Physics 25 Chapter 28 Special Relativity Dr

Physics 25 Chapter 28 Special Relativity Dr. Alward

Video Lecture 1: Mass-Energy Equivalence Video Lecture 2: Relativistic Energies Video Lecture 3: Time Dilation Video Lecture 4: Length Contraction Video Lecture 5: Relative Velocity

Mass-Energy Equivalence According to Albert Einstein’s Special Theory of Relativity, energy added to an object causes its mass to increase; similarly, all or part of the mass of an object can undergo a transformation into energy (heat, light, sound, potential energy). This “equivalence” of mass and energy is expressed in the following equation:

2 Δm = Q /c

Example:

Gain in energy causes a gain in mass, as the example below illustrates:

600 J of light energy Q is absorbed by a 10- kg object. What is the new mass of the object?

Δm = Q /c2 = 600 /(3.0 x 108)2 = 6.67 x 10-15 kg m = 10.000 000 000 000 067 kg

The most important example of a conversion of mass into energy occurs in the interior of the sun where two different isotopes of hydrogen, 2 3 “deuterium” (1H ) and “tritium” (1H ), combine (“fuse”) to form helium.

The mass of the product (the helium nucleus) is less than the mass of the reactants (the hydrogen isotopes) by the amount 3 x 10-29 kg:

Δm = -3 x 10-29 kg

Q = (Δm) c2 = (-3 x 10-29) (3 x 108)2 = -2.70 x 10-12 J

The created energy is electromagnetic energy that’s radiated away.

The electron’s “antiparticle” is the positron. If they encounter each other, they will annihilate each other and, in their place will appear electromagnetic energy.

The reverse can occur:

Under the right conditions, electromagnetic energy can disappear, and in its place will appear the equivalent amount of mass in the form of an electron-positron pair.

Total Energy

According to Einstein, the total energy of an object moving with speed v is:

E = mc2 / (1 - β2)1/2

β = v/c, where c = 3.00 x 108 m/s.

Note that when the object is at rest, β = 0, and E = mc2, which is called the “rest-mass energy.”

Relativistic Kinetic Energy

Subtract the rest portion (mc2) of the energy from the total energy, and what’s left is the motion part of the energy kinetic energy.

K = E - mc2 = mc /(1-β2)1/2 - mc2

Note that if v = 0, then K = 0, as expected for an object that’s not moving.

We often refer to this kinetic energy as the “relativistic” kinetic energy to alert the reader to the fact that this kinetic energy is the true kinetic energy equation, while the other kinetic energy equation, ½ mv2, is called the “classical” kinetic energy.

Example:

(a) What is the rest-mass energy of an electron (in MeV)?

Electron Mass: 9.11 x 10-31 kg

E = mc2 = 9.11 x 10-31 (3.0 x 108)2 = 8.20 x 10-14 J = 8.20 x 10-14 J / 1.6 x 10-19 J/eV = 0.51 x 106 eV = 0.51 MeV

(b) What the total energy of an electron moving at 0.9995 c?

β = 0.9995 E = mc2 / (1-β2)1/2 = 0.51 / (1 - 0.99952)1/2 = 16.13 MeV

K = E - mc2 = 16.13 - 0.51 = 15.62 MeV

(d) What is the kinetic energy using the “classical” equation?

K = ½ mv2 = ½ (9.11 x 10-31)(0.9995 x 3.0 x 108)2 = 4.10 x 10-14 J = 4.10 x 10-14 J/ 1.6 x 10-19 J/eV = 0.26 x 106 eV = 0.26 MeV

At very high speeds the classical kinetic energy equation is wildly inaccurate, as we see above, but at lower speeds the classical kinetic energy equation is extremely accurate, as we will see on the next page.

