# Action of metric tensor on Levi-Civita symbol

• I
• Baela
In summary, in SR a metric tensor raises or lowers indices of a tensor, while in general relativity a covariant Levi-Civita tensor exists which is negative when the metric is negative.
Baela
We know that a metric tensor raises or lowers the indices of a tensor, for e.g. a Levi-Civita tensor. If we are in ##4D## spacetime, then
\begin{align}
g_{mn}\epsilon^{npqr}=\epsilon_{m}{}^{pqr}
\end{align}
where ##g_{mn}## is the metric and ##\epsilon^{npqr}## is the Levi-Civita tensor.

The Levi-Civita symbol, which we can denote by ##\varepsilon^{npqr}##, is not a tensor. It obeys the relation
\begin{align}
\varepsilon^{npqr}=\varepsilon_{npqr}.
\end{align}

What happens if the metric tensor is multiplied with the Levi-Civita symbol ##\varepsilon^{npqr}##?
\begin{align}
g_{mn}\varepsilon^{npqr}=\,?
\end{align}

Well, what's the relation between the epsilon symbol and -tensor?

vanhees71
For a generic spacetime in GR, (2) is not correct. I think already in SR you have a minus there to rectify (2).

Baela said:
is not a tensor.
should be replaced by is a tensor. The four uppers are just replaced by four downers by being acted on with four gs.
Baela said:
What happens if the metric tensor is multiplied with the Levi-Civita symbol ?
Your Eq. (2). g just raises or lowers indices.
The LC tensor just depends on the order of the super or subscripts.
It is called an idempotent tensor because it has the same value in any system.
I am just talking about special relativity.

The components of the Levi-Civita tensor are
$$\epsilon^{\alpha \beta \gamma \delta}=\frac{1}{\sqrt{-g}} \Delta^{\alpha \beta \gamma \delta},$$
where ##g=\mathrm{det}(\hat{g})## and ##\Delta^{0123}=1## and totally anti-symmetric under exchange of its arguments, i.e., the usual Levi-Civita symbol.

The covariant components of the Levi-Civta-tensor thus are
$$\epsilon_{\alpha \beta \gamma \delta}=-\sqrt{-g} \Delta^{\alpha \beta \gamma \delta}.$$
This somewhat confusing extra sign comes from the fact that ##g<0##.

Also note that the convention of this sign is a matter of convention, i.e., it can be the opposite (e.g., in MTW). So you have to carefully check, which convention is used in any reference you read ;-).

dextercioby
I am just talking about special relativity.

vanhees71
Meir Achuz said:
I am just talking about special relativity.
Actually, you're talking about special relativity only for the specific case of a cartesian coordinate system. More general curvilinear systems (e.g., polar coordinates) require the full tensor machinery of the metric, its determinant and the Christoffel symbols, even in special relativity.

vanhees71
That's why they are not usually used in SR.

In pseud-Cartesian (Lorentzian) coordinates then you have ##\epsilon^{\alpha \beta \gamma \delta}=\Delta^{\alpha \beta \gamma \delta}## and ##\epsilon_{\alpha \beta \gamma \delta}=-\Delta^{\alpha \beta \gamma \delta}##, because then ##-g=1##.

Meir Achuz said:
I am just talking about special relativity.
I'm not sure the thread in general is restricted to SR.

Meir Achuz said:
That's why they are not usually used in SR.
I'm not sure that's necessarily the case. Rindler coordinates are fairly commonly used, and they are non-inertial.

vanhees71 and renormalize
In heavy-ion hydrodynamics often Milne coordinates are used, because they are convinient to describe "Bjorken flow" ;-).

usually

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