Lab Notes for a Scientific Revolution (Physics) http://jayryablon.wordpress.com Tue, 01 Jul 2008 00:08:15 +0000 http://wordpress.org/?v=MU en Foldy-Wouthuysen, continued http://jayryablon.wordpress.com/2008/06/30/foldy-wouthuysen-continued/ http://jayryablon.wordpress.com/2008/06/30/foldy-wouthuysen-continued/#comments Tue, 01 Jul 2008 00:08:15 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=247

Just for the heck of it, I did a calculation of what happens to the mass matrix M\equiv \beta m during the transformation from the Dirac-Pauli representation to the Newton-Wigner representation via Foldy-Wouthuysen.  This is shown in:

http://jayryablon.files.wordpress.com/2008/06/foldy-wouthuysen.pdf

Not sure where to go from there, but I’ll be away the rest of the week on vacation, so I’ll take another look when I return.

Interested in any further thoughts anyone may have.

Jay

]]> http://jayryablon.wordpress.com/2008/06/30/foldy-wouthuysen-continued/feed/ Jay R. Yablon Might Foldy-Wouthuysen Transformations Contain a Hidden Fermion Mass Generation Mechanism? http://jayryablon.wordpress.com/2008/06/29/might-foldy-wouthuysen-transformations-contain-a-hidden-fermion-mass-generation-mechanism/ http://jayryablon.wordpress.com/2008/06/29/might-foldy-wouthuysen-transformations-contain-a-hidden-fermion-mass-generation-mechanism/#comments Mon, 30 Jun 2008 01:43:57 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=242

I have been looking over the following three links for the Foldy-Wouthuysen transformation from the Dirac-Pauli to the Newton-Wigner representation of Dirac’s equation:

The first shows the calculation itself of this transformation:

I: http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf.

The second, an excellent and lucid exposition of the physics (why this is of interest), is to be found at:

II: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.3209&rep=rep1&type=pdf.

The third, dealing with Zitterbewegung motion and the velocity operator in the Dirac-Pauli representation, is at:

III: http://en.wikipedia.org/wiki/Zitterbewegung.

What I would like to discuss, for the purpose of getting your reactions as to whether I am on a sensible track, is the possibility that a mechanism for generating fermion mass may be hidden in all of this.

I say this in particular because in the Dirac-Pauli representation, the velocity operator is given by:

v^{k} =\alpha ^{k} (1)

where \alpha ^{k} = \gamma ^{0} \gamma^{k} , see reference III.  Further, the eigenvalues of this velocity operator constrain the velocity of the Fermion of be the speed of light, see reference II in the middle of page 3.  This means that the fermion must be massless and luminous, in the Dirac-Pauli representation.  Why this is so, has long been a mystery, and is thought not to make any sense, for obvious reasons.

Now, transform into the Newton-Wigner representation via Foldy-Wouthuysen.  The velocity operator in Newton-Wigner now takes the classical form:

v^{k} =dx^{k} /dt   (2)

where x^{k}  is the position operator.  But even more importantly, Newton-Wigner permits a range of eigenvalues less than the speed of light, and so, the fermions permitted by Newton-Wigner are massless and sub-luminous.

Following this to its logical conclusion, this seems to suggest that somewhere hidden in the Foldy-Wouthuysen transformation, we have gone from a fermion which is massless and luminous, to one which has a finite, non-zero rest mass and travels at sub-luminous velocity.  It seems, then, that it would be important to specifically trace how the velocity operator (1) of the Dirac-Pauli representation with \pm c eigenvalues transforms into the velocity operator (2) of Newton-Wigner which allows a continuous, sub-luminous velocity spectrum, and at the same time, to trace through how the rest mass goes from necessarily zero (with decoupled chiral components), to non-zero with chiral couplings.

By doing so, perhaps one would find a mechanism for generating fermion masses.

One contrast to make here: think about how vector boson masses are generated.  One starts with a Lagrangian in which the boson mass term is omitted entirely.  Then, via a well-knows technique, one breaks the symmetry and reveals a boson mass.  Perhaps the mystery of luminous velocity eigenvalues in the Dirac-Pauli representation is telling us a similar thing: Start out with a Dirac-Pauli Lagrangian in which the mass of the fermion is zero, i.e., without a mass term.  Then, the +/- c velocity eigenvalues make sense.  Transform that into the Newton-Wigner representation.  Somewhere along the line, a mass must appear, because a subliminous velocity appears.

I will, of course, try to pinpoint how this all happens, if it does indeed happen.  But I would for now like some reactions as to the tree up which I am barking.

Thanks,

Jay.

]]> http://jayryablon.wordpress.com/2008/06/29/might-foldy-wouthuysen-transformations-contain-a-hidden-fermion-mass-generation-mechanism/feed/ Jay R. Yablon A New Lab Note: Commutation of Linear Rest Mass with Canonical Position http://jayryablon.wordpress.com/2008/06/19/a-new-lab-note-commutation-of-linear-rest-mass-with-canonical-position/ http://jayryablon.wordpress.com/2008/06/19/a-new-lab-note-commutation-of-linear-rest-mass-with-canonical-position/#comments Fri, 20 Jun 2008 04:49:41 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=240

It has been awhile since my last blog entry, but if you want to check out some my recent wanderings through physicsland, check out sci.physics.foundations, relativity, and research.

Here, I would like to show a rather simple calculation, which may cast a different light on how one needs to think about the canonical commutation relationship \left[x_{j} ,p_{k} \right]=i\eta _{jk} ;\; j,k=1,2,3.  I would very much like your comments in helping me sort this through.  You may download this in pdf form at http://jayryablon.files.wordpress.com/2008/06/linear-mass-commutator-calculation.pdf.

