Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
Zero Point Energy Does Not Imply Planck's Radiation Law, And Does Not Replace the Quantum Assumption
H. Vic Dannon [email protected] March 1999 Revised July 2019 Abstract Zero point energy was introduced by Planck in 1912 in his second radiation law. In the derivation of his Radiation Law, Planck assumed that the radiation energy of his oscillator includes Zero Point Energy 1 ω , 2 even at absolute temperature zero. Planck was unaware of his assumption. He believed that quantized energy implied the Zero Point Energy. Actually, Planck's Zero Point Energy is an assumption independent of the well-established fact that energy is quantized. In 1913, Einstein-Stern proposed that Zero Point Energy implies Planck's radiation law with no need for assuming quantized energy.
1 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
We show here that Einstein-Stern Dipole Model that assumes Zero Point Energy fails to produce Planck's Radiation Law, and is inconsistent with it. In 1916, Einstein gave up quietly on that claim. His final treatment of Radiation does not mention Zero Point Energy. But in 1969 that claim was revived in a paper by Boyer. Boyer never supplied any derivation of the radiation law, with or without quantum assumptions. We show here that his proposed equation leads to a Ricatti type differential equation, with infinitely many solutions, that are inconsistent with the radiation law.
Keywords: Planck’s radiation law, Zero Point energy, Quantum hypothesis, Bose statistics, Entropy radiation law. Electromagnetic Radiation, Heat radiation, Blackbody Radiation, Radiation Quantum, PACS 12.90.+b.
2 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
Contents
1. Planck's Zero Point Energy 2. Einstein-Hopf Radiator 3. Einstein-Stern Zero Point Energy 4. Zero Point Energy of the Solution to Einstein-Stern Radiator 5. Einstein-Stern Dipole Model has Infinitely Many Solutions 6. Removing Zero Point Energy from the Einstein-Stern Radiator is Inconsistent with Planck's Radiation Law 7. Adding Zero Point Energy to the Einstein Model is Inconsistent with Planck's Radiation Law 8. Einstein's Giving Up on Zero Point Energy 9. Boyer's Revival of Einstein's Dead Radiator 10. Boyer Radiator has infinitely many solutions 11. Removing the Zero Point Energy from Boyer's Model is Inconsistent with Planck's radiation law 12. Adding Zero Point Energy to Boyer's Model is Inconsistent with Planck's Radiation Law
3 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
1.
Planck’s Zero Point Energy
In 1912, Planck derived his second radiation law that includes Zero Point Energy by assuming that Zero Point Energy. Unaware of his assumption, he believed that his quantum hypothesis implied the zero point energy. Had he utilized his 1901 derivation to obtain his second radiation law, as we do here, he might have noticed that assumption. His 1912 derivation masked the assumption of his Zero Pint Energy. Thus, he was unable to refute Einstein’s claim that zero point energy alone without quantum hypothesis can imply the first radiation law. Planck was not aware of the necessity of both the ZPE and the quantum hypothesis in his models for the derivation of his second radiation law that includes Zero Point Energy.
1.1 Planck's Assumption of Zero Point Energy
Planck’s assumptions in his 1912 paper define a harmonic oscillator with evenly spaced energy levels differing by ω . The probability that the oscillator will radiate its energy is
p The probability that it will retain it is
4 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
q= 1 − p The probability that the oscillator’s energy is between 0 and ω is . p1 The probability that the energy is between ω and 2ω is = ... p2 p 1 q The probability that the energy is between2ω and 3ω is = 2 p3 p 1 q ...... We have = + + + + + 1p1 p 2 p 3 .... pn .... = + + 2+ + n− 1 + p1(1 q q .... q ....)
1 = p . 1 1 − q Hence, = − = p1 1 q p and = n−1 pn pq The oscillator’s entropy is
= − − − − − s kp1log p 1 kp 2 log p 2 kp 3 log p 3 kp 4 log p 4 .... = − − − 2 2− 3 3 − kplog p kpq log pq kpq log pq kpq log pq .... logp+ log q log p+ 2log q log p+ 3log q
= − + + 2+ − + + 2 + − kqq(1 ...)log ppk (1 2 qq 3 ...) pqq log ... 1 d =p( q+ q2+ q 3 + ...) 1−q dq 1 −1 1−q 1 = −klog p− k pq log q (1− q )2
5 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
1 − p = −klog p− k log(1− p ) p
1 = k−log p− log(1− p )+ log(1− p ) p 1 1 = k− log(1− p )+ log − 1 p p 1 1 1 = k− log p− log(1− p )− log p + log − 1 p p p 1 1 1 1 = k log− − 1 log − 1. p p p p To obtain the Radiation Law with Zero Point Energy, the oscillator must have at the first energy level, between 0, and ω , an average energy of 1 ω with probability , that is, 1 ω 2 p1 2 p1 at the second energy level, between ω , and 2ω , an average energy of 3 ω with probability , that is, 3 ω 2 p2 2 p2 at the third energy level, between 2ω , and 3ω , an average energy of 5 ω with probability , that is, 5 ω 2 p3 2 p3 ...... the average energy of the oscillator must be
= 1ω + 3ω + 5 ω + u2 p1 2 p 2 2 p 3 ...
