Lectures on String Theory

– 11 –
The action has local gauge symmetry which is now
δx µ = ξẋ µ
δe = ∂ τ (ξe)
The first symmetry transformation is generated by the constraint p 2 + m 2 :
δ η x µ = η{p 2 + m 2 , x µ } = 2ηp µ = η e ẋµ = ξẋ µ , η ≡ eξ .
The second transformation for e is easily derived from e 2 = − 1 ẋ 2 . Thus, in our
m 2
new formulation we have the set of fields (x µ , e) and reparametrization symmetry
which acts on them and leaves the action invariant. This reparametrization freedom
can be used to put e = 1 which results into the following equation
m
ẍ µ = 0
This is not the end however, because there is eom for e which now reads as ẋ 2 = −1.
Complete eoms are
ẍ µ = 0 , ẋ 2 = −1
Thus, the relativistic particle moves freely in Minkowski space over time-like geodesics.
Space-like and light-like straight lines are excluded by the constraint ẋ 2 = −1.
In the case of the massless particle we can set e = 1 and get eoms
ẍ µ = 0 , ẋ 2 = 0
In both, the massive and massless cases, the constraints are integral of motions: they
are preserved in time due to the dynamical equation ẍ µ = 0.
Finally, we treat the relativistic particle in the so-called first order (the Hamiltonian)
formalism. To this end we have to represent the initial Lagrangian in the
form
L = p µ ẋ µ + L rest
where p µ = 1 eẋµ and express in L rest the derivatives ẋ µ via p µ . In doing so we obtain
the phase-space Lagrangian
L = p µ ẋ µ − e 2 (p2 + m 2 )
We clearly see that the auxiliary field e we introduced in our second formulation
plays here the role of the lagrangian multiplier to the constraint p 2 + m 2 = 0. By
using the gauge freedom we can fix the gauge e = 1 and the physical Hamiltonian
m
becomes in this case
H = 1
2m (p2 + m 2 )