential energy is converted to kinetic energy, then the resulting speed will depend on the direction. Energy is still conserved, but it is anisotropic in relation to the anisotropy of the OWSOL.