← IA Dynamics and Relativity – Full Text
These are Zixuan’s notes for Part IA – Dynamics and Relativity at the University of Cambridge in 2026. The notes are not endorsed by the lecturers or the University, and all errors are my own.
The latest version of this document is available at academic.micfong.space. Please direct any comments to my CRSid email or use the contact details listed on the site.
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1 The Structure of the Newtonian Universe
To set up the arena we are going we work in, we require
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a three dimensional space that can be endowed with a Cartesian reference frame [i.e. an origin and some axes], such that points in space are labelled as
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a time parameter that can be labelled, in an arbitrary reference frame, by a real number .
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a point particle which is an idealised object that is completely determined by its position at a given time as .
Examples include electron, tennis ball, planet depending on the context.
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the velocity which is the vector
From results in IA Vector Calculus, the velocity vector is tangent to the trajectory of the particle.
Recall that in Cartesian coordinates,
[We will discuss other coordinate systems later in the course.]
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the acceleration which is the vector
The above structure is not enough to write down Newton’s equations.
Consider a “free” particle that does not experience any forces. [e.g. the particle is alone in deep space far away from any other matter] The position of this particle is , we need to consider which reference frame we are using.
The particle may be at rest in a frame , but moving in a complicated way with respect to another frame .
In an inertial frame, we may write for a free particle,
The law of inertia is an improved version of Newton’s 1st law.
This is a true statement about the world, but not an obvious one. [In antiquity, it was believe that the natural state of an object is to be at rest, and a force is required to keep it moving.]
A Galilean transformation between two reference frames and is given by
where is a rotation and/or a reflection, is a constant translation, and is a constant velocity, called a boost.
It is easy to see that under a Galilean transformation.
Galilean invariance restricts the type of forces that are possible (see Example Sheet 1).
Galilean relativity implies that the laws of physics make reference to no special point, direction, time or velocity. All these things can only be defined relatively. [e.g. one cannot be “at rest”, but only at rest with respect to something.]
Acceleration is not relative. If one is accelerating in an inertial frame, then they will be accelerating in all other inertial frames, with the same magnitude.
Remark. The Galilean transformations form the Galilean group, often supplemented with time translations:
Laws of physics are also invariant under time translations, i.e. all inertial frames have the same time, called absolute time.
2 Forces
Once there is more than one particle in the universe, there will be interactions between the particles. In Newtonian physics, these are described by forces.
2.1 Newton’s Second Law
In an inertial frame,
where is the momentum of the particle and is the net force acting on the particle.
The momentum is defined to be , where is the inertial mass.
The mass is an additional property of particles. It could change with time, but we generally assume it is constant unless otherwise specified.
The force depends on the interaction, but can only depend on and at the current time.
The above implies that
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Newton’s second law can be written as a second order ODE for .
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given and at for all particles, Newton’s equations uniquely determine for all future times.
2.2 Conservative Forces and Gravity
Conservativeforces form an important class of forces that can be written as
for some potential (also called potential energy) .
The gravitational potential energy of a particle of mass at due to a particle of mass at is
where .
To take the gradient,
and
Hence
This gives
If we let , this is the familiar inverse square law
Sometimes we write , where is the gravitational potential
Near the surface of the Earth, take the centre of the Earth, and , where is the radius of the Earth and is the height above the surface. Then
Thus, near the surface of the Earth, we approximate the gravitational potential as
The force is then
which is a constant force near the surface of the Earth.
This force leads to the simplest example of motion due to a force.
Newton’s 2nd law gives
where . Consider the -component,
which gives
where and are the initial position and velocity at .
2.3 Conservation of Energy
Conservative forces have a conserved energy
We can check that this is conserved:
Suppose we throw an object into space and want it to never fall back down. The minimal velocity this object must have is called the escape velocity.
As the object is thrown,
To not fall back, the object must be able to reach without the velocity going to zero. Hence
Therefore we require
Hence the escape velocity is
The mass is cancelled out, since the gravitational mass (that appears in the inverse square law) is the same as the inertial mass (that appears in Newton’s 2nd law).
It is useful to write , where is the kinetic energy and is the potential energy.
Conservative forces have the property that the work done by the force as a particle moves along a trajectory , where the work done is defined as
only depends on the endpoints of the trajectory, not on the path itself.
Proof. Let the trajectory go from at to at . Then
Proof (Direct). Using results from IA Vector Calculus, we have
2.4 Electromagnetic Forces
Forces that depend on the velocity typically don’t have a conserved energy, such as friction. However, the Lorentz force is an exception.
Electromagnetic fields and exert the following force on a particle with charge
In this section, we shall restrict to static electromagnetic fields, i.e., and do not depend on time. Then
where is the electric potential.
We claim that the conserved energy is
To check this,
The velocity-dependent force is orthogonal to the trajectory of the particle, so it does no work.
Electric forces are similar to gravitational ones. The potential at due to another particle of charge at is
where is the permittivity of free space, approximately .
Like gravity, this leads to an inverse square law for the electric force, called Coulomb’s law. However, charges can be positive or negative, but mass is always positive. Hence, gravity dominates for large objects while electric foces tend to cancel out overall.
For magnetic forces, a charged particle in a magnetic field obeys
This is a vector differential equation. The most direct way to solve it is to write out components.
