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Homework Help: Another Integral

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Sove the differential equation below. Limits, t= 0 to t [itex]\rightarrow \infty[/itex]

    2. Relevant equations

    a0 is a given value (say, 9.8 m/sec2) at t = 0

    dv/dt = a0[itex]\sqrt({c^2 - v^2/c^2})[/itex]

    3. The attempt at a solution

    sin-1(v/c) = a0t/c

    v/c should go from 0 to 1. My solution does not come up with that.
  2. jcsd
  3. Aug 30, 2010 #2
    If your equation is relating physical quantitites, then, by dimensional analysis, you must have:

    [c^{2}] = [\frac{v^{2}}{c^{2}}]

    [c] = [v]^{1/2}

    and then (assuming t is time):

    \frac{[v]}{\mathrm{T}} = \frac{\mathrm{L} \, \mathrm{T}^{-2}}{[v]^{1/2}}


    [v]^{3/2} = \mathrm{L} \, \mathrm{T}^{-1}

    [v] = \left(\mathrm{L} \, \mathrm{T}^{-1}\right)^{\frac{2}{3}}

    and, the dimension of c:

    [c] = \left(\mathrm{L} \, \mathrm{T}^{-1}\right)^{\frac{1}{3}}

    I don't know of any physical quantity with these dimensions. Can you please explain?
  4. Aug 30, 2010 #3
    Let's start over again:

    a0 is a given value (say, 9.8 m/sec2) at t = 0

    dv/dt = a0[itex]\sqrt({c^2 - v^2/c^2})[/itex]

    dv/[itex]\sqrt({c^2 - v^2})[/itex] = a0dt/c

    sin-1 (v/c) = a0t/c

    Now, something I have done is wrong... what?
  5. Aug 30, 2010 #4
    try getting rid of the inverse sine by taking the sine of both sides. everything youve done looks valid so far. what do you mean it doesnt work?
  6. Aug 30, 2010 #5
    OK -

    (v/c) = sin (a0t/c)

    t starts at zero and goes to infinity. As it does, v/c starts at zero and ends at c (when t is infinite.) It is supposed to be a monotonic increasing function. The one depicted above by me is not. As t advances, the sine oscillates from -1 ...0 ... +1 ... 0 ... -1 etc.

    We have the v/c restricted to -1 ... +1 but this is not considtent with the advancing t's and it is not ever supposed to decrease.
  7. Aug 30, 2010 #6
    Where did you get this formula:

  8. Aug 31, 2010 #7
    Was afraid you would ask that...

    That is a formula I concocted to mimick a particle of mass mstarting at ground zero acceleration (9.8 m/sec2) being pushed indefinitely over t forever (t --> infinity.) This takes into account the relativistic momentum with the infamous
    m = m0/(sqrt{(1 - v^2/c^2)}[/itex]) equation we learned in high school (used to be called "relativistic mass" but now that term is verboten.

    Basicallly it states that at a given instant, a particle which had an initial acceleration (a0) of 9.8 m/sec2 or 1 g now has a reduced acceleration of
    a0[itex]\sqrt{(1 - v^2/c^2)}[/itex] depending on the achieved velocity. The dv/dt represents the acceleration at that point in time.

    When you integrate or solve the diff eq you do get the restriction of |v/c| [itex]\leq[/itex] |1| which is true but it should NOT be periodic as once a force is applied, there should be no reduction in |v/c|.
  9. Aug 31, 2010 #8
    Oh, then it's incorrect. You should use a transformation law that relates the acceleration of a particle with respect to a stationary reference frame to the acceleration of the particle with respect to a co-moving inertial reference frame.

    You should tak that [itex]a_{0}[/itex] is actually the proper acceleration measured by this co-moving frame, then you will get the correct equation of motion.

    So, do you know the acceleration transformation law in relativistic dynamics?
  10. Aug 31, 2010 #9
    I knew there was something wrong with what I did. No, off the top of my head I do not know this transformation of which you speak. I have Taylor/Wheeler's 1960s book of Spacetime Physics as well as French's book of Special Relativity (1979.)

    If you could point the way and maybe supply me with a small simple numeric example that would be most helpful.

    This is precisely why I posted this question on this thread.

    I am not afraid of hyperbolic functions but straight algebraic ones are more intuitive.

  11. Aug 31, 2010 #10
    Don't leave me now, Dickfore! I need you to finish what you were saying as that will be a learning jump for me...

  12. Aug 31, 2010 #11
    Ok, do you know Calculus? If you do, you can derive everything starting from Lorentz transformations? Do you know them?
  13. Aug 31, 2010 #12
    Yes, I do know calculus and I do know the Lorentz transformations. It's putting them together that is the problem and that is where my initial error started.

    That;s what I need the help in - choosing the right path. If I had known the right path and been able to derive v in terms of t, I wouldn't have been asking the question.

    x' = [itex]\gamma[/itex](x - vt)
    t' = [itex]\gamma[/itex](t - vx/c2)

    where v = the relative velocity of S' to S.

