# What is the meaning of spin two?

As the title suggests, what is the meaning of spin two?

I kind of understand spin half for electrons. I can kind of understand spin one for other particles. However I'm not sure how something could have spin 2.

• Related: Graviton Commented Jul 11, 2013 at 13:00
• "I'm not sure how something could have spin 2" I think the phrase is quite vague. It's like asking "I'm not sure how apples can fall down" - Nature doesn't care what you think..! Commented Jul 11, 2013 at 13:04
• Possible duplicates: physics.stackexchange.com/q/1/2451 and 10 links therein. Related: physics.stackexchange.com/q/14932/2451 Commented Jul 11, 2013 at 13:07

For a more intuitive, less rigorous, understanding of "spin", fall back to primitive geometric objects and, in particular, how they behave under a coordinate rotation.

A scalar, a number, is unchanged (invariant) under a coordinate rotation so think of this as a "spin 0" object. Indeed, when we quantize a scalar field, the quanta have zero angular momentum; the quanta are spin 0 particles.

A vector, however, is covariant under a coordinate rotation. Importantly, if the coordinate system is rotated "once around", the vector is unchanged so think of this as a "spin 1" object; the vector rotates at the same rate as the coordinate system. More technically, to transform a vector, apply the transformation once.

As you might anticipate, when a vector field is quantized, the quanta have 1 unit of angular momentum; the quanta are spin 1 particles.

Now, consider a rank 2 tensor (think of an outer product of two vectors as an example). To transform this object, the coordinate transformation must be applied twice (both vectors get the transformation).

When the coordinate system is rotated through half a rotation, the rank 2 tensor is unchanged so think of this object as a "spin 2" object; the rank 2 tensor rotates twice the rate of the coordinate system.

Now, you probably already see where this is going. When this tensor field is quantized, the quanta have 2 units of angular momentum; the quanta are spin 2 particles.

• Hi you have discussed spin-0 particle as the quanta of scalar field, spin-1 particle as the quanta of vector field and spin-2 particle as the quanta of tensor field. But what about spin-1/2 particles? What do we call the field of spin-1/2 particles? Commented Oct 23, 2014 at 7:44
• @user22180, spinor fields: mathworld.wolfram.com/SpinorField.html Commented Oct 23, 2014 at 11:02
• Wow, never thought that a rank $n$ tensor is invariant under rotation by $\frac{2\pi}n$! Commented Feb 19, 2015 at 11:09

Spin-2 means that the spin is equal to 2 in the same sense in which spin-1 means that the spin is equal to 1 or spin-1/2 means that the spin is equal to 1/2. So it's hard to believe that you could understand the words spin-1/2 and spin-1 but not spin-2. It's like knowing how to drink half a liter of water, one liter of water, but be unable to drink 2 liters of water. Well, in this water example, it's actually more plausible.

The spin $\vec J$ is the intrinsic angular momentum. The intrinsic means "innate", the part of the angular momentum that exists even when the particle is at rest. The angular momentum is the quantity that is conserved whenever the laws of physics obey the rotational symmetry. The classical rotating object has $\vec J = \sum_i \vec r_i \times \vec p_i$ summed over the mass points.

In quantum mechanics, the total angular momentum of a particle is linked to the eigenvalue of $\vec J\cdot \vec J$ and it may be shown that the eigenvalues have the form $j(j+1)\hbar^2$ where $2j=0,1,2,3,\dots$ So the spin is either integer or half-integer.

Equivalently, any projection of the spin, most commonly talked about as $j_z$, has eigenvalues that go from $j_z=-j$ to $j_z=+j$ with the spacing one. The multiplication of all the (half)integer values by $\hbar$ is understood everywhere. The maximum allowed $j_z$ for a particle is also equal to the spin $j$.

Massless particles move by the speed of light so they don't have any rest frame. In their case, we usually talk about the spin with respect to one particular axis only, the axis of the direction of motion $\vec p$, because the rotations around this axis are unbroken. If it is so, the rotation symmetry is just $U(1)\sim SO(2)$ and the individual values of $j_z$ may exist in isolation from all the other values filling the interval between $-j_z$ and $+j_z$. The maximum positive value of $j_z$ is still referred to as the spin.

The electron is massive and has $j=1/2$, with the allowed $j_z=\pm 1/2$. The photon is massless and has $j=1$ with $j_z=\pm 1$. Roughly speaking, this "one" comes from the one index of the potential $A_\mu$. Similarly, the graviton has spin 2, $j=2$, roughly speaking because its field $g_{\mu\nu}$ has two indices. The allowed projections are just $j_z=\pm 2$. The photon's $j_z=0$ and the graviton's $j_z=-1,0,1$ are rendered unphysical by the gauge symmetries and diffeomorphisms, respectively.

There are also many massive particles such as nuclei and atoms that have $j=2$. These massive particles allow $j_z=-2,-1,0,1,2$.

• Does this mean that the Higgs has no intrinsic angular momentum? Commented Jul 11, 2013 at 23:41
• Yes, the Higgs is a spinless particle - equivalently, an excitation of a scalar field. Commented Jul 12, 2013 at 4:52
• Does -1 then mean that I owe a glass of water and have to pay it back? Commented Jul 14, 2013 at 11:28

Excellent photo of these two at this link : https://skullsinthestars.com/2016/03/29/1975-neutrons-go-right-round-baby-right-round/

## 1/1 - one full rotation equals original state one time (spin 1)

Excellent gif of this at this link: https://en.m.wikipedia.org/wiki/Spin-%C2%BD#/media/File%3ASpin_One-Half_(Slow).gif

1/2 - one full rotation is half way to the original state (spin 1/2)

Top number is how many time the original state is present. Bottom number is how many rotations.