Example:

Calculate and compare the classical and relativistic kinetic energies (in joules) of a proton traveling at the “non- relativistic” speed, v = 0.01c (β = 0.01). v = 0.01 (3.0 x 108) = 3.0 x 106 m/s

Classical: K = ½ mv2 = ½ (1.67 x 10-27) (3.0 x 106)2 = 7.515 x 10-15 J

Relativistic:

mc2 = 1.67 x 10-27 (3.0 x 108)2 = 1.503 x 10-10 J

K = mc2 / (1- β2)1/2 - mc2 = 1.503 x 10-10 (1 - 0.012)-1/2 - 1.503 x 10-10 = 7.516 x 10-15 J

Even at speeds as large as 3.0 x 106 m/s, the classical kinetic energy equation is still accurate to within one- hundredth of one percent.

Classical versus Relativistic Kinetic Energy

Prove: At small speeds v, the classical kinetic energy equation (K = ½ mv2) predicts kinetic energies as accurately as does the relativistic kinetic energy equation.

The “binomial theorem”:

If x << 1, then (1 - x)n  1 – nx

For example, let x = 0.0001:

(1 – 0.0001)2 = 0.9998 Compare to 1.0000 – 2 (0.0001) = 0.9998

Proof: The role of x is played by β2, while the role of n is played by -½:

(1 - β2)-1/2 = 1 - (-½) β2 = 1 + ½ β2 if β << 1

K = mc2 [ (1 - β2)-1/2 - 1] = mc2 [1 + ½ β2 - 1] = mc2 (½ β2) = mc2 (½) (v2/c2) = ½ mv2 if v << c

Example:

How much work (in MeV) would need to be done on a proton to increase its speed from βo = 0.60 to β = 0.70?

Mass of proton: m = 1.67 x 10-27 kg mc2 = 1.67 x 10-27 (3.0 x 108)2 = 1.50 x 10-10 J = 1.50 x 10-10 J / 1.6 x 10-19 J/eV = 9.39 x 108 eV = 939 x 106 = 939 MeV

K = mc2 / (1- β2)1/2 - mc2 = 939 / (1 - 0.702)1/2 - 939 = 375.9 MeV

2 2 1/2 2 Ko = mc / (1- βo ) - mc = 939 / (1 - 0.602)-1/2 - 939 = 234.8 MeV

Use the work-kinetic energy theorem:

W = K - Ko = 375.9 - 234.8 = 141.1 MeV

Time Dilation

The time required for an event to occur depends on the observer’s motion relative to the event.

Observers moving relative to an event measure a longer time than do observers stationary relative to the event.

--Albert Einstein, 1905

Times measured by observers at rest relative to the event are called the “proper time.”

To = “proper time”

Times measured by observers moving relative to the event are called the “dilated time.”

T = “dilated time” 2 1/2 = To / (1 - β )

Note: The denominator is less than 1.0, so T > To.

Note: the word “proper” does not imply it’s the “real” time, and that the times observed by moving observers are somehow flawed because of measurement errors or illusion. The dilated time measured by the moving observer is just as real for that observer as it is for the observer who is at rest relative to the event.

Example A:

An observer at rest relative to an ice cube says it took eight minutes for the ice to melt.

Proper Time: To = 8.0 minutes

How long does an observer, moving at a speed of four- tenths the speed of light say it took for the ice to melt?

β = 0.40

2 1/2 T = To / (1 - β ) = 8.00 / (1 - 0.402)1/2 = 8.73 minutes

Dilated Time: 8.73 minutes

Example B:

A runner completes a journey in 2.00 minutes, according to her. Does she measure the dilated time, or the proper time?

The runner is located wherever the event (the running) is occurring, so she measures the proper time.

To = 2.00 minutes

Suppose a passenger in a spaceship flying overhead at 0.80 c observes the runner. How long will he say it took the runner to complete her run?

T = 2.00 /(1 - 0.802)1/2 = 3.33 minutes

Example:

A hovering drone in a hypothetical universe observes an automobile drive at speed v = 40 m/s across town in 14 minutes. The automobile driver says the trip took 12.0 minutes; what is the speed of light in that universe?

