It’s that time of year again, when we count down the days to Isaac Newton’s birthday (according to the Julian calendar, anyway), and how better to mark this than with mathematics? Thus, I’ll post an equation a day until either Christmas Eve or I run out of ideas, and talk about what it means and why it’s important for physics.

Since this is, after all, a celebration of Sir Isaac, let’s kick things off with arguably his most famous equation:

i-c034f8aa5270f55ef20355f7a63ba01e-dec01_newton_ii.png

OK, it might not look familiar in this form, but this is, in fact, the full and correct statement of Newton’s Second Law (written in modern notation), which most people know as F=ma, force equals mass times acceleration. Newton himself expressed it thusly:

Lex II: Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.

It’s in Latin, because it was the 1600’s (and also because he was kind of a dick), but this translates to (according to the 1729 translation quoted by Wikipedia, anyway):

Law II: The alteration of motion is ever proportional to the motive force impress’d; and is made in the direction of the right line in which that force is impress’d.

By “motion” Newton meant what is know known as momentum, which gets the symbol “p” in physics for reasons that passeth all understanding. Thus, this translates to the equation above: The time derivative of the momentum of an object is equal to the net force acting upon it. The derivative is the mathematical expression for the “alteration of motion” as time goes by, and the net force is, well, the “motive force impress’d.”

Why is this important? Because it’s the foundation of physics as a mathematical science.

OK, if you want to be picky, it’s his second law of motion, but the first doesn’t have a convenient mathematical representation, so if you want to do anything quantitative, you start with the second law. And this is the absolute cornerstone of what’s now known as classical mechanics: It tells you that if you know the forces acting on an object, then you can predict the resulting change in its momentum. Given the momentum, you can predict the future position, and with that, you can do anything you want.

The F=ma form that everybody learns in grade school (in an ideal world, anyway) is an approximation to the full expression, assuming an object with constant mass and low velocity. If you want to know how high you can throw a baseball, or how quickly you can stop a car, this is the form to use.

But the full equation encompasses much more than that. It works for systems whose mass is changing– for example, a rocket, which propels itself through the air by burning fuel and expelling the hot exhaust out the back of the rocket. As it goes along, the mass of the rocket decreases, so you can’t use F=ma to find the acceleration (unless you do it in tiny little steps), but you can use the derivative form with no problem. So, rocket science starts with this equation right here.

The full expression also works for any speed you like. If you’re talking about something like a baseball or a car, moving at speeds very slow compared to the speed of light, you can get away with defining momentum as mass times velocity. If you look at things moving at really high speeds, though, like a proton in a particle accelerator, or a really clever dog with access to alien propulsion technology, you need to use the full relativistic definition of momentum, which is more complicated. But Newton’s second law, in the derivative form above, will still work: if you know the forces that act on an object moving at speeds close to the speed of light, you can still predict its future momentum, and thus its future position.

So there’s a whole lot packed into that tiny little equation. There’s a lot behind it, too, since in order to come up with that, not only did Newton have to shake off thousands of years of Aristotelian thinking about the motion of objects (a process started by Galileo Galilei, who died the year Newton was born), but he needed to invent vector calculus in order to make it work.

So, as we start our countdown to Sir Isaac’s birthday, take a moment to appreciate the power and beauty of his second law. And come back tomorrow to see the next equation of the season.

Comments

  1. #1 Kenneth Cavness
    December 1, 2011

    “…but he needed to invent vector calculus in order to make it work.” — and this is the part that I find so truly awesome.

  2. #2 claudnine9
    December 2, 2011

    I always thought “p” is used for momentum because in some languages it is called “impulse” (or a variation thereof) – “i” and “m” are already taken, so “p” it is….

  3. #3 Frank Wappler
    December 2, 2011

    Chad Orzel wrote (December 1, 2011 10:26 AM):
    > [Newton’s] second law of motion, but the first doesn’t have a convenient mathematical representation

    Arguably:

    Limit{ t –> t0, t < t0 }_[ (p[ t ] - p[ t0 ]) / (t - t0) ] ==
    Limit{ t --> t0, t > t0 }_[ (p[ t ] – p[ t0 ]) / (t – t0) ].

    p.s.

    > I’ll post an equation a day until either Christmas Eve or I run out of ideas, and talk about what it means and why it’s important for physics.

    Looking forward to a discussion of

    d/dt[ -d/dx[ … ] ] == -d/dx[ d/dt[ … ] ]

  4. #4 Frank Wappler
    December 2, 2011

    Oops — did the preview eat my “&.l.t.;”?
    Formula should have been:

    Limit{ t –> t0, t < t0 }_[ (p[ t ] – p[ t0 ]) / (t – t0) ] ==
    Limit{ t –> t0, t > t0 }_[ (p[ t ] – p[ t0 ]) / (t – t0) ].