Suppose is constant and WLOG along the -axis, i.e., . The equations become
Solution 1. Using and ,
which is a 2nd order equation for . This gives
where
which is called the cyclotron frequency.
This gives
Substituting back into gives
where and are arbitrary.
The period is the time to complete one cycle,
Solution 2. Alternatively, let . Then we can write as
Solving gives
where and are complex constants.
Set and . Then taking real and imaginary parts recovers
2.5 Motion in One Dimension
Problems can often be reduced to one-dimensional motion, such as
If is independent of velocity, then it can always be written in terms of a potential, by setting
where is an arbitrary reference point. This gives
The following energy is then conserved:
Keeping constant gives a 1st order ODE for , which is easy to integrate:
This equation tells us how long it takes to move from to , if it has energy .
More often than not the integral is not analytically solvable, but it is still useful for qualitative analysis of motion.
From , we get . This restricts the range of where the particle can be.
The points where are called turning points. Typically, particles bounces off the potential at turning points and turn around.
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Within region , the particle bounces back and forth between turning points (bounded motion).
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Within region , the particle bounces off the turning point and escapes to (unbounded motion).
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A special case occurs when and . These are equilibrium points, where the particle can remain at rest if placed there.
We have and , giving .
We shall show that motion close to equilibrium points are especially simple.
Let be the equilibrium point, the Taylor expansion about gives
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If , this is the potential for a simple harmonic oscillator,
Solving this gives
where is the amplitude and is the phase. The angular frequency is
If is small enough, we can neglect higher order terms in the Taylor expansion, and the oscillating solution is valid. The point is called a stable equilibrium.
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If , then we get an unstable equilibrium point. The solution is
where
If , the exponential growth means the particle moves far away from the equilibrium point, so the approximation breaks down.
The case corresponds to rolling the particle up the potential with just enough energy to reach the top as .
If , we need to include higher order terms in the Taylor expansion.
2.6 Dimensional Analysis
Dimensional analysis is a way to obtain information about solutions to equations without solving them. At a mathematical level, dimensional analysis is the ability to rescale variables to remove certain constants from equations.
We will derive this equation for a pendulum later:
Suppose we release the pendulum from rest at some angle . We want to find the period of oscillation where
We can remove from the equation by rescaling
We are effectively writing and by chain rule,
This equation does not depend on , and so the solution is some function . Therefore, the period of may depend on the initial angle but can’t depend on . Hence,
giving
Hence,
Therefore, without solving the equation, we have found that the period of a pendulum is proportional to .
However, in general, a necessary rescaling may not be obvious. Thus, associating dimensions to all constants and variables is a form of bookkeeping that accounts for how these quantities appear in Newton’s equations.
The basic dimensions are
- length ,
- time ,
- mass .
Then we have
There can be other dimensions (such as charge), depending on the problem.
The fundamental principles of dimensional analysis are
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,
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all arguments of nontrivial functions (i.e. involving sums of different powers) must be dimensionless.
We first list all the dimensions of the relevant quantities:
Then we let
This gives
Therefore, the period is
2.7 Friction
When objects move through a medium (e.g. air or water), microscopic forces between the object and the medium cause momentum to be carried off into the medium and lost.
There are two important properties:
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Friction does not conserve energy, since momentum is lost to the medium, in the form of heat.
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Friction is irreversible. Energy is lost by the object, and energy is lost by the object but not regained. [If one every doubts about the sign of a friction force, it should slow the object down.] Moreover, friction forces must change signs under , hence friction forces must depend on velocity.
There are two common cases of friction forces:
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Linear drag. .
Linear drag depends on viscous effects [see IB Fluid Dynamics], such as a spoon in honey. In this case, objects move the medium with them.
Example 2.10Stokes’ law for a spherical object of radius stats that
where is the viscosity of the medium.
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Quadratic drag. .
Quadratic drag is the more intuitive case.
As an object bumps into molecules, the rate of collisions is proportional to the speed , and each collision imparts a momentum change proportional to . Hence the force is proportional to .
The number of collisions depends on the density of the medium and the cross-sectional area of the object, so . We can also see this by dimensional analysis:
and are called coefficients of friction. Both linear and quadratic drag are typically present. For a spherical object, which term dominates depends on the Reynolds number [see IB Fluid Dynamics], where
2.7.1 Terminal Velocity
Consider a particle falling with quadratic friction under gravity. Consider the -component of the motion. We have
The velocity starts at , then increases. Initially, is dominated by . Eventually, the two forces balance, giving a terminal velocity.
so heavier objects have higher terminal velocities.
We can also consider the timescale when the object reaches its terminal velocity. With dimensional analysis, we write
So we have
Therefore,
Note that, solving the equations gives the following graph.
We can also have motion in different directions to gravity, so that
2.7.2 Damping
Friction damps small oscillations about an equilibrium point. For small oscillations, the linear drag dominates. Thus, in a 1D system:
where is the natural frequency of oscillations without friction, and is the damping coefficient. Solving this gives
where and taking the real part gives the damped oscillations.
The three cases are
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, underdamped, decaying oscillations.
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, overdamped, exponential decay.
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, critical damping: .
3 Central Forces
An important class of potentials only depend on the distance to the origin, such that
The force points towards (or away from) the origin, so
Recall that
Thus,
We will study the motion of a particle under a central force, particularly in polar coordinates.
