1 / 255

# Rotational Mechanics & Special Relativity - PowerPoint PPT Presentation

The Department of Physics. Part IA Natural Sciences Tripos 2013/14. Rotational Mechanics & Special Relativity. Lecture 1. Point your browser at: www-teach.phy.cam.ac.uk/teaching/handouts.php. Course materials. This space is for your own notes. Text-book references:

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Rotational Mechanics & Special Relativity' - sonya-beck

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 1

• www-teach.phy.cam.ac.uk/teaching/handouts.php

• Text-book references:

• MO: Mansfield & O’Sullivan

• TM: Tiplar & Mosca

• You can also find these slides, in colour, on the Physics Teaching website:

• www-teach.phy.cam.ac.uk/teaching/handouts.php

• 7

• 6

### Mechanics in Rotational Motion

The centre of mass

• Suppose that we have a distribution of masses, all within one rigid body. Where should we put the fulcrum such that there is no rotation when suspended in a uniform gravitational field? This point is the ‘centre of weight’, but it also has special properties when there is no gravitational field and so is usually called the centre of mass. For the simple case of a uniform light rigid rod of length l connecting two point masses m1 and m2, it is obvious where the centre of mass must be:

• l2

• l1

• m1

• C

• m2

• Centre of mass

• x

• x1

• x2

• x0

We require that the total turning moment about C is zero, so we define its position to be such that m1l1 = m2l2. Since l1 = x0 – x1, and l2 = x2 – x0, we can write this as

where M = m1+m2.

• We can generalise this to a one-dimensional rigid body of N masses as follows:

• For a three-dimensional rigid body, this expression must be satisfied in each dimension simultaneously, and we may write:

• Now we can express this in vector form, writing the position vector of the centre of mass as R = (x0, y0, z0), and of each contributing mass as ri = (xi, yi, zi):

• back

• Hence, for continuous bodies, we can replace the sum over i elements with an integral:

• In 1 dimension, dm is l dl, where lis the mass per unit length and l is the length variable.

• In 2 dimensions, dm is a da, where ais the mass per unit area and a is the area variable.

• m continuous body using integration rather than summation. Write the summation for 2

• m1

• m3

• In 3 dimensions, dm is v dv, where vis the mass per unit volume and v is the volume variable.

• We can also use the principle of superposition to work out what happens if the rigid body is ‘lumpy’. The centre of mass of a composite system of several lumps of mass can be calculated by first finding the centre of mass of each lump separately, and then finding the centre of mass of the whole considering each lump as a point mass at its individual respective centre of mass.

• m2

• m1

• =

• +

• +

• m3

• For discrete masses: continuous body using integration rather than summation. Write the summation for

• For continuous bodies:

• y

• δx

• x0

• b

• x

• x

• h

• Example 1: find the centre of mass of a lamina shaped like an isosceles triangle

• Divide the lamina into thin strips, each having a width of δx. Take moments about the origin. The moment of the element shown is xδm, where δm is the mass of the element. The area of the element is 2yδx, and the area of the lamina is bh/2.

• y

• δx

• x0

• b

• x

• x

• Let δx tend to zero, and integrate to find the total moment. Then

• h

• Example 1: find the centre of mass of a lamina shaped like an isosceles triangle

• Divide the lamina into thin strips, each having a width of δx. Take moments about the origin. The moment of the element shown is xδm, where δm is the mass of the element. The area of the element is 2yδx, and the area of the lamina is bh/2.

• y continuous body using integration rather than summation. Write the summation for

• (m1+m2)g

• l1

• l2

• m2

• m1

• m2g

• m1g

• x

• x1

• x2

• x0

The lever balance

• We find experimentally, that the lever is balanced only when m1l1 = m2l2

• This is the case even though the sum of all the forces is zero wherever we put the fulcrum, i.e.

• We conclude that a secondcondition is necessary for equilibrium, that is that “the sum of all the turning moments must be equal to zero”.

• This is expressed in the law of the lever:

• where li is the perpendicular distance and fiis the force.

• The two conditions which must both be satisfied zero wherever we put the fulcrum, i.e.simultaneously for a body in static equilibrium are:

• The vector sum of all the external forces acting on the body is exactly zero.

• The sum of all the turning moments about an axis through any point is exactly zero.

• MO 149

• TM 397

Circular motion zero wherever we put the fulcrum, i.e.

• When a particle changes its position from P to P1 relative to a fixed point, O, we can describe the change in position using position vectors. The incremental vector δris added to the position vector, r, describing the position P, to find the new position P1:

• This constitutes a complete description of the change in position. Note, however, that the

• P1

• r + δr

• δr

• P

• O

• r

• change has two parts: a change in the zero wherever we put the fulcrum, i.e.radial distance from O, and a change in the angle about an axis OQ perpendicular to the plane of the triangle OPP1:

• Here, we shall be concerned with the rotational aspects of the motion, i.e. things to do with

• Q

• P1

• r + δr

• δr

• θ

• P

• O

• r

• changes in the angle, zero wherever we put the fulcrum, i.e.θ. Note that this always requires you to define an axis about which the rotation takes place, in this case the axis OQ. The rotation can be in two senses, clockwise and counter-clockwise (or anti-clockwise). We adopt the right-handed convention to describe the positive direction of the axis of rotation (as shown by the arrow from O to Q). When looking along the axis in the direction of the arrow, the positive rotation is clockwise. If you look along the axis in the other direction, such that the arrow is pointing directly at you, rotation is anticlockwise.

• Q

• θ

• θ

• O

• For very small angles, the segment of the circle AB (on the surface of the sphere of radius a centred on O) becomes approximately straight, and can therefore be represented by vector s of length a.Three small rotations can be performed such that the segments

• form a closed vector triangle. The rotation axes are perpendicular to each rotation, so straight lines proportional to the angles also form a closed triangle, i.e. they add up as vectors.

• Q

• a2

• B

• a3

• s

• a1

• A

• a

• O

• θ

The moment of a force as a vector surface of the sphere of radius

• We have already met the concept of the “moment of a force”, or the “turning moment”. We defined this as the force times the perpendicular distance of its point of action from the axis of rotation. We can represent the moment as a vector which, just as in the case of the angular velocity etc., points in a clockwise sense along the axis of rotation, and has a magnitude equal to the magnitude of the turning moment.

• We consider a force, F, acting at point P which is described by position vector r from origin O:

• G surface of the sphere of radius

• F

• Q

• θ

• r

• G

• P

• O

G = rF

• Q

• rsin(θ)

• θ

• rsin(θ)

• Force Facts at point P,

• which is at position

• vector r from point O.

• The moment of the force about O is G = Frsin() about the axis through O perpendicular to both r and F, clockwise looking in the direction of the arrow.

• We define the moment vector G=rF

• F surface of the sphere of radius

• a

• r1

• F

• r2

• O

• Example 2: find a vector expression for the moment of a couple.

• A couple is a combination of two equal and opposite forces which are not in line with each other.

• The forces act at position vectors r1 and r2 with respect to an arbitrary origin, O.

• F surface of the sphere of radius

• a

• r1

• F

• r2

• O

• Example 2: find a vector expression for the moment of a couple.

• A couple is a combination of two equal and opposite forces which are not in line with each other.

• The forces act at position vectors r1 and r2 with respect to an arbitrary origin, O.

• The moment of the couple is independent of the origin O, and depends only on the vector forces and the position vector of the point of action of one of them with respect to the other.

Angular acceleration of a rigid body surface of the sphere of radius

• Newton’s second law gives us the relationship between the linear force applied to a body of mass m and its acceleration. We have seen that the rotational equivalent of the force, F, is the moment of the force (or torque), G, and the rotational equivalent of the linear acceleration, , is the angular acceleration, . Can we find an equivalent to Newton’s second law for rotational mechanics, and if so, what is the equivalent of the mass, m ?

• Q surface of the sphere of radius

• v = rω

• ω

• m

• O

• r

• To answer this question, consider a particle of mass m rotating about an axis OQ:

• Its linear speed is v = rω, but let us suppose that its speed is increasing, i.e. it has an acceleration in the direction of v. N2 tells us that there must therefore be a force acting on the particle in this

• direction of magnitude surface of the sphere of radius F where

• The magnitude of the moment of this force about OQ is directed along OQ. For this particle we may therefore write

• Now we can consider a rigid body rotating about OQ as made up of the sum of N such elementary particles. Let the ith such particle be a distance ri from OQ and have a mass of mi. Then summing over the whole body, we have:

• where surface of the sphere of radius Gext is the total external vector moment acting on the body about OQ.

• If G is the rotational equivalent of F, and is the rotational equivalent of , then we must conclude that the rotational equivalent of the mass of the body is:

• We call this quantity the moment of inertia, I, and then the rotational equivalent of N2 is

• MO 154 surface of the sphere of radius

• TM 293

Moment of inertia

• In linear mechanics, the mass is measure of a body’s reluctance to change its state of linear motion. The larger the mass, the slower the rate of change of velocity for a given applied force. In rotational mechanics, the moment of inertia takes the place of the mass, and it is a measure of a body’s reluctance to change its state of angular motion. The larger the moment of inertia, the slower the rate of change of angular velocity for a given applied moment of a force. Like the mass, the moment of inertia is a scalar quantity, usually

• given the symbol surface of the sphere of radius I. (Do not confuse this with the vector impulse, .) Thus, for a rigid body which can be thought of as being composed from N particles:

• In terms of I, we can write the rotational equivalent of N2 as

• flywheel surface of the sphere of radius

• G

• ω

• motor

• Example 3: an electric motor is attached to the axis of a massive flywheel of moment of inertia 70 kg m2. When an electric current is switched on, the motor applies a constant torque of 150 N m. What is the rotation rate of the flywheel after 30 s?