  5. #5 Neil Bates
    December 2, 2011

    Sure, and f = ma is of course only the classical limit because of SRT (instead, proper force = restmass times proper acceleration, then transform to observers’ frame.) As for dp/dt: even that is tricky because if the mass of “the object” changes, it can be misleading. (Chad mentioned the issue but didn’t catch this problem, see below.) For example, if “the rocket” is losing mass at a rate dm/dt (negative value) because of exhaust, then even classical dp/dt:

    f “=” dp/dt = ma + v dm/dt.

    Well of course that is absurd, the force needed to accelerate can’t vary with “velocity” which is not even an absolute. But the “mass loss” is actually a “reallocation” as some particles leave the rocket, not a literal intrinsic change in mass. If the exhaust is included, it works out.

    However, things still get tricky in SRT for various reasons, there is for example the correction to momentum and mass-energy due to stress in a material. (Think: if you start squeezing a rod simultaneously in its rest frame, no momentum added – but in another frame, those forces didn’t begin to be applied simultaneously. Ahhh ….) There are complicated formulas but a simple version for the obscure (!) “augmented” momentum and energy of a compressed or stretched elastic rod, oriented parallel to observed velocity:

    p_aug = f_elastic * length * gamma*v/c²
    E_aug = f_elastic * length * gamma*v²/c²

    F_elastic is positive in compression there is a shear-force version as well. You get these by using the simultaneity difference in the observing frame, and taking the effect of the longer-applied force in the rear as seen moving. The extra energy is *in addition to* any energy ks²/2 added in rest frame by compression.

    It’s amazing how many people haven’t heard about this. I haven’t seen my own simple versions around, you have to dig to get the equivalents. But they are needed to get full conservation in cases of forces applied to extended bodies – the bane of simple understanding in SRT.

    Relativistic dynamics of extended bodies is a bear, I wrote about in my previous peer-review papers (see blog link.) It even leads to quarrels, I kid you not, about how to solve paradoxes like “the right-angle lever paradox” and regarding how to interpret the “energy current” etc. The RALP can be solved with the stress corrections, the AJP claim of Nickerson and McAdory regarding internal torques is IMHO not apt. For more and related. The Wikipedia article regarding the RALP does not IMHO properly address the issue (you know, they aren’t perfect …)

  6. #6 Frank Wappler
    December 2, 2011

    Neil Bates wrote (December 2, 2011 8:31 AM):
    > […] “augmented” momentum and energy of a compressed or stretched elastic rod, oriented parallel to observed velocity:

    > p_aug = f_elastic * length * gamma*v/c²
    > E_aug = f_elastic * length * gamma*v²/c²

    Setting

    f := k Δx,

    E := Sqrt[ E^2 – P^2 c^2 ] gamma,

    P := Sqrt[ E^2 – P^2 c^2 ] gamma v/c^2,

    e := Sqrt[ e^2 – p^2 c^2 ] gamma,

    p := Sqrt[ e^2 – p^2 c^2 ] gamma v/c^2, and

    Sqrt[ E^2 – P^2 c^2 ] := Sqrt[ e^2 – p^2 c^2 ] + 1/2 f Δx,

    I get

    P – p == 1/2 f Δx gamma v/c^2 ,

    (which may be just a factor 1/2 off “p_aug“),

    but

    E – e == 1/2 f Δx gamma …

  7. #7 Neil Bates
    December 2, 2011

    Frank, tx for digging in but you are confusing two different issues. Yes there is the extra movement given by compressing the rod, but you still have to take into account the relativity of simultaneity, why didn’t you? The ks stuff just gives the ordinary internal stress energy, which I already noted was “additional to” the augmented forms.

    Note that the ks part has little to do with the momentum, which is based on force * time, altho it does affect the final “L” to which we apply. As for *ordinary* (seen in rest frame) energy, moving the force over the total distance is the same ks when both directions are taken into account.

    Actually I did make a mistake, in the sense that I should have written “proper length” L0, and tx Chad for allowing HTML for subscript to work) instead of “length” up there. (BTW, those are worked out in the literature, even if not common – don’t think I just cooked it up, do you really want to disagree with the consensus? Would be proud of you in principle, but pick your battles carefully.)

    So, just to have the formula again corrected:

    p_aug = γ f_elastic * L0 v/c²
    E_aug = γ f_elastic * L0 v²/c²

    Note also, the augmented terms do not include “k” in the formula, they come from the intrinsic effect of RoS. So you can consider them the limit case of a rod with high k subjected to relatively low forces (ie, ΔL_0 < < L0). Then, please get the point, tx.