I.  A Known Square Mass Commutation Calculation

 Consider a particle of mass m as a single particle system.  Consider canonical coordinates x_{\mu } , and that at least the space coordinates x_{j} ;\; j=1,2,3 are operators.  If we require that the mass m must commute with all operators, then we must have \left[x_{\mu } ,m\right]=0, and by easy extension, \left[x_{\mu } ,m^{2} \right]=0.  It is well known that the commutation condition \left[x_{\mu } ,m^{2} \right]=0, taken together with the on-shell mass relationshipm^{2} =p^{\sigma } p_{\sigma } and the single-particle canonical commutation relationship \left[x_{j} ,p_{k} \right]=i\eta _{jk} ;\; j,k=1,2,3, where diag\left(\eta _{\mu \nu } \right)=\left(-1,+1,+1,+1\right) is the Minkowski tensor, leads inexorably to the commutation relationship:

\left[x_{k} ,p_{0} \right]=-ip_{k} /p^{0} =-iv_{k}    (1.1)

where v_{k} is the particle velocity (in c=1 units) along the kth coordinate.  I leave the detailed calculation as an exercise for the reader not familiar with this calculation, and refer also to the sci.physics.research thread at http://www.physicsforums.com/archive/index.php/t-142092.html or http://groups.google.com/group/sci.physics.research/browse_frm/thread/d78cbfecf703ff6a.

 I would ask for your comments on the following calculation, which is totally analogous to the calculation that leads to (1.1), but which is done using the linear mass m rather than the square mass m^{2} , and using the Dirac equation written as m\psi =\gamma ^{\nu } p_{\nu } \psi , in lieu of what is, in essence, the Klein Gordon equation m^{2} \phi =p^{\sigma } p_{\sigma } \phi that leads to (1.1).

2.  Maybe New?? Linear Mass Commutation Calculation

 Start with Dirac’s equation written as:

m\psi =\gamma ^{\nu } p_{\nu } \psi .  (2.1) 

Require that:

\left[x_{\mu } ,m\right]=0   (2.2)

 Continue to use the canonical commutator \left[x_{j} ,p_{k} \right]=ig_{jk} .  Multiply (2.1) from the left by x_{\mu } noting that \left[\gamma ^{\nu } ,x_{\mu } \right]=0 to write:

x_{\mu } m\psi =\gamma ^{\nu } x_{\mu } p_{\nu } \psi =\gamma ^{0} x_{\mu } p_{0} \psi +\gamma ^{j} x_{\mu } p_{j} \psi .  (2.3) 

This separates into:

 \left\{\begin{array}{c} {x_{0} m\psi =\gamma ^{0} x_{0} p_{0} \psi +\gamma ^{j} x_{0} p_{j} \psi } \\ {x_{k} m\psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} x_{k} p_{j} \psi } \end{array}\right. .  (2.4)

  Now, use the canonical relation \left[x_{j} ,p_{k} \right]=i\eta _{jk} to commute the space (k) equation, thus:

 \begin{array}{l} {x_{k} m\psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} x_{k} p_{j} \psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} \left(p_{j} x_{k} +i\eta _{jk} \right)\, \psi } \\ {=\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} p_{j} x_{k} \psi +i\gamma _{k} \psi } \\ {=\gamma ^{0} x_{k} p_{0} \psi +mx_{k} \psi -\gamma ^{0} p_{0} x_{k} \psi +i\gamma _{k} \psi } \end{array} .  (2.5)

In the final line, we use Dirac’s equation written as mx_{\mu } \psi =\gamma ^{\nu } p_{\nu } x_{\mu } \psi =\gamma ^{0} p_{0} x_{\mu } \psi +\gamma ^{j} p_{j} x_{\mu } \psi , and specifically, the \mu =k component equation \gamma ^{j} p_{j} x_{k} \psi =mx_{k} \psi -\gamma ^{0} p_{0} x_{k} \psi .

 If we require that \left[x_{\mu } ,m\right]=0, which is (2.2), then (2.5) reduces easily to:

 \gamma ^{0} \left[x_{k} ,p_{0} \right]\psi =-i\gamma _{k} \psi ,  (2.6)

Finally, multiply from the left by \gamma ^{0} , and employ \gamma ^{0} \gamma _{k} \equiv \alpha _{k} and \gamma ^{0} \gamma ^{0} =1 to write:

\left[x_{k} ,p_{0} \right]\, \psi =-i\alpha _{k} \psi .  (2.7) 

If we contrast (2.7) to (1.1) written as \left[x_{k} ,p_{0} \right]\phi =-iv_{k} \phi , we see that the velocity p_{k} /p^{0} =v_{k} has been replaced by the Dirac operator \alpha _{k} , that is, v_{k} \to \alpha _{k} .

3.  Questions

 Here are my first set of questions:

 1)  Is the calculation leading to (2.7) correct, and is (2.7) a correct result, or have I missed something along the way?

2)  If (2.7) is correct, has anyone seen this result before?  If so where?

3)  Now use the plane wave \psi =ue^{ip^{\sigma } x_{\sigma } } so that we can work with the Dirac spinors u\left(p^{\mu } \right), and rewrite (2.7) as:

\left\{\begin{array}{c} {\left(\alpha _{k} -\lambda \right)\, u=0} \\ {\lambda =i\left[x_{k} ,p_{0} \right]} \end{array}\right.  

The upper member of (3.1) is an eigenvalue equation.  Reading out this equation, I would say that the commutators \lambda =i\left[x_{k} ,p_{0} \right] are the eigenvalues of the Dirac \alpha _{k} matrices, which are:

{\bf \alpha }=\left(\begin{array}{cc} {0} & {{\bf \sigma }} \\ {{\bf \sigma }} & {0} \end{array}\right) and {\bf \alpha }=\left(\begin{array}{cc} {-{\bf \sigma }} & {0} \\ {0} & {{\bf \sigma }} \end{array}\right) ,  (3.2)

in the respective Pauli/Dirac and Weyl representations, and that the u are the eigenvectors associated with these eigenvalues \lambda =i\left[x_{k} ,p_{0} \right].  Am I wrong?  If not, how would one interpret this result?  Maybe the commutators \left[x_{j} ,p_{k} \right]=i\eta _{jk} can be discussed in the abstract, but it seems to me that the commutators \lambda =i\left[x_{k} ,p_{0} \right] can only be discussed as the eigenvalues of the matrices \alpha _{k} with respect to the eigenstate vectors u.  This, it seems, would put canonical commutation into a somewhat different perspective than is usual.

