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Cross product integral

by daudaudaudau
Tags: cross, integral, product
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daudaudaudau
#1
Apr14-08, 05:25 PM
P: 296
Hi.

I have an integral of a cross product:


[tex]\int a\times b(t)dt[/tex]

a does not depend on t. How can I move it outside? Is this legal:

[tex]a\times\int b(t)dt[/tex]

?
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Hurkyl
#2
Apr14-08, 05:26 PM
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Have you tried proving it from the definition of the integral? Or from the definition of the cross product?
daudaudaudau
#3
Apr14-08, 05:34 PM
P: 296
Yes I tried using the definition of the cross-product. I don't think what I suggested is legal.

[tex]\int a\times b(t) dt=\int |a||b(t)|\sin \theta_1dt[/tex]

[tex]a\times\int b(t) dt=|a|\left|\int b(t)dt\right|\sin \theta_2[/tex]

where [tex]\theta_1[/tex] is the angle between a and b(t) and [tex]\theta_2[/tex] is the angle between a and [tex]\int b(t) dt[/tex]

Hurkyl
#4
Apr14-08, 05:43 PM
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Cross product integral

(I'm assuming from the formulas that you are using that you're only working in two dimensions?)

Hrm. I was thinking more about the component-wise formula for the cross product than the angle & magnitude version; it's much more friendly in purely algebraic contexts.


It turns out that your conjecture is true -- in fact, there's a bigger principle: you can always pull a linear transformation through an integral sign. (A linear transformation is one satisfying T(u+v)=T(u)+T(v) and T(au) = a T(u), where u and v denote vectors, and a denotes a number) The function (a x __) is, in fact, a linear transformation.

It might be easier to abstractly prove the general theorem, rather than directly prove this particular case! (I think it would be an instructive exercise to undertake, when you have the time)
daudaudaudau
#5
Apr16-08, 06:07 AM
P: 296
Thank you, Hurkyl.
HallsofIvy
#6
Apr16-08, 07:24 AM
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Thanks
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[itex]|u||v|sin(\theta)[/itex] is the length of the cross product, not the cross product itself.
JonF
#7
Apr16-08, 10:27 AM
P: 617
Quote Quote by Hurkyl View Post
It might be easier to abstractly prove the general theorem, rather than directly prove this particular case! (I think it would be an instructive exercise to undertake, when you have the time)
Hint: If i remember correctly you prove the general theorem nearly the exact same way you proved you can pull constants out of integrals.

If you are having trouble with a proof, but you know a proof for a similar concept, it is often helpful to go back and look at the proof you know.
jquark
#8
Apr26-08, 04:54 PM
P: 18
i just now happened to have to integrate a cross product. this particular one is from celestial mechanics.

the cross product of the sum of the system's m*v cross a, i had to think a minute.

then i remembreed because i was doing some of these. that the derivative of the cross product follows Liebnitz' rule....

d ( u x v ) = du x v + u x dv.

so the antiderivative would be something that had produced x x a. d(x x v) has an x x a term in it, what of the other term?

dx * v = v * v. putting two of the same thing in the cross product determinant formula duplicates a line in the matrix rendering the determinant zero.

so, the integral of x * a, is x * v.

q x d. (eyecross!)

ok. i didnt know it either, and i had opened IE to look for it. then i did the antiderivative, and then i did look for it and found my home for goodies like this right on the search results, so i came in :)

let me try to get the latex to work.

(oh, summing the system's values is just that sum. placing the masses in doesnt matter to the integral, it was the cross product i had to think up.

