Lagrangian Methods in 1D Flashcards
What can we use calculus of variations for?
Instead of guessing the path, we can find function which extremises an integral.
What is the equation for the action when trying to find the minimal path?
A = integral from t0 to t1 of L(x(t), v(t), t) dt
How can we show the path varying in the variables?
x(t) -> x(t) + a(t) and hence v(t) -> v(t) + a’(t)
What do we do to the end points of the path and what does this mean for the Lagrangian?
We fix the end points: a(t0) = a(t1) = 0, so L(x, v, t) -> L(x+a, v+a’, t)
Using Taylor expansion, what does the first order of the Lagrangian for very small a(t) become?
L(x,v,t)+a(t)dL/dx + a’(t)dL/dv
What does the equation for the action become?
The new Lagrangian term subbed in so: integral from t0 to t1 of L(x,v,t)+a(t)dL/dx + a’(t)dL/dv dt
What is 𝛿A, the added on part?
New part without old part of the Action equation: integral from t0 to t1 of a(t)dL/dx + a’(t)dL/dv dt
What can we do to simplify 𝛿A?
Integrate the second term (involving a’(t)) by parts, and then use a(t0) = a(t1) = 0, so this integrated part is zero.
What is 𝛿A after the integration by parts? What do we do with this?
𝛿A = integral from t0 to t1 of a(t)*(dL/dx - d/dt(dL/dx’))
With this, for A to be stationary, we need 𝛿A = 0, so part inside the brackets = 0.
What is the Euler-Lagrange equation?
dL/dx - d/dt * dL/dx’ = 0
How can we use the E-L equation in the mass under gravity problem?
Find the different terms in the E-L equation using differentiation: find that it is equal to -mg-ma = 0, so a = -g, as expected from Newtons second law.
What do we start with for the alternative derivation of the E-L equation?
An alternative path which differs from the optimum path: x(α,t) = x(0,t)+α*Ν(t), where Ν(t) is an arbitrary function
What do we do after starting the alternative derivation of the E-L equation?
- Find the new Lagrangian is: L(x+αΝ, v+αΝ’, t)
- Know that for all Ν(t), path minimised for α = 0 as x(0,t) is optimum path (dA/dα at α = 0 = 0)
What is the integral from t0 to t1 of L(t,x,v) dt equal to?
= f(x,v)
How do we find dA/dα for the alternative derivation of the E-L equation?
- dA/dα = d/dα*integral from t0 to t1 of L dt = df/dα
- Use chain rule so df/dα = df/dx * dx/dα + df/dv * dv/dα
What do we do with df/dα after using the chain rule?
- Sub in the Lagrangian integral again and rearrange so integral is on outside of brackets
- Use equations for x and v to differentiate
- Integrate by parts
What do we get after integrating by parts?
- dA/dα = integral from t0 to t1 of Ν*(dL/dx - d/dt * dL/dv)
- if dA/dα at α = 0 = 0 for any Ν(t), we must have: dL/dx - d/dt * dL/dv = 0
- The E-L equation
How do we use the E-L equation in a 1D potential, potential energy V(x)? (find Newtons second law)
- L = T-V = 1/2 * m*v^2 - V(x)
- Apply E-L: dL/dx = -dV/dx = force due to potential, d/dt(dL/dv) = dρ/dt
- F = dρ/dt = Newtons second Law
For a swinging pendulum, what do we make the Lagrangian a function of and why?
L(θ, θ’), since the length is fixed, we can describe the position using only θ, don’t need x,y.
What is the potential energy V equal to for the pendulum?
V = const - mgLcosθ
What is the kinetic energy T equal to for the pendulum?
1/2 * m*v^2 = 1/2 * m(Lθ’)^2, where v = (Lθ’)^2
What do we do once we have found the Lagrangian for the simple pendulum?
Sub it into the E-L equation and find the differentials, then rearrange to find an equation for simple harmonic motion: θ’’ = -g/L * θ
For a rollercoaster, what can we use instead of x,y and why?
Can use arc length s(t), since it is a fixed path and x&y aren’t independent.
What is the Lagrangian equal to for the roller coaster?
L = 1/2 * m*s’^2 - mgh(s) = L(s,s’)
What do we do with the Lagrangian for the roller coaster problem?
Sub it into the E-L equation, and find s’’ = -g * dh/dm which is the equation of motion, so if we know h(s), we get s(t).
What is the generalised version of the E-L equation?
dL/dq 0 d/dt * dL/dq’ = 0, where q(t) is a generalised coordinate and may be x(t), θ(t) etc, and q’ is the related velocity.
