Special Relativity: Derivations

Transcription

1 Special Relativity: Derivations Exploring formulae in special relativity Introduction: Michelson-Morley experiment In the 19 th century, physicists thought that since sound waves travel through air, light waves must travel through a medium as well. They called this medium Aether. It was a medium that was hypothesized to be what light require in order to move through the vacuum of space. How else could light waves move through unless there was some medium there? Aether, as they assumed, is an invisible nothingness that permeates space. Its only physical property is that it only allows light to propagate through. Once precise measurements of the speed of light became possible, testing the predicted effect of the aether on the speed light became possible as well. The earth orbits the sun at about km/h. If the light travels through aether, then as the earth moves through the aether, the speed of light should be different when going with the aether or going perpendicular through it.however, relative to the speed of light, the earth s orbit around the sun is rather slow. Therefore the level of precision needed for this experiment must be extremely high. Nevertheless, in 1877, Albert A. Michelson and Edward W. Morley, two American scientists, devised an apparatus that would detect even tiny differences in speed between two beams of light. It was called the interferometer. Light from one source is split into two directions through a half silver two-way mirror. These beams are reflected between other mirrors and recombined into a single beam. When two light beams recombine, an interference pattern will occur from the superposition of the waves. The experiment was repeated many times, and each time the direction of the light beam. Therefore it is expected that the interference pattern, resulting from the two light beams sent out from different direction, will be different, showing that the light beams in each case are of different speeds. However, Michelson and Morley never detected such a difference, and the existence of aether was never proved. The issue remained unsolved for many years until Einstein came along, and proposed, altogether with his theory of relativity, that the speed of light is always the same regardless of the speed of the light source.

2 Special relativity In an attempt to explain and simplify the theory of special relativity, I will only rely on two postulates to derive the properties of space-time with what I believe to be a systematic approach and eventually we shall derive arguably the most famous equation in physics, E = mc 2. Postulates: 1. The laws of physics are the same in all inertial (non accelerating) frames of reference 2. The speed of light in vacuum is a constant, and has a value c in all inertial frames of reference The theory of special relativity is a generally accepted theory regarding the relationship between space and time. The first postulate implies that in any reference frame, the physical law v=x/t, for instance, remains true. It does not imply, however, that the distances or the times measured would be the same in different frames (don t assume time is always a constant!). The implications of the second postulate also contradict what many have thought to be true. Consider a simple experiment (I) below. A person A travelling on a train at speed v relative to the platform fires a bullet. In A s frame, the speed of the bullet was measured to be w. Another person, B, who is standing on the platform, measures the speed of the bullet. From everyday experience, we would assume that the speed of the bullet from B s frame would be equal to v+w. Consider however, another scenario, where A fires a laser (light) beam instead of a bullet, travelling at speed c in A s frame. Many would assume that B will measure the speed of the laser beam to be v + c. But that is not the case! According to the second postulate, the speed of light (say, its speed in air is roughly equal to its speed in vacuum) has a value c in all inertial frames. Therefore, the speed of the laser beam measured by B will still be equal to c, not v + c. Thus, we can conclude that the rule for combining velocities breaks down when considering the speed of light, and we cannot simply add them to arrive at the solution. Note, also, that the velocity v+c cannot be achieved, as nothing can travel faster than the speed of light (later on).

3 Lorentz interval We will start with a little bit of basic mathematics. Consider another example (II) where there are three observers. One observer, A, is standing on the train travelling at speed v relative to the platform. Here, A observed a laser beam being fired vertically from the bottom of the train to the top where it is detected by the receiver. B, who is standing on the platform, and C, who is on another train travelling in opposite direction, also observed this. Let the speed of A relative to C be v. Denote E1 to be the event when the laser was fired, and E2 to be the event when it is detected. For A, let t be the time interval between the two events. The vertical height of the train, as v=s/t, is ct, which is the spatial distance between E1 and E2. B, however, observes the two events differently, as shown below. Let the time interval between E1 and E2 be t. As for C, as the train is moving in opposite direction, C s observation will be somewhat similar to B, but with a larger distance (as in C s frame, A moves faster). Denote the time interval of the two events t. In B frame, B must measure the speed of light beam to be c, therefore the spatial separation between the two events is ct. Then we can construct a vertical line down from where E2 took place, forming a right angled triangle. Knowing that in B s frame, the train travels at speed v horizontally, then the horizontal distance is equal to v t = x and the vertical distance is ct, as from before, as the train only moves horizontally, we can regard that vertically, the train is stationary with respect to the platform, therefore the train and the platform is vertically in a similar inertial frames. As for C, the same concept applies as shown in the diagram. From Pythagoras theorem, (ct ) 2 (v t ) 2 (ct) 2 = 0 and also (ct ) 2 (v t ) 2 (ct) 2 = 0. Generalising this