Just as Dirac’s equation reveals some features that cannot be seen strictly from the Klein Gordon equation, the calculation here seems to reveal some features about the canonical commutators that the usual calculation based on \left[x_{\mu } ,m^{2} \right]=0 and m^{2} =p^{\sigma } p_{\sigma } cannot, by itself, reveal.

I’d appreciate your thoughts on this, before I proceed downstream from here.

Thanks,

Jay.

]]> http://jayryablon.wordpress.com/2008/06/19/a-new-lab-note-commutation-of-linear-rest-mass-with-canonical-position/feed/ Jay R. Yablon How Precisely can we Measure an Electron’s Heisenberg Uncertainty? (or, How Certain is Uncertainty?) http://jayryablon.wordpress.com/2008/05/08/how-precisely-can-we-measure-an-electrons-heisenberg-uncertainty-or-how-certain-is-uncertainty/ http://jayryablon.wordpress.com/2008/05/08/how-precisely-can-we-measure-an-electrons-heisenberg-uncertainty-or-how-certain-is-uncertainty/#comments Fri, 09 May 2008 01:03:51 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=229

   In a May 24 post Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?, I set forth the hypothesis that the anomalous magnetic moment first characterized by Schwinger, may in fact be a manifestation of the Heisenberg uncertainty relationship, and in particular, that the excess of the uncertainty over \hbar/2 may in fact originate from the same basis as the excess of the intrinsic spin magnetic moment g-factor g, over the Dirac value of 2.  This hypothesis is most transparently written as \Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} =\left(1+\frac{\alpha}{2\pi } +\ldots \right)\frac{\hbar }{2} , where \alpha is the running electromagnetic coupling for which \alpha \left(\mu \right)\to 1/137.036 at low probe energy \mu .  I also pointed out that a crucial next step was to employ a calculation similar to that shown at http://en.wikipedia.org/wiki/Uncertainty_principle#Wave_mechanics, but for a non-Gaussian wavefunction.

   I have now concluded a full calculation along these lines, of the precise uncertainty associated with a particle wavefunction of the general form \psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-V'\left(x\right)} .  (The primes are a convenience used in calculation where we define A\equiv A'+A'*, etc. when calculating expected values, to take into account the possibility of the wavefunction parameters being imaginary.)  While I refer to V'\left(x\right) as an “intrinsic potential,” it is perhaps better to think about this simply as an unspecified, completely-general polynomial in x, which renders the wavefunction completely general.  I have linked an updated draft of my paper which includes this calculation in full and applies it to the hypothesis set forth above, at Heisenberg Uncertainty and the Schwinger Anomaly. While the calculation is lengthy (and took a fair bit of effort to perform, then cross-check), the essence of what is contained in this paper can be summarized quite simply.  So I shall lay out a brief summary below, using the equation numbers which appear in the above-linked paper.

   The essence of the results demonstrated in this paper is as follows.  Start with the generalized non-Gaussian wavefunction:

\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-V'\left(x\right)}   (4.1)

Calculate its uncertainty by calculating its Fourier transform \psi (p) (see (6.1)), by calculating each of its variances (\Delta x)^2 (5.4) and (\Delta p)^2 (7.4), and then by multiplying these together and taking the square root to arrive at the uncertainty.  The calculation is lengthy but straightforward, and it leads to the bottom line result:

 \Delta x\Delta p=\frac{\hbar }{2} \sqrt{1-2A'\left(\frac{dV'}{dB'} \right)^{2} +4B'\frac{dV'}{dB'} } =\frac{\hbar }{2} \sqrt{1-4A'V'\frac{d^{2} V'}{dB'^{2} } +4V'} .   (8.5)

   It is important to emphasize that (8.5) is a mathematical result that is totally independent of the hypothesized relationship of the uncertainty to the intrinsic spin.  So, if you ever been dissatisfied with the inequality of the Heisenberg relationship \Delta x\Delta p\ge {\tfrac{1}{2}} \hbar and wondered what the exact uncertainty is for a given wavefunction, you will find this calculated with precision in sections 4 through 8, and the answer is (8.5) above.  The upshot is that (8.5) above is the precise uncertainty for a wavefunction (4.1) with A’, B’ and V’ all real.  We cannot give a position and momentum with precision, but we can give an uncertainty with precision.  The reasons for having A’, B’ and V’ be real are developed in the paper, but suffice it to say that A’, B’ real is necessary to avert a divergent uncertainty, and if V’ were imaginary rather than real, the uncertainty would always be exactly equal to \hbar/2 .

   Now, with the result (8.5) in hand, we return to the original hypothesis which, if it is true, would require that:

\frac{\Delta x\Delta p}{\hbar /2} =\sqrt{1+4B'\frac{dV'}{dB'} -2A'\left(\frac{dV'}{dB'} \right)^{2} } =\sqrt{1+4V'-4A'V'\frac{d^{2} V'}{dB'^{2} } } =\frac{\left|g\right|}{2} =1+\frac{a}{2\pi } +\ldots    (9.1)

   Using the series expansion for \sqrt{1+x} , we then make the connection:

V'\equiv \alpha /4\pi     (9.5)

   Now, it behooves us to return to the wavefunction (4.1), and use (9.5) to write: 

\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-\frac{\alpha }{4\pi } } ,    (9.6) 

and to rewrite the uncertainty relationship (9.1) as:

\frac{\Delta x\Delta p}{\hbar /2} =\sqrt{1+\frac{1}{\pi } B'\frac{d\alpha }{dB'} -\frac{1}{8\pi ^{2} } A'\left(\frac{d\alpha }{dB'} \right)^{2} } =\sqrt{1+\frac{\alpha }{\pi } -A'\frac{\alpha }{4\pi ^{2} } \frac{d^{2} \alpha }{dB'^{2} } } =\frac{\left|g\right|}{2} =1+\frac{\alpha }{2\pi } +\ldots (9.7)

