Physics Notes - Herong's Tutorial Notes

∟Introduction of Lagrangian

∟Other Proofs of the Lagrange Equation

This section provides two other proofs to show that the Lagrange Equation is equivalent to the Hamilton Principle.

Here are two other proofs to show that the Lagrange Equation is equivalent to the Hamilton Principle.

Melanie's Proof

Melanie Ganz gives the following proof in lecture notes "Introduction to Lagrangian and Hamiltonian Mechanics" at http://image.diku.dk/ganz/Lectures/Lagrange.pdf.

Introduce a small variation of x(t) as xs(t): 

xs(t) = x(t) + s*x(t)                       (MG.1)
  # s is small parameter
  # x(t) is the true position function

or:
  xs = x + s*x
    # (t) omitted

Its velocity function will be xs'(t): 

xs'(t) = x'(t) + s*x'(t)                    (MG.2)
  # x'(t) is the true velocity function

or:
  xs' = x' + s*x'
    # (t) omitted

This will result a new Action value Ss: 

Ss = S[xs(t)]

or:
  Ss = ∫ L(xs,xs',t)dt                       (MG.3)

or:
  Ss = ∫ L(x+s*x,x'+s*x',t)dt                (MG.4)
    # MG.1 and MG.2 applied

Now we expand the Lagrangian function with its first order partial derivatives, because the parameter s is very small. 

L(x+s*x,x'+s*x',t) = L(x,x',t) + s*x*∂L/∂x + s*x'*∂L/∂x'      (MG.5)

Apply MG.5 to MG.4. We have: 

Ss = ∫ (L(x,x',t) + s*x*∂L/∂x + s*x'*∂L/∂x')dt

or:
  Ss = ∫ L(x,x',t)dt + ∫ (s*x*∂L/∂x + s*x'*∂L/∂x')dt       (MG.6)

According the Hamilton's Principle, x(t) is a stationary point on the Action S[x(t)]. So the small variation x(t) + s*e(t) should result the same Action S value: 

S[xs(t)] - S[x(t)] = 0
  # Since x(t) is a stationary point and s is small

or:
  Ss - S = 0

or:
  ∫ L(x,x',t)dt + ∫ (s*x*∂L/∂x + s*x'*∂L/∂x')dt - S = 0
    # MG.6 applied

  ∫ L(x,x',t)dt + ∫ (s*x*∂L/∂x + s*x'*∂L/∂x')dt - ∫ L(x,x',t)dt = 0
    # Since S = ∫ L(x,x',t)dt

or:
  ∫ (s*x*∂L/∂x + s*x'*∂L/∂x')dt = 0                       (MG.7)

Further reduction of MG.7 requires the rule of integration by parts: 

s*x*∂L/∂x'| = ∫ (s*x'*∂L/∂x')dt + ∫ (s*x*(∂L/∂x')')dt
  # Between t1 and t2

or:
  ∫ (s*x'*∂L/∂x')dt = s*x*∂L/∂x'| - ∫ (s*x*(∂L/∂x')')dt        (MG.8)
  # Between t1 and t2

Applying MG.8 to MG.7, we have: 

∫ (s*x*∂L/∂x)dt + s*x*∂L/∂x'| - ∫ (s*x*(∂L/∂x')')dt = 0
  # Between t1 and t2

s*x*∂L/∂x'| + ∫ s*x*(∂L/∂x - (∂L/∂x')')dt = 0                (MG.9)
  # Between t1 and t2

Now here is a big jump. According to Melanie Ganz's lecture notes, for equation MG.9 to be true for any s, the following must be true. 

∂L/∂x - (∂L/∂x')' = 0

or:
  ∂L/∂x - d(∂L/∂x')/dt = 0
    # Since ()' = d()/dt

or:
  d(∂L/∂x')/dt = ∂L/∂x                                     (MG.10)

Equation MG.15 is exactly the Lagrange Equation G.5. So we have approved that Hamilton's Principle is equivalent to Lagrange Equation.

David's Proof

David Morin gives the following proof in the book "Introduction to Classical Mechanics" at https://scholar.harvard.edu/david-morin/classical-mechanics.

