Recently I was studying quantum electrodynamics and got a small question about tensors.
The Lagrangian of QED has this thing $\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$
And I can't understand the difference between $F_{\mu \nu}$ and $F^{\mu \nu}$
Also there is unusual derivative with such notation $\partial ^{\mu}$ and usual $\partial _{\mu}$
So my question is what does this notation mean?
Thank you!
It would be digestible if we can translate everything from tensor calculus to vector calculus, please see most typical results below:
Regard zero rank tensors as numbers, first rank as vectors, second rank as matrices , etc.
Lazy notations
$$\mathbf{a}=\sum_i a_i \mathbf{e}_i=a_i \mathbf{e}_i$$
$$\mathbb{AB}=\sum_j a_{ij} b_{jk}=a_{ij} b_{jk}=c_{ik}$$
$$\mathbf{0}=0\, \mathbf{e}_{i}$$
$$\mathbb{I}=\delta_{ij}$$
$$\mathbf{a}^T \mathbf{b}= \begin{pmatrix} a_1 & a_2 & a_3 \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}= \mathbf{a} \cdot \mathbf{b}= \delta_{ij} a_i b_j=a_i b_i$$
$$\mathbb{A} \mathbb{B}= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{pmatrix} =a_{ij} b_{jk}$$
$$\epsilon_{ijk}=\frac{(i-j)(j-k)(k-i)}{2} \quad \quad i,j,k\in \{ 1,2,3 \}$$
$$\det \mathbb{A}= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=\frac{\epsilon_{ijk} \epsilon_{pqr} a_{ip} a_{jq} a_{kr}}{3!}$$
$$\mathbf{a} \times \mathbf{b}= \begin{pmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}= \begin{pmatrix} 0 & b_3 & -b_2 \\ -b_3 & 0 & b_1 \\ b_2 & -b_1 & 0 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}= \epsilon_{ijk} a_i b_j \mathbf{e}_k$$
$$\mathbf{a}^T \times \mathbf{b}^T= \begin{pmatrix} a_1 & a_2 & a_3 \end{pmatrix} \begin{pmatrix} 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \end{pmatrix}= \begin{pmatrix} b_1 & b_2 & b_3 \end{pmatrix} \begin{pmatrix} 0 & a_3 & -a_2 \\ -a_3 & 0 & a_1 \\ a_2 & -a_1 & 0 \end{pmatrix}$$
$$g^{\mu \nu}=g_{\mu \nu}= \begin{pmatrix} 1 & \mathbf{0}^T \\ \mathbf{0} & -\mathbb{I} \\ \end{pmatrix}$$
$$x_\mu= \begin{pmatrix} ct \\ -\mathbf{r} \end{pmatrix}= g_{\mu \nu} x^{\nu}$$
$$x^\mu= \begin{pmatrix} ct \\ \mathbf{r} \end{pmatrix}= g^{\mu \nu} x_{\nu}$$
$$\partial^\mu = \frac{\partial}{\partial x_\mu}= \begin{pmatrix} \dfrac{1}{c} \partial_t \\ -\nabla \end{pmatrix}$$
$$\partial_\mu = \frac{\partial}{\partial x^\mu}= \begin{pmatrix} \dfrac{1}{c} \partial_t \\ \nabla \end{pmatrix}$$
$$\partial^\mu = g^{\mu \nu} \partial_\nu$$
$$\partial_\mu = g_{\mu \nu} \partial^\nu$$
$$\square^2= \partial^\mu \partial_\mu= \partial_\mu \partial^\mu= \nabla^2-\frac{1}{c^2} \frac{\partial^2}{\partial t^2}$$
