This chapter discusses how length, mass, and time measurements are affected for objects that travel at very very high speeds (close to the speed of light). The speed of light in vacuum is c = 3.00x108 m/s = 3.00x105 km/s = 1.86x105 miles/s. Those who drive their vehicle about 20,000miles per year, will have an odometer reading of about 186,000 miles in 9 years. Light travels this distance in one second.
Einstein classified two studies of relativity: "Special Relativity," and "General Relativity."
In "Special Relativity," the relative motion of non-accelerating objects is studied. In other words, constant velocity motion relative to a reference frame which motion is also at a constant velocity is studied.
In "General Relativity," the relative motion of accelerating objects is studied where mass, force and energy calculations play important roles.
Before starting the special relativity, it is necessary to review the topics of relative motion as well as inertial frames. An inertial frame is one that has a constant velocity. This means that not only the speed of the frame remains constant, but also, it keeps the same direction and its motion is along a perfectly straight line. Earth itself is not really an inertial frame, because it turns about its own axis as well as going around the Sun, the combination of at least two curvy motions; however, since such radii of curvature are so great, its motion very well approximates an inertial frame for short intervals of time.
I) Relative Motion in One Dimension
For simplicity, we first look at relative motion in one dimension, say the x-axis, for example. Suppose car A is moving East at 65miles/h (with respect to the origin), and, behind that, car B is moving at 55miles/h East (also, with respect to the origin). The passengers in car A see or measure the velocity of car B moving West at 10miles/h while the passengers in car B see or measure the velocity of car A moving East at 10 miles/h.
Based on the above, the relative velocity of car B with respect to A (VB/A= -10km/h) and the relative velocity of car A with respect to B (VA/B = +10km/h).
This example also emphasizes that the measurement of motion (distance, time, velocity, and acceleration) is relative and depends on who measures what!
II) Relative Motion in Two Dimensions
In the figure on the right, the position of particle P with respect to frame A is rPA. Its position with respect to frame B is rPB. Note that the position of frame B itself with respect to frame A is determined by rBA. According to vector addition, we may write: rPA = rPB + rBA.
Now suppose both of particle P and frame B are moving with respect to frame A, not necessarily at the same velocity. Taking the time derivative of both sides and setting each term equal to its corresponding velocity, results in
(d/dt)rPA = (d/dt)rPB + (d/dt)rBA. or,
vPA = vPB + vBA.
This reads as:
" The Velocity of P relative to A is equal to the velocity of P relative to B plus the velocity of B relative to A."
The order of the subscripts must be carefully written. In general,
vAB = - vBA.
A fighter jet traveling North with velocity VFG = (750 mph, N). fires a rocket Eastward at VRF = (1,800mph, East). Find VRG, the velocity of the rocket with respect to the ground (as seen by observers on the ground).
Solution: We may write: VRG = VRF + VFG
VRG magnitude is equal to (7502 + 18002)1/2 mph = 1950mph.
VRG direction is equal to tan-1(750/1800) = 23° North of East.
The point is that an observer in the jet sees the rocket traveling Eastward at 1800mph; however, the observers on the ground see the rocket moving 23° North of East at a speed of 1950mph. Again, it depends on who measures what!
A fighter jet traveling North with velocity VFG = (750 mph, N) fires a rocket that appears moving Eastward to the observes on the ground. If the rocket travels at 1800mph relative to the fighter, find the direction of the rocket relative to the East and the speed of the rocket relative to the ground.
Solution: We may write: VRG = VRF + VFG
VRF magnitude is 1800 mph as seen by the jet's occupants in the direction shown that is be calculated.
VRG magnitude is (18002 - 7502)1/2mph = 1640mph.
The direction of VRF is equal to tan-1(750/1640) = 25° South of East.
An observer in the jet sees the rocket traveling 25° South of East at 1800mph; however, the observers on the ground see the rocket moving East at a speed of 1640mph.
III) The Galilean Transformation:
In this section, it will be shown how the measurements of position, velocity, and acceleration differ in two frames that move at constant velocities relative to each other.
The figure on the right shows two frames S and S'. Frame S' moves at constant velocity u relative to frame S. Suppose that points O and O' coincide at t = 0. The positions of point P relative to the two frames are related by
r' = r - ut (1)
This vector equation is equivalent to three scalar equations in a 3-D space. Often for simplicity, we arrange for the Ox and O'x' axes to be on each other. In such case frame S' moves at constant velocity u along the x-axis of frame S, as shown in the lower figure. For this arrangement, the three components of Equation (1) become:
x' = x - ut y' = y z' = z and t' = t . (2)
Equations in (2) give the coordinates of a particle in two inertial frames moving relative to each other at constant velocity. This is called the Galilean transformation of coordinates.
