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Please advise on...

MichaelVio 45
5 hours ago
0 0

@dharr 

Please advise on solving the Planck.mw and plot the solution for h(nu) with E=ν⸱h(nu). Thus, for nu2 =9.733521364*10^16 =>h(nu2)= 3.348222989*10^(-17) eV*s and the same energy of 3.259eV for nu1 = 7.889275211*10^14 =>h(nu2)=4.135667697*10^(-15) eV*s  With the partular case the plot 

with the supposition that k = 1/2 eV/K, thus kT=>1.380649*10^(-23)*297/(1.60217*10^(-19));
=0.02559, so kT should be the energy close to 0.5 and could be assimilated to an energy E0

[1]
 

> 

restart;
> 

with(PDEtools, dchange):with(plots):
> 

with(Units):

Automatically loading the Units[Simple] subpackage
 

  
> 

kernelopts(maxdigits);

38654705646

 (1)
> 

Digits:=10;

10

 (2)
> 

params:={k= 1.380649*10^(-23)*Unit(J/K),rb=5.293*10^(-11)*Unit(m),he=4.135667697*10^(-15)*Unit(eV*s),
ec= 1.602176634*10^(-19)*Unit(C),Tq=1.765*10^(-19)*Unit(s),
c= 299792458*Unit(m/s),T=297*Unit(K),nu1=7.880979442*10^14*Unit(s^(-1)),E1=3.259*Unit(eV)};

{E1 = 0.5221493608e-18*Units:-Unit(J), T = 297*Units:-Unit(K), Tq = 0.1765000000e-18*Units:-Unit(s), c = 299792458*Units:-Unit(m/s), ec = 0.1602176634e-18*Units:-Unit(C), he = 0.6626070096e-33*Units:-Unit(m^2*kg/s), k = 0.1380649000e-22*Units:-Unit(m^2*kg/(s^2*K)), nu1 = 0.7880979442e15*Units:-Unit(1/s), rb = 0.5293000000e-10*Units:-Unit(m)}

 (3)
> 

NV:=8*Pi*Tq*nu^2*he(nu)*nu/(exp(he(nu)*nu/(k*T) - 1));

8*Pi*Tq*nu^3*he(nu)/exp(he(nu)*nu/(k*T)-1)

 (4)
> 

E:=nu*h(nu);

nu*h(nu)

 (5)
> 

ic:=h(7.889275211*10^14)=4.135667697*10^(-15);

h(0.7889275211e15) = 0.4135667697e-14

 (6)
> 

de:=diff(NV,nu)-h(nu)*nu;

24*Pi*Tq*nu^2*he(nu)/exp(he(nu)*nu/(k*T)-1)+8*Pi*Tq*nu^3*(diff(he(nu), nu))/exp(he(nu)*nu/(k*T)-1)-8*Pi*Tq*nu^3*he(nu)*((diff(he(nu), nu))*nu/(k*T)+he(nu)/(k*T))/exp(he(nu)*nu/(k*T)-1)-nu*h(nu)

 (7)
> 

dsolve({de});

{he(nu) = -k*T*LambertW((1/8)*(_C1-(Int(nu*h(nu), nu)))*exp(-1)/(Pi*Tq*nu^2*k*T))/nu}

 (8)
> 

#Thus, for k=3/2*k*T we have the solution
> 

c:=299792458;nu1:=7.880979442*10^14;c/nu1;ec:= 1.602176634*10^(-19);k:= 1/2;E1:=3.259*ec;T:=297;Tq:=1.765*10^(-19);;he := h/ec;

299792458

  

0.7880979442e15

  

0.3804000000e-6

  

0.1602176634e-18

  

1/2

  

0.5221493650e-18

  

297

  

0.1765000000e-18

  

0.6241509074e19*h

 (9)
> 

NV:=8*Pi*Tq*nu^2*he(nu)*nu/(exp(he(nu)*nu/(k*T) - 1));

27.68689003*nu^3*h(nu)/exp(0.4203036413e17*nu*h(nu)-1)

 (10)
> 

ic1:=h(9.489275211*10^16)=4.135667697*10^(-17);

h(0.9489275211e17) = 0.4135667697e-16

 (11)
> 

de:=diff(NV,nu)-h(nu)*nu;

