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Discover Article something not quite right.

  1. Jan 10, 2005 #1
    In the Sept issue of Discover they had an article about Einstein. In the article they mentioned something about Special Relativity that I always suspected but had never seen in print.

    They basically stated that the reason a beam of light always passes you at "c" (the speed of light) no matter how fast you are travelling (even if you are going 0.999 c) is because time is slowing down for you. Therefore if you are traveling at 0.9999 the speed of light your clocks have slowed down so much that you would still see the beam pass you at "c". A "stationary" observer however would see you almost keeping up with the beam. To see their diagram click go to http://public.aci.on.ca/~rallum/Discover Home.htm and see Discover.

    Well I put in the Time Dilation equations in and there is no way I can get the numbers to match. I wrote a quick letter to Discover showing them the math, but I got no reply.

    See letter : http://public.aci.on.ca/~rallum/Discover Home.htm and see letter pg 1 & 2

    If anyone can help and explain this to me it would be much appreciated.

    Richard
     
  2. jcsd
  3. Jan 10, 2005 #2

    JesseM

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    The problem is that you're not taking into account "the relativity of simultaneity", i.e. the fact that clocks which are synchronized in one reference frame will appear out-of-sync in another.

    Here's a simple example. Suppose, for the sake of making the math a bit easier, that we measure distance in units of "fivers", where a fiver is defined to be the distance light travels in 0.2 seconds, so that light is defined to have a velocity of 5 fivers/second. Suppose you see a ruler which is moving at a velocity of 3 fivers/second along your x-axis. In its own rest frame, this ruler is 40 fivers long; so in your frame its length will appear to be:
    40*squareroot(1 - v^2/c^2) = 40*squareroot(1 - 9/25) = 40*0.8 = 32 fivers long. Also, at either end of this ruler is placed a clock; using the time dilation formula, we can see that for every second on your clock, you will only see these clocks ticking 0.8 seconds forward.

    Now, say that when t=0 according to your clock, the clock on the left end of the ruler also reads t'=0. At that moment, a light is flashed on at the left end of the ruler, and you observe how long the light pulse takes to catch up with the right end. In your frame, the position of the light pulse along the x-axis at time t will be c*t, while the position of the right end of the ruler at time t will be v*t + 32. So, the light will catch up to the right end when c*t = v*t + 32 which if you solve for t means t = 32/(c - v). Plugging in c = 5 and v = 3, you get a time of 16 seconds, in your frame.

    Now the key to understanding how the ruler can also measure this pulse to be moving at c is to realize that different frames have different definitions of what it means for a pair of clocks to be "synchronized". In your frame, when the clock on the left reads t'=0, the clock on the right will not read t'=0; in your frame, the clock on the right is always 4.8 seconds behind the clock on the left, so it will read t'=-4.8. This means that after 16 seconds have passed according to your clocks, only 16*0.8 = 12.8 seconds will have elapsed on the ruler's clocks, which means the clock on the right reads -4.8 + 12.8 = 8 at the moment that the light reaches the right end. So remembering that light was emitted when the clock on the left read t'=0, the light must have taken 8 seconds to cross the ruler in the ruler's own frame; and remembering that the ruler is 40 fivers long in its own frame, the speed of the light pulse is measured to be 40/8 = 5 fivers/second. So, light does indeed have the same speed in both frames.

    The general equations for converting between coordinates (x,y,z,t) in one frame S and coordinates (x',y',z',t') in another frame S', where S measures the origin of S' to be moving at velocity v along the x-axis, are:

    x' = gamma*(x - v*t)
    y' = y
    z' = z
    t' = gamma*(t - v*x/c^2)

    and

    x = gamma*(x' + v*t')
    y = y'
    z = z'
    t = gamma*(t' + v*x'/c^2)

    where gamma = 1/squareroot(1 - v^2/c^2).

    These are known as the "Lorentz transformation" equations. You can verify that these equations work for the example above--for example, in my coordinates, the light hit the right end of the ruler at t = 16 and position x = 5*16 = 80. So with v=3 and c=5, gamma will be 1.25, so the equations say that in the ruler's coordinates, this event will happen at:

    x' = 1.25*(80 - 3*16) = 40 (the ruler is 40 fivers long in its own frame)
    t' = 1.25*(16 - 3*80/25) = 8

    which is what I got earlier. I should note that the logic is actually a bit backwards here--in relativity the fact that light moves at the same velocity in all frames isn't really a result but rather a starting assumption, you say that each observer should assign coordinates to events using a grid of rulers and clocks which are at rest relative to himself, and you synchronize clocks at different locations using the assumption that light travels at c in all directions relative to the observer (so that if light is emitted at the midpoint of two clocks, the clocks should both have the same reading at the moment the light hits them). If different observers define coordinates in this way, and if the laws of physics have the same form in each observer's coordinate system (this last part is really the empirical content of the theory of relativity), then you can show that the Lorentz transformation equations must be the correct way to translate between different observers' coordinate systems.

