Helmholtz Equation

The Helmholtz equation is named after a German physicist and physician named Hermann von Helmholtz, original name Hermann Ludwig Ferdinand Helmholtz.

This equation corresponds to the linear partial differential equation: where \[⛛^{2}\] is the Laplacian, is the eigenvalue, and A is the eigenfunction.

In mathematics, the eigenvalue problem for the Laplace operator is called the Helmholtz equation. That’s why it is also called an eigenvalue equation.


Here, we have three functions namely:

  1. Laplacian denoted by a symbol \[⛛^{2}\]

  2. The wavenumber symbolized as k

  3. Amplitude as A.


The relation between these functions is given by:


                 \[⛛^{2}\] A + \[k^{2}\] A = 0


Here, in the case of usual waves, k corresponds to the eigenvalue and A to the eigen function which simply represents the amplitude.

Helmholtz Equation Derivation

The wave equation is given by,


( \[⛛^{2}\] - 1/ \[c^{2}\] \[∂^{2}\]/\[∂x^{2}\]) u(r, t) = 0…(1)


Separating the variables, we get,

 u(r , t) = A(r) T(t)...(2)


Now substituting (2) in (1):  


\[⛛^{2}\] A/A = 1/\[c^{2}\] T. \[d^{2}\] T/ \[dt^{2}\]


Here, the expression on LHs depends on r.  While the expression on RHS depends on t.

These two equations are valid only if both the sides are equal to some constant value.

On solving linear partial differential equations by separation of variables. We obtained two equations i.e., one for A (r)  and the other for T(t).


\[⛛^{2}\] A/A = - \[k^{2}\]….(3)


And,  1/\[c^{2}\] T.  \[d^{2}\] T/\[dt^{2}\] = - \[k^{2}\] (4)


Hence, we have obtained the helmholtz equation where - \[k^{2}\] is a separation constant.      


helmholtz Equation

\[⛛^{2}\] A + \[k^{2}\] A = (\[⛛^{2}\] + \[k^{2}\])A = 0

           

Helmholtz Free Energy Equation Derivation

Helmholtz function is given by,

 F = U - TS


Here, U = Internal energy

T = Temperature

S = Entropy

Fᵢ is the initial helmholtz function and Fᵣ being the final function. 

During the isothermal (constant temperature) reversible process,  work done will be:

                     

 W   ≤    Fᵢ - Fᵣ


This statement says that the helmholtz function gets converted to the work. That’s why this function is also called free energy in thermodynamics.


Derivation:

Let’s say an isolated system acquires a δQ heat from surroundings, while the temperature remains constant.

                

So,

Entropy gained by the system = dS

Entropy lost by surroundings = δQ/T


Acc to \[2^{nd}\] law of thermodynamics, net entropy =  positive

From Classius inequality:

                                dS - δQ/T ≥ 0

                                 dS  ≥ δQ/T


Multiplying by T both the sides, we get

                       TdS  ≥  δQ

Now putting  δQ = dU + δW (\[1^{st}\]law of thermodynamics)

 

                  TdS ≥ (dU + δW)

Now, 

                         TdS ≥  dU + δW       

Or, 

                      δW   ≤ TdS - dU

Integrating both the sides:

                 w                   Sᵣ         Uᵣ

                 ∫     δW    ≤ T  ∫dS  -  ∫ dU

                 0                   Sᵢ         Uᵢ


     W   ≤   T (Sᵣ -  Sᵢ)  - (Uᵣ -  Uᵢ)

     W   ≤   (Uᵢ - TSᵢ)  -  (Uᵣ - TSᵣ)


Now,if we observe the equation.  The terms  (Uᵢ - TSᵢ) and (Uᵣ - TSᵣ) are the initial and the final helmholtz functions.

Therefore, we can say that:

              W  ≤    Fᵢ - Fᵣ

By whatever magnitude the helmholtz function is reduced, gets converted to work.


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Helmholtz Equation Thermodynamics

The Gibbs-Helmholtz equation is a thermodynamic equation. 

This equation was named after Josiah Willard Gibbs and Hermann von Helmholtz.

This equation is used for calculating the changes in GIbbs energy of a system as a function of temperature. 

Gibbs free energy is a function of temperature and pressure given by,

                                       G = G(T,P)

      And,                          G(T) = H(T) - T S(T)

                                       Here, H is the enthalpy

                                                  S = Entropy

                             Dividing LHS and RHS by T:

                                     G(T)/T = H(T)/T - T S(T)/T

Now doing the partial differentiation on both the sides:

                                   

At constant pressure,P


   \[(∂ G(T)/T)_{p}\] = - H(T)/\[^{2}\] + 1/T \[(∂ H(T)_{P}\] /∂T - \[(∂ S(T)/∂T)_{P}\]


Since 1/T  \[(∂H(T))_{p}\] /∂T and \[(∂ S(t)/∂t)_{P}\] are equal so they are cancelled out and  we get the equation as:


       \[(∂ S/∂ T)_{P}\] = \[C_{P}\] (T)/T = 1/T \[(∂H/∂T)_{P}\]

                                             \[C_{P}\](T) = \[(∂ S/∂ T)_{P}\]

                                     \[(∂ S/∂ T)_{P}\] = - H/\[T^{2}\]          

                                                           

We Get The Equation As

\[(∂ ΔG/∂ T)_{P}\] = - ΔH/\[T^{2}\]

                                

This is the Gibbs-Helmholtz equation in thermodynamics.

Applications of Helmholtz Equation

There are various applications where the helmholtz equation is found to be important. They are hereunder:

  1. Seismology:  For the scientific study of earthquakes and its propagating elastic waves.

  2. Tsunamis

  3. Volcanic eruptions

  4. Medical imaging

  5. Electromagnetism: In the science of optics, 

  6. Gibbs-Helmholtz equation:  It is used in the calculation of change in enthalpy using change in Gibbs energy when the temperature is varied at constant pressure.

  7. CHELS: A combined Helmholtz equation-least squares abbreviated as CHELS. This  method is used for reconstructing acoustic radiation from an arbitrary object.

                    


FAQ (Frequently Asked Questions)

1. What did Helmholtz discover?

Ans: A German physician and physicist named Helmholtz had interests in the physiology of senses. 

For which he revolutionized in the field of ophthalmology with the invention of the ophthalmoscope.

Ophthalmoscope is an instrument that is used to examine the inside of a human eye.

2. How is Helmholtz free energy calculated?

Ans: We know that U is the internal energy of a system.

PV = pressure-volume product.

 TS = The temperature-entropy product. Where T is the temperature above absolute zero.

 Then by Helmholtz free energy equation:

         F = U − TS, and G = H- TS.

 Where  H = U + PV. So we get that:

                                      G = U + PV - TS


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This is how we can calculate the helmholtz free energy.


3. Can Helmholtz free energy be negative?

Ans: Since we know that work done, W = Fᵢ - Fᵣ. The final helmholtz function is always lesser than the initial one. Therefore, ΔF difference between  Fᵣ and  Fᵢ  is negative.

4. What is the difference between Helmholtz free energy and Gibbs free energy?

Ans: In a closed thermodynamic system at constant temperature and pressure, Gibbs free energy is available to do a non-PV work while Helmholtz free energy is the maximum useful non-PV work that can be extracted from a thermodynamically closed system at constant temperature and volume.