Comments on: Lab Note 2, Part 2: Gravitational and Inertial Mass, and Electrodynamics as Geometry, in 5-Dimensional Spacetime http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/ Tue, 14 Jul 2009 16:04:49 +0000 http://wordpress.com/ hourly 1 By: Jay R. Yablon http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1315 Jay R. Yablon Fri, 15 Feb 2008 21:54:26 +0000 http://jayryablon.wordpress.com/?p=135#comment-1315 Hi Martin, Thanks for your comment #8. The Cornwell link is in error -- it points to one of my own posts. Can you please provide a corrected link. Thanks. Jay. PS: Update: Martin sent me a corrected link to <a href="http://www.amazon.de/Group-Theory-Physics-Infinite-Dimensional-infinite-dimensional/dp/0121898067/ref=sr_1_2?ie=UTF8&s=books-intl-de&qid=1203246874&sr=8-2" rel="nofollow"> Cornwell</a>, now fixed in #8. Jay. Hi Martin, Thanks for your comment #8. The Cornwell link is in error — it points to one of my own posts. Can you please provide a corrected link. Thanks. Jay.

PS: Update: Martin sent me a corrected link to Cornwell, now fixed in #8. Jay.

]]> By: martinbauer http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1313 martinbauer Fri, 15 Feb 2008 18:16:54 +0000 http://jayryablon.wordpress.com/?p=135#comment-1313 Hi Carl and Jay, First, there s a mistake in my first comment. You can delete the 5th sentence. The question is: are there chiral representations of the Clifford algebra in odd spacetime dimensions (p+q=n odd, no matter how many spaces, times). And the answer is simple : There are none. you can look it up in a lot of books (i.e. <a href="http://www.amazon.de/Group-Theory-Physics-Infinite-Dimensional-infinite-dimensional/dp/0121898067/ref=sr_1_2?ie=UTF8&s=books-intl-de&qid=1203246874&sr=8-2" rel="nofollow"> Cornwell</a> has an exhaustive discussion in the appendix ). You need another, fifth , 4x4 matrix, anticommuting with each $latex \gamma_i $ you already have. It is easy to check that the only possibility you have is to use some multiple of $latex \gamma_5 = a \gamma_0 \gamma_1 \gamma_2 \gamma_3 $ , a some c-number. So $latex \gamma_5 $ is part of your dirac algebra already. Now you try to define projections. To define projection operators you need a 4x4 matrix anticommuting with all $latex \gamma_i $ (also $latex \gamma_5 $ now). There is no such matrix left (= no additional degree of freedom thats not part of the algebra) . That is, because the Clifford algebra provides a basis for the space of 4x4 real matrices. If there'd be another it would be a linear-combination, plus the demand to anticommute with all the others --> it needs to be a product of all $latex \gamma_i's $, therefore $latex \gamma_6 $. you can't use the old $latex \gamma_5 $. It commutes with $latex \gamma_5 $ (jay, all your calculations fail if you put $latex \mu =5 $). Try to define a $latex \gamma_6 $ as you said above will yield $latex \gamma_6 = \gamma_5^2 = 1$ (another way: it will commute with every element of the algebra and is therefore 1 by Schurs lemma). Carl, the expert on geometric algebra must be joking with you, or he is far from beeing an expert. C(4,1) as every other clifford algebra (group) over an odd spacetime does not allow for a projective operator. To have chiral fermions you need even spacetime dimension or orbifolding the extradimension. Best, martin Hi Carl and Jay,

First, there s a mistake in my first comment. You can delete the 5th sentence.

The question is: are there chiral representations of the Clifford algebra in odd spacetime dimensions (p+q=n odd, no matter how many spaces, times). And the answer is simple : There are none.

you can look it up in a lot of books (i.e. Cornwell has an exhaustive discussion in the appendix ).

You need another, fifth , 4×4 matrix, anticommuting with each \gamma_i you already have. It is easy to check that the only possibility you have is to use some multiple of \gamma_5 = a \gamma_0 \gamma_1 \gamma_2 \gamma_3 , a some c-number. So \gamma_5 is part of your dirac algebra already.