83.06067009*nu^2*h(nu)/exp(0.4203036413e17*nu*h(nu)-1)+27.68689003*nu^3*(diff(h(nu), nu))/exp(0.4203036413e17*nu*h(nu)-1)-27.68689003*nu^3*h(nu)*(0.4203036413e17*h(nu)+0.4203036413e17*nu*(diff(h(nu), nu)))/exp(0.4203036413e17*nu*h(nu)-1)-nu*h(nu)

 (12)
> 

evalf(simplify(dsolve({de,ic1},h(nu),series)));

h(nu) = series(0.4135667697e-16-0.4060327648e71635144604154866*(nu-0.9489275211e17)+0.3287671071e143270289208309765*(nu-0.9489275211e17)^2-0.3549395354e214905433812464664*(nu-0.9489275211e17)^3+0.4310949299e286540578416619563*(nu-0.9489275211e17)^4-0.5584961425e358175723020774462*(nu-0.9489275211e17)^5+O((nu-0.9489275211e17)^6),nu = 0.9489275211e17,6)

 (13)
> 

h0(nu):=convert(%,polynom);

h(nu) = 0.3852956650e71635144604154883-0.4060327648e71635144604154866*nu+0.3287671071e143270289208309765*(nu-0.9489275211e17)^2-0.3549395354e214905433812464664*(nu-0.9489275211e17)^3+0.4310949299e286540578416619563*(nu-0.9489275211e17)^4-0.5584961425e358175723020774462*(nu-0.9489275211e17)^5

 (14)
> 

odetest(h0(nu), [de,ic1], series);

Error, (in odetest/series) numeric exception: overflow

  
> 

 

> 

evalf(c/(3*10^(-9)));1/ec;

0.9993081933e17

  

0.6241509074e19

 (15)
> 

plot(h(nu),nu=10^14..1.2*10^17);

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

  

 
> 

#All we have to do is to fit the constant k=1/2 so for lambda=3.08nm and for lambda=380.4nm we have E=3.259 eV
> 

9.733521364*10^16Hz & ;
> 

 

> 

restart; # With the particular case the plot below
> 

with(PDEtools, dchange):with(plots):
> 

with(Units):

Automatically loading the Units[Simple] subpackage
 

  
> 

kernelopts(maxdigits);

38654705646

 (16)
> 

Digits:=20;

10

 (17)
> 

params:={k= 1.380649*10^(-23)*Unit(J/K),rb=5.293*10^(-11)*Unit(m),he=4.135667697*10^(-15)*Unit(eV*s),
ec= 1.602176634*10^(-19)*Unit(C),Tq=1.765*10^(-19)*Unit(s),
c= 299792458*Unit(m/s),T=297*Unit(K),nu1=7.880979442*10^14*Unit(s^(-1)),E1=3.259*Unit(eV)};

{E1 = 0.5221493608e-18*Units:-Unit(J), T = 297*Units:-Unit(K), Tq = 0.1765000000e-18*Units:-Unit(s), c = 299792458*Units:-Unit(m/s), ec = 0.1602176634e-18*Units:-Unit(C), he = 0.6626070096e-33*Units:-Unit(m^2*kg/s), k = 0.1380649000e-22*Units:-Unit(m^2*kg/(s^2*K)), nu1 = 0.7880979442e15*Units:-Unit(1/s), rb = 0.5293000000e-10*Units:-Unit(m)}

 (18)
> 

NV:=8*Pi*Tq*nu^2*he*nu/(exp(he*nu/(k*T) - 1));

8*Pi*Tq*nu^3*he/exp(he*nu/(k*T)-1)

 (19)
> 

E:=diff(NV,nu);

24*Pi*Tq*nu^2*he/exp(he*nu/(k*T)-1)-8*Pi*Tq*nu^3*he^2/(exp(he*nu/(k*T)-1)*k*T)

 (20)
> 

solve(E=0);

{T = T, Tq = Tq, he = he, k = k, nu = 0}, {T = T, Tq = Tq, he = 0, k = k, nu = nu}, {T = T, Tq = 0, he = he, k = k, nu = nu}, {T = (1/3)*he*nu/k, Tq = Tq, he = he, k = k, nu = nu}

 (21)
> 

#Thus, for k=3/2*k*T we have the solution
> 

c:=299792458;nu1:=7.880979442*10^14;c/nu1;ec:= 1.602176634*10^(-19);k:= 1/2;E1:=3.259*ec;T:=297;Tq:=1.765*10^(-19);h := 6.62607015*10^(-34);he := h/ec;