• This answer could be a lot more useful, if you added some explanatory text rather than a collection of links. Commented Jul 10, 2019 at 5:41
• I’m guessing the idea of a full rotation and original state being obtained or not after the full rotation is pretty abstract. So imagine an object spinning 360 degrees on the horizontal axis (one full rotation), but during this horizontal rotation the object spun 180 degrees on the vertical axis. Now after one full rotation of the horizontal axis the object would not be at its original position. After another rotation horizontally the object would spin another 180 degrees vertically and therefore then be back to its original position. 1 time original state after 2 rotations. Commented Aug 18, 2019 at 4:05

A spin-2 particle just spins "harder" than a spin-1 particle.

I say harder because it's not really proper to talk of spinning faster or slower as, so far as elementary particles are pointlike objects one cannot give an extended geometry to them which is necessary to make sense of the moment of inertia that would let you figure out the angular velocity via

$$L = I\omega.$$

Hence we have to understand the spin solely through its angular momentum alone, without being able to attach a rotating reference frame to it unlike the case of, say, a spinning planet. The reason I say it as "harder" is because of an analogy of angular momentum to linear momentum: namely that linear momentum is "how much" motion is going on, under the idea that its elementary Newtonian form

$$p = mv$$

can be thought of as "there's more motion if there's more stuff (larger $$m$$) and there's more motion if that stuff is moving faster (larger $$v$$)", as it intuitively "kind of feels so" that if there is such a quantity as "how much motion is happening", then having both twice the stuff moving and the same stuff moving twice as fast should both produce twice as much "motion going on" as a baseline scenario with a known "amount of motion going on". Of course, the impressive part is that this rather intuitive quantity happens to be conserved in Newtonian mechanics, which then is what lets us generalize the quantity to its relativistic and quantum counterparts.

Hence angular momentum is the same way, though $$I$$ is no longer clearly understandable as "amount of stuff", so the relevant concept we must transfer is just the idea it measures the "amount of rotational motion". An object with 2 units of momentum thus doesn't necessarily move twice as fast as one with 1 unit of momentum, but it is moving "more" in some way, so thus we could say perhaps, searching our vocabulary, that it is moving twice as "hard". Hence we could say an object with two units of angular momentum likewise spins twice as "hard" as one with one unit of angular momentum.

Of course, here I should point out that the "2" is actually not directly the strength of the spin - the magnitude of the angular momentum is actually

$$|\mathbf{L}_S| = \sqrt{s(s+1)}\hbar,$$

where $$s$$ is the number called in your post as $$\frac{1}{2}$$, $$1$$, and $$2$$ respectively. (Why is this? It would require a more precise discussion of how that the notion of motion and rotational motion change in quantum mechanics versus in classical mechanics.) Hence a spin-2 particle actually spins

$$\frac{\sqrt{2(2+1)}\hbar}{\sqrt{1(1+1)}\hbar} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$$

times harder than a spin-1 particle, not 2 times harder.

My suspicion is that when you said you "understand" spin-1 and spin-$$\frac{1}{2}$$, what you meant is you understood something like that "spin-$$\frac{1}{2}$$ means 'matter' and spin-$$1$$ means 'radiation' or, perhaps, 'photons', or 'vector-field forces'" and you likely came across the statement that spin-$$2$$ means "tensor-field forces" viz. General Relativity and gravity. This, however, is a consequence, not the "essential meaning" of spin. Spin is really just angular momentum, except in terms of magnitude it is intrinsic to the given particle so that given any particular species it cannot be spun up nor braked, and in terms of direction - well, that gets "fuzzy", so to speak :)

Is it correct if I say, when particles with spin are placed in a magnetic field they presess. The precession depends on the strength of the magnetic field.If the particles spin(revolve) on their own axes in a magnetic field, the gyration or the precession produced per rotation of the particle is termed as spin. That is to say, Spin 1 means the particle gyrates once every revolution. Spin 0 means there is no precession when the particle rotates. Spin 1/2 means every two revolutions of the particle produce one precession and so on. In other words spin is related with precession.

An animated Illustration of basic spins using manifolds can be,

1. For spin-1/2 like for example an electron fermion particle (any arbitrary chosen point on a closed loop manifold returns to its initial exact position in space after 4π-rotations (720°) of the whole depicted manifold around its equatorial plane (i.e. horizontal xy plane):

spin-1/2

2. For integer spin-1 like for example a normal photon Boson particle (any arbitrary chosen point on a closed loop manifold returns to its initial exact position in space after a 2π-revolution (360°):

spin-1

3. For integer spin-2 like for example the hypothesized graviton Boson particle (any arbitrary chosen point on a closed loop manifold returns to its exact initial position in space after a π-revolution (180°, half circle!):

This is impossible to animate for a Boson which according to string theory as a Boson must be a closed loop manifold. The only way this hypothesized not yet discovered quantum spin-2 to be real would be IMHO what exactly in a previous answer here @The_Sympathizer mentioned "A spin-2 particle just spins "harder" than a spin-1 particle". https://physics.stackexchange.com/a/711336/183646

However, for massless Boson the theorized spin-2 graviton is theorized to be, this would be equivalent to a superluminal $$v>c$$ revolution speed spin-1 photon and impossible according to special relativity x2 "time contraction"! Instead of time dilation in the lab frame.

Relativistically impossible theorized massless Boson spin-2:

spin-2 particle (hypothesized graviton) would be equivalent to a spin-1 particle  (photon) thus a revolving circle but with an angular velocity  FTL $$c$$ speed.