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Force & acceleration equation

  1. Aug 1, 2005 #1
    While working on a problem in special relativity I ran into the following equation and am trying to understand it before going on. It was accompanied by the vague statement that at relativisitic speeds, force & acceleration aren't necessarily in the same direction.

    [tex] \vec{F} = m_{o}\gamma \vec{a} + m_{o} \vec{u} \gamma^{3}(\vec{a} \cdot \vec{u}) / c^{2}[/tex]

    [tex] \gamma \equiv \frac{1}{\sqrt{1 - u^2/c^2}}

    To understand this, I apply it to the case where
    [tex] \vec{F} = eE_{o} \hat{y}[/tex] and [tex]u(t=0) = u_{o} \hat{x}[/tex] and [tex]u_{x}(t=0)[/tex] is nearly equal to c.

    Now, looking at the x and y components, since the left side has only y component, so must the right side. Therefore the coefficients of [tex] \hat{x} [/tex] must be zero which gives me the following equation:

    [tex] a_{x} + \gamma^{2}u_{x}a_{x}/c^{2} + \gamma^{2}u_{y}a_{y}/c^{2} = 0 [/tex] with solution [tex]a_{x} = -\gamma^{2}u_{y}a_{y}/c^{2}* \frac {1}
    { (1+\gamma^{2}u_{x}/c^{2})}

    question: Is this the right approach? am I right that the x component must vanish? Notice that for the relativistic case, even though the x component F is zero, there is still an x component to the acceleration.

    I like the sign of ax because it should be negative (the x component of u must decrease as the y component increases since the speed must be less than c) and as c goes to infinity (classical case), ax goes to zero. To do this, there must be a negative x component of the acceleration.
  2. jcsd
  3. Aug 8, 2005 #2
    When I carried on with this method, I got a mess! substituting my expression for ax into the y component gave me a nonlinear differential equation involving both ux and uy and their derivatives. This is way to advanced for the book so I must be using the wrong approach.
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