Note: An assumption is made that the observers in both frames measure the same elapsed time ( t' = t ) once their clocks are synchronized, and this remains true indefinitely. This means that in Newtonian Mechanics, there is only a single universal or absolute time for all frames. The special theory of relativity has shown this assumption to be false; however, it is an excellent approximation for our everyday experiences with ordinary motions when we deal with speeds much much lower than speed of light.
To find the expressions for relative velocities, let's take the time derivative of the 3 position equations in (2). We get:
dx'/dt = dx/dt - u dy'/dt = dy/dt and dz'/dt = dz/dt, or
v'x = vx - u v'y = vy and v'z = vz .
Since the motion is in 1-D only, and there are no relative velocities in the y and z directions, we may drop the subscripts from the x equation and write the only relative velocity equation as
v' = v - u
To obtain an expression for acceleration, let's take the time derivative of this relative velocity equation. Note that u is a constant velocity and its derivative is zero (du/dt = 0); therefore,
dv'/dt = dv/dt - 0 or a' = a
This clearly states that the observers in both systems measure the same value for the acceleration of a particle.
As an example consider a train car moving to the right at velocity u in which a person throws a ball straight up relative to the train. The ball will go straight up and down back to his hands as measured in the train's frame(S') as shown. An observer on the ground frame (S) notices a parabolic path for the ball. From the point view of the observer in S, the ball has two motions: one at constant velocity u horizontally, and one up and down under gravity acceleration. The horizontal acceleration measured by both observers is zero (du/dt = 0). Also, the gravity acceleration is measured the same by both observers.
This observation does not allow us to claim that one frame is fixed and the other is moving. No experiment allows us to distinguish between inertial frames. This is expressed as the Galilean principle of relativity:
"The laws of mechanics have the same form in all inertial reference frames."
Test Yourself Questions 1:
1) Speed of sound at STP conditions is (a) 331m/s. (b) 343m/s. (c) 300m/s.
2) The speed of light is roughly (a) 100 times (b) 1000 times (c) 1000,000 times greater than speed of sound.
3) Speed of light is (a) 3.00x107m/s. (b)3.00x105km/s. (c) 1.86x106miles/s. (d) a, b, and c. click here.
4) Special relativity is about the relative motion of (a) accelerating objects. (b) objects with increasing forces. (c) non-accelerating objects.
5) General relativity is about the relative motion of (a) accelerating objects. (b) non-accelerating objects. (c) objects that move at speed of sound. click here.
6) An inertial frame is one that moves at (a) constant velocity but curves. (b) at constant velocity. (c) at constant speed along a straight line. (d) b and c.
If while you are traveling West at 70 miles/h face a car traveling East at 50miles/h,
7) you measure the velocity of the other car (a) 20miles/h West. (b) 120miles/h East. (c)120miles/h West. click here.
8) the observers in the other car measure your velocity (a) 120miles/h West. (b) 120miles/h East. (c) 20miles/h West.
A fighter jet traveling eastward at velocity VFG = (900.0 mph, East) fires a rocket southward at VRF = (2,160mph, South). Answer the following questions: (You need to draw a vector diagram and solve the problem).
9) The VRG magnitude is equal to (a) 4320miles/h. (b) 3060mples/h. (c) 2340miles/h. click here.
10) The VRG magnitude is (a) 67° South of West. (b) 67° South of East.. (c) 67° North of East..
A man holding a rifle in a train that travels North at velocity VRG = (150 mph, N) fires a bullet that moves Westward. The man must have held the rifle in the Southwest direction for the bullet to travel exactly westward relative to the ground. If the bullet leaves the rifle at 360 mph, answer the following questions:
11) VBG magnitude (bullet's speed relative to the ground) is (a) 390mph. (b) 410mph . (c) 327mph . click here.
12) The direction of vector VBG relative to East is (a) 25°. (b) 225°. (c) 205°.
13) If frame S' (that is to the right of S) is moving a constant velocity u to the right relative to S such that their x-axes coincide and also their Origins coincided at time t=0, then (a) x' = x - ut. (b) x' = x + ut. (c) x' = x ∙ ut.
