Chapter 7: Acceleration and Gravity

Transcription

1 Chapter 7: Acceleration and Gravity 7.1 The Principle o Equivalence We saw in the special theory o relativity that the laws o physics must be the same in all inertial reerence systems. But what is so special about an inertial reerence system? The inertial reerence rames are, in a sense, playing the same role as Newton s absolute space. That is, absolute space has been abolished only to replace it by absolute inertial reerence rames. Shouldn t the laws o physics be the same in all coordinate systems, whether inertial or noninertial? The inertial rame should not be such a privileged rame. But clearly, accelerations can be easily detected, whereas constant velocities cannot. How can this very obvious dierence be reconciled? That is, we must show that even all accelerated motions are relative. How can this be done? Let us consider the very simple case o a mass m on the loor o a rocket ship that is at rest in a uniorm gravitational ield on the surace o the earth, as depicted in igure 7.1(a). The orce acting on the mass is its weight w, which we write as F = w = mg (7.1) Figure 7.1 An accelerated rame o reerence is equivalent to an inertial rame o reerence plus gravity. Let us now consider the case o the same rocket ship in interstellar space ar removed rom all gravitational ields. Let the rocket ship now accelerate upward, as in igure 7.1(b), with an acceleration a that is numerically equal to the acceleration due to gravity g, that is, a = g = 9.80 m/s. The mass m that is sitting on the loor o the rocket now experiences the orce, given by Newton s second law as F = ma = mg = w (7.) That is, the mass m sitting on the loor o the accelerated rocket experiences the same orce as the mass m sitting on the loor o the rocket ship when it is at rest in the uniorm gravitational ield o the earth. Thereore, there seems to be some relation between accelerations and gravity. Let us experiment a little urther in the rocket ship at rest by holding a book out in ront o us and then dropping it, as in igure 7.1(c). The book alls to the loor and i we measured the 7-1

2 acceleration we would, o course, ind it to be the acceleration due to gravity, g = 9.80 m/s. Now let us take the same book in the accelerated rocket ship and again drop it, as in igure 7.1(d). An inertial observer outside the rocket would see the book stay in one place but would see the loor accelerating upward toward the book at the rate o a = 9.80 m/s. The astronaut in the accelerated rocket ship sees the book all to the loor with the acceleration o 9.80 m/s just as the astronaut at rest on the earth observed. The astronaut in the rocket at rest on the earth now throws the book across the room o the rocket ship. He observes that the book ollows the amiliar parabolic trajectory o the projectile that we studied in our College Physics course and that is again shown in igure 7.1(e). Similarly, the astronaut in the accelerated rocket also throws the book across the room. An outside inertial observer would observe the book moving across the room in a straight line and would also see the loor accelerating upward toward the book. The accelerated astronaut would simply see the book ollowing the amiliar parabolic trajectory it ollowed on earth, igure 7.1(). Hence, the same results are obtained in the accelerated rocket ship as are ound in the rocket ship at rest in the gravitational ield o the earth. Thus, the eects o gravity can be either created or eliminated by the proper choice o coordinate systems. Our experimental considerations suggest that the accelerated rame o reerence is equivalent to an inertial rame o reerence in which gravity is present. Einstein, thus ound a way to make accelerations relative. He stated his results in what he called the equivalence principle. Calling the inertial system containing gravity the K system and the accelerated rame o reerence the K system, Einstein said, we assume that we may just as well regard the system K as being in a space ree rom gravitational ield i we then regard K as uniormly accelerated. This assumption o exact physical equivalence makes it impossible or us to speak o the absolute acceleration o the system, just as the usual (special) theory o relativity orbids us to talk o the absolute velocity o a system But this view o ours will not have any deeper signiicance unless the systems K and K are equivalent with respect to all physical processes, that is, unless the laws o nature with respect to K are in entire agreement with those with respect to K 1 Einstein s principle o equivalence is stated as: on a local scale the physical eects o a gravitational ield are indistinguishable rom the physical eects o an accelerated coordinate system. The equivalence o the gravitational ield and acceleration ields also accounts or the observation that all objects, regardless o their size, all at the same rate in a gravitational ield. I we write mg or the mass that experiences the gravitational orce in equation 7.1 and igure 7.1(a), then F = w = mgg And i we write mi or the inertial mass that resists the motion o the rocket in igure 7.1(b) and equation 7., then F = mia = mig Since we have already seen that the two orces are equal, by the equivalence principle, it ollows that mg = mi (7.3) That is, the gravitational mass is in act equal to the inertial mass. Thus, the equivalence principle implies the equality o inertial and gravitational mass and this is the reason why all objects o any size all at the same rate in a gravitational ield. As a inal example o the equivalence o a gravitational ield and an acceleration let us consider an observer in a closed room, such as a nonrotating space station in interstellar space, ar removed rom all gravitating matter. This space station is truly an inertial coordinate system. Let 1 On the Inluence o Gravitation on the Propagation o Light, rom A. Einstein, Annalen der Physik 35, 1911, in The Principle o Relativity, Dover Publishing Co. 7-