The event is the trip; it is occurring wherever the driver is, so she measures the proper time.

The drone’s location relative to the automobile is constantly changing, i.e., the drone moving relative to the event, so it measures the dilated time, T = 14.0 minutes.

To = 12.0 minutes T = 14.0 minutes

Assume Einstein’s Special Theory holds true in the hypothetical universe.

2 1/2 T = To / (1 - β ) 14.0 = 12.0 / (1 - β2)1/2 β = 0.515 40/c = 0.515 c = 77.67 m/s

Muons

Muons are unstable particles that live for only 2.2 x 10-6 s before decaying—according to observers stationary relative to the muon. This is the proper time.

If muons are moving relative to observers, they live longer—according to the moving observers.

For example, muons are created when protons streaming toward Earth from the sun collide with molecules in the upper atmosphere; these muons move at β = 0.998 toward the ground. A hypothetical rider atop the muon would say that the muon after creation would travel the following distance through the atmosphere before decaying:

Distance = Speed x Time = (0.998)(3.0 x 108) (2.2 x 10-6) = 659 m

Earth observers measure the dilated time:

2 1/2 T = To / (1 - β ) = (2.2 x 10-6) /(1 - 0.9982)1/2 = 3.48 x 10-5 s

Earth observers say the muon in fact travels this many meters before dying:

(0.998) (3.0 x 108 m/s) (3.48 x 10-5 s) = 10, 419 m

Length Contraction

2 1/2 L = Lo ( 1 - β )

Lo = Length measured by an observer at rest relative to the object = “the proper length”

L = Length measured by an observer moving relative to the object = “the contracted length”

Both lengths are correct for their respective observers.

Contraction occurs in the dimension that is along the direction of motion. Below we show an object moving at various speeds to the right, along the x-axis. The narrowing of the object’s width occurs along the object’s x-axis.

Note: the word “proper” does not imply that the length measured by the stationary observer is the only “real” length, and all other measurement are somehow flawed because of errors by the observer, faulty equipment, or an illusion. The length measured by the moving observer is just as real for that observer as it is for the observer who is at rest relative to the object.

Distance-Speed-Time Calculations Involving Light-Years

One Light-Year (LY) is the distance light will travel in one year. For example, if an object is traveling at one-tenth the speed of light, it will take ten years to travel 1.0 LY. The table below further illustrates the point we wish to make:

Let D = Distance in LY

D (in LY)  t (in Y) 1.0 1.0 1.0 1.0 0.2 5.0 10.0 0.4 25.0 20.0 0.5 40.0

t = D/

Example

A passenger on a spaceship leaving Earth at  = 0.80 travels to a star and says its distance from Earth is 8.0 LY. The passenger is moving relative to the path connecting Earth to the star, so this distance is the contracted length, L:

L = 8.0 LY

(a) How long (in years) does the passenger say it took the spaceship to reach the star?

The passenger is always located where the event (the traveling) is occurring (wherever she is), so she measures the proper time, To. The passenger calculates the travel time To (in years, Y) as shown below:

To = L/ = 8.0 / 0.80 = 10 Y

(b) How long (in years) does the Earth observer say the spaceship took to reach the star?

The Earth observer is moving relative to where the event (the traveling) is occurring, so he measures the dilated time, T:

T = 10 /(1 – 0.802)1/2 = 16.67 Y

Example:

A stationary observer measures length of a path to be 400 m. The observer is at rest relative to the path, so he measures the proper length:

Lo = 400 m

A person runs along that path at a speed of 0.70 c: v = 0.70 (3.0 x 108 m/s) = 2.1 x 108 m/s

(a) What does the runner say is the length of the path?

The runner is moving relative to the path, so he measures the contracted length:

2 1/2 L = Lo (1 - β ) = 400 (1 - 0.702)1/2 = 286 m

(b) How much time does the runner say it took to complete the run?

