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Why was angular momentum defined as it was?

  1. May 28, 2008 #1
    Is there a particularly good reason that angular momentum was defined as the cross product of position and linear momentum vectors?
  2. jcsd
  3. May 28, 2008 #2
    If you continue with your angular velocity derivation in your other post, and calculate r'', you'll see how this term show up. Instead of using your


    term, the key is expressing things using the cross product:

    \frac{d}{dt}\left(\frac{\mathbf r}{\Vert \mathbf r \Vert}\right)
    = \left(\hat{\mathbf r} \times \frac{1}{{\Vert \mathbf r \Vert}} \frac{d \mathbf r}{dt}\right) \times \hat{\mathbf r}

    Give it a try and see where you get with it. I have a write up I did for myself of such a derivation (since I didn't like the way my mechanics book pulled all this out of a magic hat). I can post that if you want (uses Geometric Algebra though), but it's probably more instructive for you to try yourself with the hint above.
  4. Jun 2, 2008 #3
    I'm not sure about this, but I think it's just to do with the development of the rotational equivalents of the translation dynamics equations. Torque is defined as [tex]r\times[/tex][tex]F[/tex]. We need the r here because physically there is a dependence on the distance of the force from the rotation axis. And we want the rotational equivalent of [tex]\sum[/tex] F = [tex]dP/dt[/tex] which is [tex]\sum\tau = dL/dt[/tex] and so we define angular momentuml as [tex]r\times[/tex][tex]p[/tex]
    Perhaps somebody more knowledgable could clarify?
  5. Jun 2, 2008 #4


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    I agree with tonyh.

    Good ol' Newton's second law is a vector equation, and that means we can do any of the vector things to it. :smile:

    In particular, if we cross-product it with another vector, it's still valid.

    r is a vector, and the cross-product with r is [itex]\sum\tau = r\times p[/itex] .

    erm … why r and not something else?

    hmm … not sure … :rolleyes:
  6. Jun 2, 2008 #5
    Introductory Mechanics Explanation
    The cool thing about defining [itex]\vec{L} = \sum_i \vec{r_i} \times \vec{p_i}[/itex] is that it follows that [itex]d\vec{L}/dt = \vec{\tau}_{\textup{net}}[/itex] because [itex]\vec{v_i}\times \vec{p_i} = 0[/itex]. If you used any other vector other than r, you would have another term in the product rule. And defining [itex]\vec{\tau}_{\textup{net}} = \sum_i \vec{r_i}\times\vec{F_i}[/itex] is intuitive (just use a wrench or play on a see saw) but also can be shown by direct application of the translational form of Newton's Second Law to be proportional to the angular acceleration, which makes it the rotational analog of force. That implies that defining angular momentum as we do yields us a rotational analog of momentum that is conserved when the rotational analog of force vanishes.

    Intermediate Mechanics explanation
    The Lagrangian is isotropic, this symmetry is associated with a Noether current in the form of the quantity we would define as angular momentum. It just pops right out of the math that you can find in any introductory classical field theory text.
  7. Jun 2, 2008 #6
    note: small typo above (should be angular momentum not torque)

    Since jbunten was working out [itex]\mathbf{r}'[/itex] for [itex]\mathbf{r} = r\hat{\mathbf{r}}[/itex] in a different thread, I suggested doing the same thing for the acceleration. I don't know how far he/she got, but the end result, with [itex]
    \boldsymbol{\omega} = \frac{\mathbf{r} \times \mathbf{v}}{r^2}
    [/itex], is:

    you can calculate the velocity and acceleration in terms of radial components:

    \mathbf{v} = r'\hat{\mathbf{r}} + \boldsymbol{\omega} \times \mathbf{r}

    \mathbf{a} = r''\hat{\mathbf{r}} - \mathbf{r} {\lvert \boldsymbol{\omega} \rvert
    }^2 - \frac{\mathbf{r}}{r^2} \times \frac{d}{dt}(\mathbf{r} \times \mathbf{v})

    Thus for constant m you get:

    F = m\mathbf{a} = mr''\hat{\mathbf{r}} - m\mathbf{r} {\lvert \boldsymbol{\omega} \rvert
    }^2 - \frac{\mathbf{r}}{r^2} \times \frac{d}{dt}(\mathbf{r} \times \mathbf{p})

    and therefore see directly why the quantity [itex]\mathbf{r} \times \mathbf{p}[/itex] becomes significant (this is given the name angular momentum). It's derivative, which for constant m, (r x p)' = r x F, is given the name torque, and one can see how it is related to the Force on the object.