4 implies that for any frame of reference observing the two events, (ct n ) 2 [(v n t n ) 2 + (ct) 2 ] is always invariant. Note that (v n t n ) 2 + (ct) 2 is the square of the space separation between the two events, and (ct n ) 2 is the square of c x time separation of the two events. This is called the spacetime interval, or the Lorentz interval. Generally, the space-time interval is written as (Interval) 2 = [(c x time separation] 2 [space separation] 2 However, this still does not prove that the interval is always invariant, because in this thought experiment we are moving at the speed of light c, in which the interval in any frame is 0. I will come to discuss this later on. Lorentz factor From the equation (ct ) 2 (v t ) 2 (ct) 2 = 0, we can rewrite this as (ct ) 2 (v t ) 2 = (ct) 2 which solves to (t ) 2 (1- v 2 /c 2 ) = t 2. Take the square root of the equation and move the factor to the other side gives t = γt This factor γ is called the Lorentz factor. Time dilation and length contraction Where γ = 1/ 1 v2 c 2 The equation t = γt presents a significant implication. t represents the outside observer i.e. platform and t represents the true observer, i.e. train. It implies that the times measured by observers moving at different speed will not be the same! Of course, this seems absurd in our daily life, because we are always moving very slow relative to light, and the factor γ becomes roughly 1 at slow speed. However, if we are on a rocket travelling to Alpha Centauri at speed 0.6c compared to earth (III), the time t measured by observers on Earth for us to reach the star will be 1.25 times longer than the actual time t elapsed in the rocket (1/0.8 = 1.25). This effect is called time dilation (moving clocks tick slower!) Note further, that from theory of relativity, the observer in the rocket would see that it is actually Alpha Centauri that is moving towards him at speed 0.6c relative to the rocket (assuming that Alpha Centauri is stationary relative to Earth). Denote L to be the distance from Earth to Alpha Centauri measured by earth observer, and L for the distance measured by the rocket observer. From the first postulate, we may assume the equation for velocity holds true, therefore 0.6c = L /t and 0.6c = L/t. From this, L /t = L/t. Substituting t gives L = γl : the equation which represents length contraction, implying that the distances measured by observers in different frames of reference will not agree. Length contraction is the phenomenon of a decrease in length or distance as measured by an observer who is traveling at

5 any non-zero velocity relative to the object. Thus, the observer in the rocket will record a distance that is smaller than the observer on Earth, as it is moving relative to the distance between the Earth and the star. Addition of parallel velocities Now that we have grasped the idea of time dilation, the law of addition of velocities can be solved using another thought experiment (IV). A, standing on a platform, fires two successive bullets at a target held by B on a train travelling at a constant speed v. For A, the bullet was fired at speed u. We shall let the time interval, measured by a clock on the platform as observed by A, between the firing of the bullets be t g, and the time interval between the first bullets hitting the target be t t. At time 0, the target is located directly in front of the gun, being hit by the first bullet immediately after it was fired. Therefore we can equate the distance travelled by the second bullet with the distance travelled by the train: u( t t t g ) = vt t By making t g the subject gives: 1) t g = t t (1-v/u) Next, we denote the time interval between the gun firing the bullets, measured by a clock on the train to be t g, therefore the relationship between t r and t r follows from the equation of time dilation, as the train is moving away from the gun: Therefore, from 1) and 2), 2) t g = γt g 3) t g = γt t (1-v/u) Switching to B s frame, B sees A and the gun moving away at speed v. We will denote w to be the speed of the bullet as observed by B and t t to be the time interval of the bullets hitting the

6 target, and by equating the distance travelled by the second bullet with the distance travelled by the machine gun during the firing of the two bullets: By making t t the subject gives: w( t t t g ) = vt g 4) t t = t g (1 + v/w) Again we can use time dilation equation to relate t t and t t : Therefore, from 3) and 4), Multiplying 3) and 6) gives 5) t t = γt t 6) t t = γt g (1 + v/w) γ 2 = (1-v/u) (1 + v/w) After several algebraic manipulations, we arrive at the law of addition of velocities: 7) u = (v+w)/(1+ vw/c 2 ) From this expression, we can understand that for relatively low speeds, the denominator becomes approximately 1. Therefore the matter comes down to simply adding v + w. However, when one or both of the speeds tend to the speed of light, the resulting velocity u also becomes larger than the two, but yet it never becomes faster than c. The addition law of relative velocities was derived from many different ways; many required the use of Lorentz transformation. Another very simplistic way of deriving this without the use of Lorentz transformation was proposed by David Mermin in his book It s about time. However, it requires the assumption of the fact that a fraction length of an object is invariant regardless of different frames: a statement which seemed obvious but not yet proved. Therefore, I believe that this thought experiment involving moving bullets and target is the best approach, as it only relies on the properties of space-time which we have derived from the two postulates. Lorentz transformations

