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Integration of an acceleration formula involving vectors

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose a constant force F acts on a particle of mass m initially at rest.

    (a) Integrate the formula for acceleration [tex]\vec{a} = \frac{\vec F}{\gamma m} - \frac{\vec v}{\gamma mc^2}(\vec F \cdot \vec v)[/tex] where [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] to show that the speed of the particle after time t is given by [tex]\frac{v}{c} = \frac{(F/m)t}{\sqrt{(F/m)^2t^2+c^2}}[/tex]


    2. Relevant equations
    Given above

    3. The attempt at a solution
    I have taken Calc. 1-3, differential equations, and linear algebra, and yet I still do not know how to integrate the formula for the acceleration [itex]\vec a[/itex] shown above. I don't think that I've ever integrated vectors, and especially not dot products; so I have no idea where to even start. This is part of a special relativity section for an bachelor's level Astrophysics course. Any help would be greatly appreciated!
    Thanks!
     
  2. jcsd
  3. Jan 27, 2015 #2

    haruspex

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    Since F is a constant vector, it can be factored out of the integral: ##\int \vec F.\vec v = \vec F. \int \vec v##.
     
  4. Jan 27, 2015 #3
    Yes, but now I guess I would have (not sure about the dotted [itex]\vec v[/itex] 's in the integral):
    [tex]\int \vec a \, dt = \vec v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int \vec v \cdot \vec v \, dt[/tex] and I still don't know how to evaluate this with vectors (especially the dot product in the integral), and the equation I'm trying to get to (the v/c equation) has just scalars in it, so I don't know how to get there from all these vectors even if I knew how to evaluate that integral.
     
  5. Jan 27, 2015 #4

    haruspex

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    No, as you suspected, that's not valid.
    Thing to notice is that you are asked to find speed, not velocity, yet you start with acceleration as a vector.
    This suggests that the integration to be performed starts with a scalar on the left, so you have to do something before integrating. Any ideas?
     
  6. Jan 27, 2015 #5
    Well, the only thing that comes to mind would be breaking the vectors into component form, but I'm not exactly sure how to do that with this specific setup. I tried (as an attempt) just making all the vectors scalars and then integrating, but then I got an equation involving v and r (position), and the v/c equation does not involve position. Also, I'm not sure that was the correct thing to do anyway.
     
  7. Jan 27, 2015 #6

    haruspex

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    No, that's not what I had in mind. What's simple equation relating v, as a speed, to the corresponding velocity? What do you get if you differentiate?
     
  8. Jan 27, 2015 #7
    I'm not sure I understand what you mean, here's what I think you mean by the simple equation relating [itex]v[/itex] to [itex]\vec v[/itex] (correct me if I'm wrong):
    [tex]v = |\vec v|[/tex]
     
  9. Jan 27, 2015 #8

    haruspex

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    Almost. Slightly simpler (removes the modulus signs) if you square both sides: ##v^2 = \vec v.\vec v##. Now differentiate.
     
  10. Jan 27, 2015 #9
    Ok, so do you mean to replace [tex]v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int \vec v \cdot \vec v \, dt[/tex] with this [tex]v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int v^2 \, dt[/tex]? If so, how do I make the force vector just the force?
     
  11. Jan 27, 2015 #10
    Oops, sry... read your "differentiate" as "integrate" lol, let me differentiate
     
  12. Jan 27, 2015 #11
    [tex]\frac{d}{dt}[\vec v \cdot \vec v] = \frac{d}{dt}[v^2] = 2v\frac{dv}{dt}[/tex]
     
  13. Jan 27, 2015 #12

    haruspex

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    Right, but you can also perform the derivative on the left, to get an expression involving vectors, velocity and acceleration.
     
  14. Jan 27, 2015 #13
    On the left of which equation? (Also, thanks for helping me out so far)
     
  15. Jan 27, 2015 #14

    haruspex

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    Perform the derivative ##\frac d{dt} (\vec v . \vec v)##, without converting to scalars first. Use the product rule.
     
  16. Jan 27, 2015 #15
    Alright, I get
    [tex]\frac{d}{dt}[\vec v \cdot \vec v] = 2\vec a \cdot \vec v[/tex]
     
  17. Jan 27, 2015 #16

    haruspex

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    Right, so how can you use that with your starting equation?
     
  18. Jan 27, 2015 #17
    I'm sorry, but I am not sure. Are you talking about the problem's equation for [itex]\vec a [/itex]? If so, everything I try doesn't really get me anywhere.
     
  19. Jan 27, 2015 #18

    haruspex

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    Yes. What do you have to do to turn the left hand side into ##\frac d{dt} (\vec v . \vec v)##?
     
  20. Jan 27, 2015 #19
    The only thing that comes to my mind is solving [itex]\frac{d}{dt}[\vec v \cdot \vec v] = 2\vec a \cdot \vec v[/itex] for [itex]\vec a[/itex] and substituting it in the starting equation, but this involves "dividing" by a vector, which doesn't make sense. I am sorry that I can't figure this out very easily (I am easily confused).
     
  21. Jan 27, 2015 #20

    haruspex

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    No, you need to do something to both sides of the equation so that the left hand side becomes ##\vec a \cdot \vec v##. Nothing complicated.

    Btw, it seems to me you have to make use of the information that not only is the force constant but the particle starts from rest. What does that tell you about the relationship between the directions of the vectors, F, v, a?
     
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