3.1 Conservation of Angular Momentum
The most important fact about central potentials is that angular momentum is conserved. We have
where is the angular momentum, and is the linear momentum. Note that is orthogonal to both position and velocity/momentum.
is defined relative to an origin, here we are setting the origin at , which will be generalised later.
For a general force ,
where is the torque. i.e.
This is analogous to Newton’s law Eq · 8, but for rotational motion. [ can be thought of as the rotational force, and as the rotational momentum.]
For a central force, . Thus, angular momentum is conserved:
Since doesn’t change, and obeys
where the position and velocity are constrained to a plane perpendicular to . Hence, we have reduced the problem from 3D to 2D.
3.2 Polar Coordinates in the Plane
Within polar coordinates, we have
In cartesian coordinates, [see IA Vector Calculus for more details]
Note that
Important. These vectors depend on positions. We have
Hence, we must keep track of these changes when we write equations in polar coordinates.
Consider Newton’s equation [i.e. ] in polar coordinates,
We have
Thus,
Note that circular motion requires a centripetal force towards the origin.
Matching components of Eq · 124 with Eq · 139 substituted in Eq · 8, we have
These are the Newton’s equations for a central force in polar coordinates.
Hence, Eq · 142 gives
Thus is constant.
In fact, this is the magnitude of the angular momentum per unit mass:
Sometimes, is called the “angular momentum”, even though it is angular momentum per unit mass. Using the definition of in the equation for , we have
We can rewrite this as
where is the effective potential defined by
Hence we have reduced the motion to an effective one-dimensional problem in the radial direction. This is possible because of the conservation of angular momentum.
3.3 The Effective Potential
Consider , i.e. an attractive potential. The effective potential is
We have a centrifugal barrier at small due to the angular momentum term. The angular momentum prevents the particle from getting too close to the origin.
We can also see the effective potential from the conserved energy,
The centrifugal barrier is the angular kinetic energy.
Different cases arise depending on the energy .
Now, suppose instead that where . Then
Note that in there are no stable bound orbits, and the particle can fall to the origin.
Remark. Gravity in space dimensions has
Moreover, circular orbits are stable only for . Hence, our universe has a special number of dimensions for stable planetary orbits.
3.4 The Orbit Equation
We shall now see how to solve the equations of motion for a central potential. Consider
We want to derive the orbit equation for . Recall the case that
Under the change of variables ,
Then
A special case arises when , i.e. the Kepler problem.
This equation is a harmonic oscillator with a displaced centre. The solution is
Note that is largest at , and is smallest there, which is the periapsis.
We can choose axes on the plane so that . Then
where and is a constant of integration. The shape of the orbit depends on . Note that this is the equation for a conic section in polar coordinates, and is the eccentricity.
Proof. [Ellipse.]
With , we can rearrange to get
Hence we can regroup to get
where are given in terms of and . For example,
Planets in the solar system have small , so that they are close to circular. e.g.
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the largest is for Mercury, which has .
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Halley’s comet has .
Proof. [Parabola.]
With , we get
This is the equation for a parabola.
Proof. [Hyperbola.]
Note that at . Hence, the asymptotes satisfy . Hence,
We can evaluate the energy on the solution,
Hence the energy is negative for bounded orbits, zero for parabolic orbits, and positive for hyperbolic orbits. In particular, for a circular orbit,
which is the minimum of .
3.5 Kepler’s Laws
A consequence of the above are Kepler’s laws of planetary motion.
- K1
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Planets move in ellipses with the Sun at one focus.
- K2
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The line between the planet and the Sun sweeps out equal areas in equal times, and
- K3
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The period of of the orbit is proportional to .
Proof.
- K2
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We have
where
It follows from the conservation of angular momentum, for any central force.
- K3
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It is natural to consider dimensional analysis. The only parameter in Newton’s equation is . Hence the period satisfies
Hence and , so and . Thus,
Note that there is no unique radius associated to an ellipse, but taking any will do.
More precisely, starting with , we have the full period
where is the average radius of the ellipse.
3.6 Repulsive Potentials and Scattering
Given a central potential such that as , one can perform scattering experiments by sending in a particle from large and see how it moves out again.
The impact parameter is the distance of closest approach if there were no forces.
The impact parameter is related to the angular momentum (per unit mass) as
Proof. A non-interacting particle has a conserved angular momentum. The velocity does not change. At the closest point,
This must also be the angular momentum at the start. But the initial is the same for the interacting and non-interacting particles and is also conserved in the interacting case.
Rutherford scattering (1911) showed that certain scattering experiments of atoms could be explained if all the positive charge in an atom was confined to a tiny nucleus.
Scattering by a repulsive interaction for two positively charged parciules satisfies
We may reuse results from Kepler problem by setting .
In particular, the orbits are
with .
We want to find the angle through which the particle is scattered.
Note that incoming and outgoing impact parameters are equal by conservation of and . [And hence at infinity.]
Clearly
- [The particle goes to infinity at .]
Then we can get in terms of impact parameter and initial velocity,
Hence, matching the first and last expressions, we have
Note that a small leads to large angle scattering, and it allows scattering to probe very small distances.
4 Systems of Particles
There are many particles in the universe. We shall focus on of them.