14) x' = x - ut calculates the relative (a) velocity of. (b) acceleration of. (c) position of a particle in S' relative to S. click here.
15) To find an expression for relative velocities, one must take the (a) time derivate. (b) space derivative. (c) second derivative of x' = x - ut.
16) The (d /dt) of x' = x - ut, yields: (a) v' = v - u. (b) a = a' - u. (c) v' = v + u. Note: Here ( ' ) does not denote derivative.
17) Given x' = x - ut, the acceleration of a particle in frames S and S' are related by (a) a = a' - u. (b) a = a' + u. (c) a = a' . click here.
18) The accelerations being the same in both S and S' is a verification that both frames (a) are inertial. (b) move at constant velocities . (c) are indistinguishable from each other as to which one is moving. (d) a, b, and c.
19) The up-and-down trajectory of a ball in a moving train as seen by its occupants is seen by ground people as (a) up-and down. (b) parabolic. (c) like a semi-circle.
20) The assumption of everlasting synchronization of the clocks in Galilean transformation is (a) only a good approximation if objects move at low speeds. (b) good for all speeds. (c) only good for high speeds close to speed of light . click here.
Light Propagation in Free Space (Vacuum):
Recall from the chapter on waves that mechanical waves require matter for their transmission. Also, recall that the more rigid a medium, the faster mechanical waves travel in it. Before Maxwell and the introduction of electromagnetic theory, light was thought as a mechanical wave and because of its enormous speed required a medium of infinite rigidity and at the same time negligible density. These two required properties were quite contradictory to each other. Such bizarre medium was named ether. It was also assumed that the entire space was filled with ether and that ether was at absolute rest and any object's motion relative to ether would therefore be its absolute motion.
This way of thinking (the ether theory) was shattered by the Michelson -Morley experiment in 1887. Michelson developed an interferometer that showed there was no such thing as ether and therefore the thought of absolute reference frame and absolute motion became questionable.
The Michelson-Morley Experiment:
| The Interferometer:
As shown on the right, two adjacent and coherent light rays from source S travel to the right toward the mirror-glass P set at 45°. One ray that passes through P, travels toward the flat mirror M1, gets reflected back to P, and then gets reflected down toward the telescope as shown. The other ray gets reflected upward at P toward mirror M2 that reflects it back downward to P and after passing through P reaches the telescope. If the apparatus is at rest, since both rays travel the same distance, the final rays arriving at the telescope will necessarily be in phase and a certain interference pattern will be expected to be viewed by the telescope. This expectation is quite normal if the interferometer is at absolute rest.
Since Earth travels about the Sun at 19 miles/s the thought of stationary interferometer is out of question.
The interferometer is first aligned in the direction of motion of the Earth about the Sun. This means that Ray SM1 is set parallel to the direction of the Earth's motion. Now, we want to calculate the time difference between the arrival of the final rays to the telescope. If ether is present and is at absolute rest, there should be a time difference between the arrival of final rays at the telescope.
Let c be the speed of light relative to the ether, and v, the speed of Earth relative to ether. Refer to the figure above. When light goes from P to M1, distance L0 is traveled at speed c - v. The light returned from M1 travels at speed c + v. The back and forth time is therefore:
When light goes from P to M2 and back, the travel time is
The reasoning for this calculation is on the right.
The difference in travel time between the two rays is therefore,
| As light goes up to M2,
the Earth goes to the right and the velocity of light relative to the
ether is VLE or
c as shown (the hypotenuse vector).
Ether moves to the left relative to mirror M2, as shown by
VEM = -v.
Light moves straight up relative to mirror M2, VLM.
We may write the vector equation:
VLM = VLE + VEM
The magnitude of VLM is needed for travel time calculation of light going up and back down to P. Since VLE = c, and VEM = v , the magnitude of VLM from the vector triangle is
This speed appears in the denominator of T2 on the left.
According to the above calculations, with respect to a stationary ether, there must be a time delay Δt = L0v2/c3 between the arrival of final rays at the telescope causing some change in the interference patterns compared to a stationary situation. No change has been observed so far by many experimenters who have repeated this experiment with more and more precisions. Even when the apparatus is turned by 90 degrees to swap the roles of the mirrors in order to double the time delay effect, no change in the interference patterns occurs. No change in the patterns means Δt = 0. This can be true only if v = 0. This makes VEM = 0 and consequently, VLM = c. In other words, light travels at the same speed c in all directions regardless of the speed of its source.
