# Lab Notes for a Scientific Revolution (Physics)

## February 27, 2008

### Lab Note 2: Derivation of the Maxwell Stress-Energy Tensor from Five-Dimensional Geometry, using a Four-Dimensional Variation

FOR SOME REASON, THESE EQUATIONS APPEAR CORRUPTED.  I AM CHECKING WITH WORDPRESS TECHNICAL SUPPORT AND HOPE TO HAVE THIS FIXED IN THE NEAR FUTURE. FOR NOW, PLEASE USE THE LINK IN THE FIRST PARAGRAPH, AND GO TO SECTION 10 — JAY.

As mentioned previously, I have been able to rigorously derive the Maxwell tensor from a five dimensional Kaluza-Klein geometry based on Lorentz force geodesics, using a variational principle over the four spacetime dimensions of our common experience.  At the link: Kaluza-Klein Theory, Lorentz Force Geodesics and the Maxwell Tensor with QED, I have attached a complete version of this paper, which includes connections to quantum theory as well as an extensive summary not included in the version of the paper now being refereed at one of the leading journals.  This is a strategic decision not to overload the referee, but to focus on the mathematical results, the most important of which is this derivation of the Maxwell tensor.

Because this paper is rather large, I have decided on this weblog, to post section 10, where this central derivation occurs.  Mind you, there are nine sections which lay the foundation for this, but with the material below, plus the above link, those who are interested can see how this all fits together.  The key result emerges in equation (10.15) below.  Enjoy!

Excerpt: Section 10 — Derivation of the Maxwell Stress-Energy Tensor, using a Four-Dimensional Variation

In section 8, we derived the energy tensor based on the variational calculation (8.4), in five dimensions, i.e., by the variation $delta g^{{rm M} {rm N} }$.  Let us repeat this same calculation, but in a slightly different way.

In section 8, we used (8.3) in the form of ${rm L}_{Matter} =-{textstylefrac{1}{8kappa }} boverline{kappa }g^{5{rm B} } J_{{rm B} } =-{textstylefrac{1}{8kappa }} boverline{kappa }g^{{rm M} {rm N} } delta ^{5} _{{rm M} } J_{{rm N} }$, because that gave us a contravariant $g^{{rm M} {rm N} }$ against which to obtain the five-dimensional variation $delta {rm L}_{Matter} /delta g^{{rm M} {rm N} }$.  Let us instead, here, use the very last term in (8.3) as ${rm L}_{Matter}$, writing this as:

${rm L}_{Matter} equiv {textstylefrac{1}{2kappa }} R^{5} _{5} =-{textstylefrac{1}{8kappa }} boverline{kappa }left(g^{5beta } J_{beta } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)=-{textstylefrac{1}{8kappa }} boverline{kappa }left(g^{mu nu } delta ^{5} _{nu } J_{mu } +{textstylefrac{1}{4}} g^{55} g^{mu nu } boverline{kappa }F_{mu } ^{tau } F_{nu tau } right)$. (10.1)

It is important to observe that the term $g^{5beta } J_{beta }$ is only summed over four spacetime indexes.  The fifth term, $g^{55} J_{5} ={textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau }$, see, e.g., (6.8).  For consistency with the non-symmetric (9.5), we employ $g^{5beta } J_{beta } =g^{mu nu } delta ^{5} _{nu } J_{mu }$ rather than $g^{5beta } J_{beta } =g^{mu nu } delta ^{5} _{mu } J_{nu }$.  By virtue of this separation, in which we can only introduce $g^{mu nu }$ and not $g^{{rm M} {rm N} }$ as in section 8, we can only take a four-dimensional variation $delta {rm L}_{Matter} /delta g^{mu nu }$, which, in contrast to (8.4), is now given by:

$T_{mu nu } equiv -frac{2}{sqrt{-g} } frac{partial left(sqrt{-g} {rm L}_{Matter} right)}{delta g^{mu nu } } =-2frac{delta {rm L}_{Matter} }{delta g^{mu nu } } +g_{mu nu } {rm L}_{Matter}$. (10.2)

Substituting from (10.1) then yields:

$T_{mu nu } ={textstylefrac{1}{4kappa }} boverline{kappa }left(delta ^{5} _{nu } J_{mu } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F_{mu } ^{tau } F_{nu tau } right)-{textstylefrac{1}{2}} g_{mu nu } {textstylefrac{1}{4kappa }} boverline{kappa }left(g^{5beta } J_{beta } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.3)

Now, the non-symmetry of sections 8 and 9 comes into play, and this will yield the Maxwell tensor.  Because $delta ^{5} _{nu } =0$, the first term drops out and the above reduces to:

$kappa T_{mu nu } ={textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F_{mu } ^{tau } F_{nu tau } right)-{textstylefrac{1}{2}} g_{mu nu } {textstylefrac{1}{4}} boverline{kappa }left(g^{5beta } J_{beta } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.4)

Note that this four-dimensional tensor is symmetric, and that we would arrive at an energy tensor which is identical if (10.3) contained a $delta ^{5} _{mu } J_{nu }$ rather than $delta ^{5} _{nu } J_{mu }$.  One again, the screen factor $delta ^{5} _{nu } =0$ is at work.

In mixed form, starting from (10.3), there are two energy tensors to be found.  If we raise the $mu$ index in (10.3), the first term becomes $delta ^{5} _{nu } J^{mu } =0$ and we obtain:

$-kappa T^{mu } _{nu } =-{textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{mu tau } F_{nu tau } right)+{textstylefrac{1}{2}} delta ^{mu } _{nu } {textstylefrac{1}{4}} boverline{kappa }left(g^{5beta } J_{beta } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.5)

with this first term still screened out.  However, if we transpose (10.3) and then raise the $mu$ index, the first term becomes $g^{5mu } J_{nu }$ and this term does not drop out, i.e.,

$-kappa T_{nu } ^{mu } =-{textstylefrac{1}{4}} boverline{kappa }left(g^{5mu } J_{nu } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{mu tau } F_{nu tau } right)+{textstylefrac{1}{2}} delta ^{mu } _{nu } {textstylefrac{1}{4}} boverline{kappa }left(g^{5beta } J_{beta } +{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.6)

So, there are two mixed tensors to consider, and this time, unlike in section 8, these each yield different four-dimensional energy tensors.  Contrasting (10.5) and (10.6), we see that $delta ^{5} _{nu } =0$ has effectively “broken” a symmetry that is apparent in (10.6), but “hidden” in (10.5).  At this time, we focus on (10.5), because, as we shall now see, this is the Maxwell stress-energy tensor $T^{mu } _{nu } =-left(F^{mu tau } F_{nu tau } -{textstylefrac{1}{4}} delta ^{mu } _{nu } F^{sigma tau } F_{sigma tau } right)$, before reduction into this more-recognizable form.

Purposely leaving constant factors separated, the trace equation of (10.5) is then:

$kappa T=R=-2{textstylefrac{1}{4}} boverline{kappa }g^{5beta } J_{beta } -{textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.7)

and so, via the inverse equation $R^{mu } _{nu } =-kappa T^{mu } _{nu } +delta ^{mu } _{nu } kappa T$, from (10.5) and (10.7):

$R^{mu } _{nu } =-{textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{mu tau } F_{nu tau } right)+{textstylefrac{1}{2}} delta ^{mu } _{nu } {textstylefrac{1}{4}} boverline{kappa }left(-3g^{5beta } J_{beta } -{textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.8)

Note that here, traceable to the screened term, lost via $delta ^{5} _{nu } =0$, that one cannot simply glean $R^{mu } _{nu }$ from (10.5) as we were able to for (8.9).  It was necessary to use the full inverse field equation $R^{mu } _{nu } =-kappa T^{mu } _{nu } +delta ^{mu } _{nu } kappa T$.  Now, we take the trace of (10.8) to obtain:

$kappa T=R=-6{textstylefrac{1}{4}} boverline{kappa }g^{5beta } J_{beta } -3{textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.9)

Interestingly, this does not look to be the same as the trace in (10.7), yet these are the same. This means that a further relationship must subsist, and if we look closely, (10.9) is the same as (10.7), multiplied by a factor of 3.  If $x=3x$, then $x=0$, so this is an indication that the trace $kappa T=R=0$, which is characteristic of Maxwell’s tensor.