   Now, let’s get directly to the point: an electron with the wavefunction (9.6), with A' and B' real, will have the uncertainty relationship (9.7), period.  For \alpha =1/137.036, the leading uncertainty term \sqrt{1+\frac{\alpha }{\pi } } =1.00116073607, while the leading anomaly term 1+\frac{\alpha }{2\pi } =1.00116140973.  These two terms differ by just under 7 parts in 10^{-7} .  Therefore, we can state the following:

   Theorem:  For a wavefunction \psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-\frac{\alpha }{4\pi } } with A' and B' real, the uncertainty ratio \frac{\Delta x\Delta p}{\hbar /2} , to leading order in \alpha , differs from the intrinsic Schwinger g-factor g/2 by less than 7 parts in 10^{-7} .

   We have stated this as a theorem, because this is a simple statement of fact, and involves no interpretation or hypothesis whatsoever.  However, in order to sustain the broader hypothesis

\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} \ge \frac{\hbar }{2} ,   (3.4)

we do need to engage in some interpretation.

   First, we define (9.6) as the intrinsic wavefunction of a ground state electron with no orbital angular momentum and no applied external potential.  And, we define (9.7) as the intrinsic uncertainty of this intrinsic wavefunction.  Not every electron will have this wavefunction or this uncertainty or this g-factor, but this wavefunction becomes the baseline electron wavefunction from which any variation is due to extrinsic factors, such as possessing orbital angular momentum or being placed into an external potential, for example, that of a proton.  Thus, our hypothesis (3.4) is a hypothesis about the intrinsic uncertainty associated with the intrinsic wavefunction, and it says that:

   Reformulated Hypothesis:  The intrinsic uncertainty associated with the intrinsic electron wavefunction is identical with the intrinsic g-factor of the anomalous magnetic moment.

   The final section 10 of this draft paper linked above, is in progress at this time.  What I am presently trying to do, is make some sense of what appears to be a “new” type of g-factor \left|g_{{\rm ext}} \right|, emanating from an extrinsic potential (polynomial) V_{{\rm ext}}  in the wavefunction:

\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-V_{int} \left(x\right)-V_{{\rm ext}} \left(x\right)} =Ne^{-\frac{1}{2} A'x^{2} +B'x-\frac{\alpha }{4\pi } -V_{{\rm ext}} \left(x\right)}    (10.1)

This new g-factor is defined in (10.2), and is isolated in (10.3) as such:

\begin{array}{l} {\frac{\left|g_{{\rm ext}} \right|}{2} =\sqrt{1+\frac{1}{\pi } B'\frac{d\alpha +4\pi dV_{{\rm ext}} }{dB'} -\frac{1}{8\pi ^{2} } A'\left(\frac{d\alpha +4\pi dV_{{\rm ext}} }{dB'} \right)^{2} } -\sqrt{1+\frac{1}{\pi } B'\frac{d\alpha }{dB'} -\frac{1}{8\pi ^{2} } A'\left(\frac{d\alpha }{dB'} \right)^{2} } } \\ {\quad \quad =\sqrt{1+\frac{\alpha +4\pi V_{{\rm ext}} }{\pi } -A'\frac{\alpha +4\pi V_{{\rm ext}} }{4\pi ^{2} } \frac{d^{2} \alpha +4\pi d^{2} V_{{\rm ext}} }{dB'^{2} } } -\sqrt{1+\frac{\alpha }{\pi } -A'\frac{\alpha }{4\pi ^{2} } \frac{d^{2} \alpha }{dB'^{2} } } } \end{array}.   (10.3)

In section 10, I have provided my “first impression” of where this new g-factor may fit in, in relation to the Paschen-Back effect, but would be interested in the thoughts of the reader regarding what to make of the above g-factor (10.3) and where it might fit into the “scheme of things.” 

Thanks for listening, and for your thoughts.

Jay.

]]> http://jayryablon.wordpress.com/2008/05/08/how-precisely-can-we-measure-an-electrons-heisenberg-uncertainty-or-how-certain-is-uncertainty/feed/ Jay R. Yablon Heisenberg Uncertainty and Schwinger Anomaly Continued: Draft Paper http://jayryablon.wordpress.com/2008/05/01/heisenberg-and-schwinger-continued-draft-paper/ http://jayryablon.wordpress.com/2008/05/01/heisenberg-and-schwinger-continued-draft-paper/#comments Thu, 01 May 2008 06:45:20 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=226

I have been writing a paper to rigorously develop the hypothesis I presented last week, in a post linked at Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?.  I believe there is enough developed now, and I think enough of the kinks are now out, so you all may take a sneak preview.  Thus, I have linked my latest draft at:

Heisenberg Uncertainty and the Schwinger Anomaly

Setting aside the hypothesized connection between the magnetic anomaly and uncertainty, Sections 4 through 7, which have not been posted in any form previously, stand completely by themselves, irrespective of this hypothesis.  These sections are strictly mathematical in nature, and they provide an exact measure for how the uncertainty associated with a wavefunction varies upwards from \hbar/2  as a function of the potential, and the parameters of the wavefunction itself.  The wavefunction employed is completely general, and the uncertainty relation is driven by a potential V .

This is still under development, but this should give you a very good idea of where this is headed.

Of course, I welcome comment, as always.

Best regards,

Jay.

