Physics
We again start with some comments on physics Physics provides a way of looking at the world We describe physical phenomena in mathematical terms with the goal of
The description is tested with experiment Physics is an experimental sci- ence The payoff is that
• The description is either correct or incorrect
2 This mechanics course also serves as a nice introduction to the graduate text [Fetter and Walecka (2003)].
The objective is to formulate a physical law expressed as a mathematical relationship, typically reflecting the immediate evolution of a system Subsequently, this law's mathematical implications are derived and tested for accuracy.
Isaac Newton and Albert Einstein are two of the most influential figures in physics Newton developed calculus to formulate his second law of motion, while Einstein revolutionized our understanding of space and time, showing that they are relative to an observer's speed and the presence of nearby mass.
We start our discussion with homage to Newton, and give a brief review of the elements of calculus that we will need for ourIntroduction to Classical
Calculus
Consider the curve described by the functionf(x) For every smooth curve, there will be a straight-line tangent to that curve at the pointx(Fig 1.1).
Now move to a neighboring pointx+∆xwhere ∆xis a very small increment.
The function will change to f(x) + ∆f(x). tangent x θ f(x) x Δ Δ f(x)
Fig 1.1 Tangent to the curve f(x) at the point x, and increment ∆f (x) in the function when x increases by ∆x.
The angle that the tangent to the curve makes with the x-axis (theslope)
3 Maxwell and Schr¨ odinger, with their equations, are not far behind. is given by tanθ=∆f(x)
∆x ; Limit ∆x→0 (1.1) where this expression is exact in the limit that the displacement ∆xbecomes vanishingly small This quantity is called thederivative off(x) df(x) dx ≡Lim∆x→0
Given the value off(x1), one can obain the valuef(x) by stepping along the curve (Fig 1.2) f(x) =f(x1) +
Fig 1.2 From f(x 1 ) to f(x) and area under the curve.
The limit N → ∞, which implies ∆x → 0, serves to define the integral
∆x xi) dx (1.6) and as ∆x→0 we call it thedifferential dx, one has
I(x, x1) Z x x 1 df(u) du du ; integral (1.7) where we have simply re-labeled the dummy integration variable.
Stepping pastxwith a finite interval ∆xgives
The limit ∆x→0 then gives dI(x, x1) dx = df(x) dx (1.10)
The derivative of the integral with respect to its upper limit is the integrand evaluated at that upper limit.
It is evident from Fig 1.2 that as the width of each rectangle decreases, and the number of the rectangles increases, one calculates the areaunder the curvef(x)
Exactly as above, the derivative of this area with respect to xis just the function f(x) itself dA(x, x1) dx =f(x) (1.12)
Calculus connects the concept of infinitesimal change in a function, represented by the derivative df(x)/dx, to the function itself, f(x)−f(x1) This relationship extends to the area under the curve, expressed as Rx x 1f(u)du, through the process of integration, which sums these infinitesimal elements.
Example
In a vacuum, where air resistance is absent, an object dropped from rest will experience a linear increase in velocity over time Specifically, if \( z \) represents the distance fallen and \( \frac{dz}{dt} \) denotes the instantaneous velocity, the relationship can be expressed as \( \frac{dz}{dt} = gt \) Here, \( g \) is the constant acceleration due to gravity at the Earth's surface, approximately \( 9.8 \, \text{m/s}^2 \).
The total distance the object falls after a timetis then given by integrating this differential equation z=g
We can go back one step and write the rate of change of velocity, the acceleration, as dvz dt =g ; acceleration (1.16)
Integration of this relation reproduces Eq (1.13) vz = dz dt =g
Now note that Eq (1.16) reads d dt dz dt
4 The original experiment is due to Galileo, who dropped different objects from the
The second derivative, representing the derivative of the derivative, is a fundamental concept in Newton's second law regarding the motion of a body dropped from the Earth's surface Notably, as Galileo first noted, the mass of the body does not influence this relationship.
We can further manipulate these results Consider
The equation 2v z 2 = gz represents the gravitational potential, where U(z) = −gz indicates the gravitational potential per unit mass, with z measured downwards This relationship leads to the principle of energy conservation for an object that is released from rest at z = 0.
Units
In this book we use Standard International (SI) units, essentially m.k.s.
Masses are measured in kilograms (kg), distances in meters (m), and forces in newtons, where
Energy is measured in joules, where
To write Newton’s laws in three dimensions, we must employvectors, to which we next turn our attention.
We summarize the essential elements of vectors that we will need in our development.
