**14. Appendix: Mathematical derivations**

The proofs to many of the derivations below are too simple and are omitted for brevity. But the propositions are listed for purposes of reference and completeness of exposition. *Notation.* We will consistently use the following notation throughout this APPENDIX:

*E t*( ) is a real-valued function of the real-variable *t*

 *tts* is an 'interval of t' *E Et Es* () () is the 'change of *E'*  ( ) *t s P E u du* is the 'accumulation of *E'*  <sup>1</sup> ( ) *t av s E E E u du t s* is the 'average of *E* '

*Dx* indicates 'differentiation with respect to *x* '

*r* is a constant, often an 'exponential rate of growth'

## **14.1 Part I: Exponential functions**

We will use the following characterization of exponential functions without proof:

*Basic Characterization:* <sup>0</sup> ( ) *rt Et Ee if and only if D E rE <sup>t</sup> Characterization 1:* <sup>0</sup> ( ) *rt Et Ee if and only if E Pr Proof:* Assume that <sup>0</sup> ( ) *rt Et Ee* . We have that 0 0 *rt rs E Et Es Ee Ee* ,

$$\text{while } P = \underset{s}{\int} E\_0 e^{ru} du = \frac{1}{r} \Big[ E\_0 e^{rt} - E\_0 e^{rs} \Big] = \frac{\Delta E}{r} \text{ . Therefore } \Delta E = Pr \text{ .} $$

Assume next that *E Pr* . Differentiating with respect to *t*, *D E rD P rE t t* .

Therefore by the *Basic Characterization,* <sup>0</sup> ( ) *rt Et Ee* . *q.e.d* 

The Thermodynamics *in* Planck's Law 713

*e*

*<sup>E</sup> E s*

*Pr E s <sup>e</sup>*

*<sup>e</sup>* as *t s* , we apply L'Hopital's Rule.

*t*

. *q.e.d.* 

*.* Since

. *Conversely,* if ( ) <sup>1</sup> *E Eav*

*e* 

*<sup>E</sup> E s*

*av*

*Pr E s <sup>e</sup>*

*Pr r t E*

As a direct consequence of the above, we have the following interesting and important result:

*E Es DEs*

*t s* 

 and ( ) <sup>1</sup> *Pr Eav Pr E s <sup>e</sup>*

In the following we provide a direct and independent proof of *Characterization 3* .

and ( ) ( ) *<sup>s</sup>*

*t*

*s E E u du t s* . 2

*<sup>E</sup> E s e* 

*E s*( )

1 *E Eav E*

*<sup>E</sup> E s e* 

is constant with

, then by

is invariant with respect

*e* 

1 *E Eav E*

, *Theorem 2* can also be written as,

are independent of *t* , *E* .

*E*

*Theorem 2: a) For any differentiable function E t*( )*,* lim ( ) <sup>1</sup> *t s E Eav*

 *b) For any integrable function E t*( )*,* lim ( ) <sup>1</sup> *r t t s*

and <sup>0</sup>

1 0 *r t Pr*

( ) lim lim 1 ( )

<sup>2</sup> ( ) lim

*Pr E s r E s e er*

*e*

*<sup>E</sup> E s e* 

*E E t s t t*

 *is invariant with t if and only if* ( ) <sup>1</sup> *E E*

*E E t s t s E E t t E DEt e DEt E DE E*

( ) *t*

*E DEt e DEt E DE E* 

*e*

*Proof:* Since <sup>0</sup>

*e* 

*Corollary A:* <sup>1</sup> *E E*

to *t*. *q.e.d* 

*e* 

*av* 1 0 *E E E*

since 0 *E* and *E Es* ( ) as *t s* .