Label the particles by . Each particle has a momentum
and obeys Newton’s law individually,
The force on the -th particle can be external or due to the other particles:
where is the force on the -th particle due to the -th particle. The forces between particles are found to obey
due to Newton’s third law.
4.1 Centre of Mass
Consider a system of particles.
The total mass of a system is .
The centre of mass of a system is defined as
The total momentum of a system can therefore be expressed as
Therefore, the centre of mass moves as if it is a single particle of mass .
4.1.1 Conservation of Momentum
We have
We have hence shown that
Hence, the centre of mass accelerates like a point particle, subject to an external force .
In particular, implies that , so the total momentum of a system is conserved in the absence of external forces.
4.1.2 Conservation of Angular Momentum
A similar result holds for total angular momentum. Consider, about a fixed point ,
where is the total external torque on the system about . Note that the final term does not vanish in general. However, if the force comes from a potential that depends on the distance from to then
so
In this case, we have , so the total angular momentum is conserved in the absence of external torques.
It is more subtle to prove for other forces such as the Lorentz force, but it can be shown that the total angular momentum is still conserved for all known forces. So,
We often take , the centre of mass. However in general, in a general , we need to generalise the above definition (refer to Example Sheet 3).
4.1.3 Conservation of Energy
We can write
where is the position of the -th particle relative to the centre of mass. We have
Note that this is true for all time, so
The kinetic energy is
To have a conserved total energy, all forces must be conservative:
To obey N3, we must have . Hence one can show that
is a conserved quantity, i.e. . To check this,
4.2 The Two Body Problem
An important special case is the two body problem, where there are two particles with no external forces (e.g. the Earth and the Moon). We can reduce this to a single particle problem by working in terms of relative separation. Let be the relative separation.
The centre of mass is given by
So we have
By Eq · 232, the kinetic energy is
where
is called the reduced mass.
Note that
Therefore, relative separation also behaves like a single particle problem, and we can use methods already developed.
If one mass is much larger than the other, say , then i.e. in this limit, heavy object is essentially still and the lighter object moves around it.
In general, for , we cannot solve the problem analytically. However, if , we can use statistical physics to make progress (see Part II).
4.3 Rocket Equation and Variable Mass
The relation is useful when the internal forces are complicated. For example, consider a rocket that ejects fuel with speed relative to the rocket.
We shall work in the -axis, since the rocket moves in a straight line.
The process of ejection can be complicated, but the total momentum must obey
Let be the speed of the rocket at time . We have
Hence
This is called the Tsiolkovsky rocket equation. We can also write this as
where is the thrust force, coming from N3.
5 Rigid Bodies
This is a class of tractable -body problems, where the distances between the particles are fixed. In practice, this is due to very strong internal forces.
The only motions a rigid body can undergo are translations of the centre of mass, and rotations.
5.1 Angular Velocity
In three dimensions, rotations are described by an angular velocity vector . We write
where points along the axis of rotation, and is the angular speed of rotation. The direction of is determined by the right hand rule.
This is captured by the equation
We have
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orthogonal to both and
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Hence indeed . Note that .
In addition to the angular velocity, a rotation must specify a point about which the axis of ration passes, since there are infinitely many parallel axis an object can rotate about.
in the equation is the position relative to some (any) point on the axis of rotation.
5.2 Moment of Inertia
Rotation of a particle involves kinetic energy. For a single particle, we have
where is the perpendicular distance of particle from an axis of rotation.
In a rigid body, all particles rotate with the same angular velocity:
This keeps the distances between particles fixed, since
The kinetic energy of a rigid body is then
where
is the moment of inertia of the rigid body about the axis of rotation.
In Eq · 263, see that is effectively a rotational mass. The bigger is, the harder it is to rotate the body.
The angular momentum of a rigid body is
In this course, we only consider the component of along the axis of rotation, so define
Again, we can observe that is a rotational mass.
Recall that torque causes change in the angular momentum, as . If the torque is also along the axis of rotation, then we can write
and dotting with gives
Hence acts like a rotational force, causing change in the angular velocity.
To calculate the moment of inertia, we use the fact that at large , the particles are densely spaced, and the sums can be approximated by integrals.
where is the density of mass of the rigid body. We typically consider uniform density, so
which is a constant.
For example, we have
where is the perpendicular distance from to the axis of rotation.
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Consider a rotating hoop of radius . We have
Hence .
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Consider a rotating rod of length about an axis through an endpoint.
Then
Hence .
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Consider a rotating disc of radius about an axis through its centre.
Then
Hence .
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Consider a rotating sphere of radius about an axis through its centre.
Then
Hence .
5.3 Perpendicular Axis Theorem
We will now consider less symmetric axes.
Proof.
Consider any 2D body.
Then
By inspection, we have .
5.4 Parallel Axis Theorem
Let the moment of intertia through the parallel axis (not through the center of mass) be , and the moment of inertia through the centre of mass be . We have
where is the total mass of the body, and is the distance between the two axes.
Proof. To prove this, we will express all positions relative to the centre of mass. Choose an origin on the parallel axis, and let be the position of particle relative to this origin. Then
where is the position of the centre of mass, and is the position of particle relative to the centre of mass. Note
We have
Since , the middle term vanishes, and we have
by noting that and .
Consider a rotating disc about an axis through its edge.
