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Equation la plus compliquée

MessagePosté: Lundi 24 Février 2020, 18:51
par expodyssey
Bonjour!
Je suis nouveau sur Mathematex et j'aimerai savoir tout d'abord quelque chose: quelle est l'équation la plus longue (ou la plus compliquée) du monde?
j'ai trouvé ça (en LaTex) :

$$\pounds = -\dfrac{1}{4} B_{\mu\nu} B^{\mu\nu} -\dfrac{1}{8}tr(W_{\mu\nu}W^{\mu\nu}) -\dfrac{1}{2}tr(G_{\mu\nu}G^{\mu\nu})$$


$$+(\bar{\nu}_{L},\bar{e}_{L}) \tilde{\sigma}^{\mu}i D_{\mu} \begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix}+\bar{e}_{R}\sigma^{\mu}i D_{\mu}e_{R}+\bar{\nu}_{R}\sigma^{mu}i D_{\mu}\nu_{R}+(h.c.)$$

$$-\dfrac{\sqrt{2}}{\upsilon}\left[ (\bar{\nu}_{L},\bar{e}_{L})\o M^{e}e_{R}+\bar{e}_{R}\bar{M}^{e}\bar{\o}(\begin{pmatrix}\nu_{L} \\ e_{L} \end{pmatrix}\right]$$


$$-\dfrac{\sqrt{2}}{\upsilon}\left[(-\bar{e}_{L},\bar{\nu}_{L}) \o ^{\ast}M^{\nu}\nu_{R}+\bar{\nu}_{R}\bar{M}^{\nu}\o ^{T} \begin{pmatrix} -e_{L} \\ \nu_{L} \end{pmatrix}\right]$$


$$+(\bar{\upsilon}_{L},\bar{d}_{L})\tilde{\sigma}^{\mu}i D_{\mu} \begin{pmatrix} \upsilon_L \\ d_L \end{pmatrix}+\bar{\upsilon}_R\sigma^{\mu}i D_{\mu}\upsilon_R+\bar{d}_R\sigma^{\mu}i D_{\mu}d_R+(h.c.)$$


$$-\dfrac{\sqrt{2}}{\nu}\left[(\bar{\upsilon}_L,\bar{d}_L)\o M^dd_R+\bar{d}_R\bar{M}^d\bar{\o}\begin{pmatrix} \upsilon_L \\ d_L \end{pmatrix}\right]$$


$$-\dfrac{\sqrt{2}}{\nu}\left[(\bar{d}_L,\bar{\upsilon}_L)\o^{\ast}M^{\upsilon}\upsilon_R+\bar{\upsilon}_R\bar{M}^d\o^T\begin{pmatrix} -d_L \\ \upsilon_L \end{pmatrix}\right]$$


$$+\overline{(D_{\mu}\o)}D^{\mu}\o-m^2_h\left[\bar{\o}\o-\upsilon^2\div2\right]^2\div2\upsilon^2$$


Code: Tout sélectionner
$$\pounds = -\dfrac{1}{4} B_{\mu\nu} B^{\mu\nu} -\dfrac{1}{8}tr(W_{\mu\nu}W^{\mu\nu}) -\dfrac{1}{2}tr(G_{\mu\nu}G^{\mu\nu})$$
$$+(\bar{\nu}_{L},\bar{e}_{L}) \tilde{\sigma}^{\mu}i D_{\mu} \begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix}+\bar{e}_{R}\sigma^{\mu}i D_{\mu}e_{R}+\bar{\nu}_{R}\sigma^{mu}i D_{\mu}\nu_{R}+(h.c.)$$ $$-\dfrac{\sqrt{2}}{\upsilon}\left[ (\bar{\nu}_{L},\bar{e}_{L})\o M^{e}e_{R}+\bar{e}_{R}\bar{M}^{e}\bar{\o}(\begin{pmatrix}\nu_{L} \\ e_{L} \end{pmatrix}\right]$$
$$-\dfrac{\sqrt{2}}{\upsilon}\left[(-\bar{e}_{L},\bar{\nu}_{L}) \o ^{\ast}M^{\nu}\nu_{R}+\bar{\nu}_{R}\bar{M}^{\nu}\o ^{T} \begin{pmatrix} -e_{L} \\ \nu_{L} \end{pmatrix}\right]$$
$$+(\bar{\upsilon}_{L},\bar{d}_{L})\tilde{\sigma}^{\mu}i D_{\mu} \begin{pmatrix} \upsilon_L \\ d_L \end{pmatrix}+\bar{\upsilon}_R\sigma^{\mu}i D_{\mu}\upsilon_R+\bar{d}_R\sigma^{\mu}i D_{\mu}d_R+(h.c.)$$
$$-\dfrac{\sqrt{2}}{\nu}\left[(\bar{\upsilon}_L,\bar{d}_L)\o M^dd_R+\bar{d}_R\bar{M}^d\bar{\o}\begin{pmatrix} \upsilon_L \\ d_L \end{pmatrix}\right]$$
$$-\dfrac{\sqrt{2}}{\nu}\left[(\bar{d}_L,\bar{\upsilon}_L)\o^{\ast}M^{\upsilon}\upsilon_R+\bar{\upsilon}_R\bar{M}^d\o^T\begin{pmatrix} -d_L \\ \upsilon_L \end{pmatrix}\right]$$
$$+\overline{(D_{\mu}\o)}D^{\mu}\o-m^2_h\left[\bar{\o}\o-\upsilon^2\div2\right]^2\div2\upsilon^2$$

Mais je ne sais pas si on peut faire mieux merci. :wink: :wink:
:crash:
Fais marcher ta tête avant de faire marcher celle des autres :roll: :roll:
:crash: