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Module 12Paradoxes1 Module 12 Paradoxes How does one test a scientific theory? Experimental testing is certainly the preferred method. Perform an experiment.

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Presentation on theme: "Module 12Paradoxes1 Module 12 Paradoxes How does one test a scientific theory? Experimental testing is certainly the preferred method. Perform an experiment."— Presentation transcript:

1 Module 12Paradoxes1 Module 12 Paradoxes How does one test a scientific theory? Experimental testing is certainly the preferred method. Perform an experiment and show that the results are consistent with the predictions made by the theory. But one can also test a theory by trying to find conceptual flaws in it. One such conceptual flaw is a paradox, a situation where it seems that the theory gives two possible, conflicting results. If a paradox cannot be resolved, then the foundation of the theory develops a crack. Relativity theory has been tested experimentally and it has passed every test. It has also been subjected to many apparent paradoxes, perhaps more than most theories because it is a theory of relativity (it deals with the observations of two observers) and because many of its predictions are, for lack of a better word, unusual. So far, all of the paradoxes have been resolved, although some people may not be satisfied with some of the resolutions. But that is also the nature of scientific theories; you can’t satisfy everybody. We’ll look at a few apparent paradoxes and discuss their conventional resolutions. The Twins The twin paradox is arguably the most popular paradox of special relativity theory. The usual statement of the paradox goes something like this. There are two twins, A and B. B leaves Earth in a space ship, travels near the speed of light, and returns to Earth. The time for the roundtrip measured by B’s clock is a proper time interval. Thus, A’s clock will measure a longer time for the trip. So A will be older than B when B steps out of the ship. But from B’s point of view, B is stationary and the Earth and A are traveling. A’s clock will measure a proper time interval and B’s clock will measure a nonproper interval. Thus, B will be older than A when B returns. (The fact that A and B are twins isn’t important at all. It just adds dramatic effect to the scenario.)

2 Module 12Paradoxes2 The reason that this paradox is so popular isn’t so much because of its resolution but rather because of the difference in aging of the twins. This aging difference is time-travel into the future for one of the twins. We will explore this time-travel in detail because it is extremely interesting and it is supported by the theory. We will then briefly discuss the rather anticlimactic resolution of the paradox. Let’s say that B travels at 0.8c in his ship, which corresponds to a gamma factor of 1.67 between A and B. Because we are dealing with the special theory of relativity, we can’t readily handle accelerations. So we will make B go in a straight path from Earth and return along the same trajectory. We will neglect the acceleration form rest to 0.8c, the acceleration during the turning around, and the deceleration during the landing. The trip out takes 3 years measured by B on the ship’s clock. The return trip, then, also takes 3 years. So B measures 6 years for the total trip. If we take the stance that this is a proper time interval, then A will measure a dilated, nonproper time interval of 1.67 times 6 years or 10 years. Thus, A will be 10 years older than when B left, B will be 6 years older, and B will have traveled 4 years into Earth’s future as far as B is concerned. It should be noted at this point that even though the conclusion of 6 years for B being 10 years for A is correct, using the principle of time dilation here is really not appropriate. The principle strictly applies to the case where one clock, moving in a straight line in the same direction, records the proper interval while a stationary observer measures a nonproper time interval between two events that occur at different positions. In our scenario, the time measured by A is not a valid nonproper time interval. First, B changes directions during the trip and, second, A records time on one clock at one position. There is a better way to view the scenario that leads to the same final result. This better view also gives us more details about how this time-travel pans out as seen by the twins. What we do is give A and B lasers and have each of them send a pulse of light to the other at a rate of one pulse per year as measured on their respective clocks. We now want to figure out the rate at which the signals are detected. To do this, we use the relativistic Doppler shift results of Module 6. The frequencies that we are interested in comparing here are not the frequencies of the electromagnetic field oscillations (although those are shifted as well) but rather the pulse transmission and reception frequencies. For example, let’s look at twin B. B sends a pulse of light once each year. If we say that B is in frame S’, then the transmission frequency is f ‘= 1 pulse per year or 1 ppy. Now we find the Doppler-shifted reception frequency f to see when A receives these pulses in frame S.

3 Module 12Paradoxes3 On the trip out, A sees the source of the frequency moving away. We can therefore use the longitudinal Doppler shift expression [Eq. (6.27)] to find We interpret this result as meaning that A receives the first pulse form B after 3 years on A’s clock, the second pulse after another 3 years, and the third pulse after another 3 years. Thus, the three-year trip out for B is measured to be a nine-year trip for A. On the trip back, A sees the source of the frequency moving towards Earth. We use the pertinent longitudinal Doppler shift expression [Eq. (6.25)] to find We interpret this result as meaning that A receives the remaining 3 pulses form B during a single year, the tenth year on A’s clock. Thus, the three-year trip back for B is measured to be a one-year trip for A. We see right away that the 6 years for B is 10 years for A, just like the time-dilation result. But we also see that the dilation is not uniform nor symmetric. A measures the majority of the total trip time to be for the trip out. In fact, the return trip is measured by A to take less time than it does for B. The objection to using the time-dilation principle has merit. While the total trip time is dilated for A, the return trip time is actually contracted for A. If you are having a bit of trouble sorting this out, don’t feel too bad. What might help make things clearer is a pictorial representation of what is going on here. Maybe a space-time diagram will help.

