4
$\begingroup$

There are a few questions here regarding upper and lower indices of the levi-civita symbol but I haven't been able to find an answer to my exact question. I am working on a problem where I start out in Minkowski space using the metric (+---) because for some reason I thought it was a good idea. I could probably answer my question by switching to the (-+++) metric but I'm still curious about what the answer to my question is WITHOUT switching metrics i.e. I want to keep the (+---) metric.

Furthermore, in my problem, I'm committing a sin of special relativity by explicitly breaking my equations into temporal and spatial parts. My question is in this sort of situation how should I write down the cross product of two (three-)vectors, $\vec{A}$ and $\vec{B}$. To be a bit explicit about what I mean: $$ \textbf{A} = \begin{bmatrix}A^0\\A^1\\A^2\\A^3\end{bmatrix} \hspace{1 in} \vec{A} = \begin{bmatrix}A^1\\A^2\\A^3\end{bmatrix}$$

I want to calculate $\vec{A} \times \vec{B}=\vec{C}$. If I was being naive and ignoring Einstein convention and metric conventions I would just write: $$C_{i,naive} = \epsilon_{ijk} A_{j}B_{k}$$ $$\vec{C}_{naive} = \sum_i C_{i,naive} \hat{x}_i = \\(A_2 B_3 - A_3B_2)\hat{x}_1 + (-A_1 B_3 + A_3B_1)\hat{x}_2 + (A_1 B_2 - A_2B_1)\hat{x}_3$$ However, this doesn't obey Einstein notation where upper and lower indices should match and I've also lowered some indices without worrying about if I should be changing signs of things. If I wanted to make indices match maybe I would write: $$C^{i} = \epsilon^i{}_{jk} A^{j} B^{k}$$ This is where I get confused. How is $\epsilon^i{}_{jk}$ defined or how should I be defining it? On the one hand I want $\vec{C}_{Ein} = C^{i}\hat{x}_i = \vec{C}_{naive}$ in which case I would have that $$\epsilon_{ijk} = \epsilon^i{}_{ jk}$$ However, on the other hand I feel like $\epsilon^i{}_{jk} = g^{il}\epsilon_{ljk}$ but since $i$ takes the values 1 through 3 and we are in the (+---) metric this would imply that $$\epsilon^{i}{}_{jk} = -\epsilon_{ijk}$$

What is the correct way to do things here? Should I be defining the cross product differently? I know there is something about the levi-civita symbol vs. levi-civita tensor but that seems like too much detail and firepower for what I'm trying to do here. I also know there is something about a left-handed vs. a right-handed coordinate system.

$\endgroup$
  • $\begingroup$ It is in fact true that $\epsilon^{i}_{\phantom{i}jk} = -\epsilon_{ijk}$. Why is this puzzling you? $\endgroup$ – gented Nov 12 '16 at 23:33
  • $\begingroup$ Would it be true if I was in the (-+++) metric? I am confused because it seems to introduce a relative minus sign in the definition of the cross product as compared to what I would expect on Euclidean (as opposed to Minkowski) space. $\endgroup$ – jgerber Nov 12 '16 at 23:35
  • 1
    $\begingroup$ Yes, there is an additional minus sign when using covariant and contravariant indeces. The notation $(a\times b)_i= \epsilon_{ijk}a_j b_k$ is only valid in Euclidean metrics (as in general the contravariant components of a vectors differ from the corresponding covariant ones by a $g^{\mu\nu}$ multiplication coefficient). $\endgroup$ – gented Nov 12 '16 at 23:42
4
$\begingroup$

Ok. I'm going to put my response as an answer to my question since it involves some new information I've found and want to document here. Part of my confusion has stemmed from the fact that different authors use different notation regarding the transition from Levi-Civita symbols or tensor densities ($\tilde{\epsilon}$) and Levi-Civita tensors ($\epsilon$). Here are the two clearest references I have found on the subject and they illustrate the two possible conventional choices. Sean Carroll's lecture notes on geometry and spacetime ch. 2 and Christopher Pope's Electrodynamics lecture notes. I find Pope's notes to give a little bit more detail which made the difference for me.

