#$&*
course Phy 202
8/3 at 9:30
roblemA spacecraft moving at constant velocity .72 * c relative to a nonaccelerating observer keeps time by reflecting a light pulse back and forth between two mirrors oriented at a right angle to the direction of motion. The mirrors are 140 meters apart (big spaceship) and are mounted within a sealed vertical vacuum tube (vertical being perpendicular to the direction of motion) so that the light pulse will travel at its vaccuum speed.
If we assume that the laws of physics are the same in the reference frame of the spaceship as in the frame of the observer, it follows that the speed of light in a vacuum will be identical in both reference frames.
Under this assumption:
• How long does it take a pulse to travel from the lower mirror to the higher, as measured in the spaceship?
• How long does it take the pulse to travel from the lower mirror to the higher as measured by the observer outside the spaceship?
• If the pulse rate of an inhabitant of the spaceship is 59 beats/minute, as measured by that individual, then what pulse rate would be measured by the outside observer, assuming some accurate means of making the observation?
Solution
Dt=sqrt(0.4666x10^-6)^2/0.6939
=0.6724 sec
The light will be observed in the reference frame of the spaceship to travel 140 meters at 3*10^8 m/s, which will take .4666 * 10^-6 second.
In the frame of the observer outside the ship the pulse will still travel 140 meters perpendicular to the line of motion, but will also travel some distance parallel to the line of motion.
• The distance travel by the pulse will therefore be greater than the 140 meter distance observed on the spaceship.
• Since the speed of light will be the same in both reference frames, the conclusion must be that the observer outside the ship will measure a longer time interval than an observer on the ship.
If `dt' stands for the time measured by the external observer, the ship and therefore the mirrors will move distance
• distance moved by ship in time `dt': v `dt' = ( .72 * c ) `dt' = .72 ( 3 * 10^8 m/s ) `dt' = 2.16 * 10^8 m/s * `dt'.
The distance traveled by the pulse is found by the Pythagorean Theorem to be
• distance measured by outside observer = `sqrt( ( 140 m )^2 + ( 2.16 * 10^8 m/s * `dt' )^2).
The time required to travel this distance at the speed of light is
• time for pulse to travel distance = distance / velocity
= `sqrt( ( 140 m )^2 + ( 2.16 * 10^8 m/s * `dt' )^2) / ( 3 * 10^8 m/s).
This time is of course the time `dt' measured by the observer as the light travels from mirror to mirror. Thus we have
• `dt' = `sqrt( ( 140 m )^2 + ( 2.16 * 10^8 m/s * `dt' )^2) / ( 3 * 10^8 m/s).
This could be solved for the only variable `dt', which appears on both sides of the equation. However, at this point anyone with good judgment will go back and solve the problem using symbols, because carrying through all those numbers and units is going to be a real pain, and we are likely to lose sight of the meanings of the numbers.
In the generalized solution we will solve the problem using v for the speed of the spaceship, c for the speed of light and `dy for the separation of the mirrors. The result will be that any time interval `dt' measured by the outside observer will be 1 / `sqrt( 1 - v^2 / c^2 ) times the time interval `dt observed on the spacecraft.
Just to bring home the lesson about solving in symbolic form we plug on through the numerical solution:
We can solve this equation for `dt':
square both sides:
• `dt'^2 = ( ( 140 m )^2 + ( 2.16 * 10^8 m/s * `dt' )^2) / ( 3 * 10^8 m/s)^2,
multiply both sides by the denominator and factor out the `dt'^2 in the second term of the right-hand side:
• ( 3 * 10^8 m/s)^2 * `dt'^2 = ( 140 m )^2 + ( 2.16 * 10^8 m/s )^2 * `dt'^2,
collect the `dt'^2 terms on the left-hand side:
• ( 3 * 10^8 m/s)^2 * `dt'^2 - ( 2.16 * 10^8 m/s )^2 * `dt'^2 = ( 140 m )^2
factor out `dt'^2 on left:
• [ ( 3 * 10^8 m/s)^2 - ( 2.16 * 10^8 m/s )^2 ] * `dt'^2 = ( 140 m )^2
(factor out 3 * 10^8 m/s)^2 on the left:
• ( 3 * 10^8 m/s)^2 [ 1 - ( 2.16 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ] * dt'^2 = ( 140 m )^2
divide through by the coefficient of `dt'^2:
• `dt`^2 = ( 140 m ) ^ 2 / [ ( 3 * 10^8 m/s)^2 { 1 - ( 2.16 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 } ],
rearrange the right-hand side, using the knowledge that 140 m / ( 3 * 10^8 m/s) = .4666 * 10^-6 sec, the time measured on the ship:
• `dt'^2 = ( .4666 * 10^-6 sec ) ^ 2 / [ 1 - ( 2.16 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ]
evaluate [ 1 - ( 2.16 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ] :
• [ 1 - ( 2.16 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ] = .4816,
rewrite the equation:
• `dt'^2 = ( .4666 * 10^-6 sec)^2 / .4816,
take the square root of both sides and evaluate:
• `dt' = ( .4666 * 10^-6 sec) / .6939 = .6724 sec.
