Directionality of angular momentum

I was told that the sum of linear and angular momentum is conserved.

Given that angular momentum's direction as a vector is completely arbitrary (I believe there is no physical reason for choosing the cross product going the way it does (that is, perpendicular up or down, it just matters that it's perpendicular), correct me if I'm wrong), and linear momentum is often interchanged into angular momentum, surely the vector sum cannot be conserved?

Or am I wrong, and the meaning is that the magnitude is conserved?

As a sidenote, I suppose this would be a good place to ask why such an odd system was devised for calculating the direction (not magnitude) of angular momentum. Why not have the arrow pointing in the direction of motion, with its base at the point that we are considering angular momentum about? Is it useful simply to compute the conservation of it about a point, as it just involves addition?

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Did you considered that your system should be closed to have them conserved? in sense no outer forces are applied? – TMS Dec 2 '12 at 22:19
It was an implicit assumption, as with any momentum conservation questions. – Meow Dec 2 '12 at 22:21
The conserved quantities is the vectors, and thus there magnitude too, and there sum is conserved, and if interchanging happens it will always happen in away that will keep the sum intact-ed, and of course there is a reasons to take the momentum as a cross product, one of them is to make the sum conservative! – TMS Dec 2 '12 at 22:28
The problem I was having is that he seemed to imply that the value [sum of all linear momentum] could grow or shrink to [sum of all angular momentum]'s loss or gain (losses/gains are equal in magnitude). On reflection, I think he may have just been wrong, but I wanted to check here that I wasn't being an ignoramus. – Meow Dec 2 '12 at 22:31
You should be more specific, who is he? if that a book you should mention it in your question to make it clear, anyway hope you got it. – TMS Dec 2 '12 at 22:33

Linear momentum and angular momentum are each conserved separately. It is not meaningful to talk about their sum.

To talk about angular momentum using a concrete example, let's consider a wheel spinning about its axis (whose center we will call the origin). What really matters in this case is the plane in which the rotation is taking place; let's call this the x-y plane. In three dimensions, it is possible to uniquely specify a plane by giving a vector normal (i.e. perpendicular) to the plane (in the z direction, in this case). (In four dimensions, though, a plane cannot be uniquely specified using a vector; one would need to use a wedge product instead.) The magnitude of the vector is the magnitude of the angular momentum, but there are still two directions in which to choose the vector normal to the plane of rotation. By convention, we choose this using the right-hand rule. You could use a left-hand rule instead if you really wished, but nobody does it this way, and it is crucial to be consistent in the definition of the cross product (or you will find contradictions).

As an aside, any "vector" which results from the cross product of two physical vectors is actually a pseudovector. This means that if you look at the system in a mirror, the "vector" will actually point the other way. For instance, if a wheel is spinning counterclockwise (angular momentum pointing toward the observer), then a mirror placed beside the wheel would show it spinning clockwise (i.e. with the angular momentum pointing away from the observer). Physical vectors do not act this way under a mirror ("parity") transformation.

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If $x$ and $y$ are two conserved quantities, then $Ax + By$ is also a conserved quantity, for any nonzero real constants $A$ and $B$. This includes both $x+y$ and $x-y$, so it is true that their sum is conserved, no matter which way around you define the direction of angular momentum.

However, it is not a very interesting or fundamental result that the sum of linear and angular momentum is conserved, because they have different units, so their sum isn't a physically meaningful quantity. It certainly isn't true that one can be converted into the other.

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protected by Qmechanic♦Sep 22 '15 at 22:42

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