• The torque causes an angular acceleration which increases the angular velocity of the flywheel.

• flywheel surface of the sphere of radius

• G

• ω

• motor

• Example 3: an electric motor is attached to the axis of a massive flywheel of moment of inertia 70 kg m2. When an electric current is switched on, the motor applies a constant torque of 150 N m. What is the rotation rate of the flywheel after 30 s?

• The torque causes an angular acceleration which increases the angular velocity of the flywheel.

Moments of inertia of continuous bodies surface of the sphere of radius

• The expression we have obtained for the moment of inertia, I, of a rigid body involves a summation over N elemental mass contributions:

• For continuous bodies, it is more convenient to use an integral form. We imagine that the body is divided into a very large number of very small masses, δm, all joined together to make up the whole. A particular one is at distance r from the axis, so it contributes an amount δI to the total

• where surface of the sphere of radius δI = r2δm. Now go to the limit as δm tends to zero, and integrate over all the contributions to get the total moment of inertia, thus:

• The trick in applying this expression is often to express dm in terms of the distance variable (r) using the density. You should also take advantage of the spatial symmetry of the problem to simplify the expression that you must integrate. Some examples follow.

• rod surface of the sphere of radius

• δx

• x

• l

• Example 4: find an expression for the moment of inertia of a rod of length l about an axis through one end perpendicular to the rod.

• Let the rod’s density be  per unit length. Consider an element of the rod, length δx, at distance x from one end. The mass of the element is δm = δx. The contribution to the total moment of inertia from this element is δI, given by:

• Let the rod’s density be  per unit length. Consider an element of the rod, length δx, at distance x from one end. The mass of the element is δm = δx. The contribution to the total moment of inertia from this element is δI, given by:

• rod

• δx

• x

• l

• Let the disc have a surface density of  per unit area. Consider a radial element of the disc at radius r, thickness δr. This has mass δm which is equal to 2rδr. It makes a contribution, δI, to the total moment of inertia given by:

• r

• a

• δr

• Let the disc have a surface density of  per unit area. Consider a radial element of the disc at radius r, thickness δr. This has mass δm which is equal to 2rδr. It makes a contribution, δI, to the total moment of inertia given by:

• r

• a

• δr

• Note that this formula also applies to a cylinder.

• r solid disc of mass M and radius a about an axis through its centre perpendicular to the plane of the disc.i

• mi

• Ri

Parallel axes theorem

• This theorem allows us to obtain the moment of inertia, I, of a rigid body about any axis, AB, parallel to an axis, OP, through the centre of mass about which the moment of inertia of the body is I0.

• Consider the contribution from the element mi:

• B

• rigid body

• P

• centre of mass

• a

• MO 155

• TM 297

• O

• A

• zero for solid disc of mass M and radius a about an axis through its centre perpendicular to the plane of the disc.

• centre of mass

• Let I be the moment of inertia about axis AB.

• Let I0 be the moment of inertia about an axis, OP, parallel to AB and through the centre of mass.

• Let solid disc of mass M and radius a about an axis through its centre perpendicular to the plane of the disc.I be the moment of inertia about axis AB.

• Let I0 be the moment of inertia about an axis, OP, parallel to AB and through the centre of mass.

• B solid disc of mass M and radius a about an axis through its centre perpendicular to the plane of the disc.

• P

• a = l/2

• A

• O

• Example 6: find an expression for the moment of inertia of a rod of length l and mass M about an axis through its centre perpendicular to the rod.

• Let the moment of inertia about the axis through the centre be I0. We have already calculated the moment of inertia, I, about a parallel axis through one end of a rod to be Ml2/3. Thus:

• Let the moment of inertia about the axis through the centre be I0. We have already calculated the moment of inertia, I, about a parallel axis through one end of a rod to be Ml2/3. Thus:

• B

• P

• a = l/2

• A

• O

• z rod of length l and mass M about an axis through its centre perpendicular to the rod.

• y

• ri

• mi

• x

Perpendicular axes theorem

• This is another useful theorem which relates the moments of inertia about three perpendicular axes through any point in a lamina, one of which is perpendicular to the plane of the lamina.

• Consider the contribution by an elemental mass mi:

• Let the lamina be in the rod of length l and mass M about an axis through its centre perpendicular to the rod.xy plane. Then the moment of inertia about the z axis is

• We have already found the moment of inertia of the disc about the z axis to be Ma2/2. Thus

• z

• a

• y

• x

• z disc of radius a and mass M about an axis through its centre in the plane of the disc.

• a

• y

• x

• Example 7: find an expression for the moment of inertia of a disc of radius a and mass M about an axis through its centre in the plane of the disc.

• We have already found the moment of inertia of the disc about the z axis to be Ma2/2. Thus

Sounds of Pulsars disc of radius a and mass M about an axis through its centre in the plane of the disc.

• 0329+54 P=0.71452s

• 0833-45 P=0.089s

• 0531+21 P=0.033s

• 0437-4715 P=0.00575s

• 1937+21 P=0.00167s

• Single pulses are usually very variable !

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 3

• www-teach.phy.cam.ac.uk/teaching/handouts.php

• MO 103 disc of radius a and mass M about an axis through its centre in the plane of the disc.

• TM 331

• L

• P, v

• P, v

• m

• m

• θ

• θ

• r

• r

• A

• A

• B

• B

• rsin(θ)

• rsin(θ)

Angular momentum

• We define the angular momentum, L, to be the moment of the momentum about a point.

• Suppose that a particle, A, of mass m and at position vector r relative to point B, has momentum P= mv, where v is its velocity:

• The moment of the momentum (by direct analogy with the moment of a force) is given by

• L = rP = rmv = mrv

Conservation of angular momentum disc of radius a and mass M about an axis through its centre in the plane of the disc.

• The angular momentum, like the linear momentum, is conserved in an isolated system. To show this, consider a system of N interacting particles. The angular momentum of the ith particle is Li = miri vi. The total angular momentum of the system is therefore given by

• This is a vector addition of the angular momenta of all the particles about a given point. We can differentiate with respect to time to find the rate of change of L:

• Now the value of disc of radius a and mass M about an axis through its centre in the plane of the disc.Gint is zero for the following reason:

• F disc of radius a and mass M about an axis through its centre in the plane of the disc.ji

• j

• b

• rj

• i

• O

• ri

• Fij

• The internal interaction on the ith particle by the jth particle is in line and oppositely directed to the interaction on the jth particle by the ith particle by N3. The moment about O is riFij + rjFji

• = (ri – rj) Fij = bFij . This is zero since b is in line with Fij. All the internal interactions are in

• similar pairs, each of which comes to zero. Hence disc of radius a and mass M about an axis through its centre in the plane of the disc.Gint must be zero. Therefore

• and we conclude that the rate of change of the total angular momentum of a system is just the vector sum of the external moments applied to that system.

• When the system is isolated, Gext = 0, so we conclude that the total angular momentum of an isolated system is constant.

• Notes: disc of radius a and mass M about an axis through its centre in the plane of the disc.

• This statement is always true, no matter what dissipative forces there might be internally.

• This statement is always true about any axis, not just one through the centre of mass.

• The rotational equivalent of N2 is and we have just shown that . Therefore we can write

Angular impulse disc of radius a and mass M about an axis through its centre in the plane of the disc.

• We defined the impulse of a force as the integral of the force with respect to time, i.e.

• The action of the impulse was to change the linear momentum, thus:

• Now the moment of the force is rF, and the angular momentum is rP. Taking the cross product with r in the above equation we get:

• Thus we see that when a disc of radius a and mass M about an axis through its centre in the plane of the disc.moment of a force acts for a finite time, it causes the total angularmomentum of the system to change, the change being equal to the integral of the angular moment with respect to time.

Rotational kinetic energy disc of radius a and mass M about an axis through its centre in the plane of the disc.

• The translational kinetic energy for a particle of mass m moving in a straight line at speed v is mv2/2. We can obtain the equivalent rotational quantity by considering the ith particle of a rigid object in rotation about an axis at at angular speed ω. Let the particle have mass mi and beat distance ri from the axis. Then:

• Summing over the entire body ( disc of radius a and mass M about an axis through its centre in the plane of the disc.N particles), we get

• Note that, for an individual point particle, we can consider its kinetic energy either as the translational quantity mv2/2, or as the rotational quantity mr22/2. However, in general the motion of a solid object must be analysed in terms of the linear motion of its centre of mass, plus the rotation around the centre of mass, so the total KE is the sum of the translational and rotational components.

Rotational oscillations: the physical pendulum disc of radius a and mass M about an axis through its centre in the plane of the disc.

• Consider a rigid body of arbitrary shape that is suspended from, and free to rotate about, a horizontal frictionless axis (A). It is displaced slightly from equilibrium. What is the period of oscillation?

• A

• A

• l

• l y

• l

• C

• C

• y

• mg

• ω

• A equilibrium position by an amount

• l

• m

• L

• C

• M

• 2a

• Example 8: the pendulum of a grandfather clock is made from a brass disc of diameter 2a and mass M with its centre attached to the end of a thin metal rod having a mass of m so that the centre of the disk is L below the point of suspension. What is the period of the pendulum?

• The moment of inertia of the disc about an axis through its centre and perpendicular to its faceis Ma2/2. Using the parallel axes theorem we have:

• The moment of inertia of the rod is mL2/3, so the total moment of inertia is

• Now the centre of mass is at distance l below

• A equilibrium position by an amount

• l

• m

• L

• C

• M

• 2a

• the pivot such that

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 4

• www-teach.phy.cam.ac.uk/teaching/handouts.php

The general motion of a rigid body equilibrium position by an amount

• We have seen already that the linear motion of a rigid body can be analysed by considering only the linear motion of its centre of mass which moves as if it carries all the mass and is acted upon by the sum of all the external forces. This is a required condition, but is not sufficient to describe the entire motion of the body because is does not take into account the rotational motion of the body about the centre of mass. We need a second statement, added to the first, to define the general motion of a body, linear plus rotational.

