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  • $\begingroup$ What I don't understand is, that when we use normal vector notation, without the index, we would simply write the components of the electric field vector as being the derivatives of the components of the vector A and gradient, Therefore I would imply that the your last formula should be E with an upper index. But that's not the case. So do we actually define the electric field vector to be a covariant vector? $\endgroup$
    – Kubrik
    Commented Jun 10, 2023 at 17:39
  • $\begingroup$ @Kubrik $\vec{E}$ is a vector in 3-dimensional space, not in 4-dimensional spacetime. In 3-dimensional space there is no real difference between upper and lower index (because there are only $+1$ in the 3-dimensional metric). So we can just write $\vec{E}$, $\vec{A}$ or $\vec{\nabla}$ and don't need to distinguish between $E_i$ and $E^i$. But in 4-dimensional spacetime there is a difference between upper and lower indices (coming from the $+1$ and $-1$ in the metric). And that's why we need to carefully distinguish between $A^\mu$ and $A_\mu$ (with $\mu=0,1,2,3$). $\endgroup$
    – Thomas Fritsch
    Commented Jun 10, 2023 at 17:54
  • $\begingroup$ Thanks for the answer, but how would we define in the equation (1) of my question if E and A in 4-dimesional spacetime should have an upper or a lower index,, since the electric field transforms as a 2-Tensor and not as a vector? Simply by definition? $\endgroup$
    – Kubrik
    Commented Jun 10, 2023 at 19:33
  • 1
    $\begingroup$ @Kubrik $E$ is 3-vector, not a 4-vector. You can write your 3-dimensional equation (1) as $E_i=-\nabla_i\phi-\partial_t A_i$ or $E^i=-\nabla^i\phi-\partial_t A^i$. It is the same, as said in my previous comment. And don't confuse $A_i$ with $A_\mu$, they are not the same. $\endgroup$
    – Thomas Fritsch
    Commented Jun 10, 2023 at 21:17