= 1 ω + + 2 + 2 p(1 3 q 5 q ...)
6 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
= 1ωp(24+ q+ 6 q2+ ...)− 1 ω p (1+ q+ q 2 + ...) 2 2 + + 2 + 2(1 2q 3 q ...) 1= 1 1−q p
= ωp(1+ 2 q+ 3 q2 + ...) − 1 ω 2 d {q+ q2+ q 3 +...} dq
d = ωp{ q+ q2+ q 3 + ...} − 1 ω dq 2 1 −1 1−q
1 = ωp − 1 ω − 2 2 (1q ) 1 p2
1 = ω − 1 ω p 2 1 1 = ω − 1 + ω p 2
Here we note the assumption of 1 ω Zero Point Energy, which 2 was assigned the nonzero probability p . Thus,
1u 1 − 1 = − , p ω 2
1= u + 1 p ω 2 1 1 1 1 s= k log− − 1 log − 1 p p p p u1 u 1 u 1 u 1 = k + log + − − log − . ω 2 ω 2 ω 2 ω 2
7 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
∂s 1 u 1 1 1 u 1 1 = k log + + − k log − + ∂u ω ω 2 ω ω ω 2 ω
u + 1 k ω = log 2 ω u − 1 ω 2 ∂ Since s = 1 , ∂u T u + 1 k ω 1 log 2 = ω u− 1 T ω 2 u 1 + ω ω 2 = ekT u − 1 ω 2 ω 1 1 u+ ω = ekT u − ω 2 2
ω ω 1 kT+ = kT − ωe1 e 1 u 2
ω 1ekT + 1 u = ω 2 ω ekT − 1
ω 1ekT − 1+ 2 = ω 2 ω ekT − 1 1 ω = ω + . 2 ω ekT − 1
8 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
This derivation’s assumptions are equivalent to the harmonic oscillator given by the Schrödinger equation. The assumptions that guarantee the radiation law with ZPE for the harmonic oscillator model include the evenly spaced energy levels, the ZPE, and the quantum hypothesis. We see that the quantum hypothesis in Planck’s derivation of his radiation law, does not imply the zero point energy. It is the hidden assumption of ZPE in the oscillator’s average energy that leads to ZPE in the radiation law. Thus, Planck’s derivation does not establish the existence of ZPE. It only recovers what was assumed at the start.
1.2 The Likelihood of Zero Point Energy
From
1= u + 1 p ω 2
= 1+ 1+ 1 2ω 2 ekT − 1
ω kT = e , ω ekT − 1 we have
ω ekT − 1 p = ω ekT
9 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ω − = 1 − e kT Zero Point Energy is assumed with probability.
The ZPE appears in the radiation law for u()ω with probability
ω − p= 1 − e kT . Computing with
− h = (6.6260755)1034 Joul/Kelvin
ν ∼ (5)⋅ 1014 Cycles/sec, for visible light
− k = (1.380658)⋅ 1023 Joul/Kelvin
° T = 300 Kelvin, for room Temperature
hν − ∼ −79.98692773 kT ω − p= 1 − e kT
− ∼ 1 − e 80 ≈ 1. That is, Planck assumed Zero Point Energy in his derivation with almost 100% certainty.
Planck’s “discovery” of zero point energy was instrumental in prompting Einstein-Stern to look for an experimental confirmation in Eucken’s specific heat data. Their analysis precedes Mulliken’s observation of zero point energy in vibrations of isotopes. [Mulliken].
10 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
2.
Einstein-Hopf Radiator
In 1910, Einstein and Hopf utilized an oscillating dipole radiator that was used by Planck to obtain
ρ()ω , the radiation energy volume density per frequency in the frequency interval [,]ω ω + dω , in terms of u()ω , the average energy per mode in [,]ω ω + dω . ω2 ρ()()ω = u ω , π2c 3 where ω2 , π2c 3 is the mode volume density per frequency.
