Integration involving a dot product

  • Thread starter mmwave
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Main Question or Discussion Point

I am trying to do the following vector integral. When the force and the initial motion are along the same line it is easy but when they are not I can't do the integration. Hints would be appreciated.

[tex] \vec F = \frac {d} {dt} \vec p [/tex] where [tex] \vec F = (0, F_{y},0) [/tex] and [tex] \vec {p} [/tex] can be written as [tex] \vec p = m_{o} \vec u / \sqrt {1 - u^{2}/c^{2} }[/tex]
At time t = 0, [tex] \vec u = (0,u_{o},0) [/tex]

When I replace p with its equivalent in u I get into trouble because [tex]u^{2} = \vec u \cdot \vec u [/tex] and so both integrals involve both x & y components.
 

Answers and Replies

  • #2
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What you need to do is calculate the dot product, prior to integrating. By doing so you will reduce the problem of integrating vectors to a problem of integrating each component (which is a scalar) on itself.

Give the complete integral expression and i will show it to you. Beware that the integrating-differential is also a vector. For example, in the case of a 2-dimensional integration, it is a vector that is pointed perpendicular on the surface over which you integrate, directed outwards of the surface if it is closed. This is just like calculating the flux of a vector field, in the direction of the field that is perpendicular to the surface over which you integrate.

marlon
 
  • #3
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[tex]\int \vec{F} \cdot d \vec{\l} = \int (F_x,F_y,F_z) \cdot (d_x,d_y,d_z)[/tex]
Now, you just have three integrations once you calculated the dot product.

This is just an example, you know that you can also calculate a dot product by using the cosines of the angles between the two vectors. the rest is the same...

marlon
 
  • #4
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marlon said:
What you need to do is calculate the dot product, prior to integrating. By doing so you will reduce the problem of integrating vectors to a problem of integrating each component (which is a scalar) on itself.

Give the complete integral expression and i will show it to you. Beware that the integrating-differential is also a vector. For example, in the case of a 2-dimensional integration, it is a vector that is pointed perpendicular on the surface over which you integrate, directed outwards of the surface if it is closed. This is just like calculating the flux of a vector field, in the direction of the field that is perpendicular to the surface over which you integrate.

marlon
But taking the dot product of the speed doesn't separate the components in the integral. The square root term depends on the magnitude of u, and so each integral depends on all three terms. That's where I get stuck when I substitute for p.

mmave.
 
  • #5
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mmwave said:
But taking the dot product of the speed doesn't separate the components in the integral. The square root term depends on the magnitude of u, and so each integral depends on all three terms.

mmave.
That is no problem. In an integral like [tex]\int \frac{1}{\sqrt{x^2+y^2}}dx[/tex] you can treat the y variable as an ordinary constant.

It still is not clear to me what the exact integrandum is, please write down the integral completely

marlon
 
  • #6
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  • #7
rachmaninoff
HungryChemist said:
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Correct! The answer is, 23 dots.
 
  • #8
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mmwave said:
[tex] \vec F = \frac {d} {dt} \vec p [/tex] where [tex] \vec F = (0, F_{y},0) [/tex] and [tex] \vec {p} [/tex] can be written as [tex] \vec p = m_{o} \vec u / \sqrt {1 - u^{2}/c^{2} }[/tex]
At time t = 0, [tex] \vec u = (0,u_{o},0) [/tex]

When I replace p with its equivalent in u I get into trouble because [tex]u^{2} = \vec u \cdot \vec u [/tex] and so both integrals involve both x & y components.
The integral is with respect to t and u is the speed.
[tex] \int \frac {d} {dt} \vec p dt = \int \frac {m_{o} \vec u} {\sqrt {1 - u^{2}/c^{2} }} dt[/tex]
 
  • #9
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mmwave said:
The integral is with respect to t and u is the speed.
[tex] \int \frac {d} {dt} \vec p dt = \int \frac {m_{o} \vec u} {\sqrt {1 - u^{2}/c^{2} }} dt[/tex]
there is not dot product here, this is a vector integral. Just calculate the integral for each component separately and you are done. I have explained this in previous posts

regards
marlon
 
  • #10
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marlon said:
there is not dot product here, this is a vector integral. Just calculate the integral for each component separately and you are done. I have explained this in previous posts

regards
marlon
I don't think that is correct. The variable of integration is t. Each component, ux, uy, uz depends on the speed u^2 = (ux^2 + uy^2 + uz^2) where each component is a function of t. It does not split into 3 separate integrals. For the ux component:

[tex]\int \frac{m_{o}u_{x} dt } { \sqrt { 1 - (u_{x}^2 + u_{y}^2 + u_{z}^2) / c^{2}}}} [/tex]

This is not the case if you integrate wrt x or y or z.

mmwave.
 
  • #11
LeonhardEuler
Gold Member
859
1
Your right mmwave. But nothing else can be said about how to do the integral without knowing each of the components of u as a function of t.
 
  • #12
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mmwave said:
I don't think that is correct. The variable of integration is t. Each component, ux, uy, uz depends on the speed u^2 = (ux^2 + uy^2 + uz^2) where each component is a function of t. It does not split into 3 separate integrals. For the ux component:

[tex]\int \frac{m_{o}u_{x} dt } { \sqrt { 1 - (u_{x}^2 + u_{y}^2 + u_{z}^2) / c^{2}}}} [/tex]

This is not the case if you integrate wrt x or y or z.

mmwave.
in that case you need to know how the components depend on t...

marlon
 
  • #13
211
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Saddly, this was a differential equation that is supposed to give me as an answer the function [tex] \vec u(t) [/tex]. I went back to a clean sheet of paper and tried not substituting for p(t). The answer is looks trivial but isn't and you can get [tex] \vec u(t) [/tex] from [tex] \vec p(t) [/tex] and the answer checks. I can see why it can't be done with [tex] \vec u(t) [/tex] though. Thanks for the help.

mmwave.
 
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