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F=ma and e=mc^2?

  1. Mar 6, 2013 #1
    Could this two equations be used together, since they have a common variable "m"?

    For example, you could deduce that the speed of light = sqrt(a*m), and since the speed of light is a constant, if acceleration goes up, then mass must go down.

    Is this true?
  2. jcsd
  3. Mar 6, 2013 #2
    Not really. You see, in e=mc^2 , e is the rest energy of a body and m is the rest mass of the body, which is different from the mass when the body is in motion. This is derived in special relativity as a result of the speed of light being the same for all inertial observers. You can research more: look up Lorentz transformation. You'll get the derivation of E=mc^2.

    As for F=ma , this is Newtonian mechanics which is completely different from special relativity. Here, m is the mass of the body which is assumed to be unchanged even if the body is moving. Therefore, the m in each equation aren't really the same.

    (That's what I think. Maybe someone can further elaborate on this.)
  4. Mar 6, 2013 #3


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    Sure. Let's say you don't know the mass of an object. If you apply a force to it and measure the acceleration, you can get the mass. Knowing this you could use E=MC2 to find the energy content of the objects mass. In this case E=(F/A)C2

    Not quite. In both equations the M stands for mass. Mass in the equation E=MC2 is also know as "rest mass" or "invariant mass". It does not change when you change the momentum of an object. In fact, the equation is actually incomplete.
    It's true form is E2=M2C4 + P2C2, where P is the objects momentum. Increasing or decreasing the velocity of an object changes its momentum, and thus its energy content.
  5. Mar 6, 2013 #4
    No, because [itex] F = ma [/itex] is not correct in special relativity. The correct version of Newton's law is

    [tex] F_\nu = \frac{dp_\nu}{d\tau} [/tex]

    Which are four-vectors and proper time respectively.
  6. Mar 7, 2013 #5
    Drakkit, dipole is right. The method you explained works only for very small speeds compared to that of light. F=ma is no longer accurate at very high speeds. You're not taking into account the the different inertial frames of reference.

    For instance you cannot use F=ma for an object travelling at 99.99% the speed of light. Here you'll need to use four-vectors.
  7. Mar 7, 2013 #6


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    Ah, that makes sense. I was only thinking of a local reference frame where you are at low velocities, such as measuring the acceleration of a ball in a lab, not one where the object was at relativistic speeds.
  8. Mar 7, 2013 #7
    I agree with the above and would like to add that if you try to solve F=m*a in relativity then you will obtain a different mass for transverse accelerations and different for longitudinal.
    In general, E=mc**2 doesn't refer to the kinetic energy of a particle or something. It is its energy content, the energy that it will unleash if it annihilates with an antiparticle.
    It's actually different physics.
  9. Mar 7, 2013 #8
    You do not need four-vectors and proper time. The original F=dp/dt also works in special relativity. It results in

    F = [a+v·(v·a)/(c²-v²)]·E/c²
  10. Mar 7, 2013 #9
    You just have to use the right p! (or m)
  11. Mar 7, 2013 #10
    IMO, the basic premise of your argument;

    is true, despite 'math-osophical' issues.

    If you take the mass of light as 0, then your equation: 'c = sqrt (a*m)' = 0. //(unless sqrt 0 = infinity? any mathematicians there to help? BODMAS starts with B right??)

    I suppose Newton took a 'perfect' motive mass in an earth like atmosphere to be invariant, as Einstein took light in 'perfect' vacuum to be invariant.. (but that's just retroactive armchair psychology!).
  12. Mar 8, 2013 #11
    Defining force as F=dp/dt with invariant mass would be foolish and Newton wasn't a fool. But I guess he considered variable mass for open systems only because I can't imagine that he was in doubt about Galilei transformation.
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