<h2>CHAPTER 1</h2><p><b>Newton's Laws</p><br><p>The foundational equation of our subject</b></p><p>For in those days I was in the prime of my age for inventionand minded Mathematicks & Philosophy more than at any timesince.</p><p>—Newton describing his youth in his memoirs</p><br><p>Let us start with one of Newton's laws, which curiously enough is spoken as <i>F = ma</i> butwritten as <i>ma = F</i>. For a point particle moving in <i>D</i>-dimensional space with position givenby [??](<i>t</i>) = (<i>x</i><sup>1</sup>(<i>t</i>), <i>x</i><sup>2</sup>(<i>t</i>), ..., <i>x<sup>D</sup>(t)</i>), Mr. Newton taught us that</p><p><i>m d</i><sup>2</sup><i>x<sup>i</sup>/dt</i>2 = <i>F<sup>i</sup></i> (1)</p><p>with the index <i>i</i> = 1, ..., <i>D</i>. For D ≤ 3 the coordinates have traditional "names": forexample, for <i>D</i> = 3, <i>x</i><sup>1</sup>, <i>x</i><sup>2</sup>, <i>x</i><sup>3</sup> are often called, with some affection, <i>x, y, z</i>, respectively.Bad notation alert! In teaching physics, I sometimes feel, with only slight exaggeration,that students are confused by bad notation almost as much as by the concepts. I am usingthe standard notation of <i>x</i> and <i>t</i> here, but the letter <i>x</i> does double duty, as the position of theparticle, which more strictly should be denoted by <i>x<sup>i</sup>(t)</i> or [??](<i>t</i>), and as the space coordinates<i>x<sup>i</sup></i>, which are variables ranging from -∞ to ∞ and which certainly are independent of <i>t</i>.</p><p>The different status between <i>x</i> and <i>t</i> in say (1) is particularly glaring if <i>N</i> >1 particlesare involved, in which case we write <i>m d</i><sup>2</sup> <i>x<sup>i</sup> a/dt</i><sup>2</sup> = <i>F<sup>i</sup><sub>a</sub></i>or [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] with <i>x<sup>i</sup><sub>a</sub>(t)</i> for <i>a</i> =1, 2, ..., <i>N</i>. But certainly <i>t<sub>a</sub></i> is a meaningless concept in Newtonian physics. In theNewtonian universe, <i>t</i> is the time ticked off by a universal clock, while [??]<i><sub>a</sub>(t)</i> is eachparticle's private business. We will have plenty more to say about this point. Here <i>x<sup>i</sup><sub>a</sub>(t)</i>are 3<i>N</i> functions of <i>t</i>, but there are still only 3 <i>x<sup>i</sup></i>.</p><p>Some readers may feel that I am overly pedantic here, but in fact this fundamentalinequality of status between <i>x</i> and <i>t</i> will come to a head when we get to the special theoryof relativity. (I now drop the arrow on [??].) Perhaps Einstein as a student was bothered bythis bad notation. One way to remedy the situation is to use <i>q</i> (or <i>q<sub>a</sub></i>) to denote the positionof particles, as in more advanced treatments. But here I bow to tradition and continue touse <i>x</i>.</p><br><p><b>Have differential equation, will solve</b></p><p>After Newton's great insight, we "merely" have to solve some second order differentialequations.</p><p>To understand Newton's fabulous equation, it's best to work through a few examples. (Ineed hardly say that if you do not already know Newtonian mechanics, you are unlikely tobe able to learn it here.)</p><p>A priori, the force <i>F<sup>i</sup></i> could depend on any number of things, but from experience weknow that in many simple cases, it depends only on <i>x</i> and not on <i>t</i> or <i>dx/dt</i>. As physicistsunravel the mysteries of Nature, it becomes increasingly clear that fundamental forcesare derived from an underlying quantum field theory and that they have simple forms.Complicated forces often merely result from some approximations we make in particularsituations.</p><br><p><b><i>Example A</i></b></p><p>A particle in 1-dimensional space tied to a spring oscillates back and fforth.</p><p>The force <i>F</i> is a function of space. Newton's equation</p><p><i>m d</i><sup>2</sup><i>x/dt</i><sup>2</sup> = -<i>kx</i> (2)</p><p>is easily solved in terms of two integration constants: <i>x(t)</i> = <i>a</i> ω<i>t</i> + <i>b</i> sin ω<i>t</i>, withω = [square root of <i>k/m</i>]. The two constants <i>a</i> and <i>b</i> are determined by the initial position and initialvelocity, or alternatively by the initial position at <i>t</i> = 0 andd by the final position at sometime <i>t</i> = <i>T</i>. Energy, but not momentum, is conserved.