3 the observer place a book in ront o him and then release it, as shown in igure 7.(a). Since there are no orces present, not even gravity, the book stays suspended in space, at rest, exactly where the observer placed it. I the observer then took the book and threw it across the room, he would observe the book moving in a straight line at constant velocity, as shown in igure 7.(b). Let us now consider an elevator on earth where the supporting cables have broken and the elevator goes into ree-all. An observer inside the reely alling elevator places a book in ront o himsel and then releases it. The book appears to that reely alling observer to be at rest exactly where the observer placed it, igure 7.(c). (O course, an observer outside the reely alling elevator would observe both the man and the book in ree-all but with no relative motion with respect to each other.) I the reely alling observer now takes the book and throws it across the elevator room he would observe that the book travels in a straight line at constant velocity, igure 7.(d). Because an inertial rame is deined by Newton s irst law as a rame in which a body at rest, remains at rest, and a body in motion at some constant velocity continues in motion at that same constant velocity, we must conclude rom the illustration o igure 7. that the reely alling rame o reerence acts exactly as an inertial coordinate system to anyone inside o it. Thus, the acceleration due to gravity has been transormed away by accelerating the coordinate system by the same amount as the acceleration due to gravity. I the elevator were completely closed, the observer could not tell whether he was in a reely alling elevator or in a space station in interstellar space. The equivalence principle allows us to treat an accelerated rame o reerence as equivalent to an inertial rame o reerence with gravity present, igure 7.1, or to consider an inertial rame as equivalent to an accelerated rame in which gravity is absent, igure 7.. By placing all rames o reerence on the same ooting, Einstein was then able to postulate the general theory o relativity, namely, the laws o physics are the same in all rames o reerence. From his general theory o relativity, Einstein was quick to see its relation to gravitation when he Figure 7. A reely alling rame o reerence is locally the same as an inertial rame o reerence. said, It will be seen rom these relections that in pursuing the General Theory o Relativity we shall be led to a theory o gravitation, since we are able to produce a gravitational ield merely by changing the system o coordinates. It will also be obvious that the principle o the constancy o the velocity o light in vacuo must be modiied. Although the general theory was developed by Einstein to cover the cases o accelerated reerence rames, it soon became obvious to him that the general theory had something quite signiicant to say about gravitation. Since the world line o an accelerated particle in spacetime is curved, then by the principle o equivalence, a particle moving under the eect o gravity must also have a curved world line in spacetime. Hence, the mass that is responsible or causing the gravitational ield, must warp spacetime to make the world lines o spacetime curved. This is The Foundation o the General Theory o Relativity rom A. Einstein, Annalen der Physik 49, 1916 in The Principle o Relativity, Dover Publishing Co. 7-3

4 sometimes expressed as, matter warps spacetime and spacetime tells matter how to move. We will go into the details o curved spacetime in much greater detail in the next chapter. For now let us look at the problem rom a purely physical point o view. 7. The Gravitational Red Shit Although the General Theory o Relativity was developed by Einstein to cover the cases o accelerated reerence rames, it soon became obvious to him that the general theory had something quite signiicant to say about gravitation. Let us consider the two clocks A and B located at the top and bottom o the rocket, respectively, in igure 7.3(a). The rocket is in interstellar space where we assume that Figure 7.3 A clock in a gravitational ield. all gravitational ields, i any, are eectively zero. The rocket is accelerating uniormly, as shown. Located in this interstellar space is a clock C, which is at rest. At the instant that the top o the rocket accelerates past clock C, clock A passes clock C at the speed va. Clock A, the moving clock, when observed rom clock C, the stationary clock, shows an elapsed time ta, given in chapter 1 by the time dilation equation 1.64 as ta tc = (7.4) 1 v / c A And since 1 va / c is less than 1, then tc > ta, and the moving clock A runs slow compared to the stationary clock C. A ew moments later, clock B passes clock C at the speed vb, as in igure 7.3(b). The speed vb is greater than va because o the acceleration o the rocket. Let us read the same time interval tc on clock C when clock B passes as we did or clock A so the two clocks can be compared. The dierence in the time interval between the two clocks, B and C, is again given by the time dilation equation 1.64 as tb tc = (7.5) 1 v / c Because the time interval tc was set up to be the same in both equations 7.4 and 7.5, the two equations can be equated to give a relation between clocks A and B. Thus, B 7-4

5 Rearranging terms, we get t t ta tb = 1 v / c 1 v / c A B t A B = ( 1 va / c ) ( 1 B / ) A tb v c 1/ 1/ ( 1 va / c ) ( 1 vb / c ) = 1/ 1/ (7.6) But the two terms on the right-hand side o equation 7.6 can be expanded by the binomial theorem, equation 1.33, as (1 x) n = 1 nx n(n 1)x n(n 1)(n )x 3 (1.33)! 3! This is a valid series expansion or (1 x) n as long as x is less than 1. In this particular case, x = v /c which is much less than 1. In act, since x is very small, it is possible to simpliy the binomial theorem to (1 x) n = 1 nx (1.34) Hence, 1/ 1 va va ( 1 va / c ) = 1 1 = c c and 1/ 1 vb vb ( 1 vb / c ) = 1 1 = c c where again the assumption is made that v is small enough compared to c, to allow us to neglect the terms x and higher in the expansion. Thus, equation 7.6 becomes t A v A v B = 1 tb c v v 1 vv = 1 c c 4 c B A B A 4 The last term is set equal to zero on the same assumption that the speeds v are much less than c. Finally, rearranging terms, t A vb v A 1 = 1 (7.7) tb c But by Einstein s principle o equivalence, we can equally well say that the rocket is at rest in the gravitational ield o the earth, whereas the clock C is accelerating toward the earth in reeall. When the clock C passes clock A it has the instantaneous velocity va, igure 7.3(c), and when it passes clock B it has the instantaneous velocity vb, igure 7.3(b). We can obtain the velocities va and vb by the law o conservation o energy, that is, 1 mv PE = E0 = Constant = Total energy (7.8) 7-5

