# Decomposition of vector potential

Hi,

The vector potential for elctrodynamics, $A_{\mu}$, can be decomposed
$$A_{\mu}\in\mathbf{0}\oplus\mathbf{1}$$
but the $\mathbf{0}$ part we ignore. My question is: do we ignore the scalar part because of experiment, or is it ignored for mathematical reasons I am naive about?

Thanks,

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Ben Niehoff
Gold Member
The 0 part would correspond to longitudinal polarization. The photon cannot have a longitudinal polarization because it is massless. Having a mass and having a 0 polarization state are equivalent statements (because in order to have oscillations along the direction of motion, a particle must have a rest frame). And of course, the photon being massless comes from gauge invariance, and is necessary for charge conservation.

The W and Z bosons get their longitudinal polarizations by combining with the Higgs scalar; effectively the Higgs scalar gives them a third polarization state, which is the same thing as giving them mass. In fact, the photon itself is the Goldstone boson of electroweak symmetry breaking.

Thanks Ben,

For linearized gravity,
$$h_{\mu\nu}\in\mathbf{0}\oplus\mathbf{1}\oplus \mathbf{2}$$
I see how the longitudinal polarization would eliminate spin-0, but how is discarding spin-1 justified?

Thanks,

dextercioby
Homework Helper
Gauge invariance (the linearization of the diffeomorphism invariance) dismisses the 3 a-sym components and the trace.

Thanks dexter,

Now, if one wanted to form a spin-1 field and dismiss the spin-0 and spin-2 components of the (rank-2 tensor) decomposition (I understand it won't correspond to anything) what would be being discarded, e.g. for spin-0 it was longitudinal polarization. What is discarded when we eliminate spin-2 from the decomposition?

Thanks,

dextercioby
Homework Helper
Spin 1 describes electromagnetism. The e-m field is the curvature 2-form of a U(1)/gauge bundle over spacetime, hence must be antisymmetric. There's no way to 'build' electromagnetism with spin 2 fields (the symmetric components of a 2nd rank tensor).

As for linearized gravity, well, GR uses a symmetric tensor (the metric) which, through a linearization, has no spin 1 (antisymmetric components) inside. If one was to start, let's say, with GR in the vierbein-spin-connection formalism, by linearizing the vierbein one indeed obtains a 2-nd rank tensor which has no initial symmetry or antisymmetry. But the spin 1 part of the linearized vierbein doesn't propagate, it doesn't enter the field equations. The linearized action can be re-written only in terms of the symmetric components of the linearized vierbein. So there's no spin 1 in GR.

vanhees71
Gold Member
2019 Award
Masseless particles are different from massive ones in their behavior under Poincare transformations. One can understand this only by a careful study of the unitary representations of the Poincare group, see e.g. my manuscript on QFT about it:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Here, I try to give a short qualitative explanation. Let's start with massive particles. The analysis of the Poincare-group representation lead to a one-particle basis in Fock space given by mass-momentum eigenstates, and spin. The spin is determined by the behavior of the zero-momentum states under rotation. For a massive spin-1 particle these rotations (the "little group") are represented by the s=1 representation, and that means the possible values for the 3-component of the spin are $s_z=\{-1,0,1\}$. That's pretty easy to understand since we are familiar with this already from non-relativistic quantum theory of particles with spin 1 (although usually one treats only particles with spin 0 and 1/2, the former because it's the most simple case, the latter because a broad field of applications for non-relativistic quantum mechanics is atomic, molecular, and condensed-matter physics, where electrons play a very important role, and those are spin-1/2 particles).

The case of massless particles is a bit more complicated. This is due to the fact that first of all there is no restframe for a massless particle. In any inertial frame, it moves with the speed of light, which means that its four-momentum is light-like (i.e., a null vector of the Minkowski metric). Thus the three momentum cannot be 0.

Now you can always take a frame of reference, where the momentum points in a certain direction (usually one takes the z axis for this "standard momentum"). To further classify the representation, you have to figure out which Lorentz transformations keep the standard momentum fixed. Of course, that's the rotations around the direction of the standard momentum, but there are also two more independent one-parameter subgroups. All these transformations together give a group that is isomorphic to ISO(2), i.e., the group of translations and rotation in the two-dimensional Euclidean plane. This group is, contrary to the rotation group (or better said its covering, the SU(2)) not compact, and as is well known from quantum mechanics, the translations give continuous degrees of freedom, namely the "eigenvalues" of the momentum components which are the real numbers for each component. In our case, this would mean that we had an infinite number of spin-like states. A particle with such a strange property hasn't been observed ever. Thus, to discribe massless particles realistically, we have to make sure that the "translations" (in our context better named "null-rotations", because one rotates around a space-like four-momentum) are represented by the trivial representation. If one thinks in terms of classical fields, this leads precisely to the U(1) gauge property of electrodynamics, i.e., massless spin-1 particles are necessarily represented as gauge bosons! The remaining "spin-like degrees of freedom" thus come only from the rotations around the direction of the standard momentum, which is an SO(2) ~ U(1) group. Then, to get the rotations for all three momenta right, one comes to the conclusion, that one can only use the rotations around the standard-momentum direction as factors $exp(i \lambda phi)$ with $\lambda \in \{0,\pm 1/2,\pm 1,\ldots \}$. Of course $\lambda$ is the helicity of the particle. For massive particles this quantity refers to the projection of the spin to the direction of its momentum, but as is clear from the above considerations, for massless particle the concept is somewhat more subtle (see my manuscript for a quite detailed explanation).