The point is the Levi-Civita symbol with the lower indices*, $\tilde{\epsilon}_{ijk}$ is defined as an object which is anti-symmetric in its indices. i.e. $\tilde{\epsilon}_{123}=+1$ whereas $\tilde{\epsilon}_{132}=-1$. Furthermore, (as explained better in the Pope notes) the symbol takes the same value in all coordinate frames. That is if we transform from one set of coordinates $\{x^i\}$ to another set of coordinates $\{x^{'i}\}$ we have transform $\tilde{\epsilon}_{ijk}$ transforms to $\tilde{\epsilon}^{'}_{ijk}$ with the relation that $\tilde{\epsilon}^{'}_{ijk}=\tilde{\epsilon}_{ijk}$

It is then possible to prove, using some facts about determinants, that $$\tilde{\epsilon}_{ijk}=\tilde{\epsilon}^{'}_{ijk} = \left|\frac{\partial x'}{\partial{x}}\right|\frac{\partial x^a}{\partial x^{'i}}\frac{\partial x^b}{\partial x^{'j}}\frac{\partial x^c}{\partial x^{'k}}\tilde{\epsilon}_{abc}$$

That is to say $\tilde{\epsilon}_{ijk}$ transforms like a tensor density of weight +1. It turns out $\sqrt{|\text{det}(g_{ij})|}=\sqrt{|\text{det }g|}$ is a tensor density of weight -1 (as proven in the Pope notes) so by multiplying these two tensor densities you get a new tensor density of weight 0, i.e. a regular tensor (indicated by the absence of the tilde).

$$\epsilon_{ijk} = \sqrt{|\text{det }g|}\tilde{\epsilon}_{ijk}$$

Both authors agree on this much. And so far none of this would have been changed by choosing the (-+++) metric as opposed to the (+---) metric**. However, the next step is finding the upper index object. Since $\epsilon_{ijk}$ is a tensor we can raise its indices:

$$\epsilon^{ijk} = g^{ii'}g^{jj'}g^{kk'}\epsilon_{i'j'k'} = g^{ii'}g^{jj'}g^{kk'}\tilde{\epsilon}_{i'j'k'}\sqrt{|\text{det }g|} \\ =\text{det}(g^{-1}) \sqrt{|\text{det }g|} \tilde{\epsilon}_{ijk} = \frac{\text{sgn}(g)}{\sqrt{|g|}}\tilde{\epsilon}_{ijk}$$

At this point we have

$$\epsilon^{ijk} = \frac{\text{sgn}(g)}{\sqrt{|\text{det }g|}}\tilde{\epsilon}_{ijk}$$

This is where people make a convention choice. Carroll (and many others that I have seen), for example, makes the chioce that $\tilde{\epsilon}^{ijk} = \tilde{\epsilon}_{ijk}$ so that we get

$$\epsilon^{ijk} = \frac{\text{sgn}(g)}{\sqrt{|\text{det }g|}}\tilde{\epsilon}^{ijk}$$

Pope takes the convention that $\tilde{\epsilon}^{ijk} = \text{sgn}(g)\tilde{\epsilon}_{ijk}$ so that

$$\epsilon^{ijk} = \frac{1}{\sqrt{|\text{det }g|}}\tilde{\epsilon}^{ijk}$$

So we've already made at least two convention chioces. The first is how $\epsilon_{ijk}$ relates to $\tilde{\epsilon}_{ijk}$ and the second is how $\tilde{\epsilon}_{ijk}$ relates to $\tilde{\epsilon}^{ijk}$. I think we have yet another choice now of how to define the cross product. I think the responsible thing to do is to keep in mind all of the machinery up to this point and make a manifestly covariant definition of the cross product. This would look like: $$(A\times B)^i = \epsilon^{i}{}_{jk}A^jB^k=g^{ii'}\epsilon_{i'jk}A^j B^k = \pm \epsilon_{ijk}A^j B^k=\pm\tilde{\epsilon}_{ijk}A^j B^k$$

Where $\pm$ indicates the next convention choice of mostly positive or mostly negative metric signature. I think this has the disadvantage that depending on whether you take the mostly positive or mostly negative metric you get a relative minus sign in the definition of the cross product, but I think this is actually to be expected since it moves the spatial part from being a right handed to left handed coordinate system. It also has the disadvantage that you really have to remember quite a few places for negative signs. It has the advantage that is covariant so that if you do follow the rules you can do the manipulations more easily I guess.

For example @Solenodon Paradoxus gives a formula for the "correct expression" for the cross product but I'm not sure on what grounds that is the "correct expression". It follows proper Einstein convention but the symbol used is the Levi-Civita symol which is not a tensor so there's not actually a rule saying the expression SHOULD follow the Einstein convention which leaves confusion as to the motivation behind any particular definitions when there are so many choices.