$$\gamma_u=\frac{1}{\sqrt{1-\dfrac{u^2}{c^2}}}$$
$$d\tau=\frac{dt}{\gamma_u}$$
$$\frac{d\gamma_u}{d\tau}=\frac{\gamma_u^4}{c^2} (\mathbf{a} \cdot \mathbf{u})$$
$$u^\mu= \frac{dx^\mu}{d\tau}=\gamma_u \begin{pmatrix} c \\ \mathbf{u} \end{pmatrix}$$
$$a^\mu= \frac{du^\mu}{d\tau}$$
$$b^\mu= \frac{da^\mu}{d\tau}$$
$$p^\mu=m_0 u^\mu= \begin{pmatrix} \dfrac{E}{c} \\ \mathbf{p} \end{pmatrix}= \gamma_u m_0 \begin{pmatrix} c \\ \mathbf{u} \end{pmatrix}$$
$$F^\mu=\gamma_u \begin{pmatrix} \dfrac{\mathbf{F} \cdot \mathbf{u}}{c} \\ \mathbf{F} \end{pmatrix}$$
$$k^\mu= \begin{pmatrix} \dfrac{\omega}{c} \\ \mathbf{k} \end{pmatrix}$$
$$J^\mu= \begin{pmatrix} \rho c \\ \mathbf{J} \end{pmatrix}$$
$$A^\mu= \begin{pmatrix} \dfrac{\phi}{c} \\ \mathbf{A} \end{pmatrix}$$
$$K^\mu= \begin{pmatrix} \dfrac{\mathbf{E} \cdot \mathbf{J}}{c} \\ \rho \mathbf{E}+\mathbf{J} \times \mathbf{B} \end{pmatrix}$$
$${\Lambda^\mu}_\nu= \begin{pmatrix} \gamma_v & -\dfrac{\gamma_v}{c} \mathbf{v}^T \\ -\dfrac{\gamma_v}{c} \mathbf{v} & \mathbb{I}+ \left( \dfrac{\gamma_v-1}{v^2} \mathbf{v} \right) \mathbf{v}^T \\ \end{pmatrix}= {\Lambda_\nu}^\mu$$
$${\Lambda_\mu}^\nu= \begin{pmatrix} \gamma_v & \dfrac{\gamma_v}{c} \mathbf{v}^T \\ \dfrac{\gamma_v}{c} \mathbf{v} & \mathbb{I}+ \left( \dfrac{\gamma_v-1}{v^2} \mathbf{v} \right) \mathbf{v}^T \\ \end{pmatrix}= {\Lambda^\nu}_\mu$$
$${\Lambda^\nu}_\mu=({\Lambda_\mu}^\nu)^{-1}$$
$$F^{\mu \nu}=\partial^\mu A^\nu-\partial^\nu A^\mu= \begin{pmatrix} 0 & -\dfrac{1}{c} \mathbf{E}^T \\ \dfrac{1}{c} \mathbf{E}^T & \times \mathbf{B}^T \end{pmatrix}= \begin{pmatrix} 0 & -\dfrac{1}{c} \mathbf{E}^T \\ \dfrac{1}{c} \mathbf{E}^T & \mathbf{B} \times \end{pmatrix}$$
$$F_{\mu \nu}=\partial_\mu A_\nu-\partial_\nu A_\mu= \begin{pmatrix} 0 & \dfrac{1}{c} \mathbf{E}^T \\ -\dfrac{1}{c} \mathbf{E}^T & \times \mathbf{B}^T \end{pmatrix}= \begin{pmatrix} 0 & \dfrac{1}{c} \mathbf{E}^T \\ -\dfrac{1}{c} \mathbf{E}^T & \mathbf{B} \times \end{pmatrix}$$
$$G^{\mu \nu}=\frac{\epsilon_{\mu \nu \alpha \beta}}{2}F_{\alpha \beta}= \begin{pmatrix} 0 & -\mathbf{B}^T \\ \mathbf{B}^T & \times \left( -\dfrac{\mathbf{E}}{c} \right)^T \end{pmatrix}= \begin{pmatrix} 0 & -\mathbf{B}^T \\ \mathbf{B}^T & \left( -\dfrac{\mathbf{E}}{c} \right) \times \end{pmatrix}$$
$$G_{\mu \nu}=\frac{\epsilon^{\mu \nu \alpha \beta}}{2}F^{\alpha \beta}= \begin{pmatrix} 0 & \mathbf{B}^T \\ -\mathbf{B}^T & \times \left( -\dfrac{\mathbf{E}}{c} \right)^T \end{pmatrix}= \begin{pmatrix} 0 & \mathbf{B}^T \\ -\mathbf{B}^T & \left( -\dfrac{\mathbf{E}}{c} \right) \times \end{pmatrix}$$