So, setting (10.7) equal to (10.9), we obtain:

$R=-2{textstylefrac{1}{4}} boverline{kappa }g^{5beta } J_{beta } -{textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)=-6{textstylefrac{1}{4}} boverline{kappa }g^{5beta } J_{beta } -3{textstylefrac{1}{4}} boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$, (10.10)

and we find after reducing, that:

$g^{5beta } J_{beta } =-{textstylefrac{1}{2}} left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)$. (10.11)

Now, we return to the energy tensor (10.5) and shift some terms to rewrite this as:

$4kappa T^{mu } _{nu } =boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{mu tau } F_{nu tau } right)-{textstylefrac{1}{2}} delta ^{mu } _{nu } boverline{kappa }left({textstylefrac{1}{4}} g^{55} boverline{kappa }F^{sigma tau } F_{sigma tau } right)-{textstylefrac{1}{2}} delta ^{mu } _{nu } boverline{kappa }g^{5beta } J_{beta }$. (10.12)

Then, we substitute $g^{5beta } J_{beta }$ from (10.11) into (10.12), and do some further rearranging, including making use of $overline{kappa }^{2} =2kappa /hbar c$, to obtain:

$frac{16kappa }{b^{2} overline{kappa }^{2} } T^{mu } _{nu } =frac{8}{b^{2} } hbar cT^{mu } _{nu } =g^{55} left(F^{mu tau } F_{nu tau } -frac{1}{4} delta ^{mu } _{nu } F^{sigma tau } F_{sigma tau } right)$. (10.13)

If we now set $hbar =c=1$ as well as:

$b^{2} =8$ and $g^{55} =-1$, (10.14)

then (10.13) now reduces, rather fortuitously, to the Maxwell stress-energy tensor:

$T^{mu } _{nu } =-left(F^{mu tau } F_{nu tau } -{textstylefrac{1}{4}} delta ^{mu } _{nu } F^{sigma tau } F_{sigma tau } right)$,  (10.15)

in the Heaviside-Lorentz units that we have been employing from the outset.  The factor $b$ which we have employed all along is now determined to be $b^{2} =8$.  Further, because we have deduced that $g^{55} =-1$ we no longer need to straddle between a timelike and a spacelike fifth dimension: we have deduced that the fifth dimension must be spacelike.  Also, despite the five-dimensional non-symmetry that we started with, the net result is still a symmetric tensor in four-dimensions.  The stress-energy tensor is an important result, because this tensor is underpinned by extensive empirical evidence.

We can then also derive the mixed Ricci tensor corresponding to the stress-energy (10.15).  We start with (10.8), substitute (10.11), and reduce, to obtain:

$16R^{mu } _{nu } =-b^{2} overline{kappa }^{2} left(g^{55} F^{mu tau } F_{nu tau } right)+{textstylefrac{1}{4}} delta ^{mu } _{nu } b^{2} overline{kappa }^{2} left(g^{55} F^{sigma tau } F_{sigma tau } right)$. (10.16)

Clearly, this is also traceless, as it should be.  Further use of $overline{kappa }^{2} =2kappa /hbar c$ with $hbar =c=1$, and $b^{2} =8$ and $g^{55} =-1$ from (10.14), then reduces to:

$R^{mu } _{nu } =kappa left(F^{mu tau } F_{nu tau } -{textstylefrac{1}{4}} delta ^{mu } _{nu } F^{sigma tau } F_{sigma tau } right)$, (10.17)

which is summarized by the traceless field equation $-kappa T^{mu } _{nu } =R^{mu } _{nu }$, as expected.

Finally, in being able to derive the traceless equation (10.15) which among many things tells us that electromagnetic energy propagates at the speed of light, we have solved the essential riddle which concerned Einstein in [Einstein. A., Do Gravitational Fields Play an Essential Part in the Structure of the Elementary Particles of Matter, in “The Principle of Relativity,” Dover (1952).], see equations (1) versus (1a) and (3) therein, which was to find a compatibility between the field equation $-kappa T^{mu } _{nu } =R^{mu } _{nu } -{textstylefrac{1}{2}} delta ^{mu } _{nu } R$ which contains a non-zero scalar trace, and (10.15) and (10.17) above which are scalar-free.   More fundamentally, since (10.15) was derived by rigorously applying the field equation $-kappa T^{mu } _{nu } =R^{mu } _{nu } -{textstylefrac{1}{2}} delta ^{mu } _{nu } R$, we have demonstrated that Einstein’s equation, which one ordinarily applies to trace matter which can be placed at rest, is also fully compatible with, and is indeed the foundation for, the energy tensor of traceless, luminous electromagnetic radiation.