299792458

  

0.7880979442e15

  

0.3804000000e-6

  

0.1602176634e-18

  

1/2

  

0.5221493650e-18

  

297

  

0.1765000000e-18

  

0.6626070150e-33

  

0.4135667697e-14

 (22)
> 

NV:=8*Pi*Tq*nu^2*he*nu/(exp(he*nu/(k*T) - 1));

0.1834552756e-31*nu^3/exp(0.2784961412e-16*nu-1)

 (23)
> 

NV;

0.1834552756e-31*nu^3/exp(0.2784961412e-16*nu-1)

 (24)
> 

eval(eval(NV,{E(nu)=E__nu1,nu=nu1}),params);

0.2387994260e14

 (25)
> 

E:=diff(NV,nu);

0.5503658268e-31*nu^2/exp(0.2784961412e-16*nu-1)-0.5109158634e-48*nu^3/exp(0.2784961412e-16*nu-1)

 (26)
> 

solve(E=0);

0., 0.1077214207e18

 (27)
> 

evalf(subs(nu=c/(3*10^(-9)),E));

6.682779197

 (28)
> 

evalf(c/(3*10^(-9)));1/ec;

0.9993081933e17

  

0.6241509074e19

 (29)
> 

plot(E,nu=10^14..1.2*10^17);

 
> 

#All we have to do is to fit the constant k=1/2 so for lambda=3.08nm and for lambda=380.4nm we have B=3.259 eV
> 

 

> 

evalf(c/(9.993081933*10^16));evalf(c/(380*10^(-9)));evalf(subs(nu=c/(380.4*10^(-9)),E));evalf(subs(nu=c/(3.08*10^(-9)),E));

0.3000000001e-8

  

0.7889275211e15

  

0.9023714125e-1

  

9.086107279

 (30)
> 

evalf(c/(3.08*10^(-9)));evalf(c/(9.733521364*10^16));

0.9733521364e17

  

0.3079999999e-8

 (31)
> 

evalf(subs(nu=c/(345*10^(-9)),E));evalf(subs(nu=c/(315*10^(-9)),E));evalf(subs(nu=c/(10^4*10^(-9)),E));
> 

 


 

Download Planck.mw

Cap 5 Paragraph 63 Black-body Radiation (63.4) page 184 . 

 

partial solution...

MichaelVio 45
October 17 2025
0 0

@dharr 

I have the first partial solution:

The equation that fits the dimensional units is   and we assume E-=ν⸱h. The LHS has dimensions of energy J (or eV), and the RHS we have Tq=[s] and ν^3=[s^-3]. The exponential is nondimensional, and h has dimension h=[J*s] Derive h with respect to t is [J*s/s] =[J]

h =[J*s]. Deriving h with respect to t is divided by seconds [J*s/s] =[J]. So RHS=LHS [J]

Please advice 

> 

restart;
> 

with(PDEtools, dchange):with(plots):
> 

with(Units):

Automatically loading the Units[Simple] subpackage
 

  
> 

kernelopts(maxdigits);

38654705646

 (1)
> 

Digits:=10;

10

 (2)
> 

params:={k= 1.380649*10^(-23)*Unit(J/K),rb=5.293*10^(-11)*Unit(m),he=4.135667697*10^(-15)*Unit(eV*s),
ec= 1.602176634*10^(-19)*Unit(C),Tq=1.765*10^(-19)*Unit(s),
c= 299792458*Unit(m/s),T=297*Unit(K),nu1=7.880979442*10^14*Unit(s^(-1)),E1=3.259*Unit(eV)};

{E1 = 0.5221493608e-18*Units:-Unit(J), T = 297*Units:-Unit(K), Tq = 0.1765000000e-18*Units:-Unit(s), c = 299792458*Units:-Unit(m/s), ec = 0.1602176634e-18*Units:-Unit(C), he = 0.6626070096e-33*Units:-Unit(m^2*kg/s), k = 0.1380649000e-22*Units:-Unit(m^2*kg/(s^2*K)), nu1 = 0.7880979442e15*Units:-Unit(1/s), rb = 0.5293000000e-10*Units:-Unit(m)}

 (3)
> 

NV:=8*Pi*Tq*nu^2*he*nu/(exp(he*nu/(k*T) - 1));