• The additional statement is that equilibrium position by an amount there isrotation about an axis through the centre of mass of a body which is the result of the sum of the external moments of all the forces acting on the body about that axis, and the moment of inertia of the body about that axis. This second statement, added to the first, is sufficient to define the general motion of a body.

• We will clarify this with the example of a cylinder rolling down an inclined plane:

• N equilibrium position by an amount

• a

• ω

• a

• ω

• F

• v

• v

• mg

• α

• mg

• α

• Example 9: what is the acceleration of a uniform solid cylinder rolling, without slipping, down a plane inclined at angle α to the horizontal?

• The cylinder is acted on by the body force, mg, by the normal reaction force, N, and by the frictional force, F, as shown in the diagrams. The net result is (a) the centre of mass accelerates down the plane, and (b) the cylinder rolls down the plane.

• N equilibrium position by an amount

• a

• ω

• F

• v

• mg

• α

• (a) The linear motion of the centre of mass is as if it is a point mass equal to the total mass of the cylinder, acted upon by the sum of the external forces. Resolving down the slope:

• (b) The rotation about the cylindrical axis is as if acted upon by the sum of the moments of the external forces:

• where I is the moment of inertia about the cylindrical axis. We also have v = ωa. Thus

• Substituting for equilibrium position by an amount F gives us

• Now I = ma2/2, so we deduce that

• Substituting for equilibrium position by an amount F gives us

• Now I = ma2/2, so we deduce that

• N equilibrium position by an amount

• a

• In terms of energy:

• (a) Rotational KE of the cylinder is

• ω

• F

• v

• (b) The linear KE of the centre of mass is

• (c) Thus when the centre of mass has dropped through a distance h we have

• mg

• α

• s

• h

• α

• N equilibrium position by an amount

• a

• In terms of energy:

• (a) Rotational KE of the cylinder is

• ω

• F

• v

• (b) The linear KE of the centre of mass is

• (c) Thus when the centre of mass has dropped through a distance h we have

• mg

• α

• s

• h

• α

Rotating frames of reference equilibrium position by an amount

• Suppose that we have a particle, P, which is rotating at constant angular speed in a circle about O:

• y

• ω

• y

• P

• r

• x

• P

• O

• ωt

• x

• O

• r

• ω

• ω

• ω

• Oblique view

• Plan view

• This tells us that the point P has a constant acceleration (since ωandrare constant) which is of magnitude ω2r and is directed along r in the negative direction, i.e. towards O. The actual speed of the particle is constant and equal to ωr in a tangential direction, but the acceleration arises from the fact that P is constantly changing direction towards the centre of the circle, so the vector velocity is constantly changing.

• Newton’s second law of motion tells us that there must be a force associated with the acceleration.

Centripetal force (since

• The centripetal force is the force on a particle which is directed towards the axis of rotation and which is required to maintain the rotational motion of the particle. From N2:

• where is the unit vector along r. Note that the centripetal force does no work as the velocity and force are orthogonal to each other.

• m

• r

• F

• We can identify quantities in linear mechanics and rotational mechanics which behave in equivalent fashions. If you are not sure what to do in a rotational problem, think what you would do in the equivalent linear problem, and then use the table below.

• centre

• of mass

• u

• v0

• b

• B4

• a)

• b)

• Example 10: (Tripos 2000). A rod of mass M and length L lies on a smooth horizontal table. A small particle of mass m travels at speed v0 on the table at 90° to the rod. It collides with the end of the rod and sticks to it. Calculate the speed of the centre of mass of the combined rod and particle after the collision, and find the new position of the centre of mass …

• a)Linear momentum before = linear momentum after

• centre

• of mass

• u

• v0

• b

• B4

• Example 10: (Tripos 2000). A rod of mass M and length L lies on a smooth horizontal table. A small particle of mass m travels at speed v0 on the table at 90° to the rod. It collides with the end of the rod and sticks to it. Calculate the speed of the centre of mass of the combined rod and particle after the collision, and find the new position of the centre of mass, …

• a)Linear momentum before = linear momentum after

• u

• b

• b) New position of the centre of mass: moments about it sum to zero

• u

• b

• b) New position of the centre of mass: moments about it sum to zero

• After

• centre

• of mass

• u

• v0

• b

• B4

• The principle of the conservation of angular momentum may be applied about the centre of mass:

• ω

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 5

• www-teach.phy.cam.ac.uk/teaching/handouts.php

• MO 164 rotation about the centre of mass.

• TM 339

• z

• x

• y

The Gyroscope

• A gyroscope is heavy flywheel which is rapidly rotating about an axis. It is usually mounted so that it can be turned about either of the two other orthogonal axes:

• L = I

• Whilst the flywheel is stationary, there is nothing unexpected about the gyroscope, so that a moment applied about the y or z axes produces rotation about the y or z axes.

• When the flywheel is rapidly rotating about the x axis, a moment about y produces slow rotation (called precession) about z and vice-versa. This is apparently counter to expectations, but is readily understood, either in terms of angular momentum, or in terms of the forces acting on particles in the flywheel.

A precessing gyroscope unexpected about the gyroscope, so that a moment applied about the

• The gyroscope is precessing about the

• z axis. Look down from on top along –z :

• Ω

• Consider the motion of a particle (blue square) which is just coming up towards you at A, moving over the top at B, and disappearing at C

• C

• z, Ω

• ω

• B

• x, L

• The blue particle moves from A to B to C. At B, it is moving in a curved path. Therefore it must feel a force to the left

• A

• t

• tδt

• t+δt

• y

• Looking from the side along –y :

• z, 

• F

• y

• x, L

• F

• Ω unexpected about the gyroscope, so that a moment applied about the

• G

• t

• L+δL

• t+δt

• L

• L+δL

• L

• δL

Precession using vectors

• Apply a couple of moment G to a rotating flywheel:

• In time δt the moment, G, of the couple exerts a change in angular momentum of Gδt, so that δL = Gδt. Note that this change is directed along G. If G is perpendicular to L (as here) then the change in angular momentum is also perpendicular to the angular momentum.

• L then keeps constant magnitude, but is constantly changing direction in the plane of G and L, so precession Ω takes place.

• Ωδ unexpected about the gyroscope, so that a moment applied about the t

• L+δL

• δL=Gδt

• L

• Hence, there is precession about an axis perpendicular to both L and G. The rate of precession, Ω rad s1, is easily calculated. In a small time, δt, it precesses through small angle Ωδt.

• In this triangle, we can see that, as δt is small, we can write

• Gδt= ΩδtL

•  G = ΩL

• In fact, we can write this in vector form as follows:

• Ωδ unexpected about the gyroscope, so that a moment applied about the t

• L+δL

• δL=Gδt

• L

• Hence, there is precession about an axis perpendicular to both L and G. The rate of precession, Ω rad s1, is easily calculated. In a small time, δt, it precesses through small angle Ωδt.

• In this triangle, we can see that, as δt is small, we can write

• Gδt= ΩδtL

•  G = ΩL

• In fact, we can write this in vector form as follows:

• To see this, consider a gyroscope at an angle to the horizontal plane:

• Ω unexpected about the gyroscope, so that a moment applied about the

• L

• θ

• G

• In the case where the gyroscope is at an angle, first resolve the angular momentum into vertical and horizontal components:

• LV = L cos(θ) is constant.

• LH = L sin(θ) is affected by the couple and is therefore changing direction but not magnitude.

• G =Ω L sin(θ)

• G is perpendicular to the plane containingΩand L, so we may write

• G = ΩL = I Ω ω

Gyroscope examples unexpected about the gyroscope, so that a moment applied about the

• (i) Luni-solar precession

• The effect of the unequal ‘pull’ of gravity from the Moon and Sun on the non-spherical Earth applies a moment which causes the N-S axis to precess with a period of about 26,000 y. We see this a slow change in the positions of the stars with time.

• L

• N

• G

• Moon, Sun

• S

• An atom has an angular momentum and a magnetic moment. The magnetic moment subjects the atom to a couple when a magnetic field is applied which results in small changes in the energies of its electronic states. This results in the splitting of spectral lines – the Zeeman effect.

• B

• L, m

### Einstein’s theory of unexpected about the gyroscope, so that a moment applied about the Special Relativity

• MO: 193-227

• TM: 1319-1356

Frames of reference unexpected about the gyroscope, so that a moment applied about the

• A frame of reference is just a set of axes which we can use to define points (or ‘events’). We are all familiar with Cartesian frames (x,y,z), but there are others commonly used.

• We need a frame of reference in which to define positions, velocities, and accelerations. For example, a position vector r might have the coordinates (2,4,7) in one Cartesian frame. In another, the same vector might be (6,5,11). A reference frame helps us to be specific about our measurements.