*I believe this is a matter of convention. We could alternatively take the Levi-Civita symbol with upper indices to be the object which is anti-symmetric in its indices.

**but the story would maybe be different if we had chosen the upper indexed Levi-Civita symbol to be the usual anti-symmetric object instead.

edit1: for less awkward notation we could define $(A \times B)^i = \epsilon^{ijk}A_jB_k$ which would be equivalent to the definition given. For my problem I'm working on I'm thinking of contravariant components of vectors as the "physical" quantities so I'm prefering to write expression in terms of those.

$\endgroup$
1
$\begingroup$

Interesting question. The correct expression for the general case would be

$$ c^i = g_{jj'} g_{kk'} \; \frac{\varepsilon^{ijk}}{\sqrt{\left| \det g \right|}} \; a^{j'} b^{k'}, $$

or

$$ c^{i} = g^{i i'} \sqrt{\left| \det g \right|} \; \varepsilon_{i' j k} a^{j} a^{k} $$

if you wish to use the Levi-Civita symbol with lower indices.

where $g_{ij}$ is the pull-back of the spacetime metric on the 3-dimensional spatial slice (or you can start with the 3d space from the beginning), and $\det g$ is the determinant of the covariant metric tensor matrix. This expression can be proven to be a tensor, i.e. $c^i$ is a true tensor in the most general meaning of this word.

For special relativity in Minkowski coordinates, you have

$$ g_{ij} = \pm \delta_{ij}, $$

where the $\pm$ stands for the two possible conventions (-+++ and +---).

It is easy to see that in both cases you end up with

$$ c^i = \varepsilon^{ijk} a^j b^k. $$

$\endgroup$
  • $\begingroup$ 2 questions. 1) You are using $\epsilon^{ijk}$ which is NOT equal to $\epsilon_{ijk}$ based on the discussion above. Which one of these two is equal to +1 for even permutations of the indices? 2) Is that to say the definition of the cross product (using correct einstein notation) would be $c^i = -\epsilon^i{}_{jk}a^j b^k$ $\endgroup$ – jgerber Nov 13 '16 at 3:41
  • $\begingroup$ 1) no, I am using $\varepsilon^{ijk}$ which is not a tensor and is defined to be the antisymmetric object (Levi-Civita symbol). 2) No, the definition is given by the 1st formula in my answer. $\endgroup$ – Prof. Legolasov Nov 13 '16 at 6:33
  • $\begingroup$ Ok. This seems to help. But what if instead we had taken $\epsilon_{ijk}$ to be the Levi-Civita symbol rather than $\epsilon^{ijk}$. Then I guess your formula would be $c^i = g^{ii'}\epsilon_{ijk} a^j b^k$ in which case you end up with $c^i = \pm \epsilon_{ijk} a^j b^k$ depending on which metric you take. So how do we know whether we should take $\epsilon^{ijk}$ or $\epsilon_{ijk}$ to be the Levi-Civita symbol? $\endgroup$ – jgerber Nov 13 '16 at 17:02
  • $\begingroup$ @jgerber I've made an edit to include a formula for the Levi-Civita symbol with lower indices (note that it is not technically covariant since it is not a tensor). $\endgroup$ – Prof. Legolasov Nov 14 '16 at 2:52
  • $\begingroup$ I know the issue now. You have left out $\text{sgn}(\text{det }g)$ and I think this leads to you getting the wrong result in your final line. Your first two lines can be interpreted as being correct if you are using Popes convention from my answer. In that case the lower equation has no additional minus sign so has 1 total minus sign. However, the upper equation has one hidden minus sign $(\epsilon^{ijk} = \text{sgn}(\text{det }g)\epsilon_{ijk}$) resulting in 3 minus signs. Your final answer should be $c^i = \pm \epsilon_{ijk} a^j b^k$. Everywhere in this comment I use the symbol, not tensor $\endgroup$ – jgerber Nov 15 '16 at 7:47
1
$\begingroup$

In Landau book vol.II page 17, the Levi-Civita tensors are defined for the Minkowski space M$^4$ as in the metric you want. I believe that everything will work if you use the vector product definion ${\vec A}\times {\vec B}$ as $C^i=\epsilon^{0i}_{\;\;jk} A^j B^k$. And raising and lowering the spatial indices will change sign as $A_k=-A^k$. So if $ij$ indices go up we have $\epsilon^{0i}_{\;\;jk}=\epsilon^{0ijk}$. Also in M$^4$ space the norm of ${\vec A}$ will be $A_kA^k=-{\vec A}^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.