*E*

Likewise, we have ( ) lim lim ( ) <sup>1</sup> *r t r t t s t s*

respect to *t*, we have ( ) <sup>1</sup> *E Eav*

Since it is always true by definitions that

*<sup>E</sup> E s e* 

Corollary B: ( ) <sup>1</sup> *E Eav*

We first prove the following, *Lemma:* For any *E*, ( ) ( ) *<sup>t</sup>*

*Proof:* Using *Theorem 2* we have lim ( ) <sup>1</sup> *t s E Eav*

*Characterization 3*, 0 ( ) *rs Es Ee* . Since *E s*( ) is a constant,

*Theorem 2a: For any integrable function E t*( )*,* lim ( ) <sup>1</sup> *t s Pr Eav*

**14.3 Part III: Independent proof of** *Characterization 3* 

*Et E DEt*

*Proof:* We let *tts* and <sup>1</sup> ( )

*t s*

Differentiating with respect to *t* we have () () *<sup>t</sup> t s DEt E Et* .

*Theorem 1:* <sup>0</sup> ( ) *rt Et Ee if and only if* <sup>1</sup> *r t Pr <sup>e</sup> is invariant with respect to t Proof:* Assume that <sup>0</sup> ( ) *rt Et Ee* . Then we have, for *fixed s*,

$$P = \int\_{s} E\_0 e^{ru} du = \frac{E\_0}{r} \left[ e^{rt} - e^{rs} \right] = \frac{E\_0 e^{rs}}{r} \left[ e^{r(t-s)} - 1 \right] = \frac{E(s)}{r} \left( e^{r(t-s)} - 1 \right)$$

and from this we get that ( ) <sup>1</sup> *r t Pr E s <sup>e</sup>* = constant. Assume next that 1 *r t Pr <sup>C</sup> <sup>e</sup>* is constant with respect to *t,* for *fixed s*.

$$\text{Therefore, } D\_t \left[ \frac{Pr}{e^{r\Lambda t} - 1} \right] = \frac{rE(t) \cdot \left[ e^{r\Lambda t} - 1 \right] - rP \cdot \left[ re^{r\Lambda t} \right]}{\left( e^{r\Lambda t} - 1 \right)^2} = 0 \quad \text{and so, } E(t) = \left( \frac{Pr}{e^{r\Lambda t} - 1} \right) e^{r\Lambda t} = \text{C} \cdot e^{r\Lambda t}$$

where *C* is constant. Letting *t s* we get *Es C* ( ) *.* We can rewrite this as ( ) <sup>0</sup> () () *rt s rt Et Ese Ee* . *q.e.d* From the above, we have

*Characterization 2:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) ( ) <sup>1</sup> *rt s Pr E s <sup>e</sup>*

Clearly by definition of *Eav* , *av Pr r t E* . We can write 1 *r t Pr <sup>e</sup>* equivalently as 1 *Pr Eav Pr <sup>e</sup>* in the above. *Theorem 1* above can therefore be restated as, *Theorem 1a:* <sup>0</sup> ( ) *rt Et Ee if and only if* <sup>1</sup> *Pr Eav Pr <sup>e</sup> is invariant with t* 

The above *Characterization 2* can then be restated as

*Characterization 2a:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) <sup>1</sup> *Pr Eav Pr E s <sup>e</sup>* .

But if ( ) <sup>1</sup> *Pr Eav Pr E s <sup>e</sup>* , then by *Characterization 2a* , 0 ( ) *rt Et Ee* . Then, by *Characterization 1*,

we must have that *E Pr* . And so we can write equivalently ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e* . We have the following equivalence,

*Characterization 3:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e* As we've seen above, it is always true that *av Pr r t <sup>E</sup>* . But for exponential functions *E t*( ) we also have that *E Pr* . So, for exponential functions we have the following.

$$\text{Characterization 4: } E(t) = E\_0 e^{rt} \quad \text{if } and \text{ only if } \quad \frac{\Delta E}{E\_{av}} = r\Delta t$$

#### **14.2 Part II: Integrable functions**

We next consider that *E t*( )is any function. In this case, we have the following.

