Four-Vector Physics: Exploring Questions

  • Thread starter Thread starter actionintegral
  • Start date Start date
  • Tags Tags
    Physics
actionintegral
Messages
305
Reaction score
5
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?
 
Physics news on Phys.org
actionintegral said:
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?

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:
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

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

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:
Daverz said:
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

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

where u^\nu is the 4-velocity.

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}.
 
Yeah, I sort of 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:

<br /> A^\mu = (\phi, A_x, A_y, A_z)<br />

<br /> F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu<br />

<br /> \frac{dp^\mu}{d\tau} = \frac{q}{m}p_\nu F^{\mu \nu}<br />
 
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top