Introduce a small variation of x(t) as xs(t): 

xs(t) = x(t) + s*e(t)                          (DM.1)
  # s is small parameter
  # x(t) is the true position function
  # e(t) is any function with e(t1) = e(t2) = 0

or:
  xs = x + s*e
    # (t) omitted

Its velocity function will be xs'(t): 

xs'(t) = x'(t) + s*e'(t)                       (DM.2)
  # x'(t) is the true velocity function

or:
  xs' = x' + s*e'
    # (t) omitted

This will result a new Action value Ss: 

Ss = S[xs(t)]

or:
  Ss = ∫ L(xs,xs',t)dt                          (DM.3)

or:
  Ss = ∫ L(x+s*e,x'+s*e',t)dt                  (DM.4)
    # DM.1 and DM.2 applied

According the Hamilton's Principle, x(t) is a stationary point on the Action S[x(t)]. So the partial derivative of S against s should be 0. 

∂S[xs(t)]/∂s = 0                                  (DM.5)

or:
  ∂(∫ L(xs,x's,t)dt)/∂s = 0
    # Between t1 and t2

or:
  ∫ ((∂L/∂xs)*(∂xs/∂s) + (∂L/∂x's)*(∂x's/∂s) + (∂L/∂t)*(∂t/∂s))dt = 0
    # Chain rule applied

or:
  ∫ ((∂L/∂xs)*(∂xs/∂s) + (∂L/∂x's)*(∂x's/∂s) = 0
    # Since ∂t/∂s = 0

or:
  ∫ ((∂L/∂xs)*e + (∂L/∂x's)*(∂x's/∂s) = 0
    # Since ∂xs/∂s = e

or:
  ∫ (e*∂L/∂xs + e'*∂L/∂x's)dt = 0                     (DM.6)
    # Since ∂x's/∂s = e'
    # Between t1 and t2

Further reduction of DM.6 requires the rule of integration by parts: 

e*∂L/∂x's| = ∫ (e'*∂L/∂x's) + ∫ (e*(∂L/∂x's)')      (DM.7)
  # Between t1 and t2

or:
  0 = ∫ (e'*∂L/∂x's) + ∫ (e*(∂L/∂x's)')
    # Because e(t1) = e(t2) = 0

  ∫ (e'*∂L/∂x's) = - ∫ (e*(∂L/∂x's)')            (DM.8)
    # Between t1 and t2

Applying DM.8 to DM.6, we have: 

∫ (e*∂L/∂xs)dt - ∫ (e*(∂L/∂x's)')dt = 0          (DM.9)

or:
  ∫ ((∂L/∂xs - (∂L/∂x's)')*e)dt = 0            (DM.10)

Now here is a big jump. According to David Morin's book, for equation DM.10 to be true for any s, the following must be true. 

∂L/∂x - (∂L/∂x')' = 0

or:
  ∂L/∂x - d(∂L/∂x')/dt = 0
    # Since ()' = d()/dt

or:
  d(∂L/∂x')/dt = ∂L/∂x                        (DM.11)

Equation DM.11 is exactly the Lagrange Equation G.5. So we have approved that Hamilton's Principle is equivalent to Lagrange Equation.

Table of Contents

 About This Book

 Introduction of Space

 Introduction of Frame of Reference

 Introduction of Time

 Introduction of Speed

 Newton's Laws of Motion

 Introduction of Special Relativity

 Time Dilation in Special Relativity

 Length Contraction in Special Relativity

 The Relativity of Simultaneity

 Introduction of Spacetime

 Minkowski Spacetime and Diagrams

 Introduction of Hamiltonian

►Introduction of Lagrangian

 What Is Lagrangian

 Action - Integral of Lagrangian

 Action - Functional of Position Function x(t)

 Hamilton's Principle - Stationary Action

 What Is Lagrange Equation

►Other Proofs of the Lagrange Equation

 Lagrange Equation on Free Fall Motion

 Lagrange Equation on Simple Harmonic Motion

 Lagrangian in Cartesian Coordinates

 Lagrange Equations in Cartesian Coordinates

 Introduction of Generalized Coordinates

 Phase Space and Phase Portrait

 References

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