    You might also want to check out my post An illustration of relativity with rulers and clocks which gives a more detailed example of how a series of clocks which are in-sync in one frame will appear out-of-sync in the other, and vice versa; this is key to understanding some other apparent paradoxes of relativity, like how each observer can think it's the other observer's clocks which are running slower than his own.
     
    Last edited: Jan 10, 2005
  4. Jan 10, 2005 #3
    Zardoz,

    You've made the mistake that nearly all who are new to this theory make by trying to use the length contraction and time dilation formulas. They tend to let you forget that two events can be measured as being simultaneous for one observer and not for another observer if they are in relative motion.

    It's much safer to use the Lorentz transforms (from which time dilation and length contraction are derived). They say that if the observer by the side of the road says that something happens at a location x and at a time t, the cop will say it happened at x' and t' where:

    x' = gamma(x-vt) and t' = gamma(t-xv/c^2)

    In your example gamma is about 4 and v is 180,000 miles/second.

    Here's how to do this problem. There are three events of interest.

    E1: The light beam and the cop start moving forward at the same time and from the same place. Let's say the stationary observer calls these x = 0 and t= 0. The Lorentz transforms say that the cop will agree, x' = 0 and t' = 0.

    E2: The cop reaches the point we're interested in. The observer says that's x = 180,000 and it happens at t = 1. But according to the Lorentz transforms, the cop doesn't agree this time. He says x' = gamma(180,000 - 180,000*1) = 0
    E3: The light beam reaches the point we're interested in. The observer says that's at x = 186000 and it happens at t = 1. But the cop says x' = gamma(186000 - 180000*1) = 4*6000 = 24,000 and t' = gamma(1 - 186000*180000/186000^2) =4*.031 = .124 secs

    So, the cop says that in .124 secs the light beam got 24,000 miles ahead of him, which works out to just about c = 186000 miles/second for him. Where you went wrong in your calculation was in assuming that the cop will think he gets to 180000 miles at this same time. But he doesn't. The two events aren't simultaneous in his frame!
     
  5. Feb 6, 2005 #4
    Thank guys. I understand the maths, but I having a little problem visualizing the relativity of simultaneity. Have not had much quiet time to think about it. I need a quiet room to just sit and think, with piece of paper and pencil.

    You explanations were much appreciated.

    Richard
     
  6. Feb 14, 2005 #5
    Ok I see it. You are on the center part of the ruler moving at 3 "fivers" per second. You have to synchronize your clocks, and as the light from the front of the ruler clock (right) reaches you quicker than the clock on the left (back). the front clock gets put back.

    In your example you state that t' would be -4.8sec behind. Where did you calculate this? Did you just work backwards for this #? i.e you knew it had to come to 8 sec so 12.8 + x = 8 therefore x= -4.8?



    Richard
     
  7. Feb 14, 2005 #6

    JesseM

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    The basis for this number is the fact that each observer is supposed to synchronize different clocks which are at rest in their frame by making the assumption that light travels at the same speed in both directions in their own rest frame. So if I have two clocks which are at rest in my frame, and I turn on a light at the exact midpoint between these clocks, then I define these clocks to be "synchronized" in my frame if they each read the same time at the moment the light hits them.