  8. #8 Neil Bates
    December 2, 2011

    Well, sorry, it seems that double brackets for “much less than” don’t work in HTML, so that is ΔL0/L0 « 0.

    Just a test again, to see the mistake
    x << y was supposed to look like “x [much less than] y.”

  9. #9 Neil Bates
    December 2, 2011

    Well, have to adjust again, should read:
    ΔL0/L0 « 1,
    and can anyone find the “much less than” symbol directly? BTW the process cut off everything after the two angle brackets.

  10. #10 Eric Lund
    December 2, 2011

    Pro tip for Neil: This site always interprets a less than sign as the start of an HTML tag. If you want to put in a literal <, you have to use the escape sequence &lt;.

    I think, but am not 100% sure, that if the site thinks you opened an HTML tag, the next greater than sign will be interpreted as closing the tag.

  11. #11 Frank Wappler
    December 5, 2011

    Neil Bates wrote (December 2, 2011 8:31 AM):
    > The extra energy is *in addition to* any energy ks²/2 added in rest frame by compression.

    Trying to appreciate this remark more throroughly than I did the first time around I note that “E_aug” explicitly vanishes as evaluated in the rest frame, with “v = 0″.

    Now … Is it in line with what you imagine that the quantity “Sqrt[ E^2 – P^2 c^2 ]” expresses an invariant of the (compressed) system, which is evaluated equally by any given frame, where “E” denotes the entire energy and “P” the entire momentum of the (compressed) system relative to
    the frame under consideration?

    (Do you first of all suppose that “energy E” and “momentum P” as well as “speed v” of the (compressed) system may be sufficiently unambiguously evaluated for any given frame?)

    And do you consider it (sufficiently) correct to set or evaluate
    “E := Sqrt[ E^2 – P^2 c^2 ] gamma” ?

    If so, I really don’t see why to bother with “augumented terms” at all, nor indeed how to derive them …

  12. #12 zeynel
    December 18, 2011

    Chad and Kenneth Cavness: Are you sure Newton invented “vector” calculus. According to Wikipedia vector calculus was developed in the 19th century:

    http://en.wikipedia.org/wiki/Vector_calculus

  13. #13 Derek in DC
    December 23, 2011

    For those who don’t feel like searching through Google results, here’s my favorite math-symbols-on-the-web cheat sheet:
    http://barzilai.org/math_sym.htm

  14. #14 Neil Bates
    December 23, 2011

    Frank, sorry for the delay. I noticed this thread again from Derek’s post. We need the “augmented” terms because of the way stress behaves in different frames. Again, forces applied (or released) simultaneously in one frame, are not so in other ones. Also, the conventional relation then is not adequate, because it can’t represent the augmentation in the rest frame. People are still arguing over how to best deal with all this, it’s mostly-forgotten “dirty linen” of relativity theory!

    Vector calculus: I think Newton had a rough idea, to just imagine coordinates being taken care of per intuitive fx sx + … etc, but not the full development. Surely not the cross product etc.

    Derek: thanks, and also try http://www.alanwood.net/unicode/index.html, with an even bigger (basically, all) character set. It’s amazing what you can find in all that, but it’s hard to find some things (like script l.c. el used for some purposes.)

  15. #15 Frank Wappler
    February 29, 2012

    Neil Bates wrote (December 23, 2011 6:09 PM):
    > […] sorry for the delay

    Likewise.

    > […] the augmentation in the rest frame

    Does that mean “augumentation for v = 0; or for β = 0, resp.” ?

    If so, then from your definitions of December 2, 2011 11:11 AM (#7) such “augumentation” apparently evaluates to 0; both for “p_aug” and for “E_aug“.

    > forces applied (or released) simultaneously in one frame, are not so in other ones

    Sure — which is why invariants such as “Sqrt[ E^2 – P^2 c^2 ]” are much easier explicitly evaluated by the members of the former one frame, than by members of any of the latter other ones.

    And once such an invariant has been evaluated (by members of a suitable frame), and their relation to members of some particular other frame has been quantified by the number β, then it seem straightforward to evaluate
    “Sqrt[ E^2 – P^2 c^2 ] / Sqrt[ 1 – β^2 ]” and so on.

    > […] calculus: I think Newton had a rough idea, to just imagine coordinates being taken care of per intuitive fx sx + … etc

    I have no intuition at all what to make of “intuitive fx sx + … etc “.

    But someone who would dream up coordinate numbers without first considering and evaluating geometric relations evidently lacks even the roughest idea of physics.

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