The Results of the Michelson-Morley Experiment:
The results of the Michelson's experiment are:
1) There is no such thing as ether. 2) Light travels at the same speed c in all directions regardless of the velocity of its source.
The Contraction Hypothesis:
One result of Michelson's experiment is that light travels at the same speed in all directions. This result is in agreement with the way waves propagate in a uniform medium. We know that the speed of waves in a medium is a property of that medium; in other words, it is the physical properties of a medium that determines the speed at which waves are allowed to travel in it. One might think that there is nothing in vacuum to be called a medium; however, the transmission of both of the electric field and magnetic field effects are limited and dependent on two constants ε0 and μ0 . These are the permittivity and permeability of vacuum, respectively. Note that the factor 1/c2 in Maxwell's wave equation is in fact equal to the product ε0μ0.
Example: Use the values of ε0 = 8.85x10-12 Farads/m, and μ0 = 4π x 10-7 Tm/A to verify that c = 3.00x108 m/s in Metric System.
Solution: The calculation is left for students.
Important: Now, one problem or question arises here. In Michelson's experiment as soon as light waves leave slab P, they travel at speed c, but the fact that the relative speed of waves reaching M1 is c-v and after reflection the relative speed of waves arriving at slab P is c+v may not be ignored. The calculation of T1 is still valid. Also, waves going up to mirror M2 arrive at M2 not exactly at its middle, but slightly to the left of its middle because meanwhile M2 moves some distance to the right; therefore, the calculation of T2 is valid as well. With no change in the interference patterns observed by the telescope, the requirement, Δt = 0, is still valid as well and we must have T1 = T2. The question is that if light travels at the same speed horizontally and vertically, how can T1 and T2 be equal? We need to look at the expressions for T1 and T2 again.
The only way T1 and T2 can be equal is that distance Lo parallel to the direction of motion of the Earth (or the apparatus) must have shrunk to compensate for the difference. This establishes the basis for the topic of "Length Contraction." Let's look at the expressions for T1 and T2 again.
The above two are not equal the way they are; unless the L0 in T1 is replaced by a shrunk length of L0 (1- v2/c2)1/2 . This means that the length L0 in the direction of motion must have shrunk by the factor (1- v2/c2)1/2 . This was originally suggested by G. F. Fitzgerald in 1889. In 1892, H. A. Lorentz suggested that the cause of contraction must be the modification of electrical forces within the moving body.
Einstein's Postulates for Special Relativity:
As was mentioned before, special relativity deals with the relative motion of inertial frames or non-accelerating frames. Einstein used the above discussions to come up with his famous two postulates of special relativity as follows:
1) The Principle of Relativity: All physical laws have the same form in all inertial frames.
2) The Principle of Constancy of Speed of Light: The speed of light in free space is the same in all inertial frames. It does not depend on the motion of the source or the observer.
1) Event: An event is the occurrence of something at one point in space and at a single instant in time.
2) Observer: An observer is a person or an automatic device with a clock and a meter stick. Each observer can record events only in the immediate vicinity. Each observer has to rely on colleagues at other locations to record the times of distant events. An observer may see or photograph a distant event, but such observations do not count as a record of the event.
3) Reference Frame: A reference frame is a whole set of observers uniformly distributed in space as shown below on the left. All observers in a given reference frame agree on the position and the time of an event. Only one observer would be close enough to an event to record it after which the data will be communicated to other observers. The notation S, S', S", ... will be used to denote different inertial frames. The frame in which an object, such as a clock or a rod is at rest, is called its rest frame. The following notation will be used to precisely describe events in special relativity:
x : Position coordinate of an event, a point in space
Δx : x2 - x1 = L A space interval, a length
t : Time coordinate of an event, an instant in time
Δt : t2 - t1 = T A time interval, a period
Synchronization of Clocks: If the clocks are spread apart by one light second that means the distance light travels in one second, and pre adjusted to the correct number of seconds ahead of each other depending on their respective distances, we might then say that they are synchronized. Above, the figure on the right shows and explains five synchronized clocks. For example, if clock a is preset at 12 noon, then clocks b, d, e, and f must be preset at 1s past noon, 2s past noon, 3s past noon, and 4 s past noon to be synchronized.
