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We use a four-sided diamond to represent the fourdimensional biquatemion gradient 18Data Matrix Barcode barcode library on .netUsing Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications.0== - - + L i n - ,Barcode recognizer in .netUsing Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications.(3.29)Bar Code barcode library in .netusing vs .net crystal toinclude bar code in asp.net web,windows applicationand as we know from experiment that tight has spin 1, we let the wavefunction be a pure biquatemion (Le., a vector wavefunction)Control data matrix 2d barcode image for visual c#.netusing .net framework todeploy barcode data matrix in asp.net web,windows applicationF== Linifn'Control datamatrix 2d barcode size on .netto produce datamatrix 2d barcode and ecc200 data, size, image with .net barcode sdk(3.30)Data Matrix 2d Barcode generating with visual basic.netgenerate, create data matrix 2d barcode none in visual basic.net projectswhere i is the ordinary commutative imaginary number, not to be confused with iI, i2, and i 3 , the generalized noncommutative imaginary quantities of quatemion theory. Notice that 2 1 82 (3.31) 0 = 0 0 = \1 - e2 8t 2 = 0,Incoporate qrcode for .netusing barcode encoder for .net framework crystal control to generate, create qrcode image in .net framework crystal applications.where 0 is the D' Alembertian. 19 So if we write (3.24) in the form .net Framework pdf417 2d barcode printer with .netusing barcode writer for visual .net control to generate, create pdf417 image in visual .net applications.n OOF=memeF,USS Code 39 integrating in .netusing barcode maker for .net vs 2010 control to generate, create code 3/9 image in .net vs 2010 applications.it is tempting to try the factorized equation GTIN - 12 barcode library in .netgenerate, create gs1 - 12 none for .net projects(3.32)Render code 2/5 with .netuse .net framework crystal standard 2 of 5 implementation todraw industrial 2 of 5 on .netOF=pF,Attach upc code for vbusing asp.net website crystal todeploy ucc - 12 for asp.net web,windows applicationwhere Control upc-a supplement 2 size in .net upc a size in .net(3.33) (3.34)Visual Studio .NET (WinForms) Crystal ean13+2 writer for visual basicgenerate, create ean-13 none with visual basic.net projectsP==r;'Upc A integration in c#use visual studio .net (winforms) crystal upc a integrating toinclude upc code for .net c#But (3.33) is not covariant. Under a Lorentz transformation, 0 transforms as 20 a four-vector: that is, U out, where U is a biquatemion such that OU = 1.Control barcode 128 data with .net code 128 code set c data for .net(3.35)Encode gtin - 13 in .netusing barcode printing for rdlc report control to generate, create upc - 13 image in rdlc report applications.Now then, to tum OF into 0' fI, we must multiply it on the left by U, on the right by T, and F' must be obtained from F by F' = U* FT, where T is still to be determined: UOFTCreate qr code on vbgenerate, create qr code jis x 0510 none on vb projects= UO utU* FT Control gs1128 image with c#use vs .net gs1128 generating torender with c#.net=O'F'.---1(3.36)ISIn Appendix C, we only talk about the quatemion gradient (i.e., a gradient when the coefficients of the quaternion 'R = Xo + Ln inxn describing a point in four dimensional space are all real). For Minkowsky's space-time. we use biquaternions (last section of Appendix C) (i.e., use complex coefficients, with Xo = ct and Xl = ix, X2 = iy, X3 = iz in the previous expression for'R). Then it is convenient to define the biquaternion gradient as i multiplied by the gradient defined in Appendix C. 19Sometimes the D' Alembertian is written as 0 2 , perhaps because it is composed of square derivatives (i.e., second derivatives). But we feel that the superscript 2 is an unnecessary emphasis, for a square is already a square. 20See Appendix C or [506] or 9 of Lanczos's book on mechanics [386].EXTREME QUANTUM THEORY OF LIGHT WITH A TWIST Then we see that regardless of what T is, the right-hand side of (3.33) does not transform properly unless U is a quaternion with real coefficients rather than a biquaternion for U" to be equal to U. But if U* = U, the transformation can only be an ordinary spatial rotation (Le., the time component will be left unchanged).21 If our wave equation is to transform accordingly with respect to a general Lorentz transformation that also transforms the time, we must have another biquaternion multiplying F on the right-hand side of (3.33) (3.37) where=uxut,(3.38)(Le., X must transform like 0). Moreover, if we multiply (3.37) by we must recover (3.24); that is,0 from the left,OOF = JLOXF JL 2 F. (3.39) (3.40) (3.41)Then OX =JL and X must be given by Let us define A such that F = OA; then the wave equation we are looking for is OF = JL2 A,(3.42) (3.43)OA =:F. These are LanclOs's equations. 22 Instead of a single biquaternion wavefunction candidate F, we were forced by relativity to have two: F and A. To find out how they transform, we must now determine T. We want F to remain a pure biquaternion in all Lorentz frames; otherwise, this formulation will not be covariant. The simplest way for this to hold is if our F is what is called a six-vector. For instance, T cannot be U; otherwise, F will transform as the conjugate of a four-vector and will develop a scalar component in some Lorentz frames. What we want is to find a T such that for a given biquaternion 2 = Zo + E~=l inzn, the same biquaternion 21 = U* 2T in a different Lorentz frame will have its scalar component unchanged (Le., z~ = Zo must be a true scalar). The simplest solution (i.e., where F is a six-vector) turns out23 to be T = ut21 See Appendix C. 22 Actually. Lanczos's equations are more general in that the wavefunction :F does not need to be a pure biquaternion and can transform in a different way. Lanczos's equations can describe all scalar. all vector. and all pseudovector particles. 23This can be seen by noticing that (3.35) also implies that Ua = 1. Taking the complex conjugate of the former equation and of (3.35). we find that U t U 1 and U* ut = 1. Then Zo is not changed by