Question

find the value of α and β so that the equation $\large\rm { Q = q^{ α} cos βp }$ represents a canonical transformation. also find the generating function of $\large\rm { F_{3} }$ in this case.


✓Need quality answer
✓Answer according to 5 mark Question in exam
✓No spam
✓No copied answer​

Answer 1

Answer:

In Hamiltonian mechanics, a canonical transformation is a change of canonical coordinates (q, p, t) → (Q, P, t) that preserves the form of Hamilton's equations. This is sometimes known as form invariance. It needs not preserve the form of the Hamiltonian itself. Canonical transformations are useful in their own right, and also form the basis for the Hamilton–Jacobi equations (a useful method for calculating conserved quantities) and Liouville's theorem (itself the basis for classical statistical mechanics).

Since Lagrangian mechanics is based on generalized coordinates, transformations of the coordinates q → Q do not affect the form of Lagrange's equations and, hence, do not affect the form of Hamilton's equations if we simultaneously change the momentum by a Legendre transformation into

{\displaystyle P_{i}={\frac {\partial L}{\partial {\dot {Q}}_{i}}}.} P_{i}={\frac {\partial L}{\partial {\dot {Q}}_{i}}}.

Therefore, coordinate transformations (also called point transformations) are a type of canonical transformation. However, the class of canonical transformations is much broader, since the old generalized coordinates, momenta and even time may be combined to form the new generalized coordinates and momenta. Canonical transformations that do not include the time explicitly are called restricted canonical transformations (many textbooks consider only this type).re advanced mathematics such as cotangent bundles, exterior derivatives and symplectic manifolds should read the related symplectomorphism article. (Canonical transformations are a special case of a symplectomorphism.) However, a brief introduction to the modern mathematical description is included at the end of this article.

For clarity, we restrict the presentation here to calculus and classical mechanics. Readers familiar with mo

Answer 2

v=∂0F(t,q′)+∂1F(t,q′)v′. (5.1)  

We obtain a Lagrangian L′ in the transformed coordinates by composition of L with the coordinate transformation. We require that L′(t, q′, v′) = L(t, q, v), so:

L′(t,q′,v′)=L(t,F(t,q′),∂0F(t,q′)+∂1F(t,q′)v′). (5.2)  

The momentum conjugate to q′ is

p′ =∂2L′(t,q′,v′)   =∂2L(t,F(t,q′),∂0F(t,q′)+∂1F(t,q′)v′)∂1F(t,q′)   =p∂1F(t,q′), (5.3)  

where we have used

p =∂2L(t,q,v)   =∂2L(t,F(t,q′),∂0F(t,q′)+∂1F(t,q′)v′). (5.4)