    So, let's go back to that ruler example and imagine that at t=0 on your clock, the back end of the ruler is at position x=0 and the front end is at position x=32, and the ruler is moving in the +x direction at 3 fivers/second. And suppose there's a clock at the front end of the ruler and a clock at the back end, and I am sitting at the middle of the ruler (x=16 in your frame) and turn on a light at t=0 in your frame, causing beams of light to head out from that point in both directions at 5 fivers/second. At what time will the light reach both ends in your frame? Well, the position of the beam heading towards the back end is given by x(t) = 16 - ct = 16 - 5t, and the position of the back end of the ruler is given by x(t) = 3t, so they will meet when 3t = 16 - 5t, or t=2 seconds. Meanwhile, the position of the beam heading towards the front end is given by x(t) = 16 + 5t, and the position of the front end of the ruler is given by x(t) = 32 + 3t, so they will meet when 16 + 5t = 32 + 3t, or t=8 seconds. Now, we know from the time dilation formula that for every second on your clock, only 0.8 seconds go by on the clocks mounted on the ruler. At t=0 in your frame, if the ruler's back clock also read t'=0 and the ruler's front clock read t'=t0', then when the light hits the back clock it will read (2 seconds)*0.8 and when the light hits the front clock it will read t0' + (8 seconds)*0.8, so if both clocks read the same time when the light hits them, we must have 2*0.8 = t0' + 8*0.8 which means that t0' = -4.8 seconds.

    You can use the same sort of method to prove that in general, if I see two clocks which are a distance x apart in my frame and which are travelling together at velocity v, then if an observer at rest wrt these clocks synchronized them using light-signals, in my frame the back clock must be ahead of the front clock by:

    [tex]\frac{vx \sqrt{1 - v^2/c^2}}{c^2 - v^2}[/tex]

    Or in simpler form:

    [tex]\gamma \frac{vx}{c^2}[/tex]

    If you use this formula along with the formulas for Lorentz contraction and time dilation, you can get the right answers for a lot of problems in relativity without having to use the full Lorentz transformation equations.
     
    Last edited: Feb 14, 2005
  8. Feb 27, 2005 #7
    I got it all now. My original mistake was fueled by using the motorcycle analogy. I assumed the clocks (watch) on the mototcyclist and "stationary" observer were synchronized before he set off on his run.

    Thanks for your time. I can explain it to the rest of my family now. At least those of whom are interested.
     
  9. May 23, 2005 #8
    The basis for this number is the fact that each observer is supposed to synchronize different clocks which are at rest in their frame by making the assumption that light travels at the same speed in both directions in their own rest frame. So if I have two clocks which are at rest in my frame, and I turn on a light at the exact midpoint between these clocks, then I define these clocks to be "synchronized" in my frame if they each read the same time at the moment the light hits them.


    Am I correct in assuming that if both clocks were next to each other, synchronized and then place in their positions at the front and back of the motorcycle, train or whatever the result would be the same? Would the motion of placing the clocks in position have the same result as synchronizing then with an flash of light between the two? If not then I have a problem with the Michelson-Morley experiment.

    Humm
     
  10. May 24, 2005 #9

    JesseM

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    If they are next to each other and you move them apart, then you have to introduce a relative velocity between them, so time dilation will be involved. But if one clock is not accelerated while the other is accelerated to a velocity that is very small compared to light speed, then the two clocks will stay arbitrarily close to being perfectly synchronized in the frame of the clock that wasn't accelerated, which means you're staying arbitrarily close to what would happen if you synchronized two already-separated clocks using the light-signal method described above.
     
  11. Jun 8, 2005 #10
    Thx JesseM

    Yes I understand. I guess my question is; mathematically do the two scenario give you the exact same result?

    It seems to me as they must, as in the Michelson-Morkey experiment I assume they synchronized the clocks while they were next to each other. It would be silly for them to synchronized using a light flash. Trying to measure the speed of light in different directions when you calibrated the clocks using the speed of light would not make sense.

    Michelson-Morkey experiment should give the same results on a speeding train traveling close to "c" as on our earth spinning around and traveling in orbit. Therefore if you have all the clocks together, synchronized them and then separated them to different areas of the train the time dilations that occurs must be exactly equal to the result of synchronizing them using a light source in the centre. If not Michelson-Morkey experiment would not give a consistent result.

    There is a nice little article in the current Discover magazine called "If an Electron Can Be in Two Places at Once, Why Can't You?". Tries to use gravity to explain it. Worth a read if you like Quantum Mechanics.


    Thanks again

    Zardoz.
     
  12. Jun 9, 2005 #11

    JesseM

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    Yes, what I was saying was that if v is the relative velocity between the two clocks as they are moved apart, then in the limit as v/c approaches 0, this will give the same result as if you synchronized them using the light-signal method. For any finite v there will be a slight departure from the synchronization you'd get using light signals, but for the velocities between the clocks that would likely have been used in the Michelson-Morley experiment (not 'Michelson-Morkey' as you wrote it, by the way), this departure would almost certainly have been so tiny that it would not have made a difference in the measured results of the experiment.
     
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