Lecture 6
8.251 Spring 2007
Lecture 6 - Topics
• The relativistic point particle: Action, reparametrizations, and equations
of motion
Reading: Zwiebach, Chapter 5
Continued from last time.
∂P t
∂P x
+
=0
∂t
∂x
P t = µ0 ∂y/∂t
P x = −T0 ∂y/∂x
Similar to ∂µ J µ = 0, ∂ρ/∂t + � · J� = 0, Q =
�
dxρ
Free BC (Neumann BC):
P x (t, x∗ ) = 0
a
�
Py =
�
µ0 dx(∂y/∂t) =
0
�
a
t
a
�
dx∂P x /∂x = −[P x (t, x = a) − P x (t, x = 0)]
dx∂P /∂t = −
∂Py /∂t =
0
dxP t
0a
0
Conservation of momentum?
Free Relativistic Particle
Non-relativistic Action:
�
S=
�
�
1
2
dt mv
2
Calculation: dv/dt = 0
Relativistic Particles:
1
Lecture 6
8.251 Spring 2007
Everyone should agree on action. It’s a Lorentz invar.
−ds2 = −ηµν dxµ dxν
�
ds = cdt 1 − v 2 /c2 = cdτ
s = −mc2
�
P
So: s = −mc2
� tf
ti
ds
= −mc
c
�
ds
P
�
dt 1 − v 2 /c2
Check:
Lagrangian:
�
v 2
c2
1 v2
= −mc2 (1 −
− . . .)
2 c2
1
= −mc2 +
mv 2
� �� �
2
� �� �
L = −mc
2
1−
Taylor Expansion
rest energy
kinetic energy
Momentum:
∂L
P� =
∂�v
1 −2�
v
2
= −mc2 · �2 c
1−
v2
c2
mv 2
=�
2
1 − vc2
Hamiltonian:
mc2
H = p� · �v − L = . . . = �
2
1 − vc2
Parameterization
Have parameterization xµ (τ ) (the xµ ’s are functions of τ )
ds2 = −ηµν dxµ dxν
2
Lecture 6
8.251 Spring 2007
�
ds =
dxµ
dτ
�
�
−ηµν
�
tf
s = −mc
��
−ηµν
ti
�
dxν
dτ
dτ
dxµ dxν
dτ
dτ dτ
�
τ (τ ):
dxµ
dxµ dτ �
=
dτ
dτ � dτ
tf
�
�
s = −mc
dxµ dxν
dτ � dτ �
�
�
dxµ dxν
dτ �
dτ � dτ �
−ηµν
ti
�
tf
= −mc
�
−ηµν
ti
dτ �
dτ
dτ
�
�
So using a different parameter, τ � (instead of τ ) gets same action s. s is
reparameterization-invariant.
��
2
Quick calculation to find equation of motion from s = −mc
1 − vc2 dt. Should
get derivative of rel. momentum with respect to time = 0.
�
S = −mc dS
�
δS = −mc
δ(dS)
dS 2 = −ηµν dxµ dxν
dxµ dxν
(dτ )2
dτ dτ
� µ� ν
dx
dx
(dτ )2
2(dS) · δ(dS) = −2ηµν δ
dτ
dτ
(dS)2 = −ηµν
3
Lecture 6
8.251 Spring 2007
δ(dS) = −ηµν
d
dxν
(δxµ )
dτ
dτ
ds
Must vary with dxµ /dτ and dxν /dτ , but since ηµν is symmetric sufficient to
vary just dxµ /dτ and multiply by 2.
δ(dS) = −
d
dxµ
(δxµ )
dτ
dτ
dS
τf
�
�
d(δxµ )
dxµ
δS =
mc
dτ
dτ
ds
τi
�
� τf �
d
dPµ
=
(δxµ Pµ ) − δxµ
dτ
dτ
dτ
τi
�
δxµ (τi ) = δxµ (τf ) = 0
�
τf �
dS = −
τi
�
dPµ
δx (τ )
dτ
dτ
µ
Equation of Motion:
dPµ
=0
dτ
This means that Pµ constant on world-line. Constant as a function of any pa­
rameter!
dPµ
dPµ dτ
=
·
dt
dt
� ��� �dτ
��� ����
0
Therefore:
d
dτ
�
�
dxµ
= 0,
ds
d2
µ
ds2 (dx )
0
=0
�
= 0 (if τ = s. Okay because τ is arbitrary.)
But can’t assign s = τ : d2 xµ /dτ 2 �== 0.
�
�
d dxµ
� 0
=
ds dτ
4
Lecture 6
8.251 Spring 2007
Coupling to Electromagnetism
Lorentz Force Equation:
dPµ
q
dxν
= · Fµν
dS
c
ds
dPµ
q
dxν
= · Fµν
dτ
c
dτ
�
�
q
dxµ
S = −mc
dS +
Aµ (x(τ ))
dτ
c P
dτ
P
A: Nevitz-Schwartz Tensor
5