</p><br><p><b><i>Example B</i></b></p><p>We kick a particle in 1-dimensional space at <i>t</i> = 0.</p><p>The force <i>F</i> is a function of time. This example allows me to introduce the highly usefulDirac delta function, or simply delta function. By the word "kick" we mean that thetime scale τ during which the force acts is much less than the other time scales we areinterested in. Thus, take <i>F(t)</i> = <i>w</i>δ(<i>t</i>), where the function δ(<i>t</i>) rises sharply just before<i>t</i> = 0, rapidly reaches its maximum at <i>t</i> = 0, and then sharply drops to 0000. Because weincluded a multiplicative constant <i>w</i>, we could always normalize δ(<i>t</i>) by</p><p>∫ <i>dt</i> δ(<i>t</i>) = 1 (3)</p><p>As we will see presently, the precise form of δ(<i>t</i>) does not matter. For example, we couldtake δ(<i>t</i>) to rise linearly from 0 at <i>t</i> = -τ, reach a peak value of 1/τ at <i>t</i> = 0, and then falllinearly to 0 at <i>t</i> = τ. For <i>t</i> < -τ and for <i>t</i> > τ, the function δ(<i>t</i>) is defined to be zero. Takethe limit τ -> 0, in which this function is known as the delta function. In other words thedelta function is an infinitely sharp spike. See figure 1.</p><p>The δ function is somehow treated as an advanced topic in mathematical physics, but infact, as you will see, it is an extremely useful function that I will use extensively in this book,for example in chapters II.1 and III.6. More properties of the d function will be introducedas needed.</p><p>Integrating</p><p><i>d</i><sup>2</sup><i>x/dt</i><sup>2</sup> = <i>w/m</i> δ(<i>t</i>) (4)</p><p>from some time <i>t</i><sub>-</sub> < 0 to some time <i>t</i><sub>+</sub> > 0, we obtain the change in velocity <i>v</i> = <i>dx/dt</i>:</p><p><i>v</i>(<i>t</i><sub>+</sub>) - <i>v</i>(<i>t</i><sub>-</sub>) = <i>w/m</i> (5)</p><p>Note that in this example, neither energy nor momentum is conserved. The lack ofconservation is easy to understand: (4) does not include the agent administering the kick. Ingeneral, a time-dependent force indicates that the description is not dynamically complete.</p><br><p><b><i>Example C</i></b></p><p>A planet approximately described as a point particle of mass <i>m</i> goes around its sun of mass<i>M</i> >> <i>m</i>.</p><p>This is of course the celebrated problem Newton solved to unify celestial and terrestrialmechanics, previously thought to be two different areas of physics. His equation now reads</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)</p><p>where we use the notation [??] = (<i>x, y, z</i>) and <i>r</i> = [square root of [??] · [??]] = _[square root of <i>x</i><sup>2</sup> + <i>y</i><sup>2</sup> + <i>z</i><sup>2</sup>].</p><p>John Wheeler has emphasized the interesting point that while Newton's law (1) tells ushow a particle moves in space as a function of time, we tend to think of the trajectory ofa particle as a curve fixed in space. For example, when we think of the motion of a planetaround the sun, we think of an ellipse rather than a spiral around the time axis. Even inNewtonian mechanics, it is often illuminating to think in terms of a spacetime picturerather than a picture in space.</p><br><p><b>Newton and his two distinct masses</b></p><p>By thinking on it continually.—Newton (reply given whenasked how he discoveredthe law of gravity)</p><p>Conceptually, in (6), <i>m</i> represents two distinct physical notions of mass. On the left handside, the inertial mass measures the reluctance of the object to move. On the right handside, the gravitational mass measures how strongly the object responds to a gravitationalfield. The equality of the inertial and the gravitational mass was what Galileo tried to verifyin his famous apocryphal experiment dropping different objects from the Leaning Towerof Pisa. Newton himself experimented with a pendulum consisting of a hollow woodenbox, which he proceeded to fill with different substances, such as sand and water. In ourown times, this equality has been experimentally verified to incredible accuracy.