Time = Distance / Speed = 286 m / 2.1 x 108 m/s = 1.36 x 10-6 s

The event in question is the running; the runner is located wherever the running is happening, do the runner is at rest relative to the event and therefore measures the proper time:

-6 To = 1.36 x 10 s

1. Time = Distance / Speed = 400 m / 2.1 x 108 m/s = 1.90 x 10-6 s

2 1/2 2. T = To / (1 - β ) = 1.36 x 10-6 / (1 - 0.702)1/2 = 1.90 x 10-6 s

Example:

The catcher on a baseball team says the path connecting home plate to first base has a length of 27.40 m. He is at rest relative to the path, so he measures the proper length, Lo.

Lo = 27.40 m

The catcher observes a runner traveling from home plate to first base at a speed of 0.90 c. (c) What does the runner say is the length v = 0.90 (3.0 x 108) of the path? = 2.7 x 108 m/s The runner is moving relative to the path, so (a) What does the catcher say is the time it he measures the contracted length: takes the runner to reach first base? 2 1/2 L = Lo (1 - β ) 2 1/2 Time = Distance/Speed: = 27.40 (1 - 0.90 ) = 27.40 m /2.7 x 108 m/s = 11.94 m = 1.01 x 10-7 s (d) What does the runner say her speed is? The event, which is the running, is taking place at the runner’s location, so the runner Runner Measures Distance = 11.94 m -8 measures the proper time. Runner Measures Time = 4.40 x 10 s Speed = Distance / Time -8 The catcher is moving relative to the event’s = (11.94 m) / (4.40 x 10 s) 8 location (the runner’s location), so the catcher = 2.7 x 10 m/s measures the dilated time: Note that this is the same speed the catcher T = 1.01 x 10-7 s measures. The two persons disagree about distance between home plate and first base, (b) What does the runner say is the time it and they disagree about the time it takes the takes to reach first base? runner to run from home plate to first base, but they agree about the runner’s speed, The runner measures the proper time: i.e., they agree about the ratio of distance and time. The catcher says the distance is 2 1/2 longer, while the time is longer; the runner T = To / (1 - β ) -7 2 1/2 says the distance is shorter, while the time 1.01 x 10 = To / (1 - 0.90 ) -8 is shorter, but the ratios of distance over To = 4.40 x 10 s times-the speeds---are the same.

Classical Relative Velocities

Classical relative velocities are velocities that are negligibly small compared to the speed of light.

Let the velocity of an Object A relative to an Earth observer be VAO and the velocity of Object B relative to the Earth observer be VBO.

Objects moving toward Earth are negative, while objects moving away from Earth are positive.

The velocity of Object A relative to Object B is

VAB = VAO - VBO

Example: VAO = 20 m/s and VBO = -30 m/s

VAB = 20 - (-30) = 50 m/s

Relativistic Relative Velocities

At very great velocities--greater than about one-hundredth the speed of light, Newtonian physics, i.e., non-relativistic physics, is no longer reliable.

The equation that is valid for relativistic, as well as classical, velocities is shown below for Objects A and B moving relative to an observer on Earth.

AB = (AO - BO) / (1 - AO BO)

Example:

Spaceship A is moving at a velocity 0.50 c relative to observers on Earth, as it chases Spaceship B, which is moving relative to Earth at 0.30 c.

(a) What is the velocity of A as seen by Spaceship B?

AB = (AO - BO) / (1 - AO BO)

= (0.50 - 0.30 ) / [1 - (0.50) (0.30)] = 0.24

(b) Spaceship A is emitting light whose frequency is 14 fA = 6.0 x 10 Hz. What is fB, the frequency seen by Spaceship B?

Recall the Doppler Effect for Light equation discussed in Chapter 24: fo = fs (1 ± )

Here, the role of the observer is played by Spaceship B, while the source is Spaceship A. fB = fA (1 ± )

Spaceship A is approaching Spaceship B, so we use the positive sign to ensure that the observed frequency is greater than the broadcast frequency:

15 fB = 6.00 x 10 (1 + 0.24) = 7.44 x 1015 Hz