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Last modified: May 7, 2022 ✓

The Principle of Least Action in Relativity
-------------------------------------------

[time]:  dτ2 = (dt2 - dx2/c2) = proper time 
                               interval

[space]:  ds2 = c2dτ2 = (c2dt2 - dx2) = spacetime 
                                       interval

The relationship between the spacelike case and the timelike
case is:

ds2 = c2dτ2 

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 = √(gμνdxμdxν) 

We can write this as:

dτ = √(gμν(dxμ/dτ)(dxν/dτ))dt

Let's postulate that the action is:

A = -mc∫dτ

  = -mc∫√(gμν(dxμ/dτ)(dxν/dτ))dτ

So L = √(gμν(dxμ/dτ)(dxν/dτ))

For flat space we get:

       -          -  -    -
      | 1          || c2dt2 |
dτ2 = |   -1       ||  dx2  | = c2dt2 - dx2 - dy2 - dz2
      |      -1    ||  dy2  |
      |         -1 ||  dz2  |
       -          -  -    -
  
Therefore, in one spatial dimension we get:

dτ = √(c2dt2 - dx2)

   = √(c2dt2(1 - dx2/c2dt2)

   = cdt√((1 - dx2/c2dt2)

   = cdt√((1 - v2/c2)

   = cdt/γ

The action, A, is defined as:

A = -m∫dτ   where m = mass

  = -m∫dt/γ  

To satisfy 2., we must have:    
 
A = -mc2∫dt/γ  

The relativistic Lagrangian can be found by comparing this to:

A = ∫Ldt

Therefore,

L = -mc2/γ

  = -mc2√(1 - v2/c2)

For v << c we can use the binomial expansion:

L = -mc2(1 - v2/2c2)   

  = -mc2 + (1/2)mv2 

The Hamiltonian
---------------

From Lagrangian mechanics:
        .
p = ∂L/∂x = mc2v/√(1 - v2/c2)
       .
H = Σpixi - L
    i
  = mc2v2/√(1 - v2/c2) + mc2√(1 - v2/c2)

  = mc2v2/√(1 - v2/c2) + mc2(1 - v2/c2)/√(1 - v2/c2)

  = mc2/√(1 - v2/c2)

  = mc2 + mv2/2 (binomial expansion)

Rindler Space
-------------

For Rindler space in Minkowski coordinates:

       -                 -  -     -
      | (1 + 2φ)  0  0  0 ||   dT2  |
dτ2 = |    0     -1  0  0 || dX2/c2 |
      |    0      0 -1  0 || dY2/c2 |
      |    0      0  0 -1 || dZ2/c2 |
       -                 -  -     -

Therefore,

dτ2 = (1 + 2φ/c2)dT2 - dX2/c2

A = -mc2∫√((1 + 2φ/c2)dT2/dT2 - (1/c2)dX2/dT2) dT
                         .
  = -mc2∫√((1 + 2φ/c2) - X2/c2) dT
                          .
L = -mc2√(1 + (1/c2)(2φ - X2))
                           .
  = -mc2(1 + (1/2c2)(2φ - X2)  (binomial)
                    .
  = -mc2 - mφ(X) + mX2/2

Where φ = ΔRg


General Relativity
------------------

A = -m∫√(gμνdxμdxν)

  = -m∫√[gμν(dxμ/dt)(dxν/dt)]dt

  = -m∫Ldt

  = -m∫√(g00dt2 - g11dx2)dt

L = √(g00dt2 - g11dx2)
                                    .
Applying the E-L equation, d/dt(∂L/∂x) = ∂L/∂x, gives the 
geodesic equation (not proven):

∂2xμ/dτ2 = -Γσρμ(dxσ/dτ)(dxρ/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 = -mc2 - mφ(X) + mX2/2
    .    .
∂L/∂X = mX
         .
d/dt(∂L/∂X) = md2X/dt2

∂L/∂X = -m∂φ(X)/∂X

Therefore,

md2x/dt2 = -m∂φ/∂X
                                 
            = -mg

The second way:

Γrsp = (gpn/2)[∂gnr/∂xs + ∂gns/∂xr - ∂grs/∂xn]

For the x-direction this becomes:

Γ00x = (g11/2)[∂g1r/∂x0 + ∂g1s/∂x0 - ∂g00/∂x]

For v << c and we can ignore the spatial terms, ∂g1r/∂t 
and ∂g1s/∂t.  Therefore,

Γ001 = (1/2)[-∂g00/∂x]

The geodesic equation becomes:  

d2x/dτ2 = -Γ001(dx0/dτ)(dx0/dτ)

or, since dt = dτ for v << c:

d2x/dt2 = -Γ001(1)(1)   

        = (1/2)∂g00/∂x1

        = -g   [-g00 = (1 + 2ΔRg); ΔR ≡ x]

Therefore,

md2x/dt2 = -mg

 

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