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I have to prove that: $$\frac{d}{dt}\left( T+q\phi \right)=\frac{\partial}{\partial t}\left[ q\left( \phi - \vec{v}\cdot\vec{A}\right)\right] $$ Where $T=\frac{1}{2}mv^2$ is the kinetic energy and $q\phi$ es the potential energy of a particle moving with a velocity $\vec{v}$.

Using that: $$\vec{F}=-q\left[ \nabla\phi + \frac{\partial\vec{A}}{\partial t}-\vec{v}\times\left( \nabla\times\vec{A} \right) \right]$$

Now, the second equality holds because: $$ \vec{F}=\frac{d\vec{p}_{mec}}{dt}=q\left( \vec{E}+\vec{v}\times\vec{B} \right)$$

So one just have to substitute the electric and magnetic fields in terms of the potentials.

Now, I expressed the kinetic energy in terms of the velocity dot product and did some algebra and arrived to this:

$$ \frac{d}{dt}\left( T+q\phi \right)=\frac{\partial}{\partial t}(q\phi)-q\vec{v}\nabla\phi-q\vec{v} \cdot\frac{\partial \vec{A}}{\partial t}$$

Which I rearranged:

$$\frac{d}{dt}\left( T+q\phi \right)=q\frac{\partial}{\partial t}\left( \phi-\vec{v}\cdot\vec{A} \right)+q\left( \vec{A}\cdot\frac{\partial\vec{v}}{\partial t}-\vec{v}\cdot\nabla\phi \right)$$

Which is almost what I want, but how do I get rid of the second term...? Any ideas?

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First of all, the correct expression of Lorentz' force is $$\vec{F}=-q\left[ \nabla\phi + \frac{\partial\vec{A}}{\partial t}-\vec{v}\times\left( \nabla\times\vec{A} \right) \right]\:.$$ So $$\frac{d}{dt} T = \frac{m}{2} 2 \vec{v} \cdot \frac{d\vec{v}}{dt} = {\bf v}\cdot m \frac{d\vec{v}}{dt} = -q\vec{v} \cdot \left[ \nabla\phi + \frac{\partial\vec{A}}{\partial t}-\vec{v}\times\left( \nabla\times\vec{A} \right) \right]\:.$$ In other words $$\frac{d}{dt} T = -q\vec{v} \cdot \nabla\phi -q\vec{v} \cdot\left[ \frac{\partial\vec{A}}{\partial t}-\vec{v}\times\left( \nabla\times\vec{A} \right) \right]\:.$$ However $\vec{v} \cdot \left(\vec{v}\times\left( \nabla\times\vec{A} \right) \right)=0$ so that $$\frac{d}{dt} T = -q\vec{v} \cdot \nabla\phi -q\vec{v} \cdot\frac{\partial\vec{A}}{\partial t} \:.$$ This identity can be re-written as $$\frac{d}{dt} T +q\vec{v} \cdot \nabla\phi = -q\vec{v} \cdot \frac{\partial\vec{A}}{\partial t}\:,$$ that, in turn, can be re-arranged to $$\frac{d}{dt}\left(T +q \phi(t, \vec{x}(t))\right) - q\frac{\partial \phi}{\partial t} = -q\vec{v} \cdot \frac{\partial\vec{A}}{\partial t}\:,$$ namely $$\frac{d}{dt}\left(T +q \phi(t, \vec{x}(t))\right) = q\frac{\partial \phi}{\partial t} -q\vec{v} \cdot \frac{\partial\vec{A}}{\partial t}\:,$$ that is your formula $$\frac{d}{dt}\left(T +q \phi(t, \vec{x}(t))\right) = q\frac{\partial }{\partial t}\left( \phi -\vec{v} \cdot \vec{A}\right)\:,$$

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  • $\begingroup$ Ok, thank you very much. But, what if one consider that the potential, $\phi$, is independent of the position. $$\frac{d}{dt}T+q\vec{v}\cdot\nabla\phi=\frac{d}{dt}\left[T +q \phi(t, \vec{x}(t))\right] - q\frac{\partial \phi}{\partial t} $$ Would be valid?? $\endgroup$ – C. Alexander Dec 22 '17 at 18:43
  • $\begingroup$ And, in the last equality: $$ q\frac{\partial}{\partial t}\left( \phi-\vec{v}\cdot\vec{A} \right) \neq q\frac{\partial\phi}{\partial t}-q\vec{v}\cdot\frac{\partial \vec{A}}{\partial t}$$ There is a term missing, $q\vec{A}\cdot\frac{\partial \vec{v}}{\partial t}$ $\endgroup$ – C. Alexander Dec 22 '17 at 18:58
  • $\begingroup$ Well, I think you should first clarify yourself some notions of calculus before starting dealing with these sort of applications to physics. Concerning your first question: if $\phi=\phi(t)$ the identity is still valid, simply $\phi$ vanishes on both sides. Concerning the second question: There is NO missing term since $\partial v/\partial t =0$. $\endgroup$ – Valter Moretti Dec 22 '17 at 20:49
  • $\begingroup$ But that is how you started, by using the Lorentz equation assuming $\frac{\partial \vec{v}}{\partial t}\neq \vec{0}$ ....... $\endgroup$ – C. Alexander Dec 24 '17 at 4:15
  • $\begingroup$ I assumed $dv/dt \neq 0$ and not $\partial v/\partial t \neq 0$. $\endgroup$ – Valter Moretti Dec 24 '17 at 6:41

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