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Precision of physics formulas

  1. Aug 11, 2009 #1
    Are more complex physics formulas, like those involving exponents, roots, logs, trig functions, etc., approximations to a certain degree? Are even simple formulas approximations?

    If so, what kinda precision: deci-, centi-, milli-, or better?

    Is there a regulated precision for new physics formulas, if any? And which body would regulate?

    When was the last physics formula introduced? Can you provide a web link that documents the history of formulas?
     
  2. jcsd
  3. Aug 11, 2009 #2
    In general, you could say that the actual formulas have infinite mathematical precision. However, error comes into play when there is error in the actual numbers you plug into the equation.

    Look into the difference between ACCURACY and PRECISION as well.

    Let's take a classic example of launching a ball. Knowing the initial speed and angle of launch, your kenimatic equations predict an landing spot with infinite precision. However, how precisely do you know the actual initial speed?
     
  4. Aug 11, 2009 #3
    Solving a projectile problem with kinematics equations is an approximation for two reasons: first, the method assumes that the acceleration of gravity is constant at all altitudes, and, second, it assumes that the gravitational field is in the same direction everywhere, that is, the world is flat. So the equations produce an answer that is a parabola for what is more accurately an ellipse.
     
    Last edited: Aug 12, 2009
  5. Aug 12, 2009 #4

    DrGreg

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    Almost all equations come with assumptions about what can be ignored. They may ignore air resistance or other friction; they may ignore heat loss; they may ignore gravity or local variation of gravity; they may ignore relativity; they may ignore quantum theory, etc, etc. The simple harmonic motion equation for a pendulum is an approximation based on a small angle of swing.

    There is also an issue over how accurate are the numbers you feed into an equation. Obviously, an error in the input will give an error in the output, but sometimes even a tiny input error can give a huge output error. Nothing can be measured to perfect accuracy anyway.
     
  6. Aug 12, 2009 #5
    Your post suggests a complete misunderstanding of what "formulas" are. A physical theory begins with certain postulates, like "light travel at the same speed, c, in all intertial reference frames" and "physics is not dependent on the frame of reference you are in". Expressing those postulate mathematically one can then construct a mathematical theory of that system/phenomenon, and through the mathematical enterprise of proof->theorem->bigger proof -> bigger theorem a framework of RESULTS can be developed.

    Your post makes it sounds like "formulas" come by mashing together variables until they luck upon a combination that seems to work or something, which is silly.

    Take your standard kinematic equation: [itex]v_f=v_i+at[/itex] where did this COME from? Did physicists just randomly toss together variables? Absolutely not. It was derived with zero reference to the physical world, it is merely a result of calculus. The acceleration of something is the change in its velocity in time or the change in the change of position in time. So as a derivative we express that as [itex]\frac{d^2p(t)}{dt^2}[/itex] now if we are considering the case where acceleration is a constant (which is what that "formula" is) then we say [itex]\frac{d^2p(t)}{dt^2}=a[/itex] where a is a constant. We take an integral (the opposite of a derivative) we get [itex]\frac{dp(t)}{dt}=at+c[/itex] where c is a constant. So what is c? Well we can find out by looking at the case when time is zero (i.e. at the beginning. When t=0 we get [itex]\frac{dp(0)}{dt}=a(0)+c=c[/itex] so c is the the change of position with respect to time at t=0. Well the change in position with respect to time is the velocity!. So c is the initial velocity. So if we re-write [itex]\frac{dp(t)}{dt}[/itex] in the more obvious form as v(t) (the velocity at a time t) or simply as v_f (if the t we're looking at is the final t) then we get [itex]v_f=v_i+at[/itex] and that's where it comes from. Just mathematically considering the case where acceleration is constant and doing two integrals to get it in a more convenient form. There was no guess work. The result is as infinitely accurate as the assumptions. If the acceleration were truly constant this would be 100% correct ( classically anyways). However, if acceleration isn't constant then it doesn't apply. If you incorrectly assume acceleration is constant when it isn't THAT'S where you get error. However the "formula" is an infinitely precise and accurate result of the mathematical assumptions from which it was derived. But it can only be as solid as those assumptions

    This is how all of physics is done. To the best of your knowledge we have infinite accuracy when we apply our physics to a situation where the assumptoins are infinitely correct.
     
    Last edited: Aug 12, 2009
  7. Aug 12, 2009 #6

    Andy Resnick

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    Just to jump off of maverick_starstrider's post, there's also differences between a formula like "F=ma", which is a *definition*, and F = kmM/r^2, in which there is considerable research regarding the 'exactness' of the number '2'. This is also seen in scaling, where there are various critical exponents which obey simple rules (see, for example):

    http://en.wikipedia.org/wiki/Phase_transition
     
  8. Aug 12, 2009 #7
    Right, but the exactness of the 2 comes down to the correctness of the assumption that the force is spherically symmetric and conservative.
     
  9. Aug 12, 2009 #8
    In F=ma, isn't it discovered from experience that the acceleration resulting from a push is proportional to the push that caused it? I mean, being born into the world with a blank slate for a mind, someone might have thought that the inertia that must be overcome to produce acceleration depends on the present motion, or depends on the past motion, or something else.
     
  10. Aug 12, 2009 #9

    f95toli

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    Right, but how do you define "push"?
    If you want you can think of F=ma as a part of the defintion of what force is.
    Alternatively you can think of it is as the definition of inertial mass.
    Classical mechanics is not nearly as "logically consistent" as we are sometimes lead to believe. In fact, the debate over the various "interpretations" of classical mechanics was still going on when quantum mechanics came along (Mach's "The Science of Mechanics" was published around 1900).
     
  11. Aug 12, 2009 #10

    Andy Resnick

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    I think that's the argument for why it's 2.00000..., but there's research regarding the breakdown of that exponent at short distances. Not sure what the lower limit (of the exponent being 2 +/-) is.
     
  12. Aug 12, 2009 #11
    I really don't like saying the formula is 'infinitely precise and accurate result'. This doesn't make any sense. The result of the formula is never "infinitely" anything. I don't see what point that derivation of constant acceleration serves either (sorry).

    Ans: The model structure used (either experimentally or theoretically determined) is only as good as its predictive capability. You can use a statistical metric like the goodness of fit value (R^2) to quantify it. How valid are the assumptions? .....how well does it predict the data from new runs. That's the answer.
     
    Last edited: Aug 13, 2009
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