Integral of triple vector product
I'm following a Electrodynamics course and I'm currently stack in the following calculation:
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}$
that the book calculates as
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}= frac{ vec{n} times vec{n} times vec{beta}(t) }{(1-vec{n} cdot vec{beta}(t)) }$
I have been unable to reproduce that calculus without proving the reverse ( that the derivative of the right side is the integrand of the left side). I would like to calculate the integral without using a coordinate basis if possible. Some details of the notation:
$frac{d vec{beta}(t)}{d t} = vec{beta'}(t)$
$vec{n}$ is a constant vector that does not depend of the integration variable t.
This is from The Physics of Synchrotron Radiation by Albert Hofmann, section 2.8. The Fourier transform of the radiation field, page 36.
physics vector-analysis
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
asked Sep 4 '15 at 17:47
Dargor
37337
add a comment |
I'm following a Electrodynamics course and I'm currently stack in the following calculation:
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}$
that the book calculates as
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}= frac{ vec{n} times vec{n} times vec{beta}(t) }{(1-vec{n} cdot vec{beta}(t)) }$
I have been unable to reproduce that calculus without proving the reverse ( that the derivative of the right side is the integrand of the left side). I would like to calculate the integral without using a coordinate basis if possible. Some details of the notation:
$frac{d vec{beta}(t)}{d t} = vec{beta'}(t)$
$vec{n}$ is a constant vector that does not depend of the integration variable t.
This is from The Physics of Synchrotron Radiation by Albert Hofmann, section 2.8. The Fourier transform of the radiation field, page 36.
physics vector-analysis
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
asked Sep 4 '15 at 17:47
Dargor
37337
Can you give the title of the book and the page ? To see the context can be useful.
– Tony Piccolo
Sep 5 '15 at 5:34
@TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields".
– Dargor
Sep 5 '15 at 11:20
Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry.
– Tony Piccolo
Sep 5 '15 at 17:06
@TonyPiccolo Thank you anyway :)
– Dargor
Sep 5 '15 at 17:14
@PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant.
– Martin Sleziak
Sep 8 '15 at 5:24
add a comment |
I'm following a Electrodynamics course and I'm currently stack in the following calculation:
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}$
that the book calculates as
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}= frac{ vec{n} times vec{n} times vec{beta}(t) }{(1-vec{n} cdot vec{beta}(t)) }$
I have been unable to reproduce that calculus without proving the reverse ( that the derivative of the right side is the integrand of the left side). I would like to calculate the integral without using a coordinate basis if possible. Some details of the notation:
$frac{d vec{beta}(t)}{d t} = vec{beta'}(t)$
$vec{n}$ is a constant vector that does not depend of the integration variable t.
This is from The Physics of Synchrotron Radiation by Albert Hofmann, section 2.8. The Fourier transform of the radiation field, page 36.
physics vector-analysis
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
asked Sep 4 '15 at 17:47
Dargor
37337
I'm following a Electrodynamics course and I'm currently stack in the following calculation:
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}$
that the book calculates as
$int{ frac{ vec{n} times (vec{n}-vec{beta}(t)) times vec{beta'}(t) }{(1-vec{n} cdot vec{beta}(t) )^2 } dt}= frac{ vec{n} times vec{n} times vec{beta}(t) }{(1-vec{n} cdot vec{beta}(t)) }$
I have been unable to reproduce that calculus without proving the reverse ( that the derivative of the right side is the integrand of the left side). I would like to calculate the integral without using a coordinate basis if possible. Some details of the notation:
$frac{d vec{beta}(t)}{d t} = vec{beta'}(t)$
$vec{n}$ is a constant vector that does not depend of the integration variable t.
This is from The Physics of Synchrotron Radiation by Albert Hofmann, section 2.8. The Fourier transform of the radiation field, page 36.
physics vector-analysis
physics vector-analysis
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
asked Sep 4 '15 at 17:47
Dargor
37337
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
asked Sep 4 '15 at 17:47
Dargor
37337
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
edited Sep 8 '15 at 5:24
Martin Sleziak
44.7k8115271
44.7k8115271
asked Sep 4 '15 at 17:47
Dargor
37337
asked Sep 4 '15 at 17:47
Dargor
37337
asked Sep 4 '15 at 17:47
Dargor
37337
37337
Can you give the title of the book and the page ? To see the context can be useful.
