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A way to gain a formula

  1. Aug 27, 2009 #1

    ShayanJ

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    [tex]\left \begin{array} \mbox{F=ma} \\ \mbox{m=\frac{m_{0},\sqrt{1- \frac{v^2,c^2}}}} \right\{ \Rightarrow F=\frac{m_{0} a,\sqrt{1- \frac{v^2,c^2}}} [/tex]

    Is the above computation a right way to gain formulas in physics and is the gained formula from above computation a right one?
    thanks alot
     
    Last edited: Aug 27, 2009
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  3. Aug 27, 2009 #2

    CompuChip

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    The LaTeX does not render, but I assume you mean: take a classical formula (like F = ma) and replace m by the special relativistic version [itex]m_0 / \sqrt{1 - v^2 / c^2}[/itex] - does that give you a correct formula again?

    The answer would be: "no, not in general", and actually F = m a is a counterexample. Indeed, if we define [itex]\gamma = 1/\sqrt{1 - v^2/c^2}[/itex], then the correct formula is only [itex]F = \gamma m a[/itex] when the force and acceleration are perpendicular to the direction of motion. When they are parallel to it, the correct formula is [itex]F = \gamma^3 m a[/itex].
    This only follows when you derive the formula from basic principles (in this case, the relativistic postulates); you cannot find it in such an ad-hoc way.
     
  4. Aug 27, 2009 #3

    ShayanJ

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    You mean it is not a proper way for gaining a formula in all cases?
     
  5. Aug 28, 2009 #4

    Hootenanny

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    Indeed, in general one cannot simply slip a gamma term in front of any mass terms than appear to obtain a relativistically correct equation. This can be done in some cases, but you need to know a little bit about the equation you are using. For example, the classical Newton's Second law assumes constant mass. That is, we start for the general form of Newton's Second Law

    [tex]F = \frac{dp}{dt} = \frac{d}{dt}\left(mv\right)[/tex]

    and then assume that m is constant leaving

    [tex]F = \frac{dp}{dt} = m\frac{dv}{dt} = ma[/tex]

    However, if we assume that the mass is not constant, then we cannot take it outside the derivative. However, we can place a gamma inside the derivative

    [tex]F = \frac{d}{dt}\left(\gamma m\boldsymbol{u}\right)[/tex]

    If you then take the derivative and follow it through to the end, you will arrive at the correct formula the CompuChip posted.

    The moral of the story is that you need to understand the formula you are dealing with, before trying to play with it.
     
  6. Aug 31, 2009 #5

    ShayanJ

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    Another question.
    How one finds a proper coefficient for a formula.
    For example we have [tex]G[/tex] in formula [tex]F=G \frac{m1.m2}{r^2}[/tex].How the value of this [tex]G[/tex] where found?
    thanks
     
    Last edited: Aug 31, 2009
  7. Aug 31, 2009 #6

    Hootenanny

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    Newton's gravitational constant is found empirically, that is, through measurement.
     
  8. Aug 31, 2009 #7

    ShayanJ

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    But how?
    that doesn't make sense.
     
  9. Aug 31, 2009 #8

    Hootenanny

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    We take two objects of known masses, m1 and m2, and place them a distance of r apart. We then measure the force between them F. Now the only unknown in Newton's law of gravitation is the constant G, we can therefore find G by plugging the above measurements into the equation.

    See here for more information: http://www.physik.uni-wuerzburg.de/~rkritzer/grav.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
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