]]> http://jayryablon.wordpress.com/2008/05/01/heisenberg-and-schwinger-continued-draft-paper/feed/ Jay R. Yablon Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin? http://jayryablon.wordpress.com/2008/04/24/heisenberg-uncertainty-and-schwinger-anomaly-two-sides-of-the-same-coin/ http://jayryablon.wordpress.com/2008/04/24/heisenberg-uncertainty-and-schwinger-anomaly-two-sides-of-the-same-coin/#comments Thu, 24 Apr 2008 05:26:47 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=217

 In section 3 of Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?, I have posted a calculation which shows why the Schwinger magnetic anomaly may in fact be very tightly tied to the Heisenberg inequality \Delta x\Delta p\ge {\tfrac{1}{2}} \hbar .  The bottom line result, in (3.11) and (3.12), is that the gyromagnetic “g-factor” for a charged fermion wave field with only intrinsic spin (no angular momentum) is given by:

\left|g\right|=2\frac{\left(\Delta x\Delta p\right)}{\hbar /2} \ge 2   (3.11)

It is also helpful to look at this from the standpoint of the Heisenberg principle as:

\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} \ge \frac{\hbar }{2}   (3.12)

The draft paper linked above has details of how I got here.  Below, are some ways to think about this:

 First, if (3.11) is true, then the greater than or equal to inequality of Heisenberg says, in this context, that the magnitude of the intrinsic g-factor of a charged wavefunction is always greater than or equal to 2.  That is, the inequality \Delta x\Delta p\ge {\tfrac{1}{2}} \hbar becomes another way of stating a parallel inequality \left|g\right|\ge 2.  We know this to be true for the charged leptons, which have g_{e} /2=1.0011596521859, g_{\mu } /2=1.0011659203, and g_{\tau } /2=1.0011773 respectively. [The foregoing data is extracted from W.-M. Yao et al., J. Phys. G 33, 1 (2006)]

 Secondly, the fact that the charged leptons have g-factors only slightly above 2, suggests that these a) differ from perfect Gaussian wavefunctions by only a very tiny amount, b) the electron is slightly more Gaussian than the muon, and the muon slightly more-so than the tauon.  The three-quark proton, with g_{P} /2=2.7928473565, is definitively less-Gaussian the charged leptons.  But, it is intriguing that the g-factor is now seen as a precise measure of the degree to which a wavefunction differs from a perfect Gaussian.

 Third, (3.11) states that the magnetic moment anomaly via the g-factor is a precise measure of the degree to which \Delta x\Delta p exceeds \hbar /2.  This is best seen by writing (3.11) as (3.12).

Thus, for the electron, \left(\Delta x\Delta p\right)_{e} =1.0011596521859\cdot \left(\hbar /2\right), to give an exact numerical example.  For a different example, for the proton, \left(\Delta x\Delta p\right)_{P} =2.7928473565\cdot \left(\hbar /2\right).

 Fourth, as a philosophical and historical matter, one can achieve a new, deeper perspective about uncertainty.  Classically, it was long thought that one can specify position and momentum simultaneously, with precision.  To the initial consternation of many and the lasting consternation of some, it was found that even in principle, one could at best determine the standard deviations in position and momentum according to \Delta x\Delta p\ge {\tfrac{1}{2}} \hbar .  There are two aspects of this consternation:  First, that one can never have\Delta x\Delta p=0 as in classical theory.  Second, that this is merely an inequality, not an exact expression, so that even for a particle with \Delta x\Delta p\ge {\tfrac{1}{2}} \hbar , we do not know for sure what is its exact value of \Delta x\Delta p.  This latter issue is not an in-principle limitation on position and momentum measurements; it is a limitation on the present state of human knowledge.

 Now, while {\tfrac{1}{2}} \hbar is a lower bound in principle, the question remains open to the present day, whether there is a way, for a given particle, to specify the precise degree to which its \Delta x\Delta p exceeds {\tfrac{1}{2}} \hbar , and how this would be measured.  For example, one might ask, is there any particle in the real world that is a perfect Gaussian, and therefore can be located in spacetime and conjugate momentum space, down to exactly {\tfrac{1}{2}} \hbar .  Equation (3.12) above suggests that if such a particle exists, it must be a perfect Gaussian, and, that we would know it was a perfect Gaussian, if its g-factor was experimentally determined to be exactly equal to the Dirac value of 2.  Conversely, (3.12) tells us that it is the g-factor itself, which is the direct experimental indicator of the magnitude of \Delta x\Delta p for any given particle wavefunction.  The classical precision of \Delta x\Delta p=0 comes full circle, and while it will never return, there is the satisfaction of being able to replace this with the quantum  mechanical precision of (3.12), \Delta x\Delta p=\left|g\right|\hbar /4, rather than the weaker inequality of \Delta x\Delta p\ge {\tfrac{1}{2}} \hbar .

 Fifth, if (3.12) is correct, then since it is independently known from Schwinger that \frac{g}{2} =1+\frac{a}{2\pi } +\ldots , this would mean that we would have to have:

\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} =\left(1+\frac{a}{2\pi } +\ldots \right)\frac{\hbar }{2}   (3.13)

Thus, from the perturbative viewpoint, the degree to which \Delta x\Delta p exceeds {\tfrac{1}{2}} \hbar would have to be a function of the running coupling strength \alpha =e^{2} /4\pi in Heaviside-Lorentz units.  As Carl Brannnen has explicitly pointed out to me, this means that a Gaussian wavepacket is by definition non-interacting; as soon as there is an interaction, one concurrently loses the exact Gaussian.

 Sixth, since deviation of the g-factor above 2 would arise from a non-Gaussian wavefunction such as \psi (x)=N\exp \left(-{\tfrac{1}{2}} Ax^{2} +Bx\right), the rise of the g-factor above 2 would have to stem from the Bx term in this non-Gaussian wavefuction.  In this regard, we note to start, that N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} +Bx\right)dx= \sqrt{2\pi /A} \exp \left(B^{2} /2A\right), for a non-Gaussian wavefunction, versus N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} \right)dx= \sqrt{2\pi /A} for a perfect Gaussian.