The Relativity of Simultaneity:
Two spatially separated events that are simultaneous in one frame are not simultaneous in another frame that is moving relative to the first one. This effect is called the relativity of simultaneity.
To show this, consider a train (frame S') moving at a constant velocity v relative to a platform (frame S). In frame S, an observer is exactly located at O, the midpoint between A and B. In frame S', another observer is exactly located at O', the midpoint between A' and B'.
If two firecrackers explode simultaneously at points A and B, the observer at O records them simultaneously because of being equidistant from A and B. The observer at O' receives the signal from B before the signal from A. The observer in S' does not judge the two explosions to be simultaneous. This is called the relativity of simultaneity and it is reciprocal. Events simultaneous for O' will not be so for O, either.
Since simultaneous events that occur at two different locations in S are not simultaneous in S', the time interval between the events will be different in the two frames, as well. This will also affect the measurement of length. The problem with Length measurement is explained below.
| We do not have any trouble measuring
the length of a stationary rod. In order to measure the length
of a moving rod, the positions of both ends of
it must be marked against a ruler simultaneously. How
can this be done? One person alone cannot do it. We have
to rely on two measurements done simultaneously.
Even if, we do that as we are standing in frame S, the people moving with the rod (in frame S') will complain about the simultaneity of our measurements. They do not see our measurements simultaneous. Consequently, observers in frames S and S' do not agree on the length of the rod.
It is the relativity of simultaneity that leads to the length contraction introduced by Fitzgerald and Lorentz. They believed that there is a physical contraction caused by the modification of electric forces between atoms. The theory of special relativity arrives at the same results from a different point of view- a profound analysis of the process of measurement.
We are going to compare two time intervals Δt and Δt' one measured in frame S and one measured in frame S'. Let S' be the rest frame of the clock. It means the clock is at rest in S'. In this frame, if a signal is sent upward from time clock A', it will get reflected at mirror M and returns to detector B' placed very near to A'. This measurement that is done in the rest frame of the clock is called To or the proper time and is equal to:
Now, let's see how the measurement goes in frame S, in which the clock moves at velocity v. Here the time interval Δt is measured by two observers A and B at different positions. We may write:
Let's name this correction factor g on which basis the dilated time can be calculated.
and hence, T = γTo or, Δt = γΔt'
In the rest frame of the clock, frame S', the time interval for a pulse going from A to M and return to B is 2Lo/c.
In the moving frame S, the emission and detection events occur at two positions. The time interval in this case is greater than the interval recorded in the rest frame of the clock.
Since γ >1, the time interval T measured in frame S (by two clocks) is greater than the proper time, To, measured by the clock in its rest frame S'. This effect is called Time Dilation. Two spatially separated clocks, A and B, record a greater time interval between two events than the proper time recorded by a single clock that moves from A to B and is present at both events.
Time dilation is reciprocal. If Δt is the proper time for a clock in S, then the two observers in S' would measure Δt' = γΔt. If this effect were not reciprocal, there would be a way to distinguish between inertial frames. Time dilation applies to any periodic phenomenon, electronic, mechanical, or biological.
Example 3: A train moving at 0.6c takes 5μs to pass an observer on a platform. (a) What is the time interval measured in the train's frame?
Solution: The value of γ for v = 0.6c is 1.25. The observer on the platform measures a time interval of 5μs between two events at a single position, and therefore the proper time. The observers in the moving frame measure the dilated time; therefore, T = γTo = (1.25) 5μs = 6.25μs.
The length of a rod in its rest frame, S, is the distance between its ends called its "proper length, Lo ."
An observer O' in S' that moves at a velocity v relative to frame S, can measure the rod's length by recording the interval between the times at which O' passes A and B. Note that O' can not tell if he is moving to the right or the rod is moving to the left; therefore, he can measure the rod's length by two ways of thinking: (1) If he sees the rod moving to the left at velocity v and treats himself as being at rest, he records two instances while being at the same position and therefore he measures the proper time Δt' or To. In this case, since the rod is moving by and not stationary, the contracted length of it, L, is measured; therefore,
L = vTo or L = vΔt' . (1)
Now, option (2), if he sees the rod at rest and himself in motion, then he records two instances at two different positions (the rod ends) and therefore, he measures a dilated time interval, Δt, or T that corresponds to the proper length, Lo, of the rod; therefore,
Lo = vΔt . (2)
Substituting γΔt' for Δt in (2), we get::
Lo = v γΔt' (3)
Now dividing (1) by (3) yields:
L = (1/γ) Lo .
