Translation Invariance without Momentum Conservation?

Instead of the actual gravitational force, in which the two masses enter symmetrically, consider something like $$\vec F_{ab} = G\frac{m_a m_b^2}{|\vec r_a - \vec r_b|^2}\hat r_{ab}$$ where $\vec F_{ab}$ is the force on particle $a$ due to particle $b$ and the units of $G$ have been adjusted. Whenever the masses are unequal, the forces are not equal and opposite, violating Newton's third law and conservation of momentum in the process.

As momentum conservation has been violated, my understanding is that translation invariance should be violated as well by this force. But the force law still depends only on separations rather than absolute coordinates, so the physics seems to be translation invariant. What am I getting wrong?

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Why there is square of $m_b$ in your equation? –  hwlau Jan 18 '13 at 4:10
@hwlau it's postulating an alternate form for the gravitational force, just to have an example of an asymmetric force for the question. –  David Z Jan 18 '13 at 4:24
Dear Mike, off-topic. I am not 100% sure but I think that the vectors with indices have smaller arrows above the letter only, as in $\vec r_a$ (simply $\backslash{\rm vec}\,\,r\underline{ }a$) and not a big arrow above everything which you achieved by braces around $r\_a$. –  Luboš Motl Jan 18 '13 at 6:07
Thanks Lubos, I agree this looks more normal. Having thought about it, $\hat{r_{ab}}$ was probably also non-standard. –  Mike Jan 18 '13 at 6:22

Momentum conservation doesn't automatically follow from translation invariance. That only happens because of special features of physical laws, so if you want to prove that translation invariance implies conservation of momentum, you'll need to use some principles of physics to do it. Make up new laws that break those principles and you can indeed have translation invariance without momentum conservation.

The principle you need is called "least action". You need to be able to write the physical law in a way so that it's minimizing something, like, for example, light taking the fastest path between points (and minimizing travel time). This is a simple example of what to minimize; others are more complex.

In general we create a function called the action that takes as its inputs the history of the entire physical system over some time and outputs a number. Whichever motion of the system minimizes the action subject to some boundary conditions is the true motion.

Most physical laws can be written this way, including Newtonian gravity. Your law, though, can't. The reason is we can't come up with an action that makes sense. If the formula for it involves $m_a m_b^2$, well, one problem is that the universe has no way to decide which mass is which, so it would be very strange indeed!

Even if there were a way to decide the one on the left is "b", for example, we'd be stuck with that. That one on the left would always be the one that's squared. In your proposed law, we always square the "to" mass, but the action formula, even if it knows the masses are different, has no idea which one is "from" and "to", because it can only see the entire system. You couldn't get from least action to a way to treat the masses with this particular asymmetry.

So you're right. That law does violate conservation of momentum and it does have translation invariance, but the piece you're missing is that it's a strange law that doesn't obey some basic rules that real laws do.

The most-accessible introduction to these ideas is in Feynman's The Character of Physical Law, or this lecture he gave: http://www.youtube.com/watch?v=zQ6o1cDxV7o The argument of interest comes near the end, 45 or 50 minutes in.

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While I agree with the answer, I would be interested in knowing: 1) How can we prove that his postulate doesn't correspond to a Lagrangian ? (without using circular argument) 2) I am not sure what physical "sense" action has even in systems that have one. Would like hear some elaboration on that. –  Sankaran Jan 18 '13 at 4:58
Can you explain where the argument I already gave is circular? It's hand-wavy, but I think it is not logically flawed and it's obvious enough how to be more precise if you wish. –  Mark Eichenlaub Jan 18 '13 at 5:11
As for the meaning of action, that doesn't seem a necessary part of this argument to me; it's more a separate question. There's a question for it here physics.stackexchange.com/questions/9686/the-meaning-of-action –  Mark Eichenlaub Jan 18 '13 at 5:13
I mean the argument that universe cannot decide which one is sounds right but I don't see how that implies a lack of action (purely in terms of a mathematical/logic conclusion). The left-right argument seems to invoke lack of translational symmetry which is what we are after showing(?) –  Sankaran Jan 18 '13 at 5:14
No, the left-right argument has nothing to do with translational symmetry. I don't see how you get that. It's just saying there's a function L(G,m_a,m_b,r_a,r_b,v_a,v_b,t) and we look at things like $\partial L/\partial r_a$. Here m_a is a constant. If the formula has m_a^2 in it, it will be that particular mass is squared when we look at the equation of motion for a or for b. There's nothing in this about translations. –  Mark Eichenlaub Jan 18 '13 at 5:17

Your forces are always equal. It is the accelerations that are unequal in case of equal masses. The situation is similar to the Coulomb interaction. The total momentum is conserved. There is no problem here.

EDIT: As Michael Brown kindly pointed out, the forces are implied to be different. Then indeed the momentum conservation does not hold. The situation is similar to that with a known motion of a "sourcing body" $\vec{r}_b (t)$: although the force on a probe body at $\vec{r}_a$ depends only on the relative distance $|\vec{r}_a$-$\vec{r}_b(t)|$, the momentum is not conserved (neither is the energy).

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-1 because you apparently didn't notice the $m_b^2$ factor that was the focus of the question. The OP isn't asking about Newton's law applied to bodies with different masses, he is proposing a new law where the forces are unequal as a counter-example to the claim that translation invariance alone is sufficient to imply momentum conservation. –  Michael Brown Jan 18 '13 at 13:54
@MichaelBrown: Indeed, if $\vec{F}_{ab}\ne \vec{F}_{ba}$, then it is a different situation. Then it is very similar to a particle in an external force where the momentum is not conserved. –  Vladimir Kalitvianski Jan 18 '13 at 13:58
Removed the -1 now. –  Michael Brown Jan 19 '13 at 12:45