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In Sean Carroll's GR book, pg 83, between eqs. (2.69-70), the Levi-Civita symbol with raised indices is defined as $$\tilde{\epsilon}^{\mu_1\mu_2...\mu_n}=\text{sgn}(g)\tilde{\epsilon}_{\mu_1 \mu_2...\mu_n}$$ where $\text{sgn}(g) = \frac{g}{|g|}$ is the sign function of the metric determinant $g$ and $\tilde{\epsilon}_{\mu_1 \mu_2...\mu_n}$ is the Levi-Civita symbol we usually use (which is a tensor density of weight $1$).

He then said that the Levi-Civita symbol $\tilde{\epsilon}^{\mu_1\mu_2...\mu_n}$ is a tensor density of weight $-1$.

I tried to prove his statement:

$$\begin{equation} \tilde{\epsilon}^{\mu_1'\mu_2'...\mu_n'}=\frac{g'}{|g'|}\tilde{\epsilon}_{\mu_1' \mu_2'...\mu_n'} \\ = \frac{|\frac{\partial x^{\mu'}}{\partial x^\mu}| ^{-2} g} {|\frac{\partial x^{\mu'}}{\partial x^\mu}| ^{-2} |g| } | \frac{\partial x^{\mu'}}{\partial x^\mu}| \tilde{\epsilon}_{\mu_1 \mu_2...\mu_n} \frac{\partial x^{\mu_1}}{\partial x^{{\mu}'_1}}\frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}\\ = |\frac{\partial x^{\mu'}}{\partial x^\mu}|\frac{g}{|g|}\tilde{\epsilon}_{\mu_1 \mu_2...\mu_n} \frac{\partial x^{\mu_1}}{\partial x^{{\mu}'_1}}\frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}\\ = |\frac{\partial x^{\mu'}}{\partial x^\mu}| \tilde{\epsilon}^{\mu_1\mu_2...\mu_n}\frac{\partial x^{\mu_1}}{\partial x^{{\mu}'_1}}\frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}} \end{equation}$$ This shows that $\tilde{\epsilon}^{\mu_1\mu_2...\mu_n}$ is a tensory density of weight $1$. What am I doing wrong?

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  • $\begingroup$ Is there a reason you're writing your coordinates with indices down? $\endgroup$
    – J. Murray
    Mar 16, 2021 at 14:53
  • $\begingroup$ @J.Murray I wrote them wrongly... Should be indices up. Corrected them. $\endgroup$
    – TaeNyFan
    Mar 16, 2021 at 15:30

2 Answers 2

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To me, a more conceptually direct derivation goes as follows. The Levi-Civita symbol is a symbol, not a geometrical object. We can denote its value in a coordinate system $x$ as $\tilde \epsilon_{(x)\mu\nu\rho\sigma}$. On its face, this is silly - the value of $\tilde \epsilon$ does not depend on the coordinate system. However, the point here is that we can compare the (trivial) transformation behavior of $\tilde\epsilon$ with the transformation behavior we would expect from a $(0,4)$-tensor.

As stated, the actual transformation behavior of these components is trivial, i.e. $$\tilde \epsilon_{(x)\mu\nu\rho\sigma}\mapsto \tilde \epsilon_{(y)\mu\nu\rho\sigma} = \tilde \epsilon_{(x)\mu\nu\rho\sigma}$$ If it were a tensor, then it would transform as $$\tilde\epsilon_{(x)\mu\nu\rho\sigma}\mapsto \tilde \epsilon_{(y)\mu\nu\rho\sigma} = \tilde \epsilon_{(x)\alpha\beta\gamma\delta}J^\alpha_{\ \ \mu}J^\beta_{\ \ \nu}J^\gamma_{\ \ \rho}J^\delta_{\ \ \sigma}= J \tilde\epsilon_{(x)\mu\nu\rho\sigma}$$ where $J^\alpha_{\ \ \beta} \equiv \frac{\partial x^\alpha}{\partial y^\beta}$ is the Jacobian matrix and $J$ is the Jacobian determinant. Therefore, we say that the true transformation behavior is given by