    This is how I liked to view this ... the cross products are just conquences of looking at things from a radial point of view since they can be used to express the difference from the projection onto the radial direction.
    Last edited: Jun 2, 2008
  8. Jul 7, 2008 #7
    and here is a thread that almost answers my question.I think when you write the accelaration in terms of radial components the momentum adn torque terms comes automatically.
  9. Jul 7, 2008 #8
    What was your question?
  10. Jul 7, 2008 #9
    i want to know where this angular momentum or torque equaton comes from and why we use the cross product.In a lot of sites they only gave the definiton of angular momentum and torque and said that theses definitions are rotational equivelants of 2nd law.But how do they get there, i mean from pure 2nd law which doesn't say anytihng about rotation as far as i can see, how can we derive these torque equations?
  11. Jul 7, 2008 #10
    and peeter i think you derived them as a secret part of ma equation, when you trasnlate from cartesian to polar these equation which i thought coming nowhere appears, so these equations come from the transition to the polar coordinates. Am i right?
  12. Jul 7, 2008 #11
    Yes, the angular momentum and torque as cross product are terms present in F=ma expressed radially.

    In the newtonian limit it is really F=p' that is fundamental, so this radial acceleration is only valid for constant mass. I'm not sure if I've seen this mentioned in intro texts that describe conservation of angular momentum is discussed, but not where it comes from. If you have mass loss, like in a rocket ship problem, then conservation of angular momentum only applies if you continue to include all the mass loss in your "system".

    Note that I didn't derive this above, just posted the end result. There is a hint above on how to actually derive this yourself (start with r=rhat |r| and use rhat' above). It is probably more instructive for you to do that yourself (or at least attempt) than for me to post this derivation directly.
  13. Jul 7, 2008 #12
    It is not possible to derive the conservation of angular momentum from Newton's laws. It a simple thing to postulate a counter example that satisfies the Newton's laws, but violates conservation of angular momentum. So you cannot derive the torque equations either, from Newton's laws alone.
  14. Jul 7, 2008 #13
    violates conservation of angular momentum,what do you mean?if we can not derive them from newton's eqs, than they didn' completely explain the motion and angular momentum is a whole different thing?
  15. Jul 7, 2008 #14
    Peeter could you put the derivation or send it to me?I tried but my maths is weak.
  16. Jul 7, 2008 #15
    I'll use coordinates for two dimension in the example: The particle A is in position (-1,0), the particle B is in position (1,0), the particle A exerts a force F_(AB)=(0,1) to the particle B, and particle B exerts a force F_(BA)=(0,-1) to the particle A. There force vectors satisfy the Newton's third law, because they are equal in magnitude, and in opposite direction. However, the angular momentum is not conserving, because the two particles will start rotating counter clockwise on their own.

    This example proves, that if you merely assume that in some system Newton's laws are being satisfied, it is not yet possible to prove that under zero external torque, the angular momentum would be conserving.

    In order to have a complete explanation of the motion, we need to of course specify what kind of forces particles use to interact. This is something different than the Newton's laws. Newton's laws are only a set of rules which the forces must obey, but there can be more rules for the forces too. On possible restriction to the forces is that the force vector must point in the same direction as the spatial separation vector between the particles. In this case the angular momentum will be conserved.
  17. Jul 7, 2008 #16
    My counter example probably looks strange because in reality forces don't go like that usually. Well, it's purpose was to prove that you cannot derive conservation of angular momentum from the Newton's laws, in case somebody is interested in the logical structure of these basic things. ..... but anyway, indeed, conservation of angular momentum is whole different thing from the Newton's laws. It is justified to say that it is one postulate in mechanics (unless you set something else as a postulate, like rotational symmetry of Lagrange's function, or something else equivalent...)
  18. Jul 7, 2008 #17
    so you're saying that the static equilibrium conditions: sum of all forces must be zero come from newtons law and the sum of the moment must be zero comes from somewhere else.Then where is this moment conditon come from?
  19. Jul 7, 2008 #18
    dear jostpuur you're saying that it's a postulate and deal with it:), but i still believe that
    it's a cosequence of the seceond rule ,and somewhere there must be a relation between 2nd rule and rotational stuff.
  20. Jul 7, 2008 #19
    The Newton's approach is that we set a postulate, that for all forces there are corresponding counter forces. In similar spirit, we can simply set a postulate, that for all torques there are corresponding counter torques.