 Finally, to calculate this all out precisely, one would need to employ a calculation similar to that shown at http://en.wikipedia.org/wiki/Uncertainty_principle#Wave_mechanics, but for the non-Gaussian N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} +Bx\right)dx= \sqrt{2\pi /A} \exp \left(B^{2} /2A\right) rather than the GaussianN\int \exp \left(-{\tfrac{1}{2}} Ax^{2} \right)dx= \sqrt{2\pi /A} , to arrive at the modified bottom line equation of this Wiki section.  That is the next calculation I plan, but this is enough, I believe, to post at this time.

]]> http://jayryablon.wordpress.com/2008/04/24/heisenberg-uncertainty-and-schwinger-anomaly-two-sides-of-the-same-coin/feed/ Jay R. Yablon Lab Note 6: Operator Decomposition of Intrinsic Spin http://jayryablon.wordpress.com/2008/04/18/lab-note-6-operator-decomposition-of-intrinsic-spin/ http://jayryablon.wordpress.com/2008/04/18/lab-note-6-operator-decomposition-of-intrinsic-spin/#comments Fri, 18 Apr 2008 06:20:30 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=213

I’d like to lay out a nifty little mathematical calculation which allows a “decomposition” of the intrinsic spin matrices s^{i} ={\tfrac{1}{2}} \hbar \sigma ^{i} to include the position and momentum operators x^{i} , p^{i} , i=1,2,3.  To simplify matters, we will employ a Minkowski metric tensor with {\rm diag}\left(\eta _{\mu \nu } \right)=\left(-1,+1,+1,+1\right) so that raising and lowering the space indexes i=1,2,3 is simple and at will, and does not entail any sign reversal.  (This lab note is available in PDF form, with a recent update including a new section 2, at intrinsic-spin-decomposition-11.)

  We start with the general cross product for two three-vectors A and B.  Written in covariant (index) notation:

\left(A\times B\right)_{i} \equiv \varepsilon _{ijk} A^{j} B^{k} .   (1)

One can easily confirm this by taking, for example, \left(A\times B\right)_{3} \equiv A^{1} B^{2} -A^{2} B^{1} .  Now, let’s take the triple cross product \left(A\times B\right)\times C.  We can apply (1) to itself using \left(A\times B\right)^{j} \equiv \varepsilon ^{jmn} A_{m} B_{n} , to write:

\left[\left(A\times B\right)\times C\right]_{i} =\varepsilon _{ijk} \left(A\times B\right)^{j} C^{k} =\varepsilon _{ijk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} .  (2)

 The fact that the crossing of A and B takes precedence over crossing with C is retained in the fact that A_{m} B_{n} sum with \varepsilon ^{jmn} , while C^{k} alone sums into \varepsilon _{ijk} .

 Let us now expand (2) for the component equation for which i=3.  The calculation is as such:

\begin{array}{l} {\left[\left(A\times B\right)\times C\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} } \\ {=\varepsilon _{312} \varepsilon ^{123} A_{2} B_{3} C^{2} +\varepsilon _{312} \varepsilon ^{132} A_{3} B_{2} C^{2} +\varepsilon _{321} \varepsilon ^{231} A_{3} B_{1} C^{1} +\varepsilon _{321} \varepsilon ^{213} A_{1} B_{3} C^{1} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} -A_{3} B_{1} C^{1} -A_{3} B_{2} C^{2} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} B_{1} C^{1} -A_{3} B_{2} C^{2} -A_{3} B_{3} C^{3} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} \left(B\cdot C\right)} \end{array},  (3)

where we have added 0=A_{3} B_{3} C^{3} -A_{3} B_{3} C^{3} to the fourth line.  Now in the final line, we hit an impasse, because B_{3} is sandwiched between the terms we would like to form into the other dot product A\cdot C.  In order to complete the calculation, we must make an assumption that the A_{i} commute with B_{3} , i.e., that \left[A_{i} ,B_{3} \right]=0.  For now, let us make this assumption.

 Therefore, we carry out the commutation in (3), and continue along to write:

\begin{array}{l} {\left[\left(A\times B\right)\times C\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} =A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} \left(B\cdot C\right)} \\ {=B_{3} \left(A\cdot C\right)-A_{3} \left(B\cdot C\right)=B_{3} A_{j} C^{j} -A_{3} B_{j} C^{j} } \end{array}.  (4) 

Generalizing fully, we may now write (4) in two equivalent ways as:

\left\{\begin{array}{c} {\left(A\times B\right)\times C=-A\left(B\cdot C\right)+B\left(A\cdot C\right)\quad \quad } \\ {\varepsilon _{ijk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} =-A_{i} B_{j} C^{j} +B_{i} A_{j} C^{j} } \end{array}\right. .  (5) 

The reader will observe the well-known formula for the cross product.

 Now, let’s take the cross product in which A=x, B=p and C={\bf \sigma }, where x is the position operator about the center of mass, p is the momentum operator, and {\bf \sigma } are the Pauli spin matrices.  We also take into account the Heisenberg canonical commutation relationship between the position and momentum operators, that is, \left[x_{\mu } ,p_{\nu } \right]=i\hbar \delta _{\mu \nu } .  This means that we will have to be careful at the juncture between equations (3) and (4), because the position and momentum operators along the same dimension do not commute.

 So, we return to (3) with A=x, B=p and C={\bf \sigma }, to write:

\left[\left(x\times p\right)\times {\bf \sigma }\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =x_{1} p_{3} \sigma ^{1} +x_{2} p_{3} \sigma ^{2} +x_{3} p_{3} \sigma ^{3} -x_{3} \left(p\cdot {\bf \sigma }\right).  (6)