[tex]f=\int\x\times\a\dt[\tex]

[tex]f=\int\x\times\a\da[\tex]

[tex]d\(\x\times\v\)\=\dx\times\v\+\x\times\dv\[\tex]
[tex]dx\times\v=v\times\v=0[\tex]
[tex]dv=a[\tex]
[tex]int\x\times\a\dt\=dv\times\a[\tex]
trambolin
#9
Apr27-08, 10:41 AM
P: 341
Quote Quote by jquark View Post
i just now happened to have to integrate a cross product. this particular one is from celestial mechanics.

the cross product of the sum of the system's m*v cross a, i had to think a minute.

then i remembreed because i was doing some of these. that the derivative of the cross product follows Liebnitz' rule....

d ( u x v ) = du x v + u x dv.

so the antiderivative would be something that had produced x x a. d(x x v) has an x x a term in it, what of the other term?

dx * v = v * v. putting two of the same thing in the cross product determinant formula duplicates a line in the matrix rendering the determinant zero.

so, the integral of x * a, is x * v.

q x d. (eyecross!)

ok. i didnt know it either, and i had opened IE to look for it. then i did the antiderivative, and then i did look for it and found my home for goodies like this right on the search results, so i came in :)

let me try to get the latex to work.

(oh, summing the system's values is just that sum. placing the masses in doesnt matter to the integral, it was the cross product i had to think up.

[tex]f=\int{x\times a dt}[/tex]

[tex]f=\int{x\times a da}[/tex]

[tex]d(x\times v) = dx\times v+x\times dv[/tex]
[tex]dx \times v = v\times v=0[/tex]
[tex]dv = a[/tex]
[tex]\int{x\times adt} = dv\times a[/tex]
Just to show the correct code. Did not touch the elements in it, and they don't make too much sense to me...
jquark
#10
Apr28-08, 01:56 AM
P: 18
that one got away from me. i spent my 30 minutes trying to get it right and the system timed me out on editing my post. i left it without a follow up.

integral cross product seems to me to be for example int x * a dt pos * acc, that by the liebnitz rule of distributint pairs such as d(u*v) we have du * v + u * dv, that that also happens with the cross product, and i had noticed working in the conservation energies that d ( x * v ) pos times velocity, we have dx * v + x * dv. dx is v of course, and dv is a, so, that makes v * v + x * a. but, in the case of cross product, a self similar product attempt yields zero, so all we have is x * a. the integral of x * a then would be x * v.

but my example that i was working with had a nice disappearing term in it.
timewalker
#11
Oct16-11, 01:29 PM
P: 15
I saw a similar problem in electromagnetism. Consider that we want to derive the expression of magnetic dipole moment of a certain circuit which carrying uniform current. First, fix a reference point(actually this is not necessary because we will think about uniform magnetic field case) at anywhere in the space. Assume that we have a uniform magnetic field [itex]\vec{B}=\vec{B_0}[/itex]. For each differential current element, the magnetic field exerts magnetic induction.
[itex]d \vec{F} = I d \vec{l} \times \vec{B}[/itex]
Therefore, we can calculate the torque for each differential current element with respect to our reference point.
[itex]d \vec{\tau} = \vec{r} \times ( Id \vec{l} \times \vec{B} ) [/itex]
So, the total torque on circuit is its integral over whole circuit.
[itex]\vec{\tau} = \oint d\vec{\tau} = I \oint \vec{r} \times ( d \vec{l} \times \vec{B} ) [/itex]
Now, I will try to factor out [itex]\vec{B}[/itex] from the integration. Then it will give us wrong answer.
[itex]I \oint \vec{r} \times ( d \vec{l} \times \vec{B} ) = I \oint \vec{B} \times ( d \vec{r} \times \vec{l} ) \neq I \vec{B} \oint d \vec{r} \times \vec{l}[/itex]
This can be easily proved by expanding both sides.(or use the fact [itex]\vec{m} = \frac{I}{2} \oint (\vec{r} \times d \vec{l}) = I \vec{A} [/itex] and [itex]\vec{\tau} = \vec{m} \cdot \vec{B}[/itex]) Actually this is 2 times larger than the correct answer.
[itex]\oint \vec{B} \times ( \vec{r} \times d \vec{l} ) = 2 \vec{B} \times \oint (\vec{r} \times d \vec{l} ) [/itex]
Knowing [itex]\vec{A} = \frac{1}{2} \oint (\vec{r} \times d \vec{l} )[/itex] will be helpful.


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