Now you try to define projections. To define projection operators you need a 4×4 matrix anticommuting with all \gamma_i (also \gamma_5 now). There is no such matrix left (= no additional degree of freedom thats not part of the algebra) . That is, because the Clifford algebra provides a basis for the space of 4×4 real matrices. If there’d be another it would be a linear-combination, plus the demand to anticommute with all the others –> it needs to be a product of all \gamma_i's , therefore \gamma_6 .

you can’t use the old \gamma_5 . It commutes with \gamma_5 (jay, all your calculations fail if you put \mu =5 ).
Try to define a \gamma_6 as you said above will yield \gamma_6 = \gamma_5^2 = 1 (another way: it will commute with every element of the algebra and is therefore 1 by Schurs lemma).

Carl, the expert on geometric algebra must be joking with you, or he is far from beeing an expert. C(4,1) as every other clifford algebra (group) over an odd spacetime does not allow for a projective operator.
To have chiral fermions you need even spacetime dimension or orbifolding the extradimension.

Best,

martin

]]> By: Jay R. Yablon http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1311 Jay R. Yablon Thu, 14 Feb 2008 17:23:29 +0000 http://jayryablon.wordpress.com/?p=135#comment-1311 Hi Martin: I fixed up your non-printing latex equations in comment #3. I understand all you said, but don't see a problem. Maybe I am missing something, but here is how I see it: I am using a 5-D metric tensor $latex \eta _{{\rm M} {\rm N} } =\left\{\gamma _{{\rm M} } ,\gamma _{{\rm N} } \right\}$. (Though I work with curved $latex g_{{\rm M} {\rm N} } $, the Minkowski metric tensor will suffice for this discussion.) You say ``Generally $latex \gamma ^{5} $ is defined as the product of all ``lower'' gammas.'' Do I take this to mean that in 5-D, I have to define yet another $latex \gamma ^{6} \equiv -i=\gamma ^{0} \gamma ^{1} \gamma ^{2} \gamma ^{3} \gamma ^{5} $ and use this for projections? You say ``This gives a projection operator for even dimensional spacetime and 1 for odd dimensions (it's $latex \gamma ^{5} $ in 5 dimensions).'' Exactly how do you calculate that the projection operator is $latex P=\gamma _{5} ^{2} =1$ in 5 dimensions? I looked at Sundrum's lecture at http://arxiv.org/abs/hep-th/0508134 . Thanks for the link. In his (3.1), he defines $latex \Gamma_5 \equiv -i \gamma_5 $ Presumably, this is to force a spacelike signature upon the fifth dimension, i.e., + - - - -, which seems to be the penchant in 5-D theories. I just go with the flow and use the ``old'' $latex \gamma ^{5} $, which is what accounts for my + - - - + signature, i.e., for a timelike fifth dimension. As my paper shows, there are advantages that arise from a second timelike dimension, including the fact that mass and charge can be characterized by virtue of their "angle" of travel through time and we can specify the Lorentz force law as geodesic motion, etc. Does that difference perhaps account for the problem you are envisioning? It looks like it does, because, e.g., $latex 1 + \gamma_5$ and $latex 1 - \gamma_5$ are very different projection terms than $latex 1+ i\gamma_5$ and $latex 1- i\gamma_5$, and the latter get real funny when you take complex conjugates. You say ``You could use the ``old'' $latex \gamma ^{5} $ to build it,'' and in fact, I do simply use the old $latex \gamma ^{5} $ as the projection in the usual way. Why not? That is: $latex \gamma ^{5} \equiv i\gamma ^{0} \gamma ^{1} \gamma ^{2} \gamma ^{3} $ as always. I define $latex P_{L} \equiv {\tfrac{1}{2}} \left(1-\gamma _{5} \right)$ and $latex P_{R} \equiv {\tfrac{1}{2}} \left(1+\gamma _{5} \right)$, as always. Thus, when we project from a Fermion wavefunction $latex \psi $, we have $latex \psi _{L} =P_{L} \psi \equiv {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi $ and $latex \psi _{R} =P_{R} \psi \equiv {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi $. Is there a reason why I am compelled not to make these choices? The adjoint is $latex \overline{\psi }\equiv \psi ^{{\rm \dag }} \gamma _{0} $, and, of course, $latex \gamma _{5} ^{{\rm \dag }} =\gamma _{5} $. So, $latex \overline{\psi }_{L} =\psi ^{{\rm \dag }} P_{L} \gamma _{0} =\psi ^{{\rm \dag }} {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\gamma _{0} =\psi ^{{\rm \dag }} \gamma _{0} {\tfrac{1}{2}} \left(1+\gamma _{5} \right)=\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)=\overline{\psi }P_{R} $ and $latex \overline{\psi }_{R} =\psi ^{{\rm \dag }} P_{R} \gamma _{0} =\psi ^{{\rm \dag }} {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\gamma _{0} =\psi ^{{\rm \dag }} \gamma _{0} {\tfrac{1}{2}} \left(1-\gamma _{5} \right)=\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)=\overline{\psi }P_{L} $. Then, the Fermion couplings work as follows: When there is no $latex \gamma _{\mu } $ sandwiched, $latex \overline{\psi }_{R} \psi _{L} =\overline{\psi }P_{L} P_{L} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }P_{L} \psi $ and $latex \overline{\psi }_{L} \psi _{R} =\overline{\psi }P_{R} P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }P_{R} \psi $ which uses $latex P_{R} P_{R} =P_{R} $ and $latex P_{L} P_{L} =P_{L} $. Further, $latex \overline{\psi }_{R} \psi _{R} =\overline{\psi }P_{L} P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =0$ and $latex \overline{\psi }_{L} \psi _{L} =\overline{\psi }P_{R} P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =0$, which uses $latex P_{L} P_{R} =P_{R} P_{L} =0$. When there is a $latex \gamma _{\mu } $ sandwiched, then $latex \overline{\psi }_{L} \gamma _{\mu } \psi _{L} =\overline{\psi }P_{R} \gamma _{\mu } P_{L} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } P_{L} \psi $ and $latex \overline{\psi }_{R} \gamma _{\mu } \psi _{R} =\overline{\psi }P_{L} \gamma _{\mu } P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } P_{R} \psi $. Further, $latex \overline{\psi }_{R} \gamma _{\mu } \psi _{L} =\overline{\psi }P_{L} \gamma _{\mu } P_{L} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =0$ and $latex \overline{\psi }_{L} \gamma _{\mu } \psi _{R} =\overline{\psi }P_{R} \gamma _{\mu } P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =0$. All of this seems as it should be, to me. Can you please point out where I might be missing something? Thanks, Jay. Hi Martin:

