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The trivial constant force acceleration math

  1. Jun 24, 2009 #1
    I accelerate a body by a constant force:
    [tex]a = F/m = F \frac{\sqrt{1-v^2/c^2}}{m_0} [/tex]

    I simplify it by fixing F = m0 = c = 1:
    [tex]\frac{dv}{dt} = {\sqrt{1-v^2} [/tex]

    This diff equation formalizes the dependence of relativistic body acceleration on its velocity. To get the speed at time t, I solve it rearranging into

    [tex] \int{ \frac{dv}{\sqrt{1-v^2}} = t [/tex]

    , which is a handbook integral: t = arcsin v, or v = sin t. This 1) satisfies the equation and, as the Einstein's correction of Newton implies, 2) slows the initially constant acceleration down to zero as v approaches 1 and 3) precludes super-light speeds. However, sine reaches v=1 in finite amount of time while texts tell that we should approach the speed of light asymptotically in t = ∞. Oscillations is not what I expected. Where is the mistake?
     
    Last edited: Jun 24, 2009
  2. jcsd
  3. Jun 24, 2009 #2
    When a relativistic particle is accelerated, both the velocity and dynamic (not rest) mass changes. We usually use the momentum p= mv multiplied by a constant, c, to give
    pc = mvc = βmc2
    So for a constant force,

    d(pc)/dt = d(βmc2)/dt = const = d(βγm0c2)/dt = m0c2 d(βγ)/dt, where m0c2 is the rest mass in energy units.

    It is sometimes easier to use the relation

    E2 = (pc)2 + (m0c2)2
    where E is total energy, and E-m0c2 is kinetic energy.

    α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
     
  4. Jun 24, 2009 #3

    DrGreg

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    The mistake is that F = ma is not valid in relativity. But F = dp/dt is. See post #16 in "A dark part of special relativity(at least for me)" for details.
     
  5. Jun 24, 2009 #4
    Thanks, Greg. I understand the mistake now and that Bob tells me how to derive the a. The v(t) solution now turns to out to be non-trivial.
     
    Last edited: Jun 24, 2009
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