Why does the Wikipedia article use a negative sign in the 4-gradient?

[etotheipi]

Apparently the contravariant components of the 4-gradient $\partial$ are $\partial^{\mu} = \left(\partial_t, -\nabla \right)$ where $\nabla$ is the usual 3-gradient. We can use the metric to lower the index like $\partial_{\mu} = \eta_{\mu \nu} \partial^{\nu}$ and if the signature is $(+,-,-,-)$ the space signs get flipped, so we should get $\partial_{\mu} = (\partial_t, \nabla)$.

Why then in this Wikipedia article do they use $\partial_{\nu} = \left(\partial_t, -\nabla \right)$ when they write$$\partial \cdot J = \partial_{\nu} J^{\nu} = \left(\partial_t, -\nabla \right) \cdot (\rho, \vec{j}) = 0$$It's the right way round, since when we use the Minkowski inner product we recover the continuity equation $\partial_t \rho = \nabla \cdot \vec{j}$, but they put a negative sign inside $\partial_{\nu}$. So is it actually the other way around, i.e. $\partial^{\mu} = \left(\partial_t, \nabla \right)$ and $\partial_{\mu} = \left(\partial_t, -\nabla \right)$? Sorry if I missed something!

 

[PeroK]

It looks like the article gets a bit muddled. Putting the inner product symbol between two rows of numbers is bound to get conceptually cloudy in this context.

 

[etotheipi]

Ah, okay I think I get it now. Really it would be better just to say $$\partial \cdot J = \left(\partial_t, -\nabla \right) \cdot (\rho, \vec{j})$$either with all contravariant or all covariant components (in the above, all contravariant), or you can write it as the summation $$\partial \cdot J = \partial_{\mu}J^{\mu} = \partial^{\mu}J_{\mu}$$They're the same, but $\partial^{\mu}$ is an operator whilst $\left(\partial_t, -\nabla \right)$ is a tuple, and we shouldn't identify one with each other. So what they wrote is right, but maybe misleading.

 

[PeroK]

Ah, okay I think I get it now. Really it would be better just to say $$\partial \cdot J = \left(\partial_t, -\nabla \right) \cdot (\rho, \vec{j})$$or to do the summation $$\partial \cdot J = \partial_{\mu}J^{\mu} = \partial^{\mu}J_{\mu}$$They should be the same, but $\partial^{\mu}$ is an operator whilst $\left(\partial_t, -\nabla \right)$ is a tuple, and we shouldn't identify one with each other. So what they wrote is right, but maybe misleading.

$$\partial \cdot J = \partial_{\mu}J^{\mu} = \partial^{\mu}J_{\mu} = \partial_t \rho + \vec{\nabla}\cdot \vec j$$

 

[etotheipi]

$$\partial \cdot J = \partial_{\mu}J^{\mu} = \partial^{\mu}J_{\mu} = \partial_t \rho + \vec{\nabla}\cdot \vec j$$

Sounds good to me

 

[etotheipi]

Something else just occurred; how do we know whether someone is using the repeated index notation to mean$$U_{\mu}U^{\mu} = U_0 U^0 - U_1 U^1 - U_2 U^2 - U_3 U^3 \quad \left( = \eta_{\mu \nu} U^{\mu} U^{\nu} \right)$$or a literal sum$$A_{\mu} A^{\mu} = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3$$Is this just another case of needing to keep your wits about you?

 

[PeroK]

It's never this:

Something else just occurred; how do we know whether someone is using the repeated index notation to mean$$U_{\mu}U^{\mu} = U_0 U^0 - U_1 U^1 - U_2 U^2 - U_3 U^3 \quad \left( = \eta_{\mu \nu} U^{\mu} U^{\nu} \right)$$

And always this:

or a literal sum$$A_{\mu} A^{\mu} = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3$$

 

[etotheipi]

Ah, right, of course you're correct. I got muddled, it should be$$A \cdot A = \eta_{\mu \nu} A^{\mu} A^{\nu} = A^0A^0 - A^1 A^1 - A^2 A^2 - A^3 A^3$$if you contract $\eta_{\mu \nu} A^{\mu} = A_{\nu}$, you'd end up with$$\eta_{\mu \nu} A^{\mu} A^{\nu} = A_{\nu} A^{\nu} = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3$$but then, with the $(+,-,-,-)$ signature, you have $A_i = -A_i$ but $A_0 = A^0$, so $$A_{\nu} A^{\nu} = A^0A^0 - A^1 A^1 - A^2 A^2 - A^3 A^3$$and everything works out as expected! Thanks