Four-vector physics

1. Nov 17, 2006

actionintegral

I am trying to get a handle on four-vectors. I see that there is a thing called a four-potential, a four-gradient, and a four momentum.

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

2. Nov 17, 2006

robphy

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

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

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

Last edited: Nov 17, 2006
3. Nov 17, 2006

Daverz

The "four-potential" must be the electromagnetic potential $A_\mu$. 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 $\tau$. However, IIRC

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

where $u^\nu$ 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.

Last edited: Nov 17, 2006
4. Nov 17, 2006

robphy

There's a $$q$$ on the right hand side and, possibly, a conventional choice of sign. This is the Lorentz Force expression on the right.

Note: $$\frac{dp_b}{d\tau}=u^a\nabla_a p_{b}=m u^a\nabla_a u_{b}$$.

5. Nov 17, 2006

Daverz

Yeah, I sorta forgot that charge stuff. Kind of important if you want there to be any force on the body. Looking it up this time (Ohanian, Gravitation and Spacetime, 2nd ed, pp. 95-97:

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

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

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