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.(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Proof of F = -grad(U)?

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