Wednesday, April 19, 2006

Fast strings

When I explain strings to non-stringy physicist, I often start out by stating that a string is like a rubber band and has a potential energy which is proportional to its length (like a 'relativistic Hook law'). Then, all you have to do is to covariantize this statement and you arrive at the Nambu Goto action.

You can, of course read this backwards: A string with potential energy E has a length proportional to E. Now, you can often read this as an explanation of why hard string scattering behaves much better than hard scattering of particles: At high energies, the strings in fact expand and thus the interaction delocalizes.

This, however, is only semi-true: I would think of what you have in hard scattering are strings which have been accelerated so that they have large centre of mass energy. But the centre of mass energy is decoupled from the internal energy of the oscillators and thus a boosted short string will still be short although it has large (kinetic) energy. On the other hand, you can have a long string with zero kinetic energy which just happens to be very heavy. So, in general, heavy strings are long, not fast ones.

So far, this is just kinematics, but can we see this in practice? What happens to a string that runs through a linear accelerator? So, the set-up would look as follows: You start with a string which is in a low mass state which is charged under some U(1) (maybe in a KK type theory, D-branes are welcome as well). Now it feels a electric field strength (of some cavity say). This electric field is a condensate of low energy (given by the normal frequency of the cavity) photons. So, you have to compute the scattering of the string with lots and lots of low energy strings in the vector field state.

Question: Even if the individual photon has energy much less than 1/sqrt(alpha'), does this scattering excite any of the higher oscillator modes (which make the string grow)? A Feynman diagram would look somewhat like


e----x-----x-----x-----x- ... -x-----h
| | | | ... |
A A A A A

where e is the charged low energy state, A is the gauge field and h is a heavy state. Has this been done before?

3 comments:

Bee said...

But the centre of mass energy is decoupled from the internal energy of the oscillators and thus a boosted short string will still be short although it has large (kinetic) energy.

I am not sure I get the point here. If you look at the center of mass energy, it doesn't depend on the boost. Also, the extension of the string grows with the energy, see e.g. Gross and Mende, Nucl. Phys. B 303, 407 (1988).

Robert said...

Maybe my words were not well picked. I mean "the (kinetic) energy of the centre of mass" rather than "the energy in the centre of mass frame". The latter is what I called "the internal energy" or "the energy of the oscillators" and my whole point is that it is constant under a boost. But the latter is also what determines the geometric size of the string.

Re the classic Gross Mende paper: Their figure 3 shows the geometry of the saddle point that dominates high energy scattering. One can see that in the tips, the size of the strings grows before they collide.

This looks to me as if the occupation number of the first oscillator <alpha_1 alpha_(-1)> grows with target space time. How does this work from the Hamiltonian perspective?

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