A piece of rubber of length L is subject to work by hydrostatic pressure and a tensional force f

step 1 This question is basically about derivatives. All you have to do is verify that the ground state

A piece of rubber of length L is subject to work by...

Key Concepts

Conjugate Variables

In thermodynamics, conjugate variables are pairs of quantities where one is extensive (like entropy or volume) and the other is intensive (like temperature or pressure) such that their product has the dimensions of energy. Recognizing these pairs, such as temperature-entropy and pressure-volume or force-length, is essential in formulating the work terms in the differential form of energy and in applying Legendre transforms to derive new potentials.

Maxwell Relations

Maxwell relations are a set of equations derived from the equality of mixed second-order partial derivatives of thermodynamic potentials. These relations provide powerful connections between different thermodynamic quantities, enabling the determination of how one variable changes with respect to another by using experimentally accessible derivatives, and they are indispensable in the theoretical analysis of thermodynamic systems.

Thermodynamic Potentials

Thermodynamic potentials, such as internal energy, Helmholtz free energy, and Gibbs free energy, are state functions that encapsulate the energy of a system and its tendency to perform work under specific constraints. They are defined to have natural variables that correspond directly to measurable thermodynamic quantities, allowing us to determine equilibrium conditions and the response of the system to changes in its environment.

Legendre Transformation

Legendre transformation is a mathematical tool used to change the natural variables of a thermodynamic potential by replacing an extensive variable with its conjugate intensive variable. This is crucial when generating new potentials (like free energy, enthalpy, etc.) whose natural variables are more convenient for a given physical process or experimental condition, thereby simplifying the analysis of the system.

Differential Work Expression

This concept refers to expressing the infinitesimal change in internal energy (dU) in terms of contributions from entropy, volume, and other work-producing or work-consuming mechanisms. It involves accounting for the different forms of work (mechanical, pressure-volume, tension, etc.) being done on or by the system, and is fundamental in connecting the mechanical and thermal aspects of a system’s energy changes.

Transcript

00:01 This question is basically about derivatives.

00:03 All you have to do is verify that the ground state wave function, psi -1 -00, satisfies schrodinger's equation with an energy given by this expression.

00:16 So all i have to do, mostly, is calculate the second derivative of the ground -state wave function with respect to x, y and z.

00:26 So let's do it for x first.

00:31 So driving psali 100 with respect to x is equal to square root of pi r0 cubed, which suggests a constant factor with respect to x times the derivative with respect to x of e to the minus r divided by r0.

00:53 Then remember that's the derivative of an exponential is equal to itself.

01:02 Just have e to minus r0 again but now we have the derivative with respect to x of the exponent of that exponential again the r0 term is constant with respect to x and so is this minus one term here and we got minus e minus r divided by r0 divided by r0 times the square root of pi r0 cubed times the derivative of r with respect to x remember that r is equal to x squared plus y squared plus z squared to one half then this derivative will result in the following so this one half factor from the exponent will drop.

02:09 We have one half here and this we multiply the same expression elevated to one half minus one which is minus one half so we got this in the denominator and now we have the derivative of the inside so the derivative of x squared plus y squared plus z squared as we are taking the derivative with respect to x only to the first term will be affected and we got the derivative of x squared which is 2 times x.

02:48 These factors of 2 simplify and we get minus e to minus r divided by r0 divided by r0 square root of pi r0 cubed times x over r now notice.

03:06 Now note of x over r now notice that this factor here is just zi -10 -so this is equal to minus zai -10 times x divided by r -0 times r.

03:25 Now we have to derivate again.

03:30 Taking the second derivative in respect to x we get the following.

03:36 Using the product rule we get the derivative of the first times the second plus the first times the derivative of the second.

03:55 Note that r0 is constant with respect to x and we got d over d x of x divided by r.

04:10 Now we already know what is the result of this derivative.

04:15 We just we have calculated it.

04:19 So we got minus minus plus and i 1 0 times x divided by r0 times r times from this term x divided by r0 times r plus psi 1 0 divided by r 0 times times the derivative of x with respect to x which is 1 divided by r plus x times the derivative of 1 over r with respect to x which is the following to d over d x of r to the minus 1.

05:05 And then we got this is equal to minus, minus psi 1 0 0 x divided by r 0 squared r squared plus psi 1 0 0 divided by r 0 times 1 over r plus x times.

05:27 So this minus 1 factor drops from the exponent and we get an r to the minus 2.

05:34 Then you have to multiply by the derivative of the inner factor, which is the derivative of r with respect to x.

05:45 Then let this minus term minus factor come inside.

05:50 To get psi 1 -0 -0 x squared divided by x -0 times r squared plus not plus anymore minus psi -1 -0 divided by r -0 times 1 over r minus x divided by r squared times d r d r over d x d r as we had already is equal to this term here.

06:28 So it is equal to x over r.

06:31 So we get it.

06:37 Then messaging this equation, we can write it as pzai 1 -0 x squared divided by r -0 r squared...

Please give Ace some feedback