7 Continuing from the previous thought experiment (III) involving a rocket journey to Alpha Centauri, we can derive the Lorentz transformation: two equations defining how the speed of light was observed to be independent of the reference frame. In this case, observer A is on Earth and observer B is on the rocket. The clocks on the Earth and the rocket are synchronised when t = t = 0, at the instant when the rocket leaves the Earth. In Earth s frame, the rocket s speed as measured by observer A is equal to v. Therefore at one instant time t the rocket will have travelled a distance vt. Let the coordinate of Alpha Centauri in Earth frame be x. In the rocket s frame, the observer B sees the Earth and Alpha Centauri moving at speed v. Let the coordinate of Alpha Centauri at time t in the rocket frame be x ; the time which the Earth has travelled a distance vt. From the equation of length contraction, note that as the rocket is moving relative Earth, the

8 actual distance measured from the rocket to Alpha Centauri by observer B at an instant time t will be contracted as seen by observer A. Also, the distance measured from the Earth to Alpha Centauri by observer A at an instant time t will be contracted as seen by observer B. Therefore we can draw two diagrams showing the distances of each frame as shown below: From this we can conclude that x = (x /γ) + vt Hence, 8) x = γ( x vt ) As for the rocket frame; x/γ = x + vt Therefore, 9) x = γ( x + vt ) Equations 8) and 9) are the two Lorentz transformation equations for space. From these two, solving the simultaneous equations for t and t (the maths is quite complicated and dull), by eliminating x and then x, yields the following Lorentz transformation equations for time: 10) t = γ( t vx/c 2 ) 11) t = γ( t + vx /c 2 ) This method of deriving the Lorentz transformation is truly captivating, because it does not rely on the assumption of linear transformations. We have avoided the derivations of the Lorentz- Einstein transformations that start with their guessed shape, imposing the condition of linearity: the idea which I find difficult to grasp. I believe that this method is a non-traditional approach to special relativity and would be simpler for the readers to understand fully. Addition of relativistic perpendicular velocities Going back to the rocket journey (III), we can rewrite the Lorentz equations for all directions as below. The distances in the y and z direction must remain the same in both frames, as the relative motion is only in the x direction. x = γ( x - vt ) y = y z = z t = γ( t - vx/c 2 )

9 Suppose at one instant the observer in the rocket fires a bullet from the bottom vertically to the top of the rocket. The observer on Earth measured the velocity of the bullet in the y direction to be u y as measured from Earth s frame, and sees that the rocket is moving at speed v relative to him in the x direction. The velocity of the bullet (y direction) in the rocket frame is u y therefore u y = y / t. For an observer on Earth, he will see u y = y / t and u x = x / t. From this we substitute the value from the Lorentz equation which gives: u y = y / t = y / γ( t - ux/c 2 ) = y / tγ(1 + vx/tc 2 ) 12) u y = u y / γ(1 + vu x /c 2 ) Equation 12) represents the relativistic addition of velocities for perpendicular velocities (this also applies to z plane). Note that the equation also relies on the velocity in the x direction. From this, we are now able to calculate the resulting velocity from combined velocities with any given directions. Lorentz Interval proved The importance of the Lorentz transformations is that it allows us to prove arguably the most important property of space-time: the Lorentz interval. As mentioned before, the interval is an invariant regardless of reference frame and it is in the form of (Interval) 2 = [(c x time separation] 2 [space separation] 2 The interval is often denoted the symbol τ. It is important to realise that for some two events, the value of (ct) 2 x 2 may be negative, and this would be a problem as τ 2 cannot be smaller than 0. In another case, the value may also be equal to 0, as we have previously encountered. When the value of (ct) 2 x 2 > 0, the interval is timelike. When it is negative it is spacelike, and when it is equal to 0, it is light like. When this value is negative, we must multiply the expression by -1, in order for the interval to have a solution. A spacelike interval describes two events for which an inertial frame can be found such that the two events are simultaneous. It also implies that an observer cannot physically be presented at both events, because otherwise he must be travelling faster than light speed, which we shall come later on to prove that it is

10 impossible. As no information can be transmitted faster than light speed, one of the two events having spacelike interval could not have caused the other to happen, and therefore they are completely independent of one another. In this experiment, we shall let event E 1 to be the instant the rocket leaves the Earth when t = t = 0; E 2 to be an explosion on Alpha Centauri as measured at time t and t on Earth and the rocket, respectively. This way, the time and space separation between the events will be x and t for Earth frame, and x and t for rocket frame. We now show that the Lorentz interval is an invariant by substituting values from the Lorentz equations:

