Mixing
Three index entities summing to product of Kronecker deltas
Clash Royale CLAN TAG#URR8PPP
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While looking at this problem I ended up having to find entities $a_ijp$ and $b_klp,$ with $i,j,k,l,p = 1, ldots, n,$ such that
$$sum_p=1^n a_ijp , b_klp = delta_ik , delta_jl,$$
where $delta_ij$ is the Kronecker delta.
Probably $a$ and $b$ can be taken equal so that we want
$$sum_p=1^n a_ijp , a_klp = delta_ik , delta_jl.$$
Does anyone have know if this is even possible, and if so, how to choose $a_ijp$?
In the actual case, $n=3,$ but a generic solution is preferred.
How did this equation come up?
I asserted that $sigma^ij , partial_i u_j , dV$ could be rewritten as $omega wedge pi$ where $omega = alpha_kl^ij sigma_ij , dx^k wedge dx^l$ and $pi = beta_p^mn partial_m u_n , dx^p.$ This led to $$alpha_kl^ij , beta_p^mn , sigma_ij , partial_m u_n epsilon^klp , dV = sigma^mn , partial_m u_n , dV.$$
Setting $gamma^ijp = epsilon^klp , alpha_kl^ij$ then gives
$$beta_p^mn gamma^ijp = delta^mi , delta^nj.$$
For the presentation in the post the names were just changed and the summation over $p$ was made explicit.
tensors kronecker-delta
asked Jul 27 at 19:21
md2perpe
5,80511022
add a comment |Â
up vote
0
down vote
favorite
While looking at this problem I ended up having to find entities $a_ijp$ and $b_klp,$ with $i,j,k,l,p = 1, ldots, n,$ such that
$$sum_p=1^n a_ijp , b_klp = delta_ik , delta_jl,$$
where $delta_ij$ is the Kronecker delta.
Probably $a$ and $b$ can be taken equal so that we want
$$sum_p=1^n a_ijp , a_klp = delta_ik , delta_jl.$$
Does anyone have know if this is even possible, and if so, how to choose $a_ijp$?
In the actual case, $n=3,$ but a generic solution is preferred.
How did this equation come up?
I asserted that $sigma^ij , partial_i u_j , dV$ could be rewritten as $omega wedge pi$ where $omega = alpha_kl^ij sigma_ij , dx^k wedge dx^l$ and $pi = beta_p^mn partial_m u_n , dx^p.$ This led to $$alpha_kl^ij , beta_p^mn , sigma_ij , partial_m u_n epsilon^klp , dV = sigma^mn , partial_m u_n , dV.$$
Setting $gamma^ijp = epsilon^klp , alpha_kl^ij$ then gives
$$beta_p^mn gamma^ijp = delta^mi , delta^nj.$$
For the presentation in the post the names were just changed and the summation over $p$ was made explicit.
tensors kronecker-delta
asked Jul 27 at 19:21
md2perpe
5,80511022
You probably should give us some background. What is $a_ijk$ and $b_ijk$ actually.
– Sou
Jul 27 at 19:47
1
Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol
– Donald Splutterwit
Jul 27 at 20:23
@Sou. I've added some background.
– md2perpe
Jul 27 at 22:38
@DonaldSplutterwit. That's close. Perhaps some modification of $epsilon$ can be used.
– md2perpe
Jul 27 at 22:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
While looking at this problem I ended up having to find entities $a_ijp$ and $b_klp,$ with $i,j,k,l,p = 1, ldots, n,$ such that
$$sum_p=1^n a_ijp , b_klp = delta_ik , delta_jl,$$
where $delta_ij$ is the Kronecker delta.
Probably $a$ and $b$ can be taken equal so that we want
$$sum_p=1^n a_ijp , a_klp = delta_ik , delta_jl.$$
Does anyone have know if this is even possible, and if so, how to choose $a_ijp$?
In the actual case, $n=3,$ but a generic solution is preferred.
How did this equation come up?
