Question on vector identity proof in order to derive maxwell stress tensor
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In the process of deriving the Maxwell stress tensor we have proven the following vector identity
begin{align}
(vec{B} cdot vec{nabla})vec{B} + vec{B} times (vec{nabla} times vec{B}) &= (B_j partial_jB_i + epsilon_{ijk}B_jepsilon_{klm}partial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + (delta_{il}delta_{jm} - delta_{im}delta_{jl})B_jpartial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + B_jpartial_iB_j - B_jpartial_jB_i)vec{e}_i \ &= B_jpartial_iB_jvec{e}_i \ &= frac{1}{2}vec{nabla} (vec{B}^2)end{align}
I am wondering about what has happened in order to get the final "=".
Can anyone explain it exactly?
vectors physics tensors
asked Dec 23 '18 at 17:44
offlineoffline
619
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add a comment |
$begingroup$
In the process of deriving the Maxwell stress tensor we have proven the following vector identity
begin{align}
(vec{B} cdot vec{nabla})vec{B} + vec{B} times (vec{nabla} times vec{B}) &= (B_j partial_jB_i + epsilon_{ijk}B_jepsilon_{klm}partial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + (delta_{il}delta_{jm} - delta_{im}delta_{jl})B_jpartial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + B_jpartial_iB_j - B_jpartial_jB_i)vec{e}_i \ &= B_jpartial_iB_jvec{e}_i \ &= frac{1}{2}vec{nabla} (vec{B}^2)end{align}
I am wondering about what has happened in order to get the final "=".
Can anyone explain it exactly?
vectors physics tensors
asked Dec 23 '18 at 17:44
offlineoffline
619
$endgroup$
add a comment |
$begingroup$
In the process of deriving the Maxwell stress tensor we have proven the following vector identity
begin{align}
(vec{B} cdot vec{nabla})vec{B} + vec{B} times (vec{nabla} times vec{B}) &= (B_j partial_jB_i + epsilon_{ijk}B_jepsilon_{klm}partial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + (delta_{il}delta_{jm} - delta_{im}delta_{jl})B_jpartial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + B_jpartial_iB_j - B_jpartial_jB_i)vec{e}_i \ &= B_jpartial_iB_jvec{e}_i \ &= frac{1}{2}vec{nabla} (vec{B}^2)end{align}
I am wondering about what has happened in order to get the final "=".
Can anyone explain it exactly?
vectors physics tensors
asked Dec 23 '18 at 17:44
offlineoffline
619
$endgroup$
In the process of deriving the Maxwell stress tensor we have proven the following vector identity
begin{align}
(vec{B} cdot vec{nabla})vec{B} + vec{B} times (vec{nabla} times vec{B}) &= (B_j partial_jB_i + epsilon_{ijk}B_jepsilon_{klm}partial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + (delta_{il}delta_{jm} - delta_{im}delta_{jl})B_jpartial_lB_m)vec{e}_i \ &= (B_jpartial_jB_i + B_jpartial_iB_j - B_jpartial_jB_i)vec{e}_i \ &= B_jpartial_iB_jvec{e}_i \ &= frac{1}{2}vec{nabla} (vec{B}^2)end{align}
I am wondering about what has happened in order to get the final "=".
Can anyone explain it exactly?
vectors physics tensors
vectors physics tensors
asked Dec 23 '18 at 17:44
offlineoffline
619
asked Dec 23 '18 at 17:44
offlineoffline
619
edited Dec 23 '18 at 17:49
offline
asked Dec 23 '18 at 17:44
offlineoffline
619
asked Dec 23 '18 at 17:44
offlineoffline
619
asked Dec 23 '18 at 17:44
offlineoffline
619
619
add a comment |
add a comment |
1 Answer
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Well, $$B_jpartial_iB_jvec{e}_i=B_jvec{nabla}B_j=frac{1}{2}vec{nabla}left(B_jB_jright)=frac{1}{2}vec{nabla}left(vec{B}^2right).$$
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
$endgroup$
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
1
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
add a comment |
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$begingroup$
Well, $$B_jpartial_iB_jvec{e}_i=B_jvec{nabla}B_j=frac{1}{2}vec{nabla}left(B_jB_jright)=frac{1}{2}vec{nabla}left(vec{B}^2right).$$
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
$endgroup$
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
1
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
add a comment |
$begingroup$
Well, $$B_jpartial_iB_jvec{e}_i=B_jvec{nabla}B_j=frac{1}{2}vec{nabla}left(B_jB_jright)=frac{1}{2}vec{nabla}left(vec{B}^2right).$$
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
$endgroup$
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
1
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
add a comment |
$begingroup$
Well, $$B_jpartial_iB_jvec{e}_i=B_jvec{nabla}B_j=frac{1}{2}vec{nabla}left(B_jB_jright)=frac{1}{2}vec{nabla}left(vec{B}^2right).$$
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
$endgroup$
Well, $$B_jpartial_iB_jvec{e}_i=B_jvec{nabla}B_j=frac{1}{2}vec{nabla}left(B_jB_jright)=frac{1}{2}vec{nabla}left(vec{B}^2right).$$
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
edited Dec 23 '18 at 18:04
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
answered Dec 23 '18 at 17:54
J.G.J.G.
29.4k22846
29.4k22846
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
1
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
add a comment |
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
1
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
$begingroup$
Well, to be honest it remains unclear. I guess I don't see something very simple but what has been used to get the second "="?
$endgroup$
– offline
Dec 23 '18 at 17:58
1
1
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
@offline The product rule, or chain rule, whichever you prefer. Perhaps for clarity I should have worked with $partial_i$s and $vec{e}_i$s for longer.
$endgroup$
– J.G.
Dec 23 '18 at 18:04
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
$begingroup$
Ohhh, thank you! Problem solved.
$endgroup$
– offline
Dec 23 '18 at 18:07
add a comment |
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