6 The total energy per unit mass, ound by dividing equation 7.8 by m, is v PE = E0 m m The conservation o energy per unit mass when clock C is next to clock A, obtained with the aid o igure 7.4, is va mgha = E0 m m or Similarly, when the clock C is next to clock B, the conservation o energy per unit mass becomes va gha = E0 (7.9) m vb ghb = E0 m (7.10) Subtracting equation 7.9 rom equation 7.10, gives Hence, vb ghb va gha = E0 E0 = 0 m m vb va = gha ghb = gh (7.11) Figure 7.4 Freely alling clock C. where h is the distance between A and B, and gh is the gravitational potential energy per unit mass, which is sometimes called the gravitational potential. Substituting equation 7.11 back into equation 7.7, gives ta = 1 gh (7.1) tb c For a clearer interpretation o equation 7.1, let us change the notation slightly. Because clock B is closer to the surace o the earth where there is a stronger gravitational ield than there is at a height h above the surace where the gravitational ield is weaker, we will let and tb = tg ta = t where tg is the elapsed time on a clock in a strong gravitational ield and t is the elapsed time on a clock in a weaker gravitational ield. I we are ar enough away rom the gravitational mass, we can say that t is the elapsed time in a gravitational-ield-ree space. With this new notation equation 7.1 becomes t = 1 gh (7.13) tg c or gh t = tg (7.14) 7-6

7 Since (1 gh/c ) > 0, the elapsed time on the clock in the gravitational-ield-ree space t is greater than the elapsed time on a clock in a gravitational ield tg. Thus, the time elapsed on a clock in a gravitational ield is less than the time elapsed on a clock in a gravity-ree space. Hence, a clock in a gravitational ield runs slower than a clock in a ield-ree space. Thus, equation 7.14 gives the slowing down o a clock in a gravitational ield. Compare this to equation.4, the time dilation ormula, which shows the slowing down o a moving clock. t = t 0 1 v / c (.4) Equation.4 says that a clock on earth reads a longer time interval t than the clock at rest in the moving rocket ship t0. Or as is sometimes said, moving clocks slow down. Thus, i the moving clock slows down, a smaller time duration is indicated on the moving clock than on a stationary clock. Example 7.1 A clock in a gravitational ield. A clock in a gravitational ield on the earth ticks o a time interval o 10.0 hr. What time would elapse at a height o 1,000,000 km above the surace o earth. Solution The time elapsed in the gravitational ree area is ound rom equation 7.14 as gh t = tg 9 (9.80 m/s )( m t = 10.0 hr 8 ( m/s ) t = hr The time in the gravity ree space is greater than the time elapsed in the gravitational ield area. But as you can see, the dierence is still very small. To go to this Interactive Example click on this sentence. We can ind a urther eect o the slowing down o a clock in a gravitational ield by placing an excited atom in a gravitational ield, and then observing a spectral line rom that atom ar away rom the gravitational ield. The speed o the light rom that spectral line is, o course, given by c = λν = λ T (7.15) where λ is the wavelength o the spectral line, ν is its requency, and T is the period or time interval associated with that requency. Hence, i the time interval t = T changes, then the wavelength o that light must also change. Solving or the period or time interval rom equation 7.15, we get T = λ c (7.16) 7-7

8 Substituting T rom 7.16 or t in equation 7.13, we get gh T = Tg λ λg gh = 1 c c λ gh = λ g (7.17) (7.18) where λg is the wavelength o the emitted spectral line in the gravitational ield and λ is the wavelength o the observed spectral line in gravity-ree space, or at least arther rom where the atom is located in the gravitational ield. Because the term (1 gh/c ) is a positive number, it ollows that λ > λg (7.19) That is, the wavelength observed in the gravity-ree space is greater than the wavelength emitted rom the atom in the gravitational ield. Recall rom College Physics that the visible portion o the electromagnetic spectrum runs rom violet light at around nm to red light at 70.0 nm. Thus, red light is associated with longer wavelengths. Hence, since λ > λg, the wavelength o the spectral line increases toward the red end o the spectrum, and the entire process o the slowing down o clocks in a gravitational ield is reerred to as the gravitational red shit. A similar analysis in terms o requency can be obtained rom equation 7.14 as, and since equation 7.17 becomes Solving or ν gives Using the binomial theorem ν T gh = Tg (7.14) 1 T = (7.0) ν 1 1 gh = ν ν 1 ν = = g 1 g gh νg 1 gh c 1 c 1 gh gh 1 = (7.1) (7.) (7.3) Substituting equation 7.3 into equation 7.3 gives ν gh = ν g (7.4) 7-8

