Talk:Jefimenko's equations

message: The answer is that the Jefimenko equations as written require the Lorenz gauge condition to be equivalent to Maxwell's laws.  All you have to do is try to calculate the divergence of the equation for the electric field and you will find that the continuity equation won't cancel out the terms correctly.  —Preceding unsigned comment added by JoeRoser (talk • contribs) 05:39, 24 December 2008 (UTC)
replies: text: They require other conditions as well, see the JRSpriggs 2006 post at the bottom of this page. -- Army1987 – Deeds, not words. 15:24, 6 January 2009 (UTC)
thread_title:  Do the Jefimenko equations need to be supplemented with the continuity equation or the Lorenz gauge condition to be equivalent to Maxwell's laws? 

message: Would it not be more appropriate to call t_r = 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)
replies: text: 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)
thread_title:  Retarded versus Advanced time 

message: * What units are you using? Is it International System of Units or Centimeter gram second system of units?
* 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)
thread_title:  Suggestions and questions 

message: What does Jefimenko mean exactly in the last quotation? How does this come out of the equations? Also, what do these equations say differently about electric and magnetic fields than Maxwell's? Diagrams would be helpful for the less mathematically adept. JKeck (talk) 21:10, 20 January 2011 (UTC)
thread_title:  Explanation needed 

message: In the "Discussion" the notion that "electric and magnetic fields can cause each other" is mentioned as an often used explanation of electromagnetic waves. In an early version of the text this was followed by "Jefimenko's equations show otherwise". This can not be justified. Jefimenko's equations do nothing else than computing E and B.
Also, Jefimenko's own statement which is quoted: "...neither Maxwell's equations nor their solutions indicate an existence of causal links between electric and magnetic fields." is clearly not denying that the fields can cause each other. He only says that Maxwell's equations don't show that they cause each other. In the same way his own equations don't show that they do not cause each other.  — Preceding unsigned comment added by Jos.Bergervoet (talk • contribs) 13:11, 18 December 2011 (UTC)
thread_title:  Causality 

message: Several formulas use  \mathrm{d}^3 \mathbf{r}', at the end of integral expressions, that is not correct. Even if changed to  (\mathrm{d} \mathbf{r}') ^3, it is still incorrect because it should be a scalar volume and this is a vector, could author check and correct this, please.  — Preceding unsigned comment added by 109.121.109.168 (talk) 20:12, 19 November 2021 (UTC)
replies: text: Indeed a volume is a scalar and not a vector so I replaced  \mathrm{d}^3 \mathbf{r}' by the more common dV'.  — Preceding unsigned comment added by Dodeluc (talk • contribs) 15:44, 20 November 2021 (UTC)
thread_title:  Errors