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Display information for equation id:math.2555.69 on revision:2555

* Page found: Das ideale Fermigas (eq math.2555.69)

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TeX (original user input): 

\begin{align}

& U\approx \frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( {{E}_{F}} \right)}^{\frac{5}{2}}}{{\left[ 1-\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right] \\

& \frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( {{E}_{F}} \right)}^{\frac{5}{2}}}\approx \frac{3}{5}\bar{N}{{E}_{F}} \\

& {{\left[ 1-\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]\approx 1+5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \\

& \Rightarrow U\approx \frac{3}{5}\bar{N}{{E}_{F}}\left[ 1+5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right] \\

\end{align}

TeX (checked): 

{\begin{aligned}&U\approx {\frac {2}{5}}\left({\frac {V}{{h}^{3}}}\right)4\pi {\frac {\left(2s+1\right)}{2}}{{\left(2m\right)}^{\frac {3}{2}}}{{\left({{E}_{F}}\right)}^{\frac {5}{2}}}{{\left[1-{\frac {{\pi }^{2}}{12}}{{\left({\frac {kT}{{E}_{F}}}\right)}^{2}}\right]}^{\frac {5}{2}}}\left[1+{\frac {5}{2}}{\frac {{\pi }^{2}}{4}}{{\left({\frac {kT}{{E}_{F}}}\right)}^{2}}\right]\\&{\frac {2}{5}}\left({\frac {V}{{h}^{3}}}\right)4\pi {\frac {\left(2s+1\right)}{2}}{{\left(2m\right)}^{\frac {3}{2}}}{{\left({{E}_{F}}\right)}^{\frac {5}{2}}}\approx {\frac {3}{5}}{\bar {N}}{{E}_{F}}\\&{{\left[1-{\frac {{\pi }^{2}}{12}}{{\left({\frac {kT}{{E}_{F}}}\right)}^{2}}\right]}^{\frac {5}{2}}}\left[1+{\frac {5}{2}}{\frac {{\pi }^{2}}{4}}{{\left({\frac {kT}{{E}_{F}}}\right)}^{2}}\right]\approx 1+5{\frac {{\pi }^{2}}{12}}{{\left({\frac {kT}{{E}_{F}}}\right)}^{2}}\\&\Rightarrow U\approx {\frac {3}{5}}{\bar {N}}{{E}_{F}}\left[1+5{\frac {{\pi }^{2}}{12}}{{\left({\frac {kT}{{E}_{F}}}\right)}^{2}}\right]\\\end{aligned}}

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MathML (73.772 KB / 7.238 KB) :

U ≈ 2 5 ⁢ ( V h 3 ) ⁢ 4 ⁢ π ⁢ ( 2 ⁢ s + 1 ) 2 ⁢ ( 2 ⁢ m ) 3 2 ⁢ ( E F ) 5 2 ⁢ [ 1 - π 2 12 ⁢ ( k ⁢ T E F ) 2 ] 5 2 ⁢ [ 1 + 5 2 ⁢ π 2 4 ⁢ ( k ⁢ T E F ) 2 ] 2 5 ⁢ ( V h 3 ) ⁢ 4 ⁢ π ⁢ ( 2 ⁢ s + 1 ) 2 ⁢ ( 2 ⁢ m ) 3 2 ⁢ ( E F ) 5 2 ≈ 3 5 ⁢ N ¯ ⁢ E F [ 1 - π 2 12 ⁢ ( k ⁢ T E F ) 2 ] 5 2 ⁢ [ 1 + 5 2 ⁢ π 2 4 ⁢ ( k ⁢ T E F ) 2 ] ≈ 1 + 5 ⁢ π 2 12 ⁢ ( k ⁢ T E F ) 2 ⇒ U ≈ 3 5 ⁢ N ¯ ⁢ E F ⁢ [ 1 + 5 ⁢ π 2 12 ⁢ ( k ⁢ T E F ) 2 ] absent 𝑈 2 5 𝑉 superscript ℎ 3 4 𝜋 2 𝑠 1 2 superscript 2 𝑚 3 2 superscript subscript 𝐸 𝐹 5 2 superscript delimited-[] 1 superscript 𝜋 2 12 superscript 𝑘 𝑇 subscript 𝐸 𝐹 2 5 2 delimited-[] 1 5 2 superscript 𝜋 2 4 superscript 𝑘 𝑇 subscript 𝐸 𝐹 2 absent 2 5 𝑉 superscript ℎ 3 4 𝜋 2 𝑠 1 2 superscript 2 𝑚 3 2 superscript subscript 𝐸 𝐹 5 2 3 5 ¯ 𝑁 subscript 𝐸 𝐹 absent superscript delimited-[] 1 superscript 𝜋 2 12 superscript 𝑘 𝑇 subscript 𝐸 𝐹 2 5 2 delimited-[] 1 5 2 superscript 𝜋 2 4 superscript 𝑘 𝑇 subscript 𝐸 𝐹 2 1 5 superscript 𝜋 2 12 superscript 𝑘 𝑇 subscript 𝐸 𝐹 2 absent ⇒ absent 𝑈 3 5 ¯ 𝑁 subscript 𝐸 𝐹 delimited-[] 1 5 superscript 𝜋 2 12 superscript 𝑘 𝑇 subscript 𝐸 𝐹 2 {\displaystyle{\displaystyle\begin{aligned} \par&\displaystyle U\approx\frac{2% }{5}\left(\frac{V}{{{h}^{3}}}\right)4\pi\frac{\left(2s+1\right)}{2}{{\left(2m% \right)}^{\frac{3}{2}}}{{\left({{E}_{F}}\right)}^{\frac{5}{2}}}{{\left[1-\frac% {{{\pi}^{2}}}{12}{{\left(\frac{kT}{{{E}_{F}}}\right)}^{2}}\right]}^{\frac{5}{2% }}}\left[1+\frac{5}{2}\frac{{{\pi}^{2}}}{4}{{\left(\frac{kT}{{{E}_{F}}}\right)% }^{2}}\right]\\ \par&\displaystyle\frac{2}{5}\left(\frac{V}{{{h}^{3}}}\right)4\pi\frac{\left(2% s+1\right)}{2}{{\left(2m\right)}^{\frac{3}{2}}}{{\left({{E}_{F}}\right)}^{% \frac{5}{2}}}\approx\frac{3}{5}\bar{N}{{E}_{F}}\\ \par&\displaystyle{{\left[1-\frac{{{\pi}^{2}}}{12}{{\left(\frac{kT}{{{E}_{F}}}% \right)}^{2}}\right]}^{\frac{5}{2}}}\left[1+\frac{5}{2}\frac{{{\pi}^{2}}}{4}{{% \left(\frac{kT}{{{E}_{F}}}\right)}^{2}}\right]\approx 1+5\frac{{{\pi}^{2}}}{12% }{{\left(\frac{kT}{{{E}_{F}}}\right)}^{2}}\\ \par&\displaystyle\Rightarrow U\approx\frac{3}{5}\bar{N}{{E}_{F}}\left[1+5% \frac{{{\pi}^{2}}}{12}{{\left(\frac{kT}{{{E}_{F}}}\right)}^{2}}\right]\\ \par\end{aligned}}}

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MathML (9.433 KB / 645 B) :

U≈25(Vh3)4π(2s+1)2(2m)32(EF)52[1−π212(kTEF)2]52[1+52π24(kTEF)2]25(Vh3)4π(2s+1)2(2m)32(EF)52≈35N¯EF[1−π212(kTEF)2]52[1+52π24(kTEF)2]≈1+5π212(kTEF)2⇒U≈35N¯EF[1+5π212(kTEF)2]

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