In our universe energy is conserved. In a parallel universe could energy divided by distance be a conserved quantity instead of energy itself? So in this universe if two objects were to move away from each other at a constant rate the total energy of the system would increase at a constant rate. What kinds of effects would having energy divided by distance as a conserved quantity have on this universe?

  • $\begingroup$ What do you mean by it exactly? What happens in systems of three and more particles? What about fields? Without this, I can only try try to derive one possible set of equations of motion for two particles. You have a conserved quantity that can serve as energy, so we can hope for Hamiltonian and quantization. $\endgroup$
    – BartekChom
    May 13, 2016 at 4:36
  • $\begingroup$ If a quantity isn’t conserved, then it doesn’t worth to be referred as to “energy”. $\endgroup$ May 16, 2016 at 18:18

3 Answers 3


The conservation of energy is a result of Nother's Theorem and the time invariance of the laws of physics. This distance distortion would mean that the laws of physics are fundamentally different at different points in your universe.

What kind of effects? The effects of this would be so stupendously altering that we might not even recognize it as a universe. I would feel comfortable stating "If you can imagine it, it won't happen in such universe." In general, all world building questions that are phrased "What if fundamental law of physics _______ was changed to _____," have the same answer: "if the universe could even exist, you would not recognize anything about it."

The best thought might be that you could identify ways to move in your universe or stretch spacetime which nullify this effect completely, so that you locally experience exactly the same laws of energy that we are used to. These small islands of sanity would exist in a sea of truly alien behavior.

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    $\begingroup$ +1 for all world building questions that are phrased "What if fundamental law of physics _______ was changed to _____," have the same answer: "if the universe could even exist, you would not recognize anything about it.". I'd give +infinity if the laws of this universe allowed it. $\endgroup$ May 12, 2016 at 23:32
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    $\begingroup$ @Lostinfrance The only ones that annoy me more than that are 4d questions. We know nothing about a fourth dimension of space! $\endgroup$ May 13, 2016 at 4:18
  • $\begingroup$ That's what this example question was meant to illustrate. $\endgroup$
    – JDługosz
    May 13, 2016 at 16:20
  • 1
    $\begingroup$ @JDługosz Thank you! I knew that question was somewhere, but I completely forgot where to find it! $\endgroup$
    – Cort Ammon
    May 13, 2016 at 17:10

TL;DR: You are describing quantum chromodynamics. Have fun!

When you write "two objects move away from each other at constant speed", this is what happens in our universe without the influence of forces. When there are forces between them involved, their speed will decrease, so that the sum of potential and kinetic energy remains constant.

So in our universe, as in yours, a constant speed of separation implies that energy is being added. In our universe, the gravitational potential would be $U = -\frac{c}{r}$, where $c$ is a constant and $r$ is the distance. Note that it's defined as negative and increases to zero at infinite distance. It follows that the change in potential energy over increasing distances is proportional to $\frac{1}{r^{2}}$.

What you are looking for, is to replace gravitation with the strong nuclear force. In the case of this force, the energy grows linearly with distance as you try to separate two quarks. Exactly as you prescribe in your universe.


It seems that I have managed to derive an example of equations of motion for two particles in non-relativistic case. Probably I should explain something more, but I do not know, what is not understandable.

Indices: $a \in \{1, 2\}$ (which of two particles), $i \in \{1,2,3\}$ (dimension)

As always $r = |\vec x_1 - \vec x_2|$ and $E = V(\vec x_1, \vec x_2) + E_k$

Equations of motion: $$d \dot{\vec x}_a = \ddot{\vec x}_a dt = \frac{{\vec F}_a}{m_a} dt + d_2 \dot{\vec x}_a$$ where $F_{a i} = - \frac{\partial V}{\partial x_{a i}}$ and $d_2$ is the part of differentiation that differs from our world.

We start from your assumption $D=\frac{E}{r} = const$ and my clarification $d_2 \dot{x}_{a i} = \kappa \dot{x}_{a i}$.

Differentiation: $$0 = dD = d\frac{E}{r} = \frac{dE}{r} - \frac{E dr}{r^2} \implies dE = E \frac{dr}{r}$$ and $$dE=d_2 E_k = d_2 \sum_{a} \frac{m_a v_a^2}{2} = \sum_a m_a v_a dv_a = \kappa \cdot 2 E_k$$ that is $$dE = E \frac{dr}{r} = \kappa \cdot 2 E_k \implies \kappa = \frac{E dr}{2 E_k r}$$ Thus equations of motion are finally $$\ddot{\vec x}_a = \frac{{\vec F}_a}{m_a} + \frac{\kappa}{dt} \dot{\vec x}_a = \frac{{\vec F}_a}{m_a} + \frac{E \dot r}{2 E_k r} \dot{\vec x}_a$$

For example for $V = 0 \implies (\vec F_a = 0 \wedge E = E_k)$ we have $$\ddot{\vec x}_a = \frac{\dot r}{2 r} \dot{\vec x}_a$$

Note that we have $(r^2)^\cdot = 2 r \dot r = 2 (\dot{\vec x}_1-\dot{\vec x}_2) \cdot (\vec x_1-\vec x_2) \implies \dot r = \frac{2 (\dot{\vec x}_1-\dot{\vec x}_2) \cdot (\vec x_1-\vec x_2)}{2 r}$

If I did not do anything wrong, one can write a simulation. Generalisation to more particles and other complications can be harder, but we could assume $D = E \rho = const$ for example for $\rho = \sum_{\text{all pairs}} \frac{1}{|\vec x_a-\vec x_b|}$


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