Separation of Linear and Angular Momentum - Explained

[Cyrus]

I was wondering if someone could take the time to explain why linear and angular momentum are two separte things that can never be converted from one form to another. Thanks

 

[robphy]

Here's one distinction: angular momentum depends on the choice of origin.

Here's another: angular momentum is a cross-product.

 

[pervect]

I was wondering if someone could take the time to explain why linear and angular momentum are two separte things that can never be converted from one form to another. Thanks

Look at the units

linear momentum : mass * velocity
angular momentum : mass * velocity * distance

 

[Cyrus]

Look at the units

linear momentum : mass * velocity
angular momentum : mass * velocity * distance

but if i divide angular momentum by the lever arm, would it not be like converting it into angular momentum, i know this is something you can't do, just wondering why.

 

[geometer]

but if i divide angular momentum by the lever arm, would it not be like converting it into angular momentum, i know this is something you can't do, just wondering why.

Basically, the problem is you have two totally different characteristics of a particle's motion. It's like asking if you can convert a fish into a bicycle.

You can't convert angular momentum into linear momentum, but you can find the magnitude of the instantaneous linear momentum of a particle if the position vector of the particle with respect to some origin about which it has angular momentum is perpendicular to its linear momentum vector. This is what you have described above.

The instantaneous angular momentum (L) is defined as the vector cross product of the particle's instantaneous linear momentum vector with its instantaneous postion vector.

The magnitude of this vector (L) then is the magnitude of p times the magnitude of r times the sine of the angle between the two vectors. If the angle between p and r is pi/2 then sin(pi/2) = 1. So the magnitude of the angular momentum is p x r, and if you then divide by the magnitude of r, you find p. But, again, you have not converted angular momentum to linear momentum, you have just found the instantaneous linear momentum of the particle that had angular momentum L at that instant.

 

[Cyrus]

I see thanks for that insight.

 

[Fredrik]

Both conservation laws can be derived from symmetries. If the lagrangian of a physical system is invariant under translations in space, the total linear momentum will be a constant. If the lagrangian is invariant under rotations, the total angular momentum will be a constant.

So you can say that the two quantities are different because they're related to two different types of isometries on R^3.