*<sup>t</sup> rs ru rt rs rt s rt s*

*rr r* 

*<sup>e</sup>* = constant. Assume next that

where *C* is constant. Letting *t s* we get *Es C* ( ) *.* We can rewrite this as

*Pr E s <sup>e</sup>*

*<sup>e</sup> is invariant with t* 

*<sup>e</sup>* , then by *Characterization 2a* , 0 ( ) *rt Et Ee* . Then, by *Characterization 1*,

*Pr E s <sup>e</sup>* .

*<sup>E</sup> E s*

1 *r t Pr*

*E E <sup>e</sup> E s P E e du e e e <sup>e</sup>*

*<sup>e</sup> is invariant with respect to t* 

0 0 ( ) ( )

( ) 1 1

and so, ( ) <sup>1</sup>

1 *r t Pr <sup>C</sup>*

*r t Pr E t e Ce*

 

*e*

*<sup>e</sup>* equivalently as 1 *Pr Eav*

*<sup>E</sup> E s*

. We have

*e* 

*<sup>E</sup>* . But for exponential functions *E t*( ) we

*<sup>e</sup>* is constant

*rt rt*

*Pr <sup>e</sup>* in

*Pr*

 <sup>2</sup> () 1 <sup>0</sup>

*av*

. We can write

*Pr*

we must have that *E Pr* . And so we can write equivalently ( ) <sup>1</sup> *E Eav*

also have that *E Pr* . So, for exponential functions we have the following.

We next consider that *E t*( )is any function. In this case, we have the following.

*e* 

> *av Pr r t*

> > *av <sup>E</sup> r t*

*E* 

*Pr r t E*

*r t r t*

*Theorem 1:* <sup>0</sup> ( ) *rt Et Ee if and only if* <sup>1</sup> *r t*

0

*s*

and from this we get that ( ) <sup>1</sup> *r t*

with respect to *t,* for *fixed s*.

( ) <sup>0</sup> () () *rt s rt Et Ese Ee* . *q.e.d* From the above, we have

Clearly by definition of *Eav* ,

But if ( ) <sup>1</sup> *Pr Eav*

the following equivalence,

*Pr E s*

*D*

Therefore,

*Proof:* Assume that <sup>0</sup> ( ) *rt Et Ee* . Then we have, for *fixed s*,

*Pr E s*

1 1

*Pr rE t e rP re*

*<sup>t</sup> r t r t*

*Characterization 2:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) ( ) <sup>1</sup> *rt s*

the above. *Theorem 1* above can therefore be restated as,

*Characterization 2a:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) <sup>1</sup> *Pr Eav*

*Characterization 3:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) <sup>1</sup> *E Eav*

The above *Characterization 2* can then be restated as

*Theorem 1a:* <sup>0</sup> ( ) *rt Et Ee if and only if* <sup>1</sup> *Pr Eav*

As we've seen above, it is always true that

*Characterization 4:* <sup>0</sup> ( ) *rt Et Ee if and only if* 

**14.2 Part II: Integrable functions** 

*e e*

*Theorem 2: a) For any differentiable function E t*( )*,* lim ( ) <sup>1</sup> *t s E Eav <sup>E</sup> E s e b) For any integrable function E t*( )*,* lim ( ) <sup>1</sup> *r t t s Pr E s <sup>e</sup>*

*Proof:* Since <sup>0</sup> *av* 1 0 *E E E e* and <sup>0</sup> 1 0 *r t Pr <sup>e</sup>* as *t s* , we apply L'Hopital's Rule.

$$\lim\_{t \to s} \frac{\Delta E}{e^{\Delta E / \overline{E}} - 1} = \lim\_{t \to s} \frac{D\_t E(t)}{e^{\Delta E / \overline{E}} \cdot \left[ \frac{D\_t E(t) \cdot \overline{E} - D\_t \overline{E} \cdot \Delta E}{\overline{E}^2} \right]}$$

$$=\lim\_{t\to s} \frac{\overline{E}^2 \cdot D\_t E(t)}{e^{\Lambda E \sqrt{E}} \cdot \left[D\_t E(t) \cdot \overline{E} - D\_t \overline{E} \cdot \Lambda E\right]} = E(s)$$