ω = 2πν ,
8πν2 ρ(ν )= 2πρ (ω )= u (ν ), c3 dρ()()ν dρ ω = 4π2 , dν dω The dipole is at thermal equilibrium with an electromagnetic radiation field. The dipole has mass m , and charge
11 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
, qe and it oscillates at frequency ω in the z direction of the electromagnetic field. Einstein-Hopf showed [EinsteinHopf, equation 9] that the dipole is retarded at random speeds
υ()t c , by a random force 3cσ 1 dρ R()()()()ν υ t= ρ ν − ν υ t , 10πν 3 dν where σ , is the radiation-reaction constant. That force decreases the dipole’s kinetic energy. Over the time period τ = ∆t , a random recoil impulse
∆ ≡ P()∆ t , due to emission or absorption from the radiation field, increases the dipole’s kinetic energy. The thermal equilibrium is attained at the momentum balance between the recoil and the retardation in the mean.
mυ()()()()() t+ ∆ t= mυ t+ P∆ t− Rν υ t∆ t (9)
Squaring both sides, and keeping terms to first order of ∆t leads to the stochastic (energy times mass) equation
12 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon m2υ 2()()() t+ ∆ t= m 2υ 2 t+ P 2 ∆ t
+2mtPtυ ()(∆ )− 2 mtRυ2 ()()ν ∆ t− 2( PttR∆ )()()υ ν ∆ t Assuming that P(∆ t )υ ( t )= 0, and using the equipartition theorem
mυ2()() t+ ∆ t= mυ 2 t= kT , the average of the squared momentum equation is
P2(∆ t )= 2 kTR (ν )∆ t
3cσ 1 dρ = 2kTρ (ν ) − ν ∆ t . 10πν 3 dν Einstein-Hopf showed [ref. 3, equation 15] that
c4σ∆ t P2()()∆ t = ρ 2 ν . 40π2ν 3 Hence,
σ ρ 4σ∆ 3cρ ν − 1 ν d∆ = c t ρ 2 ν , 2kT ( ) t ( ) 10πν 3 dν 40π2ν 3 ρ 3 ρ ν − 1 ν d = c ρ 2 ν . 3kT ( ) ( ) 3 dν 8πν2 They confirmed by substitution that
8πν2 ρ()ν = kT . c3 is a solution. Hence, u()ν = kT . is the average energy per mode.
13 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
This is the classical Rayleigh-Jeans spectrum that blows at
ω ≈ ∞ , and led to the ultra-violet catastrophe, and the black body radiation problem. So Einstein-Hopf had to let it go.
14 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
3.
Einstein-Stern Zero Point Energy
In 1913, Einstein-Stern examined specific heat measurements by Eucken that convinced them of the existence of Zero Point Energy [EinsteinStern]. They utilized the Einstein-Hopf radiator which was earlier examined by Planck, and discarded by him since it was difficult to augment that radiator with the quantum hypothesis. After Planck's 1912 assumption of Zero Point Energy, Einstein- Stern tried to revive the Einstein-Hopf radiator by adding to it a zero point energy term. They derived [EinsteinStern] hcσ∆ t P2()()∆ t = ρ ν , 5π and noticed that equating it with 3cσ 1 dρ 2kTρ (ν ) − ν ∆ t 10πν 3 dν leads to Wien’s law. They decided to revive the Einstein-Hopf model,
σ ρ 4σ∆ 3cρ ν − 1 ν d∆ = c t ρ 2 ν , 2kT ( ) t ( ) 10πν 3 dν 40π2ν 3
hcσ∆ t by adding ρ()ν to its Right Hand Side. They obtained 5π
15 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
σ ρ 4σ∆ σ∆ 3cρ ν − 1 ν d∆ = c tρ 2 ν + hc t ρ ν , 2kT ( ) t ( ) ( ) 10πν 3dν 40π2ν 3 5π ρ 3 ρ ν − 1 ν d = c ρ 2 ν + νρ ν . 3kT ( ) ( ) h ( ) 3 dν 8πν2 Einstein-Stern substituted into their modified equation
πν2 ν ρ ν = 8 h , () ν c3 h ekT − 1 and confirmed that it satisfies the equation. That is, ν ν = h . u() hν ekT − 1 This is Planck's Radiation Law WITHOUT Zero Point Energy. Planck's Radiation Law with Zero Point Energy is
hν 1 u()ν = + hν . hν 2 ekT − 1 The assumed zero point energy 1 ω that modifies the Einstein- 2 Stern Radiator, disappears from the final u()ω Paradoxically, The Einstein-Stern Radiator that included Zero Point Energy, produced a Radiation Law with NO Zero Point Energy
They speculated that the hνρ()ν in their equation meant Zero
Point Energy of ν , while Planck had 1 ν , and Eucken h 2 h experiments on specific heats supported 1 ν . 2 h
16 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
They wrote [EinsteinStern,p.143] “…with the method of calculation sketched here, the zero point energy must be set equal to hν in order to arrive at the Planck radiation formula. Future investigation must show whether the discrepancy between this assumption and the assumption underlying the investigation on hydrogen disappears if the calculation is more rigorous…”
In fact, their equation assumes 1 ν zero point energy. 2 h
17 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
4.