• We use this concept of a ‘frame of reference’ widely in Physics. The ‘laboratory frame’, for example, is often the one in which you, the observer, are situated. You must imagine a set of axes fixed to the floor, and you are standing stationary at the origin. It is often helpful, however, to change your viewpoint to another frame, say one which is moving at a steady speed through the laboratory frame parallel to the x axis. For example, you can imagine standing on the platform of a station (the ‘laboratory frame’) watching someone run past you at 10 mph. Coming into the station is a train,

• also moving at Physics. The ‘laboratory frame’, for example, is often the one in which you, the observer, are situated. You must imagine a set of axes fixed to the floor, and you are standing stationary at the origin. It is often helpful, however, to change your viewpoint to another frame, say one which is moving at a steady speed through the laboratory frame parallel to the 10 mph parallel to the runner. If you were to transform your point of view into the train’s frame (the ‘moving frame’) you would see the runner apparently running on the spot, not making any progress at all relative to the train. We make wide use of the concept of a frame of reference in the theory of special relativity

• y

• y’

• yP’

• P

• 3

• S’

• S

• x

• 5

• 30°

• xP’

• x’

• Example 1: a particle, P, is situated at position vector (5,3) in a two-dimensional coordinate frame S. What are its coordinates measured in frame S’ which has the same origin as S but is rotated clockwise by 30 degrees relative to S?

• y

• y’

• yP’

• P

• The new coordinates are therefore

• 3

• S’

• S

• x

• 5

• 30°

• xP’

• x’

• Example 1: a particle, P, is situated at position vector (5,3) in a two-dimensional coordinate frame S. What are its coordinates measured in frame S’ which has the same origin as S but is rotated clockwise by 30 degrees relative to S?

• 1 by angle

• 2

• 3

• 4

• 5

• S

• metres

• Before coming on the the development of Einstein’s Special Theory of Relativity, it will be useful to examine in general terms some assumptions we have been making about the frames of reference in which we have been thinking about physics. Thus:

• (a) A frame of reference is just a set of calibrated axes or coordinate system against which we can measure positions, velocities, accelerations, and times. Typically, we might

• y

• S

• x

• (x,y,z,t)

• (x,y,z,t )

• (x,y,z,t )

• y

• y

• S

• x

• S

• x

• say that an ‘event’ occurs at a particular instant in space and time defined in one coordinate system, S, by (x,y,z,t). The same event, seen in another coordinate system, S, is defined by the coordinates (x,y,z,t ).

• (b) The geometry of the space in which we are working obeys the axioms derived from the postulates of Euclid: in this space, two parallel lines meet at infinity, the sum of the angles in a triangle is 180 degrees, etc. We refer to this as ‘Euclidean space’.

• (c) The distance between two events in space seen in one frame is the same when viewed in any other frame.

• Note: do not confuse frame names, S, S’ etc, with the variable s

• sAB

• y

• B

• yB

• y

• A

• yA

• x

• x

• S

• xA

• xB

• Thus if events A and B occur in Cartesian frame S at positions (xA,yA,zA,tA) and (xB,yB,zB,tB), then the space interval, sAB, between the two events viewed in S is given by Pythagoras’ Theorem.

• Thus by angle s2AB= (xB xA)2 + (yB yA)2 + (zB zA)2.

• Now view the same two events from the point of view of another frame, S.

• y

• y

• y

• B

• A

• sAB

• S

• x

• S

• x

• x

• The two events occur in S at coordinates of (xA,yA,zA,tA) and (xB,yB,zB,tB). Again the space interval, sAB, between the two events, viewed in S, is given by Pythagoras’ Theorem, i.e.

• s2AB= (xB xA)2 + (yB yA)2 + (zB zA)2.

• Our assumption is that s2AB= s2AB.

• y by angle

• x

• y

• y

• B

• y

• A

• x

• S

• S

• x

• x

• Our assumption is that s2AB= s2AB.

• (d) The by angle time interval is the same in any frame. Thus tAB= (tB tA)= tAB= (tB tA). In fact we have a strong notion that time and space are absolute quantities. We think that we can define a point in ‘absolute’ space and ‘absolute’ time, and that space and time are the same for everyone, no matter how they are moving with respect to each other. These ideas obviously work very well in everyday life, but need closer examination.

• (e) We can express the transformation between coordinates seen in one inertial Cartesian

• x by angle A

• y

• v

• A

• xA

• vtA

• S

• x

• frame and those in another using the ‘Galilean’ transformation. Suppose that frame S' is moving along the positive x axis of frame S at constant speed v, and their origins coincide att = t = 0.

• An event, A, occurs in S at (xA,yA,zA,tA), and in Sat (xA,yA,zA,tA ).

• MO 173

• TM 1322

• y

• x

• S

• We see xA = xA vtA, which is the only coordinate affected, so the Galilean transformation is:

• x = x  vt

• y= y

• z= z

• t= t

• This is the transformation which applies to all the Newtonian Physics you’ve done so far. As we shall see, it only works for transformations between frames in which v << c, the speed of light.

• (f) Note that the by angle quantitiesx, x, etc. are really intervals between the two events: (i) the origins coinciding with each other, and (ii) event A. Even here, we are expressing the transformation between space and time intervals, not between absolute space and absolute time positions. We could therefore equally well write:

• x = x  vt and equally x = x’ + vt’

• y= y y = y’

• z= z z = z’

• t= t t= t’

• where  denotes the interval between events.

• Example 2: an observer in a high-speed train, travelling at 575 km h‒1(currently the speed record held by the TGV) measures the time between his passing two signals as precisely 3 s. What is the distance between the two signals measured by a second observer on the track using a tape measure?

• The observer in the train is present at both events, so he measures a space interval of zero.

• The Galilean transformation gives:

• ΔxAB

• A

• B

• S (track frame)

• v

• A,B

• Δx’AB = 0

• Δt’AB= 3

• S’ (train frame)

• Example 2: an observer in a high-speed train, travelling at 575 km h‒1 (currently the speed record held by the TGV) measures the time between his passing two signals as precisely 3 s. What is the distance between the two signals measured by a second observer on the track using a tape measure?

• The observer in the train is present at both events, so he measures a space interval of zero.

• The Galilean transformation gives:

• ΔxAB

• A

• B

• S (track frame)

• v

• A,B

• Δx’AB = 0

• Δt’AB= 3

• S’ (train frame)

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 6

• www-teach.phy.cam.ac.uk/teaching/handouts.php

Problems with classical physics 575 km h

• Nineteenth-century physicists sought mechanical explanations for everything. Even Maxwell’s electromagnetic theory was based on mechanical interactions. In particular, it was thought that light waves, like all other known waves, must travel in a medium. Thus water waves travel on the surface of water; sound waves travel through the air; waves on my mother’s washing line travelled down the line (waves on a string). What about light, radio etc.? The work of Huygens, Young,

Problems with classical physics

• 1D wave equation for waves on a string, or water waves, or sound waves:

• Nineteenth-century physicists sought mechanical explanations for everything. Even Maxwell’s electromagnetic theory was based on mechanical interactions. In particular, it was thought that light waves, like all other known waves, must travel in a medium. Thus water waves travel on the surface of water; sound waves travel through the air; waves on my mother’s washing line travelled down the line (waves on a string). What about light, radio etc.? The work of Huygens, Young,

• represents the amplitude of the wave, that is the displacement of a point in the medium, and c is the speed of the wave. For example, for waves on a string:

• James Clerk Maxwell showed that, for electromagnetic waves (light, radio, x-ray etc.) the equation was

Problems with classical physics 575 km h

• Nineteenth-century physicists sought mechanical explanations for everything. Even Maxwell’s electromagnetic theory was based on mechanical interactions. In particular, it was thought that light waves, like all other known waves, must travel in a medium. Thus water waves travel on the surface of water; sound waves travel through the air; waves on my mother’s washing line travelled down the line (waves on a string). What about light, radio etc.? The work of Huygens, Young,

• and Fresnel seemed to account for the properties of light in terms of waves. Therefore, (it was thought) there must be a corresponding medium for them? They made the hypothesis that there was indeed such a medium, and they called it the luminiferous aether. Thus, light travelled at 3108 m s1 through this medium which was all-pervading. This hypothesis could be tested by looking for effects caused by motion through the aether.One such piece of evidence was supplied by Bradley in 1725 who observed stellar aberration.

Stellar aberration aether

• Imagine you are in vertically-falling rain. Now get on your bike – the rain appears to come down at you from an angle to the vertical.

This was easily measured by Bradley, and appeared to show evidence for a stationary aether.

• c

• aether wind

• v

• Earth’s orbital velocity through the stationary aether

• One of the most-famous attempts to measure motion through the aether was the Michelson-Morley experiment. The Earth moves at about 30 km s1 in its orbit around the Sun, which is an appreciable fraction of the speed of light, 300,000 km s1. M & M set up an optical interferometer which would have been easily sensitive enough to detect this motion. The principle was that a coherent light beam was divided into two parts, and each part sent along perpendicular paths as follows:

• B

• v

• d

• C

• A

• S

• d

• T

• O

• This is a simplified diagram of a MM interferometer. You will meet this again in more detail next year.

• A beam of light from a coherent light source, S, is split by the half-silvered mirror, A, into two beams, one travelling towards B, and the other towards C. The beams are reflected by the fully-silvered mirrors B and C. That from B passes through A to T, and that from C is reflected at A towards T. The two beams combine in the telescope, T, and interfere to produce an interference pattern, which is measured by the observer O.

• Let us suppose that the apparatus is moving through a stationary aether from right to left, at speed v, so that there is an ‘aether wind’

• blowing from left to right parallel to AC. The two light paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:

• AC is with the flow, so tAC = d/(c+v);

• CA is against the flow, so tCA = d/(cv).

• tACA = d/(c+v) + d/(cv).

Now consider the path ABA:

Both AB and BA are across the wind. The light gets ‘blown’ to the right, so the path from A to B is slightly against the flow and takes longer.

• v paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:

• aether wind

• B

• The speed from B to A is the same as that from A to B, so the total time of flight is

• v

• c

• A

• paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:the time difference between ACA and ABA is

• Now, if x is much less than 1 we can use the first two terms of the binomial expansion of

• ( paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:1+x)n1+nx, so that

• With the apparatus used by MM, this corresponded to a shift in the fringe pattern of about half a fringe, very easily seen if it existed.