I asserted that $sigma^ij , partial_i u_j , dV$ could be rewritten as $omega wedge pi$ where $omega = alpha_kl^ij sigma_ij , dx^k wedge dx^l$ and $pi = beta_p^mn partial_m u_n , dx^p.$ This led to $$alpha_kl^ij , beta_p^mn , sigma_ij , partial_m u_n epsilon^klp , dV = sigma^mn , partial_m u_n , dV.$$
Setting $gamma^ijp = epsilon^klp , alpha_kl^ij$ then gives
$$beta_p^mn gamma^ijp = delta^mi , delta^nj.$$
For the presentation in the post the names were just changed and the summation over $p$ was made explicit.
tensors kronecker-delta
asked Jul 27 at 19:21
md2perpe
5,80511022
While looking at this problem I ended up having to find entities $a_ijp$ and $b_klp,$ with $i,j,k,l,p = 1, ldots, n,$ such that
$$sum_p=1^n a_ijp , b_klp = delta_ik , delta_jl,$$
where $delta_ij$ is the Kronecker delta.
Probably $a$ and $b$ can be taken equal so that we want
$$sum_p=1^n a_ijp , a_klp = delta_ik , delta_jl.$$
Does anyone have know if this is even possible, and if so, how to choose $a_ijp$?
In the actual case, $n=3,$ but a generic solution is preferred.
How did this equation come up?
I asserted that $sigma^ij , partial_i u_j , dV$ could be rewritten as $omega wedge pi$ where $omega = alpha_kl^ij sigma_ij , dx^k wedge dx^l$ and $pi = beta_p^mn partial_m u_n , dx^p.$ This led to $$alpha_kl^ij , beta_p^mn , sigma_ij , partial_m u_n epsilon^klp , dV = sigma^mn , partial_m u_n , dV.$$
Setting $gamma^ijp = epsilon^klp , alpha_kl^ij$ then gives
$$beta_p^mn gamma^ijp = delta^mi , delta^nj.$$
For the presentation in the post the names were just changed and the summation over $p$ was made explicit.
tensors kronecker-delta
asked Jul 27 at 19:21
md2perpe
5,80511022
edited Jul 27 at 22:38
asked Jul 27 at 19:21
md2perpe
5,80511022
asked Jul 27 at 19:21
md2perpe
5,80511022
asked Jul 27 at 19:21
md2perpe
5,80511022
5,80511022
You probably should give us some background. What is $a_ijk$ and $b_ijk$ actually.
– Sou
Jul 27 at 19:47
1
Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol
– Donald Splutterwit
Jul 27 at 20:23
@Sou. I've added some background.
– md2perpe
Jul 27 at 22:38
@DonaldSplutterwit. That's close. Perhaps some modification of $epsilon$ can be used.
– md2perpe
Jul 27 at 22:39
add a comment |Â
You probably should give us some background. What is $a_ijk$ and $b_ijk$ actually.
– Sou
Jul 27 at 19:47
1
Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol
– Donald Splutterwit
Jul 27 at 20:23
@Sou. I've added some background.
– md2perpe
Jul 27 at 22:38
@DonaldSplutterwit. That's close. Perhaps some modification of $epsilon$ can be used.
– md2perpe
Jul 27 at 22:39
You probably should give us some background. What is $a_ijk$ and $b_ijk$ actually.
– Sou
Jul 27 at 19:47
You probably should give us some background. What is $a_ijk$ and $b_ijk$ actually.
– Sou
Jul 27 at 19:47
1
1
Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol
– Donald Splutterwit
Jul 27 at 20:23
Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol
– Donald Splutterwit
Jul 27 at 20:23
@Sou. I've added some background.
– md2perpe
Jul 27 at 22:38
@Sou. I've added some background.
– md2perpe
Jul 27 at 22:38
@DonaldSplutterwit. That's close. Perhaps some modification of $epsilon$ can be used.
– md2perpe
Jul 27 at 22:39
@DonaldSplutterwit. That's close. Perhaps some modification of $epsilon$ can be used.
– md2perpe
Jul 27 at 22:39
add a comment |Â
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You probably should give us some background. What is $a_ijk$ and $b_ijk$ actually.
– Sou
Jul 27 at 19:47
1
Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol
– Donald Splutterwit
Jul 27 at 20:23
@Sou. I've added some background.
– md2perpe
Jul 27 at 22:38
@DonaldSplutterwit. That's close. Perhaps some modification of $epsilon$ can be used.
– md2perpe
Jul 27 at 22:39