9 Where now the requency observed in the gravitational-ree space is less than the requency emitted in the gravitational ield because the term 1 gh is less than one. The change in requency per unit requency emitted, ound rom equation 7.4, is ν νg = gh c νg νg ν = gh νg c ν gh = ν c g (7.5) The gravitational red shit was conirmed on the earth by an experiment by R. V. Pound and G. A. Rebka at Harvard University in 1959 using a technique called the Mossbauer eect. Gamma rays were emitted rom radioactive cobalt in the basement o the Jeerson Physical Laboratory at Harvard University. These gamma rays traveled.5 m, through holes in the loors, up to the top loor. The dierence between the emitted and absorbed requency o the gamma ray was ound to agree with equation 7.5. Example 7. Gravitational requency shit. Find the change in requency per unit requency or a γ-ray traveling rom the basement, where there is a large gravitational ield, to the roo o the building, which is.5 m higher, where the gravitational ield is weaker. Solution The change in requency per unit requency, ound rom equation 7.5, is = ν gh = ν c g ( 9.80 m/s )(.5 m) 8 ( m/s) = As you can see, the change in requency per unit requency is very small. To go to this Interactive Example click on this sentence. The experiment was repeated by Pound and J. L. Snider in 1965, with another conirmation. Since then the experiment has been repeated many times, giving an accuracy to the gravitational red shit to within 1%. Further conirmation o the gravitational red shit came rom an experiment by Joseph Haele and Richard Keating. Carrying our atomic clocks, previously synchronized with a reerence clock in Washington, D.C., Haele and Keating lew around the world in On their return they compared their airborne clocks to the clock on the ground and ound the time dierences associated with the time dilation eect and the gravitational eect exactly as predicted. Further tests with atomic clocks in airplanes and rockets have added to the conirmation o the gravitational red shit. 7-9

10 7.3 The Gravitational Red Shit by the Theory o Quanta The eect o acceleration and gravitation on the time recorded on a clock was derived in the last section by observing how a clock slows down in a gravitational ield. The eect is o course known as the gravitational red shit. A remarkably simple derivation o this red shit can also be obtained by treating light as a particle, a photon, in a gravitational ield. Let an atom at the surace o the earth emit a photon o light o requency νg. As you recall rom your course in Modern Physics, a photon o light has the energy E = hν, where h is Planck s constant = ( J s) and ν is the requency o the light associated with the photon. Let this particular photon have the energy Eg = hνg (7.6) The subscript g is to remind us that this is a photon in the gravitational ield. Let us assume that the light source was pointing upward so that the photon travels upward against the gravitational ield o the earth until it arrives at a height y above the surace, as shown in igure 7.5. (We have used y or the height instead o h, as used previously, so as not to conuse the height with Planck s constant h.) As the photon rises it must do work against the gravitational ield. When the photon arrives at the height y, its energy E must be diminished by the work it had to do to get there. Thus Because the gravitational ield is weaker at the height y than at the surace, the subscript has been used on E to indicate that this is the energy in the weaker ield or even in a ield-ree space. The work done by the photon in climbing to the height y is the same as the potential energy o the photon at the height y. Thereore, W = PE = mgy (7.8) Substituting equation 7.8 and the values o the energies back into equation 7.7, gives E = Eg W (7.7) Figure 7.5 A photon in a gravitational ield. hν = hνg mgy (7.9) But the mass o the emitted photon is m = Eg = hνg c c Placing this value o the mass back into equation 7.9, gives or hν = hνg hνg c gy ν gy = ν g (7.30) Equation 7.30 says that the requency o a photon associated with a spectral line that is observed away rom the gravitational ield is less than the requency o the spectral line emitted by the atom 7-10

11 in the gravitational ield itsel. Since the requency ν is related to the wavelength λ by c = λν, the observed wavelength in the ield-ree space λ is longer than the wavelength emitted by the atom in the gravitational ield λg. Thereore, the observed wavelength is shited toward the red end o the spectrum. Note the equation 7.30 is the same as equation 7.4 that we derived in the last section rom a dierent point o view. The slowing down o a clock in a gravitational ield ollows directly rom equation (7.30) by noting that the requency ν is related to the period o time T by ν = 1/T. Hence 1 1 gy = 1 T T g But by the binomial theorem, Thus, T = T Tg 1 gy/c gy = Tg 1 1 gy gy 1 = c T gy = Tg (7.31) Equation 7.31 is identical to equation Finally calling the period o time T an elapsed time, t, we have gy t = tg (7.3) which is identical to equation 7.14, which shows the slowing down o a clock in a gravitational ield. We can also show that the slowing o a clock in a gravitational ield is identical to the slowing down o a clock by the Lorentz transormation. Consider the term gy/c in equations 7.31 or 7.3. From the law o conservation o energy we have and hence Replacing this in equation 7.3 gives Using the binomial theorem in reverse with x = v /c and n = 1/, we get PE = KE 1 mgy = mv gy = v t = tg v c 1 1 nx = (1 x) n (7.33) 1/ 1 v 1 v v 1 = 1 1 c = = c 1 v / c (7.34) Replacing this value in equation 7.3 we get 7-11

