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The Principle of Least Action in Relativity
-------------------------------------------
[time]: dτ^{2} = (dt^{2} - dx^{2}/c^{2}) = proper time
interval
[space]: ds^{2} = c^{2}dτ^{2} = (c^{2}dt^{2} - dx^{2}) = spacetime
interval
The relationship between the spacelike case and the timelike
case is:
ds^{2} = c^{2}dτ^{2}
The proper distance, ds, is analogous to the proper time. The
difference is that the proper distance is defined between two
spacelike separated events (or along a spacelike path), while
the proper time is defined between two timelike separated events
(or along a timelike path).
Consider the world-line for a single point particle.
The action principle obeys the following rules:
1. It should be an integral over the trajectory (multiplied by
any constant).
2. It has to have units of energy multiplied by time.
3. It should be Lorentz invariant. This ensures that all the
laws of physics derived from it will also be Lorentz
invariant.
4. It is required to be stationary (which allows use of the
Euler-Lagrange equation).
Relativistic Action
-------------------
World lines are always time-like (TL) curves in spacetime. We
would like to make the value of the action independent of the
parameter chosen to calculate it (the coordinate time, t). For
this purpose we choose τ (the proper time) since all Lorentz
observers agree on its value. This process is referred to as
'parameterization of the world line'. Starting with:
dτ = √(g_{μν}dx^{μ}dx^{ν})
We can write this as:
dτ = √(g_{μν}(dx^{μ}/dτ)(dx^{ν}/dτ))dt
Let's postulate that the action is:
A = -mc∫dτ
= -mc∫√(g_{μν}(dx^{μ}/dτ)(dx^{ν}/dτ))dτ
So L = √(g_{μν}(dx^{μ}/dτ)(dx^{ν}/dτ))
For flat space we get:
- - - -
^{ } | 1 || c^{2}dt^{2} |
dτ^{2} = | -1 || dx^{2}^{ } | = c^{2}dt^{2} - dx^{2} - dy^{2} - dz^{2}
^{ } | -1 || dy^{2}^{ } |
^{ } | -1 || dz^{2}^{ } |
- - - -
Therefore, in one spatial dimension we get:
dτ = √(c^{2}dt^{2} - dx^{2})
= √(c^{2}dt^{2}(1 - dx^{2}/c^{2}dt^{2})
= cdt√((1 - dx^{2}/c^{2}dt^{2})
= cdt√((1 - v^{2}/c^{2})
= cdt/γ
The action, A, is defined as:
A = -m∫dτ where m = mass
= -m∫dt/γ
To satisfy 2., we must have:
A = -mc^{2}∫dt/γ
The relativistic Lagrangian can be found by comparing this to:
A = ∫Ldt
Therefore,
L = -mc^{2}/γ
= -mc^{2}√(1 - v^{2}/c^{2})
For v << c we can use the binomial expansion:
L = -mc^{2}(1 - v^{2}/2c^{2})
= -mc^{2} + (1/2)mv^{2}
The Hamiltonian
---------------
From Lagrangian mechanics:
.
p = ∂L/∂x = mc^{2}v/√(1 - v^{2}/c^{2})
.
H = Σp_{i}x^{i} - L
^{i}
= mc^{2}v^{2}/√(1 - v^{2}/c^{2}) + mc^{2}√(1 - v^{2}/c^{2})
= mc^{2}v^{2}/√(1 - v^{2}/c^{2}) + mc^{2}(1 - v^{2}/c^{2})/√(1 - v^{2}/c^{2})
= mc^{2}/√(1 - v^{2}/c^{2})
= mc^{2} + mv^{2}/2 (binomial expansion)
Rindler Space
-------------
For Rindler space in Minkowski coordinates:
- - - -
_{ } | (1 + 2φ) 0 0 0 || dT^{2}^{ } |
dτ^{2} = | 0 -1 0 0 || dX^{2}/c^{2} |
_{ } | 0 0 -1 0 || dY^{2}/c^{2} |
_{ } | 0 0 0 -1 || dZ^{2}/c^{2} |
- - - -
Therefore,
dτ^{2} = (1 + 2φ/c^{2})dT^{2} - dX^{2}/c^{2}
A = -mc^{2}∫√((1 + 2φ/c^{2})dT^{2}/dT^{2} - (1/c^{2})dX^{2}/dT^{2}) dT
^{ }.
= -mc^{2}∫√((1 + 2φ/c^{2}) - X^{2}/c^{2}) dT
^{ }.
L = -mc^{2}√(1 + (1/c^{2})(2φ - X^{2}))
^{ }.
= -mc^{2}(1 + (1/2c^{2})(2φ - X^{2}) (binomial)
.
= -mc^{2} - mφ(X) + mX^{2}/2
Where φ = ΔRg
General Relativity
------------------
A = -m∫√(g_{μν}dx^{μ}dx^{ν})
= -m∫√[g_{μν}(dx^{μ}/dt)(dx^{ν}/dt)]dt
= -m∫Ldt
= -m∫√(g_{00}dt^{2} - g_{11}dx^{2})dt
L = √(g_{00}dt^{2} - g_{11}dx^{2})
.
Applying the E-L equation, d/dt(∂L/∂x) = ∂L/∂x, gives the
geodesic equation (not proven):
∂^{2}x^{μ}/dτ^{2} = -Γ_{σρ}^{μ}(dx^{σ}/dτ)(dx^{ρ}/dτ)
Equations of Motion
-------------------
The equations of motion can be obtained in 2 ways. Apply
the Euler-Lagrange equation directly or solve the above
geodesic equation.
The first way:
.
L = -mc^{2} - mφ(X) + mX^{2}/2
. .
∂L/∂X = mX
.
d/dt(∂L/∂X) = md^{2}X/dt^{2}
∂L/∂X = -m∂φ(X)/∂X
Therefore,
md^{2}x/dt^{2} = -m∂φ/∂X
= -mg
The second way:
Γ_{rs}^{p} = (g^{pn}/2)[∂g_{nr}/∂x^{s} + ∂g_{ns}/∂x^{r} - ∂g_{rs}/∂x^{n}]
For the x-direction this becomes:
Γ_{00}^{x} = (g^{11}/2)[∂g_{1r}/∂x^{0} + ∂g_{1s}/∂x^{0} - ∂g_{00}/∂x]
For v << c and we can ignore the spatial terms, ∂g_{1r}/∂t
and ∂g_{1s}/∂t. Therefore,
Γ_{00}^{1} = (1/2)[-∂g_{00}/∂x]
The geodesic equation becomes:
d^{2}x/dτ^{2} = -Γ_{00}^{1}(dx^{0}/dτ)(dx^{0}/dτ)
or, since dt = dτ for v << c:
d^{2}x/dt^{2} = -Γ_{00}^{1}(1)(1)
= (1/2)∂g_{00}/∂x^{1}
= -g [-g_{00} = (1 + 2ΔRg); ΔR ≡ x]
Therefore,
md^{2}x/dt^{2} = -mg