We have
5.5 Motion of Rigid Bodies
We will now consider the cases where CoM moves as the body rotates. We have
where the term will capture the rotation about the CoM if
The velocity of the body is
For the kinetic energy, we have shown in Eq · 232 that
Then the full energy of the body is , where we have previously shown that
Consider a nice case, where
then,
where is the -component of the centre of mass. Hence a rigid body is just like a point particle with mass , located at the centre of mass.
Consider a rigid rod pendulum of mass .
We will carry out calculation in two ways:
-
the end point (pivot) is fixed, so there is only rotational motion about this point, giving
in which case and .
-
the CoM is moving with . Hence
To check that the angular velocity about CoM is the same as that about the pivot, we have
Hence,
To understand the motion, consider the energy
Imposing gives
No-slip rolling occurs when the friction between the ball and the ground is so strong that the relative velocity between the point of contact and the ground is zero. Compare:
The kinetic energy is given by, with and noting that ,
Important. Because there is no relative velocity between the point of contact and the ground, no work is done by the friction force, so the energy is conserved.
The only role of rolling is to impose the no-slip condition.
In the case where the ball rolls down a slope,
The conserved energy is
By imposing , we have
Now consider a ball rolling on a horizontal surface, where we will demonstrate the conservation of energy. We have
Then
where is the friction force, and we have used the no-slip condition .
5.6 Normal Forces
Objects on a table does not fall through the table, because the table exerts a normal force on the object that pushes it away. [Microscopic origin of normal forces is electrostatic repulsion and the Pauli exclusion principle.]
At an angle,
Normal force does not prevent the object from sliding. Sliding is prevented by the dry friction force . Once the object starts moving, we typically have
where is the coefficient of friction.
Normal forces also product elastic bounces:
Normal force on impact is in the direction normal to the surface at point of impact.
Note that by conservation of total momentum,
where is the momentum of the Earth or the wall after the collision.
By conservavtion of energy,
where is the mass of the ball, and is the mass of the Earth or the wall. Clearly .
Substituting Eq · 324 into Eq · 325 gives
If , then , and we have
Hence we can conclude that
The change in momentum over a short time is called an impulse .
6 Rotating Reference Frames
Rotating reference frames (RRFs) are important examples of non-inertial frames.
6.1 Newton’s Equations in a Rotating Reference Frame
An inertial frame has Cartesian axes , , , and a rotating frame has axes , , .
From the perspective of the inertial frame, the axes rotates with angular velocity .
In the two frames, the position of a particle is, repsectively,
We wish to find in terms of and . We have
where means the derivatives of components of with respect to in the frame .
The difference between the two time derivatives is just the relative velocity of the two frames.
For Newton’s second law, we need to find the acceleration,
i.e.
In the inertial frame, we have
Hence,
A free particle does not move in a straight line in the rotating frame.
Consider the rotating frame of the earth. We have
We shall neglect the small wobbling of the Earth, so assume , and hence no Euler force.
6.2 Centrifugal Force
We have
It points away from the axis of rotation, as shown in the following diagram.
For the size of the force,
The centrifugal force is conservative, with
Hence, potential energy is lowered by moving away from the axis of rotation.
Consider a hanging string on the Earth.
Rather than hanging vertically downwards, the pendulum hangs at an angle to the vertical. We wish to find .
The forces acting on the particle satisfy
[The string is short compared to , so it does not matter whether we use at the top or the bottom of the string.]
To hold the string together, there must be a force exerted by the molecules on the string that balances the other forces, which is the tension.
The net force on the particle is zero, so
We have 2 equations (for and ) and 2 unknowns ( and ), so we can solve for :
At the equator (), the gravity is a bit weaker, but .
When , , so the effect is very small.
6.3 Coriolis Force
In Eq · 342, we have
where is the velocity of the particle in the rotating frame.
Note that this is similar to Lorentz force with , so moving particles will turn in circles.
Coriolis force is responsible for the formation of hurricanes.
When a low pressure region forms, air particles move in, and the Coriolis force bends them:
Each molecule of air in bent clockwise in the northern hemisphere [by the right hand rule with going into the plane], which leads to an anticlockwise swirling motion.
In the southern hemisphere, the Coriolis force bends particles anticlockwise, leading to a clockwise swirling motion.
Motion along the Earth’s surface is not in general perpendicular to the axis of rotation . Hence, the effect of Coriolis force is typically weaker near the equator. There are empirical observations that hurricanes do not form within near the equator.
[ can be substantial near the equator if moves along the equator, but it pushes particles vertically, and it need to compete with gravity, which is much stronger.]
Consider dropping a ball from a tower on the euqator. We will consider where the ball lands.
Initially,
As the ball falls, the distance to the axis decreases, so the angular velocity must increase to conserve angular momentum.
At the foot of the tower, , which must give , and hence the ball rotates faster than the Earth, so it lands slightly east of the foot of the tower.
In the rotating frame,
[We can neglect the centrifugal force since it does not affect the horizontal motion.]
Integrating once gives
where is the initial position of the ball. Subsituting Eq · 358 into Eq · 357 gives
The last term acts in the vertical direction and is small, so we can neglect it. Hence, integrating twice gives
Consider the following right-handed set of basis:
Then, substituting back into Eq · 360 gives
Clearly, is negative at positive , so the ball lands slightly east of the foot of the tower.
6.4 Foucault’s Pendulum
Foucault’s pendulum demonstrates the rotation of the Earth.