12 t = t g 1 v / c (7.35) But the time elapsed on the clock at rest in the gravitational ield, tg, is the same as the time elapsed to on a clock at rest in the moving coordinate system and t the time elapsed on the clock in the gravity ree space is the same as the time elapsed t on the moving coordinate system. Hence equation 7.35 becomes t = t0 (7.36) 1 v / c But this is exactly the Lorentz transormation equation or time dilation. This is very signiicant here, because it is derived on the basis o the gravitational red shit by the theory o quanta and not on inertial motion, yet the results are identically the same. Beore leaving this section we should point out that or the general case, the value or the acceleration due to gravity g that we have been using in all these equations is not a constant but varies with the mass o the main object and the distance that the second object is located with respect to the main object. Recall that the acceleration due to gravity comes rom Newton s second law F = ma = mg and Newton s law o universal gravitation Gmem F = r where me is the mass o the earth, and r is the distance rom the center o the earth to the location o the second mass. Equating the two values o F gives Gmem mg = r Thereore the acceleration due to gravity is determined rom Gm g = e (7.37) r So or any particular problem dealing with time dilation and gravitational acceleration, you can determine g by equation 7.37 i necessary. 7.4 An Accelerated Clock and the Lorentz Transormation Equations An extremely interesting consequence o the gravitational red shit can be ormulated by invoking Einstein s principle o equivalence discussed in at the beginning o this chapter. Calling the inertial system containing gravity the K system and the accelerated rame o reerence the K' system, Einstein stated, we assume that we may just as well regard the system K as being in a space ree rom a gravitational ield i we then regard K as uniormly accelerated. Einstein s principle o equivalence was thus stated as: on a local scale the physical eects o a gravitational ield are indistinguishable rom the physical eects o an accelerated coordinate system. Hence the systems K and K' are equivalent with respect to all physical processes, that is, the laws o nature with respect to K are in entire agreement with those with respect to K'. Einstein then postulated his theory o general relativity, as: The laws o physics are the same in all rames o reerence. 7-1

13 Since a clock slows down in a gravitational ield, equation 7.3, using the equivalence principle, an accelerated clock should also slow down. Replacing the acceleration due to gravity g by the acceleration o the clock a, equation 7.3 becomes ay t = ta (7.38) Note that the subscript g on tg in equation 7.3 has now been replaced by the subscript a, giving ta, to indicate that this is the time elapsed on the accelerated clock. Notice rom equation 7.38 that t > ta indicating that time slows down on the accelerated clock. That is, an accelerated clock runs more slowly than a clock at rest. In chapter 1 we saw, using the Lorentz transormation equations, that a clock at rest in a moving coordinate system slows down, and called the result the Lorentz time dilation. However, nothing was said at that time to show how the coordinate system attained its velocity. Except or zero velocity, all bodies or reerence systems must be accelerated to attain a velocity. Thus, there should be a relation between the Lorentz time dilation and the slowing down o an accelerated clock. Let us change our notation slightly and call t the time t in a stationary coordinate system and ta the time interval on a clock that is at rest in a coordinate system that is accelerating to the velocity v. Assuming that the acceleration is constant, we can use the kinematic equation v = v0 ay Further assuming that the initial velocity v0 is equal to zero and solving or the quantity ay we obtain ay = v (7.39) Substituting equation 7.39 into equation 7.38, yields Using the binomial theorem in reverse with x = v /c and n = 1/, we get v t = ta 1 nx = (1 x) n (7.40) Equation 7.40 becomes 1/ v 1 v v 1 = 1 1 c = = c 1 v / c t = t a 1 v / c (7.41) (7.4) But this is exactly the time dilation ormula, equation.4, ound by the Lorentz transormation. Thus the Lorentz time dilation is a special case o the slowing down o an accelerated clock. This is a very important result. Thereore, it is more reasonable to take the slowing down o a clock in a gravitational ield, and thus by the principle o equivalence, the slowing down o an accelerated clock as the more basic physical principle. The Lorentz transormation or time dilation can then be derived as a special case o a clock that is accelerated rom rest to the velocity v. 7-13

14 Example 7.3 Comparison o time dilation by Lorentz s time dilation equation and the slowing down o an accelerated clock. (a) A rocket ship is moving at a speed o 1610 km/s = m/s. A clock on the rocket ship ticks o a time interval o 1.00 hr. Using the Lorentz time dilation equation, ind the time elapsed on the clock on earth. (b) To arrive at the speed o 1610 km/s, the rocket ship accelerates at 9.80 m/s. How ar must the rocket travel to arrive at this velocity. A clock on the rocket ticks o a time interval o 1.00 hr. Find the time recorded on the earth clock using the equation or time dilation o an accelerated clock. Compare the results o the Lorentz time dilation equation and the results o time dilation or an accelerated clock. Solution a. The time elapsed on the clock on earth is ound by the Lorentz time dilation ormula equation.4 as t0 t = 1 v / c 1.00 hr = ( m/s) /( m/s) = hr As we would expect with time dilation, this relatively large speed o 1610 km/s = 3,600,000 mph, the dierence in the clocks is very small. That is, a dierence o hr = seconds between the moving clock and the stationary clock in a period o time o 1.00 hours. b. To ind the dierence in time using the concept o an accelerated clock we start with equation 7.38 ay t = ta Now in order to compare the accelerated clock with the Lorentz clock we have to know the acceleration o the rocket that gave it its speed, and the distance the rocket moved during this acceleration so it could attain its speed. We can obtain this simply rom the kinematic equation v = v0 ay (7.43) We assume the rocket starts rom rest so v0 = 0, and equation 7.43 becomes v = ay (7.44) Since the velocity v o the Lorentz rocket must be the same as the velocity v o the accelerated rocket we solve equation 7.44 or ay as ay = v / (7.45) Now we have to pick a reasonable value or the acceleration a, and then we can solve or the distance the rocket has to move to acquire the velocity v. That is, y = v a (7.46) 7-14

