Einstein Summation Convention Question 2

[Athenian]

Homework Statement

Solve for $\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k$

Relevant Equations

Refer Below $\longrightarrow$

Below is my attempted solution:

$$\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k$$
$$\Rightarrow (\delta_{ik} \, \delta_{jm} - \delta_{im} \, \delta_{jk}) \epsilon_{ijm} \, a_k$$
$$\Rightarrow \delta_{ik} \, \delta_{jm} \, \epsilon_{ijm} \, a_k - \delta_{im} \, \delta_{jk} \epsilon_{ijm} \, a_k$$
$$\Rightarrow a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}$$

From here on out, I am not sure what I am doing. Therefore, any guidance or assistance would be greatly appreciated.

$$\Rightarrow a_m \epsilon_{ijm} - a_m \epsilon_{ijm} = 0$$

Perhaps, my biggest question is would eliminating the Kronecker delta and getting $a_m$ be mathematically legal?

Also, I have heard from someone else that the above equation results to zero because of anti-symmetry. How is one able to determine that?

Thank you for the help!

 

[George Jones]

Also, I have heard from someone else that the above equation results to zero because of anti-symmetry. How is one able to determine that?

If $A$ and $B$ are general, is it always true that

$$A_{ij} B_{ij} = A_{ji} B_{ji}?$$

If so, why? If not, why not?

 

[Athenian]

Unfortunately, I am not quite sure what you mean by "general" as I have only seen the Kronecker delta possessing two subscripts thus far in my studies. However, if I am to apply the above equation using "Kronecker delta" logic, I would say yes as they will both ultimately equate to 0.

I apologize if this doesn't properly answer your question. Nonetheless, I really appreciate the helping hand.

 

[George Jones]

Unfortunately, I am not quite sure what you mean by "general" as I have only seen the Kronecker delta possessing two subscripts thus far in my studies.

Think of $A$ and $B$ as being matrices ($3\times3$, if you like). The Kronecker delta is the component version of the identity matrix. By "general", I mean that $A$ and $B$ do not possesses any special symmetry or antisymmetry properties. I am trying to illustrate and import aspect of the Einstein summation convention.

 

[Athenian]

If $A$ and $B$ are general, is it always true that

$$A_{ij} B_{ij} = A_{ji} B_{ji}?$$

If so, why? If not, why not?

Hmmm, despite thinking for a bit to find a better answer, I still only came up with an assumption of "no" since matrix multiplication is not communicable.

While I did try to prove my guess by writing something along the lines of $A_{ij} \, B_{ik} \, \delta_{jk} = A_{ij} \, B_{ij}$, I didn't think that was going very well in the end ...

 

[Gaussian97]

Hmmm, despite thinking for a bit to find a better answer, I still only came up with an assumption of "no" since matrix multiplication is not communicable.

While I did try to prove my guess by writing something along the lines of $A_{ij} \, B_{ik} \, \delta_{jk} = A_{ij} \, B_{ij}$, I didn't think that was going very well in the end ...

Try to write the sum explicitly for $i,j=1,2,3$.

 

[Athenian]

Wrote the sum explicitly down and that definitely helped me better visualize what's going on.

Anyway, I take back my answer. They are equal to each other. In short, $A_{ij}$ and $A_{ji}$ are the same because one is simply moving the $\Sigma$ (i.e. sum) for $i$ and $y$ back and forth (or in different orders). However, in the end, though, the resulting sum would be the same.

Would this logic, I can thus say that the below equation is correct:

$A_{ij} B_{ij} =A_{ji} B_{ji}$

 

[George Jones]

Wrote the sum explicitly down and that definitely helped me better visualize what's going on.

Anyway, I take back my answer. They are equal to each other. In short, $A_{ij}$ and $A_{ji}$ are the same because one is simply moving the $\Sigma$ (i.e. sum) for $i$ and $y$ back and forth (or in different orders). However, in the end, though, the resulting sum would be the same.

Would this logic, I can thus say that the below equation is correct:

Now suppose that $A$ is symmetric and that $B$ is antisymmetic.

 

[Athenian]

I see. Then, using my logic from earlier, $B_{ij}$ would be symmetric while $-B_{ji}$ would be antisymmetric.

Therefore, $A_{ij} \, B_{ij} = - A_{ij} \, B_{ji}$, right?

Provided that I am correct, I still don't exactly see how this solution method helps me solve my above equation. Are there dots I am supposed to be connecting here?

Thank you for your help!

 

[George Jones]

I see. Then, using my logic from earlier, $B_{ij}$ would be symmetric while $-B_{ji}$ would be antisymmetric.

I am not sure what what you mean, here. $A$ symmetric means $A_{ij} = A_{ji}$ always, i.e,, both inside and outside sums. The Kronecker delta is symmetric. $B$ antisymmetric means $B_{ij} = -B_{ji}$ always, i.e,, both inside and outside sums. The Levi-Civita symbol is antisymmetric with respect to any pair of indices.

Where we stand: for any $A$ and $B$, $A_{ij} B_{ij} = A_{ji} B_{ji}$, because of the implied sums.

Now suppose that $A$ is symmetric and that $B$ is antisymmetic. Using the properties I give above, rewrite $A_{ij} B_{ij}$.

 

[nrqed]

Homework Statement:: Solve for $\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k$
Relevant Equations:: Refer Below $\longrightarrow$

Below is my attempted solution:

$$\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k$$
$$\Rightarrow (\delta_{ik} \, \delta_{jm} - \delta_{im} \, \delta_{jk}) \epsilon_{ijm} \, a_k$$
$$\Rightarrow \delta_{ik} \, \delta_{jm} \, \epsilon_{ijm} \, a_k - \delta_{im} \, \delta_{jk} \epsilon_{ijm} \, a_k$$
$$\Rightarrow a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}$$

As I said in another thread, when you see a Kronecker delta, use it to eliminate one of the indices that appears in it. For example here what do you get if you do the sum over m in both terms?

