Is it possible to solve cross products using Einstein notation?

Question

I'm considering a case where I have an equation of the form $\mathbf{x}\times\mathbf{b}+\mathbf{c}=0$; I wish to solve for $\mathbf{x}$ given that $\mathbf{b}\perp \mathbf{c}$. It was in the context of studying guiding center drifts in plasmas.

An easy way to do this is to set up a coordinate system with $\mathbf{b}=b\hat{\mathbf{z}}$ and $\mathbf{c}=c\hat{\mathbf{y}}$. Then you can get three simultaneous equations to uniquely specify the individual components of $\mathbf{x}$. There's no loss of generality because we're given that $\mathbf{b}$ and $\mathbf{c}$ are orthogonal.

But this should be possible with Einstein notation without assuming a coordinate system, because the previously stated assumptions about the directions of $\mathbf{b}$ and $\mathbf{c}$ don't add any new information: they just serve as a way to express the orthogonality constraint. I tried this out; I got stuck when I reached this step: $$\epsilon_{ijk}x_jb_k=-c_i.$$ Ordinarily, I'd try to divide by $b_k$, because this is an equation with just scalars, but that doesn't work because we're taking a summation over the dummy index $k$. My next idea was to take a dot product of both sides and exploit orthogonality. I'd multiply both sides by $c_k$ or $b_i$, but I don't see how that would help (and multiplying by $c_k$ would be nonsensical on the RHS).

I think my problem can be generalized to this: "What do you do when you are trying to solve a series of equations in Einstein notation for a vector that is involved in a cross product?"

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As a clarification in response to an answer posted, it seems I glossed over a component of the argument made through the method that assumes a cartesian coordinate system: we neglect the component of $x$ along $\mathbf{b}$ (which is trivial) and focus on the magnitude of $\mathbf{x}$ that is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$. This is the value that I'm interested in computing through Einstein notation.

But I think the real conceptual question is about the approach to use when you're solving for the components of a vector that appears through a dummy summation index. In ordinary vector notation, I'd separate the equation into individual components. That doesn't scale well to higher dimensional vector spaces, and even in 3 dimensions, it doesn't work out in Einstein notation. What do we do?

Answer 1

You painted yourself into an impossible notational corner, by using a terrible and misleading name for your unknown! Call, it, instead, v, so $$\mathbf{v}\times\mathbf{b}+\mathbf{c}=0 ; \qquad \epsilon_{ijk}v_jb_k=-c_i.$$ It is then evident, with your choice of coordinate system, that your unknown vector $$ \mathbf{v}= \mu\mathbf{x}+\rho\mathbf{y} + \lambda \mathbf{z}, $$ plugs into your equation to yield $$ 0= \rho b\mathbf{x}+(c-\mu b) \mathbf{y} ,\implies \rho=0, ~~ \mu=c/b. $$ $\lambda$ is arbitrary, since it was projected out.

The comment by @J.G. already has your answer, which I rewrite in abstract index notation, $$ v_i=\mu \epsilon_{ijk}\frac{c_jb_k}{bc}+\rho c_i/c +\lambda b_i/b ~, $$ as you have effectively defined an orthonormal basis.

Plugging into your equation, you have, by inspection, (4), $$ 0=\left (1-\frac{\mu b}{c}\right )c_i + \epsilon_{ijk} \frac{\rho}{c} c_j b_k. $$ But this is a vanishing combination of two orthogonal vectors, so their coefficients must vanish, hence $\rho=0$ and $\mu = c/b$.

Answer 2

If $\mathbf{\vec X}$ satisfies $\mathbf{\vec X}\times\mathbf{\vec b}+\mathbf{\vec c}=\vec 0$, then so does $(\mathbf{\vec X}+\lambda \mathbf{\vec b})$.

So, the solution is not unique.

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UPDATE

Given $\mathbf{\vec V}\times\mathbf{\vec b}+\mathbf{\vec c}=\vec 0$, which I will write as $$-\mathbf{\vec c}=\mathbf{\vec V}\times\mathbf{\vec b}$$ $$-c_i =\epsilon_{ijk}V_j b_k ,$$ consider this operation "$\vec b \times \square$ "
and apply the BAC-CAB rule $\vec A\times(\vec B\times \vec C)= \vec B(\vec A\cdot \vec C)- \vec C(\vec A\cdot \vec B)$.

• See https://en.wikipedia.org/wiki/Triple_product#Vector_triple_product
• See https://en.wikipedia.org/wiki/Levi-Civita_symbol#Three_dimensions_2

I'll leave a “trail of breadcrumbs”: $$\begin{align} \epsilon_{lmi} b_m \left[-c_i \right] &= \epsilon_{lmi} b_m \left[ \epsilon_{ijk}V_j b_k \right] \\ "\vec b\times (-\vec c)" &=(-1)^2\epsilon_{ilm} b_m \left[ \epsilon_{ijk}V_j b_k \right] \\ &= (-1)^2\epsilon_{ilm} \epsilon_{ijk}V_j b_k b_m \\ &= (-1)^2\left[ \delta_{lj}\delta_{mk}- \delta_{lk}\delta_{jm}\right] V_j b_k b_m \\ &= (-1)^2\left[ V_l b_m b_m - V_m b_l b_m \right] \\ &=" \vec V (b^2) - \vec b(\vec V\cdot \vec b) "\\ &=" b^2 \left( \vec V - \hat b(\vec V\cdot \hat b) \right) "\\ \end{align} $$

So, $$ \vec V_{\bot \vec b}\equiv \left(\vec V - \hat b(\vec V\cdot \hat b)\right)= \frac{\vec b\times(-\vec c)}{b^2},$$ which I think agrees with the result from @CosmasZachos .

Answer 3

your equation is

$$\vec x\times\vec b+\vec c=0\quad \text{or}\\ -\vec b\times \vec x=-\vec c$$

with the components of vector b

$$ -\underbrace{\left[ \begin {array}{ccc} 0&-b_{{z}}&b_{{y}}\\ b_{ {z}}&0&-b_{{x}}\\ -b_{{y}}&b_{{x}}&0\end {array} \right]}_{\mathbf B} \vec x=-\vec c$$

to solve this equation for $\vec x$ the determinate of the matrix $~\mathbf B~$ must be unequal zero, which is not ($\det(\mathbf B)=0~$) , thus you can't obtain the solution for $~\vec x$