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Thomas Fritsch
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You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

In the above I have adopted the convention to use greek indices ($\alpha,\mu,\nu\in\{0,1,2,3\}$) for 4 dimensions and latin indices ($i\in\{1,2,3\}$) for 3 dimensions.

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

In the above I have adopted the convention to use greek indices ($\alpha,\mu,\nu\in\{0,1,2,3\}$) for 4 dimensions and latin indices ($i\in\{1,2,3\}$) for 3 dimensions.

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Thomas Fritsch
  • 40.1k
  • 13
  • 74
  • 138

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potentialelectromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index loweringindex lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

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Thomas Fritsch
  • 40.1k
  • 13
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  • 138

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

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Thomas Fritsch
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  • 138
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Source Link
Thomas Fritsch
  • 40.1k
  • 13
  • 74
  • 138
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