Why is carbon dioxide non-polar every explanation keeps using the symmetry argument but I want to know what is fundamentally cancelling out because as far as I can tell there should be a positive middle without two negatives on the outside?
[Physics] Why is carbon dioxide a non-polar molecule
dipole-momentmoleculesphysical-chemistrysymmetry
Related Solutions
$\vec\mu$ is just the electric dipole moment. However, a molecule can be polar with $\vec\mu=0$, as polarity has to do with charge separation, so a particle with any form of multipole moment is polar.
Molecules like methane, carbon dioxide, and perchlorate have $\vec\mu=0$, but have some level of charge separation, making them polar (these have quadrupole moments, not sure about higher order moments).
Actually, all molecules are polar by this definition, just that many aren't polar enough for this to matter. Generally, when we call a molecule "polar", we are talking about only $\vec\mu$.
Assume that the Lagrangian density
$$\tag{1} {\cal L} ~=~ {\cal L}(\phi(x), \partial \phi(x), x) $$
does not depend on higher-order derivatives $\partial^2\phi$, $\partial^3\phi$, $\partial^4\phi$, etc. Let
$$\tag{2} \pi^{\mu}_{\alpha} ~:=~ \frac{\partial {\cal L}}{ \partial (\partial_{\mu}\phi^{\alpha})} $$
denote the de Donder momenta, and let
$$\tag{3} E_{\alpha}~:=~ \frac{\partial {\cal L}}{ \partial \phi^{\alpha}} - d_{\mu} \pi^{\mu}_{\alpha} $$
denote the Euler-Lagrange equations. Let us for simplicity assume that the infinitesimal local quasi-symmetry$^1$ transformation
$$\tag{4} \delta_{\varepsilon} \phi^{\alpha}~=~ Y^{\alpha}(\varepsilon) ~=~Y^{\alpha}\varepsilon + Y^{\alpha,\mu} d_{\mu}\varepsilon $$
is vertical$^2$ and that it does not depend on higher-order derivatives of the infinitesimal $x$-dependent parameter $\varepsilon$. [It is implicitly understood that the structure coefficients $Y^{\alpha}$ and $Y^{\alpha\mu}$ are independent of the parameter $\varepsilon$. If the theory has more that one symmetry parameter $\varepsilon^a$, $a=1, \ldots m$, we are just investigating one local symmetry (and its conservations law) at the time.] The bare Noether current $j^{\mu}(\varepsilon)$ is the momenta times the symmetry generators
$$\tag{5} j^{\mu}\varepsilon + j^{\mu,\nu}d_{\nu}\varepsilon ~=~j^{\mu}(\varepsilon) ~:=~ \pi^{\mu}_{\alpha}Y^{\alpha}(\varepsilon) ,$$
$$\tag{6} j^{\mu}~:=~ \pi^{\mu}_{\alpha}Y^{\alpha}, \qquad j^{\mu,\nu}~:=~ \pi^{\mu}_{\alpha}Y^{\alpha,\nu}. $$
(Again, it is implicitly understood that the structure coefficients $j^{\mu}$ and $j^{\mu\nu}$ are independent of the parameter $\varepsilon$, and so forth.) That the infinitesimal transformation (4) is a local quasi-symmetry$^1$ implies that variation of the Lagrangian density ${\cal L}$ wrt. (4) is a total space-time divergence
$$ d_{\mu} f^{\mu}(\varepsilon) ~=~ \delta_{\varepsilon} {\cal L} ~\stackrel{\begin{matrix}\text{chain}\\ \text{rule}\end{matrix}}{=}~ \frac{\partial {\cal L}}{ \partial \phi^{\alpha}} Y^{\alpha}(\varepsilon) + \pi^{\mu}_{\alpha}d_{\mu}Y^{\alpha}(\varepsilon) $$ $$\tag{7} ~\stackrel{\begin{matrix}\text{Leibniz'}\\ \text{rule}\end{matrix}}{=}~ E_{\alpha}Y^{\alpha}(\varepsilon) + d_{\mu} j^{\mu}(\varepsilon). $$
Here$^3$
$$ \tag{8} f^{\mu}(\varepsilon) ~=~ f^{\mu}\varepsilon + f^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} f^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon $$