$$\tilde\epsilon_{(x)\mu\nu\rho\sigma}\mapsto \tilde \epsilon_{(y)\mu\nu\rho\sigma} = J^{-1}\tilde \epsilon_{(x)\alpha\beta\gamma\delta}J^\alpha_{\ \ \mu}J^\beta_{\ \ \nu}J^\gamma_{\ \ \rho}J^\delta_{\ \ \sigma}$$

The fact that the power of $J$ out front is $-1$ means that $\tilde \epsilon$ is a tensor density of weight $-1$.


We now turn our attention to the symbol with upstairs indices. In this case, we note that the actual transformation behavior is now $$\tilde \epsilon_{(x)}^{\mu\nu\sigma\rho}\mapsto \tilde \epsilon_{(y)}^{\mu\nu\sigma\rho} = \mathrm{sgn}(J)\epsilon_{(x)}^{\mu\nu\sigma\rho}$$ whereas its expected transformation behavior (if it were a tensor) would be $$\tilde \epsilon_{(x)}^{\mu\nu\sigma\rho}\mapsto \tilde \epsilon_{(y)}^{\mu\nu\sigma\rho} =\epsilon_{(x)}^{\alpha\beta\gamma\delta}(J^{-1})^\mu_{\ \ \alpha}(J^{-1})^\nu_{\ \ \beta}(J^{-1})^\rho_{\ \ \gamma}(J^{-1})^\sigma_{\ \ \delta}=J^{-1} \tilde\epsilon_{(x)}^{\mu\nu\rho\sigma}$$

Therefore, we can write that the true transformation behavior is

$$\tilde \epsilon_{(x)}^{\mu\nu\rho\sigma} \mapsto \tilde \epsilon_{(y)}^{\mu\nu\rho\sigma} = \mathrm{sgn}(J) J\epsilon_{(x)}^{\alpha\beta\gamma\delta}(J^{-1})^\mu_{\ \ \alpha}(J^{-1})^\nu_{\ \ \beta}(J^{-1})^\rho_{\ \ \gamma}(J^{-1})^\sigma_{\ \ \delta}$$

The fact that the power of $J$ is now $+1$ means that the weight is now $+1$; the fact that we also have a $\mathrm{sgn}(J)$ means that $\tilde \epsilon^{\mu\nu\rho\sigma}$ is in fact a pseudotensor density of weight $+1$. Carroll does not discuss this because he assumes that the coordinate transformation in question is orientation-preserving, which implies that $\mathrm{sgn}(J)=+1$ and the distinction between tensors and pseudotensors (and their respective densities) can be ignored.


In summary, we contrast the actual transformation behavior of the $\tilde \epsilon$ symbols with their expected transformation behavior if they were genuine tensors to determine in what way the actual transformation behavior differs. If it transforms like a tensor except for an additional factor of $J^w$, we call it a tensor density of weight $w$. If it additional picks up a factor of $\mathrm{sgn}(J)$, we call it a pseudotensor density.

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  • $\begingroup$ Excellent! Very Helpful! $\endgroup$
    – Daren
    Feb 5, 2023 at 4:32
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I believe the answer lies in plain sight. Carroll defines weight as the power of det(∂x'/∂x). Apply (2.66) in Carroll's book and replace |M| by det(∂x'/∂x). Multiply sgn(g') to LHS and sgn(g) to RHS. The equation still holds since they are equal. Then you get an equation similar to (2.67) and the first item of RHS is 1/det(∂x'/∂x) QED. Carroll uses sgn(g) to connect Levi-Civita symbol and tensor in (2.70). Shultz defines Levi-Civita symbols with upper/lower indices the same, see Geometrical Method in Mathematical Physics.

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