# Nature of conservation laws

1. Jul 26, 2011

### QuArK21343

It seems to me that there is a difference between, say, the law of conservation of linear momentum or energy and the law of conservation of angular momentum. The first two are valid in any frame of reference and their invariance is a direct consequence of the relativity principle, whereas angular momentum may be conserved for observer 1 and not for observer 2, if they compute it around two different points (both observers are inertial). How can it be so? Is there a reason? And is it true that energy and momentum must be conserved for all inertial observers?

2. Jul 26, 2011

### xts

Really??? It is not so... Try to find an example.
Angular momentum is conserved for both observers 1 and 2. The total angular momentum may have different value for each of them, but from each one's perspective it is conserved during any evolution of isolated system.

As momentum conservation is a consequence of translational symmetry and energy conservation is consequence of symmetry of time shift, the angular momentum results from rotational symmetry.

Not both but all three conservation laws are true for all inertial observers.

3. Jul 26, 2011

### Petr Mugver

QuArK21343, can you give an example of a system whose energy, linear and angular momentum are conserved in one inertial system, but not in a different inertial system?

4. Jul 26, 2011

### QuArK21343

I believe I never said that the conservation laws are valid for one inertial observer but not for another. To verify the equivalence of all inertial observer, as far as dynamics is concerned, we can limit ourselves to the principle of conservation of momentum and the definition of force. If they are invariant, all other consequences of these two still hold in a new inertial frame. Take the principle of conservation of momentum. In an isolated system of two particles:

$m_1v_1+m_2v_2=const$

For an observer O' moving with constant velocity v, $v_1'=v_1-v$ and $v_2'=v_2-v$. Substituting, we find the total momentum is still a constant. Also, differentiating $v_1=v_1'+v$, we see that $a_1=a_1'$, so acceleration is not a relative concept in newtonian mechanics. Therefore, both inertial observers measure the same force on particle 1. From here, we can verify that also energy is a constant in both frames, that is

$\frac{1}{2}m_1v_1^2+U=\frac{1}{2}m_1'v_1'^2+U'$

where $U=U'$, if the potential energy depends only on the relative distance, as it usually is.

What I am saying is that, in an inertial frame, angular momentum may be conserved around one point but not around another. (Let's speak about only one observer now, since it is clearer) To a beginner like me, it seems strange and rather peculiar. So I was asking if there is some deep reason I still don't know.

Please, feel free to correct me if I am wrong.

Last edited: Jul 26, 2011
5. Jul 26, 2011

### mikeph

angular momentum around all points is separately conserved. So I believe this is incorrect ^

6. Jul 26, 2011

### QuArK21343

I believe angular momentum conservation depends on the choice of the fixed point around which you compute it. Let the angular momentum $L$ of a system of $n$ particles be:

$L=\sum m_i r_i \times v_i$

where $r_i$ is computed relative to a point, moving with velocity $v_0$ in our inertial frame, in general different from the origin. The rate of change of angular momentum is:

$\dot L=\sum m_i \dot r_i\times v_i + m_i r_i\times \dot v_i$

Of course, $\dot r_i= v_i - v_0$ and therefore $\dot L=\sum r_i\times F_i -v_0\times M v_{CM}$

where the forces $F_i$ are only the external ones, since if we assume the internal forces are central, their contribution is exactly zero and where $v_{CM}$ is the velocity of the center of mass of the system. So angular momentum is conserved if the RHS of the previous equation is zero. In particular, the term $v_0\times M v_{CM}$ is zero if the point is stationary, if the center of mass is stationary, if the point is moving in the same direction as the center of mass or if the point is the center of mass. Let's suppose this first condition is met. Then, the moment of external forces is zero if the system is isolated (and conservation is guaranteed for any point) or if, for example, the forces are central with respect to the fixed point or, in general, if the moment of external forces is anyway zero with respect to this point. If we choose another point, the moment of forces may not be zero. Now, if the resultant of all forces is zero, then the moment is independent of the point chosen, but in general this is not true.

Returning to my original question, isn't that weird that conservation of angular momentum depends on the point chosen? Second part of my questions is: given that energy and momentum are conserved for all inertial frames in newtonian mechanics, is it true in all fields of physics? But I think xts partly answer this second part.

7. Jul 27, 2011

### mikeph

I believe you're wrong. Conservation doesn't depend on the point chosen, that's what I'm trying to say. Angular momentum depends on the point, but for any fixed point, that angular momentum is conserved.

8. Jul 27, 2011

### QuArK21343

That would explain everything, but could you show it to me? And could you then spot the error in my previous post?

9. Jul 27, 2011

### mikeph

I think you go wrong when you introduce external forces. Nobody said angular momentum is conserved if you have a general, external torque being applied.

10. Jul 27, 2011

### QuArK21343

Could you be more precise? I can't understand what is wrong with my introduction of external forces.

Also, think about this example. Let's consider a particle going around the sun in a circular orbit. If I compute the angular momentum relative to the sun S, then I get a constant vector of magnitude $mvr$. Angular momentum is conserved! Now take a point P on the circular orbit: angular momentum is changing all the time. It is zero when the particle passes through point P, since the position vector is zero. It is not zero, for sure, in any other position. Angular momentum is not conserved! This is so because when I choose to compute angular momentum around S the external torque is zero, since the force is central. But if I choose another less special and uninteresting point, it is different because now the torque is not zero.