are some functions with
$$\tag{9}f^{\mu,\nu\lambda}~=~f^{\mu,\lambda\nu}. $$
The full $\varepsilon$-dependent Noether current $J^{\mu}(\varepsilon)$ is defined as$^3$
$$\tag{10} J^{\mu}\varepsilon + J^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} J^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon ~=~J^{\mu}(\varepsilon) ~:=~ j^{\mu}(\varepsilon) - f^{\mu}(\varepsilon), $$
where
$$\tag{11}J^{\mu,\nu\lambda}~=~J^{\mu,\lambda\nu}. $$
Eqs. (7) and (10) imply the $\varepsilon$-dependent off-shell Noether identity
$$ \tag{12} d_{\mu} J^{\mu}(\varepsilon) ~=~ -E_{\alpha}Y^{\alpha}(\varepsilon) . $$
The $\varepsilon$-dependent off-shell Noether identity (12) is the key identity. Decomposing it in its $\varepsilon$-independent components leads to the following set (13)-(16) of identities,
$$ \tag{13} d_{\mu}J^{\mu} ~=~-E_{\alpha} Y^{\alpha} , $$
$$ \tag{14} J^{\mu} + d_{\nu} J^{\nu,\mu}~=~-E_{\alpha} Y^{\alpha,\mu} ,$$
$$ \tag{15} J^{\nu,\lambda}+J^{\lambda,\nu}+d_{\mu}J^{\mu,\nu\lambda} ~=~0 , $$
$$ \tag{16} \sum_{{\rm cycl}.~\mu,\nu,\lambda}J^{\mu,\nu\lambda} ~=~0, $$
in accordance with Noether's second theorem. Eq. (13) is just the usual off-shell Noether identity, which can be derived from the global symmetry alone via Noether's first theorem (where $\varepsilon$ is $x$-independent). As is well-known, the eq. (13) implies an on-shell conservation law
$$ \tag{17} d_{\mu}J^{\mu}~\approx~ 0, $$
or more explicitly written as
$$ \tag{18} \frac{d Q}{dt}~\approx~ 0,\qquad Q~:=~\int_{V} \! d^3V ~J^0. $$
(Here the $\approx$ sign denotes equality modulo Euler-Lagrange equations $E_{\alpha}\approx 0$. We have assume that the currents $J^i$, $i\in\{1,2,3\}$, vanish at the boundary $\partial V$.)
The remaining eqs. (14)-(16) may be repackaged as follows. Define the second Noether current ${\cal J}^{\mu}(\varepsilon)$ as$^4$
$$ \tag{19} {\cal J}^{\mu}\varepsilon + {\cal J}^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} {\cal J}^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon ~=~ {\cal J}^{\mu}(\varepsilon)~:= ~ J^{\mu}(\varepsilon)+ E_{\alpha} Y^{\alpha,\mu}\varepsilon. $$
It satisfies an $\varepsilon$-dependent off-shell conservation law
$$ d_{\mu} {\cal J}^{\mu}(\varepsilon) ~\stackrel{(12)+(19)}{=}~ -E_{\alpha}Y^{\alpha}(\varepsilon)+d_{\mu}(E_{\alpha} Y^{\alpha,\mu}\varepsilon)$$ $$ \tag{20}~\stackrel{(13)+(14)}{=}~ - \varepsilon d_{\mu}d_{\nu} J^{\nu,\mu}~\stackrel{(15)}{=}~\frac{\varepsilon}{2}d_{\mu}d_{\nu}d_{\lambda} J^{\lambda,\mu\nu}~\stackrel{(16)}{=}~0 . $$
One may introduce a so-called superpotential ${\cal K}^{\mu\nu}(\varepsilon)$ as$^3$
$$ {\cal K}^{\mu\nu}\varepsilon+{\cal K}^{\mu\nu,\lambda}d_{\lambda}\varepsilon~=~{\cal K}^{\mu\nu}(\varepsilon)~=~-{\cal K}^{\nu\mu}(\varepsilon) $$ $$~:=~ \left(\frac{1}{2} J^{\mu,\nu}-\frac{1}{6}d_{\lambda}J^{\mu,\nu\lambda}\right)\varepsilon+ \frac{1}{3} J^{\mu,\nu\lambda}d_{\lambda}\varepsilon-(\mu\leftrightarrow \nu)$$ $$ \tag{21}~\stackrel{(14)+(16)}{=}~ \left( J^{\mu,\nu}+\frac{1}{3}d_{\lambda}(J^{\lambda,\mu\nu}-J^{\mu,\nu\lambda})\right)\varepsilon+ \frac{1}{3}\left( J^{\mu,\nu\lambda}-J^{\nu,\mu\lambda}\right)d_{\lambda}\varepsilon$$
A straightforward calculation