I fixed up your non-printing latex equations in comment #3. I understand all you said, but don’t see a problem. Maybe I am missing something, but here is how I see it:

I am using a 5-D metric tensor \eta _{{\rm M} {\rm N} } =\left\{\gamma _{{\rm M} } ,\gamma _{{\rm N} } \right\}. (Though I work with curved g_{{\rm M} {\rm N} } , the Minkowski metric tensor will suffice for this discussion.)

You say “Generally \gamma ^{5} is defined as the product of all “lower” gammas.” Do I take this to mean that in 5-D, I have to define yet another \gamma ^{6} \equiv -i=\gamma ^{0} \gamma ^{1} \gamma ^{2} \gamma ^{3} \gamma ^{5} and use this for projections?

You say “This gives a projection operator for even dimensional spacetime and 1 for odd dimensions (it’s \gamma ^{5} in 5 dimensions).” Exactly how do you calculate that the projection operator is P=\gamma _{5} ^{2} =1 in 5 dimensions?

I looked at Sundrum’s lecture at http://arxiv.org/abs/hep-th/0508134 . Thanks for the link. In his (3.1), he defines \Gamma_5 \equiv -i \gamma_5 Presumably, this is to force a spacelike signature upon the fifth dimension, i.e., + – - – -, which seems to be the penchant in 5-D theories. I just go with the flow and use the “old” \gamma ^{5} , which is what accounts for my + – - – + signature, i.e., for a timelike fifth dimension. As my paper shows, there are advantages that arise from a second timelike dimension, including the fact that mass and charge can be characterized by virtue of their “angle” of travel through time and we can specify the Lorentz force law as geodesic motion, etc. Does that difference perhaps account for the problem you are envisioning? It looks like it does, because, e.g., 1 + \gamma_5 and 1 - \gamma_5 are very different projection terms than 1+ i\gamma_5 and 1- i\gamma_5, and the latter get real funny when you take complex conjugates.

You say “You could use the “old” \gamma ^{5} to build it,” and in fact, I do simply use the old \gamma ^{5} as the projection in the usual way. Why not? That is: \gamma ^{5} \equiv i\gamma ^{0} \gamma ^{1} \gamma ^{2} \gamma ^{3} as always. I define P_{L} \equiv {\tfrac{1}{2}} \left(1-\gamma _{5} \right) and P_{R} \equiv {\tfrac{1}{2}} \left(1+\gamma _{5} \right), as always. Thus, when we project from a Fermion wavefunction \psi , we have \psi _{L} =P_{L} \psi \equiv {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi and \psi _{R} =P_{R} \psi \equiv {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi . Is there a reason why I am compelled not to make these choices?

The adjoint is \overline{\psi }\equiv \psi ^{{\rm \dag }} \gamma _{0} , and, of course, \gamma _{5} ^{{\rm \dag }} =\gamma _{5} . So, \overline{\psi }_{L} =\psi ^{{\rm \dag }} P_{L} \gamma _{0} =\psi ^{{\rm \dag }} {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\gamma _{0} =\psi ^{{\rm \dag }} \gamma _{0} {\tfrac{1}{2}} \left(1+\gamma _{5} \right)=\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)=\overline{\psi }P_{R} and \overline{\psi }_{R} =\psi ^{{\rm \dag }} P_{R} \gamma _{0} =\psi ^{{\rm \dag }} {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\gamma _{0} =\psi ^{{\rm \dag }} \gamma _{0} {\tfrac{1}{2}} \left(1-\gamma _{5} \right)=\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)=\overline{\psi }P_{L} . Then, the Fermion couplings work as follows:

When there is no \gamma _{\mu } sandwiched, \overline{\psi }_{R} \psi _{L} =\overline{\psi }P_{L} P_{L} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }P_{L} \psi and \overline{\psi }_{L} \psi _{R} =\overline{\psi }P_{R} P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }P_{R} \psi which uses P_{R} P_{R} =P_{R} and P_{L} P_{L} =P_{L} . Further, \overline{\psi }_{R} \psi _{R} =\overline{\psi }P_{L} P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =0 and \overline{\psi }_{L} \psi _{L} =\overline{\psi }P_{R} P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =0, which uses P_{L} P_{R} =P_{R} P_{L} =0.

When there is a \gamma _{\mu } sandwiched, then \overline{\psi }_{L} \gamma _{\mu } \psi _{L} =\overline{\psi }P_{R} \gamma _{\mu } P_{L} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } P_{L} \psi and \overline{\psi }_{R} \gamma _{\mu } \psi _{R} =\overline{\psi }P_{L} \gamma _{\mu } P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } P_{R} \psi . Further, \overline{\psi }_{R} \gamma _{\mu } \psi _{L} =\overline{\psi }P_{L} \gamma _{\mu } P_{L} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1-\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right){\tfrac{1}{2}} \left(1-\gamma _{5} \right)\psi =0 and \overline{\psi }_{L} \gamma _{\mu } \psi _{R} =\overline{\psi }P_{R} \gamma _{\mu } P_{R} \psi =\overline{\psi }{\tfrac{1}{2}} \left(1+\gamma _{5} \right)\gamma _{\mu } {\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =\overline{\psi }\gamma _{\mu } {\tfrac{1}{2}} \left(1-\gamma _{5} \right){\tfrac{1}{2}} \left(1+\gamma _{5} \right)\psi =0.

All of this seems as it should be, to me. Can you please point out where I might be missing something?

Thanks,

Jay.