To take the next step, we want to place p_{3} in front of the x_{i} .  In so doing, we can commute p_{3} with x_{i} for i=1,2.  But, for i=3, we must employ x_{3} p_{3} =p_{3} x_{3} +i\hbar .  Therefore, (6) now becomes:

 \begin{array}{l} {\left[\left(x\times p\right)\times {\bf \sigma }\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =p_{3} x_{1} \sigma ^{1} +p_{3} x_{2} \sigma ^{2} +\left(p_{3} x_{3} +i\hbar \right)\sigma ^{3} -x_{3} \left(p\cdot {\bf \sigma }\right)} \\ {=p_{3} \left(x\cdot {\bf \sigma }\right)-x_{3} \left(p\cdot {\bf \sigma }\right)+i\hbar \sigma _{3} =p_{3} x_{j} \sigma ^{j} -x_{3} p_{j} \sigma ^{j} +i\hbar \sigma _{3} } \end{array},  (7)

lowering the index on i\hbar \sigma ^{3} with {\rm diag}\left(\eta _{ij} \right)=\left(+1,+1,+1\right).  Now all of a sudden, i\hbar \sigma ^{3} has made an unexpected appearance.  Generalizing (7), we may write:

\left\{\begin{array}{c} {\left[\left(x\times p\right)\times {\bf \sigma }\right]=-x\left(p\cdot {\bf \sigma }\right)+p\left(x\cdot {\bf \sigma }\right)+i\hbar {\bf \sigma }\quad \quad \quad } \\ {\varepsilon _{ijk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =-x_{i} p_{j} \sigma ^{j} +p_{i} x_{j} \sigma ^{j} +i\hbar \sigma _{i} } \end{array}\right. ,  (8 )

 This is the also the well-known formula for the triple-cross product, but with an additional term i\hbar {\bf \sigma } emerging from the canonical commutation relationship.  In fact, moving terms, equation (8 ) gives us a way to decompose the intrinsic spin matrix so as to contain the position and momentum, and as we shall also see, orbital angular momentum operators.

 First, we rewrite (8 ) as:

\left\{\begin{array}{c} {i\hbar s=\left[\left(x\times p\right)\times s\right]+x\left(p\cdot s\right)-p\left(x\cdot s\right)\quad \quad \quad } \\ {i\hbar s_{i} =\varepsilon _{ijk} \varepsilon ^{jmn} x_{m} p_{n} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} } \end{array}\right. ,   (9) 

where we have multiplied through by {\tfrac{1}{2}} \hbar and then set s_{i} \equiv {\tfrac{1}{2}} \hbar \sigma _{i} .  This decomposes the intrinsic spin matrix into an expression involving itself, as well as the position and momentum operators.

 Now, using the definition (1) but with A=x and B=p, let’s introduce the orbital angular momentum operator :

l^{j} \equiv \left(x\times p\right)^{j} \equiv l^{j} \equiv \varepsilon ^{jmn} x_{m} p_{n}   (10)

It is easy to see, for example, that l^{3} =x_{1} p_{2} -x_{2} p_{1} .  Using (10), we now rewrite (9) as:

\left\{\begin{array}{c} {i\hbar s=\left(l\times s\right)+x\left(p\cdot s\right)-p\left(x\cdot s\right)\quad \; \; } \\ {i\hbar s_{i} =\varepsilon _{ijk} l^{j} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} } \end{array}\right. ,   (11) 

We see that part of this decomposition includes the cross-product l\times s of the orbital angular momentum with the intrinsic spin.  We may also multiply the lower equation (11) through by \varepsilon ^{mni} and then employ the commutation relationship \left[s^{m} ,s^{n} \right]=i\hbar \varepsilon ^{mni} s_{i} , to write:

\left[s^{m} ,s^{n} \right]=\varepsilon ^{mni} \varepsilon _{ijk} l^{j} s^{k} +\varepsilon ^{mni} x_{i} p_{j} s^{j} -\varepsilon ^{mni} p_{i} x_{j} s^{j} =l^{m} s^{n} +\varepsilon ^{mni} x_{i} p_{j} s^{j} -\varepsilon ^{mni} p_{i} x_{j} s^{j} .  (12) 

Note, we have also made use of \varepsilon ^{mni} \varepsilon _{ijk} =\delta ^{mni} _{ijk} . 

 Equation (11) allows us to decompose the total spin S for a Dirac field \psi , as follows: WORDPRESS DOES NOT LIKE THE INTEGRALS — NEED TO FIX

\left\{\begin{array}{c} {S=\int \left(\overline{\psi }s\psi \right)\, d^{3} x =-{\tfrac{i}{\hbar }} \int \left(\overline{\psi }\left[\left(l\times s\right)+x\left(p\cdot s\right)-p\left(x\cdot s\right)\right]\, \psi \right)\, d^{3} x\quad \; \;  } \\ {S_{i} =\int \left(\overline{\psi }s_{i} \psi \right)\, d^{3} x =-{\tfrac{i}{\hbar }} \int \left(\overline{\psi }\left[\varepsilon _{ijk} l^{j} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} \right]\psi \right)\, d^{3} x } \end{array}\right.  (13)

See Ohanian, H., What is spin?,  equation (18).

More to follow . . .

]]> http://jayryablon.wordpress.com/2008/04/18/lab-note-6-operator-decomposition-of-intrinsic-spin/feed/ Jay R. Yablon John Archibald Wheeler RIP http://jayryablon.wordpress.com/2008/04/14/john-archibald-wheeler-rip/ http://jayryablon.wordpress.com/2008/04/14/john-archibald-wheeler-rip/#comments Mon, 14 Apr 2008 05:19:31 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=207

I just read that John Archibald Wheeler passed away this past day. 

With Wheeler’s passing, we have lost a giant in the world of physics.  Those who have followed my work know that I have been very heavily influenced by Wheeler’s view that all of nature must have a geometrodynamic foundation.  While his dream has not yet been realized, and many have abandoned his dream in favor of other avenues, there remain a few of us quixotic die-hards who will not give up the ghost on Wheeler’s approach, because it is difficult to see how God might have done anything other than to rest nature upon geometry.  In fact, if I were to sum up in one sentence the central thrust of my research, it is to show that Wheeler was right, and that his dream of realizing a geometrodynamic foundation for all of nature has been prematurely abandoned.

