# The Electromagetic Tensor and Minkowski Metric Sign Convention

I am trying to figure out how to switch between Minkowski metric tensor sign conventions of (+, -, -, -) to (-, +, +, +) for the electromagnetic tensor $$F^{\alpha \beta}$$. For the convention of (+, -, -, -) I know the contravariant and covarient forms of the electromagnetic tensor are: $$F^{\alpha \beta} = \begin{bmatrix} 0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\ \frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\ \frac{E_{y}}{c} & B_{z} & 0 & -B_{x} \\ \frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \\ \end{bmatrix}$$ and $$F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} = \begin{bmatrix} 0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\ -\frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\ -\frac{E_{y}}{c} & B_{z} & 0 & -B_{x} \\ -\frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \\ \end{bmatrix}.$$

Now for the convention of (-, +, +, +) are the contravariant and covariant forms of the electromagnetic tensor just switched from above along with signs?:

$$F^{\alpha \beta}= \begin{bmatrix} 0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\ -\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\ -\frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\ -\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\ \end{bmatrix}$$ and $$F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} = \begin{bmatrix} 0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\ \frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\ \frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\ \frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\ \end{bmatrix}~?$$

Basically, I am trying to figure out how to switch between the two sign conventions.

The electromagnetic tensor is defined by the Lorentz force equation, which gives the four-force acting on a charged particle as $$f^a=qF^a{}_bv^b$$. The definition of the upper-index four-force and four-velocity have nothing to do with the choice of signature, so the components of the mixed-index electromagnetic tensor $$F^\mu{}_\nu$$ do not depend on the choice of signature. The forms $$F_{\mu\nu}$$ and $$F^{\mu\nu}$$ do have components that depend on the signature, and they can be found, if required, from the components of $$F^\mu{}_\nu$$.
• If $η_{μν} =diag(+1,−1,−1,−1)$ and $\overline{η}_{μν}=diag(−1,+1,+1,+1)$ then wouldn't with corresponding Lorentz force laws be of the form $\ddot x^{\mu} = \eta_{\mu v}F^{\mu v} \dot x^{\lambda}$ As the trajectories $x^{μ}$,$\overline{x}^{μ}$ should agree we can equate the terms $η_{νλ}F^{μν}=\overline{η}_{νλ}\overline{F}^{μν}$. So I actually think that all of the components of the $F^{\mu v}$ are opposite to one another in when changing sign conventions instead of just the first column and row signs being changed. Is this right? – Jay Apr 29 at 0:20
• @Jay: I oversimplified in my original answer. It's only the mixed-index form of F that's the same in both signatures. I've edited appropriately. However, I don't think your comment is correct, and I'm not really sure what you mean. Your expression $\ddot{x}^\mu=\eta_{\mu\nu}F^{\mu\nu}\dot{x}^\lambda$ is ungrammatical, because the free index $\mu$ appearing on the left is contracted on the right, and the free index $\lambda$ doesn't appear on the left. People who use $+---$ and $-+++$ don't differ in the way they write the forms of equations, only in how they related those to components. – Ben Crowell Apr 29 at 1:22
• So I actually think that all of the components of the $F^{\mu\nu}$ are opposite to one another in when changing sign conventions instead of just the first column and row signs being changed. Yes, that seems right, since $F^{\mu\nu}=F^\mu{}_\lambda g^{\lambda\nu}$. Every element of the metric is sign-flipped when we change signatures, and the $F^\mu{}_\lambda$ are invariant. – Ben Crowell Apr 29 at 1:28
• @BenCrowell Unfortunately MTW are not very accurate on that point. It's true that they define ${F^\mu}_\nu$ in mixed form but they fail to tell that the e.m. tensor is antisymmetric. Of course to speak of antisymmetry for a mixed tensor is nonsense, but something could be said anyway. Contract your equation with $v_a$ finding $v_af^a=q\,v_a\,{F^a}_b=q\,v^a\,F_{ab}\,v^b$. As to LHS, since $f^a=m\,dv^a/d\tau$ it must vanish because of $v_a\,v^a=1$ (or $-1$ depending on signature). So $v^a\,F_{ab}\,v^b=0$ showing antisymmetry. – Elio Fabri Apr 29 at 13:10
• @BenCrowell Note that later (§4.2) $\bf F$ is defined as an antisymmetric two-form and 4-force becomes a 1-form (co-vector). Not really a clean job IMHO. – Elio Fabri Apr 29 at 13:12
I use this way: $$$$\tag{1} F_{ab} = \partial_a \, A_b - \partial_b \, A_a,$$$$ where $$$$\tag{2} A^a = (\phi, \, A_x, \, A_y, \, A_z), \qquad\qquad A_a = (\phi, - A_x, - A_y, - A_z).$$$$ Then, we have: \begin{align} E_i &= \Big( -\, \vec{\nabla} \, \phi - \frac{\partial \vec{A}}{\partial t} \Big)_i, \tag{3} \\[12pt] B_i &= (\vec{\nabla} \times \vec{A})_i. \tag{4} \end{align} (1) and sign convention (2) implies $$$$\tag{5} F_{0 i} = \partial_0 \, A_i - \partial_i \, A_0 \equiv E_i.$$$$ Also: $$F^{0 i} = -\, E_i$$.