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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.

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Related: Graviton –  Glen The Udderboat Jul 11 '13 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..! –  Waffle's Crazy Peanut Jul 11 '13 at 13:04
    
Possible duplicates: physics.stackexchange.com/q/1/2451 and 10 links therein. Related: physics.stackexchange.com/q/14932/2451 –  Qmechanic Jul 11 '13 at 13:07

2 Answers 2

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 $j_z=-2,-1,0,1,2$.

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Does this mean that the Higgs has no intrinsic angular momentum? –  Jitter Jul 11 '13 at 23:41
    
Yes, the Higgs is a spinless particle - equivalently, an excitation of a scalar field. –  Luboš Motl Jul 12 '13 at 4:52
    
Does -1 then mean that I owe a glass of water and have to pay it back? –  Jitter Jul 14 '13 at 11:28

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.

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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? –  user22180 2 days ago
    
@user22180, spinor fields: mathworld.wolfram.com/SpinorField.html –  Alfred Centauri 2 days ago

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