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Spin of a p-form |
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| Dec11-07, 05:11 AM | #1 |
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Spin of a p-form
Spin of a 1-form (e.g. electromagnetic field) is 1.
What is the spin of a p-form? (My first guess is p, but I know that higher p-forms appear in physically interesting supersymmetric theories, whereas spins higher than 2 do not appear there.) |
| Dec12-07, 11:38 AM | #2 |
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I guess one important point here is that it doesnīt make much sense to talk abount spins higher than 2 in any elementary particle field as for example is pointed out in weinberg vol. 1 ( the chapter about infrared behaviour), as they seem to have any interactions only in the very high energy regiem.
But anyways i guess that your guess is right that for every spinor index you get 1/2 spin. And for every lorentz index you get spin 1 as worked out in weinberg :) The thing in susy theories is that your forms are not scalar valued, they are grassmann and scalar valued i guess thatīs the reason isnīt it ? |
| Dec13-07, 04:38 AM | #3 |
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| Dec20-07, 10:16 PM | #4 |
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Spin of a p-form
p forms do not have spin-p. I'm actually not sure what their spin is, but a spin-s field is represented by a SYMMETRIC tensor of rank-s. And forms are antisymmetric. So they don't have the right number of degrees of freedom.
Higher p-forms only exist in higher dimensions: in simply connected D=4, you can only write down 1-forms, which are equivalent to 1-tensors (EM field). You start to have higher p-form fields in D=10,11 Supergravity theories, for example. But as you say, these objects are not higher spin. I'm not an expert on these things, but I believe you can describe these higher p-form fields in terms of branes in string theory, where the usual "point particle" description fails, and all bets are off. But as I said, I'm not an expert... |
| Dec21-07, 04:26 AM | #5 |
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| Dec21-07, 10:44 AM | #6 |
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However, it unfortunately doesn't answer your question - rather, these solitons are the things that COUPLE to the form-fields, not the form-fields themselves. Let me try again: in higher dimensions (where nontrivial form fields exist), the reps of the Lorentz group are not given by a single half-integer, and so talking about "spin" of these objects is misleading. I think that's the right way to think about it - the little group allows for more complicated representations than just the usual ones in D=4. Again, I'm not sure I'd be willing to invest money in this, but maybe someone with more experience than me can work out the bugs. As far as susy is concerned: what is important is that the form fields are bosonic (all indices are vector-indices). |
| Dec21-07, 01:57 PM | #7 |
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I was just reading John Terning's excellent text "Modern Supersymmetry" where he talks about supergravity in chapter 15. There, he looks at the sugra spectrum in D=10,11 dimensions and gets p-forms, as he should. But later he looks at what these fields like like to a four-dimensional observer (someone living on a 3-brane, for instance). He finds that the massless modes (lowest KK-states) coming from the p-form fields are all spin-1 or spin-0. There are several of them(!) but they all have these lower spins. This is nothing more than the KK-paradigm that usually happens.
So I stand by what I said above, that you can't talk about spin in the usual way for higher-dimension fields. But when looking at these things from the low-energy (4D) point of view, you see these objects as spin-1 and spin-0. The only question left is what does the 4D picture look like for form fields. Once again, you can have 0-form fields (s=0) and 1-form fields (s=1). 3(4)-form fields are dual to 1(0)-form fields, so they're also s=1(0). The only question is: what about 2-form fields. My GUESS is that these are all exact - that is, they can be written as total exterior derivatives (coboundaries). This is true if the 2nd cohomology group of Minkowski space is trivial. Is it???? |
| Jan2-08, 04:23 AM | #8 |
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I think that the notion of spin is a physical (not purely mathematical) property NOT uniquely determined by its kinematic geometric properties. (For example, the electromagnetic field F=dA is a 2-form, but its spin is 1. A less familiar example is a symmetric second rank tensor which, in general, contains components of spin 0, 1, and 2. If this tensor is further restricted to satisfy the Einstein equation, then only spin 2 components remain.) Instead, spin is determined by dynamics, i.e. by the Lagrangian. More precisely, spin of a field is determined by its contribution to the total conserved angular momentum, which is determined by the Lagrangian. Since p-forms in supersymmetric theories couple in the same way as the electromagnetic field (the Lagrangian is quadratic in curvature F), which is very different from that of gravitational field (linear in curvature R), I conjecture that all p-forms in supersymmetric theories have spin 1.
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| Jan2-08, 04:41 AM | #9 |
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One of the interesting things about spin-1/2 and massless spin-1 is that both have 2 degrees of freedom.
And I believe that they both have the same geometric phase (or Berry phase); that is, when you send them through a series of direction changes and back to their original orientation, in such a way that the path encloses a spherical area of S ster radians, they both pick up a complex phase of exp( i S / 2 ). This is one of the things that makes me wonder if they really are that different from each other. |
| Jan2-08, 06:07 AM | #10 |
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| Jan10-08, 06:48 AM | #11 |
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http://lanl.arxiv.org/abs/hep-th/9912164 It is true that dynamics is important, but it is not true that all p-forms in supersymmetric theories have spin 1. Instead, their spin may be 1 or 0, or they may carry no dynamical degrees of freedom at all. In a specific supersymmetric 4-dimensional theory discussed there we have: 3-form: no dynamical degrees of freedom 2-form: s=0 1-form: s=1 The 4-form does not appear in this supersymmetric theory, but a 4-form in 4 dimensions is dual to a scalar, which means that in any 4-dimensional theory a 4-form is either nondynamical or has spin s=0. |
| Jan10-08, 07:43 AM | #12 |
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"D=11 is the maximal number of dimensions of a supersymmetric theory that does not contain massless fields with spins higher than 2." Does the word "spin" in that sentence referres only to the 4-dimensional view of the higher dimensional theory? It makes sense if one visualizes spin as a rotation in the 2-plane normal to the spatial direction of particle motion. Only in 3 space dimensions such a plane is unique. |
| Jan11-08, 11:07 AM | #13 |
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But I'm not sure what you meant by your earlier post of "physical" vs "mathematical" spin. Spin is the quantum number that tells you what representation of the rotation group you are in. It is "physical" in the sense that ANY quantum number is physical! But that quantum number is only defined in D=4, since in higher dimensions, the representation is no longer simply a half-integer. This thought shouldn't bother anyone too much: in 3 dimensions, you can have "anyons" which have "spin" any real number! To CarlB's post: D=4 is a funny world, indeed! It is where massless degrees of freedom always have only 2 components, irregardless of their spin. It is also the "critical dimension" where the Wilson-Fisher fixed point of condensed matter becomes the trivial fixed point. Maybe G-d just loves the number 4! ![]() Comments??? Complaints??? |
| Jan14-08, 04:17 AM | #14 |
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| Jan14-08, 09:33 AM | #15 |
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| Jan17-08, 05:45 AM | #16 |
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Which page is it? Thanks. |
| Jan17-08, 09:04 PM | #17 |
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