 

[Athenian]

I am not sure what you mean, here. $A$ symmetric means $A_{ij} = A_{ji}$ always, i.e,, both inside and outside sums. The Kronecker delta is symmetric. $B$ antisymmetric means $B_{ij} = -B_{ji}$ always, i.e,, both inside and outside sums. The Levi-Civita symbol is antisymmetric with respect to any pair of indices.

Where we stand: for any $A$ and $B$, $A_{ij} B_{ij} = A_{ji} B_{ji}$, because of the implied sums.

Now suppose that $A$ is symmetric and that $B$ is antisymmetic. Using the properties I give above, rewrite $A_{ij} B_{ij}$.

My mistake here. In that case, to answer the question $\Longrightarrow A_{ij} B_{ij} = -A_{ji} B_{ji}$ assuming that $A_{ij}$ is symmetric and $B_{ij}$ is anti-symmetric.

 

[Athenian]

As I said in another thread, when you see a Kronecker delta, use it to eliminate one of the indices that appears in it. For example here what do you get if you do the sum over m in both terms?

Thanks a lot for the help. I figured out the logic behind solving the question. Essentially, if I assume $j \neq m$, then both Kronecker deltas will go to zero - making the entire solution $0$. However, if I make $j = m$, then both Kronecker deltas will go to one. However, the Levi-Civita will then become $\epsilon_{ijj}$ or $\epsilon_{imm}$. And, in this case, the Levi-Civita would also go to zero. Therefore, regardless whether $j = m$ or $j \neq m$, the entire solution will go to $0$.

Is this reasoning accurate?

 

[nrqed]

Thanks a lot for the help. I figured out the logic behind solving the question. Essentially, if I assume $j \neq m$, then both Kronecker deltas will go to zero - making the entire solution $0$. However, if I make $j = m$, then both Kronecker deltas will go to one. However, the Levi-Civita will then become $\epsilon_{ijj}$ or $\epsilon_{imm}$. And, in this case, the Levi-Civita would also go to zero. Therefore, regardless whether $j = m$ or $j \neq m$, the entire solution will go to $0$.

Is this reasoning accurate?

That's completely correct.

In general, when you have a Kronecker delta, it is very nice because you can do the sum over one of its indices trivially, so you should always do that and get rid of all the Kronecker deltas this way. In this problem, if you had not realized the point you just made, you would have ended up with Levi-Civita symbols with two indices equal, which would have told you the result is zero.

 

[Athenian]

Thank you for confirming my explanation. It really helped and I sincerely appreciate it!

 

[George Jones]

As @nrqed has explianed, it is very important to know how to eliminate sum using the Konecker delta. It also can be useful to know how to use symmetry properties.

I can thus say that the below equation is correct:

$$A_{ij} B_{ij} = A_{ji} B_{ji}$$

$A_{ij} B_{ij} = -A_{ji} B_{ji}$ assuming that $A_{ij}$ is symmetric and $B_{ij}$ is anti-symmetric.

Putting these two things together gives
$$\begin{align}
A_{ij} B_{ij} &= -A_{ij} B_{ij} \nonumber \\
2A_{ij} B_{ij} &= 0 \nonumber \\
A_{ij} B_{ij} &= 0 \nonumber
\end{align}$$

for any symmetric $A$ and antisymmetric $B$ (or vice versa). In particular, it can be used in both terms of $a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}$.

The first term contains $\delta_{jm} \, \epsilon_{ijm}$, with $\delta_{jm}$ symmetric in $j$ and $m$, and $\epsilon_{ijm}$ antisymmetric in $j$ and $m$, so $\delta_{jm} \, \epsilon_{ijm} = 0$.

The second term contains $\delta_{im} \, \epsilon_{ijm}$, with $\delta_{im}$ symmetric in $i$ and $m$, and $\epsilon_{ijm}$ antisymmetric in $i$ and $m$, so $\delta_{im} \, \epsilon_{ijm} = 0$.

 

[nrqed]

As @nrqed has explianed, it is very important to know how to eliminate sum using the Konecker delta. It also can be useful to know how to use symmetry properties.Putting these two things together gives
$$\begin{align}
A_{ij} B_{ij} &= -A_{ij} B_{ij} \nonumber \\
2A_{ij} B_{ij} &= 0 \nonumber \\
A_{ij} B_{ij} &= 0 \nonumber
\end{align}$$

for any symmetric $A$ and antisymmetric $B$ (or vice versa). In particular, it can be used in both terms of $a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}$.

The first term contains $\delta_{jm} \, \epsilon_{ijm}$, with $\delta_{jm}$ symmetric in $j$ and $m$, and $\epsilon_{ijm}$ antisymmetric in $j$ and $m$, so $\delta_{jm} \, \epsilon_{ijm} = 0$.

The second term contains $\delta_{im} \, \epsilon_{ijm}$, with $\delta_{im}$ symmetric in $i$ and $m$, and $\epsilon_{ijm}$ antisymmetric in $i$ and $m$, so $\delta_{im} \, \epsilon_{ijm} = 0$.

To the OP: make sure you understand George's example with an the product of arbitrary symmetric $A$ and antisymmetric $B$, it can be very useful.

 

[Athenian]

Wow, thank you so much for the explanation, @George Jones! This was incredibly insightful and I have managed to connect all the conceptual dots to understanding the symmetry properties.

I really appreciate the help and I'll be sure to practice more similar problems.