Since γ > 1 ; therefore, L < Lo and the effect of length contraction is apparent from this equation.
Example 4: The length of a train that moves at 0.6c relative to the ground is 640m as measured by the people on the ground. Calculate (a) the actual length of the train. How long does it take the train to pass a light pole at the station (b) in the ground frame, and (c) in the train frame.
Solution: (a) The 640m is the contracted length, L, measured by the ground observers. The value of γ for v = 0.6c is 1.25. We may calculate the proper length by Lo = γL = (1.25)(640m) = 800m.
(b) The travel time of the train measured by the ground people is L = v To ; 640m = (0.6c)(To) ; To= 3.56μs
(c) The travel time of the train measured by observers on the train is Lo = v T ; 800m = (0.6c)(T) ; T = 4.44μs
The Relativistic Doppler Effect:
For sound waves, the observed frequency depends on the source frequency according to the following equation:
For sound there is a medium (air) that serves as an absolute reference frame. For Electromagnetic waves (signals) traveling in vacuum, there is no absolute reference frame and therefore relativistic Doppler effect must be considered. The frequency of the waves emitted from a moving frame, S', and arrived at S (treated to be at rest), depends on the relative velocity of S and S', only. There are two extreme cases:
1) Source and observer are receding and the direction of v is completely along the line of sight: :
Suppose S' is emitting E&M signals at period To in its own frame while moving away from S at a velocity v to the right. The signals recorded by S will have a longer period because of time dilation. The thought that S records time dilated signals of period Δt = γTo is not completely correct. This is not the whole story! Not only S receives dilated signals, but also there will be a time delay for each Δt equal to d/c associated distance d traveled by S' during each Δt. Since d as measured by S is
d = vΔt = vγTo, the period of signals recorded by S is:
T = Δt + d/c = γTo + vγTo/c = γ(1 + v/c)To
Clocks tick slower in a moving frame relative to another frame treated to be at rest. If To is the period of signals emitted in S', then γTo is the dilated period. S however records
T = γTo + d/c where d = vΔt = vγTo, the distance traveled by S' during each Δt as measured by S.
Since γ = ( 1 + v/c )-1/2 ( 1 - v/c )-1/2 ; therefore,
This equation is used when source and observer move away from each other such that the direction of v lies completely along the line of sight. For approach, the sign of v must be changed.
2) The direction of v is perpendicular to the line of sight:
In this case, the transverse Doppler effect is used and involves the effect of time dilation, only.
Example 5: A spaceship approaching Earth at 0.8c emits E&M pulses that are 1.20 seconds apart in the ship's fame. Calculate the frequency recorded here on Earth.
Solution: The period for the approaching pulses is T = [(c-v)/(c+v)]0.5 To = 0.40s. The frequency is therefore f = 1/T = 2.5Hz
The Twin Paradox:
Let's see how time elapses for a twin that one stays here on the Earth and the other makes a round trip to a distant point in space at high speed and back. Suppose twin A stays here and twin B travels at 0.8c. If twin B emits 15 pulses at intervals of 6 minutes while going away and 15 pulses of the same period on his way back, the total time for his trip according to his calculation (proper time) is To = 180 minutes. Twin A however, received longer period pulses one way and shorter period pulses on B's return and his total calculated time for B's trip is (dilated time) T = 300 minutes.
For v = 0.8c, the value of γ = 5/3. As we see the relation T = γTo holds true for this calculation 300 = (5/3) 180 !!! The details of calculations is shown below:
( B )
Going away trip: Δt'1 =(15)(6min) = 90min.
Return trip: Δt'2 =(15)(6min) = 90min.
Total time for twin B = 180min.
( A )
Δt1 =(15)(6min)[(c+v)/(c-v)]0.5 = 270min.
Δt2 =(15)(6min)[(c-v)/(c+v)]0.5 = 30min.
Total time for twin A = 300min.
|For v = 0.8c, the value of γ = 5/3. This verifies that 300 = (5/3) 180 or, T = γTo|
The situation is not symmetric. One reason is that the going away time as calculated by A is much longer than the return trip. Also, twin A remains at rest and relaxed while twin B undergoes modified inter-atomic electric forces that account for length contraction. Moreover, it is twin B that undergoes changes in acceleration at the return point. Note that at the return point, special relativity is not sufficient to account for switching to accelerated frames; however, general relativity also confirms that the twin paradox is true.