4 Module 12Paradoxes4 Check out the space-time diagram in Figure The diagram is drawn such that A is seen to be at rest at the origin of frame S. Thus, we see the world line of A going right on top of the ct axis. Each marked division represents one year on A’s clock. We have B at rest at the origin of frame S’, with this origin moving to the right during the trip out at a constant speed of 0.8c. Thus, B’s outgoing world line corresponds to the ct’ axis which is tilted clockwise at 38.7  from the vertical. A division of 1 year on B’s world line is 2.13 times longer than a division of 1 year on the world line of A. (You should confirm this angle of 38.7  and this factor of 2.13 using what you learned in Module 8!) All we need to do to make B’s world line for the return trip is to draw a straight line tilted 38.7  counterclockwise from vertical. The orange lines are the world lines of the laser pulses sent out by B. These lines are are directed towards A and tilted 45  from vertical since they are light world lines. We see that B sends a pulse after each year of the ship’s clock. We see that the first pulse is received by A after 3 years on the Earth clock, just as we deduced earlier. And, sure enough, the second pulse is received after 6 years, the third after 9 years, and the final 3 pulses all arrive during the last year. The last pulse is received by A as B lands  ct (c-years) x outgoing world line of B returning world line of B turnaround point Fig 

5 Module 12Paradoxes5 You might be thinking that we could also show what happens to the signals received by B on a space-time diagram. You would be right! Look at Figure Now the light world lines are drawn emanating from A after each year on A’s clock. We see that the first pulse is not received until 3 years after launch on B’s clock, at the turnaround point. On the return trip, B receives nine pulses at a constant frequency with the tenth pulse received as B lands. This is consistent with the Doppler shifted pulse frequencies observed by B. On the trip out, the observed frequency is 1/3 ppy, and on the trip back it is 3 ppy. So a three-year trip out means B receives one pulse and a three-year trip back means nine more pulses are received. ct (c-years) x outgoing world line of B returning world line of B turnaround point Fig. 12-2

6 Module 12Paradoxes6 As if things aren’t strange enough, let’s look at how each twin observes the aging of the other to occur. Each knows that the other is sending one pulse each year so each can track the age of the other by counting the received pulses. We look at Figure 12-1 for A’s perspective and Figure 12-2 for B’s perspective. Let’s say that the ship leaves when the twins are 20 years old. Table 12.1 summarizes the observed ages. Notice that the aging looks identical for both twins on the trip out. Each says on their twenty-third birthdays, “I’m 23 and my twin is 21.” These observations are consistent with the principle of time dilation and, for the trip out, we can invoke time dilation since the motion is only in one direction. A sees B moving to the right and measures a longer time interval for the trip out than B does. But B sees A moving to the left and measures a longer time interval for the trip out than A. True time dilation is symmetric. It is only after the turn back home that the aging starts to look different for each twin. Both turn 24 on the same day if you ask B, but A would say that B is still 21 on A’s twenty-fourth birthday! However, both agree that A is 30 and B is 26 when the ship lands, and that B has traveled four years into Earth’s future. Table Observed Ages of Twins Twin A A’s AgeMy AgeB’s AgeMy Age Twin B What an interesting and bizarre trip this has turned out to be. But, wait. We still have not resolved the paradox. We have analyzed the trip with A at rest in frame S and B in frame S’. But can’t we say that B is in frame S and A is in S’ so that B ages 10 years and A ages 6 years? The answer is no, this symmetry does not exist. What breaks the symmetry? The answer commonly proffered is that B must accelerate to return to Earth. A does not accelerate. It is this acceleration which breaks the symmetry and resolves the paradox. You might argue, however, that we could think of B as stationary and the Earth with A on it accelerating as it goes out and returns to B’s ship. Thus, who can really say which twin is accelerating? But this can be determined experimentally. Give each twin an accelerometer before launch. Only one will record an acceleration and that will be the one on B’s ship. You can’t have it both ways here; B will be younger when the ship lands.