since 0 *E* and *E Es* ( ) as *t s* . Likewise, we have ( ) lim lim ( ) <sup>1</sup> *r t r t t s t s Pr E s r E s e er* . *q.e.d. Corollary A:* <sup>1</sup> *E E E e is invariant with t if and only if* ( ) <sup>1</sup> *E E <sup>E</sup> E s e Proof:* Using *Theorem 2* we have lim ( ) <sup>1</sup> *t s E Eav <sup>E</sup> E s e .* Since 1 *E Eav E e* is constant with respect to *t*, we have ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e* . *Conversely,* if ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e* , then by *Characterization 3*, 0 ( ) *rs Es Ee* . Since *E s*( ) is a constant, 1 *E Eav E e* is invariant with respect to *t*. *q.e.d* 

Since it is always true by definitions that *av Pr r t E* , *Theorem 2* can also be written as, *Theorem 2a: For any integrable function E t*( )*,* lim ( ) <sup>1</sup> *t s Pr Eav Pr E s <sup>e</sup>* As a direct consequence of the above, we have the following interesting and important result:

Corollary B: ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e* and ( ) <sup>1</sup> *Pr Eav Pr E s <sup>e</sup>* are independent of *t* , *E* .

#### **14.3 Part III: Independent proof of** *Characterization 3*

In the following we provide a direct and independent proof of *Characterization 3* . We first prove the following,

*Lemma:* For any *E*, ( ) ( ) *<sup>t</sup> Et E DEt t s* and ( ) ( ) *<sup>s</sup> E Es DEs t s t*

*Proof:* We let *tts* and <sup>1</sup> ( ) *s E E u du t s* .

Differentiating with respect to *t* we have () () *<sup>t</sup> t s DEt E Et* .

The Thermodynamics *in* Planck's Law 715

( ) () () ( ) ( ) ( ) ( )

 , or as *D t <sup>s</sup>* 

 

*E s* and therefore 0 ( ) *rs Es Ee* . *q.e.d.*

I am indebted to Segun Chanillo, Prof. of Mathematics, Rutgers University for his encouragement, when all others thought my efforts were futile. Also, I am deeply grateful to Hayrani Oz, Prof. of Aerospace Engineering, Ohio State University, who discovered my posts on the web and was the first to recognize the significance of my results in Physics. Special thanks also to Miguel Bayona of The Lawrenceville School for his friendship and help with the graphics in this chapter. And Alexander Morisse who is my best and severest

Frank, Adam (2010), *Who Wrote the Book of Physics?* Discover Magazine (April 2010)

Physics Society Newsletter, Winter 2001/2002 Vol 1 Issue 4 http://www.oufusion.org.uk/pdf/FusionNewsWinter01.pdf

presented at Cambridge University, England, March 27, 2008,

Vibration, 329 (2010) 2565–2602, doi:10.1016/j.jsv.2009.12.021

Ragazas, C. (2010) *A Planck-like Characterization of Exponential Functions*, knol

Keesing, Richard (2001). *Einstein, Millikan and the Photoelectric Effect,* Open University

Öz, H., Algebraic Evolutionary Energy Method for Dynamics and Control, in: *Computational* 

Öz, H., Evolutionary Energy Method (EEM): An Aerothermoservoelectroelastic

Öz , H., *The Law Of Evolutionary Enerxaction and Evolutionary Enerxaction Dynamics* , Seminar

Öz , Hayrani; John K. Ramsey, *Time modes and nonlinear systems,* Journal of Sound and

Planck, Max (1901) *On the Energy Distribution in the Blackbody Spectrum,* Ann. Phys. 4, 553,

*Nonlinear Aeroelasticity for Multidisciplinary Analysis and Design*, AFRL, VA-WP-TR-

Application,: *Variational and Extremum Principles in Macroscopic Systems*, Elsevier,

 

= constant.

*DEs DEs E Es E E Es DEs Es E Es E t E*

*s s <sup>s</sup>*

*Ds <sup>t</sup>* 

Differentiating (A4) above with respect to *s*, we get <sup>2</sup> *D D tD ss s*

( ) *DEs <sup>s</sup> <sup>r</sup>*

. Working backward, this gives *D r <sup>s</sup>*

   2

. (A4)

 .