Example 6: To verify that the twin paradox is not symmetric, solve the above problem by assuming that A emits pulses and B receives them. Find the number of pulses twin B receives while (a) receding from A, and (b) approaching A.
Solution: The solution is left for students. Answer: (a) 5 pulses (b) 45 pulses
Test Yourself Questions 2:
1) Light waves are (a) mechanical (b) electromagnetic (c) transverse (d) b and c. click here.
2) Because is high speed, light transmission requires a medium of (a) high rigidity (b) low density (c) both a and b (d) neither a nor b.
3) The ether theory meant a medium of (a) high density (b) low density (c) high rigidity (d) low rigidity (e) a & d (f) b & c.
4) The ether medium was thought as (a) relatively fixed (b) absolutely at rest (c) moving with light (d) a and b . click here.
5) The motion of light in ether was thought as its (a) absolute motion (b) relative motion (c) its rest frame motion.
6) Michelson-Morley experiment (a) proved (b) just verified (c) shattered the ether theory.
7) Michelson's experiment disqualified the (a) absolute reference frame (b) absolute motion (c) a & b. (d) just b.
8) Standing in a train that travels East at 25m/s, if you throw a ball Eastward at 5m/s, the speed of the ball relative to Earth (if treated as an absolute frame) is (a) 30m/s East (b) 20m/s East (c) 30m/s West. click here.
9) Standing in a train that travels East at 25m/s, if you throw a ball Westward at 5m/s, the speed of the ball relative to Earth (if treated as an absolute frame) is (a) 30m/s East (b) 20m/s East (c) 30m/s West.
10) Standing in a train that travels East at 25m/s, if you throw a ball Eastward at 5m/s that hits a wall in front of you and bounces back at the same speed, the return speed of the ball relative to Earth (if treated as an absolute frame) is (a) 30m/s East (b) 20m/s East (c) 30m/s West.
11) Question (9) thinking may be used to conclude that the speed at which light travels the horizontal distance Lo from P to M1, in Michelson's apparatus as shown above, is (a) c-v (b) c+v (c) neither a nor b. click here.
12) Question (9) thinking may be used to conclude that the speed at which light travels the horizontal distance Lo from M1 to P, in Michelson's apparatus as shown above, is (a) c-v (b) c+v (c) neither a nor b.
13) In Michelson's experiment as shown above, the actual velocity direction of the ray that goes upward, VLE, must be (a) pure North (b) pure East (c) Northeast.
14) The reason for the Northeast direction of VLE in Question 13 is that (a) the bounced off rays from slab P want to go straight up (b) the slab that is the source for bounced off rays, has itself an Eastward motion (c) both a and b. click here.
15) In Michelson's experiment as shown above, the Northeast VLE = c has two components, VLM and VME = v. If we think of an ether wind that neutralizes the component VME = v, then the ether wind must have a velocity equal to (a) VEM = -v (b) VEM = +v (c) neither a nor b.
16) In Michelson's experiment as shown above, the actual upward speed of rays, VLM is (a) v (b) c (c) (c2 - v2)1/2 .
17) In Michelson's Experiment the arrived final rays at the telescope show a change in interference patterns (a) only for motion in the Earth direction (b) only when the roles of the arms are swapped (c) in none of many trials in different directions and with different precisions. click here.
18) In Michelson's Experiment, since no change in interference patterns is observed in any of the cases, one may conclude that (a) the travel time for the horizontal and vertical rays, as shown in the above figure, must be equal (b) the travel times may not be equal (c) ether theory was not bad at all.
19) In order for T1 and T2 in Michelson's experiment to be equal, (a) the apparatus must be at absolute rest (b) answer (a) is not possible because there is no such thing as absolute rest (c) the arm along the direction of motion must shrink by the factor (c2 - v2)1/2 (d) all of a, b, and c.
20) The suggestion that the arm along the direction of motion in Michelson's experiment must contract by the factor (c2 - v2)1/2 was made by (a) Einstein (b) Fitzgerald (c) Lorentz. click here.
21) The reason for length contraction of Lo along the direction of motion suggested by Lorentz was (a) the collision of the apparatus with air molecules (b) vacuum pressure (c) the modification of electric forces in between atoms and molecules of the moving body.
22) According to Einstein's first postulate, (a) the Galilean transformation is true in all inertial frames (b) ether theory is true for non-inertial systems (c) all physical laws have the same form in all inertial frames.