• However, paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:no shift was ever seen, despite the experiment being repeated with first AC then AB parallel to the Earth’s orbital velocity, and at six month intervals (just to check that the aether wasn’t coincidentally moving at the same velocity as the Earth when the experiment was first done).

• Although this experiment is often cited as evidence that the aether does not exist, Einstein was probably not aware of it when he formulated his theory of special relativity.

• not accelerated paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:

Classical electromagnetism

• Newtonian physics appears to operate in accord with the Galilean transformation, i.e. we can transform the (x,y,z,t) coordinates of events seen in one inertial frame into those seen in another using this transformation.

• However, Einstein was aware that Maxwell’s laws of classical electromagnetism (which you will come to next year) did not transform in the same way. In particular, the speed of radio or light waves was predicted to be given

• by the expression , where ε0 and μ0

• are constants associated with paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:the vacuum. This seems to say that the speed of light in a vacuum is independent of the motion of the source or the observer, since there is no meaning to motion relative to a vacuum. No need for an aether – it just doesn’t exist. He postulated that this was not some quirk of electromagnetism, but that its consequences applied to the whole of physics. He formulated the theory of relativity – special relativity (1905) applying to un-accelerated frames, and general relativity (1916) which is about gravity. Here we do the special theory.

• NB! paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:

Einstein’s postulates

• MO 194

• TM 1321

• Einstein’s ideas about electromagnetism and the nature of physical laws may be summarised in his two postulates:

• All of the laws of physics are the same in every inertial (un-accelerated) frame.

• The speed of light in a vacuum is the same for all observers.

• The first of these postulates is quite consistent with the Newtonian mechanics that you have already met in school. You made the assumption that

• Newton’s laws were true and obeyed in any inertial frame, and were consistent with the Galilean transformation. However, the new thing is that all the laws of Physics, including electromagnetism and anything else that you can mention, are the same in every inertial frame.

• Turning that around, the first postulate states that, if you are in an inertial frame, there is no internal experiment which you can do which can distinguish between that frame and any other inertial frame – all are equivalent. For example, there is no such thing as an absolute rest frame.

• When I am at rest in my inertial frame, that state of rest is the same as any other in any inertial frame no matter how fast it is moving relative to mine  a revolutionary concept for people seeking an all-pervading aether.

• The second postulate really follows on from the first. If there is no internal experiment I can do to tell which particular inertial frame I am in, then the speed of light  in a vacuum must be the same for me as for anyone else. This has far-reaching implications for space and time.

What is time? is the same as any other in any inertial frame no matter how fast it is moving relative to mine

• Time is a notion or a concept. We know that it is not a substance. We know that it always moves in one direction, i.e. ‘time passes’ or ‘time advances’. We assume that its rate of flow is uniform, and that it is universal.

• We measure time in terms of intervals, that is we identify events and then we measure how many ‘ticks’ (assumed equally spaced) there are between the events using a machine – a clock – that has been designed to produce ticks at as uniform a rate as possible.

• Time is the most-accurately measured of all physical quantities by many orders of magnitude.

How is time measured? is the same as any other in any inertial frame no matter how fast it is moving relative to mine

• Time intervals are measured using processes which are assumed to be exactly periodic: the swinging of a pendulum; the oscillation of an escapement mechanism; the rotation of the Earth; the orbit of the Earth around the Sun; the vibration of an excited atom.

• Since 1967, we have used atomic clocks to measure time. The SI second is defined to be exactly 9,192,631,770 cycles of vibration in an atomic clock controlled by one of the characteristic frequencies of caesium 133.

• Notes: is the same as any other in any inertial frame no matter how fast it is moving relative to mine

• Atomic time is now independent of astronomy.

• We keep times consistent with astronomy by inserting leap seconds up to twice per year at midnight on June 30th or December 31st.

• The pulses received from highly regular ms pulsars may supersede atomic clocks in the future.

• The time scale disseminated by radio (UTC) – e.g. the ‘time pips’ on the (analogue) BBC – is an average over many atomic clocks in many countries.

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 7

• www-teach.phy.cam.ac.uk/teaching/handouts.php

• MO 195 is the same as any other in any inertial frame no matter how fast it is moving relative to mine

• TM 1324

Time dilation

• One of the consequences of Einstein’s postulates is that identical clocks appear to run at different rates depending on their relative motion. To see how this comes about, we will consider a special kind of clock. Its ‘pendulum’ is a photon reflected back and forth between mirrors. Since the speed of light is directly involved with the mechanism of this clock, we can see quite easily how Einstein’s postulates, especially the second one, affect the performance of the clock. His first postulate tells us that all clocks, whatever the mechanism, must be affected in

• mirrors

• photon

• clock face

• B ‘speed’ of our inertial frame relative to absolute rest by comparing this clock with one having a different mechanism):

• First view the situation in the upper clock’s rest frame. Let events A, B, C be a photon leaving the base mirror, arriving at the upper mirror, and arriving back at the base mirror respectively. In the clock’s rest frame, S, the photon takes a time 2h/c to travel from A to B and back to C (where c is the speed of light). Let this time interval be . Then

• h

• A

• C

• rest frame S

• back

• Now let’s look at the ‘speed’ of our inertial frame relative to absolute rest by comparing this clock with one having a different mechanism):same events as seen in the lower frame, S. The time interval between events A and C in this frame is tAC. Remember that the photon still travels at speed c, but has further to go, so takes longer:

• B

• h

• A

• D

• C

• v

• In the triangle ‘speed’ of our inertial frame relative to absolute rest by comparing this clock with one having a different mechanism):ABD, by Pythagoras’ theorem, we have

• But in the moving clock’s rest-frame, S, we have

• So substituting for h gives us

• Rearranging, we find ‘speed’ of our inertial frame relative to absolute rest by comparing this clock with one having a different mechanism):

• where .

• Note that is greater than one for all speeds such that 0  v  c, and is undefined for speeds of c or greater. This equation therefore shows us that the time interval between events A and C is shortest in the rest frame of the clock, and that the time interval measured between the same two events viewed in a frame in which that clock is moving is longer.

• Notes ‘speed’ of our inertial frame relative to absolute rest by comparing this clock with one having a different mechanism):

• We are forced to conclude that the rate of the passage of time depends on relative motion.

• The shortest time interval between two events is measured by a clock which is present at both events.

• Such a time interval is called a proper time interval, and such a clock measuresproper time.

• The time interval measured in another moving frame, using two clocks each of which is at only one of the events, is always longer.

• (e) This is sometimes summarised in the statement ‘moving clocks run slow’. Be careful with this statement as it can cause confusion. Always ask: “Which clock was present at both events?” (It measures the shortest time interval.)

• (f) This true for everykind of clock, not just light-clocks (remember Einstein P1).

• (g) The effect is tiny in every day life: 70 mph for 6 years causes a 1s shift

• (h) The effect is larger for space travellers: at four-fifths of c, the shift is from 3 s to 5 s

• Two quotations on the relativity of time: clocks run slow’. Be careful with this statement as it can cause confusion. Always ask: “Which clock was present at both events?” (It measures the shortest time interval.)

• Einstein 1905: “Thence we conclude that a clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.”

• Rosalind 1599: “Time travels in divers paces with divers persons.”

• (As you like it, Act 3 Scene 2, by William Shakespeare)

• MO 204 clocks run slow’. Be careful with this statement as it can cause confusion. Always ask: “Which clock was present at both events?” (It measures the shortest time interval.)

• TM 1335

• We have seen that the time interval between two events measured by the captain of a space ship is shorter than the time interval between the same two events measured by observers at rest relative to the Earth.

• Consider the following train of events:

• A

• B

• v

• C

• E

• A rocket leaves Earth (event clocks run slow’. Be careful with this statement as it can cause confusion. Always ask: “Which clock was present at both events?” (It measures the shortest time interval.)A) and travels to a distant point at very high speed. It turns around (event B) and travels back to Earth, arriving (event C) several years after leaving. We know from what we have done already that the time interval ΔtABbetween events A and B, and the time interval ΔtBC, between events B and C, measured by the rocket captain, will be shorter than the corresponding intervals, ΔtABand ΔtBC, measured on the Earth. (Remember that only the rocket captain’s clock is present at all the events.)

• Thus we conclude (correctly) that the twin brother of the rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• But now consider the same events as seen by the rocket captain. She sees the Earth receding at high speed v behind her on the outward leg, and then approaching at high speed v on the return leg. According to the rocket captain, her brother left on Earth has been travelling relative to her, and so will be younger than her when she gets back to Earth. This is the twin paradox.

• The twin paradox arises rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey. through sloppy thinking. Actually there is no paradox because the two situations are not symmetrical. Just ask yourself the question: “Who’s clock was present at all the events, A, B, and C ?” Also, the twin brother left on Earth remains the whole time in a single inertial frame of reference. The twin sister in the rocket changes from one inertial frame to another mid-course. On the way out, the frame travels at speed v; on the way back it travels at speed –v. We therefore need to analyse the situation carefully. We will return to this a bit later.

• MO 205 rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• TM 1330

Simultaneity

• Another consequence of Einstein’s postulates is that if two events are simultaneous in one frame, they are not necessarily simultaneous when viewed in another moving frame. Consider the following example: a first observer is at rest relative to, and exactly half-way between, two flashing beacons, one blue and one red. He observes that a flash from the red beacon arrives at exactly the same instant as one from the blue beacon. He therefore deduces that the two flashes were emitted by the beacons at the same moment in his frame.

