Relativistic Boats: an explanation of special relativity. Brianna Thorpe, Dr. Michael Dugger

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1 height Relativistic Boats: an explanation of special relativity Brianna Thorpe, Dr. Michael Dugger Time Dilation Relativity is all about your point of view. We are working with the speed of light and some things can be difficult to visualize, but in the end it just comes down to where you're standing. We refer to this as your reference frame. Depending on your reference frame, you're going to see things a bit differently. Time will get all mixed up with position. What one might usually think of as a simple measured quantities suddenly is not quite so simple in relativity. But however strange things get, it is important to remember that no matter your reference frame the physical laws remains the same. If we use relativistic equations and the correct transformations between reference frames, we don't need to use two different equations to find the same thing. All we need to do is think about where we're standing. So, let's pretend you're on a really fast boat. You're sitting below deck, and for some reason you've got a laser pointer in your hands and a mirror right above your head. Of course, you do what any good scientist would do and you shine the laser pointer into the mirror. The beam goes up, hits the mirror, and goes straight back down. You're able to measure the time it takes for the beam to go up and down, and you get some number. I'm on the shore, and your really fast boat passes by me just as you decide to shine the laser pointer into the mirror. Well, I feel like doing some science, so I watch your beam of light with a timer in my hand. I see the light leave the laser, go in a diagonal path up towards the mirror, hit the mirror, and go in a diagonal path back down. In your reference frame the light's path was different than in mine. You were moving such that the light moves in the vertical direction, and I was standing still watching the light move with the boat. The path of the light which you measured looked like it was going straight up and down, but the path of the light I measured looked as though it was going diagonal up to the mirror, then diagonal back down. From what we ve seen, the light looks like it had taken two different paths depending on the reference frame of the observer. If we were to draw out the paths for each reference frame, we would see that the straight up and down path seems shorter than the diagonal path. That is not too strange. All we are doing is comparing two different ways to observe the path that the light went. The way you take is more direct; the distance you measure is simply twice the distance between the laser pointer and the mirror. But the distance I measure is a lot longer. While the up and down distance between the laser pointer and the mirror remains constant, the up diagonal path, and the down diagonal path, have formed the legs of a triangle of which the up and down distance is the height. To make this clearer, please see Fig. 1 and 2. In Fig. 1, the path of light as seen by you on the boat is graphically shown, while in Fig. 2, the path as seen by me on the shore is illustrated. Light hits mirror

2 height Figure 1: The path of light seen by you on the boat (as explained in the text). Light hits mirror Figure 2: The path of light seen by me on the shore (as explained in the text). Light leaves laser pointer So, while the distance (height) between the laser pointer and the mirror remains the same, the distance which the light travels changes depending on the reference frame. The speed at which the light travels is still constant. If you put those together, the changing distance and the constant speed of light, time has to be messed with. If he doesn t, the speed at which the light travels would have to change depending on the reference frame, and that doesn t happen. Light always travels at the same speed in a vacuum. The units for velocity are meters per second. That means that velocity is directly proportional to the distance travelled and inversely proportional to the time it took to go that distance. If you want to keep your velocity constant and have different distances, you need to make your time shorter if the distance is shorter, your time longer if the distance is longer. In the boat example given earlier, we need to keep the velocity of light constant, but the paths for the light within each frame was different, so we need to change the time it took to travel. Let s look at each situation separately. First, in your situation the distance which the light travels is twice the distance between the laser pointer and the mirror. The distance is equal to 2D, in which D is the distance between the laser pointer and the mirror. If the distance as measured by me, is double the distance as measured by you, then the time I measure must be twice the time you measured in order to keep the velocity of the light constant. So, looking at all this, we can see that the time which it takes for your light to go up and down is equal to t 0 = 2D c, where t 0 represents the time which you measure, and c represents the speed of light. My situation is much the same. Instead of caring about the distance between the laser pointer and the mirror, we now care about the diagonal length which the light travels from when it leaves the laser pointer to when it hits the mirror. We ll use the variable L for this. Now when we look at the diagram for my situation, we can see that the distance between laser pointer and mirror forms a triangle (triangle A) with the light traveling the diagonal length. The diagonal length is the hypotenuse of the triangle, the distance D is the vertical leg, and half of the distance that I observe you to move during the measurement is the horizontal leg of the triangle. The distance you travel from the time when I start the timer to when I stop can be measured by multiplying your velocity by the time which I measure. The