Had we used the result of our symbolic solution we would have obtained the same result directly:
• `dt' = `dt / `sqrt( 1 - v^2 / c^2 ) = .4666 sec / `sqrt ( 1 - ( .72 * c / c ) ^ 2 )
= .4666 sec / `sqrt( 1 - `b^2 ) = .4666 sec / .6939 = .6724 sec.
The time `dt' measured by the 'outside' observer will be 1 / `sqrt( 1 - v^2 / c^2) = 1 / `sqrt ( 1 - ( .72 * c / c ) ^ 2 ) = 1.441 times as great as that observed on the spacecraft. Note that this is the ratio `dt' / `dt = .6724 sec / ( .4666 sec).
Generalized Solution
If the velocity of the spacecraft is symbolized by v, the speed of light by c and the separation of the mirrors by `dy, we see that the time required in the spaceship frame is
• `dt = `dy / c.
If the time measured by the outside observer is `dt', then the distance `dx' moved by the spacecraft as the light travels from mirror to mirror is
• `dx' = v * `dt',
and the distance `ds' moved by the light pulse is the hypotenuse of a right triangle with legs `dy and `dx':
• `ds' = `sqrt( `dy^2 + `dx' ^ 2 ) = `sqrt ( `dy^2 + ( v * `dt' ) ^ 2 ).
The time required for light to move this distance, moving as it must at velocity c, is
• `dt' = `ds' / c = `sqrt ( `dy^2 + ( v * `dt' ) ^ 2 ) / c .
We thus have the equation
• `dt' = `sqrt ( `dy^2 + ( v * `dt' ) ^ 2 ) / c ,
which we solve for `dt.
We square both sides:
• `dt' ^ 2= ( `dy^2 + ( v * `dt' ) ^ 2 ) / c^2
then divide through the right-hand side by c^2:
• `dt' ^ 2= `dy^2 / c^2 + ( v * `dt' ) ^ 2 / c^2 = (`dy / c)^2 + ( v / c * `dt' ) ^ 2
We observe that `dy / c = `dt, the time interval measured in the rocket frame. Using this fact and subtracting the second term on the right from both sides we have
• `dt`^2 - ( ( v / c ) `dt' ) ^ 2 = `dt^2.
Factor out `dt'^2 on the left-hand side:
• ( 1 - ( v / c ) ^ 2 ) * `dt'^2 = `dt^2
then dividing by the coefficient of `dt'^2 and taking square roots we have
• `dt' = `dt / `sqrt( 1 - ( v / c ) ^ 2 ).
This could be written as
• `dt' = `dt * [ 1 / `sqrt( 1 - ( v / c ) ^ 2 ) ],
and we see that the time measured by the outside observer will be equal to the product of the time measured by the observer on the rocket and the factor 1 / `sqrt( 1 - ( v / c ) ^ 2 ).
This factor 1 / `sqrt( 1 - ( v / c ) ^ 2 ) is called the 'time dilation factor'.
• When v = 0, ( v / c ) ^ 2 = 0 and it is easy to see that this factor is 1. That is, if the rocket isn't moving relative to the observer, everyone measures time the same.
• If v is very much smaller than c, as it is for anything we can move around in, v / c is very small and (v / c ) ^ 2 is very very small, practically 0, so that the factor is again near zero.
• If v is a significant proportion of the speed of light, then (v / c ) ^ 2 ceases to be negligible and makes the denominator of the factor less than 1, making the factor itself greater than 1. The larger v is, the greater the effect and the greater and the greater the time dilation factor will become.
• As v approaches c, (v / c ) ^ 2 approaches 1 and the denominator approaches 0, so the factor grows without bound. That is, as we approach the speed of light there is no upper bound on the time dilation factor. If the ship is moving fast enough a second of its time could correspond to a million years of our time. Again, there is no limit.
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oblem
If a spacecraft whose length is observed by its pilot to be 23 meters passes an observer at a relative velocity of .78 * c, where c is the speed of light, then provided the spacecraft's length is oriented in its direction of motion what will be the length of the craft as measured by the observer?
Solution
Length= 23(sqrt(1-0.78^2)
=14.4m
At relative velocity v any observer will measure the dimensions of objects in the other frame of reference to be decreased in the direction of motion. This follows easily from the fact that time is dilated, as seen in the preceding problem, but the details of this reasoning are not included here.
In the present situation the pilot is presumed to be at rest with respect to the spacecraft. The length as measured by the pilot is called the 'rest length' of the spacecraft. The observer in this situation is outside the spacecraft observing its passage.
Assuming that the spacecraft is oriented with its length in the direction of motion, its length as detected by the observer will therefore be less than that observed by the pilot who is presumably at rest with respect to the rocket.
The 'length contraction factor' will be `sqrt ( 1 - v^2 / c^2 ). In this case v = .78 * `c, so the observed length will be
• observed length = rest length * length contraction factor = 23 meters * `sqrt( 1 - ( .78 * c )^2 / c^2 ) = 23 meters * `sqrt( 1 - .78^2) = 23 meters * .6257796 = 14.39293 meters.
