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Proof of F = -grad(U)?

  1. Sep 16, 2008 #1
    Hi, all. A friend challenged me to prove that F = -grad(U), and now that I did, I'm thinking of submitting it to the university I'm applying to. Before I do so, I want to see if this is right.

    Theorem: For a particle in a conservative force field, F = -grad(U).

    Proof: In a conservative force field, the potential energy is depends only on the coordinates of the particle, and not on its velocity. Using the Euler-Lagrange equation,

    [tex]\frac{\partial L}{\partial x_{i}} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_{i}} = 0[/tex]

    [tex]\frac{\partial}{\partial x_{i}}[T(\dot{x}_{i}) - U(x_{i})] - \frac{d}{dt} \frac{\partial}{\partial \dot{x}_{i}}[T(\dot{x}_{i}) - U(x_{i})] = 0[/tex]

    [tex]\frac{\partial U}{\partial x_{i}} + \frac{d}{dt} \frac{\partial T}{\partial \dot{x}_{i}} = 0[/tex]

    [tex]\frac{\partial U}{\partial x_{i}} + \frac{d}{dt} \frac{\partial}{\partial \dot{x}_{i}}(\frac{1}{2} m \sum \dot{x}_{i}^{2}) = 0[/tex]

    [tex]\frac{\partial U}{\partial x_{i}} + \frac{d}{dt} m\dot{x}_{i} = 0[/tex]

    [tex]\frac{\partial U}{\partial x_{i}} = - m\ddot{x}_{i} = -F_{i}[/tex]

    [tex]\textbf{F} = -\nabla \textbf{U}.[/tex] QED.

    Does this seem like a good proof?

    One thing that I'm worried about is if I overlooked that F = -grad(U) is actually a requirement for the Euler-Lagrange equation to work, and that through this "proof" I might just be making a redundant statement that really doesn't prove anything. Please tell me this isn't true.

    Also, is there anything I can do to make it cleaner or more professional? Thanks!
  2. jcsd
  3. Sep 17, 2008 #2
    Hmm I have always thought that the Euler-Lagrange equations can be derived using [tex]F=\dot{p}=-\nabla U[/tex], albeit a little crudely. The way I remember seeing it goes something like this, with the same assumption of a curlless field,

    For one dimension,

    \dot{p} = -\frac{\partial U}{\partial x}

    Since L = T - U, and only T and U have explicit dependence on dx/dt and x respectively, then it is correct to say that

    [tex]\dot{p}=\frac{\partial L}{\partial x}[/tex]

    [tex]\because \frac{\partial T}{\partial\dot{x}} = \frac{\partial}{\partial\dot{x}}(\frac12m\dot{x}^2)=m\dot{x}=p,[/tex]

    [tex]\frac{d}{dt}\frac{\partial T}{\partial\dot{x}}=\frac{\partial L}{\partial x}[/tex]

    [tex]\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=\frac{\partial L}{\partial x}[/tex]


    This I think is essentially your proof in reverse.
    Last edited: Sep 17, 2008
  4. Sep 17, 2008 #3


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    The potential energy is originally introduced by
    [tex]\Delta U=-\int{\bf F}\cdot{\bf dr}[/tex]
    from which F=-grad U follows in one step.
    Your proof just verifies that the Lagrangian includes this necessary feature.
    Last edited: Sep 17, 2008
  5. Sep 17, 2008 #4


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    I think what you have is fine. But, is not anything ground-breaking. It does show that you are familiar with the formalism of Lagrangians and partial derivatives and how to apply them, etc. And that's good if you are applying to a university.

    Proofs that the Lagrangian formulation of mechanics and Newton's laws are equivalent under certain circumstances are given in many textbooks, so if you want to make your proof look more professional you could go get a classical mechanics textbook from the library and have a look through.
  6. Sep 17, 2008 #5
    Yeah, I know it's definitely not new - it's probably been done since Lagrange's time - but one of my teachers recommended showing it to the admissions office.

    I have noticed a couple of mistakes, though. First, the [tex]U[/tex] in the last line shouldn't be a vector - potential energy is a scalar quantity, and either way, the gradient of a vector is wrong. Also, when I wrote out the summation in the fourth line, I used [tex]i[/tex] as the index of summation. I think this is wrong, since I'm using it as the index for [tex]x[/tex]. Should I use a different index?

    hyperon, that's interesting. I had never seen the Euler-Lagrange equations derived that way - I usually see it through Hamilton's Principle and the calculus of variations.
  7. Sep 19, 2008 #6


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    As hyperon pointed out, in a mathematically consistent framework, the choice of assumptions and derived statements is a matter of taste. But since it is a matter of taste, I think it is reasonable to derive F=-gradU, if the "forces" are not functions of "velocities", provided the assumptions are clearly stated, and F=-gradU is not stated as an assumption.

    Perhaps you should start from a slightly more general form of Lagrange's equation which can handle both conservative and non-conservative forces. Then show that they reduce to the form of Lagrange's equation you started from, if the "forces" are not functions of "velocities".

    Take a look at lectures 9 and 10 of How's and Deyst's Aerospace Dynamics lectures:
    http://ocw.mit.edu/OcwWeb/Aeronautics-and-Astronautics/16-61Aerospace-DynamicsSpring2003/LectureNotes/ [Broken]

    Another way might be to start from Newton's laws, go through the work-kinetic-energy theorem, and define a "conservative force" to be one in which the work done is path independent.
    Last edited by a moderator: May 3, 2017
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