We can rewrite the above as follows,

and so, ( ) <sup>1</sup>

Using (A1), this can be written as

**15. Acknowledgement** 

From (A1), we then have that ( )

critic of the Physics in these results.

2002 -XXXX, 2002, pp. 96-162.

http://talks.cam.ac.uk/show/archive/12743

2005, pp. 641-670.

1901

Therefore, <sup>2</sup> 0 *Ds*

**16. References** 

( ) *DEs <sup>s</sup> E Es t E* .

Rewriting, we have ( ) ( ) *<sup>t</sup> Et E DEt t s* . Differentiating with respect to *s* we have () () *<sup>s</sup> t s DEs E Es* . Rewriting, we have ( ) ( ) *<sup>s</sup> E Es DEs t s* . *q.e.d. Characterization 3:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e Proof:* Assume that 0 ( ) *rt Et Ee* . From,

$$P = \int\_{s}^{t} E\_0 e^{rt} d\mu = \frac{E\_0}{r} \left[ e^{rt} - e^{rs} \right] = \frac{E\_0 e^{rs}}{r} \left[ e^{r\Lambda t} - 1 \right] = \frac{E(s)}{r} \left[ e^{r\Lambda t} - 1 \right]$$

we get, ( ) <sup>1</sup> *r t Pr E s <sup>e</sup>* . This can be rewritten as, ( ) <sup>1</sup> *Pr Eav Pr E s <sup>e</sup>* . Since *E Pr* , this can further be written as ( ) <sup>1</sup> *E Eav <sup>E</sup> E s e* .

*Conversely*, consider next a function *E s*( ) satisfying

$$E(s) = \frac{\Delta E}{e^{\tilde{s}} - 1}, \text{ where } \begin{cases} \Delta E = E(t) - E(s) \\ \Delta t = t - s \\ \tilde{\xi} = \frac{\Delta E}{\overline{E}} \\ \overline{E} = \frac{1}{\Delta t} \int\_{s}^{t} E(u) du \end{cases} \text{ and } t \text{ can be any real value.}$$

From the above, we have that () () () () <sup>1</sup> ( ) () () *E Et Es Es Et e E s Es Es* .

Differentiating with respect to *s*, we get 2 () () () ( ) ( ) *s s <sup>s</sup> Et DEs DEs e D <sup>e</sup> E s E s* 

$$\text{and so, } D\_s \underline{\mathfrak{E}} = -\frac{D\_s \underline{\mathbf{E}}(\mathbf{s})}{E(\mathbf{s})} \tag{A1}$$

From the above *Lemma* we have

$$D\_s \overline{E}(s) = \frac{\overline{E} - E(s)}{t - s} \tag{A2}$$

Differentiating *<sup>E</sup> E* with respect to *s* we get,

$$D\_s \xi = \frac{-D\_s E(\mathbf{s}) \cdot \overline{E} - \Delta E \cdot D\_s \overline{E}(\mathbf{s})}{\overline{E}^2} \tag{A3}$$

and combining (A1), (A2), and (A3) we have

$$-\frac{D\_s E(\mathbf{s})}{E(\mathbf{s})} = \frac{-D\_s E(\mathbf{s}) \cdot \overline{E} - \frac{\Delta E}{\Delta t} (\overline{E} - E(\mathbf{s}))}{\overline{E}^2} = -\frac{D\_s E(\mathbf{s})}{\overline{E}} - \frac{\Delta E}{\Delta t} \cdot \frac{\left(\overline{E} - E(\mathbf{s})\right)}{\overline{E}^2}$$

We can rewrite the above as follows,

$$\frac{D\_s E(\mathbf{s})}{E(\mathbf{s})} - \frac{D\_s E(\mathbf{s})}{\overline{E}} = D\_s E(\mathbf{s}) \left( \frac{\overline{E} - E(\mathbf{s})}{E(\mathbf{s}) \cdot \overline{E}} \right) = \frac{\Delta E}{\Delta t} \cdot \frac{\left( \overline{E} - E(\mathbf{s}) \right)}{\overline{E}^2}$$