• blue beacon rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• first observer

• B

• C

• A

• red beacon

• First observer’s frame

• The beacons and first observer are stationary in this frame. The flashes arrive at the same instant (event C), and since the distances are the same, they must have left the two beacons at the same instant (events A and B are deduced to be simultaneous in this frame).

• v rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• v

• v

• Now consider the same circumstances from the point of view of a second observer who is moving at speed v relative to the first observer:

• blue beacon

• first observer

• v

• B

• C

• A

• red beacon

• Second observer’s frame

• In the second observer’s frame, the beacons and first observer are all moving to the left at speed v.

• The second observer rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.must agree that the two photons arrive at the same time at the first observer’s position (event C) since this is an event, and the nature of the event cannot be changed by relative motion. But the first observer is moving at speed v to the left relative to the photons in the two flashes, and so the red photons have to make up extra distance whilst the blue photons have less distance to travel (remember that both sets of photons travel at the same speed, c, in any frame). The second observer must conclude that the red photons left (event B) before the blue photons (event A) since they arrive together (event C) and the reds have

• further to go than the blues rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey..Thus, we have the first observer saying that A and B must have been simultaneous, whilst the second observer says that B happened before A. This is not a contradiction as we allow time to be relative, just as space is relative. We can think of space and time together as making up ‘spacetime’, and that two events, points in spacetime, can be considered as being joined by a vector called a ‘four vector’. All we are doing when we change frames is that we are viewing the four vector from a different point of view.

• I am turning rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.right

• I am turning left

An example in Euclidean space …

• Who is correct? Both are. From the red car’s point of view, the road is on the left. From the green car’s point of view, the road is on the right. We are used to this and so don’t find it strange. So it is with events in space time. Whether one event happens before another depends on your point of view – i.e. which inertial frame you are in.

• MO 201 rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• TM 1326

Length contraction

• A third consequence of Einstein’s postulates is that the length of a rigid bar measured in its rest frame is always greater than the length of the same bar measured in a frame moving parallel to its length.

• To see this, consider the following. A spaceship travels from one interplanetary beacon to another.

• A

• B

• v

• L0

• Event rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.A is that of passing the first beacon, and event B is that of passing the second. The beacons are L0 apart in their rest frame (measured with a tape-measure). An observer at rest with respect to the beacons finds the time interval between A and B is ΔtAB = L0/v.

• The captain of the spaceship measures the distance between the beacons as L, and the time interval as ΔtAB = L/v. Now ΔtAB = γΔtAB(the captain measures a proper time interval).So we must conclude that .

• This means that the captain of the spaceship measures a rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.shorter distance between the two interplanetary beacons than is measured by a tape-measure (or rigid bar) fixed between them. This result applies to all measurements of length, no matter how they are made. The length of an object appears to be contracted in the frame in which it is moving along its length, and rulers are longest when measured in their rest frames.

• What about lengths perpendicular to the motion? The answer is ‘no change’. We can see this by considering a relativistic train running on a section

• of straight track at a steady speed rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.v. When viewed in the rest-frame of the track, the train runs smoothly by at speed +v. When viewed in the rest-frame of the train, the track runs smoothly by at speed v. In neither case is the train seen to come off the track. Therefore, we must conclude that lengths perpendicular to the motion do not change. If they did, an observer could test whether he is ‘fixed’ or ‘moving’ by observing whether the train stayed on the track or came off it, and that would violate Einstein’s first postulate.

Summary rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• Einstein’s postulates lead to the following conclusions:

• Time dilation:

• Simultaneity: events simultaneous in one frame are not necessarily so in another.

• Length contraction parallel to motion:

• No length contraction perpendicular to motion.

Summary rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• Einstein’s postulates lead to the following conclusions:

• Time dilation:

• Simultaneity: events simultaneous in one frame are not necessarily so in another.

• Length contraction parallel to motion:

• No length contraction perpendicular to motion.

• Measuring the speed of light rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• The Department of Physics rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 8

• www-teach.phy.cam.ac.uk/teaching/handouts.php

Lorentz transformation rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• MO 198

• TM 1322

• We have already seen how the Galilean transformation transforms the coordinates of events between different inertial frames which are moving with respect to each other on the basis of classical Physics, i.e. before Einstein proposed his revolutionary postulates. We now need to revise that transformation to incorporate the effects of Special Relativity. The new formulation is called the Lorentz Transformation, and it can be used to solve relativistic problems in a straightforward manner.

• v rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• A

• x

• S'(moving frame)

• S(laboratory frame)

• We consider an event, A, which occurs at position (x,y,z,t) as measured in the laboratory frame S.

• The same event is also viewed in another frame, S’, moving at speed v along x, with its axes parallel to those of S, and such that x = x’ = 0 at t = t’ = 0.

• v rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

• measured in S

• vt

• A

• r

• x

• S'(moving frame)

• S(laboratory frame)

• At time t (measured in S), the origin of the moving frame is at position x = vt in S:

• Imagine that a rigid ruler is fixed to the y’ axis of S’ and is of just the right length so that its far end is instantaneously at event A. Its length in S is r.

• We see that rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.x = r + vt, so r = x ‒ vt.

• Event A occurs at position (x,y,z,t ) measured in the moving frame S. Now we know that a ruler is longest when measured in its rest frame, and is contracted by a factor γ when measured in a frame in which it is moving. In this case, the rigid ruler has a length of r in S and r0 in S’ (it is stationary in this frame) such that r0= γr. But r0 is the x’-coordinate of event A measured in S’.

•  x =  (x – vt) . (1)

• We could have done all this the other way around, i.e. start in frame S’ and then move to S. We do

• not need to start again, however, as all we need to do is to switch the dashes and replace v with v, so we get

• x =  (x + vt ) . (2)

• To find the way the times transform, we can substitute in equation (2) for x from equation (1):

• x =  (x + vt ) =  ( (x – vt) + vt )

• After a bit of algebraic manipulation, we get

• t =  (t – vx/c2).

• Replacing v with –v, and switching the dashes:

• t =  (t + vx/c2).

• Back

• Now motion do not change. Collecting everything together, we have:x, t etc. are really intervals between the two events ‘the origins coincide’ and event A. We can therefore equally well write:

• Notes: motion do not change. Collecting everything together, we have:

• This set of equations is called the Lorentz Transformation. Use it - it will help you to solve relativity problems and get the correct answer.

• The sets (ct,x,y,z) and (ct,x,y,z ) are examples of 4-vectors. The LT can be written:

• Δ relativity, all of which transform using the same xAB

• A

• B

• S (track frame)

• v

• A,B

• Δx’AB = 0

• Δt’AB= 1.198×10‒6

• S’ (train frame)

• Example 3: advances in technology enable the high-speed train of Ex. 2 to travel at 0.8 c. An observer in the train measures the time between his passing two signals as 1.198 μs. What is the distance between the two signals measured by a second observer on the track using a tape measure?

• The observer in the train is present at both events, so he measures a space interval of zero.

• The Lorentz transformation gives:

• Δ relativity, all of which transform using the same xAB

• A

• B

• S (track frame)

• v

• A,B

• Δx’AB = 0

• Δt’AB= 1.198×10‒6

• S’ (train frame)

• Example 3: advances in technology enable the high-speed train of Ex. 2 to travel at 0.8 c. An observer in the train measures the time between his passing two signals as 1.198 μs. What is the distance between the two signals measured by a second observer on the track using a tape measure?

• The observer in the train is present at both events, so he measures a space interval of zero.

• The Lorentz transformation gives:

• t relativity, all of which transform using the same 

• v

• t

• A

• B

• A

• x0/c

• tA = ?

• x0

• xA = ?

• B

• xB = ?

• x0/2c

• 2x0

• x

• S

• x

• S

• Example 4: (part of Ex 11 in book). The coordinates of two events measured in frame S are A: xA = x0, tA = x0 /c; B: xB = 2x0, tB = x0 /(2c). (a) What is the speed and direction of travel of an inertial frame S in which both events occur at the same time? (b) When do the events occur as measured in S ?

• (i) Events are already identified in the question.

• (ii) Draw diagrams:

• (iii) Write down the intervals (remember that the origins coinciding is event 0, so x, t etc are intervals already)

• Event A: xA= x0, tA = x0 /c,

• Event B: xB = 2x0, tB = x0 /(2c)

• Example 15: ( relativity, all of which transform using the same NST – Physics Part IA 1999). Twins Alice and Bob go travelling in space. They each carry a clock to record how much they age during the trip. Alice leaves Earth and travels at a steady speed of 5c/13 to a space-station which is 1 light year away. Bob leaves Earthat the same time as Alice, but travels at a speed of 5c/13 in the opposite direction. When Alice reaches the space-station, she immediatelyturns around and travels towards Bob at a new speed of 12c/13, eventually catching up with him. Which twin is older, and by how much, when they meet in space?

• Bob leaves Earth

• turns around

• meet in space

• Event C

• Event B

• Event A

• Example 15: (NST – Physics Part IA 1999). Twins Alice and Bob go travelling in space. They each carry a clock to record how much they age during the trip. Alice leaves Earth and travels at a steady speed of 5c/13 to a space-station which is 1 light year away. Bob leaves Earthat the same time as Alice, but travels at a speed of 5c/13 in the opposite direction. When Alice reaches the space-station, she immediatelyturns around and travels towards Bob at a new speed of 12c/13, eventually catching up with him. Which twin is older, and by how much, when they meet in space?