3 horizontal distance I measure is equal to v t where v represents your velocity and t represents the change in the time I measure. The horizontal leg of triangle A is equal to 1 v t. With this information, 2 we can now find out everything we want to know about our triangle just by using the Pythagorean Theorem. The theorem says that the vertical leg of a triangle squared plus the horizontal leg of a triangle squared is equal to the hypotenuse of the triangle squared. So, the length which the light travels that I perceive is equal to ( 1 2 v t)2 + D 2. Next, we can figure out what D equals just by realizing that the D we re using in my reference frame is the exact same as the D which you measure in your reference frame. Now we get that the length which the light travels as I perceive in my reference frame is ( 1 2 v t)2 + ( 1 2 t 0c) 2. The strange thing about this situation is that it looks like we re using two different times in the equation, and we are. t is the time which I perceive, and t 0 is the time which you perceive. The fact that we re using two different times seems strange, so let s figure out why we do this. We have calculated the time it took, in your reference frame, for the light to travel the distance D. If your time is also equal to mine, then the fact that we re using different times must be really pointless. But if they re different then we have figured out something really cool. To figure it all out, let s just rearrange the equation to find out what the time I perceive is equal to. Starting with Pythagorean theorem: 2L 2 = (v t) 2 + ( t 0 c) 2. And since L is equal to cδt, then we can write (c t) 2 = (v t) 2 + ( t 0 c) 2, and rearrange to obtain t 2 (c 2 v 2 ) = t 0 2 c 2. Isolating Δt we obtain the result of interest: t = t 0 (1 ( v c )2. After all this we see that the change in the time which I perceive is actually a velocity dependence on the time which I perceive. To see this, just mess with the values of velocity and see what happens. In fact, I will measure a longer time than you measure. While both times are valid, we call your time the proper time since you only need one clock for your measurements. I need two clocks because I m measuring the time at two different places. My time being longer can also be seen on the diagrams. Our light is going at the same speed but the path of the light which I see is longer than the path of the light which you see. So, of course I m going to measure a longer time. My light has to go a longer distance. This measuring of different times is called time dilation. Going broader than the analogy, this means that when an event is timed in a moving reference frame, that time will be shorter than that measured in a stationary reference frame.

4 Length Contraction Length contraction follows the same idea as time dilation. Reference frames make a difference, and things you think ought to be the same are actually not the same. Let s say that you are still on the boat and I m on the shore. The boat is still moving really fast, maybe 99% of the speed of light. We both want to measure the length of the boat. When you measure the length of the boat, all you need to do is pull out a ruler and measure the boat. This is because, in your frame, the boat is at rest and the world is moving past you at a rapid pace. Because the thing being measured is at rest in your reference frame, the measurement you take will be called the proper length. As you re passing the shore, you also notice that some mark on the shore will pass you in a certain amount of time. Since the marker first passes the front of the boat, then passes the back of the boat, you need to use two different clocks. The first clock records the time at which the marker passed the front of the boat, and the second clock records the time at which the marker passed the back of the boat. Because you need to use two different clocks, the time which you measure is not considered proper time. You know your relative velocity and the time it takes for that mark to pass, so you can express the length which you measure through the following equation: L 0 = v t. Now we get to my measurements. I m standing still on the shore, right by the mark. In my reference frame, the boat is moving past me near the speed of light. Since what I want to measure is moving with respect to me, the length which I find will not be the proper length. The time which I measure, however, will be considered the proper time since I can use a stationary clock to measure the time. I can express the boat s length as L = v t 0. The equation for length contraction is just a combination of those we use for our respective measurements of length. The length which you measure is L o = v t, where L o the proper length. The length which I measure is L = v t o. If we combine these we can see the rate at which the length you measure changes with respect to the length which I measure. The ratio of the lengths is given by L L 0 = v t 0 v t = 1 γ, where γ 1 (1 ( v c )2. Which may be simplified to yield: L = L o γ. Going back to the equation which we found for time dilation, we let the variable γ represent the 1 to term. Because t =, we can substitute t in our above equation with that. For the equation 1 v2 c 2 1 v2 c 2 of length contraction the velocities cancel out, as do the proper times, leaving us with the inverse of γ. Four Vectors Now we ve got two equations. The first, t = t 0 (1 ( v c )2, gives the time which a viewer moving relative to another perceives, in terms of the time of a stationary viewer. The second, L = L o, puts the γ

