Talk:Jefimenko's equations

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[edit] Retarded versus Advanced time

Would it not be more appropriate to call tr = t − R / c the "advanced" time since it is EARLIER than t? t + R / c would be the retarded time since it is later. Would it not? JRSpriggs 06:02, 25 April 2006 (UTC)

The whole thing is called retardation because the electric and the magnetic fields at a point are not affected by changes in the charge and current density at a remote point immediately, but only at a later time. It is a standard terminology to call the time that needs to be substituted in the integral the retarded time. Yevgeny Kats 11:30, 25 April 2006 (UTC)

[edit] Suggestions and questions

  • Are there similar equations for the vector potential and voltage? If so, what are they and what gauge (such as the Lorenz gauge condition) do they assume?
Assuming the Lorenz gauge condition and Gaussian units...
\phi(\vec{r},t)=\int \frac{\rho(\acute{\vec{r}},t_r)}{R}\mathrm{d}^3\acute{r}
\vec{A}(\vec{r},t)=\int \frac{\vec{J}(\acute{\vec{r}},t_r)}{Rc}\mathrm{d}^3\acute{r}
  • Please provide a proof that Jefimenko's equations are Lorentz invarient and so consistent with special relativity. It is clear that they should be, if they are going to be equivalent to Maxwell's equations. But it is not obvious to me that they are. Of course, they cannot be consistent with general relativity without modification because they implicitly assume a uniform metric for space-time.
  • Please provide a proof that they are equivalent to Maxwell's equations as claimed.
Here's an abbreviated part of the proof (the easy part) in Gaussian units...
\vec{B} = \int \vec{J} \times \frac{\vec{R}}{R^3c} + \frac{\partial \vec{J}}{\partial t} \times \frac{\vec{R}}{R^2c^2} dr^3 (Jefimenko's equation)
\vec{B} = \int \vec{J} \times \frac{\vec{R}}{R^3c} - \frac{\partial \vec{J}}{\partial t} \times \frac{\nabla t_r}{Rc} dr^3 (gradient of the retarded time inserted)
\vec{B} = \int \vec{J} \times \frac{\vec{R}}{R^3c} + \frac{1}{Rc} \nabla \times \vec{J} dr^3 (from the chain rule)
\vec{B} = \int \nabla \left( \frac{1}{Rc} \right) \times \vec{J} + \frac{1}{Rc} \nabla \times \vec{J} dr^3 (gradient of 1/R inserted)
\vec{B} = \int \nabla \times \left( \frac{\vec{J}}{Rc} \right) dr^3 (using the identify for the curl of a scalar times a vector)
\nabla \cdot \vec{B} = 0 (divergence of a curl is zero)
\vec{E} = \int \rho \frac{\vec{R}}{R^3} + \frac{\partial \rho}{\partial t} \frac{\vec{R}}{R^2c} - \frac{\partial \vec{J}}{\partial t} \frac{1}{Rc^2}dr^3 (Jefimenko's equation)
\vec{E} = \int - \rho \nabla \left( \frac{1}{R} \right) + \frac{\partial \rho}{\partial t} \frac{\vec{R}}{R^2c} - \frac{\partial \vec{J}}{\partial t} \frac{1}{Rc^2}dr^3 (gradient of 1/R inserted)
\vec{E} = \int - \rho \nabla \left( \frac{1}{R} \right) - \frac{\partial \rho}{\partial t} \frac{\nabla t_r}{R} - \frac{\partial \vec{J}}{\partial t} \frac{1}{Rc^2}dr^3 (gradient of the retarded time inserted)
\vec{E} = \int - \rho \nabla \left( \frac{1}{Rc} \right) - \frac{\nabla \rho}{R} - \frac{\partial \vec{J}}{\partial t} \frac{1}{Rc^2}dr^3 (from the chain rule)
\vec{E} = \int - \nabla \left( \frac{\rho}{R} \right) - \frac{\partial \vec{J}}{\partial t} \frac{1}{Rc^2}dr^3 (from the identity for the gradient of a product of scalars)
\nabla \times \vec{E} = \int - \nabla \times \frac{\partial \vec{J}}{\partial t} \frac{1}{Rc^2}dr^3 (the curl of a gradient is zero)
\nabla \times \vec{E} = \int - \frac{\partial}{\partial t} \nabla \times \left( \frac{\vec{J}}{Rc^2} \right) dr^3 (curl and time differentiation commute)
\nabla \times \vec{E} = -\frac{1}{c}\frac{\partial \vec{B}}{\partial t} (expression for the magnetic field in the above step inserted)
  • I think you should mention that the equivalence of Jefimenko's equations plus the continuity equation with Maxwell's equations depends on some assumptions, including: (1) there are no cosmological electric or magnetic fields (i.e. none which extend infinitely in space), (2) there is no electromagnetic radiation coming in from the infinite past (and infinite distance), and (3) the integrals can be integrated, i.e. the distribution of charges and currents drops off rapidly enough to zero as the distance from the origin increases. JRSpriggs 05:53, 5 September 2006 (UTC)