Zero Point Energy of the Solution to Einstein-Stern Radiator
The solution to the Einstein-Stern Radiator has indeed 1 ν Zero 2 h Point Energy. To add Zero Point Energy to the Einstein-Hopf Radiator,
σ ρ 4σ∆ 3cρ ν − 1 ν d∆ = c t ρ 2 ν , 2kT ( ) t ( ) 10πν 3 dν 40π2ν 3 we assume that ν → ν + 1 ν . u()() u2 h Then, 8πν2 8πν 2 4πhν 3 ρ()()()()ν = uν → ( uν + 1 hν ) = ρ ν + c3 c 32 c 3
dρ(ν ) 16πν 8πν2 du = u()ν + dν c3 c 3 dν
16πν 8πν2 du → ( ν + 1ν ) + + 1 u() 2 h 2 h c3 c 3 dν
dρ(ν ) 16πν 8πν2 = + 1hν + 1 h dν c32 c 3 2 ρ ν π ν2 = d( )+ 12 h dν c3 Thus, for the left hand side of the Einstein-Hopf model
18 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
3cσ 1 dρ 2kTρ (ν ) − ν ∆ t → u()()ν → uν + 1 hν 10πν 3 dν 2
σ π ν3 ρ ν π ν 2 3c + 4 h− 1 d ( )+ 12 h ∆ = 2kTρ (ν ) ν t 10πν c3 3 dν c 3 3cσ 1 dρ (ν ) = 2kTρ (ν ) − ν ∆ t . 10πν 3 dν That is, the Left hand Side of the model does not changed when
ν → ν + 1 ν u()() u2 h For the Right Hand Side of the Einstein-Hopf Model,
2 4σ∆ 4σ∆ πν 2 2 c tρ 2 ν → c t 8 ν + 1 ν ()()1 (u h ) 2 3u()()ν → uν + hν 2 3 3 2 40π ν 2 40π ν c
To first order of h ,
2 4σ∆ πν 2 4σ∆ πν 2 πν 2 c t 8 2 c t 8 8 = u()()ν + uν hν 40π2ν 3 c 3 40π 2ν 3 c 3 c 3 ρ 2()ν ρ()ν
c4σ∆ t cσ∆ t = ρ 2()()ν + ρ ν hν 40π2ν 3 5πν which is the Right Hand Side of the Einstein-Stern modified model.
Thus, the Einstein-Stern Model indicates 1 ν Zero Point Energy. 2 h However, the Einstein-Stern Model fails because it has infinitely many solutions.
19 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
5.
Einstein-Stern Dipole Model Has
Infinitely Many Solutions
In ω , the Einstein-Stern equation is
2 3 dρ π c 2 kT 3ρ (ω )− ω = ρ (ν )+ ωρ (ω ). dω ω2 Substituting ω2 ρ()()ω = u ω π2c 3
dρ 2ω ω2 du (ω ) = u()ω + dω π2c 3π 2 c 3 dω we have
ω2ω ω 2 ω 2du ( ) kT3 u (ω )− ω ( u (ω )+ ) = π2c 3π 2 c 3π 2 c 3 dω ω2 du()ω u()ω − 2 3 π c dω
2 π2 3 ω 2 ω 2 c 2 = u()()ω + ω u ω ω2 π 2c 3 π 2 c 3 ω2 2 ω + ω ω u()() u π2c 3 du()ω 2 kT u()()()ω − = uω + ω u ω . dω The equation is satisfied by
20 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ω ω = , u() ω ekT − 1 as well as infinitely many other solutions It is a Bernoulli differential equation that is reduced to a first order equation by the Lie group transformation
1 u()ω = . y()ω Then, ω 1+ ω y '= 1 1+ 1 yy2 kT y 2 kT y
1 ω y+ ω y' = + y kT kT ω 1 ωy'+ y (1− ) = kT kT For the homogeneous equation,
ω ωy'+ y (1− )= 0 , kT
y '= − 1 y kT ω
ω logy= − logω + log C kT 1 ωy ω log = C1 kT
ω ωy = ekT C1
21 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ω C y()ω = 1 ekT ω By the variation of parameter method, we substitute
ω C ()ω y()ω = 1 ekT . ω into 1 ω y+ ω y' = + y kT kT We obtain
ω ω ω ω CCCC()()'()()ω ω ω ω 1kT+ − 1 kT+ 1 kT+ 1 kT = eω e e e ω ω2 ω ω kT
ω 1 ω C ()ω = + 1 ekT , kT kT ω
ω 1 C'()ω ekT = , 1 kT
ω 1 − C'()ω = e kT 1 kT
ω 1 − C()ω = − ekT + C 1 0
ω ω − ω = 1− 1 kT+ kT y() e C0 e ω
ω 1 1 = − + C ekT ω ω 0 1 u()ω = , ω 1 1 − + C ekT ω ω 0
22 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ω = , ω − + kT 1 C0 e Denoting = , CC 0 ω = . ω CekT − 1 ω2 ω ρ ω = . () ω π2c 3 CekT − 1 Can we determine C by imposing the boundary conditions ρ(ω )≈ 0, at ω ≈ 0 , and at ω ≈ ∞ ?
