SAT Physics Subject Test

Chapter 16 Modern Physics

SPECIAL RELATIVITY

The SAT Physics Subject Test may ask a question or two on Einstein”s theory of special relativity. Since you aren”t expected to have an in-depth mastery of this subject, we”ll simply state some relevant facts from this theory and present some examples.

Let”s begin with the two postulates of special relativity.

An inertial reference frame is one in which Newton”s first law holds. Given one inertial reference frame, any other reference frame that moves with constant velocity relative to the first one will also be inertial. For example, a person standing at a train station considers herself to be in an inertial reference frame. If she places her suitcase on the ground next to her and exerts no forces on it, it stays at rest. Now, if a train moves past the station, traveling on a smooth track in a straight line at constant speed, then a passenger on the train considers himself to be in an inertial reference frame, too. If he places his suitcase on the floor and exerts no forces on it, it stays at rest (relative to him). Now, if the train is speeding by at, say, 40 m/s, then the person standing at the train station would say that the man”s suitcase on the train is moving at a speed of 40 m/s, but the man on the train would say that his suitcase is at rest. So, while it”s true that these two observers will naturally disagree about the velocity of the suitcase on the train, they won”t disagree about physics laws, such as conservation of momentum. This is the essence of Postulate 1. Two people playing a game of billiards on a smoothly moving train (that is, one that travels in a straight line at constant speed) don”t have to “adjust” their shots to account for the motion of the train.

Now let”s look at Postulate 2 more closely.

1) The Relativity of Velocity

Let”s say that the man in the train speeding by the platform at 40 m/s throws a ball down the aisle, parallel to the direction of motion of the train, at a speed of 5 m/s as measured by him. But as measured by the woman standing on the platform, the ball would be moving at a speed of 40 + 5 = 45 m/s. It seems clear that we”d just add the velocities.

This simple addition of velocities does not, however, extend to light or even to objects moving at speeds close to that of light. Imagine a spaceship moving at a speed of c toward a planet. If the spaceship emitted a light pulse toward a planet, the speed of that light pulse, as measured by someone on the planet, would not be c + c = c; Postulate 2 says that the speed of light would be c, regardless of the motion of the spaceship. Since we haven”t been able to travel at speeds even remotely approaching that of light, this result seems very strange.

The correct, relativistic formula for the “addition” of velocities—that is, the one that follows from the theory of relativity—looks like this. Imagine that a reference frame, S_me, is moving with velocity u past you. Now if an object moves at velocity v (parallel to u), as measured by me in my reference frame, then its speed as measured by you would not be v_you = u + v, but would instead be

At normal, everyday speeds, u and v are so small compared with the speed of light that the fraction uv/c² is negligibly small, nearly zero. In this case, the denominator in the formula above is nearly 1, and it becomes the formula v_you = u + v. It”s only when u or v is close to the speed of light that the difference becomes measurable.

Let”s apply the correct formula to our previous example about the spaceship emitting a pulse of light. With you on the planet and me on the spaceship, u = c. Naturally, I measure the speed of the light pulse to be v = c, and you would measure its speed to be

just as Postulate 2 says you must.

Question 14-15

An enemy spaceship of the Empire is traveling toward the planet Ceti Alpha VI at a speed of 0.4c. The ship emits a beam of antiprotons at the planet that travel at a speed of 0.5c relative to the ship.

14. How fast are the antiprotons traveling relative to the planet?

15. If the Empire ship emits a pulse of red light just before its blast of antiprotons, with what speed does the red light travel, relative to the planet?

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14. With u = 0.4c and v = 0.5c, the speed of the antiprotons relative to the planet is

15. As stated in Postulate 2 of the Theory of Special Relativity, the speed of light through empty space is always measured to be c, regardless of the motion of the source (or observer).

2) The Relativity of Time

Let”s go back to me on my spaceship, passing by your planet at a speed of c. Let”s say, as I”m standing in the ship, I sneeze, and then, 1 second later, I sneeze again. If you were holding a stopwatch and measured the time between my sneezes, you would not get 1 second. Time is relative. The time as measured by you would not agree with the time as measured by me.

To see how these time intervals would be related, imagine the following. In my reference frame, let me call the time between my sneezes ∆T₁. If my velocity past you is v, then the time that you”d measure between my sneezes would be

Proper Time and
Dilated Time

∆ T₁ can be referred to
as the proper time, which
is the time measure in
the frame at rest with
respect to an event. In this
case, the sneezing is the
event, so the proper time
is measured by a clock at
rest with respect to the
person sneezing. ∆T₂, a
time measured in a different
inertial frame, can be
referred to as the dilated
time.

∆ T₂ = γ • ∆T₁

where γ is the relativistic factor

Because the denominator of this fraction is never greater than 1, the value of is never less than 1. So unless my spaceship was standing still relative to you, the time that you”d measure between my sneezes would always be longer than the time I”d measure. Seems strange? Of course, because, once again, we have no understanding of moving at speeds close to the speed of light, so we don”t have any experience for time dilation. In fact, even at v = c (which is about 67 million miles per hour!) the relativistic factor γ is still only 1.005. But highly sensitive atomic clocks have been flown on commercial jet airplanes, and it”s been found, upon landing, that they are a little slow relative to their synchronized counterparts that didn”t fly. Furthermore, the time difference has been shown to be just what the relativistic formula above predicts.

