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New to relativity

  1. Jan 25, 2009 #1
    Im just starting to learn about relativity and find it very interesting and fun. Anyways I have a question to see if Im on the right track when it comes to taking this stuff in.

    If a man who is stationary times a car moving and the car itself has a clock that measures itself moving. And they both measure out 1 complete second, will the man whos stationary since he moves faster in times dimension think that the cars speed is slower compared to the clock which is in the car because his second happens much faster than the cars.

    Anyways i did some number playing and calculated that if the cars clock measures that the car moved 100 m in that 1 second than the stationary man will think that the car only moved 99.999964 m or something which is a slightly smaller distance obviously. So are they both equally right because everyones reference of time is relevant to their movement in spatial dimensions? So it all depends on the observer's speed in spatial dimensions as to what the speed of time will be.

    im hoping im on the right track here because it all seems to make sense right now. Thanks in advance for any clarification.
     
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  3. Jan 25, 2009 #2

    tiny-tim

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    Hi cam875! :smile:

    The man thinks the car's speed is, say 60 mph.

    The car thinks the man's speed is also 60 mph.

    The man thinks the car moves 88 feet in 1 second.

    The car thinks the man moves 88 feet in 1 second. :wink:
     
  4. Jan 25, 2009 #3

    JesseM

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    Yes, if two inertial observers each measure the speed of the other one in their own rest frames, they'll both get the same answer for the other one's speed, it's reciprocal in this sense.
     
  5. Jan 25, 2009 #4
    im sorry but i am now completely lost with this, why would the car think the man is moving at 60 mph and vice versa. and is my answer thatg I provided to my question i made up even correct. Thanks for the reply.
     
  6. Jan 25, 2009 #5

    JesseM

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    Your answer isn't right, if you take into account the combined results of length contraction, time dilation, and the relativity of simultaneity you'll find that each one measures the same speed for the other one in their own rest frame (you can also just use the Lorentz transformation to find the coordinates that each observer assigns to points on the other observer's path, and from this show that each calculates the same distance/time for the other observer). I gave an illustration of two ruler-clock systems moving alongside each other at 0.866c (which means that each measures the other's rulers shrunk by a factor of two and the other's clock ticks to be lengthened by a factor of 2), if you look at that it might give you a better sense of how each one's view of the other's speed will be symmetrical.
     
  7. Jan 25, 2009 #6
    so in the end if they both measure the same speed where does time dilation come from? or am i now confusing all this new stuff flooding my brain lol.
     
  8. Jan 25, 2009 #7

    JesseM

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    Well, if each one is carrying a clock, then each will measure the other one's clock to be running slower, so there is time dilation. If you look at that thread with the illustration I made, and take note of the clock times, you can see that each individual clock on one ruler (like the red clock at the middle or the left or right green clocks) is ticking at half speed in the frame of the other ruler.

    It might help to do the math for a similar example. Say we have two observers A and B who each have a ruler which is 20 light-seconds long in their own rest frame, and each one has 2 clocks attached to either end of their own ruler which are synchronized in their own frame. Suppose the two rulers approach each other and pass side-by-side, and each observer notes when any pair of clocks pass right next to each other. In A's frame, if B is approaching from the right, and B's ruler is shorter than A's ruler in this frame, then first B's left clock will pass next to A's right clock (event 1), then later B's right clock will pass next to A's right clock (event 2), then B's left clock will pass next to A's left clock (event 3), and finally B's right clock will pass next to A's left clock (event 4). It may help to draw a little diagram of this.

    Let's say that in A's frame, B is approaching at 0.8c. That means that in this frame B's ruler will be shrunk by a factor of [tex]\sqrt{1 - 0.8^2}[/tex] = 0.6, so in A's frame B's ruler is only 20*0.6 = 12 light-seconds long. Suppose also that when event 1 occurs, both A's right clock and B's left clock (which are passing next to each other at this moment) read t=0 seconds. B's right clock is 12 light-seconds away from A's right clock at this moment, so if B is moving at 0.8c B's right clock will reach A's right clock 12 l.s./(0.8 l.s./s) = 15 seconds later in A's frame, so A's clock reads t=15 seconds at the time of event 2. And since A's ruler is 20 l.s. long while B's is 12 l.s. long in this frame, at the moment their right clocks line up, the left clock of B will be 20 - 12 = 8 light seconds away from the left clock of A. So, it'll take another 8/0.8 = 10 seconds for the two left clocks to line up (event 3), meaning A's clock reads t=15+10=25 seconds at the time of event 3. At this moment B's right clock is 12 light-seconds away from A's left clock, so it takes another 12/0.8=15 seconds for those clocks to line up (event 4), so A's left clock reads t=25+15=40 seconds at the moment of event 4.

