MEDICINE AND MATHEMATICS

ABSTRACT

equilibrium temperature when two phases coexist.

Along a phase transition line, the pressure and To find out the dependence of pressure on temperature are not independent of each other, since the system is univariant, that is, only one intensive parameter can be varied independently.

When the system is in a state of equilibrium, i.e., thermal, mechanical and chemical equilibrium, the temperature of the two phases has to be identical, the pressure of the two phases has to be equal and the chemical potential also should be the same in both the phases.

Representing in terms of Gibbs free energy, the criterion of equilibrium is:

at constant T and P

or,

Consider a system consisting of a liquid phase at state 1 and a vapour phase at state 1’ in a state of equilibrium. Let the temperature of the system is changed from T1 to T2 along the vaporization curve.

For the phase transition for 1 to 1’:

or

or

In reaching state 2 from state 1, the change in the Gibbs free energy of the liquid phase is given by:

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]]>The J0 Bessel function

The equation for J0 Bessel is the zeroth order Bessel equation

.

The “standard form” of differential equations is often specified as having the coefficient of the highest order derivative cancelled through. Thus in standard form the equation would be written

.

Our procedure for the series solution of this equation is to take the assumed series

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]]>EXPLORING OF ELLIPTICAL PROPERTIES FOR MEDICINE

ABSTRACT

The equation for J0 Bessel is the zeroth order Bessel equation

.

The “standard form” of differential equations is often specified as having the coefficient of the highest order derivative cancelled through. Thus in standard form the equation would be written

.

Our procedure for the series solution of this equation is to take the assumed series

and substitute it into the equation. This involves using the first and second derivatives

and

In the expression for the second derivative we have (as in the sine/cosine case above) shifted the dummy summation variable by 2 so that the sum expression contains xn explicitly.

So far we have left the sum for the first derivative unchanged. The point here is that what the differential equation contains is , and the expression for this must be written as a sum in xn. For this reason we shift the dummy variable in the series for the first derivative by 2:

1 INTRODUCTION

This may be tidied into

,

which gives a recurrence relation for the coefficients:

.

We may now build up the coefficients from the term. Starting from we find

.

Now putting gives

and so on. We see that the coefficients of all the even powers of x are given in terms of and we obtain the solution to the ODE as

.

The series specifies the J0 Bessel function:

.

So the solution to the ODE which we have discovered is a constant times the J0 Bessel function

.

Thus far this is quite good; we have discovered a new function which solves the above differential equation. But it is a second order differential equation and therefore, as with the previous SHO equation, there should be two independent solutions. Where is the other solution?

When we examined the solution of the wave equation for a drumhead we found the separated radial equation took the form of the zeroth order Bessel equation. And at that stage we simply noted that Mathematica gave, as independent solutions to that equation, the two zeroth order Bessel functions J0(x) and Y0(x). We plotted the functions and the behaviour of the functions in the vicinity of x = 0 gives us an important clue about the “other” solution.

J0(x) and Y0(x) Bessel functions

The J0(x) function goes to 1 as x goes to 0. This we see on the plot and we have discovered this in the series solution. The Y0(x) function, on the other hand, looks as if it is heading for minus infinity as x goes to 0. That is the problem.

Recall the point made when we introduced the power series method. A series

will only work when the function is “well behaved”. This is OK for J0(x), but going off to infinity is an example of “bad behaviour”; then a simple power series won’t work. We will

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]]>APPLICATIONS OF BESSEL FUNCTIONS IN LAGRANGIAN AND MEDICINE

ABSTRACT

The J0 Bessel function

The equation for J0 Bessel is the zeroth order Bessel equation

.

The “standard form” of differential equations is often specified as having the coefficient of the highest order derivative cancelled through. Thus in standard form the equation would be written

.

Our procedure for the series solution of this equation is to take the assumed series

and substitute it into the equation. This involves using the first and second derivatives

and

In the expression for the second derivative we have (as in the sine/cosine case above) shifted the dummy summation variable by 2 so that the sum expression contains xn explicitly.

So far we have left the sum for the first derivative unchanged. The point here is that what the differential equation contains is , and the expression for this must be written as a sum in xn. For this reason we shift the dummy variable in the series for the first derivative by 2:

so th

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]]>MEDICINE JOURNAL AND MATHEMATICS

ABSTRACT When the system is in a state of equilibrium, i.e., thermal, mechanical and chemical equilibrium, the temperature of the two phases has to be identical, the pressure of the two phases has to be equal and the chemical potential also should be the same in both the phases.

Representing in terms of Gibbs free energy, the criterion of equilibrium is:

at constant T and P

or,

Consider a system consisting of a liquid phase at state 1 and a vapour phase at state 1’ in a state of equilibrium. Let the temperature of the system is changed from T1 to T2 along the vaporization curve.

For the phase transition for 1 to 1’:

or

or

In reaching state 2 from state 1, the change in the Gibbs free energy of the liquid phase is given by:

Similarly, the change in the Gibbs free energy of the vapour phase in reaching the state 2’ from state 1’ is given by:

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]]>CAVALIERI AND CLAPEYRON RELATION WITH LAGRANGIAN

ABSTRACT Along a phase transition line, the pressure and temperature are not independent of each other, since the system is univariant, that is, only one intensive parameter can be varied independently.

When the system is in a state of equilibrium, i.e., thermal, mechanical and chemical equilibrium, the temperature of the two phases has to be identical, the pressure of the two phases has to be equal and the chemical potential also should be the same in both the phases.

Representing in terms of Gibbs free energy, the criterion of equilibrium is:

at constant T and P

or,

Consider a system consisting of a liquid phase at state 1 and a vapour phase at state 1’ in a state of equilibrium. Let the temperature of the system is changed from T1 to T2 along the vaporization curve.

For the phase transition for 1 to 1’:

or

or

In reaching state 2 from state 1, the change in the Gibbs free energy of the liquid phase is given by:

Similarly, the change in the Gibbs free energy of the vapour phase in reaching the state 2’ from state 1’ is given by:

Therefore,

Or

Where the subscript sat implies that the derivative is along the saturation curve.

The entropy change associated with the phase transition:

Hence,

Which is known as the Clapeyron equation

Since is always positive during the phase transition, sat will be positive or negative depending upon whether the transition is accompanied by expansion ( >0) or contraction ( > )and hence = = =RT/P.

The Clapeyron equation becomes:

or

which is known as the Clausius-Clapeyron equation.

Assume that is constant over a small temperature range, the above equation can be integrated to get,

or +constant

Hence, a plot of lnP versus 1/T yields a straight line the slope of which is equal to –(hfg/R).

Kirchoff Equation

Kirchoff relation predicts the effect of temperature on the latent heat of phase transition.

Consider the vaporization of a liquid at constant temperature and pressure as shown in figure. The late

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]]>KLEIN GORDON EQUATION AND LAGRANGIAN

ABSTRACT

temperature are not independent of each other, since the system is univariant, that is, only one intensive parameter can be varied independently.

When the system is in a state of equilibrium, i.e., thermal, mechanical and chemical equilibrium, the temperature of the two phases has to be identical, the pressure of the two phases has to be equal and the chemical potential also should be the same in both the phases.

Representing in terms of Gibbs free energy, the criterion of equilibrium is:

at constant T and P

or,

For the phase transition for 1 to 1’:

or

or

Similarly, the change in the Gibbs free energy of the vapour phase in reaching the state 2’ from state 1’ is given by:

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]]>SOME FOUNDATIONS OF KLEIN GORDON EQUATION

ABSTRACT

The Clapeyron equation becomes:

or

which is known as the Clausius-Clapeyron equation.

Assume that is constant over a small temperature range, the above equation can be integrated to get,

or +constant

Hence, a plot of lnP versus 1/T yields a straight line the slope of which is equal to –(hfg/R).

Kirchoff Equation

Kirchoff relation predicts the effect of temperature on the latent heat of phase transition.

Consider the vaporization of a liquid at constant temperature and pressure as shown in figure. The latent heat of vaporization associated with the phase change 1 to 1’ is ( – ) at temperature T. When the saturation temperature is raised to (T+dT), the latent heat of vaporization is ( – ). The change in latent heat,

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