Banking Exams Preparation, Bank PO/Clerical Entrance India, Banking Career, Ask Banking Experts

SMARTY

Newtonian Mechanics

Inertial frames:

Newton?s first law defines a class of inertial frames. Inertial frames are reference frames for which the trajectories for force-free motion are solutions to dr/dt = 0. With respect to inertial frames Newton?s second law has the form

F = dp/dt. (r = coordinate, F = force, p = mv momentum)

Let F_ik be the force that particle i exerts on particle k. Newton?s third law states that F_ik= -F_ki.

Newton?s laws are well suited for the study of unconstrained mechanical systems. Constraints, such as requiring a particle to follow a given curve in space, tell us that there are external forces, but do not tell us what these forces are. The forces are only known in terms of their effect on the motion.

Conservation laws are very important tools in solving mechanics problems.

	For a system of particles momentum is conserved if F_ext= 0; F_ext= 0 Û P = constant.
	Angular momentum (L = r´p) is conserved if the torque t_ext= 0; t_ext= 0 Û L = constant.
	Energy E = T + U is conserved if all forces are conservative; òF× dr = 0 Û T + U = constant.

Formulas:

Laws:
Newton's 2nd law:	F = dp/dt
Newton's third law:	F_ik= -F_ki
Forces:
Static and kinetic friction:	f_s£ m_sN, f_k= m_kN
Gravity:
Uniform circular motion:	F = mv²/r
Hooke's law:	F = -kr, F_x= -kx
Concepts:
Work:	W = F×d
Kinetic energy:	K = (1/2)mv²
Work-kinetic energy theorem:	W_net= DK = (1/2)m(v_f²-v_i²)
Elastic potential energy:	U = (1/2)kx²
Gravitational potential energy:
Conservative systems:	E = K + U, F_x= -dU/dx
Power:	P = F·v or P = dW/dt
Momentum:	p = mv
Impulse:	I = Dp = F_avgDt
Angular momentum:	L = r´p
Torque	t = r´F
Angular momentum and torque:	, , dL = tdt

Impulsive forces

Forces and torques that act so powerfully but so briefly that they produce finite changes in linear and angular momentum while the system undergoes negligible displacement are said to be impulsive.

Linear Impulse: dp = Fdt, Dp = òFdt, Dp = F_avgDt
The integral of force over time as Dt approaches 0 is called the impulse of the force.

Angular impulse: dL = tdt, DL = òtdt, DL = t_avgDt
The integral of torque over time as Dt approaches 0 is called the angular impulse of the torque.

Collisions

In collisions, it is assumed that the colliding particles interact for such a short time, that the impulse due to external forces is negligible. Thus the total momentum of the system just before the collision is the same as the total momentum just after the collision.

	Elastic collision: momentum is conserved, mechanical energy is conserved
	Inelastic collisions: momentum is conserved, mechanical energy is not conserved

Lagrangian Mechanics

Constraints

Imposing constraints on a system is simply another way of stating that there are forces present in the problem that cannot be specified directly, but are known in term of their effect on the motion of the system. Holonomic constraints are constraints of the form

f_m(r₁,r₂,r₃,¼,r_n,t) = 0, m = 1,2,3, ? ,k.

They reduce the number of degrees of freedom of the system; k equations of constraints reduce the number of the degree of freedom of an n-particle system from 3n to 3n - k, if the constraints are holonomic.

If the constraints are holonomic, then the forces of constraints do no virtual work.

Consider a virtual displacement of the system, i.e. an infinitesimal change in the coordinates of the system, denoted by dr_i, consistent with the constraints imposed on the system at a given instant t. The work done by the force in the virtual displacement dr_iis called the virtual work. For holonomic constraints, the forces of constraints are perpendicular to the virtual displacements and do no virtual work.

Generalized coordinates

Any set of independent quantities q₁, q₂, ? , q_s, which completely define the position of the system with s degrees of freedom, are called generalized coordinates of the system, and the derivatives are called generalized velocities.
Examples:

	A particle is constraint to move in the x-y plane, the equation of constraint is z = 0, the constraint is holonomic. Possible generalized coordinates for the system with two degrees of freedom are x, y; r, f; ? .
	A particle is constraint to move on a circle in the x-y plane, the equations of constraints are z = 0, x²+ y²- r²= 0. The constraints are holonomic. Possible generalized coordinates for the system with 1 degree of freedom are f ; f³, ? .

The generalized coordinates q₁, ? , q_s can be expressed in terms of the Cartesian coordinates the system.

q₁ = q₁(r_1,r_{2, ? ,}r_n), ? , q_n = q_n(r_1,r_{2, ? ,}r_n).

These equations, together with the equation of constraints, can be inverted to find the r?s in terms of the q?s.

