Degrees of freedom in the infinite momentum frame

+ 2 like - 0 dislike
1375 views

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?

asked Dec 24, 2012
retagged Apr 13, 2014

+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-12 15:30 (UCT), posted by SE-user twistor59
answered Dec 26, 2012 by (2,500 points)
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 ...

This post imported from StackExchange Physics at 2014-03-12 15:30 (UCT), posted by SE-user Dilaton
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).

This post imported from StackExchange Physics at 2014-03-12 15:30 (UCT), posted by SE-user twistor59
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).

This post imported from StackExchange Physics at 2014-03-12 15:30 (UCT), posted by SE-user twistor59

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification