Lectures on String Theory

– 8 –
2. Relativistic particle
Consider relativistic particle of mass m moving in d-dimensional Minkowski space:
η µν = (−1, +1, +1, . . . , +1).
Action
∫ s
S = −α ds
s 0
Note that ∫ s
s 0
ds has maximum along straight lines, this explains the sign “-” in front
of the action.
Embedding x µ ≡ x µ (τ):
If x µ = (cτ, ⃗x) then
Thus, the action is
ds =
√
− dxµ
dτ
ds = √ c 2 − ⃗v 2 ,
√
dx µ
dτ dτ ≡ dx
−η µ dx ν
µν
dτ dτ dτ
⃗v = d⃗r
dτ
∫ √
τ1
S = −αc 1 − ⃗v2
τ 0
c dτ 2
The Lagrangian in the non-relativistic limit
√
( )
L = −αc 1 − ⃗v2
c dτ = −cα 1 − ⃗v2 + . . . = −αc + α⃗v2
2 2c 2 2c + . . .
To get the standard kinetic energy one has to identify
α = mc
In what follows we will work in units in which c = 1.
The action is invariant under reparametrizations of τ:
Let us show this
δx µ = ξ(τ)∂ τ x µ as long as ξ(τ 0 ) = ξ(τ 1 ) = 0
δ( √ 1
−ẋ µ ẋ µ ) =
2 √ 1
−ẋ µ ẋ µ (−2ẋν δẋ ν ) = −√ −ẋµ ẋ µ ẋν ∂ τ (ξẋ ν ) =
1
[
]
= −√ ẋ ν 1
ẋ −ẋµ ẋ µ ν ˙ξ + ξẋνẍ ν = −√ −ẋµ ẋ µ ẋν ẋ ν ˙ξ + ξ∂τ ( √ −ẋ µ ẋ µ )
= √ −ẋ µ ẋ µ ˙ξ + ξ∂ τ ( √ −ẋ µ ẋ µ ) = ∂ τ (ξ √ −ẋ µ ẋ µ )