negative angular momentum?

[h4tt3n]

In this wikipedia article the orbital element the "specific relative angular momentum vector" h is defined as:

 

h = r cross v

 

where r is the position vector and v is the velocity vector. In two dimensions this is the normal dot product of r and v:

 

h = r.x * v.y - r.y * v.x

 

The trouble is that counterclockwise orbits have negative h values. Is this correct?

 

Cheers,

Mike

[Bignose]

In 2 dimensions, there is no such thing as a cross product. The resulting vector is always perpendicular to both of the two input vectors -- impossible in just 2-D.

[h4tt3n]

Well yes, I do realise that the 2d cross product isn't clearly defined and expected a reply similar to yours for the same reason. None the less, the normal dot product or perpdot product (or whatever you like to call it) is a very useful substitute for the 3d cross product. So far I haven't come across a 3d math/physics problem where the cross product couldn't be replaced by the normal dot product in its 2d version.

 

Keeping the thread on rail, if not by using the above method, how do you then calculate h in a 2d case?

 

Cheers,

Mike

[proton]

In this wikipedia article the orbital element the "specific relative angular momentum vector" h is defined as:

 

h = r cross v

 

where r is the position vector and v is the velocity vector. In two dimensions this is the normal dot product of r and v:

 

h = r.x * v.y - r.y * v.x

 

The trouble is that counterclockwise orbits have negative h values. Is this correct?

 

Cheers,

Mike

I don't understand what you've written. What are x and y. And why do you asser that "In two dimensions this is the normal dot product of r and v"? I.e. where did this relationship come from?

[Royce]

This is correct.

 

Say you have two vectors r= [1, 2, 0] , v= [3, 2, 0]. the k(z) component is zero because there is none, hence 2d.

 

Now we want the cross product between r and v.

 

To take the cross product you write

i j k

1 2 0

3 2 0

Rewrite the first two lines

i j k i j

1 2 0 1 2

3 2 0 3 2

 

Multiply diagonally from top left to bottom right, subtract when you come back multiplying from top right to bottom left. This is a conveniant way to get the determinant. This is where the negative in your equation comes from.

 

You will get this: (0i + 0j +2k) -(0j +0i +6k) =-4k

 

This means that in a coordinate system following the right hand rule the direction of your angular momentum is clockwise which can be easily seen by drawing a 2-d representation of your vectors.

 

(z points out of the page)

 

z.---------------y +

| \ r

| \ /

| /

x+ v pointing to the +x axis

 

As you can visually see the velocity of the object is is clockwise. (Srry bad pic)

 

Thus h value indicates counterclockwise movement, while -h values indicate clockwise. This is all based on the coordiante system we chose, which is the based on the right hand rule. It could very well be the opposite on another coordinate system.

 

 

A good example to think of is a wrench on a bolt in the x y plane. Its mouth is wrapped around the z axis, the centre of the bolt. The length of the shaft is the radius, which we will say lies directly on the + y axis. We apply a force on the end of it down perpendicular to the y axis in the -x direction.

 

We know from experience that this bolt is going to tighten, righty-tighty. Its going to move in the negative z direction. But in a twisting motion, clockwise.

 

Mathamatically, it tells us the same thing. We have y times -x and we get -xy

a negative torque. Thus we know clockwise motion.

Edited May 20, 2009 by Royce
Spacing error

[h4tt3n]

Thanks for the very comprehensive reply Royce.

 

For information, I originally picked the 2d cross product equation up at the mathworld site.

 

So, summing up, is it mathematically / physically correct to say that in some cases an orbiting body can have negative angular momentum? I suppose this would have some relevance for the 3d case aswell, since all two-body cases can be simplified to 2d if you align the orbital plane with the frame of reference (z axis zeroes out).

 

Cheers,

Mike

[Royce]

I wouldn't go around saying that you have a negative angular momentum. All the negative sign indicates is the direction. So if h=-4Nms, from calculations, I would write my final anwser as, 4Nms clockwise.

[h4tt3n]

That makes good sense. Thanks for the help.

 

Cheers,

Mike