For ω ≈ 0 , ω 0 ekT≈ e kT = 1, and 03 1 ρ()ω ≈ . π2c 3 C − 1 For any C ≠ 1, we get ρ(ω )≈ 0.
03 1 0 If C = 1, then ≈ . But by L'hospital, 2 3 π c 0 0 ekT − 1 ω3 d 3ω2 ω π2 3 2 3 d c = π c d ω ω (ekT− 1) e kT dω kT
kT ω2 = 3 π2c 3 ω ekT
23 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
2 ≈ kT 0 = 32 3 0 π c 0 ekT Thus, for any arbitrary constant C , ω ≈ 0⇒ ρ (ω )≈ 0 . For ω ≈ ∞ ,
ω ∞ ekT≈ e kT = ∞ , and ∞3 1 ρ()ω ≈ . π2c 3 C∞ − 1 For any C ≠ 0 , we get ∞3 ∞ 1 ≈ . π2c 3 C∞ − 1 ∞ By L'hospital, 3 ω3 d 6 ω π2 3 2 3 d c = π c 3 ω 3 ω d (CekT− 1) C e kT dω kT
3 ≈ 6kT 1 = . 0 C2π 2 c ∞ ekT Thus, Einstein-Stern dipole model has infinitely many solutions.
24 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
6.
Removing the Zero Point Energy
From Einstein-Stern Radiator is Inconsistent with Planck's
Radiation Law
To be consistent with the radiation law, the removal of the 1 ν 2 h Zero Point Energy from Einstein's dipole model should reduce the solution to a radiation law without ZPE. Einstein-Stern Radiator solution is already a Radiation Law without ZPE, and the removal of the ZPE from the model contradicts the Radiation Law. Removing the zero point energy from Bernoulli's equation, gives
1 u()'()()ω − ω uω = u2 ω kT The Lie Transformation 1 u = y reduces the equation to
1+ ω y '= 1 1 , yy2 kT y 2
25 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
1 y+ ω y ' = . kT C y()ω = 1 ω solves the homogeneous equation, and we substitute
C ()ω y()ω = 1 ω into the equation. Then,
ω ω ω CCC1()'()()+ ω 1− ω 1 = 1 , ω ω ω2 kT
1 C '()ω = , 1 kT 1 CC()ω = ω + . 1kT 0 C ()ω 1 1 y= 1 = + C ω kT ω 0 ω + = kTC0 kTω kTω u()ω = . ω + kTC0
ω2 kTω ρ()ω = 2 3 ω + π c kTC0 For ω ≈ 0 , For any ≠ , we get C 0 0
ω2 kT ω ρ()ω ≈ 2 3 + π c 0 kTC 0
26 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
2 ≈ 0 0 = 2 3 0 π c C0 If = , then C 0 0 02 kT ω ρ(ω )≈ = 0 . π2c 3 ω Thus, for any , C 0 ω ≈ 0⇒ ρ (ω )≈ 0 . For ω ≈ ∞ , ω2 kTω ρ()ω = . 2 3 ω + π c kTC0 ω2 ≈ kT π2c 3 kTC 1 + 0 ω
∞2 ≈ kT ≈ ∞ π2 3 c + kTC0 1 ∞ That is, for any , C 0 ω ≈ ∞ ⇒ ρ()ω ≈ ∞, indicating the ultra violet catastrophe, and the classical Rayleigh- Jeans spectrum. Thus, the removal of the zero point energy destroys the model, and reveals its inconsistency with the radiation law.
27 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
7.
Adding Zero Point Energy to
Einstein-Stern Radiator is
Inconsistent with Planck's
Radiation Law.
Had Planck assumed his zero point energy to be ω rather than 1 ω , his radiation law would have been 2 ω ω = + ω . u() ω ekT − 1 showing the added zero point energy.
The addition of ω zero point energy in the Einstein-Stern model gives the Bernoulli equation du()ω 2 kT u(ω )− = u (ω )+ 2ω u (ω ) dω Then, 1 ω u(ω )− ω u '(ω )= u2 (ω )+ 2 u (ω ), kT kT with infinitely many solutions
ω ω = 2 u() 2 ω Ce kT − 1
28 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
The factor 2 ω e kT further corrupts Planck’s radiation law. The solutions have no zero point energy because
ω 2 T≈ 0⇒ ekT ≈ ∞ ⇒ u (ω )≈ 0
The additional zero point energy does not show up in u()ω , but generates a nonphysical radiation law.
29 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
8.
Einstein’s Giving Up on Zero
Point Energy
Einstein-Stern erroneously believed that their model did not use the quantum hypothesis, and hopped that the Radiation Law resulted from the Zero Point Energy. They speculated that the radiation law solves the differential equation that represents the model, and that Zero Point Energy may replace the quantum hypothesis [EinsteinStern,p145]: “The assumption of zero point energy opens a way for deriving Planck’s radiation formula without recourse to any kind of discontinuities. Nevertheless, it seems doubtful that the other difficulties can also be overcome without the assumption of quanta.”