15 We will assume a value o a o 9.80 m/s. This acceleration is the same as the acceleration due to gravity and an astronaut could easily take it or the long time necessary or the rocket to acquire the velocity desired. We can change the acceleration value later to some other value i we wish. The distance y that the rocket must move to attain this velocity is ound rom equation 7.46 as ( m/s) v y = = a 9.8 m/s y = m The value o ay that we need or equation 7.46 is ound as ay = (9.80 m/s )( m) = m s We can now determine the time t rom equation 7.38 as ay t = ta 1 ( m s t = 1.00 hr 8 ( m/s ) t = hr The dierence in the value o t ound with the Lorentz time dilation equation and the time t ound with the accelerated clock time dilation equation is Lorentz time dilation = E00 hr Accelerated time dilation = E00 hr And is essentially zero. Hence the accelerated time dilation is the same as the Lorentz time dilation. Thereore we can assume that a Lorentz clock is equivalent to a gravitational clock and an accelerated clock. To go to this Interactive Example click on this sentence. Just as the slowing down o a clock in a gravitational ield can be attributed to the warping o spacetime by the mass, it is reasonable to assume that the slowing down o the accelerated clock can also be thought o as the warping o spacetime by the increased mass, due to the increase in the velocity o the accelerating mass. The Lorentz length contraction can also be derived rom this model by the ollowing considerations. Consider the emission o a light wave in a gravitational ield. We will designate the wavelength o the emitted light by λg, and the period o the light by Tg. The velocity o the light emitted in the gravitational ield is given by cg = λg (7.47) We will designate the velocity o light in a region ar removed rom the gravitational ield as c or the velocity in a ield-ree region. The velocity o light in the ield-ree region is given by Tg c = λ T (7.48) 7-15

16 where λ is the wavelength o light, and T is the period o the light as observed in the ield-ree region. I the gravitating mass is not too large, then we can make the reasonable assumption that the velocity o light is the same in the gravitational ield region and the ield-ree region, that is, cg = c. We can then equate equation 7.47 to equation 7.48 to obtain λg = λ Tg T Solving or the wavelength o light in the ield-ree region, we get λ = T Tg λg Substituting the value o T rom equation 7.31 into this we get Tg gy λ = 1 λ g T g gy λ = λ g (7.49) Equation 7.49 gives the wavelength o light λ in the gravitational-ield-ree region. By the principle o equivalence, the wavelength o light emitted rom an accelerated observer, accelerating with the constant acceleration a through a distance y is obtained rom equation 7.49 as ay λ = 0 1 λ a (7.50) where λ0 is the wavelength o light that is observed in the region that is not accelerating, that is, the wavelength observed by an observer who is at rest. This result can be related to the velocity v that the accelerated observer attained during the constant acceleration by the kinematic equation v = v0 ay Further assuming that the initial velocity v0 is equal to zero and solving or the quantity ay we obtain ay = v Substituting this result into equation 7.50 we obtain v λ0 = λ a c (7.51) Using the binomial theorem in reverse as in equation 7.41, equation 7.51 becomes v 1 = c 1 v / c λ = 0 λ a 1 v / c 7-16

17 Solving or λa we get λa 0 1 v / c = λ (7.5) But λ is a length, in particular λa is a length that is observed by the observer who has accelerated rom 0 up to the velocity v and is usually reerred to as L, whereas λ0 is a length that is observed by an observer who is at rest relative to the measurement and is usually reerred to as L0. Hence, we can write equation 7.5 as L L0 1 v / c = (7.53) But equation 7.53 is the Lorentz contraction o special relativity. Hence, the Lorentz contraction is a special case o contraction o a length in a gravitational ield, and by the principle o equivalence, a rod L0 that is accelerated to the velocity v is contracted to the length L. (That is, i a rod o length L0 is at rest in a stationary spaceship, and the spaceship accelerates up to the velocity v, then the observer on the earth would observe the contracted length L.) Hence, the acceleration o the rod is the basic physical principle underlying the length contraction. It is physically the acceleration that gives the object its velocity that is used in the original Lorentz equations. Example 7.4 Comparison o length contraction by Lorentz-Fitzgerald Contraction and the length contraction equation caused by an acceleration. a. Using the Lorentz-Fitzgerald contraction equation. What is the length o a meter stick at rest on earth, when it is observed by an astronaut moving at a speed o v = m/s = 3,600,000 miles/hr. b. Length contraction or an accelerated rod. To arrive at the speed o m/s, the rocket ship accelerates at 9.80 m/s. How ar must the rocket travel to arrive at this velocity. Find the length contraction o the meter stick caused by this acceleration. c. Compare the results o the Lorentz-Fitzgerald length contraction and the accelerated length contraction. Solution a. The length o the meter stick is measured in the rame where it is at rest, and is called its proper length and is denoted by L0. The length o the meter stick as measured by the astronaut in the moving rame is denoted by L. The Lorentz-Fitzgerald contraction o a meter stick or a speed o m/s is ound by the Lorentz equation as L = L0 1 v c ( m/s) = (1.00 m) 1 8 ( m/s) = (1.00 m) = m 6 As we would expect with length contraction, this relatively large speed o m/s, gives a very small dierence in the inal length. b. A meter stick is at rest on the earth. A rocket ship now accelerates away rom the earth. What is the length contraction caused by the acceleration o the rocket ship away rom the earth where the 7-17

