# homogeneity of space for a free particle means no position dependence?

In Landau and Lifshitz, chapter 1, the argument is made that given the homogeneity of space, the motion of the particle cannot depend on its position.

i'm having a hard time understanding this. the solution to the lagrangian would be the motion of the particle as a function of time, or q(t). thus, wouldn't the precise position of the particle in any inertial frame depend on where i pick the center of my axis to be / my initial conditions?

Peter: photon is right, a geometrical symmetry is whether the Lagrangian is independent of the related vector that defines that symmetry. See for example the treatments of those in Lagrangians at http://home.strw.leidenuniv.nl/~icke/ps/SymmetryLagrangian.pdf.

For the Lagrangian not to depend on x say, it means space is the same, or more specifically the physics that the Lagrangian describes, is independent of the value of x. [note: the Lagrangian is T-V, with T kinetic energy and V potential, and since normally (not always) depends on x, y and z, if V is independent of x it means no force in the x direction] in Newtonian physics, The result is that the momentum in the x direction is conserved, consistent with is no force in the x axis direction. If you want full homogeneity then it's the same in all 3 directions, and total momentum, as a vector, is conserved. There is no force.

In Newtonian physics a symmetry means no physical condition depends on that coordinate that is a symmetry. Of course things can be more general.

Noether's theorem has generalized the results to show that for every symmetry there is a conserved physical variable. See the wiki article at https://en.m.wikipedia.org/wiki/Noether's_theorem

I can understand trying to interpret a geometrical concept, such as symmetry in the x direction, in terms of some physical entity. it just means nothing in the physics depends on that entity. And the Lagrangian describes all the physics of a situation, so it is reflected in it, as being independent of it.

In General Relativity (GR) it is broader, and with a more geometrical interpretation. If the metric (or equivalently, the spacetime) is independent of x, spacetime is symmetric or the same along that axis, and one can define a conserved quantity, the relativistic momentum in that direction. One has to be careful in GR, because one can make any coordinate transformation and the equations of GR still hold, so coordinates can be changed (e.g., one can take the origina of x axis to be uniformly moving in that direction, or accelerating, or something rotated around the x axis. But when there is such a symmetry, it is a preferred direction physically and one can make physical conclusions. One can say that there exists a coordinate system where the metric, and then all the gravitational physics, is the same along that axis. Thus, the universe is (in the big), homogeneous and isotropic, and that leads to a preferred spatial coordinate system (and really a lot more, it leads to the semi-unique Robertson Walker spacetime).

• Hi Bob thank you for the clear comment. I think I understand your point. A symmetry (homogeneity of space) implies that the physics that the Lagrangian describes is independent of the value of x. Commented Jan 1, 2017 at 7:58
• Yes. You got it. You're welcome. It gets a little more complex mathematically when it's a more complicated symmetry (eg, spherical symmetry, better to transform coordinates to spherical coordinates, and careful since sin($\theta$) can appear anyway), and later with so called internal and gauge symmetries, but there's math ways of getting those also. Commented Jan 1, 2017 at 8:06
• One more question if you will -- if a particle has a velocity v along the x axis in inertial frame A, the solution to the Lagrangian is x(t) = vt, whereas if I were to choose a different frame with a different point of origin, B, x(t) = vt + x0. While the physics is the same, the exact solution to the Lagrangian depends on the position/velocity of the frame I chose. What am I missing here? Commented Jan 1, 2017 at 8:10
• It does, but in a trivial way. In fact, exactly the way you posed it. In one coordinate frame it's your first equation, in the other the second. Since you know the two frames different by $x_0$ they are the same exact physics. You could have gotten the two equations from solving the two Lagrangians, at the end transform to the same coordinate frame for both, and you can compare the two, and they are the same answer (transform at the end to the first frame and the $x_0$ goes away. So you can solve in any frame, but to compare you have to have both in one frame). Commented Jan 1, 2017 at 8:29
• So that was transforming by changing the zero x point. That's translational invariance. You could also have done a Galilean transformation to a moving frame at velocity v, and your solution would have been x = 0. That's moving to another inertial frame going at speed v. Newton's equations and classical physics (pre-relativity) are valid in any inertial frame you choose, i.e. Going at a constant velocity wrt each other. That's called Galilean relativity. Commented Jan 1, 2017 at 8:33

If you consider a free particle, the Lagrangian is just $$L=\tfrac{1}{2}m\dot q^2$$ and the equation of motion is $$m\ddot q=0$$ If you shift your coordinate system as $$q\rightarrow q+a$$ your equations of motion stay the same, hence the solution also will not depend on the shift $a$.

• thank you for your reply. I was not clear -- this step occurs after the homogeneity of space imposes the constraint that the Langrangian is only a function of v, not x. my question is regarding why the homogeneity of space causes the langrangian to not depend on x. Commented Dec 31, 2016 at 22:58
• This is what homogeneity means, by definition. Commented Dec 31, 2016 at 23:09