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Background

As per convention, Carroll defines the Levi-Civita symbol $\tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n}$ as the sign of the permutation of $01\dots(n-1)$ on page 82. He states the Levi-Civita symbol has these components in any right-handed coordinate system (not necessarily orthogonal): they're defined not to change under coordinate transformations.

However, Carroll then goes on to derive how the components transform under coordinate transformations (seemingly contradicting what he just said before).

Given any $n \times n$ matrix $M^{\mu}{}_{\mu'}$, the determinant obeys $$ \tilde{\epsilon}_{\mu_1' \mu_2' \dots \mu_n'} |M| = \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n} M^{\mu_1}{}_{\mu_1'}M^{\mu_2}{}_{\mu_2'} \dots M^{\mu_n}{}_{\mu_n'} \tag{2.66}$$

By setting $M^{\mu}{}_{\mu'} = \frac{\partial x^{\mu}}{\partial x^{\mu'}}$, we obtain the transformation law:

$$\tilde{\epsilon}_{\mu_1' \mu_2' \dots \mu_n'} = \left| \frac{\partial x^{\mu'}}{\partial x^{\mu}} \right| \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n} \frac{\partial x^{\mu_1}}{\partial x^{\mu_1'}} \frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}} \dots \frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}} \tag{2.67}$$

revealing the symbol is a tensor density of weight 1 - i.e. the components do change under a general coordinate transformation?

Again, on page 89, to show that the volume element $d^n x$ is a tensor density (of weight 1), Carroll states "under a coordinate transformation [the Levi-Civita symbol] stays the same, while the one-forms change" - although I don't think this statement actually affects his derivation. I can outline this in more detail if necessary.

Question 1

Just by definition of the determinant, I think we should have:

$$ \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n} |M| = \tilde{\epsilon}_{\nu_1 \nu_2 \dots \nu_n} M^{\nu_1}{}_{\mu_1}M^{\nu_2}{}_{\mu_2} \dots M^{\nu_n}{}_{\mu_n} \tag{A}$$

where both Levi-Civita symbols belong to the same coordinate system.

Then, if the symbol has the same components in any coordinate system, $\tilde{\epsilon}_{\mu_1' \mu_2' \dots \mu_n'} = \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n}$, allowing us to recover equation 2.66 by relabelling of dummy indices. Is this the correct way to interpret equation 2.66?

Question 2

Many resources online suggest the symbol is only invariant under orthogonal transformations, but Carroll seems to suggest the components are the same in any right-handed coordinate system (which I don't think necessarily has to be orthogonal?). But if this is the case, I'm quite confused what exactly equation 2.67 is showing.

Indeed if $|M| = 1$, and the symbol transformed like a tensor (not density), we see that $\tilde{\epsilon}_{\mu_1' \mu_2' \dots \mu_n'} = \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n}$. But, as stated in my first question, I think we're using $\tilde{\epsilon}_{\mu_1' \mu_2' \dots \mu_n'} = \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n}$ to get to equation 2.67 (how the symbol actually transforms).

Wikipedia states:

Under the ordinary transformation rules for tensors the Levi-Civita symbol is unchanged under pure rotations, consistent with that it is (by definition) the same in all coordinate systems related by orthogonal transformations.

Under a general coordinate change, the components of the permutation tensor are multiplied by the Jacobian of the transformation matrix. This implies that in coordinate frames different from the one in which the tensor was defined, its components can differ from those of the Levi-Civita symbol by an overall factor. If the frame is orthonormal, the factor will be ±1 depending on whether the orientation of the frame is the same or not.

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  • $\begingroup$ I can't write an answer right now but look at 2.67 more closely - it is perfectly consistent with the LC symbol being invariant. You have two inverse determinants, which cancel. $\endgroup$
    – Javier
    Commented Aug 31, 2020 at 15:39

1 Answer 1

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There's no contradiction. Under any coordinate transformation that preserves the right-handedness of the coordinate system, the components of the Levi-Civita symbol stay the same. Another way to show this is to explicitly compute how the components change (as Carroll did in (2.67)). If you plug in actual values for the terms in that expression, you'll find that everything cancels out, so the components indeed stay the same. Another way of thinking about this is that since you already know the Levi-Civita components stay the same, (2.67) is merely a restatement of the definition of the determinant.)

In general, it's a bad idea to trawl for online sources because conventions will differ. Here, there are two different concepts: the Levi-Civita symbol, and the Levi-Civita tensor. The Levi-Civita symbol is the one whose components are all $0$ or $\pm 1$, and contraction with it yields the orientation of an ordered set of basis vectors. The Levi-Civita tensor is a true tensor, which yields the volume spanned by an ordered set of basis vectors. The Levi-Civita symbol doesn't scale with volume, which is why its transformation has an extra power of the determinant.

The Levi-Civita symbol and tensor coincide for orthogonal bases, since the determinant is $1$. So if you define it there, then there are many different ways to extend the definition to general bases:

  • Extend the definition by multilinearity, which yields the Levi-Civita tensor. (Depending on the source, this might also be called the Levi-Civita symbol.)
  • Define the components to always be $0$ or $\pm 1$, which yields the Levi-Civita symbol. (Depending on the source, this might also be called the Levi-Civita tensor, even though it isn't one.)
  • Declare that the definition only works for orthogonal bases and refuse to define it more generally. (This is common for people who work in $\mathbb{R}^3$ who don't often need nonorthogonal bases.)

Because every source will have different conventions on nomenclature, Wikipedia (which itself is compiled inconsistently from many sources) is not reliable.

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  • $\begingroup$ Thank you for your answer! This really helps to clear up my confusion. "[Since we know the components stay the same], (2.67) is merely a restatement of the definition of the determinant" - I think this is what I was suggesting may be the case under the "question 1" heading. This makes sense. $\endgroup$
    – Shrey
    Commented Aug 31, 2020 at 18:48

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