Turning our attention next to field energy defined in Equation (
68) we obtain the following expression for fourth order term in
:
in which we are reminded that there are no field contributions which are first order in
. According to Equation (
A6)
, and according to Equation (A65)
, hence the above equation simplifies as follows:
The field energy can be clearly partitioned to electric field and magnetic field contributions:
We begin with evaluating
by using
of Equation (
A5) and
of Equation (A56). Obtaining:
We show in
Appendix D (see also [
17]) that:
hence:
Taking into account Equation (
14) we obtain:
hence:
using the mechanical momentum Equation (
21) we thus obtain:
Turning our attention to the magnetic part of the field energy we notice that a fourth order correction of the magnetic field
is needed, this can be calculated according to Equation (
61) as:
Taking into account Equation (A185) the fourth order correction to the magnetic field is thus:
Hence for a static current:
We shall find it convenient to label the different terms of the magnetic field of the fourth order:
Furthermore, thus the magnetic energy can also be partitioned:
We shall start by evaluating
. Using Equations (
A13) and (A208) we obtain:
Using a well known identity from vector analysis we may write:
Let us look at the integral expression
This is an expression of the type described in Equation (A305) of
Appendix B with
and
. According to
Appendix A the expression in Equation (A213) can be expressed as a surface integral. Assuming that our system is contained in an infinite sphere we have according to Equations (A316) and (A22):
The following asymptotic expressions will now come in handy (see Equations (A72) and (A84)):
Inserting Equations (A215) and (A216) into Equation (A214) and taking the limit will yield:
However, according to Equation (A129) a closed loop integral over a constant is null hence:
Furthermore, Equation (A212) simplifies to:
Now:
Taking into account Equation (A19), we have:
Plugging Equation (A221) into Equation (A219) and using Gauss theorem we obtain:
Let us perform the surface integral on an infinite sphere as usual and look at the integral:
Now Equation (A216) takes the asymptotic form:
Using Equations (A215) and (A224) in Equation (A223) and taking the limit:
However, according to Equation (A129) a closed loop integral over a constant is null hence:
and thus Equation (A222) simplifies
Taking into account the definition Equation (
14) this is simplified to the form:
Next we turn our attention to
(see Equation (A210)), using Equations (
A13) and (A209) we obtain:
However, this integral is the same as the integral given in Equation (A211) with the indices 1 and 2 interchanged. It immediately follows that
is equal to right hand side of Equation (A228) with the indices 1 and 2 interchanged, thus:
However, according to Equation (
14):
Thus, we obtain:
Furthermore, Equation (A210) simplifies to the form:
this is to be expected as the energy terms should not depend on acceleration but only on velocity. We now turn our attention to
defined in Equation (A210). Using Equations (
A13) and (A209) we obtain:
Using a well known identity from vector analysis we may write:
Let us look at the integral expression
This is an expression of the type described in Equation (A305) of
Appendix B with
and
. According to
Appendix A the expression in Equation (A202) can be expressed as a surface integral. Assuming that our system is contained in an infinite sphere we have according to Equations (A316) and (A22):
Now:
Taking into account the asymptotic expressions in Equations (A215) and (A25) the limit of Equation (A237) takes the following form:
However, according to Equation (A129) a closed loop integral over a constant is null hence:
Furthermore, Equation (A235) simplifies to:
Now:
Taking into account Equation (A19), we have:
Plugging Equation (A242) into Equation (A241) and using Gauss theorem we obtain:
Let us perform the surface integral on an infinite sphere as usual and look at the integral:
Using Equations (A215) and (A224) in Equation (A245) and taking the limit we obtain:
However, according to Equation (A129) a closed loop integral over a constant is null hence:
and thus Equation (A244) simplifies to:
Taking into account the definition in Equation (
14) this is simplified to the form:
Hence:
Taking into account the mechanical momentum Equation (
21) and the mechanical energy Equation (
22) this can be written as:
Finally we turn we turn our attention to