Effective potential in a central field

  • #1
pbilous
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Hi, I am confused by a point which should be relatively simple. When we consider classical motion of a particle in a central field U(r), we write the total energy E = T + U, where T is the kinetic energy. The kinetic energy contains initially r, r' and φ' (where ' denotes the time derivative). We replace φ' via the angular momentum L, which is an integral of motion due to the rotational symmetry. The kinetic energy takes then the form T=mr'2/2 + L2/(2mr2). Finally we attribute the latter term to the potential energy and introduce the effective potential Ueff = U + L2/(2mr2). The energy E remains the same under such term redistiribution and we obtain effectively 1D motion in potential Ueff.

Now the question. If we consider the Lagrangian L = T - U instead of energy E =T + U and carry out the same procedure, then such "redistribution" of the terms would lead to different effective potential Veff = U - L2/(2mr2). Why is then the way with E "better" than the one with L?

Alternatively one could formulate it as follows. When we shift the term L2/(2mr2) to the potential energy Ueff = U + L2/(2mr2), the total energy E = T + U remains the same, but at the same time the Lagrangian L = T - U changes by L2/(mr2). Doesn't it describe already a completely different system? Why wouldn't we choose to preserve L instead of E and introduce Veff = U - L2/(2mr2)?
 

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  • #2
vanhees71
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The trick is to use the conservation laws, and the conserved quantities are usually energy (Lagrangian doesn't depend explicitly on time) and the canonical momenta of cyclic coordinates (which come also from symmetries, i.e., the choice of the variables takes into account some symmetry like rotational symmetry around an axis or even around a point). If you have enough conserved quantities "first integrals" you can express the conserved energy in terms of only one variable, and then the entire motion is effectively a one-dimensional motion of this variable in the corresponding effective potential. All other variables can be then found by simple integration after the one-dimensional problem is solved.

The reason to use the energy rather than the Lagrangian in this context is simply that energy is often a conserved quantity.
 
  • #3
pbilous
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Dear vanhees71, thanks for your comment. I perfectly agree with you that the important difference between the energy and the Lagrangian in this context is that the energy is the first integral and is conserved, whereas the Lagrangian is changing somehow with the motion of the particle.

However, I still don't see why it is necessary to restrict only to first integrals (your last sentence). I mean, we can express φ' via the angular momentum and substitute into the Lagrangian as well. Then we redistiribute the terms such that we get the "wrong" effective potential. The Lagrangian is not conserved, but I don't see in this fact any problem why we couldn't carry out the described procedure.

On the other hand. Even if we stick to energy E = T + U = mr'2/2 + M2/(2mr2) + U and introduce the "correct" effective potential Ueff = U + M2/(2mr2) (where M is the angular momentum), something strange happens to the Lagrangian. Indeed, the Lagrangian for the obtained effective 1D motion is L = mr'2/2 - Ueff = mr'2/2 - M2/(2mr2) - U. This differs from the initial Lagrangian by M2/(mr2), although the system described is exactly the same. This difference M2/(mr2) does not look like a full time-derivative, so these Lagrangians describe really two different systems. Do you have any idea how to resolve this contradiction?
 
  • #4
vanhees71
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The Lagrangian formalism is only form invariant under point transformations, i.e., under diffeomorphisms transforming a set of generalized coordinates ##q## to another set ##q'##, i.e., you cannot expect when doing transformations involving the canonical momenta to get a valid Lagrangian. For that you need the Hamiltonian formalism, where you are free to use the larger class of canonical transformations, which is a transformation of the phase-space variables ##(p,q)##.
 

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