# Derivation of Jefimenko's Equation in Jackson's EMT book

I have been trying to understand the derivation of Jefimenko's equation in Jackson on p.246-247 which can be seen in the photographs attached. First of all I did not fully comprehend the transformation between the two square bracket notations one with the ∇' inside the square brackets and the one outside it. How is the minus/plus selected in Eqs.(6.53 and 6.54). Furthermore, after the transformation Eq.(6.54) and substitution into Eq.(6.52) integration by parts is applied to the first term(the one with the curl) and there are actually two terms and one is neglected without any justification. I would appreciate it if someone could provide the justification for that. The transformation is as follows:

$$\nabla'\times(\frac{\vec{J}}{R})=\frac{\vec{J}\times(\nabla'R)+R\nabla'\times \vec{J}}{R^2}$$ Hence

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$

Yet apparently the first term on the right hand side is neglected in Eq.(6.56), is there a justification for it?

EDIT After substitution you get the term on the LHS. We transform this term obeying vector calculus identities(I proved it in tensor notation and checked the result in Griffiths btw.), then we arrive at two terms and in the integration the first term on the RHS that is $\nabla'\times(\frac{\vec{J}}{R})$ neglected in 6.56. I wonder how you can justify this neglection.

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$

• I have answered the first question you asked. But I cannot understand how the integration by parts thing works. Can you show how the integration by parts is done and what the two terms are? Apr 19 '15 at 5:25
• @gautam1168 Thanks for your answer. For the integration by parts for the magnetic field you substitute 6.54 into 6.52, a.k.a referred to as the preliminary solution in the text. I am editing the question to specify the term you get. Apr 19 '15 at 9:42
• I can't seem to figure it out either. Lets see if someone else can answer this. I'll keep trying in the meanwhile. Apr 19 '15 at 12:12
• @gautam1168 You can check my latest answer if you are interested. Apr 23 '15 at 21:30

The square bracket transformation

This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a function of primed time coordinate as well. So the gradient operator has to applied using chain rule. $$=\left \lbrace \sum_i \left(\frac{\partial x_i'}{\partial x_i'}\frac{\partial }{\partial x_i'} + \frac{\partial x_j'}{\partial x_i'}\frac{\partial }{\partial x_j'} + \frac{\partial x_k'}{\partial x_i'}\frac{\partial }{\partial x_k'} + \frac{\partial t'}{\partial x_i'}\frac{\partial }{\partial t'}\right) \hat{i}\right\rbrace[\rho(x_i',x_j',x_k',t')]_{ret}\\ =\left \lbrace \sum_i \left(\frac{\partial }{\partial x_i'} + 0+ 0 + \frac{\partial t'}{\partial x_i'}\frac{\partial }{\partial t'}\right) \hat{i}\right\rbrace[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ So the primed spacial derivatives are taken while keeping the primed t constant. $$=\left \lbrace \left(\frac{\partial }{\partial x_i'} \hat{i}+\frac{\partial }{\partial x_j'} \hat{j}+\frac{\partial }{\partial x_k'} \hat{k}\right) + \left(\frac{\partial t'}{\partial x_i'} \hat{i}+\frac{\partial t'}{\partial x_j'} \hat{j}+\frac{\partial t'}{\partial x_k'} \hat{k}\right)\left(\frac{\partial }{\partial t'}\right) \right \rbrace[\rho]_{ret}\\ =[\nabla'\rho]_{ret} + \nabla'(t')\left[\frac{\partial \rho}{\partial t'}\right]_{ret}$$

Where the $\nabla'$ acts while keeping $t'$ fixed and the derivative wrt $t'$ is taken while keeping primed spatial coordinates fixed. So,

$$[\nabla' \rho]_{ret} = \nabla'[\rho]_{ret} - \left[\frac{\partial \rho}{\partial t'}\right]_{ret} \nabla'(t-R/c)$$

• Any more ideas? Apr 19 '15 at 19:15

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$

Thanks to Prof. Y. F. Chen I was able to figure it out. While in the integral the first term on the RHS can be converted into a surface integral as below:

$$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$

Since the integration takes place in whole space and the current source vanishes at infinity the surface integral of this term, hence the volume integral, is zero. The justification is as follows:

$$\int(\partial_iA_j-\partial_jA_k)d^3x^{'}=\int (n_iA_j-n_jAi)d^2x^{'}$$