Lagrangian Mechanics

Assume a system has n independent generalized coordinates {q_i}. Assume that the generalized applied forces

are given by

,

with U some scalar function, i.e. the generalized applied forces are derivable from a potential. Then the equations of motion may be obtained from Lagrange?s equations,

,

where L = T - U is the Lagrangian of the system. L is a function of the coordinates and the velocities.

If not all the forces acting on the system are derivable from a potential, then Lagrange's equations can be written in the form

,

where L contains the potential of the conservative forces and Q_j represents the generalized forces not arising from a potential.

Define the generalized momentum or conjugate momentum or canonical momentum through

.

If the Lagrangian does not contain a given coordinate q_j then the coordinate is said to be cyclic and the corresponding conjugate momentum p_j is conserved.

The Hamiltonian H of a system is given by

.

H is a function of the generalized coordinates and momenta of the system. The equations of motion can be obtained from Hamilton?s equations,

.

	If the Lagrangian does not explicitly depend on time, then the Hamiltonian does not explicitly depend on time and H is a constant of motion. [If H does explicitly depend on time, H = H(t), then H is not a constant of motion.]
	If the generalized coordinates do not explicitly depend on time, then H = T + U = E, the total energy of the system. [If the generalized coordinates do explicitly depend on time, then H is not the total energy of the system.]
	So only if Lagrangian does not explicitly depend on time and the generalized coordinates do not explicitly depend on time, then H = T + U = E and the energy is a constant of motion.

Lagrange multipliers

Assume you have chosen coordinate for a system that are not independent, but are connected by m equations of constraints of the form

.

Then the equations of motion can be obtained from

(n equations), and

(m equations).

We have m + n equations and m + n unknowns, the n coordinates and the m l?s. The l_l are called the undetermined Lagrange multipliers.

is the generalized force of constraint associated with the coordinate q_k

Small oscillations

Let

.

Then solutions of the form q_j = Re(A_jeⁱ^w^t) can be found. We can find the w² from det (k_ij-w ²T_ij) = 0. For a system with n degrees of freedom, n characteristic frequencies w_a can be found. Some frequencies may be degenerate.
For a particular frequency w_a we solve

to find the A_j_a.

[While the secular equation det(k_ij-w ²T_ij) = 0 can in principle always be solved, it is often simpler to find the normal modes by using physical insight and noting the symmetries of the system.]

The most general solution for each coordinate q_j is a sum of simple harmonic oscillations in all of the frequencies w_a .

Rigid body motion

A rigid body is a system of particles in which the distances between the particles do not vary. To describe the motion of a rigid body we use two systems of coordinates, a space-fixed system X, Y, Z, and a moving system x, y, z, which is rigidly fixed in the body and participates in its motion.

Let the origin of the body-fixed system be the body?s center of mass (CM). The orientation of the axes of that system relative to the axes of the space-fixed system is given by three independent angles. The vector R points from the origin of the spaced-fixed system to the CM of the body. Thus a rigid body is a mechanical system with six degrees of freedom.

Let r denote the position of an arbitrary point P in the body-fixed system. In the space fixed system its position is given by r + R, and its velocity is

v = d(R + r)/dt = dR/dt + dr/dt = V + W ´ r.

Here V is the velocity of the CM and W is the angular velocity of the body. The direction of W is along the axis of rotation and W = df/dt.

The kinetic energy of the body is

T = (1/2)Sm_iv_i² = (1/2)Sm_i(V + W ´ r)².

We rewrite

T = (1/2)MV² + (1/2)Sm_i(W²r_i² - (W×r_i)²), M = Sm_i, Sm_ir_i = 0.

We find T = T_CM+ T_rot, i.e. the kinetic energy is the sum of the kinetic energy of the motion of the CM and the kinetic energy of the rotation about the CM. In component form we write

.

Here

is the inertia tensor. The W_i are the components of W along the axis of the body fixed system. For a continuous system

.

By appropriate choice of the orientation of the body-fixed coordinate system the inertia tensor can be reduced to diagonal form. The directions of the axes x_i are then called the principal axes of inertia and the diagonal components of the tensor are then called the principal moments of inertia. Then

.

Definitions:

Asymmetrical top:
Symmetrical top:
Spherical top:

Let L denote the angular momentum about the CM of the body.

,

which in component form yields

.

If x₁, x₂, and x₃ are the principal axes of inertia, then

L₁ = I₁W₁, L₂ = I₂W₂, L₃ = I₃W₃.

The Lagrangian of a rigid body is

.

The equations of motion

The equations of motion of a rigid body are dP/dt = F, where P = total momentum and F = external forces, and dL/dt = t , where L = angular momentum about CM and t = total torque produced by external forces.

Let A be an arbitrary vector and dA/dt its rate of change with respect to the space fixed axes, d'A/dt its rate of change with respect to the body fixed axes.

dA/dt = d'A/dt + W ´ A.