7 Module 12Paradoxes7 Faster Than Light Many paradoxes put forth involve scenarios where it looks like something travels faster than light. We have seen that relativity theory predicts that superluminal speeds are not possible, hence the paradox. In all of these cases one can show that no object is traveling faster than light and that no information is traveling faster than light. We’ll take a look at a few scenarios. Scenario 1 Let’s take a rod of length r and rotate it at an angular speed of  as shown in Figure The piece of the rod at its end has a tangential velocity of r v  Fig (12.1) We can make this speed be faster than that of light if (12.2) To do this at a reasonable angular speed would take a very long rod but the math says it can be done. But the physics does not! The rod is made of matter. As the chunk of rod at the end speeds up, its inertial mass will increase and its angular acceleration will decrease. The rod will warp and bend as it spins and will definitely break. (Even if we could find a miracle material that would not break, no section of the rod can ever surpass c.)

8 Module 12Paradoxes8 Scenario 2 The previous paradox was simple to resolve. Matter can not reach a speed of c, period. But what if we replace the rod with a beam of light? Consider Figure Here we have a laser rotating at angular speed . The beam travels a distance of r whereupon it hits a curved wall. The tangential speed of the spot on the wall is r  Fig A B (12.3) We can make this speed be faster than that of light if the beam length satisfies (12.4) So, for instance, if the laser rotates at 5000 rpm,  = 523 rad/s, and r must be greater than 574 km. Again, that’s a long distance, but it should work, right? The laser spot should move from A to B at a speed faster than c, right? It seems so. Since this is a beam of light and not a rod, we don’t have any inertial mass to prevent this from being so. Is this paradox irresolvable? Is the theory of special relativity threatened? No. Look closely at what is happening. The photons travel from the laser to point A at c. They travel from the laser to point B at c. They are not moving from A to B along the wall. It’s okay if this spot moves faster than c; we’re not violating relativity here. But doesn’t that mean we could somehow send a signal from A to B at a speed faster than light? Nope. Remember that the light is coming from the laser and it is confined to speed c. Any information must be encoded in the laser light and it simply cannot travel faster than light. (You’re welcome to try to think of a way that you could use this set-up to send information from A to B with the laser spot. If you come up with a way that works, contact the Nobel Prize headquarters.)

9 Module 12Paradoxes9 Scenario 3 Let’s try a rod again to send information faster than light. This time, though, we’ll keep the rod’s speed under c. Suppose we want to transmit information along the x-axis by means of a long rod as shown in Figure The rod is tilted at angle  with respect to the x-axis. The intersection point of the rod and the axis is labeled P. Notice that this intersection point will move to the right if the rod is moved downward. Thus, we can use the intersection point to send a signal along the x- axis. Point P begins at the transmitter. All we have to do is to set our detector farther out along on the axis and, at some later time, point P will hit the detector. Let’s now figure out how fast we can move point P. Receiver x y t = 0 t > 0 P yy uyuy  Transmitter Fig xx Suppose the rod moves downward at speed u y. During a time of  t point P moves to the right a distance of  x. Thus, its velocity along the x-axis is (12.5)

10 Module 12Paradoxes10 During this same time the rod moves downward a distance of (12.6) We can relate the ratio of the distances to angle  with (12.7) Combining the previous three equations gives the speed of the intersection point P as (12.8) Interesting…the rod can move less than the speed of light but we can easily obtain v > c by choosing an appropriate angle. For instance, if u y = 0.6c and  = 25 , then v = 1.3c. It seems that we can transmit our signal faster than light can travel. Hold on. Alas, we cannot. While it is true that this mathematical intersection point can travel faster than c, there is no way to transmit information with it. For example, suppose you want to signal the detector when a light goes on at the transmitter. When the light turns on, the rod begins to move. It must accelerate to reach its speed and this takes time. Point P will not beat the light signal to the receiver. Good try, but this paradox has been resolved. You may want to try and replacing the rod with a laser beam as we did in Scenario 2. The beam would form the angle  with the x-axis and a downward movement of the laser would result in the laser spot on the x-axis moving to the right faster than light. But just as in Scenario 2, there is no way to send information with this laser spot. In fact, all we have done here is to change from the circular geometry of Scenarios 1 and 2 to a linear geometry. Scenarios can be concocted where points of intersection of light with walls or axes can move faster than c but there is no way to send information with these intersection points.

11 Module 12Paradoxes11 The conclusion that information cannot travel faster than light is commonly accepted. However, quantum theory does predict what may be a violation of this rule. When certain particles that have the same possible quantum states interact and then separate, they seem to be able to transmit information about their states instantaneously over long distances. These particles exhibit what is called quantum entanglement. This principle is key to the design of proposed quantum computers where information about many different quantum states is transmitted extremely fast. Is information really traveling faster than light? Some would say yes. Others would say no; rather space-time is being warped by quantum effects and that distances between these particles are being shrunk so that it just appears that information is traveling faster than c. In the end, it may not matter which view is correct. What we do know is that without quantum effects, information has a speed limit and that limit is c.


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