$$\square^2 A^{\mu}=\partial_\lambda F^{\lambda \mu}=\mu_0 J^{\mu} $$
$$ \begin{pmatrix} \dfrac{1}{c} \partial_t & \nabla^T \end{pmatrix} \begin{pmatrix} 0 & -\dfrac{1}{c} \mathbf{E}^T \\ \dfrac{1}{c} \mathbf{E}^T & \times \mathbf{B}^T \end{pmatrix}=\mu_0 \begin{pmatrix} \rho c & \mathbf{J}^T \end{pmatrix}$$
$$\partial_\mu G^{\mu \nu}=0 $$
$$ \begin{pmatrix} \dfrac{1}{c} \partial_t & \nabla^T \end{pmatrix} \begin{pmatrix} 0 & -\mathbf{B}^T \\ \mathbf{B}^T & \times \left( -\dfrac{\mathbf{E}}{c} \right)^T \end{pmatrix}= \begin{pmatrix} 0 & \mathbf{0}^T \end{pmatrix}$$
$$x'^\mu={\Lambda^\mu}_\nu \, x^\nu$$
$$x'_\mu={\Lambda_\mu}^\nu \, x_\nu$$
$$\partial'^\mu={\Lambda^\mu}_\nu \, \partial^\nu$$
$$\partial'_\mu={\Lambda_\mu}^\nu \, \partial_\nu$$
$$F'^{\mu \nu}={\Lambda_\mu}^\alpha {\Lambda_\nu}^\beta F^{\alpha \beta}$$
$$F'_{\mu \nu}={\Lambda^\mu}_\alpha {\Lambda^\nu}_\beta F_{\alpha \beta}$$
$$G'^{\mu \nu}={\Lambda_\mu}^\alpha {\Lambda_\nu}^\beta G^{\alpha \beta}$$
$$G'_{\mu \nu}={\Lambda^\mu}_\alpha {\Lambda^\nu}_\beta G_{\alpha \beta}$$
$$x^\mu x_\mu=x_\mu x^\mu=c^2 t^2-r^2=s^2$$
$$ds^2= g_{\mu \nu} dx^\mu dx^\nu= g^{\mu \nu} dx_\mu dx_\nu= dx^\mu dx_\mu= dx_\mu dx^\mu$$
$$u^\mu x_\mu=\gamma_u(c^2 t-\mathbf{u} \cdot \mathbf{r})$$
$$u^\mu u_\mu=c^2$$
$$a^\mu u_\mu=0$$
$$a^\mu a_\mu+b^\mu u_\mu=0$$
$$p^\mu p_\mu=m_0 c^2$$
$$F^\mu p_\mu=0$$
$$F^\mu F_\mu=-\gamma_u^2 \left[ F^2-\frac{(\mathbf{F} \cdot \mathbf{u})^2}{c^2} \right]$$
$$F^\mu x_\mu=\gamma_u \mathbf{F} \cdot (\mathbf{u} t-\mathbf{r})$$
$$F^\mu=\frac{dp^\mu}{d\tau}=m_0 a^\mu$$
$$k^\mu k_\mu=\frac{\omega^2}{c^2}-k^2$$
$$k^\mu x_\mu=\omega t-\mathbf{k} \cdot \mathbf{r}$$
$$J^\mu J_\mu=\rho^2 c^2-J^2$$
$$\partial_\mu J^\mu=0$$
$$\partial_\mu A^\mu=0$$
$$A^\mu A_\mu=\frac{\phi^2}{c^2}-A^2$$
$$J^\mu A_\mu=\rho \phi-\mathbf{J} \cdot \mathbf{A}$$
$$J^\mu K_\mu=0$$
$$K^\mu=F^{\mu \nu} J_\nu$$
$$F^\mu=qF^{\mu \nu} u_\nu$$
$$F^{\mu \nu} F_{\mu \nu}=2 \left( B^2-\frac{E^2}{c^2} \right)$$
$$F^{\mu \nu} G_{\mu \nu}=-\frac{4\mathbf{E} \cdot \mathbf{B}}{c}$$
$$m_0 a^\mu-qF^{\mu \nu} u_\nu=\frac{1}{4\pi \epsilon_0} \frac{2q^2}{3c^3} \left( b^\mu+\frac{a^\lambda a_\lambda}{c^2} u^\mu \right)$$
Every four-vectors can be transformed using Lorentz matrix.
For easy clarification, we may use $E_{\mu \nu}$, $B_{\mu \nu}$ instead of $F_{\mu \nu}$, $G_{\mu \nu}$ and may extend the ideas to $D_{\mu \nu}$, $H_{\mu \nu}$, $P_{\mu \nu}$ and $M_{\mu \nu}$.
Besides four-vectors, we can use quaternions instead, that is
$$ct+x i+y j+z k \in \mathbb{H}$$