This is basically the same question as this one. I have the same problem with the sign. In the Dirac equation $(i\gamma^{\mu}\partial_{\mu}-m)\psi = 0$, the term $i\gamma^{\mu}\partial_{\mu}$ is: $$i\gamma^{\mu}\partial_{\mu} = \sum_{j=0}^{3}i\gamma^{j}\partial_{j}$$ However, the Einstein summation convention is being used. The answer of the linked question says that this is because $\partial_{\mu}$ is contravariant; however, I have seen many times $\partial_{\mu}$ being used the sign convention. For instance, the D'Alembertian is: $$\square^{2} = \partial_{0}^{2}-\partial_{1}^{2}-\partial_{2}^{2}-\partial_{3}^{2} = \partial^{\mu}\partial_{\mu} = \partial_{\mu}\partial^{\mu}$$ So, what's the difference here? When do I have to change the signs and when do I not?


2 Answers 2


We use the metric $[\eta]=\mathrm{diag}(+,-,-,-).$

Note first that

$$X^\mu Y_\mu=X^0Y_0+X^1 Y_1+X^2Y_2+X^3Y_3 \tag{1},$$

but also

$$X^\mu Y_\mu=\eta^{\mu\nu}X_\mu Y_\nu=\eta^{00}X_0Y_0+\eta^{11}X_1Y_1+\eta^{22}X_2Y_2+\eta^{33}X_3Y_3, \tag{2}$$

which, using the components of the metric gives

$$X^\mu Y_\mu=X_0Y_0-X_1Y_1-X_2Y_2-X_3Y_3. \tag{3}$$

Note the position of the indices in $(3)$ compared to $(1)$. We have both indices down in $(3)$ at the cost of introducing factors of $\pm1$ from the Minkowski metric.


The superscript "$2$" on the „Box”/D'Alembertian is superfluous, because the operator is second order in derivatives.

Here is the deal: the "flat" 4D gradient noted by $\partial_{\mu}$ is made up of 4 differential operators $\left( \frac{\partial}{\partial x^0}, \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial x^3}\right)$, where $x^\mu = \{x^0, x^1, x^2, x^3\} =\{t,x,y,z\}$. If the Greek index on the gradient is "downstairs", we call it "covariant", not "contravariant" as you say in your text.

The contraction of the covariant gradient is necessarily done with the "upper index Gamma matrices" to properly obey not only Einstein's notation, but also to ensure that no 4D index "floats", i.e. is left free. The flat space-time Minkowski metric used is $g_{\mu\nu} = \left(g^{-1}\right)^{\mu\nu} = \text{diag} (+1,-1,-1,-1)$ and allows you to simultaneously raise the Greek index in the gradient and lower the Greek one in the Gamma matrices (equation $(1)$ below).

The Box/D'Alembertian is in the mostly minus metric that you propose exactly the one you wrote.

So if everything is done correctly, there is really no minus sign appearing anywhere explicitly.

$$\gamma^{\mu}\partial_{\mu} = \left(g^{\mu\alpha}\gamma_{\alpha}\right)\left(g_{\mu\sigma}\partial^{\sigma}\right) = \delta^{\alpha}_{\sigma} \gamma_{\alpha}\partial^{\sigma} = \gamma_{\sigma}\partial^{\sigma} \equiv \gamma_{\mu}\partial^{\mu} \tag{1} $$

The only minuses you may see are only on the individual components themselves (because the metric has 3 minuses in the spatial part), but remember that components here are summed over (i.e. the Greek indices are not free). I invite you to expand $(1)$ in individual sums of products:

$$ \gamma^{\mu}\partial_{\mu} = \gamma^{0}\partial_{0}+\gamma^{1}\partial_{1}+\gamma^{2}\partial_{2}+\gamma^{3}\partial_{3} \tag{2}$$

Now in order to reach the 5th term of $(1)$ use the metric.

$$ \gamma^{0} = g^{00}\gamma_{0}, ~ \text{etc.} \tag{3}$$

The "non-covariant form of the D'Alembertian" is addressed by Charlie in his reply.


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