    The Lagrange's and Noether's approach reveals something else. If the Lagrange's function is invariant under translations, then it implies that the total momentum is conserved, which is equivalent with the existence of counter forces. If the Lagrange's function is invariant under rotations, then it implies that the total angular momentum is conserved, which is equivalent with the existence of the counter torques.

    Notice, translation invariance does not imply rotational invariance.

    Good luck on your search :wink:
  21. Jul 7, 2008 #20
    Jostpuur, Can you point me to a reference (preferably online) that discusses this in more detail. My mechanics texts do not really discuss this in a way that I find satisfactory. My assumption was that Newton's laws were more fundamental (in the v << c limit) than angular momentum conservation, and that angular momentum was only conserved under specific circumstances (like constant and non-relativistic mass).
  22. Jul 7, 2008 #21
    Not really... I don't remember very well where I have learned this myself. I don't think I've ever read that claim from anywhere. I've only thought about it myself.
  23. Jul 7, 2008 #22
    I looked, but can't find a derivation in my notes that does not use Geometric Algebra (I'm sure I did one, but it must be on paper somewhere). I've attached what I do have typed up... you can probably get an idea how to directly approach the same radial acceleration decomposition using traditional cross products. Like I said, the key is exploiting a cross product expansion of \rcap', and taking derivitives of r = rcap |r|. If you can't derive the rcap' expression yourself just use what I posted above (my old Salus/Hille Calculus text did that rcap' derivation but it was a bit involved and indirect).

    Attached Files:

  24. Jul 7, 2008 #23
    Not exactly sure how much this will help but when applying a force to something so that it rotates a general path, the direction of displacement tells the direction of the effect the force had. So the sin(theta) added in can have several meanings that add up to that (such a case would be torque). The same concept could be applied to angular momentum.

    With the cross product, we see that two vectors interact within a plan to induce another vector in a direction perpendicular to the plane. Also take notice that you could take the cross product of any two interacting vectors...although the procedure may or may not hold much meaning to it.
  25. Jul 8, 2008 #24


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    Newton's third law implies conservation of angular momentum

    Hi jostpuur! :smile:

    Newtonian mechanics essentially does not recognise force at a distance, except as allowed by Newton's third law (action and reaction).

    When equal but opposite forces are applied perpendicular to opposite ends of a rigid rod, you might say that Newton's second law only requires that the centre of the rod remains stationary, and says nothing about the rotation.

    But that fails to treat the ends of the rod as separate bodies. Each end feels a tension force along the rod (it must do, or the rod wouldn't be rigid), and Newton's second law then makes the end of the rod go in a circle.

    Although the applied forces are not equal and opposite (and so do not obey Newton's third law), the tension forces are. :smile:

    All central forces (such as gravity and electric charge) comply with Newton's third law, and they can easily be proved to conserve angular momentum.

    By contrast, a uniform magnetic field (which is perpendicular-to-central), for example, sends charged bodies in circles, and that obviously does not conserve angular momentum about a distant axis!
    Yes … obviously, we could define forces in a way that violates Newton's second and third law, which your example does.

    Forces which don't meet at a point must be augmented by connecting forces obeying Newton's third law.

    Thus all forces can be grouped according to the point at which they meet.

    And, at each such point, forces which meet at a point can be proved to conserve angular momentum about any axis:
    ∑mv´ = 0 implies ∑mrxv´ = 0​

    Angular momentum can only fail to be conserved in Newtonian mechanics if there is a field which does not obey Newton's third law. :smile:
  26. Jul 8, 2008 #25
    The Newton's laws were not being violated in my example.
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