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Is it reasonable to ask if the negative four-gradient of the four-potential is equal to the time derivative of the four-momentum?

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- #1

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Is it reasonable to ask if the negative four-gradient of the four-potential is equal to the time derivative of the four-momentum?

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No. Transcribing what you said: [tex]-\nabla_a A_b \stackrel{?}{=} \frac{d}{dt} p_a [/tex]. The indices don't balance.actionintegral said:

Is it reasonable to ask if the negative four-gradient of the four-potential is equal to the time derivative of the four-momentum?

(Note the four-potential [tex]A_a[/tex] refers to the electromagnetic potential, which decomposes in an observer's coordinate system into the scalar potential [tex]\phi[/tex] and the vector potential [tex]\vec A[/tex].)

By "time derivative", do you mean derivative with respect to proper-time?

Last edited:

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The "four-potential" must be the electromagnetic potential [itex]A_\mu[/itex]. Try writing down your second sentence as a tensor equation, it doesn't make sense, and the time derivitive of the momentum is not covariant, you need to use the proper time [itex]\tau[/itex]. However, IIRC

[itex]

\frac{dp_\mu}{d\tau} = (\partial_\mu A_\nu - \partial_\nu A_\mu)u^\nu

[/itex]

where [itex]u^\nu[/itex] is the 4-velocity of a body. But note what the space part of this equation reduces to in the rest frame of the body.

[itex]

\frac{dp_\mu}{d\tau} = (\partial_\mu A_\nu - \partial_\nu A_\mu)u^\nu

[/itex]

where [itex]u^\nu[/itex] is the 4-velocity of a body. But note what the space part of this equation reduces to in the rest frame of the body.

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- #4

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There's a [tex]q[/tex] on the right hand side and, possibly, a conventional choice of sign. This is the Lorentz Force expression on the right.Daverz said:The "four-potential" must be the electromagnetic potential [itex]A_\mu[/itex]. Try writing down your second sentence as a tensor equation, it doesn't make sense, and the time derivitive of the momentum is not covariant, you need to use the proper time [itex]\tau[/itex]. However, IIRC

[itex]

\frac{dp_\mu}{d\tau} = (\partial_\mu A_\nu - \partial_\nu A_\mu)u^\nu

[/itex]

where [itex]u^\nu[/itex] is the 4-velocity.

Note: [tex]\frac{dp_b}{d\tau}=u^a\nabla_a p_{b}=m u^a\nabla_a u_{b} [/tex].

- #5

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[itex]

A^\mu = (\phi, A_x, A_y, A_z)

[/itex]

[itex]

F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu

[/itex]

[itex]

\frac{dp^\mu}{d\tau} = \frac{q}{m}p_\nu F^{\mu \nu}

[/itex]

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