714 Thermodynamics – Interaction Studies – Solids, Liquids and Gases

*e* 

*E E <sup>e</sup> E s P E e du e e e <sup>e</sup>*

*<sup>t</sup> rs ru rt rs r t r t*

*rr r* 

0 0

*<sup>e</sup>* . This can be rewritten as, ( ) <sup>1</sup> *Pr Eav*

. Differentiating with respect to *s* we have

( ) 1 1

*<sup>e</sup>* . Since *E Pr* , this can

*t s* . *q.e.d.*

*E Es DEs*

*Pr E s*

and *t* can be any real value.

( ) () () *E Et Es Es Et*

(A1)

() () () ( ) ( )

 

(A2)

*E s E s*

(A3)

 2 2

*s s <sup>s</sup> Et DEs DEs e D <sup>e</sup>*

*E s Es Es*

.

( ) ( ) *<sup>s</sup> E Es DEs*

*t s*

2 ( ) ( ) *s s <sup>s</sup> DEs E E DEs <sup>D</sup> E*

( ) ( ) ( ) ( ) ( )

*DEs DEs E E Es DEs <sup>E</sup> E Es <sup>t</sup> E s E E E t* 

*<sup>E</sup> E s*

Rewriting, we have ( ) ( ) *<sup>t</sup>*

*Proof:* Assume that 0 ( ) *rt Et Ee* . From,

further be written as ( ) <sup>1</sup> *E Eav*

we get, ( ) <sup>1</sup> *r t Pr E s*

( ) <sup>1</sup> *<sup>E</sup> E s e* 

, where

and so, ( )

Differentiating *<sup>E</sup>*

*<sup>s</sup> <sup>s</sup> DEs <sup>D</sup> E s*

From the above *Lemma* we have

( )

*E*

and combining (A1), (A2), and (A3) we have

( )

*Et E DEt*

() () *<sup>s</sup> t s DEs E Es* . Rewriting, we have ( ) ( ) *<sup>s</sup>*

*Characterization 3:* <sup>0</sup> ( ) *rt Et Ee if and only if* ( ) <sup>1</sup> *E Eav*

0

*<sup>E</sup> E s e* .

 

() ()

*E Et Es tts E E*

> <sup>1</sup> ( ) *t*

From the above, we have that () () () () <sup>1</sup>

*e*

with respect to *s* we get,

*<sup>s</sup> s s E*

*s*

Differentiating with respect to *s*, we get 2

*E E u du t*

*Conversely*, consider next a function *E s*( ) satisfying

*s*

*t s*

and so, ( ) <sup>1</sup> ( ) *DEs <sup>s</sup> E Es t E* .

Using (A1), this can be written as

$$-D\_s \xi = \frac{\xi}{\Delta t} \quad \text{or as} \quad \xi = -D\_s \xi \cdot \Delta t \text{ }\tag{A4}$$

Differentiating (A4) above with respect to *s*, we get <sup>2</sup> *D D tD ss s* . Therefore, <sup>2</sup> 0 *Ds* . Working backward, this gives *D r <sup>s</sup>* = constant. From (A1), we then have that ( ) ( ) *DEs <sup>s</sup> <sup>r</sup> E s* and therefore 0 ( ) *rs Es Ee* . *q.e.d.*

### **15. Acknowledgement**

I am indebted to Segun Chanillo, Prof. of Mathematics, Rutgers University for his encouragement, when all others thought my efforts were futile. Also, I am deeply grateful to Hayrani Oz, Prof. of Aerospace Engineering, Ohio State University, who discovered my posts on the web and was the first to recognize the significance of my results in Physics. Special thanks also to Miguel Bayona of The Lawrenceville School for his friendship and help with the graphics in this chapter. And Alexander Morisse who is my best and severest critic of the Physics in these results.