Therefore

dP/dt = d'P/dt + W ´ P = F and dL/dt = d'L/dt + W ´ L = t .

Let the body fixed axes be the principal axes of inertia of the body. Then

L_i = I_iW_i and d'L_i/dt = I_idW_i/dt,

and we have Euler?s equations:

The Eulerian angles

The orientation of the body-fixed coordinate system with respect to the space-fixed coordinate system is described by three angles. These angles are often taken as the Eulerian angles, defined in the figure.

We can express the components of the angular velocity W along the body-fixed axes x, y, z in terms of the Eulerian angles and their derivatives.

Formulas

Center of mass:

Motion in a non-inertial frame

In non-inertial frames fictitious forces appear. Consider a particle moving with velocity v in a reference frame K which moves with velocity V(t) relative to the inertial frame K₀ and rotates with angular velocity W(t).

The Lagrangian of the particle is

L = (1/2)mv² + mv × (W ´ r) + (1/2)m(W ´ r)² - mW×r - U, with W = dV(t)/dt.
¶L/¶v = mv + m(W ´ r), ¶L/¶r = m(v ´ W) + m(W ´ r) ´ W - mW - ¶U/¶r.

The equations of motion are

mdv/dt = -¶U/¶r - mdV/dt + mr ´ dW/dt - 2mW ´ v - mW ´ (W ´ r).

Here

	-mdV/dt = fictitious force due to acceleration of frame
	mr ´ dW/dt = fictitious force due to non-uniform rotation of frame
	-2mW ´ v = Coriolis force
	-mW ´ (W ´ r) = Centifugal force

For a uniformly rotating frame dW/dt = 0, dV/dt, and the equations of motion are

mdv/dt = -¶U/dr - 2mW ´ v - mW ´ (W ´ r).

Motion in a central potential

Consider a particle moving in a central potential. In a central potential the energy E and the angular momentum M are conserved. The motion is in a plane. Choose this plane to be the x-y plane. Then

.

Kepler?s second law:

.

The area swept out per unit time is constant for all central potentials.

Equations of motion involving r only:

yields

.

Lagrange?s equations yield

.

Equations for the orbit:

.

From

we obtain

,

.

From

we obtain

,

or

.

The Kepler problem

Let

.

The equation for the orbit yields

.

This is the equation of a conic section. Here e is the eccentricity.

				hyperbola
				parabola
				ellipse
				circle

The ellipse and the circle are closed orbits. For closed orbits we have

semi-major axis:

semi-minor axis:

From

we find the period

.

This is Kepler?s third law.

Two interacting particles

In the CM frame of two interacting particles the Lagrangian

can be written as

.

The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass m moving in a central potential U(r).

Elastic scattering

Consider the scattering of a particle by a central potential. We define the differential scattering cross section

through the following expression:

# of particles scattered into the solid angle dW per unit time = Is(W)dW ,

where I is the intensity of the incident beam, i.e the number of beam particles per unit area per unit time. For a central potential s(W) is independent of f. We write

.

The number of particles scattered through an angle between q and q +dq per unit time is

.

We define the impact parameter b through

,

where M is the angular momentum and v₀ is the incident speed at infinite distance. Once E and b are fixed, the scattering angle is uniquely determined.

.

In a central potential the motion is in a plane and M and E are constant.

From

we find

.

Let q be the angle between the incident and the scattered direction and f₀ be the angle between r( z=-¥) and r_min. Then

and

determines u_max. We have

q =p-2f₀ for a repulsive potential,

q =2f₀-p for an attractive potential; (or q=p-2f₀, q < 0).

If

then

.

Rutherford?s formula

Frame transformations

The number of particles scattered into a detector is the same in the laboratory and in the CM frame. Therefore

Special theory of relativity

An inertial fame is a reference frame in which all relative accelerations due to external forces are eliminated.

Postulates

I. In vacuum, light propagates with respect to any inertial frame and in all directions with the universal speed c. This speed is a constant of nature.
II. The laws of nature are the same in all inertial reference frames.

The Lorentz transformation

Consider two reference frames K and K?. Assume that the coordinate axes in the two frames are parallel and that the origins of the coordinates coincide at t = t? = 0. Assume that K? is moving with velocity

with respect to K. The Lorentz transformation gives the coordinates of a space-time point (x₀,x₁,x₂,x₃)=(ct,x,y,z) in K in terms of its coordinates (x'₀,x'₁,x'₂,x'₃)=(ct',x',y', z') in K? and vice versa.

.

Since 0 £ b £ 1, we may write b=tanhB, where B is the boost parameter or the rapidity.

.

Then

and

,

reminiscent of a rotation. We define as a 4-vector any set of 4 quantities which transform like (x₀,x₁,x₂,x₃) under a Lorentz transformation; (a₀,a₁,a₂,a₃) is a 4-vector if

,

or

.

