MichaelVio

https://perso.crans.org/sylvainrey/Biblio%20Physique/Physique/Physique%20statistique/%5BLandau%2C%20Lifshitz%5D%2005%20Statistical%20Physics%20Part.1.pdf

page 184

I hope it will work

All good!

Vio

Document PDF reattached...

Answered: MichaelVio 45

October 20 2025

0 0

@dharr

If you search on Google, ^[1] Landau & Lifshitz Vol 9 Statistical Physics Cap 5 Paragraph 63 Black-body Radiation (63.4) page 184. You will find the document, but I reattached it

LandauLifsh.pdf

For 1, " you have both he(nu) and h(nu), but I assume you mean to solve the de that led to it, assuming he(nu)=h(nu)." It was a mistake. I apologize, there is only h(nu) I want to calculus h(nu) and Plot it !!!!

I reattached the Planck1.mw

For 2, "I am assuming that when you say 'fit, ' you mean find the single value of k*T that makes those two conditions true." Yes, I want the value of k*T for that

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

Planck1.mw

I look froward to your's reply!

I check and please advice me...

Answered: MichaelVio 45

October 20 2025

0 0

@dharr

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) I want to solve the equation he(nu) = -k*T*LambertW(((_C1 - Int(nu*h(nu), nu))*exp(-1))/(8*Pi*Tq*nu^2*k*T))/nu

plot the solution for k=1/2 Planck.mw

2) and if it is possible to solve the k that fit 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

The calculus is according to ^[1] Landau & Lifshitz Vol 9 Statistical Physics Cap 5 Paragraph 63 Black-body Radiation (63.4) page 184.

Landau&Lifsh.pdf

Please advise on...

Answered: MichaelVio 45

October 19 2025

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);

(1)

>	Digits:=10;

(2)

>	params:={k= 1.38064910^(-23)Unit(J/K),rb=5.29310^(-11)Unit(m),he=4.13566769710^(-15)Unit(eVs), ec= 1.60217663410^(-19)Unit(C),Tq=1.76510^(-19)Unit(s), c= 299792458Unit(m/s),T=297Unit(K),nu1=7.88097944210^14Unit(s^(-1)),E1=3.259Unit(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:=8PiTqnu^2he(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);

(5)

>	ic:=h(7.88927521110^14)=4.13566769710^(-15);

(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=3/2kT we have the solution

>	c:=299792458;nu1:=7.88097944210^14;c/nu1;ec:= 1.60217663410^(-19);k:= 1/2;E1:=3.259ec;T:=297;Tq:=1.76510^(-19);;he := h/ec;

(9)

>	NV:=8PiTqnu^2he(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(9.48927521110^16)=4.13566769710^(-17);

(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;

(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);

(16)

>	Digits:=20;

(17)

>	params:={k= 1.38064910^(-23)Unit(J/K),rb=5.29310^(-11)Unit(m),he=4.13566769710^(-15)Unit(eVs), ec= 1.60217663410^(-19)Unit(C),Tq=1.76510^(-19)Unit(s), c= 299792458Unit(m/s),T=297Unit(K),nu1=7.88097944210^14Unit(s^(-1)),E1=3.259Unit(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:=8PiTqnu^2henu/(exp(henu/(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=3/2kT we have the solution

>	c:=299792458;nu1:=7.88097944210^14;c/nu1;ec:= 1.60217663410^(-19);k:= 1/2;E1:=3.259ec;T:=297;Tq:=1.76510^(-19);h := 6.62607015*10^(-34);he := h/ec;

(22)

>	NV:=8PiTqnu^2henu/(exp(henu/(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);

(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);

(27)

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

(28)

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

(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/(9.99308193310^16));evalf(c/(38010^(-9)));evalf(subs(nu=c/(380.410^(-9)),E));evalf(subs(nu=c/(3.0810^(-9)),E));

(30)

>	evalf(c/(3.0810^(-9)));evalf(c/(9.73352136410^16));

(31)

>	evalf(subs(nu=c/(34510^(-9)),E));evalf(subs(nu=c/(31510^(-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...

Answered: 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);

(1)

>	Digits:=10;

(2)

>	params:={k= 1.38064910^(-23)Unit(J/K),rb=5.29310^(-11)Unit(m),he=4.13566769710^(-15)Unit(eVs), ec= 1.60217663410^(-19)Unit(C),Tq=1.76510^(-19)Unit(s), c= 299792458Unit(m/s),T=297Unit(K),nu1=7.88097944210^14Unit(s^(-1)),E1=3.259Unit(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:=8PiTqnu^2henu/(exp(henu/(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=3/2kT we have the solution

>	c:=299792458;nu1:=7.88097944210^14;c/nu1;ec:= 1.60217663410^(-19);k:= 1/2;E1:=3.259ec;T:=297;Tq:=1.76510^(-19);h := 6.62607015*10^(-34);he := h/ec;

(7)

>	NV:=8PiTqnu^2henu/(exp(henu/(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);

(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);

(12)

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

(13)

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

(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/(9.99308193310^16));evalf(c/(38010^(-9)));evalf(subs(nu=c/(380.410^(-9)),B));evalf(subs(nu=c/(3.0810^(-9)),B));

(15)

>	evalf(c/(3.0810^(-9)));evalf(c/(9.73352136410^16));

(16)

>	evalf(subs(nu=c/(34510^(-9)),B));evalf(subs(nu=c/(31510^(-9)),B));evalf(subs(nu=c/(10^4*10^(-9)),B));

(17)

>

>	k:=0.4640872608;

Download phot.mw

dim and units...

Answered: 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!...

Answered: 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);

(1)

>	Digits:=30;

(2)

>	params:={k= 1.380649000010^(-23)Unit(J/K),rb=5.29310^(-11)Unit(m), ec= 1.60217663410^(-19)Unit(C),Tq=1.76510^(-19)Unit(s), c= 299792458Unit(m/s),T=297Unit(K),a=1,nu1=7.88097944210^15Unit(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:=7.88097944210^15;c/nu1;ec:= 1.60217663410^(-19);k:=1.380649000010^(-23);E1:=3.259;T:=297;Tq:=1.76510^(-19);

(4)

>	de:=nudiff(h(nu),nu)+h(nu)=(((nuh(nu)-Tq^28Pinuh(nu)/(exp(nuh(nu)/(kT)-1)))(exp(nuh(nu)/(kT)-1))^2)/(Tq^28Pinu)/(exp(nuh(nu)/(kT)-nuh(nu)/(kT))((exp(nuh(nu)/(kT)))-nuh(nu)/(kT)(exp(nuh(nu)/(kT)))-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});

(6)

>	invals:=eval({f1=nu1Tq,E1=E1/(kT),f2=1e18Unit(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(7.88097944210^15)=4.13566769710^(-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));

(8)

>	E(nu):= (3.996510^(-15)ln(nu)+8.941594733*10^(-20))

(9)

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

>	E(nu):= -(3.996510^(-30)ln(1/nu)-8.941594733*10^(-20));

(10)

>	plot(E(nu),nu=c10^14..c10^18);

Download fot.mw

To fit in the new vision...

Answered: 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.docx plgOld.mw

You are right!...

Answered: 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...

Answered: 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...

Answered: 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...

Answered: 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

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About phot1.mw...

corection about Tq=rb/c...

equation and dsolve...

Link for Landau Lifshits Vol 5 Part 1...

Document PDF reattached...

I check and please advice me...

Please advise on...

partial solution...

dim and units...

plot!...

To fit in the new vision...

You are right!...

Details in document attached...

Dimensionless energy ok...

dimensional...

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