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Lenny Susskind explains in this video at about 40min, as an extended object (for example a relativistic string) is boosted to the infinite momentum frame (sometimes called light cone frame), it has no non-relativistic degrees of freedom in the boost direction. Instead, these degrees of freedom are completely determined by the (non-relativistic) motions in the plane perpendicular to the boost direction.

I dont see why this is, so can somebody explain to me how the degrees of freedom are described in this infinite momentum frame?

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Without seeing the quote/context I can only imagine that it means something like: if you take, say, a cube moving at close to c in the z direction, then (in the frame in which it's moving) its z extent gets Lorentz contracted to virtually zero, so it is effectively now a square in the xy plane and has only the degrees of freedom that a square in the xy plane has.

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  • $\begingroup$ Thanks twistor59, I first similarly thought about it as you explain here. But Lenny Susskind explained it in this lecture and as somebody in the audience asked him he made quite a fuzz about it, saying that why the degrees of freedom in the boosted direction is still not exactly understood, it was the start for people to think about holography, etc ... So I thought there is something else apart from what you say which I am missing ... $\endgroup$
    – Dilaton
    Commented Dec 26, 2012 at 12:24
  • $\begingroup$ I had a quick listen - I think the relevant stuff starts ~40min. He's talking about relativistic composite systems (I'm thinking, for example, of quarks in a proton confined so tightly they move at relativistic speeds). If you now boost the entire proton to speeds near c in some direction z, then the quark motions in the plane perpendicular to z can now be treated non relativistically (provided you use a time coord which has a large gamma factor). $\endgroup$
    – twistor59
    Commented Dec 26, 2012 at 13:58
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    $\begingroup$ For more info, a search on "light front dynamics" or something like that will turn up some stuff (though I can't find much at a review level). $\endgroup$
    – twistor59
    Commented Dec 26, 2012 at 14:18

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