</p><p>That the same <i>m</i> appears on both sides of the equation turns out to be one of thegreatest mysteries in physics before Einstein came along. His great insight was that thisunexplained fact provided the clue to a deeper understanding of gravity. At this point, allwe care about this mysterious equality is that <i>m</i> cancels out of (6), so that[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], withκ [equivalent to] <i>GM</i>.</p><br><p><b>Celestial mechanics solved</b></p><p>Since the force is "central," namely it points in the direction of [??], a simple symmetryargument shows that the motion is confined to a plane, which we take to be the (<i>x-y</i>)plane. Set <i>z</i> = 0 and we are left with</p><p>[??] = -κ<i>x/r</i><sup>3</sup> and [??] = -κ<i>y/r</i><sup>3</sup> (7)</p><p>I have already, without warning, switched from Leibniz's notation to Newton's dot notation</p><p>[??] [equivalent to] <i>dx/dt</i> and [??] = <i>d</i><sup>2</sup><i>x/dt</i><sup>2</sup> (8)</p><p>Since this is one of the most beautiful problems in theoretical physics, I cannot resistsolving it here in all its glory. Think of this as a warm-up before we do the heavy liftingof learning Einstein gravity. Also, later, we can compare the solution here with Einstein'ssolution.</p><p>Evidently, we should change from Cartesian coordinates (<i>x, y</i>) to polar coordinates(<i>r</i>, θ). We will do it by brute force to show, in contrast, the elegance of the formalismwe will develop later. Differentiate</p><p><i>x</i> = <i>r</i> cos θ and <i>y</i> = <i>r</i> sin θ (9)</p><p>twice to obtain first</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (10)</p><p>and then</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (11)</p><p>(Note that in each pair of these equations, the second could be obtained from the first bythe substitution θ -> θ - π/2, so that cos θ -> sin θ, and sin θ -> - cos θ.)</p><p>Multiplying the first equation in (7) by cos θ and the second by sin θ and adding, weobtain, using (11),</p><p>[??] - <i>r]<i>??]<sup>2</sup> = -κ/<i>r</i><sup>2</sup> (12)</p><p>On the other hand, multiplying the first equation in (7) by sin θ and the second by cos θand subtracting, we have</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)</p><p>I remind the reader again that we are doing all this in a clumsy brute force way to showthe power of the formalism we are going to develop later.</p><p>After staring at (13) we recognize that it is equivalent to</p><p><i>d/dt</i> (<i>r</i><sup>2</sup> [??]) = 0 (14)</p><p>which implies that</p><p>[??] = <i>l/r</i><sup>2</sup> (15)</p><p>for some constant <i>l</i>. Inserting this into (12), we have</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (16)</p><p>where we have defined</p><p><i>v(r)</i> = <i>l</i><sup>2</sup>/2<i>r</i><sup>2</sup> - κ/<i>r</i> (17)</p><p>Multiplying (16) by [??] and integrating over <i>t</i>, we have</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]</p><p>so that finally</p><p>1/2 [??]<sup>2</sup> + <i>v(r)</i> = ε (18)</p><p>with ε an integration constant.</p><p>This describes a unit mass particle moving in the potential <i>v(r)</i> with energy ε. Plot <i>v(r)</i>.Clearly, if ε is equal to the minimum of the potential <i>v</i><sub>min</sub> = -κ<sup>2</sup>/2<i>l</i><sup>2</sup>, then [??] = 0 and r staysconstant. The planet follows a circular orbit of radius <i>l</i><sup>2</sup>/κ. If ε > <i>v</i><sub>min</sub> the orbit is elliptical,with <i>r</i> varying between <i>r</i><sub>min</sub> (perihelion) and <i>r</i><sub>max</sub> (aphelion) defined by the solutions toε = <i>v(r)</i>. For ε > 0 the planet is not bound and should not even be called a planet.</p><p>We have stumbled across two conserved quantities, the angular momentum <i>l</i> and theenergy ε per unit mass, seemingly by accident. They emerged as integration constants,but surely there should be a more fundamental and satisfying way of understandingconservation laws. We will see in chapter II.4 that there is.