A vector can be characterized by its cartesian components (Fig 2.1)
Fig 2.1 A vector ~ v in a cartesian coordinate system, with components (v x , v y , v z ).
• Vectors can beadded by adding their components
~a+~b= (ax+bx, ay+by, az+bz) (2.2)
• Vectors can bemultiplied by a scalarthrough multiplication of their
The scalar product (“dot product”) of two vectors is given by
~aã~b≡axbx+ayby+azbz ; scalar product (2.4)
• The square of thelengthof a vector is
• It follows from the above that (see Fig 2.2)
From the law of cosines
We can alternatively express the vector ~v as a linear combination of orthonormal unit vectors 1
~v=vxxˆ+vyyˆ+vzzˆ ˆ x 2 = ˆy 2 = ˆz 2 = 1 ˆ xãyˆ= ˆxãzˆ= ˆyãzˆ= 0 (2.9)
1 In this work, a hat consistently denotes a unit vector.
The vector product (“cross product”) of two vectors is given by
~a×~b≡(aybz−azby) ˆx+ (azbx−axbz) ˆy+ (axby−aybx) ˆz
; vector product (2.10) This is most conveniently written as adeterminant
~a×~b≡det ˆ x yˆ zˆ axay az bx by bz
= det ay az by bz ˆ x−det axaz bx bz ˆ y+ det axay bx by ˆ z (2.11)
The last line is obtained through an expansion in minors.
The cross product has the following properties:
• The dot product of~a×~bwith either of its constituentsvanishes
The vector \( \mathbf{a} \times \mathbf{b} \) is derived from the definition in Equation (2.10) or from the observation that the expressions result in a determinant with two identical rows as shown in Equation (2.11) Consequently, this vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), as illustrated in Figure 2.3.
• Thedirectionof~a×~bis given by theright-hand rule Put the fingers of your right hand along~a, curl them into
~b, then your thumb points along~a×~b.
• Thelengthof~a×~bis calculated as (see Prob 2.2)
• The vector product changes sign under the interchange of its elements
• Evidently, the cross product of a vector with itself vanishes
A vector~v=vxxˆ+vyyˆ+vzzˆ can also be written as a length times a unit vector denoting the direction~v=vˆn This is seen as follows.
The square of the length of the vector is 2 v 2 =~vã~v=v x 2 +v y 2 +v z 2 (2.17) The length is therefore v= (v x 2 +v y 2 +v z 2 ) 1/2 (2.18) The quantity ˆn≡~v/vis then a unit vectordenoting the direction ˆ n≡ ~v v = 1
(v 2 x +v 2 y +v z 2 ) 1/2 (vxxˆ+vyyˆ+vzz)ˆ (2.19) The vector itself is thus written as
A vector canchangeboth because its length and its direction can change.
The infintesimal form of the change in a vector is evidently d~v= ˆn dv+v dˆn (2.21)
Since ˆnis a unit vector with ˆnãnˆ= 1, it follows that ˆ nãdˆn= 0 (2.22)
The gradient is a differential operator defined by
Here thepartial derivativesimply that the other members of the variable set
(x, y, z) are to be held fixed when evaluating the derivative The gradient of a scalar function V(x, y, z) is then
Letd~sindicate a small displacement in position d~s= ˆx dx+ ˆy dy+ ˆz dz (2.25)
The corresponding change in the function V(x, y, z) is obtained from the projection of the gradient along this displacement dV(x, y, z) =∇~V(x, y, z)ãd~s
∂zdz ; total differential (2.26) Some comments:
• This is known as thetotal differentialofV(x, y, z) ;
• It gives the small change in V under small changes in (x, y, z) ;
• It is exact in the limit as these small changes go to zero (dx, dy, dz)→0.
As an example, consider the position vector~r
SupposeV(r) is only a function ofr, then 3
2xxˆ+ 2yyˆ+ 2zzˆ (x 2 +y 2 +z 2 ) 1/2 =−dV(r) dr
Hence, when acting on a function of the radial distancer, the gradient is a simple radial derivative∇ →~ ˆr(d/dr).
3 Use the chain rule from Prob 1.6.
To understand Newton's laws, it's essential to introduce inertial coordinate systems, with the primary system being one that remains at rest relative to the fixed stars Although this concept is not absolute, it can be defined with arbitrary accuracy Observers on Earth can accurately identify stars that do not move against the background of the heavens, effectively establishing a reference frame for the observable universe where Newton's laws apply.