Thus a massless particle with spin $s \geq 1/2$ has two spin-like degrees of freedom, no matter of the value of $s$. For $s=1$ that means it has only two helicity states, namely $\lambda=\pm 1$. A massive spin-1 particle has three spin-degrees of freedom according to the possible values of $s_z \in \{0,\pm 1\}$.

Now another issue is the case of massive gauge bosons that became their mass through the Higgs mechanism. In the case of non-Abelian gauge symmetries that's the only way to build Dyson-renormalizable models with massive spin-1 fields, and the electroweak part of the standard model rests upon this discovery by Higgs, Kibble, Anderson, Brout,...

The Higgs mechanism works as follows: You introduce a multiplet of the gauge group in question and give it a negative mass and a quartic self-interaction such that the stable vacuum state is given at a finite vacuum expectation value of this field. If the symmetry under consideration is a global symmetry, you get some massless boson-degrees of freedom, the famous Nambu-Goldstone bosons and some massive particles left. This happens in the chiral limit of the light-quark part in QCD. How many Goldstone bosons and how many massive particles you get in the model depends on the dimension of the group and the dimension of the subgroup that leaves the stable-vacuum field configuration invariant and the dimension of the representation (as a real Lie group!). Take only u and d quarks, which (in the chiral limit) are assumed to be massless. Then one has a symmetry group $\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R$. To describe pions, one can introduce four real scalar fields to represent this ciral group as SO(4) and write down an SO(4)-symmetric lagrangian with negative mass-squared and a four-particle point interaction. That's the linear $\sigma$ model of chiral symmetry. The non-zero vacuum expectation value defines then a direction in the four-dimensional field space, and the remaining symmtry group are the rotations around this axis. Thus the subgroup that leaves the stable vacuum invariant is SO(3). That means, we have 3 massless Goldstone bosons (the pions in the chiral limit). Since we started with a four-dimensional representation, there's one massive boson left, which represents the $\sigma$ meson, which however is a broad resonance rather than a well-defined particle.

If the symmetry is a local gauge symmetry, you can find a special special gauge, the unitary gauge, where all those fields which would make massless Goldstone bosons in the case of a local symmetry, are absorbed into some of the gauge bosons of the group. Each of these gauge bosons become massless. The other bosonic degrees of freedom become also massive and these are the Higgs bosons of the model. In the electroweak standard model the gauge group is $G=\mathrm{SU}(2) \times \mathrm{U}_1$. The group is four-dimensional. In the version with a "minimal Higgs sector" one introduces a doubled of complex scalar fields (i.e. 4 real field-degrees of freedom). The symmetry is broken to the electromagnetic H=U(1) (which is different from the U(1) we started with!). Counting according to our rules leads to dim(representation of the gauge group to which the scalar fields belong)-dim(H)=4-1=3 would-be Goldstone bosons, which are absorbed to three of the four gauge-boson degrees of freedom, i.e., we end up with 3 massive vector bosons and one massless vector boson, which become the W/Z bosons and the photon, respectively. The photon stays massless as it must be. It is of course not a Goldstone boson as claimed in one earlier posting!

So to make sure I am clear. It is possible to write done a field equation for a rank-2 symmetric tensor field that has spin-1, it just won't correspond to E&M or Gravity, without any internal contradictions?

I asked these questions hoping to find out what is possible to write down, not what corresponds to real life, thats why this is so awkward...

Thanks again,

dextercioby
Homework Helper
According to the Lorentz group, spin 1 couldn't come from a symmetric rank 2-tensor, but only either from a vector or an antisymmetric 2-nd rank tensor.

According to the Lorentz group, spin 1 couldn't come from a symmetric rank 2-tensor, but only either from a vector or an antisymmetric 2-nd rank tensor.
But doesn't any symmetric rank-2 tensor decompose into a spin 0, 1, and 2 part under the Lorentz group? Why couldn't spin-1 come from that decomposition? and entertain the thought that the spin-0 and spin-2 parts are ruled out by experiment in this toy model.