</p><br><p><b>Orbit closes</b></p><p>One fascinating apparent mystery is that the orbit closes. In other words, as the particlegoes from <i>r</i><sub>min</sub> to <i>r</i><sub>max</sub> and then back to <i>r</i><sub>min</sub>, θ changes by precisely 2π. To verify that thisis so, solve (18) for [??] and divide by (15) to obtain <i>dr/d</i>θ = ±(<i>r</i><sup>2</sup>/<i>l</i>) [square root of (ε - <i>v(r</i>))]. Changingvariable from <i>r</i> to <i>u</i> = 1/<i>r</i>, we see, using (17), that 2(ε - <i>v(r)</i>) becomes the quadraticpolynomial 2ε - <i>l</i><sup>2</sup><i>u</i><sup>2</sup> + 2κ<i>u</i>, which we can write in terms of its two roots as <i>l</i><sup>2</sup>(<i>u</i><sub>max</sub> -<i>u)(u - u</i>min). Since u varies between <i>u</i><sub>min</sub> and <i>u</i><sub>max</sub>, we are led to make another changeof variable from <i>u</i> = <i>u</i><sub>min</sub> + (<i>u</i><sub>max</sub> - <i>u</i><sub>min</sub>) sin<sup>2</sup> ζ to ζ, so that ζ ranges from 0 to π/2. Thus,as the particle completes one round trip excursion in r, the polar angle changes by (notethat <i>u</i><sub>min</sub> = 1/<i>r</i><sub>max</sub> and <i>u</i><sub>max</sub> = 1/<i>r</i><sub>min</sub>)</p><p>[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (19)</p><p>That this integral turns out to be exactly 2π is at this stage nothing less than an apparentmiracle. Surely, there is something deeper going on, which we will reveal in chapter I.4.Note also that the inverse square law is crucial here. Incidentally, the change of variablehere indicates how the Newtonian orbit (and also the Einsteinian orbit, as we will see inpart VI) could be determined. See exercise 2.</p><p>Bad notation alert! In (1), the force on the right hand side should be written as <i>F<sup>i</sup>(x(t))</i>(in many cases). In C, the gravitational force exists everywhere, namely <i>F(x)</i> exists as afunction, and what appears in Newton's equation is just <i>F(x)</i> evaluated at the position ofthe particle <i>x(t)</i>. In contrast, in A, with a mass pulled by a spring, <i>F(x)</i> does not makesense, only <i>F(x(t))</i> does. The force exerted by the spring does not pervade all of space, andhence is defined only at the position of the particle <i>x(t)</i>, not at any old <i>x</i>. I can practicallyhear the reader chuckling, wondering what kind of person I could be addressing here, butbelieve me, I have encountered plenty of students who confuse these two basic concepts:spatial coordinates and the location of particles. I may sound awfully pedantic, but when weget to curved spacetime, it is often important to be clear that certain quantities are definedonly on so-called geodesic curves, while others are defined everywhere in spacetime.</p><br><p><b>A historical digression on the so-called Newton's constant</b></p><p>Wouldn't we be better off with the two eyes we now have plus athird that would tell us what is sneaking up behind? ... With sixeyes, we could have precise stereoscopic vision in all directionsat once, including straight up. A six-eyed Newton might havedodged that apple and bequeathed us some levity rather thangravity.</p><p>—George C. Williams</p><br><p>Physics textbooks by necessity cannot do justice to physics history. As you probably know, inthe <i>Principia</i>, Newton (1642–1727) converted his calculus-based calculations to geometricarguments, which most modern readers find rather difficult to follow. Here I want tomention another curious point: Newton never did specifically define what we call hisconstant G. What he did with <i>ma = GMm/r</i><sup>2</sup> was to compare the moon's accelerationwith the apple's acceleration: <i>a</i><sub>moon</sub> <i>R</i><sup>2</sup><sub>lunar orbit</sub> =<i>GM</i><sub>earth</sub> = <i>a</i><sub>apple</sub> <i>R</i><sup>2</sup><sub>radius of earth</sub>. But to write<i>GM</i><sub>earth</sub> = <i>a</i><sub>apple</sub> <i>R</i><sup>2</sup><sub>radius of earth</sub>, he had to prove what is sometimes referred to as the first ofNewton's two "superb theorems," namely that with the inverse square law the gravitationalforce exerted by a spherical mass distribution acts as if the entire mass were concentratedin a point at the center of the distribution. (See exercise 4.) Even with his abilities, Newtonhad to struggle for almost 20 years, the length of which contributed to the bitter priorityfight he had with Hooke on the inverse square law, with Newton claiming that he had thelaw a long time before publication. You should be able to do it faster by a factor of ~10<sup>4</sup> asan exercise.</p><p><i>(Continues...)</i>