• Identify the events:

• (B) relativity, all of which transform using the same

• B

• Bob S

• A

• C

• Alice S

• E

• S

• 1

• 1

• xAC

• In S:

• In Alice’s frame S:

Experimental evidence for relativity relativity, all of which transform using the same

• (a) Time dilation in the decay of muons

• Cosmic rays produce showers of muons at the top of the atmosphere. These have lifetimes of only about 2s, so should travel only a few hundred metres before decaying. (Their speeds are close to c.) In practice, we measure most of them at ground level after travelling through many tens of km of atmosphere. This is because the muon clocks measure proper time intervals between events ‘enter atmosphere’ and ‘decay’, (which are about 2 s), whilst the Earth-based clocks measure much longer time intervals.

• (b) relativity, all of which transform using the same Michelson-Morley experiment

• This is evidence for the absence of the aether. Jaseja, Javan, Murray & Townes (1964) showed that any effect is less that 0.1% of that expected if there was an Aether.

• (c)Magnetic effects

• The magnetic force between two parallel current-carrying wires can be calculated instead from relativistic modifications of the electrostatic forces between the charges in the wires. This demonstrates the consistency between electromagnetism and mechanics brought about by Einstein’s postulates.

• (d) relativity, all of which transform using the same GPS Clocks

• The rates of the clocks in the Global Satellite Positioning System (GPS) satellites need to be adjusted relative to those on the ground for both the time dilation of special relativity and the general relativistic effect of the difference in gravitational potential. Otherwise, the Earth-based clocks drift with respect to the satellite clocks, with corresponding growing errors in positions.

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 9

• www-teach.phy.cam.ac.uk/teaching/handouts.php

The speed of light as the ultimate speed relativity, all of which transform using the same

• The Lorentz transform shows that the speed of light places an upper limit to the speed which which two frames can move with respect to each other. As v c, so  . Then lengths observed in a frame moving with respect to the observer shrink to zero, and time intervals between events (in the moving frame) extend to infinity. Note that this applies to moving objects, or more generally to moving energy. Virtual objects, such as the point of intersection of two

• fixed

• point of intersection

• moves at speed > v

• move up at speed v

• B

• A

• S

• MO 208

• TM 1336

• It is clear that the Galilean addition of speeds formula, , will not work if either or v is a large fraction of c. For example, if = c/2 and v = 3c/4, the result would be 5c/4, an impossible result since it must be less than c. Use the Lorentz transform to get the right formula.

• To see how to do this, consider a rocket travelling at speed relative to a frame S between two markers, A & B. The markers are a distance apart fixed in this frame, and the rocket takes time to travel between them.

• v

• Clearly

• From the point of view of an observer in frame S, relative to which frame S is moving parallel to at speed v, the corresponding intervals are xAB and tAB. The speed of the rocket in this frame is therefore given by

• B

• A

• S

• S

• Dividing through by infinity.tABand cancelling the gammas, we get

• What about motion at an angle? This time, the rocket travels at velocity u in frame S, so we must first resolve u into x and y components:

• B infinity.

• A

• v

• u

• S

• S

• Thus, the ‘vector’ velocity addition formula is

• v

• Guard

• S’ (train frame)

• Guard

• S (track frame)

• Example 6: further advances in technology enable the guard of the high-speed train of Ex. 2 to travel at 0.5 c from the back of the train towards the front. With what speed would an observer on the track see the guard moving when the train is travelling at a speed of 0.8 c?

• c. givessin(θ )

• c

• v

• θ

• c.cos(θ )

• L

• S

• S

Aberration of light

• We have already seen that Bradley measured the aberration of light, and took this to be evidence for the existence of the Aether. But aberration can also be understood in terms of relativity and the complete absence of the Aether. Here’s how. Consider a photon emitted from a light source L which is stationary in frame S’:

• What is the corresponding angle, givesθ, when the photon is viewed from frame S? We can use the addition of speeds formulae to find the components of the photon’s velocity as seen in S:

• In frame S we have cx = ccos(θ), cy = csin(θ).

• In frame givesS we have cx = ccos(θ), cy = csin(θ).

Stellar aberration gives

• In Bradley’s case, θ= 90°, and the direction of the photon was reversed (i.e. incoming photon rather than outgoing). We conclude

• c

sin( ) =cos()  

•   v/c

as Bradley measured and used as evidence, incorrectly, of the existence of the Aether.

• 

• v

• S

• S

Relativistic Doppler effect

• We are all familiar with the Doppler effect in relation to the changing pitch of wailing sirens on emergency vehicles as they pass by. Here we find a formula for the Doppler effect as applied to electromagnetic waves. We do this by considering a pulsing light source which is at rest in a frame S, and finding the time between two consecutive pulses measured in frame S, relative to which S’ is moving along the x-axis at a steady speed of v.

• Let events A and B be the emission of consecutive pulses by the light source in S’.

• v gives

• v

• xAB = ?

• tAB = ?

• O

• B

• A

• xAB

• S

• v

• xAB = 0

• tAB = T0

pulses in frame S:

• The observer in S sees the light source move between pulses, so the second pulse has further to go to reach the observer, O, than the first pulse.

• A,B

• light source

• S

• S

• back

• We use the Lorentz transform to find givesxABand tAB:

• Now the pulse from B has an extra distance, xAB, to travel before reaching the observer as compared with the pulse from A. Thus the time, T1, measured by the observer, O, between the arrival of the pulses is:

• We use the Lorentz transform to find givesxABand tAB:

• Now the pulse from B has an extra distance, xAB, to travel before reaching the observer as compared with the pulse from A. Thus the time, T1, measured by the observer, O, between the arrival of the pulses is:

• back

• If the frequency of the pulses, in the rest-frame of the light source, is 0,then the frequency, 1,measured by the observer in S is

• Note that this is the change observed when the source is moving directly away from the observer. There is a (smaller) Doppler shift even for sources moving at right angles to the line of sight because of the time dilation factor.

• v light source, is

• v

• S – space station frame

• Example 7: an observer on a space station sees two spacecraft passing each other, one coming directly towards him and the other moving directly away, both at a speed of c/2. Both are transmitting radio waves. What is the ratio of the frequencies received by the observer on the space station?

• The observer sees frequency ν1 from the spacecraft moving towards him, and ν2 from the one moving away. The frequency measured on board the spacecraft is ν0.

• v light source, is

• v

• S – space station frame

• Example 17: an observer on a space station sees two spacecraft passing each other, one coming directly towards him and the other moving directly away, both at a speed of c/2. Both are transmitting radio waves. What is the ratio of the frequencies received by the observer on the space station?

• The observer sees frequency ν1 from the spacecraft moving towards him, and ν2 from the one moving away. The frequency measured on board the spacecraft is ν0.

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 10

• www-teach.phy.cam.ac.uk/teaching/handouts.php

• y light source, is 

• y

• y

• Q

• y

• P

• x

• S

• x

• S

• x

• x

Intervals

• Consider two points (‘events’), P and Q, in Euclidean space, observed in frames S and S’:

• The interval, light source, is PQ, is the distance, Δs from P to Q, and is given by Pythagoras’ theorem:

• Note that this is invariant in Euclidean space under rotation and translation – i.e. it has the same value whether viewed in frame S or S.

• What is the equivalent interval in special relativity? Clearly, it is not the same since

• x x whilst y = y. Furthermore, we can see that it must include time.

• The ‘equivalent’ of Pythagoras in SR is light source, is

• You can show this quite easily by substituting for x and t using the Lorentz transform. The quantity s is invariant under the Lorentz transform – that is it has the same value when observed in any inertial frame. We can think of it as the ‘length’ (or norm) of the space-time four-vector. With all four dimensions it is

• The space-time interval between the pulses in the rest-frame of the light source is given by

• The corresponding interval measured in the S frame (the observer’s frame) is

• B consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• A

• C

• ct

• path of a photon

• path of a photon

• Q

• P

• C

• x

• A

• B

• S

Space-time (Minkowski) diagrams

• We can plot events and tracks. The geometry is that of Minkowski space rather than Euclidean space:

• The photon meets observer A at event P, and observer B at event Q

• If we chose the scales consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.ct and x to be the same, then the path of a photon is a straight line at 45° to the axes. The vertical lines AA and BB represent the paths through space-time of stationary observers. The straight line CC is the path of a moving observer, with v < c.

• All of these lines are called world lines.

• This diagram is for the laboratory frame, S. How do we represent the view as seen in a moving frame, S, on the same diagram? The clue is given by CC – i.e. sloping lines for moving observers.

• ct consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• path of a photon

• in both frames

• P

• calibrate the axes using the x-t invariant,e.g:

• x2 – c2t2 = 1

• ctp

• x

• 1

• xp

• 1

• Let the origins coincide at t = t = 0:

• ct

• x

• ct consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• path of a photon

• in both frames

• P

• calibrate the axes using the x-t invariant,e.g:

• x2 – c2t2 = 1

• ctp

• x

• 1

• xp

• 1

• Let the origins coincide at t = t = 0:

• ct

• x

The twin paradox consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• Let us now revisit the twin paradox, and analyse it using space-time diagrams. We will take the specific case that the rocket travels outward for a distance of 4 light years (as measured on the Earth) before returning, travelling at a steady speed of 4c/5.

• v = 4c/5

• A

• B

• C

• E

• 4 light years

The twin paradox consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• ct

• The total journey time taken according to Earth clocks is 10 years: 5 out and 5 back. The twins agree to send radio messages to each other on the anniversaries of departure. According to the Earth-based twin, the rocket will not receive the first message until Earth year 5 after departure, and then it will receive eight other messages at regular intervals over the succeeding 5 years.

• The Earth-based twin knows about special relativity, and recognises that the clocks on the rocket measure proper time and therefore appear to run more slowly than those on Earth. He calculates that the value of  is 5/3, so he expects the rocket-based twin to send just five messages, two on the outward leg, one at turn-around, and two on the return leg, since the rocket years are 5/3 Earth years long.