As the Earth rotates under the pendulum, from the point of view of someone on the Earth, it will look like the pendulum rotates.
At a general latitude,
In the basis on the Earth’s surface, we have
The tension in the string is
Since the string doesn’t break, we have
Now, for the equations of motion,
Note that we have all the quantities defined, with 4 equations (3 ODEs and 1 constraint) and 4 unknowns (, , , and ), so we can solve for the motion of the pendulum. Our strategy is as follows
-
Solve constraint for in terms of and ,
-
Substitute into the ODEs,
-
Eliminate to get 2 ODEs for and ,
-
Solve the ODEs.
The exact solution is tedious, but the upshot is that the pendulum follows an ellipse in the plane that slowly rotates, i.e.
The period of rotation is
7 Special Relativity
7.1 Basic Postulates of Relativity
Maxwell’s equations (1862) predicted the existence of electromagnetic waves. These travel in the vaccum at the speed of light,
The quantity is exact due to the definition of the metre.
However, a theory with a preferred velocity cannot be Galilean invariant, since Galilean theories can only have relative velocities. [This could be fine, such as the case of the sound in air which travels at a definitive speed, but this is relative to the air, which is at rest. In the case of light, there is no medium, so there is no reference frame in which the light is at rest.] It was assumed that light must also be travelling in some fixed medium, called the luminiferous æther.
Michelson and Morley (1881) performed an experiment that effectively showed that if one run towards a wave of light, the speed of the light does not change. This was a strong indication that there is no such thing as the æther, and that the speed of light is invariant in all inertial frames.
Einstein (1905) postulated that there was no æther, and
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The laws of physics are the same in all inertial frames. [This is Galileo’s principle of relativity.]
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The speed of light in vacuum is the same in all inertial frames.
7.2 Lorentz Transformations
From the postulates, we must change the rules for transforming between frames.
We shall start with one spatial dimension .
A frame has coordinates . A frame has coordinates , and moves at a constant speed relative to .
According to Galileo,
Instead, let’s allow for a general transformation,
By postulate (1), law of inertia is still true, so a particle experiencing no forces moves at a constant velocity in all frames. i.e. for such a particle,
We shall choose our frames to have a common origin i.e. . [We can always shift and to achieve this.]
Hence, the transformation must map all lines in the plane to lines in the plane. These are precisely linear transformations. Hence, we can write
Note that does not depend on or , but can depend on .
Consider the frame to be moving at speed relative to in . Then, the point should map to . Hence,
We will follow a few steps to determine more results about the transformation.
Consider and where the -axis is inverted in direction, i.e.
Since moves at speed relative to , moves at speed relative to . Hence, the transformation from to is given by
and so .
Another argument for is that in 3D there is no preferred direction, so can only depend on .
Step 2. We can assume that if we boost by and then by , we should get back to the original frame, i.e.
i.e. the boost by is the inverse transformation of the boost by .
We have
So
We have
and
Hence, solving for gives
See that the transformation only makes sense for . So we have derived
This is the Lorentz transformation, or the Lorentz boost. These are linear transformations, so we can invert them to get
i.e. .
For velocities , and , so these become the Galilean transformations i.e. the non-relativistic limit.
We will now explore some possibly counter-intuitive consequences of the above.
7.3 Addition of Velocities
A particle moving with speed in a frame which in turn moves at a speed in frame . Consider the speed the particle moves at in frame .
By Eq · 395 and Eq · 396, we have
Therefore, we have the relativistic formula for addition of velocities:
It is easy to check that if and are both less than , then so is . Hence, we cannot make an object move faster than light by adding velocities.
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Let . Then
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Let . Then
7.4 Spacetime Diagrams and Simultaneity
Consider the following type of diagram
We may put the axes for a frame on the spacetime diagram of :
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axis is at , which gives .
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axis is at , which gives .
where the axes are symmetrical about the light ray .
We can draw lines at constant and to get
So and are simultaneous in but not in . This is the relativity of simultaneity.
There is a direct consequence of the speed of light being the same, which is demonstrated by a famous illustration due to Einstein. Consider a train.
A light is emitted from the centre of the train. The light reaches the front and back of the train at the same time in the frame of the train. However, for an observer on the platform, the light reaches the back of the train before it reaches the front. Hence, events that are simultaneous in one frame may not be simultaneous in another frame.
[In Galilean physics, the platform observer would see light going at in one direction and in the other, reaching both ends at the same time.]
One may concern that if different frames see things at different times, we might be able to reverse cause and effect. We will see that this is not the case.
The future light cone of consists of all points that can be influenced by , and the past light cone of consists of all points that can influence .
Note that the lines of simultaneity of can be at most at as . Hence, we can make simultaneous with , but cannot make simultaneous with .
Therefore, in all frames, the future light cone of is to the future of , and nothing moves faster than light, so causality is ensured. i.e. all frames agree on what events can influence.
7.5 Time Dilation
A clock at rest in frame ticks at intervals . i.e. the ticks occur at
In frame , using Eq · 395 and Eq · 396, where
with , we have the ticks occurring at
So . Recall that , so . i.e. a moving clock runs more slowly. [ is the time interval between ticks on the moving clock as measured in , and is the time interval between ticks on the clock as measured in its rest frame .]
