option h: relativity h4 consequences of special relativity

The twin paradox

H.4.1 Describe how the concept of time dilation leads to the twin paradox.

H.4.2 Discuss the Hafele-Keating experiment.

Option H: RelativityH4 Consequences of special relativity

The twin paradox

Describe how time dilation leads to the paradox.

This has already been done in Topic H3.


EXAMPLE: The twin paradox: Suppose Einstein has a twin brother who stays on Earth while Einstein travels at great speed in a spaceship. When he returns to Earth Einstein finds that his twin has aged more than himself! Explain why this is so.

SOLUTION: Since Einstein is in the moving spaceship, his clock ticks more slowly. But his twin’s ticks at its Earth rate. The twin is thus older than Einstein on his return!

By the way, this is NOT the paradox. The paradox is on the next slide…

EXAMPLE: The twin paradox: From Einstein’s perspective Einstein is standing still, but his twin is moving (with Earth) in the opposite direction. Thus Einstein’s twin should be the one to age more slowly. Why doesn’t he?

SOLUTION: The “paradox” is resolved by general relativity (which is the relativity of non-inertial reference frames). It turns out that because Einstein’s spaceship is the reference frame that actually accelerates, his is the one that “ages” more slowly.

The twin paradox

Describe how time dilation leads to the paradox.

This has already been done in Topic H3.


The twin paradox

Discuss the Hafele-Keating experiment.

From Wikipedia: “The Hafele–Keating experiment was a test of the theory of relativity. In 1971, Hafele, a physicist, and Keating, an astronomer, took four atomic clocks aboard commercial airliners.” “They flew twice around the world, first eastward, then westward, and compared the clocks against others that remained at the United States Naval Observatory.”“When reunited, the three sets of clocks were found to disagree with one another, and their differences were consistent with the predictions of special and general relativity.”


This experiment obviously tested time dilation.

Velocity addition

H.4.3 Solve one-dimensional problems involving the relativistic addition of velocities.


Velocity addition

Solve one-dimensional problems involving the relativistic addition of velocities.

Suppose you are in Car B, shown below.Suppose car A is moving at ux = +20 m s-1 and your car B is moving at v = +40 m s-1 as shown.

As far as you are concerned, the velocity ux’ of car A relative to you is -20 m s-1 , because A is moving backwards relative to you.


A

B

velocity of A relative to Bux’ = ux - v

FYIThis is the Galilean transformation which we learned way back in Topic 2.1.

Velocity addition


Now consider two spaceships leaving Earth in opposite directions:

Suppose ux = 0.75c and v = 0.50c (to the left).

Then according to the Galilean transformation, the ux’ (the velocity of A relative to B) would be

ux’ = ux – v = 0.80c - -0.50c = 1.3c.

This is in contradiction to the tenet that c is the maximum speed anything can have.


FYIThe Galilean transformation for addition of velocities fails at relativistic speeds.

Velocity addition


At relativistic speeds we will use:


relativistic velocity addition

ux’ =ux - v

1 - u v c2x

Where ux and v are signed quantities.

EXAMPLE: Spaceship B is traveling leftward at 0.50c wrt Earth. Spaceship A is traveling right- ward at 0.80c wrt Earth. What is the speed of A relative to B?

SOLUTION: v = -0.50c, ux = +0.80c:

● ux’ = (ux - v) / (1 - uxv/c2)

= (0.8c – -0.5c) / (1 - -0.5c 0.8c / c2)

= 1.3c / (1 + 0.4) = 0.93c.

Velocity addition



●Use ux’ = (ux – v)/(1 – uxv/c2), where ux = 0.8c and v = -0.8c (since they are traveling in opposite directions.

●Then ux’ = (0.8c – -0.8c)/[1 – (0.8c)(-0.8c)/c2].

ux’ = 1.6c/[1 + 0.64].

ux’ = 0.98c.FYIMake sure you are consistent with your signs and that you get an answer that is LESS THAN c!

Velocity addition



●A Galilean transformation assumes that time and length are absolute (the same in all reference frames) and fails under conditions of relativistic speeds.●A Galilean transformation simply adds velocities in a straight-forward manner according to “intuition.”

●EXAMPLE: ux’ = ux – v.

Velocity addition



where ux = 0.9800c and v = -0.9800c (since they are traveling in opposite directions).

●Use ux’ = ux – v,

●Then ux’ = 0.98c – -0.98c = 1.960c.

Velocity addition



●Use ux’ = (ux – v)/(1 – uxv/c2), where ux = 0.98c and v = -0.98c (since they are traveling in opposite directions.

ux’ = (0.98c – -0.98c)/[1 – (0.98c)(-0.98c)/c2]

= 1.96c/[1 + 0.982]

= 0.9998c.

Velocity addition



●In (b)(i) v > c which is not possible.

●Thus the Galilean transformation is not applicable to this problem.

Mass and energy

H.4.4 State the formula representing the equivalence of mass and energy.

H.4.5 Define rest mass.

H.4.6 Distinguish between the energy of a body at rest and its total energy when it is moving.

H.4.7 Explain why no object can ever attain the speed of light in a vacuum

H.4.8 Determine the total energy of an accelerated particle.


Mass and energy

State the formula representing the equivalence of mass and energy.

●Recall E = mc2. It has been slightly changed:

●We used this formula when we looked at mass defect in nuclear energy problems. It is the energy of a mass m0 in its rest frame.


Equivalence of mass and energy

E0 = m0c2

PRACTICE: A nuclear power plant converts about 30 kg of matter into energy each year. How many joules is this? How many watts?SOLUTION: ●E0 = m0c2 = 30(3108)2 = 2.71018 J.●P = E0/t = 2.71018/[365243600] = 8.61010 W.

Much of this energy is wasted in conversion to electrical power.

