Is absolute motion inconsistent with homogenous space? I read that homogenous space implies conservation of momentum (according to Noether's theorem) Conservation of momentum is kinda the statement that bodies continue moving with constant velocity unless a force is applied.
Suppose we instead lived in a universe where a bunch of frames could be described as absolutely at rest. Suppose anything moving relative to these frames would inevitably start slowing down relative to these frames, and would eventually also come to rest. I guess this implies that physics experiments would produce such results in constant velocity frames from which you could deduce that you were moving.
Though how does this imply that space is not homogenous in this universe? All the origins of these rest frames have equal preference here
 A: After the comment by @ZeroTheHero I realized my answer is wrong. Here is the explanation of what went wrong. I will delete the answer later
Conservation of ordinary momentum is implied by homogeneity of space and time, not just space. The most easiest way to see it is by the fact that you can simulate decelerating particle by adding a time dependent term to free particle lagrangian.
E.g. In one dimensional case (for simplicity):
$$L=\frac{mv^2}{2}-ma_0vt.$$
This leads to EoM:
$$ma=ma_0,$$
where $a_0$ is acceleration you want the particle to have, yet the lagrangian is still invariant under spatial translation.
The conserved momentum is this one:
$$p=mv-ma_0t$$
and this momentum does not imply that free bodies continue moving with constant velocities..
The problem with this statement is that we can just transform to accelerated frame and we are back in ordinary newtonian physics. Thus I did not show anything new, I just rewritten the old physics in an accelerated frame.
Edit
To explain further:
When we deal with symmetries, the newtonian formulation of mechanics is not the best way to do it. Much better way is to use lagrangian mechanics. Here we simple declare function called lagrangian $L$ and we say, that particle will move between points $A$ and $B$ in such a way as to make the action given by this integral along the curve $\gamma(t)$ from $A$ to $B$:
$$S=\int_{\gamma(t)} L dt$$
minimal. Thus to find real movement of the particle, you just need to find which curve from $A$ to $B$ minimizes the integral. This leads to certain equations called equations of motion (EoM) or Euler-Lagrange equations.
Since in mechanics the system is determined once the position and velocities of all particles is known, the lagrangian can in general depend on velocities and positions of the particles and the time. Thus $L=L(\dot{x}_i^j,x_i^j,t)$, where lower index is giving component of the vector and upper which particle we are talking about. In this case, the equations by which particles will move will be given by Euler-Lagrange equations in the form:
$$\frac{d}{dt}\frac{\partial{L}}{\partial \dot{x}_i^j}=\frac{\partial{L}}{\partial x_i^j}.$$
So to recapitulate. In lagrangian mechanics we can give just one function and from the demand of minimalization of action we can arrive at some equations, which for the appropriate lagrangian are the same as traditional newton formula $m\vec{a}=\vec{F}.$
Now, consider some symmetry. For example symmetry under spatial translation (homogeneity of space). This means transformation $t'=t$ and $\vec{x}'=\vec{x}+\vec{\epsilon}$, where $\vec{\epsilon}$ can be any constant vector. If physics should be the same, it means EoM should remain the same and this will happen if the action does not change under the transformation. This requirement of action not being changed by (differentiable) transformation leads to Noether's theorem.
In simple case of lagrangian not being dependent on certain coordinate $x$ with $v=\dot{x}$, the EoM gives:
$\frac{d}{dt}\frac{\partial{L}}{\partial v}=0$ and it means, that the generalized momentum defined as $p\equiv\frac{\partial{L}}{\partial v}$ is conserved.
For ordinary momentum being the same as this conserved current, it must hold that $mv=\frac{\partial{L}}{\partial v}$ which means that $L=mv^2/2+C$, where integration constant $C$ can depend on anything but the velocity $v$. In Newtonian mechanics, the lagrangian is $L=T-U$, where $T$ is kinetic energy and $U$ is potential energy, which in mechanics should not depend on velocities of particles, so the conservation of ordinary momentum is true. For different spatially homogenenous lagrangians, the conserved (due to homogeneity of space) generalized momentum does no longer need to be the same as ordinary momentum.
Edit2
However we could just used some random coordinates (this is the point I did not realize in my original answer), which leads to weird form of momentum. The real question then is, wheter in spatially homogenenous universe we can always find coordinates that leads to conservation of ordinary momentum? Sadly I do not have an answer for that.
A: Let's say that anything moving relative to the "absolute rest" frame decelerates at a constant rate, $a = -x \ m/s^2$, until it is at rest with respect to the "absolute rest" frame. Then, if you are in the absolute rest frame, you will see this deceleration.
If you are not in the absolute rest frame, but one that is moving at constant velocity w.r.t to that frame, you will not see this. In fact if you were originally moving at the same speed as the object (i.e. at the start of the motion you perceive the object to be at rest), you would see it accelerate.
Therefore different observers see different things, and the universe is not homogeneous.