18 rod is at rest? To ind the dierence in the length using the concept o an acceleration we start with equation 7.50 as ay λ0 = λ a (7.50) but change the length designation rom λ to L, so that equation 7.50 can be rewritten as L ay = L 0 1 a (7.54) Where La is the length that is observed by the accelerated observer who has accelerated rom 0 up to the velocity v, whereas L0 is the length that is observed by an observer who is at rest on the earth next to the rod. For this problem the rod is at rest on the earth (L0), and the rod observed by the astronaut is La. Hence we have to rearrange equation 7.54 into the orm L a L 0 = ay 1 c (7.55) Now in order to solve this equation we have to know the acceleration a o the rocket that gave it its speed, and the distance y the rocket moved during this acceleration so it could attain its speed. We can obtain this simply rom the kinematic equation as we did in Example 7.3 as v = v0 ay (7.43) We assume the rocket starts rom rest so v0 = 0, and equation 7.43 becomes v = ay (7.44) Since the velocity v o the Lorentz rocket must be the same as the velocity v o the accelerated rocket we solve equation 7.44 or ay as ay = v / (7.45) Now we have to pick a reasonable value or the acceleration a, and then we can solve or the distance the rocket has to move to acquire the velocity v. That is, y = v a (7.46) We will assume a value o a o 9.80 m/s. This acceleration is the same as the acceleration due to gravity and an astronaut could easily take it or the long time necessary or the rocket to acquire the velocity desired. We can change the acceleration value later to some other value i we wish. The distance y that the rocket must move to attain this velocity is ound rom equation 7.46 as ( m/s) v y = = a 9.8 m/s y = m The value o ay that we need or equation 7.46 is then ound as ay = (9.80 m/s )( m) = m s 7-18

19 We can now determine the length Lo rom equation 7.55 as L a = L L 0 a = ay m ( m s 1 La = m 1 8 ( m/s) c. Hence the results obtained by the Lorentz-Fitzgerald length contraction and the accelerated length contraction are the same. That is, the length observed by the accelerating astronaut is the same length that we obtained by the Lorentz- Fitzgerald contraction. The important thing here to observe is that we get the same results by using the Lorentz- Fitzgerald contraction as we do by considering a contraction as the result o gravity or an acceleration. To go to this Interactive Example click on this sentence. Thus, both the time dilation and length contraction o special relativity should be attributed to the warping o spacetime by the accelerating mass. The warping o spacetime by the accelerating mass can be likened to the Doppler eect or sound. Recall that i a source o a sound wave is stationary, the sound wave propagates outward in concentric circles. When the sound source is moving, the waves are no longer circular but tend to bunch up in advance o the moving source. Since light does not require a medium or propagation, the Doppler eect or light is very much dierent. However, we can speculate that the warping o spacetime by the accelerating mass is comparable to the bunching up o sound waves in air. In act, i we return to equation 7.30, or the gravitational red shit, and again, using the principle o equivalence, let g = a, and dropping the subscript, this becomes ν ay = ν a (7.56) Using the kinematic equation or constant acceleration, ay = v /. Hence equation 7.56 becomes Again using the binomial theorem Equation 7.56 becomes v ν = ν a 1 c v 1 v / c = c ν ν a 1 v / c (7.57) = (7.58) Equation 7.58 is called the transverse Doppler eect. It is a strictly relativistic result and has no counterpart in classical physics. The requency νa is the requency o light emitted by a light source that is at rest in a coordinate system that is accelerating past a stationary observer, whereas ν is 7-19

