1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What exactly is meant by potential energy in the Lagrangian?

  1. Oct 24, 2015 #1
    I was solving the double pendulum problem via Lagrangian methods but something bothered me quite a lot.

    (Consider the two bobs are of equal mass and the pendula are of equal length). Then the potential energies are conveniently written as ##V_1=-mgl\cos\theta## and ##V_2=-mg(l\cos\theta+l\cos\phi)##. Where ##\big\{\theta,\phi\big\}## refer to the usual coordinate angles. ##V## total would be ##V=V_1+V_2##.

    What bothers me is:

    Why isn't the potential energy written (analogously to EM) as ##\frac{1}{2}\sum m_i \Phi(\mathbf{r})## where ##\Phi## would be the gravitational voltage (and of course this includes the sun too).

    As another example the potential energy of a baseball is written in the Lagrangian formalism as ##-mgy## where ##y## is the height. Why is it valid to say that a single particle can have potential energy?

    Namely, potential energy is strictly about a system of particles. We sometimes (incorrectly) speak of the potential energy of a particle only when we're interested in the dynamics of the particle and know that we can find the force on it via the potential energy.

    So I arrived to the conclusion that the potential energy in Lagrangian mechanics more precisely refers to that quantity which would give the correct generalized force; not the actual potential energy of the system. In that sense Lagrangian mechanics isn't about potential energies but about the scalar potential in vector calculus. Which although related, aren't the same thing.

    Where am I screwing up here?

    Thanks.
     
  2. jcsd
  3. Oct 24, 2015 #2
    I don't think there should be a 1/2 there. Perhaps you meant ##V = \sum m_i \Phi(\mathbf{r}_i)## ?

    The center of masses of the two objects can be described by a small number of parameters, mainly ##\theta## and ##\phi##. You can express the potential in the additive form much like the one you gave. Then if you re-express it in terms of those parameters, it no longer looks like the additive form. But the two forms would be equivalent.


    It makes an individual contribution to the total potential energy. Why not? Are you referring to the fact that the real gravitation potential exists between two bodies and doesn't just depend on the position of one object at a time?
     
    Last edited: Oct 24, 2015
  4. Oct 24, 2015 #3
    Yes, that's exactly my point. Although one could state the scalar potential for the force field acting on the baseball that way. Which is why I say Lagrangian mechanics is really about scalar potentials and not the potential energies of the system, which is really defined on 2+ bodies.

    If you see that I'm referring to the fact that real potential exists between 2+ bodies, then you'll see why I have that factor of 2 there. Its exactly what happens with electrostatics (that factor appears there as well).
     
  5. Oct 24, 2015 #4
    I'd like to ask you a question though: How would you write the potential energy term for the lagrangian in the case of two point particles?


    Strictly speaking the potential energy of that system is ##\frac{kq_1q_2}{r}## which is in contradiction to what we've been doing with the gravitational case, since that logic applied to the electric case would say ##U=\frac{kqq'}{r}+\frac{kqq'}{r}## which is not the potential energy of the system.
     
  6. Oct 24, 2015 #5
    You were probably thinking of
    $$V=\frac{1}{2}\sum_{ij} m_im_j v(\mathbf{r}_i -\mathbf{r}_j) $$
    Which is called a pairwise additive potential. For the potential energy in the penulum problem, we are making an approximation of this. The earth has so much mass compared to the pendulum, and its center of gravity so far away compared to the lengths of the pendulum parts, that we just assume the Earth is fixed and the potential is linear. So we get a potential that looks like

    $$V\approx m_{earth}\sum_{i} m_i v(\mathbf{r}_i(\theta, \phi) -\underbrace{\mathbf{r}_{Earth}}_0) $$
    Which should give us exactly what the potentials you wrote earlier.

    Not all potentials have either of these two forms. The only thing that is guaranteed about a potential is that it is a function only of the generalized coordinates. All of this post applies just as much to Newtonian mechanics as it does to Lagrangian mechanics. It's not something special to this type of mechanics at all.
     
    Last edited: Oct 24, 2015
  7. Oct 24, 2015 #6
    It has the same form for electrostatic and graviational potential energy. which is as follows.
    $$V = k\frac{qq'}{|\mathbf{r} - \mathbf{r}' |} + qV_{ext}(\mathbf{r})+q'V_{ext}(\mathbf{r}')$$
    Like I wrote already we might make an approximation and and ignore the first term.
     
  8. Oct 24, 2015 #7
    Are you saying that we are making the approximation (in the gravitational case) that ##qV_{ext}(\mathbf{r})## would be the linear form ##mgy##? I'm starting to understand what you say now :D
     
  9. Oct 24, 2015 #8
    Yeah, that's what's happening. We are also ignoring the first term because the energy of attraction between the parts of the pendulum is negligible.
     
  10. Oct 24, 2015 #9
    But why wouldn't you include the Earth in the Lagrangian though? It seems unnatural to speak of the potential energy of a system and include the contributions of something you aren't considering (the earth in this case).

    I guess this is because in the end what you care about are the Lagrange equations of motion for the particles, and these would be the same regardless of whether you include the Earth or not.
     
  11. Oct 24, 2015 #10
    Perhaps it sounds like I'm making you repeat yourself, hehe... I've encountered many problems where they speak about the potential energy of a particle and its starting to make me think there's nothing wrong about that concept. Do you agree? Are both concepts physically sound?
     
  12. Oct 25, 2015 #11
    Some physical theories are closer to reality than others. The idea that each particle makes an independent contribution to potential energy can give good results for practical purposes. However keep in mind there is something else closer to reality. The idea is just a simplified model.

    The electrostatic model is also a simplification. For example we know that the forces are not transmitted faster than light. Since a particle can gain kinetic energy from the force, there might have been a time in-between when particle one has sent some potential energy and when particle 2 receives it. So maybe we can have energy in the spaces in between. So the potential energy could be different than some property of the classical particle like its velocity or orientation, which the particle could be carrying along with itself. It turns out the question of where exactly is the potential energy can be a puzzle!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What exactly is meant by potential energy in the Lagrangian?
Loading...