5 formatting edited Jul 10 at 17:20 Chris K 9,86933 gold badges2323 silver badges5454 bronze badges The following input text  Clear[ρ, ν, λ, T, h, c, kB] λ[ν_] := c/ν ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) DifferentialD]ν ρ[ν, T] ρ[λ, T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$$$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3λ^3 \[Pi]hπh \[DifferentialD]\[Lambda]\[DifferentialD]λ)/(c^3 (-1 + E^((\[Lambda]λ)/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals  ? The following input text  Clear[ρ, ν, λ, T, h, c, kB] λ[ν_] := c/ν ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) DifferentialD]ν ρ[ν, T] ρ[λ, T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals  ? The following input text Clear[ρ, ν, λ, T, h, c, kB] λ[ν_] := c/ν ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) $DifferentialD]ν ρ[ν, T] ρ[λ, T]  is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be (8 λ^3 πh \[DifferentialD]λ)/(c^3 (-1 + E^((λ)/(kB T))))  which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]}$ As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals? 4 Format code edit approved Jul 10 at 17:18 Rohit Namjoshi 2,33911 gold badge55 silver badges1717 bronze badges The following input text  Clear[Rho]Clear[ρ, \[Nu]ν, \[Lambda]λ, T, h, c, kB] \[Lambda][\[Nu]_]λ[ν_] := c/\[Nu]ν \[Rho][\[Nu]_ρ[ν_, T_] := (8 \[Pi]hπh/ c^3) (\[Nu]^3ν^3/(Exp[h \[Nu]ν/(kB T)] - 1)) \[DifferentialD]\[Nu]\[DifferentialD]ν \[Rho][\[Nu]ρ[ν, T] \[Rho][\[Lambda]ρ[λ, T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ? The following input text  Clear[Rho], \[Nu], \[Lambda], T, h, c, kB] \[Lambda][\[Nu]_] := c/\[Nu] \[Rho][\[Nu]_, T_] := (8 \[Pi]h/ c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu] \[Rho][\[Nu], T] \[Rho][\[Lambda], T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ? The following input text  Clear[ρ, ν, λ, T, h, c, kB] λ[ν_] := c/ν ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) DifferentialD]ν ρ[ν, T] ρ[λ, T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ? 3 Space tweaks. edited Jul 10 at 15:08 Somos 2,97511 gold badge22 silver badges1111 bronze badges The following input text  Clear[Rho], \[Nu], \[Lambda], T, h, c, kB] \[Lambda][\[Nu]_] := c/\[Nu] \[Rho][\[Nu]_, T_] := (8 \[Pi]h/ c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu] \[Rho][\[Nu], T] \[Rho][\[Lambda], T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ? The following input text  Clear[Rho], \[Nu], \[Lambda], T, h, c, kB] \[Lambda][\[Nu]_] := c/\[Nu] \[Rho][\[Nu]_, T_] := (8 \[Pi]h/ c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu] \[Rho][\[Nu], T] \[Rho][\[Lambda], T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ? The following input text  Clear[Rho], \[Nu], \[Lambda], T, h, c, kB] \[Lambda][\[Nu]_] := c/\[Nu] \[Rho][\[Nu]_, T_] := (8 \[Pi]h/ c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu] \[Rho][\[Nu], T] \[Rho][\[Lambda], T]  , which visually appears as \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} , is employed to get the expression for $$\rho [\lambda ,T]$$, which is expected to be (equivalent to) $$\frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1}$$ but turns out to be  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))  , which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} As can be told by comparison, $$\lambda$$ is literally substituted for $$\nu$$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ? 2 deleted 15 characters in body edited Jul 10 at 10:11 SOUser 14355 bronze badges 1 asked Jul 10 at 10:06 SOUser 14355 bronze badges