    Einstein derived these equations without any supposition that v < c. It was the
    [itex]\gamma[/itex] = 1/[itex]\sqrt{(1 - v^2/c^2}[/itex]) which comes along in the development of this proof that put the kibosh on v ever > c.
    Last edited: Aug 31, 2010
  14. Aug 31, 2010 #13
    Ok, now, imagine a particle is at a point with coordinates [itex](x, y, z)[/itex] at some instant [itex]t[/itex] according to the frame S. Since the particle is moving, it will, in general, have different coordinates at an infinitesimally later time [itex]t + dt[/itex]. Due to the finite speed of propagation of the particle, each of the coordinates has to change by an infinitesimal amount. Hence, the coordinates of the particle at a later instant are [itex](x + dx, y + dy, z + dz)[/itex]. By definition, we call the following ratios:

    v_{x} = \frac{dx}{dt}, \; v_{y} = \frac{dy}{dt}, \; v_{x} = \frac{dz}{dt}

    the components of the velocity of the particle (relative to the frame S).

    Using the Lorentz transformation, we may write for the infinitesimal increments of the coordinates and time in the frame S':

    dx' = (x' + dx') - x' = \gamma \, \left[(x + dx) - \beta \, c \, (t + dt) \right] - \gamma \left(x + \beta \, c \, t \right) = \gamma \, \left(dx - \beta \, c \, dt \right)

    dy' = (y' + dy') - y' = (y + dy) - y = dy

    and similarly

    dz' = dz

    and, finally:

    dt' = (t' + dt') - t' = \gamma \, \left[(t + dt) - \frac{\beta}{c} \, (x + dx) \right] - \gamma \, \left( t + \frac{\beta}{c} \, x \right) = \gamma \, \left(dt - \frac{\beta}{c} \, dx\right)

    I have written everything out, but notice that what we are doing is expressing the differentials of [itex]x', y', z', t'[/itex] as functions of [itex]x, y, z, t[/itex] according to the rules of Multi-Variate Calculus.

    Then, according to S', the components of the velocity are:

    v'_{x} = \frac{dx'}{dt'}, \; v'_{y} = \frac{dy'}{dt'}, \; v'_{z} = \frac{dz'}{dt'}

    Substituting for all of the above differentials, we get:

    v'_{x} = \frac{dx - \beta \, c \, dt}{dt - \frac{\beta}{c} \, dx}

    v'_{y} = \frac{dy}{\gamma \, \left(dt - \frac{\beta}{c} \, dx \right)}

    and, similarly for [itex]v'_{z}[/itex]:

    v'_{z} = \frac{dz}{\gamma \, \left(dt - \frac{\beta}{c} \, dx \right)}

    In all of these equations [itex]\beta = \frac{u}{c}[/itex], where [itex]u[/itex] is the relative velocity of the frame S' with respect to S and [itex]\gamma = (1 - \beta^{2})^{-1/2}[/itex].

    If you divide each term in both the numerator and the denominator by [itex]dt[/itex] and use the definitions for the components of the velocity in the frame S, then you will get:

    v'_{x} = \frac{v_{x} - u}{1 - \frac{v_{x} \, u}{c^{2}}}

    v'_{y} = \frac{v_{y} \sqrt{1 - \frac{u^{2}}{c^{2}}}}{1 - \frac{v_{x} \, u}{c^{2}}}

    v'_{z} = \frac{v_{z} \sqrt{1 - \frac{u^{2}}{c^{2}}}}{1 - \frac{v_{x} \, u}{c^{2}}}

    These are the transformation formulas for the components of the velocity. Notice that the formula for the z-component is the same as for the y-component by making the substitution [itex]y \rightarrow z[/itex] everywhere. You can get the inverse transformations (from S' to S) in two ways. Either solve the above equations for the unprimed velocity components or simply use the same formulas with the place of the unprimed and primed components interchanged and [itex]u \rightarrow -u[/itex]. You should get the same result.

    Now comes your question:
    1. Find the differential [itex]dv'_{x}[/itex] (assuming [itex]u[/itex] is constant);

    2. Find the differentials [itex]dv'_{y}[/itex] and then [itex]dv'_{z}[/itex] by analogy;

    3. Use the definition of acceleration in S' and S and the above expression for [itex]dt'[/itex] to find the acceleration components transformation formulas.
  15. Aug 31, 2010 #14
    Ill restrict to ONE dimension (say the x dimension) and work on that. Right now I have to watch Becker - my role model.
  16. Sep 2, 2010 #15
    After I differentiate 1) [I don't need to worry about 2) and 3)] - I know
    denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared - how painful does it get?

    What will I have? Where does it get me?

    Remember my question was take a mass m0 and push it with force f. With m0 at rest the initial acceleration will be a0. As m0 moves (gets a velocity v) its acceleration will decrease as v increases. They used to call it "relativistic mass" but that is a forbidden term.
    The old equation used to be: m = m0/[itex]\sqrt{(1 - v^2/c^2)}[/itex]

    Its acceleration at time t will be: a0[itex]\sqrt{(1 - v^2/c^2)}[/itex]

    This, we have the differential equation:
    a = dv/dt = a0[itex]\sqrt{(1 - v^2/c^2)}[/itex]

    That is where I got the brainy idea to solve that diff eq to get the v after a force f, which gave an initial acceleration of a0 when the object was at rest, was applied for time t. But then the solution with the arcsin answer gave a periodic function which didn't make any sense at all. One intuitively knows that even though the acceleration is getting smaller the faster the object is moving that it is still accelerating and has a limit of c if t is infinite.
  17. Sep 4, 2010 #16
    The final solution as posted by yuiop. I don't know how he got it but that is the answer.
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