Modern Physics: Mastering Physics: ± When Time Flies...It Runs More Slowly

Learning Goal: To understand length contraction and time dilation.
An inertial frame of reference is one in which Newton's laws hold. Any frame of reference that moves at a constant speed relative to an inertial frame of reference is also an inertial frame. The proper length l_0

of an object is defined to be the length of the object as measured in the object's rest frame. If the length of the object is measured in any other inertial frame, moving with speed

relative to the object's rest frame (in a direction parallel to l_0

), the resulting length

is given by the length contraction equation:

$l = l_0\sqrt{1-\frac{u^2}{c^2}} \;$ ,

where

is the speed of light. Similarly, if two events occur at the same spatial point in a particular reference frame, and an observer at rest in this frame measures the time interval between these two events, the time interval so measured is defined to be the proper time $\Delta t_0$ . When the time interval is measured in any other inertial frame, again moving with speed

relative to the first frame, the resulting time interval $\Delta t$ is given by the time dilation equation:

$\Delta t = \frac{\Delta t_0}{\sqrt{1-\frac{u^2}{c^2}}} \;$ .

Part A

Suppose that you measure the length of a spaceship, at rest relative to you, to be 400 $\rm m$ . How long will you measure it to be if it flies past you at a speed of $u = 0.75c$ ?

Express the length

in meters to three significant figures.

ANSWER:

264.6

$\rm m$

Notice that this is smaller than the proper length of the spaceship. This is why the discrepancy between measurements of moving objects and their proper length is sometimes referred to as length contraction.

Part B

The spaceship from Part A has a large clock attached to its side. This clock ran at the same rate as your watch when you were in the same reference frame. How much time $\Delta t$ will pass on your watch as 80 $\rm s$ passes on the clock attached to the ship?

Express your answer in seconds to three significiant figures.

ANSWER:

$\Delta t$ =

120.9

$\rm s$

Notice that this is larger than the time measured on the ship. This is why the difference between measurements of moving objects and measurements of proper time is referred to as time dilation.

Two spaceships, named A and B, are flying toward each other with relative speed $0.800c$ .

Part C

If the captain of ship A fires a missile, counts 10.0 $\rm s$ on his watch, and then fires a second missile, how much time $\Delta t$ will the captain of ship B measure to have passed between the firing of the two missiles?

Express your answer in seconds to three significant figures.

ANSWER:

$\Delta t$ =

16.7

$\rm s$

Part D

The captain of ship B knows that ship A uses 2-m-long missiles. She measures the length of the first missile, once it has finished accelerating, and finds it to be only 0.872 $\rm m$ long. What is the speed

of the missile, relative to ship B?

Express your answer in meters per second to three significant figures. Use $c = 3 \times 10^8 \; \rm m/s$ .

ANSWER:

$2.70{\cdot}10^{8}$

$\rm m/s$

It should be noted that the same equations apply to events in your everyday life. The reason that you don't notice them is that objects in your everyday life move so much slower than the speed of light. Now let's look at the differences in measurements between two frames moving relative to one another at a speed of 30 $\rm m/s$ (108 $\rm kph$ or about 67 $\rm mph$ ). Your calculator may not be able to store enough digits to work these problems accurately, so you may need to use the approximations from the binomial expansion:

$\frac{1}{\sqrt{1-\frac{u^2}{c^2}}} \approx 1+ \frac{1}{2} \frac{u^2}{c^2} \quad {\rm and} \quad \sqrt{1-\frac{u^2}{c^2}} \approx 1- \frac{1}{2} \frac{u^2}{c^2}$ .

Part E

What would be the difference between the time

measured by an observer moving at 30 $\rm m/s$ and the proper time t_0

for a proper time interval of 1 hour (3600 $\rm s$ )? The answer is small but nonzero. You will need to find an expression for the time difference using the approximation given in this problem before you substitute in the numbers; otherwise your calculator will just give zero.

Express your answer in seconds to three significant figures. Use $c = 3 \times 10^8 \; \rm m/s$ .

ANSWER:

$t-t_0$ =

1.800×10⁻¹¹

$\rm s$

You could not notice this small of a difference. For everday situations, it is a very good approximation to assume that all measured times are equal to their proper times.

Part F

What would be the difference between proper length l_0

and the length

measured by an observer moving at 30 $\rm m/s$ for a proper length of 5 $\rm m$ (a reasonable proper length for a car)? Use a similar procedure to the one you used in the previous part.

Express your answer in meters to three significant figures. Use $c = 3 \times 10^8 \; \rm m/s$ .

ANSWER:

$l_0-l$ =

2.500×10⁻¹⁴

$\rm m$

You could not possibly notice this difference, which is much smaller than the diameter of an atom. For everday situations, it is a very good approximation to assume that all measured lengths are equal to their proper lengths.

Modern Physics

Saturday, January 22, 2011

Mastering Physics: ± When Time Flies...It Runs More Slowly

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