Mass and energy

Define rest mass.

●It is beyond the scope of this course, but not only do time and length change with speed, but so does mass!

●We call m the relativistic mass. Recall that is the Lorentz factor, and it is instrumental in solving special relativity problems.

●We call m0 the rest mass. This is the mass of the object as measured in a reference frame in which it is at rest.


relativistic mass

m = m0 where = 1

1 - v2/c2

FYINote that as v c that . ●Thus as v c we see that m .

Mass and energy

Define rest mass.


relativistic mass

m = m0 where = 1

1 - v2/c2

PRACTICE: At CERN a proton can be accelerated to a speed such that its relativistic mass is that of U238. How fast is it going?SOLUTION: ●First, find the Lorentz factor: m = m0

238mp = mp

= 238.●Then solve for v: (1 – v2/c2)1/2 = 1/238 1 – v2/c2 = 1/2382

1 – 1/2382 = v2/c2

0.9999823 = v2/c2 v = 0.9999912c.

Make sure you can reproduce both of these graphs.

Mass and energy

Distinguish between the energy of a body at rest and its total energy when it is moving.

●Since mass increases with velocity according to m = m0, clearly the total energy of a moving mass is E = m0c2.


relativistic energy

E0 = m0c2 where = 1

1 - v2/c2E = m0c2

PRACTICE: Show that the relativistic energy E reduces to the rest energy E0 when v = 0.SOLUTION: ●If v = 0 then = 1/(1 - 02/c2)1/2 = 1/11/2 = 1.●Then E = m0c2 = 1m0c2 = m0c2

= E0.

FYIThus E = m0c2 is the total energy of the object, whether the object is moving or not.

Mass and energy

Explain why no object can ever attain the speed of light in a vacuum

●Various arguments can be presented to show that no object with a non-zero rest mass can attain the speed of light.


EXAMPLE:

Argument 1: m as v c.●If v = c then = 1/(1 - c2/c2) = 1/0 = .●Since m = m0

then m also.

●But there is not even an infinite amount of mass in the universe. (Reductio ad absurdum).

Argument 2: E as v c.●Since E = m0c2 then as so does E.

●But there is not even an infinite amount of energy in the universe. (Reductio ad absurdum).

Mass and energy

Determine the total energy of an accelerated particle.

●Recall that the acceleration of a charge e through a potential difference V produces a kinetic energy given by EK = eV.

●The total energy E of a particle of rest mass m0 is just the sum of its rest energy E0 and its kinetic energy EK = eV.


total energy of an accelerated

particle

E = E0 + EK

where m = m0 mc2 = m0c2 + eV

FYIWe can not use (1/2)mv2 = eV at relativistic speeds to find v because it assumes all of the energy eV is going into the velocity. But the mass also changes at large speeds.

Mass and energy



●Use E = m0c2 = E0. Then

3E0 = E0 = 3.

●Since = 1/(1 – v2/c2)1/2 = 3, then(1 – v2/c2)1/2 = 1/3

1 – v2/c2 = 1/9v2/c2 = 8/9

v = 0.94c.

Mass and energy



●The mass as measured by an observer at rest with respect to the mass.●The mass as measured by an observer in the rest frame of the mass.

●From the formula we see that v V.

●Thus if V is large enough, v > c, which cannot happen.

Mass and energy



●Use E = E0 + EK.

mc2 = eV m = eV/c2

m = e(5.0106 V)/c2

m = 5.0 MeV c-2

m = (1.610-19)(5106)/(3108)2

m = 8.910-30 kg.

●Alternate method…

●Then mc2 = m0c2 + eV.mc2 - m0c2 = eV.

Mass and energy



●Use E = E0 + eV. Then

mc2 = m0c2 + eV.

m0c2 = m0c2 + eV.

= 1 + eV/(m0c2)

Mass and energy



●Use = 1 + eV/(m0c2). = 1 + e(500 MV)/(938 MeVc-2c2) = 1 + 500 MeV/(938 MeV) = 1 + 500/938 = 1.53 = 1/(1 – v2/c2)-1/2 = 1.53

1 – v2/c2 = 1/1.532 = 0.427

1 - 0.427 = v2/c2

v2 = 0.573c2

v = 0.76c.

Mass and energy



●As v c, .

●Since E = m0c2 we see that

●As , E .●Since there is not an infinite amount of energy in the universe, you cannot accelerate an object with a rest mass to the speed of light.

Mass and energy



E = E0 + EK = 0.51 MeV + 6.00 MeV = 6.51 MeV.

EK = eV = e(6.00106) V = 6.00 MeV.

(For an electron, E0 = m0c2 = 0.51 MeV.)

Mass and energy



●E = m0c2 = (0.51 MeV) = 6.51 MeV● = 6.51 MeV / 0.51 MeV = 12.17.

= 1/(1 – v2/c2)-1/2 = 12.171 – v2/c2 = 1/12.172 = 1/12.172

1 - 0.0067 = v2/c2

v2 = 0.9933c2

v = 0.997c.

Mass and energy



●Rest mass energy is E0 = m0c2 and is the energy that a particle has in its rest frame.

●Total energy is E = m0c2 + EK and is the sum of the rest mass energy and the kinetic energy EK = eV.●For these problems we always assume there is no potential energy.

Mass and energy



●E0 = m0c2 = (938 MeV c-2)c2 = 938 MeV.

Mass and energy



● = 1/(1 – v2/c2)-1/2 = 1/(1 – 0.9802c2/c2)-1/2 = 5.03.

●E = m0c2 = m0c2 + eV

eV = ( - 1)m0c2

V = ( - 1)m0c2/eV = (5.03 - 1)(938 MeV)/eV = 3780 MeV.

option h: relativity h4 consequences of special relativity

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