The “other difficulties” are the incompatibility of ω Zero Point Energy with Eucken’s experiments and Planck’s derivation that indicate 1 ω . 2 In fact, the Einstein-Stern Model is based on a solution with 1 ω 2 zero point energy. However, Einstein-Stern were not aware of their assumption of quanta. Their decomposition of the electromagnetic radiation field
30 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon into modes [EinsteinHopf] is a step towards the second quantization of the radiation field, and its description as a collection of oscillators. Furthermore, their model fails to produce Planck’s radiation law. Even worse, their model that seemed to include Zero Point Energy, produced a Radiation Law with NO Zero Point Energy. In his 1916 final treatment of the radiation problem, Einstein gave up on the dipole model as a tool to derive Planck’s radiation law, and on zero point energy as a replacement for the quantum hypothesis. In fact, Zero Point Energy has to be assumed separately from the fact that energy is quantized. In his last treatment of the quantum theory of radiation [Einstein], Einstein assumes the quantum hypothesis, Wien’s law, and spontaneous emission with Bohr rules, to derive Planck’s radiation law without zero point energy, and to confirm that
P2(∆ t )= 2 kTR (ν )∆ t .
He claims that the radiation-reaction is given by
ε hν hν − n 1 − ν = m kTρ ν − νρ ν − kT R( ) pn B n e ( ) '( ) 1 e c2 S 3 He substitutes πν2 ν ρ ν = 8 h , () ν c3 h ekT − 1 and obtains
31 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ε ν − n ρ ν ν ν = hm kT () h R() pn B n e c2 S 3 kT Thus, ε 2 2− n 1 hν ν ∆ = m kT ρ ν ∆ . 2kTR ( ) t pn B n e ( ) t S 3 c The first factor on the right is the number of induced transitions
from the molecular state Zm , into the molecular state Zn , The second factor is the average momentum transfer per transition. Therefore, the product equals
P2()∆ t . And this confirms that radiation is a directional collision process. While these claims need to be checked out, Zero Point Energy is not amongst them.
32 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
9.
Boyer's Revival of Einstein's Dead
Radiator
In 1912, Planck added 1 ω photon of Zero Point Energy to his 2 Radiation Law of 1901, and proved that the average internal radiation energy per mode in the frequency interval [,]ω ω + dω at a given absolute temperature T is ω ω = + 1 ω . u() ω 2 ekT − 1 ω2 There are modes in [,]ω ω + dω , and the average radiation π2c 3 energy density in it is ω2 ω 1 ρ()ω = + ω π2 3 ω 2 c ekT − 1 In 1913, Einstein-Stern proposed that Zero Point Energy implies the radiation law with no need for the quantum hypothesis. Using the differential operator
2 3 dρ π c 2 kT 3ρ (ω )− ω − ρ (ω ), dω ω2 Einstein-Stern failed to show that Zero Point Energy alone, without quantum assumptions, implies Planck's Radiation Law.
33 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
In 1916 Einstein gave up quietly on that claim. But in 1969 it was revived in a paper by Boyer. Boyer never supplied any derivation of the radiation law, with or without quantum assumptions.
In [Boyer1], Boyer plugged Planck's ρ()ω into the Einstein differential operator, and found that
2 3 dρ π c 2 kT 3ρ (ω )− ω − ρ (ω ) = dω ω2 ω2 ω 1 = 3kT + ω π2 3 ω 2 c ekT − 1 ω2 ω d 1 −kω T + ω ω π2 3 ω 2 d c ekT − 1
2 π2 3 ω 2 ω c 1 − + ω ω2 π 2 3 ω 2 c ekT − 1
ω2 2 = − ( 1 ω ) . π2c 3 2 That confirms that ρ()ω satisfies
dρ π2 c 3ω 2 2 ρ ω − ω − ρ 2 ω = − ( 1 ω ) . kT 3 ( ) ( ) 2 dω ω2π 2c 3 If this is a derivation without quantum assumptions,
where does the quantization constant come from?
34 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
Furthermore, without solving the above equation, Boyer claimed that this is the only solution to it.
In [Boyer1,p.1379], he called ρ()ω “The solution…”, and in [Boyer2, p.1310], “a unique solution…”. He concluded that this is a derivation of Planck's Radiation Law without assuming that radiation energy is quantized. Indeed, the equation has infinitely many solutions which are inconsistent with the radiation law. And nowhere did Boyer derive Planck's Radiation Law, with or without assuming, that radiation energy is quantized.
35 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
10.
Boyer's Radiator has Infinitely
Many Solutions
Boyer's equation for u()ω ,
2 ( ω − ω ω ) − 2 ω = −( 1 ω ) kT u()'()() u u 2 is a Ricatti differential equation. Thus, the Lie group transformation
y '(ω ) u()ω = y()ω will yield a second order differential equation with no y2()ω term.
