I don't understand quantum mechanics, the example deals with the collision of two bodies, and the conservation of momentum.

So how does homogenity of space imply linear momemtum conservation?

I have thought a lot about it, but couldn't find the relation between them. It's not symmetrical to me, as the bodies undergoing collision in space have different velocities and mass. So it is possible to distinguish them.

I see no symmetry in this physical system.

  • 2
    $\begingroup$ Are you familiar with Noether's Theorem? $\endgroup$ Commented Jul 6, 2017 at 17:16
  • $\begingroup$ I read it on wikipedia several times, and it makes the same statement. But I understand physics upto high school. Moreover this question is classical physics, and noether theorem states the same statement. I fail to see the symmetry. $\endgroup$
    – koe
    Commented Jul 6, 2017 at 17:21
  • $\begingroup$ The symmetry is in the fact that things are the same whether move this direction or that direction. Whether a collision occurs moving to the right of you or whether a collision occurs moving to the left of you, as long as the two masses are the same and the two velocities are the same except for a reflection (i.e., one of the velocity components is negative in one case and positive in the other), the physical laws are the same and the result is the same, except for the propagation of the negative sign. That's the symmetry. $\endgroup$ Commented Jul 6, 2017 at 17:28
  • $\begingroup$ This is one of those cases in which, even in classical mechanics, you won't get a satisfactory explanation until you're familiar with the required mathematical machinery. In this case, the result was proven based on results from the Lagrangian formalism and the calculus of variations. I will, however, attempt an explanation as an answer. $\endgroup$ Commented Jul 6, 2017 at 17:29
  • $\begingroup$ @joshua yes, I understand it. But in this situation the two masses and velocities are not the same, so how do we see the symmetry? $\endgroup$
    – koe
    Commented Jul 6, 2017 at 17:30

2 Answers 2


Lagrangians and Actions

So the foundation for all of this theory comes from this remarkable fact: that the laws of physics can be reduced to this function, the Lagrangian, which takes as inputs all of the positions and velocities of all of your particles, and the current time, and it tells you the difference between the kinetic and potential energies for that configuration. In other words, this function $L$ defined over the space of all possible configurations, is a summary of all of the physics of the system. For a 2-particle system you have two vector positions and two vector velocities, so the Lagrangian takes the shape $L(\vec x_1, \vec x_2, \vec v_1, \vec v_2, t).$ For more particles you might have more vectors making up a configuration of the space.

More properly, there is a derived function which operates on paths $(t_0, t_1, \mathbf x(t)),$ where by $\mathbf x$ I mean a specification of all of the positions of all of the particles. This derived function is called the "action" of the path, and it is defined as $$S[t_0,t_1,\mathbf x] = \int_{t_0}^{t_1} dt~L(\mathbf x(t),~\dot{\mathbf x}(t),~ t).$$

It turns out that all of the physics reduces to this single claim:

If you fix $\mathbf x(t_{0})$ and $\mathbf x(t_{1})$ then any physical path from that starting point to that ending point must have stationary action. In other words the family of similar paths $(t_0, t_1, \mathbf x + \epsilon ~\mathbf y)$ for small $\epsilon$ and $\mathbf y$ such that $\mathbf y(t_0) = \mathbf y(t_1) = \mathbf 0$ will have the exact same numerical value for their action integral (up to second order in $\epsilon$).

This is usually called the principle of "least" action but "stationary" is a little more general than "least," so that's not quite 100% true.

In practice: the Euler-Lagrange equations

This stationary-action principle allows us to calculate the equations of motion, the so-called "Euler-Lagrange equations", for any Lagrangian we give them.

How does that work? Well, let me show you a simple example, the Lagrangian for a simple 1-D harmonic oscillator: $$L(x, v) = \frac12 m v^2 - \frac12 k x^2$$ This relates back to the equations of motion by the so-called Euler-Lagrange equations, which basically just say that you derive a "generalized momentum" by taking a derivative with respect to one of your velocity components $v$ and a "generalized force" in that direction by taking a derivative with respect to the position $x$. You then construct $\dot p = F$ as per usual in that direction. The technical way this is written is, $$\frac{d}{dt}\frac{\partial L}{\partial v_i} = \frac{\partial L}{\partial x_i}.$$ So in this case that process gives$$m \dot v = - k~x,$$ the normal Newtonian equation for a spring force.

