# I Spin of particles and relativity

1. Jul 17, 2017

### davidge

In Carrolls notes on General Relativity, it is said that the general formula for finding the spin number of a particle is $$\frac{360°}{\theta}$$ where $\theta$ is the angle of rotation, after which the state of the field describing the particle returns to its original state polarization. He argues that the graviton (if it exists) should have spin-2, because after a $180°$ rotation the polarization of the field describing the plane wave returns to its starting state.

My question is, for particles which have spin zero, it means that their fields never return to their original state?

2. Jul 17, 2017

### Staff: Mentor

This is a heuristic, which only applies to particles with nonzero spin.

I'm not sure why you put the word "polarization" here. The point is that, for particles with nonzero spin, a rotation changes the state, but there is some rotation angle, dependent on the spin, at which the state transformation induced by the rotation is the identity again (I say "again" because a rotation by an angle of zero is the identity, so the behavior of rotation transformations is periodic).

No, it means that a rotation by any angle is equivalent to the identity transformation--i.e., that rotation does nothing to the state of a spin-zero particle. So the heuristic Carroll gives, that works for particles with nonzero spin, breaks down for particles with zero spin.

3. Jul 17, 2017

### davidge

Oh, ok.

Is there a more appropriate way for finding the spin-number (maybe using quantum-theory)?

4. Jul 17, 2017

### Staff: Mentor

It's not a matter of "finding" the spin number; it's more a matter of figuring out what spin numbers are possible, based on other considerations. The basic framework for doing that, as I understand it, is quantum field theory. A quick and dirty summary would be that the possible spin numbers are derived from the representation theory of the group SU(2), which arises from the "rotational symmetry" part of Lorentz invariance.

5. Jul 17, 2017

Ok. Thanks.