23) According to Einstein's 2nd postulate, (a) the speed of light in vacuum decreases at higher frequencies (b) the speed of light in free space is the same in all inertial frames. (c) the speed of emitted light increases as the speed of its source increases (d) both b & c. click here.
24) An event is the occurrence of something (a) at one point in space (b) at a single instant in time (c) both a and b tohether.
25) Suppose clocks A and B are synchronized; A reads 10:00:45, and B reads 10:00:48, then the distance from B to A is (a) 9.00x108m (b) 3.00x108m (c) 6.00x108m
26) The relativity of simultaneity states that two simultaneous events in one frame (a) are necessarily simultaneous in other inertial frames (b) are not simultaneous in other inertial frames that move at a nonzero relative velocity (c) are simultaneous in other frames that have a constant relative acceleration.
27) Measuring the length of a moving rod (picturing it to be in S ) while you are in S' considering yourself at rest, requires a stop watch to record in instances when the rod ends pass by you. If that happens, you have measured (a) the proper length, Lo (b) the proper time, To (c) the contacted length, L (d) both b and c. click here.
28) Measuring the length of a rod while traveling in a moving frame S' and passing by the rod that is placed in frame S (treated to be at rest), requires a stop watch. If that happens, (a) the actual length, Lo is measured (b) the dilated time is measured (c) both a and b (d) neither a nor b.
29) In relativistic Doppler effect (a) an absolute reference frame must be used (b) there is no absolute reference frame because there is no medium (c) similar formulas exist as were developed for sound.
30) The relativistic Doppler effect for light depends on (a) the velocity of observer only (b) the velocity of source only (c) depends on the relative velocities between source and observer. click here.
1) Find the speed of a meter stick that is measured to be 96.0cm as passing by an observer.
2) Use the binomial expansion; (1+x)n = 1 + nx + .... to show that for ( v << c ), γ = 1 + v2/2c2. click here
3) A spaceship moving at 0.800c relative to a station has a length of 300m as measured by the observers at the station. Find its length as measured by its occupants.
4) Alpha-Centauri is the nearest star to our Sun and is 4 light-years (ly) away. How fast should a spaceship move so that the travel distance measured in the ship is 3ly? Note that ly is a measure of distance and not time.
5) How fast must a clock (frame S') move relative to frame S in order to lose (a) 30days in 1year, (b)1day in 1year, and (c) 1s in 1year as measured in S? Let each year be 365days. Hint: If v<<c, the approximation in Problem 2 may be used. click here
6) A spaceship travels at a speed of 0.397c relative to the Earth (frame S) for 365days as measured by the observers here on the Earth. Calculate (a) the proper duration as measured by the clocks in the ship's frame, and (b) the time loss in the ship's frame.
7) A train traveling at 0.6c takes 8μs to pass an observer on a platform. (a) What is the time interval measured in the train's frame? What is the train's length according to observers (b) on the train, and (c) on the platform?
8) At what speed will the measured rate of a clock be 25% of its measured rate at rest? click here
9) Spaceship B overtakes spaceship A at a relative speed of 0.25c. Observers in A measure the length of B 136m. (a) What is the proper length of B? How long does it take B to pass a given point on A as measured by observers (b) in A, and (c) in B?
10) A train moving at 0.5c relative to the ground has a measured length of 405m in the ground frame. How long does it take to pass a tree as measured (a) in the ground frame, and (b) in the train frame? click here
11) A spaceship travels the 800-km distance between two cities at a speed of 0.16c. (a) What is the elapsed time according to the ship's occupant? (b) What distance do they measure for this trip?
13) A spaceship moving at 0.8c relative to the Earth has a flashing light that flashes 90.0 times per minute. At what rate do the pulses arrive on Earth if the ship is (a) approaching, and (b) receding? click here
14) A galaxy moves away from the Earth at 0.316c. What is the natural wavelength of a spectral line whose wavelength measured in a laboratory is 589nm?
15) A police radar gun that uses 1.5-cm E&M waves is held toward a car that is approaching at 30m/s. Calculate (a) the Doppler shift in frequency received by the car, (b) the Doppler shift in frequency of waves emitted by the car and received by the gun, and (c) the total shift in frequency recorded by the gun.
16) Show that the sum [(c+v)/(c-v)]0.5 + [(c-v)/(c+v)]0.5 = 2γ. click here .