7.6 Twin Paradox [Not a Paradox]
Consider twin that stays on Earth, and that goes to Neptune and back at almost light speed. We would investigate which one is younger.
Each twin sees the other one moving while themselves staying at rest, so each one thinks the other one is younger. This seems paradoxical.
The key asymmetry is that twin needs to turn around to come back.
From the perspective of , time passes, while has passed on ‘s clock, so is younger.
From the perspective of , while travelling out and back, indeed ‘s clock runs more slowly, and passes in total.
However, the jump in the lines of simultaneity at the turnaround means that sees ‘s clock jump forward by a large amount, so is older.
7.7 Length Contraction
A rod has length at rest in frame , where the length is defined as the distance between endpoints at a fixed time.
Now consider frame .
We would like to calculate given .
We have
i.e. for we have .
Hence,
Therefore, moving objects appear shorter. This is called Lorentz contraction.
Consider a ladder of length and a barn of length .
Due to lorentz contraction, if the ladder moves fast enough, an external observer would see the ladder fit in the barn. However, the observer on the ladder would see the barn even shorter, so the ladder cannot fit in the barn.
7.8 The Invariant Interval
Consider two events and with coordinates and in frame . Let
each represent the separation in time and space between the two events.
The invariant interval is defined as
We can check that it is indeed invariant:
Hence, observers may disagree about time passed and the distance between events, but they all agree on the invariant interval. We can write
With this matrix form, the Lorentz transformation can be written as
The invariance of is equivalent to
i.e. Lorentz transformations preserve the Minkowski metric.
The Minkowski metric is not positive definite. So, we have points with
Note that, two points with are connected by a light ray. The Minkowski metric mesures distances in spacetime.
7.9 Rapidity
To make the analogy with rotations more clear, we will define rapidity as
Then,
So the Lorentz transformation can be written as
Hence, two sequential Lorentz boosts satisfy
i.e. rapidities add like angles in rotations.
In contrast, in terms of velocities,
which is consistent with the relativistic addition of velocities.
7.10 Lorentz Transformations in 4 Dimensions
The 4D Minkowski metric is
we write this with indices as
An event in spacetime is given by a 4-vector
we write this with indices as
The invariant distance between and an event is given by the inner product
The inner product of is not positive definite. We call each of the cases
The 4D Lorentz transformations are matrices such that
[In indices, this is ]
The defining feature of Lorentz transformations is that they have the inner product invariant,
Consider the number of . There are 16 entries, and since both sides of the above are symmetric, we have 10 constraints. Hence, we expect to find 6 families of Lorentz transformations.
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3 of them are rotations of the form
which satisfy .
These give 3 independent rotations about different axes. Composition of them also includes reflections.
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3 of them are Lorentz boosts along the 3 possible axes of the form
and similarly for boosts along and axes.
These matrices form the Lorentz group: .
Because the inner product is preserved by these transformations, it automatically follows that the speed of light is the same in all frames, because null 4-vectors are null in all frames.
Subgroups with are called the proper Lorentz group, denoted by .
A further subgroup are those that do reserve the time direction, called the proper orthochronous Lorentz group, denoted by . For example,
is in but not in .
7.11 Proper Time
We want to define a velocity that is a 4-vector. We need to find a time that is invariant under Lorentz transformations, and define the 4-velocity as
Given two points along a worldline, the invariant interval is the same in all inertial frames. Hence, the proper time between these points is defined as
Note that worldlines are always timelike, so is real.
All frames agree on , and they can parameterise the worldline by
[This is a Lorentzian version of the arc length.]
Along a small segment of the worldline,
We can define the 3-velocity as
hence
where here is a function of the instantaneous 3-velocity . Hence by above,
A clock following the worldline has
i.e. the proper time is the time measured by an observer following the worldline.
7.12 4-Velocity
Because is invariant and
transforms by Lorentz: , then
This also transforms as
The definition of a 4-vector implies that it transforms this way.
In particular, because of Eq · 450,
In fact,
The relativistic 4-vector incorporates the familiar 3-velocity into a nice Lorentzian object.
7.13 4-Momentum
The 4-momentum is defined as
where is the property of the particle called rest mass.
The relativistic energy and the relativistic 3-momentum are defined by
So the 4-momentum can be written as
Note that
i.e. the 4-momentum combines energy and momentum, analogously to how combines time and space.
In the absense of forces, is conserved,
Eq · 457 is a Lorentz-invariant generalisation of Newton’s First Law. It also combines conservation of energy and momentum.
From , we get
In the non-relativistic limit, , we get
The rest mass energy term is a new consequence fo relativity, where we can predict that mass leads to an energy.
Now consider the high velocity limit
where as , . So .
i.e. more and more forces are needed to accelerate the particle. In particular, any finite force cannot accelerate a particle to the speed of light.
Hence, a particle with (called a massive particle) has .
7.14 Massless Particle
In Galilean physics, the notion of a massless particle does not make sense. Our relativistic expression
suggests that massless particles should have
This imply that the 4-momentum of a massless particle lies along a light ray.
4-velocity does not exist for a massless particle, and the 4-momentum is the more fundamental concept.
Using Eq · 431 and Eq · 468, we have
where satisfies . We can interpret this as
where is the 3-velocity, so that , hence massless particles moves at the speed of light.
i.e.
along a light ray , since any points on the trajectory are lightline separated.
