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Q&A

Can I keep our universe, but without the speed limit (of light)?

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It seems like nothing can move faster than light and this is quite bothersome for interstellar travel. It takes decades in the best case to get anywhere interesting in our little Milkyway (Many thousands of years to travel an appreciable distance within it) and unless we find some loopholes traveling between galaxies seems off the table entirely.

I want to have a universe where there is no maximum speed but that is otherwise relatively similar to ours. Is this possible?

Preferably this universe would have:

  • A beginning (a big bang?)
  • galaxies
  • stars
  • planets

That will do for now. How can I make this possible with a well defined rule-set and what notable, large differences would there be between our universe and this universe?

Some things to get started:

  • Forces and all massless particles (like light) might travel at infinite speed.
  • Special relativity likely doesn't hold.
  • The big bang and what happens in the following moments are probably immensely important.
  • Can I still have quantum stuff?
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This post was sourced from https://worldbuilding.stackexchange.com/q/3646. It is licensed under CC BY-SA 3.0.

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First, the good news: You still can have quantum stuff. All the quantum weirdness already exists in non-relativistic quantum mechanics.

Without the speed limit, full relativity is, of course, out of question. Therefore let's look at the alternatives:

Preferred frame of reference

Assumptions: Matter behaves as in non-relativistic quantum mechanics, while unchanged electrodynamics holds for electromagnetic fields. This is basically the aether theory of before Einstein, except of course they didn't know about quantum mechanics yet.

Note that in this case the sped of light would be $c$ only in the preferred frame. The speed of light would not be a limit for particle movement, though.

Let's look at what would happen with atoms at high speeds. To get a feeling of what is bound to happen, let's first use a classical model of the atom, before taking a closer look at quantum mechanics.

As custom when looking at atomic physics, we approximate the nucleus as charged point particle. We will also neglect any spin/magnetic moment of the nucleus.

The fields of a moving electric point charge are

$$\vec E = \frac{q}{4\pi\epsilon_0} \frac{\gamma}{r^3\left(1+\gamma^2 \frac{v_r^2}{c^2}\right)^{3/2}} \vec r$$ $$\vec B = \frac{\vec v\times\vec E}{c^2}$$

Here $q$ is the charge (for an atom with atomic number $Z$, we have $q=Ze$ with $e$ the elementary charge), $r$ is the distance from the (moving) charge, $\vec v$ is the velocity of the charge, and $v_r$ is the radial component of that velocity. Moreover we have the gamma factor $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

If you object that the calculation on the linked page was relativistic: Electrodynamics is inherently Lorentz invariant, therefore those calculations are still valid in the hypothetical non-relativistic world, as long as we are in the preferred frame.

Now let's look at a classical electron (charge $-e$) orbiting the nucleus in a plane orthogonal to the velocity, in a circular orbit. If that electron is at radius $r$, its velocity $\vec v_e$ has the component $\vec v$ as it moves with the atom, and additionally the tangential component for orbiting.

Now let's calculate the force acting on that electron. First, the electric field causes a force $\vec F=-e\vec E$. Since we are perpendicular to the velocity, $v_r=0$, and the only correction compared to the field of a charge at rest is the gamma factor. So we get an attractive radial force with strength $$F_E = -\frac{Ze^2}{4\pi\epsilon_0} \frac{\gamma}{r^2} = \gamma F_0$$ where $F_0$ is the absolute value of the force an electron orbiting an atom at rest would experience at the same radius. The minus indicates attraction.

On the other hand, the electron moves through the magnetic field of the moving nucleus with velocity $\vec v_e$, giving rise to a Lorentz force $\vec F_B = (-e)\vec v_e\times \vec B$. Since the orbital component of $\vec v_r$ is in the direction of $\vec B$, only the component $\vec v$ due to the atom velocity would enter the formula. This gives a repulsive force of the strength $$F_B = \frac{v^2}{c^2} eE = \frac{v^2}{c^2}\gamma F_0$$ Together we therefore get $$F = F_E + F_B = -\left(1-\frac{v^2}{c^2}\right)\gamma F_0 = -\sqrt{1-\frac{v^2}{c^2}}F_0$$ We see that the attractive force is reduced as we get faster. Quantum mechanically this means the atoms will get wider in the orthogonal direction.

More importantly, we see that as soon as we reach the speed of light, the force will go to zero. In other words, the electron will stop being bound.

So in short, while this model would allow acceleration of particles beyond the speed of light, matter would disintegrate as soon as passing the speed of light. So from a practical point of view, the speed of light would still be a limit; indeed, it would not even be safe to just come close to it.

Taking the limit $c\to\infty$

Another option would be to remove the speed limit by letting it go to infinity. Indeed, the Minkowski spacetime of Special Relativity would then transform to the Galilean spacetime of Newtonian physics. The electric field would work as instantaneous force following the Coulomb law, analogous to Newtonian gravitation. There would be an absolute time, but no preferred frame.

However, when looking closer at Maxwell's equations, one recognizes that this would also mean that there are no magnetic fields.

Also, quite obviously there could be no light waves. It would be a dark universe.

As before, but adding "Newtonian light"

We can fix the last issue by simply going back to Newton's idea of light: Just postulate light particles that are emitted from light sources. Thanks to quantum mechanics, we don't even need to forego interference. To work without relativity, those light particles would need to have mass.

However this would likely give a very different universe from ours.

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