8*Pi*Tq*nu^3*he/exp(he*nu/(k*T)-1)

 (4)
> 

E:=diff(NV,nu);

24*Pi*Tq*nu^2*he/exp(he*nu/(k*T)-1)-8*Pi*Tq*nu^3*he^2/(exp(he*nu/(k*T)-1)*k*T)

 (5)
> 

solve(E=0);

{T = T, Tq = Tq, he = he, k = k, nu = 0}, {T = T, Tq = Tq, he = 0, k = k, nu = nu}, {T = T, Tq = 0, he = he, k = k, nu = nu}, {T = (1/3)*he*nu/k, Tq = Tq, he = he, k = k, nu = nu}

 (6)
> 

#Thus, for k=3/2*k*T we have the solution
> 

c:=299792458;nu1:=7.880979442*10^14;c/nu1;ec:= 1.602176634*10^(-19);k:= 1/2;E1:=3.259*ec;T:=297;Tq:=1.765*10^(-19);h := 6.62607015*10^(-34);he := h/ec;

299792458

  

0.7880979442e15

  

0.3804000000e-6

  

0.1602176634e-18

  

1/2

  

0.5221493650e-18

  

297

  

0.1765000000e-18

  

0.6626070150e-33

  

0.4135667697e-14

 (7)
> 

NV:=8*Pi*Tq*nu^2*he*nu/(exp(he*nu/(k*T) - 1));

0.1834552756e-31*nu^3/exp(0.2784961412e-16*nu-1)

 (8)
> 

NV;

0.1834552756e-31*nu^3/exp(0.2784961412e-16*nu-1)

 (9)
> 

eval(eval(NV,{E(nu)=E__nu1,nu=nu1}),params);

0.2387994260e14

 (10)
> 

E:=diff(NV,nu);

0.5503658268e-31*nu^2/exp(0.2784961412e-16*nu-1)-0.5109158634e-48*nu^3/exp(0.2784961412e-16*nu-1)

 (11)
> 

solve(E=0);

0., 0.1077214207e18

 (12)
> 

evalf(subs(nu=c/(3*10^(-9)),E));

6.682779197

 (13)
> 

evalf(c/(3*10^(-9)));1/ec;

0.9993081933e17

  

0.6241509074e19

 (14)
> 

plot(B,nu=10^14..1.2*10^17);

 
> 

#All we have to do is to fit the constant k=1/2 so for lambda=3.08nm and for lambda=380.4nm we have B=3.259 eV
> 

 

> 

evalf(c/(9.993081933*10^16));evalf(c/(380*10^(-9)));evalf(subs(nu=c/(380.4*10^(-9)),B));evalf(subs(nu=c/(3.08*10^(-9)),B));

0.3000000001e-8

  

0.7889275211e15

  

0.9023714125e-1

  

9.086107279

 (15)
> 

evalf(c/(3.08*10^(-9)));evalf(c/(9.733521364*10^16));

0.9733521364e17

  

0.3079999999e-8

 (16)
> 

evalf(subs(nu=c/(345*10^(-9)),B));evalf(subs(nu=c/(315*10^(-9)),B));evalf(subs(nu=c/(10^4*10^(-9)),B));

.1093758750

  

.1307980197

  

0.1343086142e-3

 (17)
> 

 

> 

k:=0.4640872608;


Download phot.mw

dim and units...

MichaelVio 45
October 17 2025
0 0

@dharr 

The equation that fits the dimensional units is ; where t=1/ν and we assume E-=ν⸱h(nu). The LHS has dimensions of energy J (or eV), and the RHS we have Tq^3=[s^3] and ν^3=[s^-3]. The exponential is nondimensional, and h has dimension h=[J*s] Derive h with respect to t is [J*s/s] =[J]

h =[J*s]. Deriving h with respect to t is divided by seconds [J*s/s] =[J]. So RHS=LHS [J]

I'm working on the program and will be back soon.

plot!...

MichaelVio 45
October 16 2025
0 0

@dharr 

In the case that I change the equation, like in file fot.mw

And change the equation so E(nu)=nu*h(nu) and d/d(nu)(E(nu))=nu*d/d(nu)(h(nu))+h(nu);

and the equation is:  

Could you do the plot?

