Sorry, there hasn’t been much activity on this blog lately – which is an understatement, I acknowledge! But, for now, here’s a small article by Terry Tao on CNNOpinion.

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Topological Musings

April 25, 2010 in Mathematics | Tags: CNN, mathematics, Terry Tao

Sorry, there hasn’t been much activity on this blog lately – which is an understatement, I acknowledge! But, for now, here’s a small article by Terry Tao on CNNOpinion.

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## 4 comments

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May 19, 2010 at 11:46 pm

BabaNo, its ok. Don say sorry

June 11, 2010 at 2:20 pm

zigojackoGood article you link to ðŸ™‚

March 10, 2019 at 11:54 am

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values stored at the grid to the entire domain. Both (continuous) spline interpolation as well as (discontinuous) Legendre or Lagrange interpolation are popular choices. Furthermore, similar to discontinuous Galerkin methods, multiple coefficients can be stored for each cell (which yields a local reconstruction at the expense of additional memory demands). The Christoffel symbols calculations can be quite complicated, for example for dimension 2 which is the number of symbols that has a surface, there are 2 x 2 x 2 = 8 symbols and using the symmetry would be 6.

For dimension 4 the number of symbols is 64, and using symmetry this number is only reduced to 40. Certainly there are many calculations and this is just to find the equations of geodesics (after, we must solve or analyze it).

Let and let C be the set of functions such us

y(a)=c, e y(b)=d if

reaches its minimum at some then, is a solution of differential equation

We write the Lagrangian associated with this metric, ie

And then, we calculate

,

,

n order to introduce the concept of generalized coordinates, let us first consider a few simple examples.

Consider, first, the problem of a simple pendulum moving in the x-y plane. The pendulum has a length l and moves under the action of gravity, so that its potential energy is mgh. The system is illustrated in the figure below.

We could use Cartesian coordinaets x and y to describe the location of the pendulum bob, but x and y are not independent. In fact, since the length of the pendulum is constant, they are related by

This condition would need to be imposed as a constraint on the system, which can be inconvenient. It is more natural to use the angle that the pendulum makes with respect to the vertical to describe the motion. But what would be the equation of motion for ? In order to find out what this is, we only need to express the Lagrangian in terms of . Now, the Lagrangian in terms of x and y is givn by

where we have introduced a general potential function, however, for this example, we know that the potential is g

2 FORMULATION

We see immediately that if, then , and the particle will remain in circular motion with a centripetal acceleration . For , we have

which can also be written as

The last line is in the form of a conservation law, which states that the quantity,

.

This is known as the conservation of areal velocity. The constant is usually defined such that

or

Substitution of this conservation law into the radial equation gives an uncoupled equation for r alone:

The conservation law and uncoupled radial equation would have been difficult to decude directly in terms of Cartesian coordinates, however, the Lagrangeâ€™s equations provide a powerful technique for obtaining the equations of motion for systems with more than one degree of freedom. They represent arguably the most general and useful expression of the Energy Principle. Compared with other methods of obtaining equations of motion, the method offers the following potential advantages:

â€¢ Accelerations do not have to be determined; only velocities. This considerably simplifies the kinematics. Also, since velocities are squared, difficulties with algebraic sign are often avoided.

â€¢ Required number of equations is automatically obtained.

â€¢ Same basic method is used whatever set of co-ordinates might be chosen, which allows us to work with the most convenient set of co-ordinates.

We shall first consider the following form of the equations, appropriate to non-conservative systems:

(1)

The terms in the equation are now described.

It is necessary to define a set of i independent generalised co-ordinates {qi}. These will often be distances and angles. In general, a system with n degrees of freedom will require n independent generalised co-ordinates {q1, q2,â€¦, qn}. A special class of system known as non-holonomic does exist in which the number of generalised co-ordinates required exceeds the number of degrees of freedom; we shall not discuss systems of this type.

Example: Double pendulum â€“ Generalised co-ordinates are angles q1 and q2.

Generalised forces {Qi}

When the system undergoes a virtual displacement dqi in which all other co-ordinates qj remain unchanged, the virtual work done dWi = Qidqi. This defines the ith generalised force Qi.

Total kinetic energy (T)

The total kinetic energy of the system can be expressed as a function â€˜Tâ€™ of the generalised co-ordinates qi and their rates of change with time .

Example: Double pendulum

Position vectors of masses 1 and 2:

Velocity vectors of masses 1 and 2:

Total kinetic energy:

Application of Lagrangeâ€™s equation to single degree of freedom system

We first determine the equation of motion of a mass M under the action of a single horizontal force F. First, define the generalised co-ordinate q1 to be the linear displacement of the mass.

Then the generalised force Qi = F. (To see this, note that the virtual work done by F = Fdqi. This by definition is equal to Q1dq1). Next, the kinetic energy is given by . Hence equation (1) above gives:

.

Lagrangeâ€™s Equations for Conservative Systems

In a conservative system, the generalised forces can be derived from a potential function V. Potential fuThe idea of ensemble averaging can also be expressed in terms of an average over all such microstates (which comprise the ensemble). A given macroscopic property, A, and its microscopic function , which is a function of the positions and momenta of a system, i.e. the phase space vector, are related by

where is the microstate of the th member of the ensemble.

However, recall the original problem of determining the microscopic detailed motion of each individual particle in a system. In reality, measurements are made only on a single system and all the microscopic detailed motion is present. However, what one observes is still an average, but it is an average over time of the detailed motion, an average that also washes out the microscopic details. Thus, the time average and the ensemble average should be equivalent, i.e.

This statement is known as the ergodic hypothesis. A system that is ergodic is one for which, given an infinite amount of time, it will visit all possible microscopic states available to it (for Hamiltonian dynamics, this means it will visit all points on the constant energy hypersurface). No one has yet been able to prove that a particular system is truly ergodic, hence the above statement cannot be more than a supposition. However, it states that if a system is ergodic, then the ensemble average of a property can be equatePhase space distribution functions and Liouvilleâ€™s theorem

Given an ensemble with many members, each member having a different phase space vector corresponding to a different microstate, we need a way of describing how the phase space vectors of the members in the ensemble will be distributed in the phase space. That is, if we choose to observe one particular member in the ensemble, what is the probability that its phase space vector will be in a small volume around a point in the phase space at time t. This probability will be denoted

where is known as the phase space probability density or phase space distribution function. Itâ€™s properties are as follows:

2 FORMULATION

Figure 1:

The flux through the small surface area element, dS is just . Then the total flux out of volume is obtained by integrating this over the entire surface that encloses :

which follows from the divergence theorem. is the 6N dimensional gradient on the phase space

On the other hand, the rate of decrease in the number of systems out of the volume is

Equating these two quantities gives

But this result must hold for any arbitrary choice of the volume , which we may also allow to shrink to 0 so that the result holds locally, and we obtain the local result:

But

This equation

March 10, 2019 at 11:55 am

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