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Display information for equation id:math.2117.1 on revision:2117

* Page found: Impulsbilanz (eq math.2117.1)

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TeX (original user input): 

\begin{align}
& \bar{B}\times \left( \nabla \times \bar{B} \right)=\frac{1}{2}\nabla \left( \bar{B}\cdot \bar{B} \right)-\left( \bar{B}\cdot \nabla  \right)\bar{B} \\
& \bar{B}\times \left( \nabla \times \bar{B} \right)=\nabla \cdot \left\{ \left( 1 \right)\frac{1}{2}\left( \bar{B}\cdot \bar{B} \right)-\bar{B}\otimes \bar{B} \right\}+\bar{B}\left( \nabla \cdot \bar{B} \right)=\nabla \cdot \left\{ \left( 1 \right)\frac{1}{2}\left( \bar{B}\cdot \bar{B} \right)-\bar{B}\otimes \bar{B} \right\} \\
& \bar{B}\left( \nabla \cdot \bar{B} \right)=0 \\
\end{align}

TeX (checked): 

{\begin{aligned}&{\bar {B}}\times \left(\nabla \times {\bar {B}}\right)={\frac {1}{2}}\nabla \left({\bar {B}}\cdot {\bar {B}}\right)-\left({\bar {B}}\cdot \nabla \right){\bar {B}}\\&{\bar {B}}\times \left(\nabla \times {\bar {B}}\right)=\nabla \cdot \left\{\left(1\right){\frac {1}{2}}\left({\bar {B}}\cdot {\bar {B}}\right)-{\bar {B}}\otimes {\bar {B}}\right\}+{\bar {B}}\left(\nabla \cdot {\bar {B}}\right)=\nabla \cdot \left\{\left(1\right){\frac {1}{2}}\left({\bar {B}}\cdot {\bar {B}}\right)-{\bar {B}}\otimes {\bar {B}}\right\}\\&{\bar {B}}\left(\nabla \cdot {\bar {B}}\right)=0\\\end{aligned}}

LaTeXML (experimentell; verwendet MathML) rendering

MathML (35.977 KB / 3.865 KB) :

B ¯ × ( ∇ × B ¯ ) = 1 2 ⁢ ∇ ⁡ ( B ¯ ⋅ B ¯ ) - ( B ¯ ⋅ ∇ ) ⁢ B ¯ B ¯ × ( ∇ × B ¯ ) = ∇ ⋅ { ( 1 ) ⁢ 1 2 ⁢ ( B ¯ ⋅ B ¯ ) - B ¯ ⊗ B ¯ } + B ¯ ⁢ ( ∇ ⋅ B ¯ ) = ∇ ⋅ { ( 1 ) ⁢ 1 2 ⁢ ( B ¯ ⋅ B ¯ ) - B ¯ ⊗ B ¯ } B ¯ ⁢ ( ∇ ⋅ B ¯ ) = 0 missing-subexpression ¯ 𝐵 ∇ ¯ 𝐵 1 2 ∇ ⋅ ¯ 𝐵 ¯ 𝐵 ⋅ ¯ 𝐵 ∇ ¯ 𝐵 missing-subexpression ¯ 𝐵 ∇ ¯ 𝐵 ⋅ ∇ 1 1 2 ⋅ ¯ 𝐵 ¯ 𝐵 tensor-product ¯ 𝐵 ¯ 𝐵 ¯ 𝐵 ⋅ ∇ ¯ 𝐵 ⋅ ∇ 1 1 2 ⋅ ¯ 𝐵 ¯ 𝐵 tensor-product ¯ 𝐵 ¯ 𝐵 missing-subexpression ¯ 𝐵 ⋅ ∇ ¯ 𝐵 0 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle\bar{B}\times\left(% \nabla\times\bar{B}\right)=\frac{1}{2}\nabla\left(\bar{B}\cdot\bar{B}\right)-% \left(\bar{B}\cdot\nabla\right)\bar{B}\\ &\displaystyle\bar{B}\times\left(\nabla\times\bar{B}\right)=\nabla\cdot\left\{% \left(1\right)\frac{1}{2}\left(\bar{B}\cdot\bar{B}\right)-\bar{B}\otimes\bar{B% }\right\}+\bar{B}\left(\nabla\cdot\bar{B}\right)=\nabla\cdot\left\{\left(1% \right)\frac{1}{2}\left(\bar{B}\cdot\bar{B}\right)-\bar{B}\otimes\bar{B}\right% \}\\ &\displaystyle\bar{B}\left(\nabla\cdot\bar{B}\right)=0\\ \end{aligned}}}

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MathML (experimentell; keine Bilder) rendering

MathML (5.033 KB / 469 B) :

B¯×(∇×B¯)=12∇(B¯⋅B¯)−(B¯⋅∇)B¯B¯×(∇×B¯)=∇⋅{(1)12(B¯⋅B¯)−B¯⊗B¯}+B¯(∇⋅B¯)=∇⋅{(1)12(B¯⋅B¯)−B¯⊗B¯}B¯(∇⋅B¯)=0

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