]]> By: Jay R. Yablon http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1310 Jay R. Yablon Thu, 14 Feb 2008 16:00:17 +0000 http://jayryablon.wordpress.com/?p=135#comment-1310 Hi Martin: As regards your comment #3, I was curious if you felt that Carl provided a satisfactory answer in #4? (I will try to decipher and fix your "does not parse" equations.) Just saw your new post over at http://martinbauer.wordpress.com/. Your analogy to pirates looking for treasure and not having a great map is a good one. Carl, by the way, has is own blog over at http://carlbrannen.wordpress.com/. Best regards, Jay, Hi Martin:

As regards your comment #3, I was curious if you felt that Carl provided a satisfactory answer in #4? (I will try to decipher and fix your “does not parse” equations.)

Just saw your new post over at http://martinbauer.wordpress.com/. Your analogy to pirates looking for treasure and not having a great map is a good one.

Carl, by the way, has is own blog over at http://carlbrannen.wordpress.com/.

Best regards,

Jay,

]]> By: Lab Note 2, Part 3: Gravitational and Electrodynamic Potentials, the Electro-Gravitational Lagrangian, and a Possible Approach to Quantum Gravitation « Lab Notes for a Scientific Revolution (Physics) http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1309 Lab Note 2, Part 3: Gravitational and Electrodynamic Potentials, the Electro-Gravitational Lagrangian, and a Possible Approach to Quantum Gravitation « Lab Notes for a Scientific Revolution (Physics) Thu, 14 Feb 2008 06:39:51 +0000 http://jayryablon.wordpress.com/?p=135#comment-1309 [...] Note: This Lab Note picks up where Lab Note 2, Part 2, left off, following section 7 thereof.  Equation numbers here, reference this earlier Lab [...] [...] Note: This Lab Note picks up where Lab Note 2, Part 2, left off, following section 7 thereof.  Equation numbers here, reference this earlier Lab [...]

]]> By: carlbrannen http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1307 carlbrannen Mon, 11 Feb 2008 17:58:36 +0000 http://jayryablon.wordpress.com/?p=135#comment-1307 To do the dirty deed in an odd spacetime you can add an extra degree of freedom to the Clifford algebra that is not a part of the algebra. Replace the four gamma matrices with the symbols x, y, z, t to allow me to avoid tex. Add a fifth gamma matrix s. Sure enough, the product xyzst is an object that squares to +1 and commutes with everything in the algebra. It's natural to assume that it is therefore equal to 1 but you don't have to do that. This is entirely up to the user. Personally, I think you should leave it as a non unit and that makes (1 +- xyzst)/2 perfectly good as a projection operator. An alternative construction, and one that I worked with for quite some time, is to begin with an even Clifford algebra, like the Dirac algebra with x, y, z, and t, and add a single extra degree of freedom to it. The extra degree of freedom is assumed to commute with everything else in the algebra and square to 1. That will give you an algebra that is isomorphic to that generated by {x,y,z,s,t} and it explicitly defines the (1+xyzst)/2 projection operator as non trivial. Eventually an expert on the geometric algebra versions of Clifford algebra pointed out to me that what I was doing was isomorphic to the usual C(4,1) so I quit doing that and just used C(4,1) instead. They make perfectly good projection operators. It is also easy to give a physical justification for adding an extra degree of freedom that commutes with everything else in the algebra. Suppose you wish to define a theory that has non commutative degrees of freedom for the spatial components but is Newtonian in that time is not included in with the spatial degrees of freedom. The natural number of dimensions to take is 3. For the spatial degrees of freedom you get the Pauli algebra. The Dirac algebra arose from looking for a square root for the Klein-Gordon equation. The massless Klein-Gordon (which is appropriate for theories that split the particles into massless chiral halves) arises naturally in perfectly Newtonian circumstances without spacetime. For example, earth quake waves are governed by it. In this case, the time coordinate is not a part of the geometry, so when you take a square root of it, there's no (relativistic) motivation to make the time coordinate act like the others. However, the massless Klein-Gordon equation (and any other wave equation) needs two degrees of freedom in time. So if you do not treat time like space you will instead have to split it into two equations that are distinct and coupled. They look something like this: [tex]D \psi = \phi, D\phi = \psi[/tex] where D is your Dirac operator without a gamma_0. So this means that instead of using just psi or phi, what you end up with is a vector that contains \psi and \phi as separate components. To melt them into the same object, just define a commuting vector "T" which satisfies TT = 1, and write [tex]\Psi = \psi + T\phi[/tex]. Then you can take the above two coupled linear equations and combine them into a single linear equation: [tex]D \Psi = 0[/tex] In doing this, you can use the projection operators (1 +- T)/2 to split the \Psi back into its components. It's perfectly physical, correct mathematically, and there is nothing wrong with it, rock on. To do the dirty deed in an odd spacetime you can add an extra degree of freedom to the Clifford algebra that is not a part of the algebra. Replace the four gamma matrices with the symbols x, y, z, t to allow me to avoid tex. Add a fifth gamma matrix s.