It is important to understand that this observation is not an optical illusion. This is what will be observed by any measuring instrument, including an array of high-speed and very accurate electronic sensors.
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Problem
A beam of protons, each with rest mass about 1.6 * 10^-27 kg, are accelerated to near-light-speed velocities before entering a magnetic field of .8 Tesla. The magnetic field is directed perpendicular to the direction of the beam. Without taking account of the relativistic effect on mass, what should be the radius of curvature the protons if they are moving at 2.81 * 10^8 m/s as the beam passes through the magnetic field?
• What should be the radius of curvature of the beam if we take relativistic effects into account?
• What will happen to the radius of curvature as the velocities of the protons approach the speed of light?
Solution
F= 1.6x10^-19 x 2.81x10^8 x 0.8
=3.6x10^-11 N
Acceleration= 3.6x10^-11/1.67x10^-27
=2.2x10^16
Curvature= (2.81x10^8)^2/2.2x10^16
=4.2 m
Mass= 1.67x10^-27 x ((1/sqrt(1-(2.81x10^8)/(3x10^8)^2))
=13.6x10^-27
Acceleration=10^-11/13.610^-27
=0.26x10^16
Curcature=(2.81x10^8)^2/0.26x10^16
= 34.1 m/s^2
For reasons related to time dilation and length contraction, as a massive particle approaches the speed of light an observer in the 'lab frame' will conclude that under a constant applied force the particle will accelerate at a decreasing rate. The observer will therefore conclude that the mass of the particle is increasing with its velocity.
Without presenting the details, we will simply state that the 'mass increase factor' is identical to the 'time dilation factor' 1 / `sqrt( 1 - ( v / c ) ^ 2 ), so that if m is the 'rest mass' of the object (the mass as observed when stationary with respect to the observer), its mass at velocity v with respect to the observer is m' = m * [ 1 / `sqrt( 1 - ( v / c ) ^ 2 ].
The force exerted on a proton by the given magnetic field is
• force on protons: F = q v B = (1.6 * 10^-19 C) * ( 2.81 * 10^8 m/s) * ( .8 Tesla) = 3.5968 * 10^-11 Newtons
(note that the field and the velocity are perpendicular so sin(`theta) = 1 and we can just use F = q v B ).
• This results in a centripetal acceleration aCent = F / m = 3.5968 * 10^-11 N / ( 1.67 * 10^-27 kg) = 2.153773 * 10^16 m/s^2.
• The resulting radius of curvature will be r = v^2 / aCent = ( 2.81 * 10^8 m/s ) ^ 2 / ( 2.153773 * 10^16 m/s^2) = 4.178714 m.
If we take the relativistic mass of the proton into account we find that the force F = 3.5968 * 10^-11 N acts on a mass of
• relativistic mass of proton = m * [ 1 / `sqrt( 1 - ( v / c ) ^ 2 ] = 1.67 * 10^-27 kg * [ 1 / `sqrt( 1 - { ( 2.81 * 10^8 m/s ) / ( 3 * 10^8 m/s) } ^ 2 ] = 13.61536 * 10^-27 kg.
The centripetal acceleration will thus be
• aCent = F / relativistic mass = 10^-11 N / ( 13.61536 * 10^-27 kg) = .2641723 * 10^16 m/s^2,
which results in radius of curvature
• radius of curvature = v^2 / aCent = ( 2.81 * 10^8 m/s ) ^ 2 / ( .2641723 * 10^16 m/s^2 ) = 34.06868 m.
As v approaches c, v^2 / c^2 approaches 1 and the relativistic mass approaches infinity. As a result the centripetal acceleration approaches zero, and the radius of the circle approaches ininity. Thus the path approaches a straight line.
Generalized Solution
At velocity v in magnetic field B, oriented perpendicular to v, `theta will be 90 degrees so a charge q will experience force
• force = q v B sin(`theta) = q v B.
If the charged particles have rest mass m and are not moving at relativistic speeds we will thus have centripetal acceleration
• centripetal acceleration = force / mass = q v B / m.
Setting centripetal acceleration equal to v^2 / r we can solve for r:
• q v B / m = v^2 / r
so
• r = m v / ( q B ).
If the velocity is relativistic then the mass is m / `sqrt( 1 - v^2 / c^2 ) so
• r = m v / ( `sqrt( 1 - v^2 / c^2 ) * q B ),
which is 1 / `sqrt( 1 - v^2/c^2) times the radius that woud be predicted ignoring relativistic effects.
As velocity approaches c, v^2/c^2 approaches 1 so 1 - v^2/c^2 approaches zero. Thus the denominator of the mass increase factor 1 / `sqrt( 1 - v^2/c^2) approaches 0, and the factor approaches infinity. That is, there is no limit to the increase of the mass. As the mass becomes larger and larger the centripetal acceleration aCent = F / m will approach zero, and the path of the particle will approach a straight line.
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Problem 27 does not work
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