Hence no time passes for a massless particle.
There are only two known massless particles: photons and gravitons. We only conider photons (the particles of light) in this course.
From quantum mechanics, we have for a photon,
where is the angular frequency of the photon, and is the wavelength of the photon.
Length contraction means that different observers see different wavelengths of light, and hence sees different energies. (See Example Sheet.)
7.15 Particle Physics
Particle accelerators like CERN work by colliding particles at relativistic speeds. The proper framework for this is quantum field theory, but nonetheless we can learn some things from
i.e. the 4-momentum must be the same before and after the collision.
Basic processes include particle delay and particle collisions.
7.15.1 Particle Decay
Heavy particles are often unstable and decay into lighter particles, i.e. the Higgs boson decays in seconds.
Such particles are detected by their decay products, for example,
where is the Higgs boson and is a photon.
From Large Hadron Collider,
which is about times heavier than an electron.
By conservation of 4-momentum,
Before solving this, we need to choose a frame to work in. The two canonical options are
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the “lab” frame, in which the particles are moving.
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the “centre of mass” (or “centre of momentum”) frame, in which the total 3-momentum is zero.
(2) is often more convenient. For decays, (2) is the rest frame of the unstable particle.
In the CoM frame, we have
By conservation of 4-momentum, we have
Hence,
i.e. the two photons emerge with opposite 3-momenta and equal energies. Also,
Hence, each photon carries half of the rest mass energy of the Higgs boson. i.e. rest mass energy has been converted into kinetic energy.
In the lab frame,
consider the angle . The idea is to use conservation and invariants [quantities that do not change under Lorentz tranformations, such as the inner product of 4-vectors]. Since for photons, we have
Now, since and , we have
Using this alongside Eq · 460, we can solve for given and .
7.15.2 Particle Collisions
Consider the process of two particles of mass colliding, in the centre of mass frame, scattering at an angle ,
By conservation of 4-momentum, we have
In the CoM frame, we have
Hence and . Therefore,
Considering the time component of Eq · 487, we have
i.e. any angle is allowed, but momenta and energy afterwards equal the initial momenta and energy. All momenta are in the same plane, so we can pick, for example,
In the lab frame,
where is at rest, hit by . Our objective is to find the angle .
The important step is to use invariants. We firstly have
Since we are looking for information related to , the best way to eliminate is to take the inner product with itself, so that
Since is at rest,
Hence,
We have obtained in terms of and . Using Eq · 460 eliminates the terms.
See Example Sheet 4 for Compton scattering, which describes a massless particle scattering off a massive one.
7.15.3 Particle Creation
If two particles collide with enough energy, some of that energy can be used to create a third particle. This is usually how we discover new particles.
In the CoM frame,
Let and be of mass , and the created particle be of mass . We have
Since we are in the CoM frame,
Squaring Eq · 499 gives
Note that
With variables as defined above,
Proof. is invariant, so we can work in any frame. In the rest frame of ,
Hence
Since , we get
Note that we chose the sign for the square root. The other sign leads to wrong non-relativistic limit, i.e. the kinetic energy would be negative. Recall that proper orthochronous Lorentz transformation will preserve positivity of .
Now, applying the lemma to Eq · 502, we have
Using Eq · 501, we have
i.e. the total kinetic energy of the incoming particles
must be greater than the rest mass energy of the created particle, .
7.16 Accelerated Motion in Special Relativity
We may define the acceleration 4-vector, for a massive particle,
Since , we have
Since , using chain rule , we have
We would like to consider motion for a constant acceleration, but we need to define a frame in which the acceleration is constant.
We will take an inertial frame that, at some moment in time, is instantaneously travelling at the same speed as the particle, i.e. . Hence and in that frame.
[More precisely, given that in that frame.]
To make calculation easier, we will consider motion in one spatial dimension, so that in we have
We say the acceleration is constant if is constant. [Later we will see that this follows from a constant force.]
We can get from by using inverse Lorentz transformation Eq · 395 and Eq · 396, by speed ,
Matching with the general expression for ,
We can hence solve for and hence , giving
Then
For early times , we get .
For late times , we get .
Since , we can integrate again to get
This is the equation of an hyperbola. An accelerated particle follows a hyperbola in spacetime.
The Rindler horizon separates the accessible and inaccessible regoins of sapce time. [Black holes work in a similar way: one need to accelerate not to fall in.]
Let the force 4-vector obey
Let us parameterise as
Recall that . For the spatial component of Eq · 525,
Hence is the force appearing in Newton’s law, but note that is the relativistic 3-momentum. This connects to previous discussion that particles get heavier as they speed up.
Also, if is constant [i.e. a constant force], then is constant in , i.e. indeed a constant produces a constant acceleration in the instantaenous rest frame.
Consider the time component of Eq · 525,
So is proportional to change of energy with time.
To recover a familiar equation, using Eq · 454,
Hence,
which essentially says that the change in energy is the rate of work done.
7.17 Example Lorentz Force [Non-Examinable]
Lorentz force can be written in the form
where is the electromagnetic 4-tensor,
The time component of Eq · 533 is
i.e. the electric field does work, but magnetic field does not.
The spatial component of Eq · 533 is
Therefore
i.e. the Lorentz force is relativistically invariant, but we need to account for the mixing of and under Lorentz transformations, where