I) First consider the point particle
S=m∫√−˙X2dτ.
If you choose the static gauge τ=X0
=m∫√1−˙Xj˙Xjdτ.
So now, you have an equivalent action with true degrees of freedom only. In fact, you can do the same with the light-cone gauge τ=X+
S=m∫√2˙X−−˙XI˙XIdτ.
II) Is it possible to do the same for the string? I have found no reference doing this. I have been trying replacing the conditions of the light-cone gauge. I think that the answer could be
L∼˙XI˙XI−XI′XI′.
This is eq.(12.81) in the 2nd edition of Zwiebach's book.
Additional Information: Basically you have to show that the gauge conditions n⋅X=2βα′n⋅pτ,
n⋅p=2πβn⋅P,
(where β=2 for open string and β=1 for closed string) imply ˙X2=−X′2and˙X⋅X′=0.
And with these relations you can easily reduce the action.
For the open string you can show that condition (2) implys ˙X⋅X′=0.
This can be done using the boundary conditions of the open string.1
But for the closed string I have found no way to show that. In Zwiebach page 180 there is an idea of this for open and closed strings.2
To sum up, my question is if one can deduce the same for the closed string. What would be the procedure?
1See Sundermayer, Constrained Dynamics page 218, or Hansen-Regge-Teitelboim, Constrained Hamiltonian Systems page 58.
2 But in the case of closed string the full σ=0 line is constructed by requiring that at each point its tangent be orthogonal to X′. So it is like he is imposing ˙X⋅X′=0, not deducing it.
No comments:
Post a Comment