Question

Can the Polyakov action be derived from the point particle action?

Answer 1

S0=12∫dτ(η−1X˙μX˙μ−ηm2)S0=12∫dτ(η−1X˙μX˙μ−ηm2)

by considering the Lagrangian

L0(τ)=12(η−1(τ)X˙μ(τ)X˙μ(τ)−η(τ)m2)L0(τ)=12(η−1(τ)X˙μ(τ)X˙μ(τ)−η(τ)m2)

as the Lagrangian density

L1(τ,σ)=12(η−1(τ,σ)X˙μ(τ,σ)X˙μ(τ,σ)−η(τ,σ)m2)+T(τ,σ)L1(τ,σ)=12(η−1(τ,σ)X˙μ(τ,σ)X˙μ(τ,σ)−η(τ,σ)m2)+T(τ,σ)

of the Polyakov action, and I'm guessing a potential TT would be needed as the 'point' is no longer free.

I have tried working the other direction, differentiaing the Lagrangian of the Polyakov action and performing some integration by parts to obtain

L1=∂Sp∂σ=14πα′h−−√(hγδ∂hγδ∂σ∂hαβ∂σ∂αXμ∂βXμ+hαβ∂αXμ∂βXμ+hαβ∂α∂Xμ∂σ∂βXμ)L1=∂Sp∂σ=14πα′h(hγδ∂hγδ∂σ∂hαβ∂σ∂αXμ∂βXμ+hαβ∂αXμ∂βXμ+hαβ∂α∂Xμ∂σ∂βXμ)

where XμXμ≡gμνXμXνXμXμ≡gμνXμXν.

However I am not entirely sure how to proceed from here - I guess my problem is that I expect the resulting Lagrangian to be independent of σσ but I'm unsure how to deal with this.

If anybody could point me to a paper which does something along these lines, or could elucidate as to why this is not possible then I would be very grateful

Answer 2

Explanation:

physics, action is an attribute of the dynamics of a physical system from which the equations of motion of the system can be derived through the principle of stationary action. Action is a mathematical functional which takes the trajectory, also called path or history, of the system as its argument and has a real number as its result. Generally, the action takes different values for different paths.[1] Action has the dimensions of