The "dot product"

is invariant under a Lorentz transformation.

The 4-vector (x₀,r) defines an event. The space-time interval between two events is ds.

.

In a reference frame in which two events have the same space coordinates dr=0 and

where

is the proper time interval. It is a Lorentz invariant quantity.

.

Transformation of velocities

A particle moves in K with velocity

. Its velocity in K?,

, is given by

.

It is impossible to obtain speeds greater than c.

Important 4-vectors

4-velocity:

,

a Lorentz invariant scalar.

4-vector momentum:

We define

.

Then

and

.

Relativistic collisions

In every reference frame energy and momentum are conserved.

For each component p_m of the 4-vector (p₀,p₁,p₂,p₃) we have

,

where i denotes the particles going into the collision and j denotes the particles emerging from the collision.

For transformations between reference frames we have

.

This is a consequence of the invariance of the dot product under a Lorentz transformation.

Hamilton?s equations

The Hamiltonian of a system is expresses in terms of the generalized coordinates and the generalized momenta of the system,

.

Hamilton?s canonical equations are

.

Symplectic notation:

Let

,

for a system with n degrees of freedom. Let

,

where I is the

identity matrix. Then

.

Properties of the matrix J:

.

Canonical transformations

In the Hamiltonian formalism we have 2n independent variables. We denote these variables by q and p and write H(q,p,t). We can also express H in terms of 2n different independent variables Q and P, where Q_i=Q_i(q,p,t) and P_i=P_i(q,p,t).

Canonical transformations, or contact transformations are a special class of transformations which preserve the structure of the canonical equations, i.e. for which there exists a function K(Q,P,t) such that

.

If we can find a generating function F, such that

then the transformation is canonical. F must be a function of 2n independent variables and

.

Canonical transformations in symplectic notation

Let h denote a set of canonical variables,

. Let c denote a second set of variables. Let

.

The transformation is canonical if and only if

.

Poisson brackets

Let H=H(q,p,t) and f=f(q,p,t). H is the Hamiltonian and f is an arbitrary function.

Then

.

The values of all Poisson brackets are independent of the set of canonical variables they are expressed in.

A transformation is canonical if and only if [Q,P] = 1.

Canonical transformations

In the Hamiltonian formalism we have 2n independent variables. We denote these variables by q and p and write H (q,p,t). We can also express H in terms of 2n different independent variables Q and P, where Q_i=Q_i (q,p,t) and P_i=P_i(q,p,t).

Canonical transformations, or contact transformations are a special class of transformations which preserve the structure of the canonical equations, i.e. for which there exists a function K(Q,P,t) such that

.

If we can find a generating function F, such that

then the transformation is canonical. F must be a function of 2n independent variables and

.

Canonical transformations in symplectic notation

Let h denote a set of canonical variables,

. Let c denote a second set of variables. Let

.

The transformation is canonical if and only if

.

Poisson brackets

Let H=H(q,p,t) and f=f(q,p,t). H is the Hamiltonian and f is an arbitrary function.

Then

.

The values of all Poisson brackets are independent of the set of canonical variables they are expressed in.

A transformation is canonical if and only if [Q,P] = 1.

The Hamilton-Jacobi equation

Assume we have a system with a Hamiltonian

. If we make a canonical transformation with a generating function F₂(q,P,t) such that

and

we obtain the Hamilton-Jacobi equation. The function F₂ is called Hamilton?s principal function S, and the Hamilton-Jacobi equation is

.

Since K=0,

,

i.e. all new variables are constant in time. Therefore

, where the a?s are the transformed momenta. We have

.

To solve a problem using the Hamilton-Jacobi method

	solve the Hamilton-Jacobi equation .
	The a?s are the transformed momenta. Find the transformed coordinates from
	These equations give the coordinates q as a function of 2n constants a and b. Solve for q. Solve for p by differentiating

Simplification:

If H=H(q,p), then

Here a₁ is a constant, often it is the energy. W(q,a) is called Hamilton?s characteristic function. W(q,a) is a generating function for its own canonical transformation.

To solve a problem using the Hamilton?s characteristic function

	solve .
	The a?s are the transformed momenta. Find the transformed coordinates from
	These equations give the coordinates q as a function of 2n constants a and b. Solve for q. Solve for p by differentiating

Action-angle variables

Assume a conservative system with one degree of freedom which has periodic motion. Let H(q,p)=a, p=p(q,a). Define

The integration is over one complete period of the motion. J is called the action variable, and it is taken to be the new constant momentum in Hamilton?s principal function

Now we look at the canonical transformation generated by W(q,J). J is the new constant momentum and w the new coordinate. We have

w is called the angle variable. n is the frequency associated with the periodic motion. We can find n without finding a complete solution of the problem.