Angular momentum confusion


by aftershock
Tags: angular, confusion, momentum
aftershock
aftershock is offline
#1
Dec7-12, 12:53 AM
P: 113
Imagine i have a particle in the xy plane rotating in a circle around the z-axis at a constant angular velocity. The angular momentum is r cross p which would give me a constant value at all times, there is no change and it's conserved. However if I was to move the rotating particle higher up and then draw my r vector out from the origin the direction of angular momentum would constantly change as the particle rotated.

So at one chosen origin the angular momentum is constant, but at another chosen origin it changes? If this is true, wouldn't that imply there is a torque and also that there's no torque?
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Philip Wood
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#2
Dec7-12, 02:32 AM
P: 861
Interesting. I agree with everything you say, including your conclusion. But there is nothing paradoxical about the conclusion. It's well known that torque (like angular momentum) depends upon the chosen point about which we calculate it.

Only when we have a system of particles acting on each other by central forces can we say that the sum of angular momenta about any point is constant, and that the resultant torque about any point is zero.
Rap
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#3
Dec7-12, 08:29 AM
P: 789
Quote Quote by Philip Wood View Post
Interesting. I agree with everything you say, including your conclusion. But there is nothing paradoxical about the conclusion. It's well known that torque (like angular momentum) depends upon the chosen point about which we calculate it.

Only when we have a system of particles acting on each other by central forces can we say that the sum of angular momenta about any point is constant, and that the resultant torque about any point is zero.
Right. The fact that angular momentum is not conserved at every point shows that the system described is physically unrealistic. I'm not sure that "only" when we have a system of particles acting on each other by central forces is the angular momentum constant at every point. For example, an electron in a magnetic field. But even then, if you include the mechanism that is generating the magnetic field as part of the system, angular momentum is conserved at every point.

nasu
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#4
Dec7-12, 06:15 PM
P: 1,904

Angular momentum confusion


There is nothing unrealistic.
In order to have uniform circular motion you need some centripetal force.
The torque of this force is zero in respect to the center of the trajectory but is non-zero in respect to a different point along the axis.
Rap
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#5
Dec7-12, 09:36 PM
P: 789
Quote Quote by nasu View Post
There is nothing unrealistic.
In order to have uniform circular motion you need some centripetal force.
The torque of this force is zero in respect to the center of the trajectory but is non-zero in respect to a different point along the axis.
It is unrealistic - the angular momentum depends on the coordinate system. Physics is independent of the coordinate system that you choose. By Newton's law, the particle will exert a force on the source of the centripetal force, and this will give rise to an additional angular momentum. The sum of the two will be independent of coordinate system and will be constant in time.
nasu
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#6
Dec7-12, 09:43 PM
P: 1,904
I suppose we should make sure that we mean the same thing by "unrealistic".
How will this make it unrealistic? What parts are not conform with reality?
Rap
Rap is offline
#7
Dec7-12, 10:00 PM
P: 789
Quote Quote by nasu View Post
I suppose we should make sure that we mean the same thing by "unrealistic".
How will this make it unrealistic? What parts are not conform with reality?
You will never find a free particle simply rotating in a circle. A free particle will always be traveling in a straight line at a fixed velocity. The angular momentum of such a particle is a constant and is the same in any coordinate system. If you have a particle rotating in a circle, there must be a force on that particle, and there must be a source of that force. By Newton's law, the source exerts a force on the particle, the particle exerts a force on the source. If you just consider the particle alone, its angular momentum will be a function of which coordinate system you choose. If you just consider the source alone, its angular momentum will also be a function of which coordinate system you choose. Only when you calculate the sum of the two angular momenta will it be constant in time, independent of coordinate system.

The angular momentum of a closed system (not subject to outside forces) is constant in time and independent of the coordinate system you choose. If you ever calculate the angular momentum of a system to be changing in time, then you will know that it is not closed - there is some outside force causing it. If you expand the system to include the source of that force, and there are no other forces on that system, then its angular momentum (and linear momentum, and energy) will be constant in time and independent of the coordinate system that you choose.
nasu
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#8
Dec7-12, 10:16 PM
P: 1,904
Quote Quote by Rap View Post
You will never find a free particle simply rotating in a circle. A free particle will always be traveling in a straight line at a fixed velocity.
Oh, I see what you mean.
But who implied that you will?
Who said that the particle is free? I did not for sure.
If you just look at the angular momentum of one particle, what you find out is not less realistic than what you get when looking at the system. As long as you know what are you looking at.
Rap
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#9
Dec7-12, 10:58 PM
P: 789
Quote Quote by nasu View Post
Oh, I see what you mean.
But who implied that you will?
Who said that the particle is free? I did not for sure.
If you just look at the angular momentum of one particle, what you find out is not less realistic than what you get when looking at the system. As long as you know what are you looking at.
I took the OP's statement "So at one chosen origin the angular momentum is constant, but at another chosen origin it changes?" as a question about how the physical principle of angular momentum conservation can hold if it changes with your choice of coordinate system. Maybe I was wrong.


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