• The Earth-based twin expects to find his twin sister to be four years younger than him on her return to Earth.

• 10

• 8

• 6

• 4

• 2

• x

• 0

• S (Earth)

• 2

• 4

• 6

Causality consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• ct

• ct

• Future

• Q

• photon world lines

• R

• x

• x

• P

• Elsewhere

• Elsewhere

• Past

• If one event, P, causes another, Q, then Qmust

• lie in the future cone of consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.P, as nothing can travel faster than light.The time interval, PQ, is ‘time-like’, i.e. ctPQ > xPQ. It is then possible to transform to another frame, S, in which P and Q both occur at the same place, separated only by a time interval.

• If the interval is ‘space-like’, i.e. ct < x, as in PR, it is possible to find a frame in which P and R are simultaneous, and yet another in which R occurs beforeP. It is then not possible for R to have been caused by P. The event R lies in the ‘elsewhere’ of P.

• The Department of Physics consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 11

• www-teach.phy.cam.ac.uk/teaching/handouts.php

Relativistic mechanics consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

• MO 210

• TM 1340

• Perspective:

• We know that Einstein’s postulates have altered our views of space, time, and addition of speeds etc. Therefore, we must expect to have to modify our ideas of mechanics – momentum, energy etc.

• We have certain laws which hold true for slowly-moving frames, e.g. conservation of momentum, conservation of energy, conservation of mass etc. How can we modify them so that they hold good in all frames?

• (c) The same modified laws have to hold good in the limit of small v and asymptotically approach the classical formulae.

• (d) We can interface classical conditions with relativistic conditions by considering an elastic glancing collision between two particles of equal mass, m. We can apply the formula for the classical momentum in one frame in order to deduce what the relativistic equivalent must be in the other.

• relativistic

• p, v

• θ

• We assume that the principle of conservation of momentum applies. Then conserving momentum vertically implies that

• 2psin(θ) = 2mu

• p = mu/sin(θ)

Now transform into a frame travelling through the laboratory with speed vcos(θ):

• u

• classical

• Frame S

• Frame S

• The situation is exactly reversed. Apply the relativistic formula for addition of speeds to the lower mass:

• u

• θ

• p, v

• relativistic

• with small uy = vsin(θ), ux = vcos(θ), uy = u so

• Now the interface between the purely classical (low-speed) motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• This suggests that momentum is conserved when defined using the above formula. The only modification required to the classical definition is to multiply by .

• Before motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• After

• u

• v

• at rest

• m

• m

• M

• S (laboratory)

• S (laboratory)

• v

• v

• v

• v

• at rest

• S (ZMF)

• S (ZMF)

Relativistic energy

• Consider the perfectly inelastic high-speed collision between two equal masses:

• The amalgamated mass motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude thatM is stationary in the Zero Momentum Frame, but is travelling at speed v in the laboratory frame. We conclude that the ZMF is also travelling at speed v. Using the relativistic formula for the addition of speeds and cons. of momentum:

• this is the classical KE in the ZMF

• Let’s expand this equation:

• If v << c, then we can use the first two terms of the binomial expansion to get

• v motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• v

• v

• v

• at rest

• S (ZMF)

• S (ZMF)

• This equation tells us that we need to consider mass and energy together in order to conserve energy:

• Notes motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• We deduce from this that mass and energy need to be considered together as different aspects of the same thing. Mass and energy are equivalent to each other.

• The conservation of energy is really the conservation of mass-energy.

• The mass-energy is calculated by E = γmc2.

• When the mass is at rest in a particular frame of reference, γ= 1. Therefore, the rest energy is given by E0 = mc2.

• (e) The kinetic energy, motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude thatK, is the extra energy that the mass has as a result of its motion. Therefore K =E – E0 =γmc2mc2. We can expand this for v << c as follows:

• (e) The kinetic energy, motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude thatK, is the extra energy that the mass has as a result of its motion. Therefore K =E – E0 =γmc2mc2. We can expand this for v << c as follows:

• The classical result for the KE is just an approximation in the limit of zero speed.

• The Department of Physics motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• Part IA Natural Sciences Tripos 2013/14

Rotational Mechanics

& Special Relativity

• Lecture 12

• www-teach.phy.cam.ac.uk/teaching/handouts.php

Relativistic mechanics - summary motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• The momentum of a particle is γmv.

• The energy of a particle is γmc2.

• The total momentum of a system is

• (conserved)

• The total energy of a system is

• (conserved)

• The kinetic energy of a particle is

• (not conserved)

• Before motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• u

• stationary

• A

• B

• 2m

• m, K

• S (lab)

• After

• v

• M

• S (lab)

• Example 19: A particle of mass m and kinetic energy 2mc2strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

• Apply conservation of momentum and mass-energy:

• Before motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• u

• stationary

• A

• B

• 2m

• m, K

• S (lab)

• After

• v

• M

• S (lab)

• Example 9: A particle of mass m and kinetic energy 2mc2strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

• Apply conservation of momentum and mass-energy:

• back

The E-p invariant motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• We saw earlier how we could think of x and ct as two components of a space-time four-vector, and that the ‘length’ or norm of this four-vector was invariant under the Lorentz transformation, i.e.

• In the same way, p and E/c are also components of another four vector called the energy-momentum four vector. This means that the components transform by the Lorentz transformation but it also means that there is an associated invariant quantity (i.e. its ‘length’ or ‘norm’) which remains

• the same for a given system when viewed motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude thatat any time in any inertial frame. This is very powerful and helps to simplify problems considerably. The invariant is

• E2 p2c2 = E2 p 2c2 = m2c4,

• where E is the total energy, p is the total momentum of the given system. The value of m is the total mass in a frame (if there is one) in which all the particles of the system are at rest. For a system of more than one particle, use:

The energy-momentum 4-vector motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• We saw previously that the Lorentz transform for space and time may be written as

• ct and x, y, z form the components of a 4-vector, and we can write the Lorentz transformation more generally as b = A.b, where bandb are 4-vectors and A is the transformation matrix which

• is given by motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• It can be shown that energy and momentum form a four vector, i.e. b= (E/c, px, py, pz):

• Note motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• The ‘length’ or norm of a 4-vector is invariant under the Lorentz transformation. The square of the norm is bb which in Minkowski geometry is given by

• bb = b12  b22 b32 b42

• For the E-p 4-vector this is

• (E2/c2) px2 py2 pz2

• and for the t-r4-vector it is

• (c2t2) x2 y2 z2

• Before motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• u

• stationary

• A

• B

• 2m

• m, K

• S (lab)

• After

• v

• M

• S (lab)

• Example 19 again: A particle of mass m and kinetic energy 2mc2strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

• Apply the left-hand side of the E-p invariant to ‘Before’, and the right-hand side to ‘After’:

• Before motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• u

• stationary

• A

• B

• 2m

• m, K

• S (lab)

• After

• v

• M

• S (lab)

• Example 18 again: A particle of mass m and kinetic energy 2mc2strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

• Apply the left-hand side of the E-p invariant to ‘Before’, and the right-hand side to ‘After’:

Nuclear binding energies motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• The nucleus of an atom is held together by nuclear binding forces, and work must be done against these forces to split the nucleus into its constituents parts, or into fragments containing smaller numbers of nucleons. The binding energy appears as increased mass, so that the mass of the nucleus is smaller than the sum of the masses of its constituent parts. The binding energy per nucleon, however, varies with increasing atomic mass number as follows:

Binding energy per nucleon motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

• The binding energy per nucleon increases sharply at first with atomic mass number, reaching a broad plateau with a peak at about Fe, and then falls again more slowly. We can extract some of the binding energy either in a nuclear fusion process or in a nuclear fission process.

• The fusion of two nuclei with atomic mass numbers smaller than that of Fe generally releases energy, whilst the fusion of nuclei heavier than Fe generally absorbs energy. For example, nuclear fusion of H atoms to form He is the main process powering the Sun at the present time.

• The fission into lighter fragments of a heavy nucleus with an atomic mass number larger than that of Fe generally releases energy, whilst the fission of nuclei lighter than Fe generally absorbs energy. Controlled nuclear fission is used in present-day atomic power stations to generate heat which is turned into steam to drive electricity generators. It may be possible in future to maintain the right conditions for nuclear fusion in a stable system so that electricity can be generated from water with the inert gas He as the by-product.

Splitting the atom: an atomic mass number larger than that of Fe generally releases energy, whilst the fission of nuclei lighter than Fe generally absorbs energy. Controlled nuclear fission is used in present-day atomic power stations to generate heat which is turned into steam to drive electricity generators. It may be possible in future to maintain the right conditions for nuclear fusion in a stable system so that electricity can be generated from water with the inert gas He as the by-product.the Cockcroft-Walton experiment

• protons injected here an atomic mass number larger than that of Fe generally releases energy, whilst the fission of nuclei lighter than Fe generally absorbs energy. Controlled nuclear fission is used in present-day atomic power stations to generate heat which is turned into steam to drive electricity generators. It may be possible in future to maintain the right conditions for nuclear fusion in a stable system so that electricity can be generated from water with the inert gas He as the by-product.

• lithium target

• 0 KV

• 400 KV

• 200 KV

• alpha particles

• fluorescent screen

I wish you a pleasant Easter break an atomic mass number larger than that of Fe generally releases energy, whilst the fission of nuclei lighter than Fe generally absorbs energy. Controlled nuclear fission is used in present-day atomic power stations to generate heat which is turned into steam to drive electricity generators. It may be possible in future to maintain the right conditions for nuclear fusion in a stable system so that electricity can be generated from water with the inert gas He as the by-product.