To eliminate also the y '()ω term, we use the Lie transformation
y '()ω u()()ω = α ω . y()ω Then, y'''' y y u'''= α + α − α y y y y2 Substituting into the Ricatti equation,
2 y'''''' y y y y 2 α − ωα + α − α − α = −( 1 ω ) kT kT '' y 2 y y yy2 y
2 To eliminate the (')y terms, we equate y2 kTωα = α2
36 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
Then, α()ω = kTω ,
α ' = kT
2 Then, both the (')y , and the y ' terms vanish. y2 y And the equation reduces to
2 y '' 1 −kTωα = −( ω ) , y 2
y '' 2 ()kTω 2 = ( 1 ω ) y 2
2 y '' 1 = y2 kT
2 y''− y = 0 2kT Therefore,
ω − ω ω = 2kT+ 2 kT y() C1 e C 2 e
ω − ω ω = 2kT− 2 kT y'() C1 e C 2 e 2kT
y ' u()()ω = α ω y
ω − ω 2kT− 2 kT C1 e C 2 e 2kT = kTω ω − ω 2kT+ 2 kT C1 e C 2 e
37 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ω C ekT − C = 1 ω 1 2 2 ω kT + C1 e C 2
C ω − 1 ekT + 1 = 1 ω C2 2 C ω − 1 ekT − 1 C2 C Denoting − 1 = C C2 ω CekT + 1 = 1 ω 2 ω CekT − 1
ω CekT − 1+ 2 = 1 ω 2 ω CekT − 1
ω CekT − 1 2 = 1ω + 1 ω 2 ω 2 ω kT− kT − Ce1 Ce 1 1
1 = 1 ω + ω 2 ω CekT − 1 = ω 1 + 1 ω 2 CekT − 1 This is a variation on the radiation Law, with what will turn out to be a pesky constant C that we cannot get rid of. Hence, infinitely many solutions.
38 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
We have, ω3 ρ ω = 1 + 1 . () π2 3 ω 2 c CekT − 1 Can we determine C from the boundary conditions
ρ(ω )≈ 0, at ω ≈ 0 , and at ω ≈ ∞ ?
For ω ≈ 0 , ω 0 ekT≈ e kT = 1, and 03 1 ρ ω ≈ + 1 . () 2 π2c 3 C − 1 For any C ≠ 1, we get ρ(ω )≈ 0.
03 1 0 If C = 1, then ≈ . But by L'hospital, 2 3 π c 0 0 ekT − 1 ω3 d 3ω2 ω π2 3 2 3 d c = π c d ω ω (ekT− 1) e kT dω kT
kT ω2 = 3 π2c 3 ω ekT
2 ≈ kT 0 = 32 3 0 π c 0 ekT Thus, for any arbitrary constant C , ω ≈ 0⇒ ρ (ω )≈ 0 . For ω ≈ ∞ ,
39 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
ω ∞ ekT≈ e kT = ∞ , and ∞3 1 ρ ω ≈ + 1 . () 2 π2c 3 C∞ − 1 For any C ≠ 0 , we get ∞3 ∞ 1 ≈ . π2c 3 C∞ − 1 ∞ By L'hospital, 3 ω3 d 6 ω π2 3 2 3 d c = π c 3 ω 3 ω d (CekT− 1) C e kT dω kT
3 ≈ 6kT 1 = 0 C2π 2 c ∞ ekT ω3 Although 1 + 1 ≈ ∞, because of the zero point π2 3 ω 2 c CekT − 1 energy part. In any event, the problem has infinitely many solutions.
40 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
11.
Removing the Zero Point Energy from Boyer's Model is Inconsistent with Planck's Radiation Law .
To be consistent with the radiation law, the removal of the zero point energy from the dipole model should reduce the solution to Planck’s radiation law without ZPE. Removing the zero point energy from Ricatti's equation, gives
kT( u(ω )− ω u '(ω )) − u2 (ω )= 0 The Lie Transformation y '()ω u()()ω = α ω . y()ω with α()ω = kTω reduces the equation to y '' −kTωα = 0 y
y ''= 0 = ω + y A1 A 2
A u()ω = kTω 1 ω + AA1 2
41 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
1 = kTω ω + A2 A1 A Denoting C = 2 , A1 1 = kTω ω + C
Therefore, ω3 kTω ρ()ω = . π2c 3 ω + C For ω ≈ 0 , kT ρ(ω )≈ 04 = 0 π2c 3 C and C remains undetermined. For ω ≈ ∞ , ω = 1 ω + CC 1 + ω
1 ≈ = 1 C 1 + ∞
kT ρ()ω ≈ ω3 π2c 3
≈ kT ∞3 = ∞ π2c 3 Hence, for any C , we obtain the ultra violet catastrophe. Thus, the removal of the zero point energy destroys the model, and reveals its inconsistency with the radiation law.
