How do we know the value of spin?

  • Thread starter Kyleric
  • Start date
  • #1
12
0
"Experimental evidence like the hydrogen fine structure and the Stern-Gerlach experiment suggest that an electron has an intrinsic angular momentum, independent of its orbital angular momentum. These experiments suggest just two possible states for this angular momentum, and following the pattern of quantized angular momentum, this requires an angular momentum quantum number of 1/2." http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html

So if spin is determined by experiment, how do we know the graviton has spin 2 if none have been observed?

One argument I read about is that it is a tensor field of rank 2. But how do we know that a tensor field of rank n has spin n? Is it just by inductive reasoning, because experiment showed that rank 1 has spin 1 and rank 0 has spin 0?
 

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
I'm not familiar with the technical details of all of what I'm saying here, so it's possible that some of this is inaccurate, but I'll try anyway.

The question "how do we know that a graviton has spin 2" is a bit strange, since the spin is one of the things that identifies a particle species. Spin 2 is part of the definition of "graviton".

The connection between spin 2 and gravity was made by Feynman and Weinberg, who both found (independently) that any theory of particles that (when when described in terms of the individual terms of a perturbation theory expansion) interact by exchanging spin 2 particles, must have general relativity as a low energy limit.

If you're given a field that satisfies an equation derived from a relativistically invariant Lagrangian, the way to find out what spin quantum number it corresponds to, is to use Noether's theorem to find the conserved quantities associated with invariance under rotations around each of the three mutually orthogonal spatial axes. This way you find the spin component operators, and if you compute S12+S22+S32, you should find it to be equal to s(s+1) times the identity operator, for some non-negative integer s. This number is the spin associated with the field.
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
13,046
594
Fredrik, spin is built within the field, you needn't write down a Lagrangian, field equations, Noether theorem or any other thing to find out its spin.

If I write down [itex] A_{\mu} [/itex], a covector on flat Minkowski space-time, I know that, when properly quantized, a field theory built with it describes particles of spin 1.

Einstein showed that the free limit of GR is described by a symmetric tensor field of rank 2. When quantum mechanics was developed, spin 2 was found to be described by symmetric tensor field of rank, guess what, 2. That's how gravity, when quantized in the free limit, has massless quanta of spin 2, called gravitons.

Feynman and Weinberg were not the only ones (and I'm almost sure not the first either) to show the converse of Einstein's work, namely that general relativity described by the H-E action plus the cosmological term is the only consistent self-coupling of a massless spin-2 tensor field (the so-called Pauli-Fierz field).
 
  • #4
12
0
If I write down [itex] A_{\mu} [/itex], a covector on flat Minkowski space-time, I know that, when properly quantized, a field theory built with it describes particles of spin 1.

How do you know that it describes particles of spin 1? From experiment, or is there are mathematical proof?
 
  • #5
970
3
Feynman and Weinberg were not the only ones (and I'm almost sure not the first either) to show the converse of Einstein's work, namely that general relativity described by the H-E action plus the cosmological term is the only consistent self-coupling of a massless spin-2 tensor field (the so-called Pauli-Fierz field).

So there can be no massless spin-2 particles besides the graviton?

Also is a massless spin-2 meson possible? For example the binding energy of a quark/antiquark pair could be so tight that it cancels out the rest mass and internal kinetic energy of the quarks, giving a composite mass of zero, and the spin would be 2 from the two spin 1/2's coming from each quark, and 1 from internal orbital angular momentum between the quarks? Or does this only apply to fundamental particles and not composites?
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,046
594
How do you know that it describes particles of spin 1? From experiment, or is there are mathematical proof?

There's a mathematical proof of course, based on the representation theory of the restricted Poincare group.
 
  • Like
Likes Cesar Romani
  • #7
dextercioby
Science Advisor
Homework Helper
Insights Author
13,046
594
So there can be no massless spin-2 particles besides the graviton?

No.

Also is a massless spin-2 meson possible? For example the binding energy of a quark/antiquark pair could be so tight that it cancels out the rest mass and internal kinetic energy of the quarks, giving a composite mass of zero, and the spin would be 2 from the two spin 1/2's coming from each quark, and 1 from internal orbital angular momentum between the quarks? Or does this only apply to fundamental particles and not composites?

That doesn't sound feasible to me, not even on a theoretical basis. You can't add 2 spins and orbital angular momentum to end up with pure spin. The spin of elementary particles (as per Wigner's definition) is dictated by the symmetry of the spacetime they could live in and its reconciliation with quantum mechanics.
 
  • #9
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
Fredrik, spin is built within the field, you needn't write down a Lagrangian, field equations, Noether theorem or any other thing to find out its spin.
Ah, yes, it temporarily slipped my mind. Thanks for the reminder.
 
  • #10
970
3
Here's how I look at it, although I've got it wrong somewhere because I don't get the right results:

1) Law of physics must be the same (i.e., Lorentz invariant) in any uniform velocity frame
2) Euler-Lagrangian equations are Lorentz invariant, provided the Lagrangian (i.e., Lagrangian density) is too.
3) So how to form scalar Lagrangians from fields? Contraction of two vectors is obvious, but it seems that the Lorentz group is not fundamental: it is the product of two SU(2)s. So instead of the vectors, a direct product of spinors |L>x|R> should be the building blocks. You form your Lorentz invariants by transforming a field |L>x|R> and taking the dot product with a similarly transformed adjoint field: (<L2|x<R2|).(|L1>x|R1>)=(<L2|L1>)(<R2|R1>). However, irreducible representations transform by themselves, so you can just use the irreducible parts to construct your fields and take the dot product with just those. The irreducible parts of SU(2)xSU(2) are just the irreducible parts of |j/2>x|k/2> with j and k integral.
4) So the spin is just what you can get from the direct product of SU(2)'s, and that is always half integer or integral. So for example if you set j=1 and k=1, you get: 1/2x1/2=0+1, so a spin 1 particle would have 3 components (in an irreducible representation) and a spin 0 particle would be a scalar with 1 component. However, a spin 2 particle would have 5 components, which is not the same as a symmetric 2nd rank tensor. So something is clearly wrong.

Also I believe each of the |L> and |R> fields are Grassman, but I don't remember what effect that has or why.
 
Last edited:
  • #11
dextercioby
Science Advisor
Homework Helper
Insights Author
13,046
594
RedX said:
4) So the spin is just what you can get from the direct product of SU(2)'s, and that is always half integer or integral. So for example if you set j=1 and k=1, you get: 1/2x1/2=0+1, so a spin 1 particle would have 3 components (in an irreducible representation) and a spin 0 particle would be a scalar with 1 component. However, a spin 2 particle would have 5 components, which is not the same as a symmetric 2nd rank tensor. So something is clearly wrong.

You're mixing dimensions. Indeed, in euclidean 3d space, spin 2 is described by an SU(2) spinor with 5 components. That corresponds to a traceless, symmetric (2,0) tensor field in R^3.

In 4 dimensions, however, spin 2 is described by an SL(2,C) spinor with 9 components. They correspond to a traceless, symmetric (2,0) tensor field in M_4.
 

Related Threads on How do we know the value of spin?

Replies
38
Views
3K
Replies
15
Views
451
  • Last Post
2
Replies
35
Views
35K
  • Last Post
2
Replies
27
Views
4K
Replies
5
Views
6K
  • Last Post
Replies
1
Views
628
Top