$$ d_{\nu}{\cal K}^{\mu\nu}(\varepsilon) ~\stackrel{(15)+(21)}{=}~J^{\mu,\nu}d_{\nu}\varepsilon -\varepsilon d_{\nu}\left(J^{\nu,\mu}+d_{\lambda}J^{\lambda,\mu\nu}\right)$$ $$ \tag{22}+\frac{\varepsilon}{3}d_{\nu}d_{\lambda}\left(J^{\lambda,\mu\nu}-J^{\mu,\nu\lambda}\right) +\frac{1}{3}\left( J^{\mu,\nu\lambda}-J^{\nu,\mu\lambda}\right)d_{\nu}d_{\lambda}\varepsilon ~\stackrel{(14)+(16)+(19)}{=}~{\cal J}^{\mu}(\varepsilon)$$
shows that ${\cal K}^{\mu\nu}(\varepsilon)$ is the superpotential for the second Noether current ${\cal J}^{\mu}(\varepsilon)$. The existence of the superpotential ${\cal K}^{\mu\nu}(\varepsilon)=-{\cal K}^{\nu\mu}(\varepsilon)$ makes the off-shell conservation law (20) manifest
$$ \tag{23}d_{\mu}{\cal J}^{\mu}(\varepsilon)~\stackrel{(22)}{=}~d_{\mu}d_{\nu}{\cal K}^{\mu\nu}(\varepsilon)~=~0. $$
Moreover, as a consequence the superpotential (22), the corresponding second Noether charge ${\cal Q}(\varepsilon)$ vanishes off-shell
$$ \tag{24}{\cal Q}(\varepsilon)~:=~\int_{V} \! d^3V ~{\cal J}^0(\varepsilon) ~=~\int_{V} \! d^3V ~d_i{\cal K}^{0i}(\varepsilon) ~=~\int_{\partial V} \! d^2\!A_i ~{\cal K}^{0i}(\varepsilon)~=~0, $$
if we assume that the currents ${\cal J}^{\mu}(\varepsilon)$, $\mu\in\{0,1,2,3\}$, vanish at the boundary $\partial V$.
We conclude that the remaining eqs. (14)-(16) are trivially satisfied, and that the local quasi-symmetry doesn't imply additional non-trivial conservation laws besides the ones (13,17,18) already derived from the corresponding global quasi-symmetry. Note in particular, that the local quasi-symmetry does not force the conserved charge (18) to vanish.
This is e.g. the situation for gauge symmetry in electrodynamics, where the off-shell conservation law (20) of the second Noether current ${\cal J}^{\mu}=- d_{\nu}F^{\nu\mu}$ is a triviality, cf. also this and this Phys.SE posts. Electric charge conservation follows from global gauge symmetry alone, cf. this Phys.SE post. Note in particular, that there could be a nonzero surplus of total electric charge (18).
--
$^1$ An off-shell transformation is a quasi-symmetry if the Lagrangian density ${\cal L}$ is preserved $\delta_{\varepsilon} {\cal L}= d_{\mu} f^{\mu}(\varepsilon)$ modulo a total space-time divergence, cf. this Phys.SE answer. If the total space-time divergence $d_{\mu} f^{\mu}(\varepsilon)$ is zero, we speak of a symmetry.
$^2$ Here we restrict for simplicity to only vertical transformations $\delta_{\varepsilon} \phi^{\alpha}$, i.e., any horizontal transformation $\delta_{\varepsilon} x^{\mu}=0$ are assumed to vanish.
$^3$ For field theory in more than one space-time dimensions $d>1$, the higher structure functions $f^{\mu,\nu\lambda}=-J^{\mu,\nu\lambda}$ may be non-zero. However, they vanish in one space-time dimension $d=1$, i.e. in point mechanics. If they vanish, then the superpotential (21) simplifies to ${\cal K}^{\mu\nu}(\varepsilon)=J^{\mu,\nu}\varepsilon$.
$^4$ The second Noether current is defined e.g. in M. Blagojevic and M. Vasilic, Class. Quant. Grav. 22 (2005) 3891, arXiv:hep-th/0410111, subsection IV.A and references therein. See also Philip Gibbs' answer for the case where the quasi-symmetry is a symmetry.
Best Answer
Yes it is about the partial charges on the atoms but the dipole is a vector not just the charge distribution which is positive in the center and negative on the O's but the vectors cancel