# Particles mass determined by SO(D-2) vs SO(D-1)

I've recently come across this statement that massless particles arise from $SO(D-2)$ symetry and massive particles from $SO(D-1)$.

I would have guessed that it would be the exact opposite way, but that doesn't seem so. Could someone point me in the right direction what exactly I should be reading to understand this?

As I'm a mathematician who has only recently been trying my way through String Theory, I've been wondering, why is this? Is there some intuitiv way of thinking about this? (This was only briefly mentioned in the ST book by Becker, Becker, Schwarz)

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I'm not entirely sure that I've understood what you're referring to, but it sounds like the following. To figure out the allowed representations for a massive particle in $D$-dimensional space-time dimensions, we can boost to its rest frame. Then we have the remaining $SO(D-1)$ rotations that leave us in the rest frame of the particle, and so its the respresentation of $SO(D-1)$ that determine the properties of the particle. So in $D = 4$ our massive particles are labelled by representations of $SO(3)$, which is good old angular momentum.

On the other hand if we have a massless particle, we can't boost to its rest frame since it doesn't have a rest frame. So what we have to look at instead are the transformations that leaves it's direction fixed, that is the transformations that fix a null ray. This is $SO(D-2)$ (if we ignore the fact that one of the directions is not compact, I think its really $SO(D-3,1)$ or whatever it's called), and so it is the representations of $SO(D-2)$ that determine the states of massless particle. In $D = 4$ we have that our massless particles are labeled by representations of $SO(2)$. So when we say the photon has angular momentum 1 we are not talking representation of $SO(3)$. It does not have a state with $m = 0$ - a longitudinal polarization - even though the $l=1$ represenation of $SO(3)$ has three states $m=-1,0,1$.

I hope this is what you were looking for.

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Thanks! This is what I was looking for –  Michael Aug 8 '11 at 18:27
Nice explanation such that though not having much (technical) clues about ST (yet?) even I can understand it ;-) ... +1 –  Dilaton Aug 9 '11 at 21:16
It has to do with the fact that massless particles do not have a rest frame. If the particle is massive, we can view it in the frame where its 3-momentum is zero. Its symmetry group then becomes $SO(D-1)$ (the group that arises in this way is known as the "little group"). Then the particles are classified according to the representations of this little group.
If the particle is massless, the simplest form the 4-momentum can take through Lorentz transformations $(P_0,P_1,0,0)$ where $P_0$ is the time component of the 4-vector. There is no frame in which the particle is not moving. So then the little group becomes the group of transformations of $D-2$ Euclidean space, in the normal $D=4$ case this is translations and rotations. I've also seen this group written as $ISO(D-2)$. So the particles now arise from representations of this group.