Sure enough, the product xyzst is an object that squares to +1 and commutes with everything in the algebra. It’s natural to assume that it is therefore equal to 1 but you don’t have to do that. This is entirely up to the user. Personally, I think you should leave it as a non unit and that makes (1 +- xyzst)/2 perfectly good as a projection operator.

An alternative construction, and one that I worked with for quite some time, is to begin with an even Clifford algebra, like the Dirac algebra with x, y, z, and t, and add a single extra degree of freedom to it. The extra degree of freedom is assumed to commute with everything else in the algebra and square to 1. That will give you an algebra that is isomorphic to that generated by {x,y,z,s,t} and it explicitly defines the (1+xyzst)/2 projection operator as non trivial.

Eventually an expert on the geometric algebra versions of Clifford algebra pointed out to me that what I was doing was isomorphic to the usual C(4,1) so I quit doing that and just used C(4,1) instead. They make perfectly good projection operators.

It is also easy to give a physical justification for adding an extra degree of freedom that commutes with everything else in the algebra. Suppose you wish to define a theory that has non commutative degrees of freedom for the spatial components but is Newtonian in that time is not included in with the spatial degrees of freedom. The natural number of dimensions to take is 3. For the spatial degrees of freedom you get the Pauli algebra.

The Dirac algebra arose from looking for a square root for the Klein-Gordon equation. The massless Klein-Gordon (which is appropriate for theories that split the particles into massless chiral halves) arises naturally in perfectly Newtonian circumstances without spacetime. For example, earth quake waves are governed by it. In this case, the time coordinate is not a part of the geometry, so when you take a square root of it, there’s no (relativistic) motivation to make the time coordinate act like the others. However, the massless Klein-Gordon equation (and any other wave equation) needs two degrees of freedom in time. So if you do not treat time like space you will instead have to split it into two equations that are distinct and coupled. They look something like this:

[tex]D \psi = \phi, D\phi = \psi[/tex]

where D is your Dirac operator without a gamma_0. So this means that instead of using just psi or phi, what you end up with is a vector that contains \psi and \phi as separate components. To melt them into the same object, just define a commuting vector “T” which satisfies TT = 1, and write [tex]\Psi = \psi + T\phi[/tex]. Then you can take the above two coupled linear equations and combine them into a single linear equation:

[tex]D \Psi = 0[/tex]

In doing this, you can use the projection operators (1 +- T)/2 to split the \Psi back into its components. It’s perfectly physical, correct mathematically, and there is nothing wrong with it, rock on.