For all ordinary speeds, where v is very, very small compared to c, the value of γ is negligibly greater than 1, so ∆T₂ ≈ ∆T₁, and we don”t notice a difference for ordinary time intervals. But, as v gets closer and closer to c, γ gets bigger and bigger. For example, for passengers on a spaceship moving at, say, v = 0.99c relative to the earth, the value of γ is about 50. So, if someone on the ship says that they”ve been on the ship for 2 years (as measured by them), we here on Earth would say that the elapsed time is 50 × 2 = 100 years.

3) The Relativity of Length

Let”s once again go back to me on my spaceship, passing by the earth at a speed of c. Let”s say I measure the length of my ship to be 100 meters. If you, on earth, were watching my ship fly by, you would not measure its length to be 100 m. Length is relative. The length as measured by you would not agree with the length as measured by me.

Proper Length and
Contracted Length

L₁ can be referred to as
the proper length, which
is the length measured
in the frame at rest with
respect to the object. In
this case, the ship is the
object, so the proper
length is measured by a
ruler at rest with respect
to the ship. The length L₂,
measured in a different inertial
reference frame,
can be referred to as the
contracted length.

To see how these lengths would be related, imagine the following. In my reference frame, let me call the length of my spaceship L₁. If my velocity past you is v, then the length that you”d measure for my ship will be

where γ is the relativistic factor. Because γ is greater than 1, the length you”d measure would be shorter than what I”d measure. This is known as length contraction; lengths that are parallel to the velocity v are shortened by a factor of γ.

Questions 16-18

A type of radioactive, subatomic particle, a μ lepton (also known as a muon, for short) is created by cosmic rays from space entering the earth”s atmosphere. Suppose that a muon is traveling downward toward the earth at a speed of 0.99c, relative to the earth.

16. At rest a muon typically lives for about 0.2 microsecond before it decays. How long will the moving muon survive, as measured by observers here on earth?

17. How far does the muon travel, as measured by Earth observers?

18. How far does the muon travel, as measured by the muon itself?

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16. As measured on Earth, the time interval between the moment when the muon is created and the moment it decays is

T₂ = γ • ∆T₁ = γ • (0.2 microsec)

In this case, the relativistic factor is

So, as measured by observers on Earth, the muon survives for

∆T₂ = γ • ∆T₁ ≈ 7 • (0.2 microsec) = 1.4 microsec

17. Since the speed of the muon is 0.99c, and we measure its lifetime as 1.4 microseconds, the distance it travels, as measured by us, is

D = vT = (0.99c)(1.4 × 10^–6 s) = (0.99)(4.2 × 10² m) ≈ 420 m

18. In the reference frame of the muon, it”s the earth that”s rushing up toward the muon at a speed of 0.99c. Since, in the muon”s frame of reference, it only lives for 0.2 microsecond, the muon measures its distance of travel to be

d = vt = (0.99c)(0.2 × 10^–6 s) = (0.99)(0.6 × 10² m) ≈ 60 m

Notice that since the muon is moving relative to the earth, it measures distances as being shorter than we do. So, from our point of view, the muon lives longer as it travels a proper distance of 420 m. From the muon”s point of view, it lives its ordinary, proper lifetime, 0.2 microsecond, but only travels 60 m. These points of view are both correct (remember, distances and time are relative) and they”re compatible. In one point of view, the time stretched by a factor of γ = 7, while in the other point of view it was the distance that shortened by a factor ofγ = 7. (The time interval between two events is said to be proper if the two events occur at the same location in the reference frame of the observer. An object”s length is said to be proper if the object is at rest in the reference frame of the observer.)

Relativistic Energy

You”ve undoubtedly seen the formula E = mc², probably the most famous formula in physics. What it says is that mass (m) and energy (E) are equivalent, and the formula tells us how much energy is equivalent to a given amount of mass. This energy is called rest energy because an object resting, say, on your desk, has energy—in fact, is energy—simply by virtue of the fact that it exists and has mass. Because c², the square of the speed of light, is such a big number, a small amount of mass is equivalent to a huge amount of energy. When an exothermic nuclear reaction takes place, the total mass of the product nuclei is always less than the total mass of the original nuclei. The “missing mass,” ∆ m, has been converted to energy, ∆E, in accordance with the equation ∆E = (∆m)c². This is the basis for how the world”s nuclear power plants generate energy. The fission of a neutron with a uranium-235 nucleus creates five product particles (a barium-140 nucleus, a krypton-93 nucleus, and 3 neutrons) whose total mass is less than the total mass of the original neutron and uranium-235 nucleus by about 3 × 10^–28 kg. Multiplying this ∆m by the factor c²= 9 × 10¹⁶ m²/s², we get ∆E = 2.7 × 10^–11 J.

But what about kinetic energy? Up to now, we”ve been using the formula KE = mv² for the kinetic energy of an object of mass m moving with speed v. But if v is close to c (in which case we say that the object is moving at a relativistic speed), then this formula won”t work. The correct formula is

KE = (γ − 1)mc²

where γ is the usual relativistic factor. It can be shown that when v is very small compared with c, this formula becomes the familiar KE = mv².

An object”s total energy, E_total, is now defined as the sum of its rest energy, E_rest = mc², and its kinetic energy.

If we plot the kinetic energy of a particle of mass m as a function of its speed v, we get the following graph:

Notice that as v approaches c, the object”s kinetic energy approaches infinity. This shows why it”s impossible for a massive particle to move at the speed of light (or at any speed greater than c). The force accelerating a particle would have to do more and more work to increase the particle”s kinetic energy, and it would take an infinite amount of work to push a particle to the speed of light (where its kinetic energy would be infinite). The universe has a speed limit: Light moves at speed c, and everything else moves at a speed less than c.