– Tony Piccolo
Sep 5 '15 at 5:34
@TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields".
– Dargor
Sep 5 '15 at 11:20
Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry.
– Tony Piccolo
Sep 5 '15 at 17:06
@TonyPiccolo Thank you anyway :)
– Dargor
Sep 5 '15 at 17:14
@PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant.
– Martin Sleziak
Sep 8 '15 at 5:24
add a comment |
Can you give the title of the book and the page ? To see the context can be useful.
– Tony Piccolo
Sep 5 '15 at 5:34
@TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields".
– Dargor
Sep 5 '15 at 11:20
Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry.
– Tony Piccolo
Sep 5 '15 at 17:06
@TonyPiccolo Thank you anyway :)
– Dargor
Sep 5 '15 at 17:14
@PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant.
– Martin Sleziak
Sep 8 '15 at 5:24
Can you give the title of the book and the page ? To see the context can be useful.
– Tony Piccolo
Sep 5 '15 at 5:34
Can you give the title of the book and the page ? To see the context can be useful.
– Tony Piccolo
Sep 5 '15 at 5:34
@TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields".
– Dargor
Sep 5 '15 at 11:20
@TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields".
– Dargor
Sep 5 '15 at 11:20
Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry.
– Tony Piccolo
Sep 5 '15 at 17:06
Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry.
– Tony Piccolo
Sep 5 '15 at 17:06
@TonyPiccolo Thank you anyway :)
– Dargor
Sep 5 '15 at 17:14
@TonyPiccolo Thank you anyway :)
– Dargor
Sep 5 '15 at 17:14
@PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant.
– Martin Sleziak
Sep 8 '15 at 5:24
@PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant.
– Martin Sleziak
Sep 8 '15 at 5:24
add a comment |
1 Answer
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If you expand out the integrand you see one term is zero, since the cross product of two parallel vectors is zero:
begin{align}
I &= int frac{vec{n}times(vec{n}-vec{beta(t)})timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}-vec{n}timesvec{beta(t)}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt
end{align}
This gets you halfway there. I'll try to come back to this to finish it soon!
answered yesterday
SabrinaChoice
246
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you expand out the integrand you see one term is zero, since the cross product of two parallel vectors is zero:
begin{align}
I &= int frac{vec{n}times(vec{n}-vec{beta(t)})timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}-vec{n}timesvec{beta(t)}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt
end{align}
This gets you halfway there. I'll try to come back to this to finish it soon!
answered yesterday
SabrinaChoice
246
add a comment |
If you expand out the integrand you see one term is zero, since the cross product of two parallel vectors is zero:
begin{align}
I &= int frac{vec{n}times(vec{n}-vec{beta(t)})timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}-vec{n}timesvec{beta(t)}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt
end{align}
This gets you halfway there. I'll try to come back to this to finish it soon!
answered yesterday
SabrinaChoice
246
add a comment |
If you expand out the integrand you see one term is zero, since the cross product of two parallel vectors is zero:
begin{align}
I &= int frac{vec{n}times(vec{n}-vec{beta(t)})timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}-vec{n}timesvec{beta(t)}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt
end{align}
This gets you halfway there. I'll try to come back to this to finish it soon!
answered yesterday
SabrinaChoice
246
If you expand out the integrand you see one term is zero, since the cross product of two parallel vectors is zero:
begin{align}
I &= int frac{vec{n}times(vec{n}-vec{beta(t)})timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}-vec{n}timesvec{beta(t)}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt\
&= int frac{vec{n}timesvec{n}timesvec{beta'(t)}}{(1-vec{n}cdot vec{beta(t)})^2}dt
end{align}
This gets you halfway there. I'll try to come back to this to finish it soon!
answered yesterday
SabrinaChoice
246
answered yesterday
SabrinaChoice
246
answered yesterday
SabrinaChoice
246
answered yesterday
SabrinaChoice
246
246
add a comment |
add a comment |
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Can you give the title of the book and the page ? To see the context can be useful.
– Tony Piccolo
Sep 5 '15 at 5:34
@TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields".
– Dargor
Sep 5 '15 at 11:20
Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry.
– Tony Piccolo
Sep 5 '15 at 17:06
@TonyPiccolo Thank you anyway :)
– Dargor
Sep 5 '15 at 17:14
@PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant.
– Martin Sleziak
Sep 8 '15 at 5:24