0
$\begingroup$

As far as I'm aware from my course notes and what I've found online, the four vector gradient expands as $$ \begin{align*} \partial_\mu x_\mu &= \left( \frac{1}{c} \frac{\partial}{\partial t}, -\frac{\partial}{\partial x^1}, -\frac{\partial}{\partial x^2}, -\frac{\partial}{\partial x^3} \right)\left( x_0, x_1, x_2, x_3 \right)\\ &= \left(\partial_0 x_0, -\vec{\nabla}\vec{x} \right) \end{align*} $$ where $c=1$,

and the four vector divergence expands as

$$\begin{align*}\partial_\mu x^\mu &= \left( \frac{1}{c} \frac{\partial}{\partial t}, -\frac{\partial}{\partial x^1}, -\frac{\partial}{\partial x^2}, -\frac{\partial}{\partial x^3} \right) \cdot \left( x^0, x^1, x^2, x^3 \right)\\ &= \left( \frac{1}{c} \frac{\partial}{\partial t}, - \vec{\nabla} \right) \cdot \left( x^0, \vec{x} \right)\\ &= \partial_0 x^0 + \vec{\nabla} \cdot \vec{x} \end{align*}$$

In the latter case, where does the negative sign caused by the metric signature go?

$\endgroup$
  • 3
    $\begingroup$ Comment to the post (v3): The raised and lowered positions of indices are applied inconsistently in various places. For starters, we should have $\partial_{\mu}= \frac{\partial}{\partial x^{\mu}}$ and $\partial^{\mu}= \frac{\partial}{\partial x_{\mu}}$. $\endgroup$ – Qmechanic Jul 15 '18 at 11:26
  • $\begingroup$ I just noticed the raised index of $x_0$ in the gradient expansion, but can't see any others. I've only recently learned this notation. Can you tell me which others are incorrect? $\endgroup$ – perilousGourd Jul 15 '18 at 11:33
  • $\begingroup$ @user3949312 look at the indices for the expression $\partial_\mu x^\mu$. $\endgroup$ – Rumplestillskin Jul 15 '18 at 11:35
  • $\begingroup$ These definition seem quite fishy. Have you computed these yourself or taken these from your notes? $\endgroup$ – Rumplestillskin Jul 15 '18 at 11:44
  • 2
    $\begingroup$ Also just FYI it’s not called a Laplacian operator either :) $\endgroup$ – Rumplestillskin Jul 15 '18 at 12:02
1
$\begingroup$

The divergence of the four-vector defining position $x^\mu = (x^0, x^1,x^2,x^3)$ will give you the dimension of the spacetime and is given by the scalar product or $\partial_\mu$ with $x^\mu$. It is defined as

$$ \partial_\mu x^\mu = \frac{\partial}{\partial x^0} x^0 + \left( \frac{\partial}{\partial x^i} x^i \right) = 4, $$

where we sum over repeated indices.

Or, for an arbitrary four-vector $a^\mu = (a^t, \mathbf{a})$ where $\mathbf{a}$ is the Cartesian spatial components is given by

$$ \partial_\mu a^\mu = \frac{\partial}{\partial t} a^t + \mathbf{\nabla} \cdot\mathbf{a}. $$

Your confusion regarding the minus sign is that you will get one minus sign from the four-vector scalar product and the other comes from the spatial components of $\partial_\mu$. Hope that clears it up.

$\endgroup$
  • 1
    $\begingroup$ Sorry, but that's not correct or at least bad notation, you are mixing up upper and lower indices. $\endgroup$ – Photon Jul 15 '18 at 12:07
  • $\begingroup$ Thank you! I thought the minus signs in the scalar product and the spatial components were from the same process just being applied at different times. Can I ask why the metric signature affects the scalar product operation and the four-vector derivative components but not the components of the $x$ four-vector or the gradient operation? (Any entry-level resources would be great.) $\endgroup$ – perilousGourd Jul 15 '18 at 12:08
  • $\begingroup$ It got better but, I fixed some more index positions and signs. :) $\endgroup$ – Photon Jul 15 '18 at 12:31
  • $\begingroup$ @Photon looks better after both contributions. Cheers. $\endgroup$ – Rumplestillskin Jul 15 '18 at 12:33
  • $\begingroup$ Actually, the original notation is completely fine in Cartesian coordinates $\endgroup$ – Rumplestillskin Jul 15 '18 at 12:35
2
$\begingroup$

Your first expression is not well-defined. According to the Einstein summing convention, you sum over double indices, where one index is upper and the other one is lower. Such indices are called silent. Such an expression is then explicitly written as:

$$x^\mu y_\mu = x^0y_0 + x^1y_1 + x^2y_2 + x^3y_3 = x^0y^0 - x^1y^1 - x^2y^2 - x^3y^3$$

Note the sign change when passing from a lower to an upper index (works only in Minkowski metric).

Expressions like $x^\mu y^\mu$ or $x_\mu y_\mu$ are not well-defined, don't use them!

Indices which only appear once (on each side of an equation if an equation is considered) are called free indices. Example:

$$x^\mu y^\nu = ?$$

Since they are free, you need to specify some index values to evaluate the expression explicitly, say:

$$\left(x^\mu y^\nu\right)_{\mu=2\nu=3} = x^2y^3$$

You need to be careful when derivatives are involved. Derivatives are "naturally" defined with lower indices:

$$x^\mu = (x^0, \vec x),\qquad x_\mu = (x^0,-\vec x) $$

but

$$\partial_\mu = (\partial_0, \vec \nabla),\qquad \partial^\mu = (\partial^0,-\vec \nabla) $$

where $\vec x = (x^1, x^2, x^3)$, $\vec \nabla = (\partial_1, \partial_2,\partial_3)$ and the metric convention is that $x_0=x^0$ and $x_i=-x^i$.

Therefore,

$$x^\mu y_\mu = x^0 y^0 - \vec x\vec y$$

as above, and also

$$\square = \partial^\mu \partial_\mu = (\partial_0)^2 - {\vec \nabla}^2 $$

but

$$\partial_\mu x^\mu = \partial_0 x^0 + \vec \nabla \vec x$$

$\endgroup$
  • $\begingroup$ I'm confused as to why the last expression has $\partial_0 x^0$ as opposed to $\partial^0 x^0$, as the latter was how $\partial_\mu$ was defined earlier. Also, what is the difference between raised and lowered indices for the first element of a four-vector? Does the number that is that component change in any way? $\endgroup$ – perilousGourd Jul 15 '18 at 12:21
  • $\begingroup$ I found this information very helpful in general, though, so thank you. $\endgroup$ – perilousGourd Jul 15 '18 at 12:22
  • $\begingroup$ Sorry, a typo, I corrected it! But in the metric convention I chose, it holds $\partial^0=\partial_0$ and $\partial^i = -\partial_i$, so it was technically not wrong but very confusing. $\endgroup$ – Photon Jul 15 '18 at 12:28
  • $\begingroup$ Found and fixed another typo (in the definition of $\square$). $\endgroup$ – Photon Jul 15 '18 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.