Science > Physics > equation about scalar and vectorial product... explain me please :)
| Topic: |
Science > Physics |
| User: |
"stefjnoskynov" |
| Date: |
02 Jun 2005 11:42:35 AM |
| Object: |
equation about scalar and vectorial product... explain me please :) |
Can someone explain me the following relation (possibility in a
intuitive and not rigorous way)?
\del\cdot(AxB)=B\cdot\del x A -A\cdot \del x B
in latex language
where A and B are two vectors,
\del is id/dx+jd/dy+kd/dz
\cdot is the scalar product and
x is the vectorial product
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| User: "Zigoteau" |
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| Title: Re: equation about scalar and vectorial product... explain me please :) |
04 Jun 2005 03:29:28 AM |
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Hi, Stefjnoskynov,
Can someone explain me the following relation (possibility in a
intuitive and not rigorous way)?
Is your difficulty the minus sign?
\del\cdot(AxB)=B\cdot\del x A -A\cdot \del x B
in latex language
where A and B are two vectors,
\del is id/dx+jd/dy+kd/dz
\cdot is the scalar product and
x is the vectorial product
I think a lot of things become more elegant when you use tensor
notation. The vector product AxB in 3D becomes epsilon_ijk*A_j*B_k,
where epsilon is the unit antisymmetric tensor, defined by:
if ijk is an even permutation of 123 then epsilon_ijk=1 else
if ijk is an odd permutation of 123 then epsilon_ijk=-1 else
epsilon_ijk=0.
I am using the Einstein summation convention whereby summation is
implied over the repeated suffixes j and k.
Then your expression becomes
del_i*(epsilon_ijk*A_j*B_k) = epsilon_ijk*(del_i*A_j)*B_k +
epsilon_ijk*A_j*(del_i*B_k)
There is no negative sign here because the tensor notation allows you
to write down the terms following epsilon in any order. Translating to
your notation, the second term changes sign for the same reason that
BxA = - AxB. In your notation in the second term, to get del and B in
the right order for the cross product, you have to swap i and j, which
is an odd-parity permutation of ijk.
It takes a while to get used to tensor notation, but it's worth it. It
is explicit, it allows you to write down expressions with several
matrices and vectors which are either impossible or very clumsy in
standard vector notation, and it makes identities like yours much more
obvious.
Cheers,
Zigoteau.
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| User: "" |
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| Title: Re: equation about scalar and vectorial product... explain me please :) |
02 Jun 2005 07:43:43 PM |
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stefjnoskynov wrote:
Can someone explain me the following relation (possibility in a
intuitive and not rigorous way)?
\del\cdot(AxB)=B\cdot\del x A -A\cdot \del x B
in latex language
where A and B are two vectors,
\del is id/dx+jd/dy+kd/dz
\cdot is the scalar product and
x is the vectorial product
Dunno about the "intuitive" part, but what exactly is there to
"explain" here? You write out the gradient of the vector field on the
left that'll come to d/dx(AyBz-AzBy)+d/dy( ... etc, distribute the
differential operators over the parentheses (d/dx(p+q) = dp/dx+dq/dx)
collect the terms that contain Bx (and the terms that contain By and
Bz) and you should be left with what's on the right side.
If there's anything more to this than a little algebra, you might want
to tell us what exactly you're looking for.
cordially
Y.T.
--
Remove YourClothes before you email me.
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| User: "stefjnoskynov" |
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| Title: Re: equation about scalar and vectorial product... explain me please :) |
03 Jun 2005 12:34:38 PM |
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In article <1117759422.991309.208650@g47g2000cwa.googlegroups.com>,
ytyourclothes@p.zapto.org says...
Dunno about the "intuitive" part, but what exactly is there to
"explain" here? You write out the gradient of the vector field on the
left that'll come to d/dx(AyBz-AzBy)+d/dy( ... etc, distribute the
differential operators over the parentheses (d/dx(p+q) = dp/dx+dq/dx)
collect the terms that contain Bx (and the terms that contain By and
Bz) and you should be left with what's on the right side.
Yes, I was not attempt to describe what I want to know. \del is a
differential operator that can assume different forms like rotor or
gradient or what I wrote some time ago.
I try to describe better my question:
\del is a vectorial operator and so It is considerable like a "normal
vector" in a appropriate vector space. Now Can I see this equation only
by geometrical arguments?
I hope to be more understandable, sorry for my english :)
If there's anything more to this than a little algebra, you might want
to tell us what exactly you're looking for.
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