20 the requency o light observed by the stationary observer. Notice that the transverse Doppler eect comes directly rom the gravitational red shit by using the equivalence principle. Thus the transverse Doppler eect should be looked on as a requency shit caused by accelerating a light source to the velocity v. It is important to notice here that this entire derivation started with the gravitational red shit by the theory o the quanta, then the equivalence principle was used to obtain the results or an accelerating system. The Lorentz time dilation and length contraction came out o this derivation as a special case. Thus, the Lorentz equations should be thought o as kinematic equations, whereas the gravitational and acceleration results should be thought o as a dynamical result. Time dilation and length contraction have always been thought o as only depending upon the velocity o the moving body and not upon its acceleration. As an example, in Wolgang Rindler s book Essential Relativity, 3 he quotes results o experiments at the CERN laboratory where muons were accelerated. He states that accelerations up to g (!) do not contribute to the muon time dilation. The only time dilation that could be ound came rom the Lorentz time dilation ormula. They could not ind the eect o the acceleration because they had it all the time. The Lorentz time dilation ormula itsel is a result o the acceleration. Remember, it is impossible to get a nonzero velocity without an acceleration. Summary o Basic Concepts Equivalence principle On a local scale, the physical eects o a gravitational ield are indistinguishable rom the physical eects o an accelerated coordinate system. Hence, an accelerated rame o reerence is equivalent to an inertial rame o reerence in which gravity is present, and an inertial rame is equivalent to an accelerated rame in which gravity is absent. The general theory o relativity The laws o physics are the same in all rames o reerence (note that there is no statement about the constancy o the velocity o light as in the special theory o relativity). Gravitational red shit Time elapsed on a clock in a gravitational ield is less than the time elapsed on a clock in a gravity-ree space. This eect o the slowing down o a clock in a gravitational ield is maniested by observing a spectral line rom an excited atom in a gravitational ield. The wavelength o the spectral line o that atom is shited toward the red end o the electromagnetic spectrum. Photon A small bundle o electromagnetic energy that acts as a particle o light. The photon has zero rest mass and its energy and momentum are determined in terms o the wavelength and requency o the light wave. An Accelerated Clock and the Lorentz Transormation Equations The Lorentz time dilation is a special case o the slowing down o an accelerated clock. Thereore, it is more reasonable to take the slowing down o a clock in a gravitational ield, and thus by the principle o equivalence, the slowing down o an accelerated clock as the more basic physical principle. The Lorentz transormation or time dilation can then be derived as a special case o a clock that is accelerated rom rest to the velocity v. The Lorentz contraction is a special case o contraction o a length in a gravitational ield, and by the principle o equivalence, a rod L0 that is accelerated to the velocity v is contracted to the length L. Hence, the acceleration o the rod is the basic physical principle underlying the length contraction. It is physically the acceleration that gives the object its velocity that is used in the original Lorentz equations. 3 Springer-Verlag, New York, 1979, Revised nd edition, p

21 Summary o Important Equations Slowing down o a clock in a gravitational gh ield t = tg (7.14) Gravitational red shit o wavelength λ gh = λ g (7.18) Gravitational red shit o requency ν gh = ν g (7.4) Change in requency per unit requency ν gh = (7.5) ν c g Slowing down o an accelerated clock ay t = ta (7.38) t = t a (7.4) 1 v / c Length contraction in a gravitational ield gy λ = λ g (7.49) Length contraction in an acceleration ay λ0 = λ a (7.50) L L0 1 v / c = (7.53) Slowing down o a clock in a gravitationield gy T = Tg (7.31) L a L 0 = ay 1 c (7.55) Slowing down o a clock in a gravitational gy ield t = tg (7.3) Questions or Chapter 7 1. When light shines on a surace, is momentum transerred to the surace?. Could photons be used to power a spaceship through interplanetary space? 3. Which photon, red, green, or blue, carries the most (a) energy and (b) momentum? 4. Ultraviolet light has a higher requency than inrared light. What does this say about the energy o each type o light? *5. Why could red light be used in a photographic dark room when developing pictures, but a blue or white light could not? Problems or Chapter 7 1. A photon has an energy o 5.00 ev. What is its requency and wavelength?. Find the mass o a photon o light o nm wavelength. 3. Find the momentum o a photon o light o nm wavelength. 4. Find the wavelength o a photon whose energy is 500 MeV. 5. What is the energy o a 650 nm photon? 6. Find the energy o a photon o light o nm wavelength. 7. A radio station broadcasts at 9.4 MHz. What is the energy o a photon o this electromagnetic wave? 7-1

22 7.3 The Gravitational Red Shit 8. One twin lives on the ground loor o a very tall apartment building, whereas the second twin lives 00 t above the ground loor. What is the dierence in their age ater 50 years? 9. The lietime o a subatomic particle is s on the earth s surace. Find its lietime at a height o 500 km above the earth s surace. 10. An atom on the surace o Jupiter (g = 3.1 m/s ) emits a ray o light o wavelength 58.0 nm. What wavelength would be observed at a height o 10,000 m above the surace o Jupiter? Additional Problems *11. Using the principle o equivalence, show that the dierence in time between a clock at rest and an accelerated clock should be given by ax tr = ta where tr is the time elapsed on a clock at rest, ta is the time elapsed on the accelerated clock, a is the acceleration o the clock, and x is the distance that the clock moves during the acceleration. *1. A particle is moving in a circle o m radius and undergoes a centripetal acceleration o 9.80 m/s. Using the results o problem 11, determine how many revolutions the particle must go through in order to show a 10% variation in time. 13. The pendulum o a grandather clock has a period o s on the surace o the earth. Find its period at an altitude o 00 km. Hint: Note that the change in the period is due to two eects. The acceleration due to gravity is smaller at this height even in classical physics, since g = GM (R h) To solve this problem, use the act that the average acceleration is g = GM R(R h) and assume that gh t = tg 14. Compute the ractional change in requency o a spectral line that occurs between atomic emission on the earth s surace and that at a height o 35 km. Interactive Tutorials 15. Properties o a photon. A photon o light has a wavelength λ = 40.0 nm, ind (a) the requency ν o the photon, (b) the energy E o the photon, (c) the mass m o the photon, and (d) the momentum p o the photon. 16. Gravitational red shit. An atom on the surace o the earth emits a ray o light o wavelength λg = 58.0 nm, straight upward. (a) What wavelength λ would be observed at a height y = 10,000 m? (b) What requency ν would be observed at this height? (c) What change in time would this correspond to? To go to these Interactive Tutorials click on this sentence. To go to another chapter, return to the table o contents by clicking on this sentence. 7-