5 length which a viewer moving relative to another measures in terms of the length which a stationary viewer measures. Both of these equations depend on something called proper time or proper length. We decided what each of these are by determining which viewer will use one clock or one ruler to make measurements. But, frankly, it s just easy to mix up what s proper and what s not when you re deep in the bowels of relativity. It d be really nice if we could forget about all this proper stuff. Fortunately, we have a tool that can take proper out of the equation, allowing us to solve for anything as long as we re given some measurement of time or length. This tool is called a four vector. Before getting too far into this, let s get some housekeeping stuff out of the way. Below are definitions, equations, and some variables to make life easier. β = v, where v represents the velocity of the object we care about, and c represents the speed c of light γ = 1 1 ( v c )2 = 1 1 β 2 t = t 0 (1 ( v c )2 = Δt 0 1 (β 2 ) = t 0γ L = L o γ Proper time: The time measured in the frame where the observer is at rest (T 0 ) Proper length: The length measured in the frame where the object being measured is at rest (L 0 ) With that out of the way, it s time to talk about what dimensions we re working in. We care about four dimensions: time, space in the x direction, space in the y direction, and space in the z directions. The x direction deals with side to side movement, the y with up and down, and the z with back and forth. With these dimensions we can describe the motion of any object. We could work with separate equations for each of these, but it s simpler to be able to put them all in one place. This place we put them in is called a matrix. This allows us to, in a way, see a chart of the magnitude of a quantity in each of the four directions. Below is a four-vector matrix equation for relating spacetime from one frame to another. ct γ βγ 0 0 ct x βγ γ 0 0 x [ ] =[ ] [ ] y y z z In this matrix equation, c represents the speed of light and t represents time. Therefore, the quantity ct is that of position. While speed of light multiplied by time spits out a position, what we really care about is the time. So, when looking at ct, think of this as a way to get time instead of a position. The quantities ct, x, y, and z are referred to as boosted quantities. This means that they are the result of the measured quantities (ct, x, y, and z) after undergoing a transformation due to one of the frames having a velocity relative to another (boost).

