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Physics Formulas To Proof

  1. Sep 21, 2008 #1
    zE=kE=mv² / 2,
    F=kp1 q2/d²,

    Could you proof the formulas above?and also Galileo's h=gt²/2
    Can you proof them?
  2. jcsd
  3. Sep 21, 2008 #2
    Physical formulae can be "Proved" by measuring them against real world, however all of the equations you have listed have their limits, and only apply in certain (although the vast majority of common) situations.

  4. Sep 21, 2008 #3


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    What you are asking here is extremely vague and could be based on a misunderstanding, or even simply based on semantics.

    One cannot prove something in physics the way one can in mathematics. This is because physics is a description based on our understanding of the physical world. And this understanding changes all the time because we keep learning more and more of our world, and thus, causes refinement in what we know.

    Now, if you're asking of one can DERIVE such equations, then you will need to provide the STARTING POINT of such derivation, i.e. how far back, or at what level do you want to use as the starting point? Note that in many cases in physics, the equations are purely phenomenological in nature, i.e. they are not derived from any starting point. Classical Coulomb's Law (one of the equation you listed) is one such example. The most general form of this equation cannot be classically derived. It can be presented in various forms depending on the problem (such as for a point charge), but the general form is phenomenological.

    When you provide an equation, especially when it is either non-standard, or have been adapted to a particular situation, please provide the definitions for all the symbols involved. While practically everyone knows what "F=ma" is, "fx", "brE", and "zE" are all meaningless.

  5. Sep 21, 2008 #4
    Although all the posts above me are right (i.e. in physics something is proven if it works against observation), I guess a few of those formulas stem from more fundamental ideas, which cannot be proven mathematically (either because they're defined to be true or they work experimentally).

    pE=mgh: The definition of work is the line integral Sum F(dot)dr. In the simple case of a constant force in one direction (i.e. gravity), moving an object against the force would result in W = F (Integral) dr = Fd=mad (F can be taken out of the integral because it is constant), and when talking about gravity we know a=g and d=h, so W=mgh.

    kE=mv2/2 : Let's say we have an object moving in the +ve x direction with mass m and a speed v0 and energy E (which we dont know yet). Now let's say we apply a constant force of 1N against the object (-x direction) until it stops. Then F=ma=-1. Integrate this a few times to find the distance it takes for it to stop: mv=-t+C. We know at t=0, the initial velocity of the mass was v0, so C=mv0. We want v=0 (when the object stops), so t=C=mv0. Integrate again you get md=-1/2t2+mv0t, plug in t=mv0: md=-1/2(mv0)2+(mv0)2=1/2(mv0)2, so d=1/(2m)(mv0)2=1/2mv02. By the definition of work used before, W = F (Integral) dr = Fd=(1)(1/2)mv02. So if it took us 1/2mv02 units of energy to stop the object, it (by conservation of energy) but have had energy of 1/2mv02. This derivation makes sense, although it may not be the standard way of doing it since I just made it up...

    F=Kq1q2/d^2: This comes straight from Maxwell's equations (Gauss's Law) applied to a sphere or a point. I'm not gonna try to do it here just because (as I found out from the above derivation), doing math with a keyboard really sucks. What you do is just use Gauss's law on a sphere or point of charge q and find the E fiend around the object. It gets pretty easy since, because of spherical symmetry, you know the field must point radially it or out of the sphere.

    F=GMm/r^2: Again, (although I havent done this one), I'm pretty sure it's just Gauss's Law but applied to a gravity field. Note that both F=GMm/r^2 and F=Kq1q2/d^2 are examples of the "inverse square law", which, in general, described the intensity of something coming from a point particle (or sphere) with respect to distance. (Note that 1/r^2 term is *always* there).
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