42 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
12.
Adding Zero Point Energy to
Boyer's Model is Inconsistent with
Planck's Radiation Law
Had Planck assumed his zero point energy to be ω rather than 1 ω , his radiation law would be 2 ω ω = + ω . u() ω ekT − 1 This does not solve Boyer’s model with ω zero point energy. The Ricatti equation
kT( u()'()()()ω − ω uω ) − u2ω = ω 2 leads to 2 y''− y = 0, kT and to the infinitely many solutions
ω ω = 2 + ω u() 2 ω Ce kT − 1 The factor 2 ω e kT further distinguishes the solutions from Planck’s radiation law.
43 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
The solutions are non-physical, and inconsistent with the radiation law. Boyer’s claim that zero point energy implies the radiation law fails because the dipole's infinitely many solutions are different from Planck’s radiation law, and inconsistent with it. Nowhere does Boyer derive Planck's Radiation Law without quantum assumptions. The quantum hypothesis has no substitute, and no replacement for the derivation of the Radiation Law.
44 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
References
[Boyer1] Boyer, T., "Derivation of the Blackbody Radiation Spectrum without Quantum Assumptions", Physical Review, Vol. 182, No 5, (1969) p.1374
[Boyer2] Boyer, T., "Classical Statistical Thermodynamics, and Electromagnetic Zero Point Radiation" Physical review, Vol. 186, No 5, (1969) p.1304
[Dannon1] Vic Dannon, "Zero Point Energy Does Not Imply the Radiation Law" Abstract Submitted for the NEF99 Meeting of the American Physical Society, on November 6, 1999.
[Dannon1] H. Vic Dannon, "Zero Point Energy: Planck Radiation Law" Gauge Institute Journal of Math and Physics, Volume 1, No. 3, August 2005.
[Dannon2] H. Vic Dannon, "Zero Point Energy: Thermodynamic Equilibrium and Planck Radiation Law" Gauge Institute Journal of Math and Physics, Volume 1, No. 4, November 2005.
[Dannon3] H. Vic Dannon, "Zero Point Energy and the Charge- Radiation Equation in Bohr's Atom" Gauge Institute Journal of Math and Physics, Volume 8, No. 4, November 2012.
[Dannon4] H. Vic Dannon, "The Neutron as a Collapsed-Hydrogen Atom: Zero Point Energy & Nuclear Binding Energy, X Rays & Gamma Rays, Nuclear Forces & Bonding, Neutronic Electrons Orbitals, and Neutron Stars" Gauge Institute Journal of Math and Physics, Volume 11, No. 3, August 2015.
45 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
[Dannon5] H. Vic Dannon, "The Casimir Effect Fallacy: Casimir's Zero Point Energy has Infinite Mass, is not Conserved, and is Falsely based on Euler’s Formula. The Fallacy No Journal Accepted, that Resonates to No End in Journals" Gauge Institute Journal of Math and Physics, Volume 11, No. 4, November 2015.
[Dannon6] H. Vic Dannon, "Zero Point Energy, and Electromagnetic Entropy" Gauge Institute Journal of Math and Physics, Volume 14, No. 3, August 2018.
[Dannon7] H. Vic Dannon, "American Journal of Physics Claims that a Harmonically Oscillating mass has Zero Point Energy" Gauge Institute Journal of Math and Physics, Volume 14, No. 3, August 2018.
[Einstein] Einstein, A.{Verh.d. Deutsch. Physikal. Gesellschaft., 13/14 (1916) p.318}. Document 38 in The collected papers of Albert Einstein Vol. 6, p.220. Princeton 1997.
[EinsteinStern]Einstein, A. and Stern, O. {Annalen der physik 40 (1913):p.551-60}. Document 11 in The collected papers of Albert Einstein Vol. 4, p.137, Princeton 1996.
[EinsteinHopf] Einstein, A. and Hopf, L.{Annalen der physik 33(1910):p.1105-1115}. Document 8 in The collected papers of Albert Einstein Vol. 3, p.220. Princeton 1993.
[Mulliken] Mulliken, R. S. Nature, 114 (1924) p.349.
[Planck1] Planck, M., {Annalen der physik 37 (1912):p.642}.
46 Gauge Institute Journal Volume 15, No 3, August 2019 H. Vic Dannon
Also in, Planck, M. "The theory of heat radiation", Dover 1959.
[Planck2] Planck, M. “On the theory of the law of the distribution of energy in the normal spectrum” {Annalen der physik 4 (1901):p.553}. Planck’s original papers in quantum physics, Wiley, 1972.
[Planck3] Planck, M. {Verh.d. Deutsch. Physikal Gesellschaft, 2 (1900);p.202}. #15 in Early concepts of energy in atomic physics, Dowden, 1979
47