6 This four vector matrix allows us to find all these boosted quantities through some pretty simple calculations. All you need to do is multiply everything through by multiplying the horizontal quantities from the matrix by the vertical quantities in the right-most vector. If you do that, you get the following equations for the boosted quantities: ct = γct ± βγx + 0y + 0z = γct βγx x = βγct + γx + 0y + 0z = βγct + γx y = 0ct + 0x + 1y + 0z = y z = 0ct + 0x + 0y + 1z = z In the equation for ct, you can rearrange the variables to find the boosted time: t = γct βγx c Now we have some nice equations for the stuff we care about. We don t need to care about what s proper and what s not. All we need to do is measure the quantities we see, use some variables to calculate the boosted quantities. Invariant Quantities We ve now gotten things broken down into a few equations for boosted quantities. While the definitions are much simpler than all the proper stuff we were dealing with before, we re now faced with a new problem. Everything we are working with is undergoing a transformation. It isn t very pretty or easy to deal with a bunch of changing quantities, so, to make life easier we re going to find a quantity which is constant whether you re standing still or travelling near the speed of light. We ll call this the invariant quantity. In order to find an invariant quantity, let s play around with some possibilities. First, we re going to look at velocity. Velocity is the time derivative of position, so v = dx, where v represents velocity, x dt represents position, and t represents time. First thing you can notice is that this is a mess. Both time and position are changing, so this guy s a mess. We won t be able to get a good four vector out of him. In order to get a good four vector, we re going to rewrite velocity as U u = dxu dt, where U represents velocity and T represents proper time. The exponent u allows us to look at these quantity in the x, y, and z dimensions. So, now we have a nice four vector for velocity. Playing around with it, we get the following: t = γt, therefore dt = γdt and dt = 1 γ dt We can write our new equation for velocity in terms of non-proper time by applying a transformation. This gives us U u = γ dxu. Using our four vector matrix, we can write the dt following: o U 0 = cγ, U x = cβ x γ, U y = cβ y γ, U z = cβ z γ

7 o U u = (γc, γβ c) = (γc, γv ) Now that we have something for velocity, we can use that velocity to find the momentum of whatever object we care about. Momentum is the mass of an object multiplied by its velocity, written as p = mv, where p represents momentum, m represents mass, and v represents velocity. Since we have a four vector for velocity, we can write our momentum as p u = mu u. Since this is now a four vector, the following can be written: p x = mu x, p y = mu y, p z = mu z or p x = γmv x, p y = γmv y, p z = γmv z Putting this in another form we get p = γmv = γmcβ. In non relativistic situations, γ goes to 1, so γβ = β, which lets us take momentum to what we re familiar with: p = mv. Taking the integral of this gets you kinetic energy, which is E k = 1 2 mv2, where E k represents kinetic energy. This can get us to work, which is written as W = 1 2 mv f mv i 2 = E kf E ki. So, work done on a particle is the change in the particle s kinetic energy. If the particle started at rest, this is simply W = 1 2 mv f 2. But what we care about is relativistic situations, so let s work through that same mathematical argument with our equation for relativistic momentum. For this we re going to use p = mcγβ = mc ( β = β (1 β 2 ) 1 2 p p 2 +m 2 c 2. ). With this, we can find β in terms of momentum and mass: The above equation for β comes in handy when finding the work done on the particle: W = vdp = c βdp = c dp. p 2 +m 2 c 2 But since, p d dp (p2 + m 2 c 2 ) 1 2 = 1 ( 1 p 2 p 2 +m 2 c2) 2p = p 2 +m 2 c 2, then W = c (p 2 + m 2 c 2 ) 1 2 = (p 2 c 2 + m 2 c 4 ) 1 2 mc 2. That last equation for work is equal to the total energy of the particle minus the rest energy of the particle. The rest energy is equal to mc 2 and the total energy is equal to (p 2 c 2 + m 2 c 4 ) 1 2. We will let the total energy be represented by E. So, E 2 = p 2 c 2 + m 2 c 4. Since p = mcγβ, this can be rewritten as E 2 = m 2 c 4 γ 2 β 2 + m 2 c 4. That can be simplified to: E 2 = m 2 c 4 (γ 2 β 2 + 1) = m 2 c 4 ( β2 1 β 2 + 1) = m2 c 4 ((β 2 + (1 β2 ) 1 β 2 = m2 c 4 1 ( 1 β 2) = m2 c 4 γ 2. So that

8 E = mc 2 γ. That equation for E is our p 0 multiplied by the speed of light. The momentum four vector can now be written as p u = ( E, p ). Now, after some more algebra: c p u p u = E2 c 2 p2 = γ 2 m 2 c 2 γ 2 m 2 c 2 β 2 = γ 2 m 2 c 2 (1 β 2 ) = m2 c 2 (1 β 2 ) 1 β 2 = m 2 c 2. This shows that our invariant quantity for our four vector is the mass of the particle and is not dependent upon frame.