Now consider any frame moving with aconstant velocity~vR relative to this primary system (Fig 3.1). v fixed stars
Fig 3.1 Inertial coordinate system moving with constant velocity ~ v R relative to the fixed stars.
Since~vR does not change in the primary frame d~vR dt = 0 ; in primary frame (3.1)
Newton's laws can be expressed through the rate of change of momentum, represented as d(m~v0)/dt, in a primary inertial coordinate system When considering an object's velocity in a moving frame, denoted as ~v, its corresponding velocity in the primary frame can be derived accordingly.
The rate of change of velocity, or acceleration, is the same in any frame moving with constant velocity with respect to the fixed stars Hence, such a frame is againinertial.
Thus, although formulated in the primary inertial coordi- nate system, Newton’s laws hold in any inertial frame mov- ing with constant velocity relative to the fixed stars.
Note that underlying these arguments in newtonian physics, we have assumed thattimeis absolute and does not change from frame to frame — time is the same in any inertial frame 1
1 This assumption was later invalidated by Einstein in his special and general theories of relativity.
1) Newton’sfirst law states that
In an inertial frame of reference, an object will either stay at rest or continue moving at a constant velocity unless a force acts upon it Consequently, when the net force is zero, the object maintains a constant velocity and its momentum remains unchanged, defined by the equation \( \mathbf{p} = m \mathbf{v} \).
In this article, we explore the concept of force (F~), defined as a push or pull, and introduce the idea of inertial mass (mI), with further details to follow.
2) Newton’ssecond law, and this is really the heart of the matter, states
In an inertial frame, application of a force alters the mo- mentum in an amount specified by the quantitative relation d~p dt = d(mI~v) dt =F~ ; Newton’s second law (4.1)
It is the physicist’s job to classify the forces; Newton’s second law then tells how a particle moves Let us give some examples offorces
• When a one-dimensional spring is displaced from its equilibrium length by a distance x (which may have either sign), there is a restoring force
F =−κx ; spring (4.2) where κis thespring constant This constant is readily measured in the laboratory through a static expansion or compression of the spring For our purposes, Eq (4.1) now provides an appropriate
15 definition of theinertial massmI It describes how an object moves under this spring force.
• When an object is dropped in thez-direction at the surface of the earth (see chapter 2), there is a gravitational force
FG=mGg ; gravity at earth’s surface (4.3)
It has been known since Galileo that mI≡mG ; equivalence principle (4.4)
The implication being, as we have seen, that the mass then drops out of Newton’s second law of motion for the particle.
The equivalence principle is a remarkable relationship that suggests the inertial mass, which governs a particle's motion under any force, is identical to the gravitational mass, which determines how it is influenced by gravity This intriguing concept raises questions about the fundamental nature of mass and its role in the universe.
We will assume that Eq (4.4) holds (all existing experimental ev- idence supports this), and we henceforth simply denote the mass bym.
• Newton’s law of gravitational attraction states that the force F~21 exerted on particle 1 with massm1located at position~r1, by par- ticle 2 with massm2 located at position~r2, is given by
|~r1−~r2| 3 ; Newton’s gravity (4.5) HereGis Newton’s constant
3) Newton’sthird law of motion states that
To each action, there is an equal and opposite reaction.
Thus, in the third example above,
F~12=−F~21 ; Newton’s third law (4.7) and these forces act along the line joining the particles.
1 It was again Einstein who recognized the deep implications of this relation; it is a starting point for his theory of general relativity.
We give some examples of the application of Newton’s laws.
Consider the one-dimensional problem of a massmon a spring with spring constant κand extensionx(t) Newton’s second law in any inertial frame states that md 2 x(t) dt 2 =−κx(t) ; spring (5.1)
This article discusses a second-order, linear, homogeneous differential equation, highlighting that the functions which remain unchanged after two differentiations are sines and cosines Consequently, the general solution to this differential equation can be expressed as x(t) = A cos(ωt + φ), where ω² is defined as κ/m.
The mass oscillates back and forth with an angular frequencyω=p κ/m.
The system forms asimple harmonic oscillator.
The amplitude and phase of the solution (A, φ) are determined by the two initial conditions x(0) osφ ; initial conditions dx(t) dt t=0
Newton's second law results in a second-order differential equation of motion, necessitating the provision of both the initial position and initial velocity of a particle to accurately determine its future position.
Projectile motion on Earth's surface can be analyzed within a short time frame as an inertial frame of reference In this context, the z-direction is defined as upward, while the x-direction runs parallel to the surface The velocity of the projectile is then examined within these directional parameters.