> 

restart;
> 

with(PDEtools, dchange):with(plots):
> 

with(Units):

Automatically loading the Units[Simple] subpackage
 

  
> 

kernelopts(maxdigits);

38654705646

 (1)
> 

Digits:=30;

30

 (2)
> 

params:={k= 1.3806490000*10^(-23)*Unit(J/K),rb=5.293*10^(-11)*Unit(m),
ec= 1.602176634*10^(-19)*Unit(C),Tq=1.765*10^(-19)*Unit(s),
c= 299792458*Unit(m/s),T=297*Unit(K),a=1,nu1=7.880979442*10^15*Unit(s^(-1)),E1=3.259*Unit(eV)};

{E1 = 0.522149360725642886832993280287e-18*Units:-Unit(J), T = 297*Units:-Unit(K), Tq = 0.176500000000000000000000000000e-18*Units:-Unit(s), a = 1, c = 299792458*Units:-Unit(m/s), ec = 0.160217663400000000000000000000e-18*Units:-Unit(C), k = 0.138064900000000000000000000000e-22*Units:-Unit(m^2*kg/(s^2*K)), nu1 = 7880979442000000.000000000*Units:-Unit(1/s), rb = 0.529300000000000000000000000000e-10*Units:-Unit(m)}

 (3)
> 

c:=299792458;nu1:=7.880979442*10^15;c/nu1;ec:= 1.602176634*10^(-19);k:=1.3806490000*10^(-23);E1:=3.259;T:=297;Tq:=1.765*10^(-19);

299792458

  

7880979442000000.000000000

  

0.380400000033396864176212883463e-7

  

0.160217663400000000000000000000e-18

  

0.138064900000000000000000000000e-22

  

3.259

  

297

  

0.176500000000000000000000000000e-18

 (4)
> 

de:=nu*diff(h(nu),nu)+h(nu)=(((nu*h(nu)-Tq^2*8*Pi*nu*h(nu)/(exp(nu*h(nu)/(k*T)-1)))*(exp(nu*h(nu)/(k*T)-1))^2)/(Tq^2*8*Pi*nu)/(exp(nu*h(nu)/(k*T)-nu*h(nu)/(k*T))*((exp(nu*h(nu)/(k*T)))-nu*h(nu)/(k*T)*(exp(nu*h(nu)/(k*T)))-1)));

nu*(diff(h(nu), nu))+h(nu) = 0.127723473498619951824413776992e37*(nu*h(nu)-0.782941437942341091303183058694e-36*nu*h(nu)/exp(243871061146125264521.757765153*nu*h(nu)-1))*(exp(243871061146125264521.757765153*nu*h(nu)-1))^2/(nu*(exp(243871061146125264521.757765153*nu*h(nu))-243871061146125264521.757765153*nu*h(nu)*exp(243871061146125264521.757765153*nu*h(nu))-1.))

 (5)
> 

dsolve({de,h(nu1)=E1/nu1});

h(nu) = RootOf(-ln(nu)+ln(7880979442000000)+Int(-(1562500000000000000000000000000000000000000000000/3991358546831873494512930531)*(243871061146125264521757765153*_a*exp((243871061146125264521757765153/1000000000)*_a)-1000000000*exp((243871061146125264521757765153/1000000000)*_a)+1000000000)/(_a*(391470718971170545651591529347*exp((243871061146125264521757765153/1000000000)*_a-1)-500000000000000000000000000000000000000000000000000000000000000000*exp((243871061146125264521757765153/500000000)*_a-2))), _a = _b .. 407374999999999999999999999998338018199/125000000000000000000000000000000000000)-(Int(-(1562500000000000000000000000000000000000000000000/3991358546831873494512930531)*(243871061146125264521757765153*_a*exp((243871061146125264521757765153/1000000000)*_a)-1000000000*exp((243871061146125264521757765153/1000000000)*_a)+1000000000)/(_a*(391470718971170545651591529347*exp((243871061146125264521757765153/1000000000)*_a-1)-500000000000000000000000000000000000000000000000000000000000000000*exp((243871061146125264521757765153/500000000)*_a-2))), _a = _b .. _Z)))/nu

 (6)
> 

invals:=eval({f1=nu1*Tq,E1=E1/(k*T),f2=1e18*Unit(s^(-1))*Tq},params);