Quantum field theory must certainly be credited for its astounding success in describing nature.  But, that does not mean that geometrodynamics cannot work.  It just means we have not yet been able to find out the way in which QFT and geometrodynamics are compatible and, indeed, inseparable.  One day, we will have a recognized theory of quantum geometrodynamics, and when we do, Wheeler will be recognized as the visionary who laid out the program, and who kept Einstein’s dream of a geometric foundation of nature alive for successive generations of physicists, even as many went their separate ways.

Though Wheeler was the third name on Misner, Thorne & Wheeler (and all one needs to say is MTW, and every student of physics knows exactly what book that means), he was the visionary author.  With all the encyclopedic calculation developed in MTW, it is Wheeler who wrote the “perspective” pieces, and always kept a forward eye on where physics should, and will, one day end up. 

Wheeler can teach many lessons to those who become so bogged down in physics calculation or dogma or pedagogy, that they miss or forget that the central aim of physics study is to understand how God created nature, and that the process of uncovering this understanding, fundamentally, is creative, and human.  All else is secondary. 

Farewell to one of the great souls of physics, who in this past day, has returned to his creator, and is now undoubtedly asking that creator directly, all of the questions he asked when he walked among us on this earth.  In his memory, let us rededicate ourselves to keeping alive Wheeler’s geometrodynamic program.

Another WordPress author paid his own triubute at http://mogadalai.wordpress.com/2008/04/13/john-archibald-wheeler-rip/.  I am sure there will be more in the coming days from all over the scientific world.

Jay.

]]> http://jayryablon.wordpress.com/2008/04/14/john-archibald-wheeler-rip/feed/ Jay R. Yablon Lab Note 5: The Central Role in Physics, of the Dirac Anticommutator g^uv=(1/2){gamma^u,gamma^v} http://jayryablon.wordpress.com/2008/04/13/lab-note-5-the-central-role-in-physics-of-the-dirac-anticommutator-guv12gammaugammav/ http://jayryablon.wordpress.com/2008/04/13/lab-note-5-the-central-role-in-physics-of-the-dirac-anticommutator-guv12gammaugammav/#comments Mon, 14 Apr 2008 04:38:56 +0000 Jay R. Yablon http://jayryablon.wordpress.com/?p=204

   I would like to take a break from my current work on Kaluza-Klein, and focus on the central importance to physics of the Dirac anticommutator relationship \eta ^{\mu \nu } \equiv {\tfrac{1}{2}} \left(\gamma ^{\mu } \gamma ^{\nu } +\gamma ^{\nu } \gamma ^{\mu } \right)\equiv {\tfrac{1}{2}} \left\{\gamma ^{\mu } ,\gamma ^{\nu } \right\}, when generalized to a non-zero gravitational field in the form g^{\mu \nu } \equiv{\tfrac{1}{2}} \left\{\Gamma ^{\mu } \Gamma ^{\nu } +\Gamma ^{\nu } \Gamma ^{\mu } \right\}\equiv {\tfrac{1}{2}} \left\{\Gamma ^{\mu } ,\Gamma ^{\nu } \right\}.  In particular, when g^{\mu \nu } \ne \eta ^{\mu \nu } but rather include a gravitational field g^{\mu \nu } (x)=\eta ^{\mu \nu } +\kappa h^{\mu \nu } (x), then also the \Gamma ^{\mu } \ne \gamma ^{\mu } , but rather include a “square root” gravitational field h^{\mu } (x) which may be defined as \Gamma ^{\mu } (x)\equiv \gamma ^{\mu } +\kappa h^{\mu } (x).  Combining all the foregoing, this means that \kappa h^{\mu \nu } \equiv {\tfrac{1}{2}} \kappa \left[h^{\mu } \gamma ^{\nu } +\gamma ^{\mu } h^{\nu } +h^{\nu } \gamma ^{\mu } +\gamma ^{\nu } h^{\mu } \right]+{\tfrac{1}{2}} \kappa ^{2} \left[h^{\mu } h^{\nu } +h^{\nu } h^{\mu } \right].

   We also note that in perturbation theory, non-divergent perturbative effects are, in the end, captured in a correction to the vertex factor given by \overline{{\rm u}}(p)\gamma ^{\mu } {\rm u(p)}\to \overline{{\rm u}}(p)\left(\gamma ^{\mu } +\Lambda ^{\mu } \right){\rm u(p)} operating on a Dirac spinor {\rm u(p)}.  That is, the bare vertex \gamma ^{\mu } becomes the dressed vertex \gamma ^{\mu } +\Lambda ^{\mu } .  By then associating the perturbative \Lambda ^{\mu } with \kappa h^{\mu } just specified,  we raise the possibility that gravitational and perturbative descriptions of nature may in some way be interchangeable.  More to the point: when we consider perturbative effects in particle physics, we may well be considering gravitational effects without knowing that this is what we are doing.  The inestimable benefit of gravitational theory over  perturbation theory is that it is non-linear and exact.  The inestimable benefit of perturbation theory over gravitational theory is that we know something about how to achieve its renormalization.  Perhaps by developing this link further via the vitally-central physical relationship g^{\mu \nu } \equiv {\tfrac{1}{2}} \left\{\Gamma ^{\mu } ,\Gamma ^{\nu } \right\}, we can infuse the exact, non-linear character of gravitational theory into perturbation theory, and the renormalizability of perturbation theory into gravitational theory.  Recognizing that “Lab Notes” is in the nature of a scientific diary, this, in any event, is the starting point for this lab note.

   Now, there are two main directions in which to exploit the connection g^{\mu \nu } \equiv {\tfrac{1}{2}} \left\{\Gamma ^{\mu } ,\Gamma ^{\nu } \right\}, and both need to be considered.  First, we may start with the metric tensor g_{\mu \nu } from a known, exact solution to Einstein’s equation, calculate its associated \Gamma ^{\mu } , and then employ \Gamma ^{\mu } in the Dirac equation, in the form 0=\left(\Gamma ^{\mu } \left(i\partial _{\mu } +eA_{\mu } \right)-m\right)\psi .  Using the Schwarzschild solution a