11. Jul 27, 2011

### mikeph

Your two body system is flawed, the sun and Earth will both orbit a common central point (somewhere between the two) called the barycentre. So the computation of the angular momentum from S cannot be done, because S moves. You're computing angular momentum of a single object when there are two objects in the system moving under constant mutual interaction.

It's like me saying "the momentum of a snooker ball is not conserved when it strikes another ball".

12. Jul 28, 2011

### QuArK21343

What I was trying to say is that, in general, whenever you have a central force, and a body moving under its action, the centre of force is special, because angular momentum is always conserved around it, but it is in no way clear if angular momentum is conserved around any other point.

Anyway, here is another one: let's imagine that a particle is accelerating along a line. First, we compute its angular momentum around a point P along the line. Then its position vector and velocity are parallel and angular momentum is equally zero. Angular momentum is conserved. Now, we take a point P' not along the line. The arm of the force is always the same throughout the motion of the particle, but its speed v is increasing. So the magnitude of its angular momentum is mvb, which is changing all the time. Angular momentum is not conserved.

13. Jul 28, 2011

### mikeph

Accepted. Angular momentum isn't conserved, in general, if you have forces acting on the body...

Again, the particle is accelerating so you've got an external force. I don't get where this is going. You're simply thinking of situations in which angular momentum of a particle is not conserved because of the action of external forces. Nobody is debating this.

14. Jul 28, 2011

### Philip Wood

The fundamental law of conservation of momentum is that in any closed system of particles acting on each other by central forces, the sum of angular momenta of bodies about any point is constant.

What's wrong with your last counterexample is that you're not dealing with a closed system. You have a force acting from outside, otherwise your particle wouldn't be accelerating.

Confusingly, for a non-closed system angular momentum may be conserved about a special point. Consider a single planet orbiting around the centre of mass (CM) of the star-planet system. The planet's angular momentum is conserved about the CM. Here we're considering the angular momentum of just the planet, which is clearly acted upon by an external force, but since this is always directed to the CM, the planet's angular momentum is conserved about this point (only). Note that this is is not a case of the P of C of AM, as defined earlier. If we wanted to apply the fundamental P of C of AM we'd need to consider a system including both the star and the planet. Then we'd find that $\sum$AM is conserved about any point.

Last edited: Jul 28, 2011
15. Jul 28, 2011

### QuArK21343

Thanks for your reply. Can I ask you what is a closed system in this context? Is it a system on which there are no external forces? Or is it a system on which there is no external torque? What I know is that the requirement for the conservation of angular momentum is that the external torque be zero, so I guess a closed system is one with no external torque on it. Is this right?

16. Jul 28, 2011

### Philip Wood

I think that for the fundamental P of C of AM to apply we need no external forces to act.

If we say 'no external torque', we would have to state about what point $\sum$G was zero. This would be a special point, and angular momentum of the system would be conserved about that point. We're back to planets' AM being conserved only about the CM of the stellar system.

17. Jul 28, 2011

### QuArK21343

I see: when I say P of C of AM I ask that the total torque must be zero, because whenever the total torque of external forces is zero (if we assume the strong principle of action-reaction!) AM is conserved (if all other conditions listed above are met about the term $v_0\times Mv_{CM}$); on the contrary, you only ask that there are no external forces acting on the system. This is a particular case, but it guarantees that conservation of AM is true about all points.

This is exactly my point. There is something about C of AM that doesn't arise in the P of C of momentum or energy. What I find counterintuitive is the fact that AM may be conserved about only one point or many points but not about all points, and so in problems we must carefully choose our pole to exploit this principle of conservation. It is not something we can count in only one way. My questions: is there a reason for that? Maybe does it depend on the fact the torque, as a vector, is unlike force or velocity?

18. Jul 28, 2011

### Philip Wood

But if you state the P of C of AM as applying to a closed system, meaning one whose particles experience no forces from outside, then $\sum$AM is conserved about any point. That's fine isn't it?

Whose idea was it to weaken the requirement to that of no external torque acting?

19. Jul 28, 2011

### Oudeis Eimi

When computing the angular momentum from a point moving with respect to the origin, you need to use the velocities of the particles as seen from that point, otherwise you're mixing up quantities as seen from different inertial systems.

If we have $R_i$ and $r_i$ as the positions as seen from the origin and the point O, respectively, both considered inertial, with the point O at coordinates $r_0$ moving with velocity $v_0$ (as seen from the origin), then

$r_i = R_i - r_0$, and
$v_i = V_i - v_0$.

Denoting $L_\mathrm{orig}$ the angular momentum as seen from the origin, which we assume is conserved, and $R_\mathrm{CM}$ and $V_\mathrm{CM}$ the position and velocity of the centre of mass as seen from the origin, we have:

$L = \sum_i m_i r_i \times v_i = \sum_i m_i (R_i - r_0) \times (V_i - v_0) = L_\mathrm{orig} + M r_0 \times v_0 - M R_\mathrm{CM} \times v_0 - M r_0 \times V_\mathrm{CM}$,

$\dot L = \dot L_\mathrm{orig} -M V_\mathrm{CM} \times v_0 - M v_0 \times V_\mathrm{CM} = \dot L_\mathrm{orig} -M V_\mathrm{CM} \times v_0 + M V_\mathrm{CM} \times v_0 = \dot L_\mathrm{orig} = 0$,

where in the last equation we've made use of the antisymmetry of the vector product.

20. Jul 28, 2011

### magison

thanks Oudeis Eimi