]]> By: martinbauer http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1304 martinbauer Thu, 07 Feb 2008 11:07:56 +0000 http://jayryablon.wordpress.com/?p=135#comment-1304 Hi Jay, You will not be able to build these projection operators in odd spacetimes. Generally $latex \gamma_5$ is defined as the product of all "lower" gammas. This gives a projection operator for even dimensional spacetime and 1 for odd dimensions (it's $latex \gamma_5^2 $ in 5 dimensions). So $latex P_L = 0, P_R = 1$ or vice versa. You could use the "old" $latex \gamma_5$ to build it, but then again you need $latex \left[\gamma _{5} ,\gamma _{\mu } \right]=0$ to get kinetic terms in the lagrangian which couple left handed fermions to left handed , but you already have $latex \left\{\gamma _{5} ,\gamma _{\mu } \right\}=0$. In your case, its even worse. Cause $latex \left(P_{L} P_{R} \right)={\tfrac{1}{4}} \left(1-\gamma _{5} ^{2} \right)={\tfrac{1}{4}} \left(1-g_{55} \right)$... You can look it up at sundrums lecture : http://arxiv.org/abs/hep-th/0508134 theres is some supersymmetry book also, but i cant remember which one right now. will tell you later. martin Hi Jay,

You will not be able to build these projection operators in odd spacetimes. Generally \gamma_5 is defined as the product of all “lower” gammas. This gives a projection operator for even dimensional spacetime and 1 for odd dimensions (it’s \gamma_5^2 in 5 dimensions).

So P_L = 0, P_R = 1 or vice versa. You could use the “old” \gamma_5 to build it, but then again you need \left[\gamma _{5} ,\gamma _{\mu } \right]=0 to get kinetic terms in the lagrangian which couple left handed fermions to left handed , but you already have \left\{\gamma _{5} ,\gamma _{\mu } \right\}=0.

In your case, its even worse. Cause \left(P_{L} P_{R} \right)={\tfrac{1}{4}} \left(1-\gamma _{5} ^{2} \right)={\tfrac{1}{4}} \left(1-g_{55} \right)…

You can look it up at sundrums lecture : http://arxiv.org/abs/hep-th/0508134
theres is some supersymmetry book also, but i cant remember which one right now. will tell you later.

martin

]]> By: Jay R. Yablon http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1303 Jay R. Yablon Wed, 06 Feb 2008 23:57:31 +0000 http://jayryablon.wordpress.com/?p=135#comment-1303 Hi Martin: I presume the chiral projection operators you refer to are 1 +/- gamma^5? But I am not sure why these would be thought to disappear in an odd-dimensional spacetime. Would you please send me or point out, something that pinpoints the root of the problem you have in mind, and I'll see if perhaps some solution seems apparent. Thanks, Jay. email to: jyablon@nycap.rr.com Hi Martin:

I presume the chiral projection operators you refer to are 1 +/- gamma^5? But I am not sure why these would be thought to disappear in an odd-dimensional spacetime.

Would you please send me or point out, something that pinpoints the root of the problem you have in mind, and I’ll see if perhaps some solution seems apparent.

Thanks,

Jay.
email to: jyablon@nycap.rr.com

]]> By: martinbauer http://jayryablon.wordpress.com/2008/02/06/lab-note-2-part-2-gravitational-and-inertial-mass-and-electrodynamics-as-geometry-in-5-dimensional-spacetime/#comment-1298 martinbauer Wed, 06 Feb 2008 16:19:12 +0000 http://jayryablon.wordpress.com/?p=135#comment-1298 Hi Jay, by tag surfing i stumbled over your blog and i will most likely return. Very interesting ideas and you are certainly able to present them arrestingly. I had not the time to read the whole entry, but you use the 5th gamma matrix to get a 5-dimensional representation of the clifford algebra. We use the same approach in the Randall-Sundrum Mmodel. The problem we encounter is that because of this in odd dimensional spacetime there is no projection operator. So we are not able to define chiral fermions. We deal with this by orbifolding the extra dimension. I don't want to explain this further. Just to draw your attention on the problem... you probably already thought about. keep up the good work. -martin Hi Jay,

by tag surfing i stumbled over your blog and i will most likely return. Very interesting ideas and you are certainly able to present them arrestingly.

I had not the time to read the whole entry, but you use the 5th gamma matrix to get a 5-dimensional representation of the clifford algebra. We use the same approach in the Randall-Sundrum Mmodel. The problem we encounter is that because of this in odd dimensional spacetime there is no projection operator. So we are not able to define chiral fermions. We deal with this by orbifolding the extra dimension. I don’t want to explain this further. Just to draw your attention on the problem… you probably already thought about.

keep up the good work.

-martin

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