Notice that the above Lagrangian does not conserve momentum. Notice also that to the mapping $x \mapsto x + \epsilon y$ does not preserve this Lagrangian. Let's rectify that.

Suppose we (still in 1D) now have two masses separated by a spring whose rest length is $\ell$:

$$L(x_1, x_2, v_1, v_2) = \frac12 m_1 v_1^2 + \frac12 m_2 v_2^2- \frac12 k (|x_1 - x_2| - \ell)^2$$

Now we have a freedom which does preserve the form of the Lagrangian, $x_{1,2}\mapsto x_{1,2}+ \epsilon~y.$ Notice that it's the same $\epsilon$ and the same $y$, which is important in the two cancelling when we get to the $x_1 - x_2$ expression.

The continuous translation symmetry of space was not present in the earlier equation which had a specially significant $x=0$ position; but neither $x_1=0$ nor $x_2=0$ has any intrinsic meaning to the above Lagrangian; it only depends on $x_1 - x_2.$

Noether's theorem

Here is Emmy Noether's theorem in the simplest non-calculusy way that I can explain it: that stationary-action principle has an astonishing property of coordinate invariance. You can write out that action integral over with whatever coordinates $c_i$ you want, as long as you make sure to compute corresponding time derivative coordinates $s_i = \dot c_i$ and re-express all of the $x_i$ and $v_i$ in terms of the new coordinates. And then you can get the self-same Euler-Lagrange equations to deal with the equations of motion in the new coordinates.

Now one set of alternate coordinates would be $c_1 = x_1 - x_2$, $c_2 = x_1 + x_2$, that would describe this system. And we're done.

What, we're done? Yes! We're done! We saw that the potential energy only depends on $c_1$ not on $c_2$, which means the generalized force in the $c_2$ direction is going to be 0. So the generalized momentum in the $c_2$ direction must be conserved, and that's conservation of momentum. Taking it more carefully, of course, we write $s_1 = v_1 - v_2,$ and $s_2 = v_1 + v_2$, and then we use these to find that $v_1 = (s_1 + s_2)/2$ while $v_2 = (s_2 - s_1)/2$, so the same Lagrangian in the new coordinates is

$$L(c_1, c_2, s_1, s_2) = \frac18 m_1 (s_1 + s_2)^2 + \frac18 m_2 (s_2 - s_1)^2 - \frac12 k (|c_1| - \ell)^2,$$and thus the momentum in this direction which is conserved must be $\partial L/\partial s_2 = \frac14 m_1 (s_1 + s_2) + \frac14 m_2 (s_2 - s_1),$ which we can see is just $m_1 v_1 + m_2 v_2$ from our original coordinates.

So Noether's theorem essentially says that a continuous symmetry in the Lagrangian must always have some "direction" that it's symmetric about; and that "direction" by virtue of this symmetry must have zero generalized-force, which means that the generalized-momentum must be constant in that direction.


It's not the symmetry of the masses or the velocities, which matters. It's the symmetry of space itself. If you assume an empty space without masses, the space poses a continuous symmetry. An other way to put it is to realize, that the origin is not uniquely defined, but that we can put the origin of our spacial coordinate system $\vec r = (x, y, z)$ where ever we like to.

Note: For Noether's theorem to work, we need a continuous symmetry. A discrete symmetry won't do the job.

  • $\begingroup$ I understand that. But how is it momentum of two collision bodies conserved in this case? how does it imply the momentum conservation? $\endgroup$
    – koe
    Commented Jul 6, 2017 at 18:46
  • $\begingroup$ Let's say a momentum is created to the right side, so what? space will still remain symmetrical , as both the bodies are assymetrical. $\endgroup$
    – koe
    Commented Jul 6, 2017 at 18:47
  • $\begingroup$ I fear, I don't have a simple picture to describe this fact. So, I agree with #probably_someone: This is derived from a mathematical formalism, which unfortunately is not very descriptive. $\endgroup$
    – Semoi
    Commented Jul 6, 2017 at 21:36

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