# Could a universe with a relativity that is a cross between Galilean and Lorentzian Relativity be self consistent?

I was thinking of a universe in which there is a relativity that is between Galilean and Lorentzian Relativity. In this universe there would be effects of length contraction and time dilation but it would not be according to the Lorentz Factor. In this universe there would be no cosmic speed limit but the faster an object would move the harder it would be to accelerate. There would be no special reference frames and the laws of physics would be the same in all reference frames moving with constant velocity.

Could this universe be self consistent?

• This is a question for SE Physics, not World Building. physics.stackexchange.com – MichaelK Aug 4 '16 at 8:27
• But it is not going to be welcomed at physics though... – Hohmannfan Aug 4 '16 at 8:48
• @MichaelKarnerfors, do we have a threshold on how hard the physics is allowed to be before we turn it away? It's still an imaginary universe not a real one. – Separatrix Aug 4 '16 at 13:04
• @MichaelKarnerfors, "Effects of events or world elements, including biology, technology and magic" but not physics for some reason. That seems an inexplicable oversight. – Separatrix Aug 4 '16 at 13:11
• @MichaelKarnerfors, I argue that he is creating a setting, it's an entire universe with different physics and therefore on topic A setting might not be a planet; it can be larger than a multiverse or smaller than a village. – Separatrix Aug 4 '16 at 13:16

No, the only physical consistent models where the relativity postulate works (i.e., no special reference frames and the laws of physics would be the same in all reference frames moving with constant velocity) are Galilean and Lorentzian relativity. The proof is given by David Morin here (section 11.10).

To ensure relativity, the transformation between reference frames must be of the form \begin{align}x' =& A_v(x+vt)\\ t' =& A_v\left[t+\frac{x'}{v}\left(1-\frac{1}{A_v^2}\right) \right] \end{align} where $A_v$ could depend on the velocity $v$. Now we define $V_v$ as $$V_v = \frac{v}{\sqrt{1-\frac{1}{A_v^2}}}$$ With this change of variable, the relativistic transformations read \begin{align}x' =& \frac{x+vt}{\sqrt{1+\frac{v^2}{V_v^2}}}\\t' =& \frac{t+\frac{vx}{V_v^2}}{\sqrt{1+\frac{v^2}{V_v^2}}} \end{align} which look a lot like the usual Lorentz transformations, except for the fact that $V_v$ could depend on $v$.

But we need one more requirement: the relativity transformation must form a group: if you transform from reference frame RF1 to RF2, and then from RF2 to RF3, the you can also transform directly from RF1 to RF3. \begin{align}x'' =& \frac{x'+wt'}{\sqrt{1+\frac{w^2}{V_w^2}}} = \frac{1}{\sqrt{1+\frac{v^2}{V_v^2}}}\frac{(x+vt)+w(t+vx/V_v^2)}{\sqrt{1+\frac{w^2}{V_w^2} }}\\=& \frac{1}{\sqrt{1+\frac{v^2}{V_v^2}}\sqrt{1+\frac{w^2}{V_w^2} } }\left[x\color{red}{\left(1+\frac{vw}{V_v^2}\right)}+t(v+w)\right]\\ t'' =& \frac{t'+\frac{wx'}{V_w^2}}{\sqrt{1+\frac{w^2}{V_w^2}}} = \frac{1}{\sqrt{1+\frac{v^2}{V_v^2}}}\frac{t+\frac{vx}{V_v^2}+\frac{w}{V_w^2}(x+vt) }{\sqrt{1+\frac{w^2}{V_w^2}}}\\=& \frac{1}{\sqrt{1+\frac{v^2}{V_v^2}}\sqrt{1+\frac{w^2}{V_w^2} } }\left[t\color{red}{\left(1+\frac{vw}{V_w^2}\right)}+x\left(\frac{v^2}{V_v^2}+\frac{w^2}{V_w^2}\right)\right] \end{align}

See the parts in red? They should be equal in a relativity transformation. This means that $V_v^2 = V_w^2$, and therefore, $V_v^2$ is a constant. Which values can it take?

• $V^2=0$: Transformations between different reference frames are not possible.
• $V^2<0$: The relativity transformations are rotations, and the periodicity produces problems with casuality.
• $V^2=+\infty$: Galilean relativity.
• $V^2$ some finite positive value: Lorentzian relativity, with speed of light $c=\sqrt{V^2}$.
• That is just beautiful, really beautiful. – a4android Aug 4 '16 at 13:01
• Have you read this? – JDługosz Aug 5 '16 at 0:37
• @JDługosz Yes! Orthogonal is the case $V^2<0$, and causality can be a little troublesome, as seen in The arrows of time – Bosoneando Aug 5 '16 at 8:50
• I don’t follow the step where you say that the red part must be equal in order to form a group. I understand that you want to ensure that adding two velocities works right, so the double-prime transform should be equal to the transform of some velocity. So I suppose you skipped sweeping it up to be in the same form as the single transform? – JDługosz Aug 5 '16 at 11:00

See my answer on Physics.SE. The familiar special relativity is the inescapable answer, from basic symmetry of space and time. You can have Galilean where there is no speed limit and nothing to make it harder and harder to accelerate either; you can have Special Relativity exactly as it is, and you can have Egan’s Orthogonal universe.

What you describe is contradictory. If an object becomes harder to accelerate at speed, but there’s no special invarient speed, what’s it measuring speed against? That contradicts the lack of preferred frames.

No it wouldn't be self-consistent. There needs to be an invariant velocity to make Lorentzian relativity work. Because that's what no cosmic speed limit effectively means. It's the equivalent of setting the speed of light as infinity. Therefore, no time dilation, no length contraction. Pure Galiliean relativity. It can't be a cross between Galilean and Lorentzian relativity.

Without Lorentzian relativity electromagnetism becomes pure nonsense.

If the faster an object's motion, then harder it is to accelerate it, then this would default to standard Eisteinian special relativity as its velocity approached some limiting velocity (because this is basic Lorentzian relativity). Although this sounds closer to special relativity as imagined by Walther Ritz which retained the principle of relativity, but rejected the principle of invariance. Note: Einstein worked on a version of relativity similar to Ritz's, but abandoned in favour of special relativity.

The main problem is that the OP will need to choose a value for an invariant velocity. This would simply resemble the physics of our universe with a different value for lightspeed.

The other alternative is that material objects will require stronger and stronger forces to accelerate them until the value of the accelerating force exceeds any reasonable force that can be meaningfully generated.