Generalised potential and generalised momentum

Generalised potential and generalised momentum

Generalised potential and generalised momentum

  • In today’s article, we are going to study about ‘Generalised potential and generalised momentum’. We are going to discuss whole expression for generalised potential and derivation for its standard formula. Along with that we will also discuss generalised momentum and cyclic co-ordinate briefly. So, take your notebooks in you hand and get ready to study physics in an easy and sorted way.

Generalised potential

Lagrangian for a charge particle moving in an electromagnetic field

In general the Lagrange’s equation can be written as-

Gk = (δ/δt)(δL/δqk^•) – δT/δqk  …(1)

Where Gk= -δv/δqk

For a conservative system-

Gk = (δ/δt)(δL/δqk^•) – δL/δqk  …(2)

L= T-V

Above equation is called Lagrange’s equation. We can put the value of L in equation (2) and can obtain a expression for generalised force. This force is obtained by a function U(qk, qk^•).

Hence, Gk= (δ/δt)(δU/δqk^•) – δU/δqk …(3)

Where  U(qk, qk^•) is called the velocity dependence potential or generalised potential.

In this case of a charge in electromagnetic field. The Maxwell’s first equation are-

div.B =0 and ∇×E+ δB/δt= 0

Where E and B are electric field and magnetic field. Let us assume that the charge is moving with velocity v in an electric field E and magnetic induction B. Therefore the force acting on charge can be given by following expression,

F= q(E + V×B)  …(4)

By Maxwell’s first equation we know that,

∇.B = 0

B= ∇×A

Where A is called magnetic vector potential. Now substituting the value of B in Maxwell’s equations, we get

∇×E + δB/δt= 0

∇×E + (δ/δt) ∇×A =0

And hence,

∇(E + (δA/δt)) =0

The vector quantity (E + δA/δt) as the gradient of scalar function Φ.

E + δA/δt = -∇Φ

E= -∇Φ- δA/δt  …(5)

B= ∇× A            …(6)

By substituting the values from equation (5) and (6) into equation (4) we get,

F= q[( -∇Φ- δA/δt)+ v×(∇× A)]

∴ [ v×∇×A= ∇(v.A)- (∇.v)A)]

F= q[ ( -∇Φ- δA/δt)+ ∇(v.A)- (∇.v)A))   …(7)

Let a function A= A(x,y,z,t)

dA/dt = (δA/δx).(δx/δt)+  (δA/δy).(δy/δt)+  (δA/δz).(δz/δt)+  (δA/δt)

dA/dt = (δA/δx).Vx+  (δA/δy).Vy+  (δA/δz).Vz+  (δA/δt)

Consequently, dA/dt= (∇.v)A+ δA/δt

dA/dt- δA/δt= (∇.v)A   …(8)

Substituting equation (8) into (7) we get,

F= q[(-∇Φ-δA/δt)+ ∇(v.A)-dA/dT+ δA/δt]

F= q[(-∇Φ+ ∇(v.A)-dA/dT]

The x-component of the force Fx is-

Fx= q[-(δ/δx( Φ- v.A)- δA/δt]    …(9)

-δAx/δt= (d/dt). (δ/δVk) Φ- v.A

So putting this value in equation (9),

Fx= q{ -(δ/δx( Φ- v.A)+  (d/dt). (δ/δVk) Φ- v.A}

We define a generalised potential U is given by-

U=  q -(δ/δx( Φ- v.A)

Fx = [-δ/δx + d/dt (δ/δVx)]   …(10)

The lagrange’s equation is,

qk= x

qk^• =x^•

Gk= Fx

Gk= Fx = (d/dt( δT/δVx)- δT/δx]   …(11)

Substituting Fx from equation (11) in equation (10), we get the lagrange’s equation

[d/dt {δ/δx (T-U)]- δ/δx(T-U)}]= 0

d/dt (δL/δx)- δL/δx =0

Hence, L= T-U

L= T-qΦ+ qv.A   …(12)

And equation (12) gives the Lagrangian for a charge particle moving in an electromagnetic field.

Generalised momentum and cyclic co-ordinate

In generalised momentum, we take a simple example of single particle, moving with velocity x^• along with x- axis.

The kinetic energy of particle is-

T= mx^• /2    …(1)

The derivative of T with respect to x^• is given by,

δT/δx^• =mx^•= P

δT/δx^• =P

Where P represent the momentum.

If V is not a function of velocity x^•. It means

V= V(x)

δV/δx^•= 0

Then, the momentum P can be written also as,

P= δ/δx(T-V) =δL/δx^•   …(2)  (∴ L= T-V)

Similarly, for a system a set of generalised coordinate qk and the generalised velocity qk^•, we define the generalised momentum corresponding to the generalised coordinate qk as,

Pk= δL/δx^•   …(3)

This is called conjugate momentum or canonical momentum.

For a conservative system of Lagrangian equation are given as

d/dt (δL/δqk^•) – δL/δqk^•= 0   …(4)

Substitute the value from equation (3) into (4),

dPk/dt- δL/δqk=0

Pk^•= δL/δqk   …(5)

Where the expression for Lagrangian L of a system if a coordinate qk does not appear explicity. Then,

δL/δqk=0   (6)

Which shows that, from equation (5).

Pk^•= d/dt( δL/δqk^•)= 0  ….(7)

And hence the integration-

Pk= δL/δqk^• = constant

The lagrangian function does not contain a coordinate qk explicitly. The generalised momentum Pk is a constant of motion. The coordinate qk is called cyclic.


So our article is finished and after completely reading this article, one can easily tell what is generalised potential and how it can be expressed in its mathematical form.

We can now also derive the expression for generalised momentum and cyclic co-ordinate in easiest way.

So one can say that they got a detailed information about generalised potential and generalised momentum.

IT was an article based on ‘basic concepts and mathematical form of generalised potential and generalised momentum’. We have already discussed some basic terms and important topics related to today’s article. If you haven’t checked that article you can visit it via given link below.

D’ Alembert’s principle and Lagrange’s equation

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