11 Relativistic mass and Momentum Now we have derived the law of addition of velocities, it may be shown that the classical law of momentum poses a problem. Consider two rocks with the same mass colliding with speed 0.9c. Suppose that the rocks stick together after the impact, it would be obvious that both of the rocks would collide and remain at rest in Earth s frame (total momentum = 0). However, if an observer is in one of the rock s frame, whose speed is 0.9c relative to a ground observer, he would measure the second rock speed to be 0.994c (from relativistic addition law). In accordance with observer in Earth s frame, the observer moving with the rock should observe the rocks after the collision to move at 0.9c relative to him. However, from the classical momentum law, mu 1 + mu 2 = 2mv, and this results in v being equal to 0.497c, instead of 0.994c. Therefo re, we must conclude that the equation for momentum only holds true when v<<c, and must account for this obvious contradiction. Consider a similar scenario when two balls which have the same mass m 0 when at rest in the same frame. The two balls are then observed by A to be travelling both at speed w but in opposite direction, colliding and sticking after the impact. As the two balls are moving at the same speed relative to B, the masses of the two balls as measured by B can be considered to be equal (m w ). We will not assume that the masses measured are equal in different frames. From the conservation of momentum, we can conclude that the total momentum is zero, and the two balls would remain at rest after the impact. Observer B who is stationary relative to the left ball sees the right ball moving at speed v and

12 collides with the left ball at rest. After the collision, the two masses stick and would be observed to travel at speed w (principle of relativity). From the equation of conservation of momentum; From the law of addition of velocity; m v v = (m 0 + m v )w v = 2w / (1+w 2 /c 2 ) Solving the simultaneous equations does seem complicated and dull. Nevertheless, the workings are shown below, substituting both v and w. Simplifying m v and w into single terms Making w the subject from the equation

13 Substituting for w (finally..) The final result gives m v = γm 0. This is a very important result, as it implies that the mass of an object is dependent on its velocity. The term gamma increases without limit as the velocity approaches c, signifying that the mass of the ball tends to infinity as the object approaches the speed of light. For the relativistic momentum, multiplying both sides by v gives the following equation: m v v = γm 0 v Going back to our contradictory experiment with rocks, we now apply the true relativistic momentum equation. γ'm 0 (0.994c) = γ(2m 0 ) v 13.4(0.994c) = 2γv = γv v 2 = (3.96c 2 /4.96) v = 0.9c This answer now agrees with observer in another frame and therefore this equation for momentum is no longer contradictory. Relativistic Energy The equation for momentum is no longer mv, therefore the force law, defined as the rate of change of momentum, must also be written as F = γm 0 v / t. This also has an effect on the work done on an object, expressed as F dx

14 Suppose that a varying force is exerted on an object with rest mass m 0 with initial velocity zero, and the all of the work done on the object is converted to its kinetic energy. We can write the equation as K.E. = F dx For classical mechanics, the kinetic energy may be derived from the integral as shown below, and the result, as expected, is mv 2 /2.

15 The mass, however, isn t constant with its speed, therefore instead we must differentiate with respect to the true law of momentum:

16 This gives the formula for relativistic kinetic energy, which looks nothing similar to the classical one. However, when the speed of an object is very much smaller than the speed of light, the formula can be shown to approximate to mv 2 /2. Note that the value γm 0 c depends on the speed of the object, whereas m 0 c does not. We know that the total energy of an object must equal to the sum of its rest energy and its kinetic energy. Therefore we may consider rest energy, which would not depend on its speed, to be mc 2, and we have arrived at the infamous formula relating mass and its rest energy: E = m 0 c 2 The formula can also be shown to equal to Rest energy 2 = Total energy 2 (momentum x c) 2 which is written as E 2 p 2 c 2. The rest energy must remain the same in all inertial frames (because it is when at rest in that frame). Thus we have arrived at another invariant, The derivations of properties of space-time from the two postulates do not require a lot of mathematical background. However, it does prove to consist of algebraic manipulations and imaginative thought experiments. Although the sequence of these derivations may seem unconventional and systematic, it seemed to be the approach which poses the least confusion and assumptions. This, however, shows only the formulas for space-time properties with a few explanations to their effects. The reader may then be tempted to apply the concept to different thought experiments and problems, which would be very helpful and improve their understanding of these derived formulae.

17 Reference: Mr Hearn Spacetime physics by Taylor Wheeler Black holes: Introduction to General Relativity by Taylor Wheeler It s about Time by David Mermin