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The variational principle clearly gives

$$\frac{\partial \rho}{\partial t} + \overrightarrow{\nabla}\cdot \mathbf{J} = 0.$$

So the sign is positive. However in my lecture notes it is claimed that the Noether current, conserved under symmetries, is

$$J^{\mu} = \Sigma_j \frac{\partial L}{\partial(\partial _{\mu}\psi_j)}\delta \psi_j$$

And it is conserved in that its four-divergence is zero:

$$\partial _{\mu}J^{\mu}=0.$$

However I thought that covariant componenets had switched sign, so in fact this would read $$\frac{\partial \rho}{\partial t} - \overrightarrow{\nabla}\cdot \mathbf{J} = 0,$$ which is certainly not necessarily true.

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  • $\begingroup$ Both expressions are fine, but they have $\boldsymbol J_1=-\boldsymbol J_2$. The actual sign depends on conventions (e.g., the metric signature). $\endgroup$ – AccidentalFourierTransform Nov 12 '18 at 21:09
  • $\begingroup$ @AccidentalFourierTransform but surely here we are working within a single framework, and this is a clear inconsistency? $\endgroup$ – Meep Nov 12 '18 at 21:12
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The 4-gradient in Minkowski space $(w,x,y,z)$ is $$\partial_{\mu} = \begin{bmatrix}\frac{\partial}{\partial w}& \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\end{bmatrix}$$having all-positive signs, corresponding to the 4-position, $$r^\mu = \begin{bmatrix}w\\ x \\ y \\ z\end{bmatrix}.$$You are correct that in general when one transposes these one inserts some minus signs either on the $w$ component ($-+++$ convention) or on the $x,y,z$ components ($+---$ convention), but these minus signs do not exist "naturally" on the 4-gradient or the 4-position components.

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  • $\begingroup$ I see. It is because of the nature of the derivative. $\endgroup$ – Meep Nov 12 '18 at 21:19
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When you have your typical invariant $x^\mu x_\mu = x_\mu x^\mu = t^2-x^2-y^2-z^2$ you are only "subtracting" because you are upping or lowering the index of one 4-vectors, so you get things like, $x_\mu = (t,-\vec x) $ while $x^\mu = (t,\vec x)$, and there's your minus sign.

In your case you are not changing any index from upper to lower or vice versa, since by definition $\partial_\mu = (\partial_t,\nabla)$ and $J^\mu = (\rho, \vec J)$, one is a lower index and the other upper already, so there's no reason for any minus sign.

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  • $\begingroup$ But I thought that the metric function (which is invertible) provides and isomorphism between tangent and dual spaces, thereby meaning that neither is more 'correct' than the other. You speak of the $x$ as if the covariance or contravariance of its indices is not established, whereas it is for $\partial$ and $J$. I am pretty sure this is not the case and have come across $\partial ^{\mu}$ and $J_{\mu}$ just as well? $\endgroup$ – Meep Nov 12 '18 at 21:31
  • $\begingroup$ I'm just going by the usual conventions, you ultimately have to check your notes, but in my example and the conventions I'm using $\partial^\mu$ would be $(\partial_t, -\nabla)$ and $J_\mu = (\rho,-\vec J)$. Neither is more correct, but if you want to derivate by an upper index variable (like in this case, because that's how the $J^\mu$ was chosen), that turns into a lower index four vector $$\partial_\mu = \frac{\partial }{\partial x^\mu}$$ If you had chosen $J_\mu$ you have to derivate a lower index and thus the differential is upper, $$\partial^\mu = \frac{\partial}{\partial x_\mu}$$ $\endgroup$ – luci Nov 12 '18 at 21:56
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$\partial_\mu=(\frac{1}{c}\frac{\partial}{\partial t},\nabla)$ but $\partial^\mu=(\frac{1}{c}\frac{\partial}{\partial t},-\nabla)$

and $j^{\mu}=(c\rho,\mathbf{j})$

so the rest is obvious.

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  • $\begingroup$ I thought the covariant and contravariant had different signs (so you could get one from the other using the Monkowski metric), and yet here the contravariant J and the derivatives have opposite signs? $\endgroup$ – Meep Nov 12 '18 at 21:15

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