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Derive the Lagrangian of a free nonrelativistic particle

  1. Apr 1, 2008 #1
    As we know, the variational principle can be used as the fundamental principle of mechanics. Without knowing Newton's laws, the Lagrangian could be derived from symmetric consideration. As the simplest case, the Lagrangian of a free particle could be derived from Galileian invariance, or the homogeneity of space and time. But I never saw a rigrious proof. I think we just "guess" the form of the Lagrangian for some reason rather than prove it.
    In Landau's famous "Mechanics", he said that the Lagragian of a free particle must be a function of v^2 only. This may be a cool argument of Landau's style, full of physics intuition, but not a rigrious proof. Is it possible to prove that "if space and time are homogeneous, the Lagrangian must be of the form L(v)+df(x,t)/dt"?
  2. jcsd
  3. Apr 4, 2008 #2
    The proof is very simple, and I believe Landau gives a rigorous proof. We first assume the Newton-Laplace Principle of Determinacy, which states that if the positions and velocities of all the particles in the system are known, the future motions of all these particles can be predicted. From this, we deduce that the Langrangian of any system can only be a function of position, velocity and time:[tex]L(\vec{x},\vec{v},t)[/tex]. If we assume that space is homogeneous, then the gradient of L (in other words, the partial derivative of L with respect to [tex]\vec{x}[/tex] must be equal to zero. Therefore, L has no explicit dependence on position, and can thus only be a function of velocity and time. But due to the homogeneity of time, the partial derivative of L with respect to t is zero, and thus the Lagrangian can have no explicit dependence on time. Therefore, L can only be dependent on the velocity [tex]\vec{v}[/tex]. But that is not enough to complete the proof. We also need to assume that space is isotropic, meaning that it does not change when you rotate it. As a consequence of this, the Lagrangian cannot depend on the direction of any vector, specifically the velocity vector. Therefore, we have proved that the Lagrangian can only depend on the magnitude of [tex]\vec{v}[/tex].
  4. Apr 4, 2008 #3
    Thank you, lugita15.
    But this is just the same as what Landau said, without more sophisticated details. We know that if [itex]L(\mathbf x+\delta \mathbf x, \mathbf v, t)-L(\mathbf x, \mathbf v, t)=df(\mathbf x,t)/dt[/itex], then we get the same equation of motion. So generally, the Lagrangian could be a function of x and t. But I come to realize that the homogeneity of x is "defined" to be that L does not contain x. This is just the same meaning as in the discussion of conservation laws.
    Without the homogeneity being mathematically well-defined, we can not get a rigrious explanation.
  5. Apr 4, 2008 #4
    As I said before, the mathematical definition of homogeneity of space is that the partial derivative of any (physically meaningful) quantity with respect to position is zero. Similarly, the definition of homogeneity of time is that the derivative of any (physically meaningful) quantity with respect to position is zero. This is a rigorous mathematical definition of homogeneity: the partial derivative with respect to the variable that is homogeneous must be equal to zero.
  6. Apr 4, 2008 #5
    Yes, such properties are "suggested" by Noether's theorem. So the form of Lagrangian is defined, not proved. Also we need not define "any (physically meaningful) quantity" have such properties. All the other conservation laws are corollaries of certain symmetries of Lagrangian.
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