Gravity provides the downward force F~G = −mgz, and since the unitˆ vectors do not change with time, Newton’s second law reads m dvx dt xˆ+dvz dt zˆ
The components are then equated to give dvz dt =−g dvx dt = 0 (5.6)
When a projectile is launched from the origin (0, 0) with an initial velocity \(v_0\) at an angle \(\phi\), its horizontal and vertical components of velocity can be expressed as \(v_{0x} = v_0 \cos \phi\) and \(v_{0z} = v_0 \sin \phi\) By integrating Newton's second law, we derive the vertical velocity \(v_z\) and position \(z\) as follows: \(v_z = v_{0z} - gt\) and \(z = v_{0z}t - \frac{1}{2}gt^2\).
2gt 2 (5.8) and for (vx, x) vx=v0x x=v0xt (5.9)
Suppose we ask how long it takes for the projectile to reach its maximum height h, and how high that is After that time th, one hasvz= 0 Hence th = v0z g = v0 g sinφ h= v 2 0z 2g = v 2 0
To determine the time it takes for a projectile to return to Earth and the distance it travels, we can establish that at time \( t_d \), the projectile reaches the ground (where \( z = 0 \)) and covers a horizontal distance \( x = d \) The relationship can be expressed as \( t_d = \frac{2v_0 \sin \phi}{g} \) and \( d = v_{0x} t_d = \frac{v_0^2 \sin 2\phi}{g} \), illustrating how the initial velocity and launch angle influence the projectile's flight path.
Note thattd= 2th, so that it takes just twice as long to complete its jouney as it does to reach its maximum height.
We can ask for the angle of inclination that will give the maximum range for a given muzzle velocity This is evidently achieved for sin 2φ= 1 ; maximum range φ= 45 o (5.12)
Consider two particles, one with (m1, ~r1) and a second with (m2, ~r2), in- teracting through a force F~ that depends only on the relative separation
~r1−~r2 Newton’s second law for the pair in an inertial frame reads 2 m1 d 2 ~r1 dt 2 =F~21 m2d 2 ~r2 dt 2 =F~12 (5.13)
Introduce the center-of-mass, and relative coordinates for the pair
If the two Eqs (5.13) are added together, and Newton’s third law in
Eq (4.7) is invoked, one has
The center-of-mass (C-M) moves as afree particle of massM with no force.
1 Note that t = 0 with (x, z) = 0 similarly provides a solution for the motion!
Multiply the first of Eqs (5.13) by m2/(m1 +m2), the second by m1/(m1+m2) and subtract, and again make use of Newton’s third law.
This gives àd 2 ~r dt 2 =F~21 ; relative motion à≡ m1m2 m1+m2
In the C-M frame, which is again inertial, the problem has been reduced to that of a single particle ofreduced massà, located at~r, moving under a force F(~r~ ).
According to Newton's gravitational force, the relationship can be expressed as d²~r/dt² = −(GM)~r/r³, illustrating the fundamental principles of Newton's gravity The gravitational parameters are encapsulated in the constant (GM), which is essential for analyzing the relative motion within a binary gravitational system based on observational data.
In the previously discussed examples, we can introduce a constant of motion, specifically energy, which serves as a first integral of Newton's second law This allows us to derive the particle's velocity from its position or the other way around, eliminating the need to explicitly integrate the second-order differential equation again We will now demonstrate how this is achieved in the three examples examined.
Multiply Newton’s second law for the spring in Eq (5.1) bydx(t)/dt m dx dt d 2 x dt 2
(6.1) This can be rewritten in the following form md dt
2x 2 ; potential energy (6.3) Since its time derivative vanishes, the total energy in the oscillator, which is the sum of the kinetic and potential energies E=T +V, is a constant
Substitution of the solution for x(t) in Eq (5.2) into these relations yields
Energy in a spring system oscillates between kinetic energy, associated with the motion of the mass, and potential energy, linked to the spring's expansion or compression, while the total energy remains constant throughout the motion This total energy is directly proportional to the square of the amplitude.
Workis done on an object by a force moving through a distance 1 dW =F dx ; work (6.6)
When I stretch the spring, I must exert a force just infinitesimally greater than κx It follows that the work done in stretching the spring a total distance xis thepotential energystored in the spring
Projectile motion on Earth's surface offers a brief inertial frame, with the z-direction pointing upwards and the x-direction aligned along the surface The projectile's velocity is a key factor in this motion.
It follows just as in section 1.3 that the total energy of the particle is