{f1 = 0.139099287151300000000000000000e-2, f2 = .176500000000000000000000000000*Units:-Unit(1/s), 0.522149360725642886832993280287e-18*Units:-Unit(J) = 794775788275222237076.408556634}

 (7)
> 

sol:=dsolve(eval({de,h(7.880979442*10^15)=4.135667697*10^(-15)},{a=1} union invals),h(nu),numeric,stiff=true);

Error, (in dsolve/numeric/SC/firststep) unable to evaluate the partial derivatives of f(x,y) for stiff solution

  
> 

odeplot(sol,nu=eval(f1..f2,invals),labels=[nu/Tq,E/k/T]);

Error, (in Units:-Simple:-=) the following expressions imply incompatible dimensions: {nu = 0.139099287151300000000000000000e-2 .. .176500000000000000000000000000*Units:-Unit(1/s)}

  
 
> 

Diff(sol(nu),nu);
dchange({nu=f/Tq},%,[f],'params'=Tq); # chain rule
dsdnu:=simplify(eval(%,sol(f)=sol_f));

Diff(sol(nu), nu)

  

0.176500000000000000000000000000e-18*(diff(sol(f), f))

  

0.

 (8)
> 

 E(nu):= (3.9965*10^(-15)*ln(nu)+8.941594733*10^(-20))

0.399650000000000000000000000000e-14*ln(nu)+0.894159473300000000000000000000e-19

 (9)
> 

plot(E(nu),nu=10^14..10^18);        

 
> 

E(nu):= -(3.9965*10^(-30)*ln(1/nu)-8.941594733*10^(-20));        

-0.399650000000000000000000000000e-29*ln(1/nu)+0.894159473300000000000000000000e-19

 (10)
> 

plot(E(nu),nu=c*10^14..c*10^18);

Download fot.mw

To fit in the new vision...

MichaelVio 45
October 14 2025
0 0

Please fit the new vision to the file plg.mw to fulfill the work, and check with the old work.

Please apply the same method to verify it for the Gamma ray.

Download plg.mw

PGam.docxplgOld.mw

You are right!...

MichaelVio 45
October 13 2025
0 0

You are right, Professor Et is for the same energy, depending on the period of oscillation. Is the same energy with dependence on (t=1/nu), like the attached file...

Download plm4.mw

Details in document attached...

MichaelVio 45
October 13 2025
0 0

PEdoc.docx

Details are provided in the attached document for the proper equation.

Please advice.

Dimensionless energy ok...

MichaelVio 45
October 13 2025
0 0

It's ok, "I would propose writing e=E/kT as a dimensionless energy." YES

Thanks for the dimensional analysis.

Could we move on?

I'm looking forward to your suggestions!

dimensional...

MichaelVio 45
October 13 2025
0 0

Tq is in seconds

rb in meter 

E in Joule

c in m/s

V = volume in m^3

N*V=J*s^3/m^3

k constant Boltzmann

si ec electron change 

And we can have as an Initial condition Nu 1 with the energy that I specify; or a boundary condition

Nu2:= 9.993081933*10^16 => E:= 0.

Please advice 

initial condition: nu1:=c/(380.4*10^(-9...

MichaelVio 45
October 12 2025
0 0

initial condition:

nu1:=c/(380.4*10^(-9));E(nu1)=3.259*1.6*10^(-19);

thus: E(7.880979442*10^14) = 5.214400000*10^(-19);

@acer And on plkG.mw the T9 does no...

MichaelVio 45
March 20 2025
0 0

@acer And on plkG.mw the T9 does not plot...

at file Aud2e.mw the command plot(E(nu),nu=10..60000);#???? neither plot([Re,Im](E(nu)),nu=10..6*10^4);#????

Aud2e.mwPlanckML.mwplkG.mw

Wish you well

@acer I tried writing again E(nu):=...

MichaelVio 45
March 20 2025
0 0

@acer I tried writing again E(nu):=% it dosent work

please help

ok...

MichaelVio 45
March 20 2025
0 0

@acer ok

how do I close the post?

yes...

MichaelVio 45
March 20 2025
0 0

@acer Yes it's ok, please take a look at 3 maple files where I put a question mark ("????") at the plots that are inconsistent....Aud2e.mwPlanckML.mwplkG.mw

@acer  Yes, you are right.... that...

MichaelVio 45
March 19 2025
0 0

@acer 

Yes, you are right.... that is the issue  E(0):=3.259*ec;

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