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samalkhaiat
But doesn't any symmetric rank-2 tensor decompose into a spin 0, 1, and 2 part under the Lorentz group?
No, rank-2 symmetric tensor decomposes into rank-2 traceless symmetric tensor (i.e., spin-2 object) and a scalar (i.e., spin-0 object which in this case is just the trace of the original tensor)
Why couldn't spin-1 come from that decomposition?
$$T_{ab}= T_{ab}-(1/n)\eta_{ab}T + (1/n)\eta_{ab}T$$

There is no spin-1 part in this decomposion.

Ben Niehoff
Gold Member
Is "spin 1" even the correct way to characterize a rank 2 antisymmetric Lorentz tensor? In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. According to the Wiki page:

http://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group

a generic 2-form field corresponds to the $(1,0) \oplus (0,1)$ representation. I guess this is a variety of spin 1. (Generally in higher dimensions one doesn't speak of "spin 1", "spin 2", etc., because the representation theory needs something more specific than that).

Jfy, in string theory one has a massless rank 2 tensor field $E_{\mu\nu}$ with no imposed symmetry properties. We decompose this field into three parts:

$$E = \phi \oplus B \oplus G$$

where $\phi$ is the trace part, B is the antisymmetric part, and G is the symmetric traceless part. Each of these components has different equations of motion. They also have gauge symmetries (since they are massless).

One can devise gravity theories that make use of these additional pieces. $\phi$ is called the dilaton, and appears together with G in scalar-tensor theories, such as Brans-Dicke. It is possible to interpret the B field as a kind of torsion, which we don't typically use in gravity theories, but we could, I suppose.

dextercioby
Homework Helper
But doesn't any symmetric rank-2 tensor decompose into a spin 0, 1, and 2 part under the Lorentz group? Why couldn't spin-1 come from that decomposition? [...]
No, the FULL non-symmetric rank 2 tensor decomposes into spin 0,1 and 2. But I've told you that essentially there's no non-symmetric rank 2 tensor, not even the linearized vierbein. Antisymmetric is electromagnetism, symmetric and traceless is linearized gravity (Pauli-Fierz field).

Thank you everyone again,

I will work hard the next couple of hours and try and get this deeply understood. I can tell by the redundancy and declarative nature of the posts there isn't any ambiguity here. I just have one more thing to clear up then. In my Maggiore book Modern Introduction to Relativisitic Quantum Field Theory there is a part in Chapter 2 page 23 where it says the following:
Since we have identified the trace $S$ with a $\mathbf{0}$ and $A^{\mu\nu}$ with $\mathbf{1}\oplus\mathbf{1}$, comparison with eq. (2.41) shows that the nine components of a symmetric traceless tensor $S^{\mu\nu}$ decompose, from the point of view of spatial rotations, as
$$S^{\mu\nu}\in\mathbf{0}\oplus\mathbf{1}\oplus \mathbf{2}.$$
This has been my reference book for QFT recently and this quote led me to believe that spin-1 could be found in rank-2 symmetric tensors. Where is my misinterpretation coming from?

EDIT: not to push this too far either, but here is another line that seems confusing:
In general, a symmetric tensor with $N$ indices contains angular momenta up to $j=N$.
This is on page 24.

EDIT2: Here are pages 22-24
http://www.scribd.com/doc/56578591/12/Decomposition-of-Lorentz-tensors-under" [Broken]

Thanks again,

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Ben Niehoff
Gold Member
Unfortunately, it does not let me quote text that is inside your quote box. However, I think the key phrase is "from the point of view of spatial rotations". So it looks like he is taking the symmetric, traceless part of a Lorentz tensor and then looking at how it transforms under just spatial rotations. Without context, I'm not sure why he's doing that.

dextercioby
Homework Helper
I see what you mean. You've got us on a wrong track. We've explained you from the point of view of the Lorentz group only, not from SO(3). Indeed, as Willard Miller in his group theory book nicely treats it, finite dimensional group representations of the Lorentz group (even the irreducible ones) can be further reduced/decomposed with respect to the spatial rotation group only. The theory of reduction indeed allows for a <spin> of 1 to enter the decomposition, but that <1> is not from the Lorentz group, but from SO(3).

Componentwise: 16 = 1 + 6 + 9 from the Lorentz group perspective only.
Futher we decompose wrt SO(3): 9 = 1 + 3 + 5. The 9s are identified through an isomorphism of representation spaces, since they are both 9-dimensional.

Great,

Thanks a lot Dexter and Ben. So when we say that the electromagnetic field decomposes as spin-1 under the Lorentz group, we are not even referring to the same spin-1 from, say, the decomposition of a symmetric rank-2 tensor under SO(3)? That is, they do not correspond to the same thing physically, they just got stuck with the same name?

Thanks,

dextercioby
A 4-vector which is irreducible under SO(1,3) is reducible under SO(3), as $\mathbb{4=1+3}$. Formula 2.41 contains the decomposition of a tensor product of 2 4-vectors under SO(3), while 2.36 contains the same decomposition, but under SO(1,3), where the 4 vectors are irreducible to start with.