    Now we can figure out what B's clocks read at each of these passing-events. In A's frame, both of B's clocks are ticking slow, by a factor of 0.6. Also, because of the relativity of simultaneity, if B's clocks are synchronized and a distance L apart in B's own frame, in a frame where they are moving at speed v they'll be out-of-sync by vL/c^2, with the front clock's time being behind the back clock's time by this amount. In A's frame, v=0.8c, and the distance between the clocks L in B's frame is 20 l.s., so the two clocks are out of sync by 16 seconds in A's frame; this means at the moment B's left clock reads t'=0 seconds (event 1), B's right clock already reads t'=16 seconds in A's frame. 15 seconds later in A's frame, each of B's clocks has only ticked forward by 15*0.6=9 seconds, so at the moment of event 2 (when B's right clock lines up with A's right clock), B's left clock reads t'=0+9=9 seconds and B's right clock reads t'=16+9=25 seconds. 10 seconds later at the time of event 3 (A's left clock lining up with B's left clock) in A's frame, each of B's clocks has only ticked forward by an additional 10*0.6=6 seconds, so now B's left clock reads t'=9+6=15 seconds and B's right clock reads t'=25+6=31 seconds. Finally, 15 seconds later at the time of event 4 (A's left clock lining up with B's right clock) in A's frame, each of B's clock has ticked forward by another 15*0.6=9 seconds, so B's left clock now reads t'=15+9=24 seconds and B's right clock reads t'=31+9=40 seconds.

    So, looking only at the clocks that line up, we have:
    event 1: A's right clock reads t=0, B's left clock reads t'=0
    event 2: A's right clock reads t=15, B's right clock reads t'=25
    event 3: A's left clock reads t=25, B's left clock reads t'=15
    event 4: A's left clock reads t=40, B's right clock reads t'=40

    Now, look at what B will conclude about this in his own frame, where B's clocks are synchronized and running at normal speed, and the distance between them is 20 light-seconds rather than 12. Event 1 is A's right clock passing B's left clock, and event 2 is A's right clock passing B's right clock; these events are 25 seconds apart according to B's clocks, and this is the time it took A's right clock to pass from the left end to the right end of B's 20 light-second ruler, so A's right clock must be traveling at 20 l.s./25 s=0.8 l.s./s=0.8c in B's frame. Likewise, event 3 and event 4 represent A's left clock passing from one side of B's ruler to the other, and these events happened 40-15=25 seconds apart in B's frame too, so B also concludes the left clock was moving at 0.8c.
     
  9. Jan 26, 2009 #8

    tiny-tim

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    Hi cam875! :smile:

    (btw, which book are you using to learn relativity from?)

    It's the word "if" that I don't like. :cry:

    There are two basic principles of relativity …

    i] if A measures B's speed as v, then B measures A's speed as v

    ii] everyone measures the speed of light as the same.​

    That's what you start with

    you don't examine time dilation length contraction simultaneity and so on, and then come to the conclusion that the speeds are the same …

    you start with i] and ii], and then (if absolutely necessary) you come to conclusions about time dilation length contraction simultaneity and so on (as JesseM :smile: demonstrates).
     
  10. Jan 26, 2009 #9
    umm alright starting to make sense, im using the book "the elegant universe" to learn the ideas of relativity not the equations or anything yet. So basically from something I read in that book it sounds like it says that no matter how fast you move light will always travel at c faster than you, but that concept doesnt make sense to me.
     
  11. Jan 26, 2009 #10
    Hi Cam875. I think what you need to get your head around is that
    with respect to you

    The with respect to you part is important, because somebody who is standing still will also measure the lights speed as being c.

    The reason for this is that velocities don't just add up the way they seem to.
     
  12. Jan 26, 2009 #11
    Hello Cam: you posted.....
    That is correct....light appears to all inertial observers, no matter what their own speed, as "c"; The speed of light is fixed for all inertial observers: "c". (In special relativity, no gravity.)

    And you are also correct: it does NOT make intuitive sense! It is NOT "logical" using classical approaches: that's one of the things that makes relativity so incredible!! And one of the things that causes people to say that Einstein is one of the greatest scientists of all time.

    By all means post quotes,and pages, from the Elegant Universe....I have it on my bookshelf and have read it twice...so far, parts four or five times.

    But this stuff takes time to sink in...a casual reading while watching TV, for example, is not likely to make much of a dent in anyone's brain the first time around.
     
  13. Jan 26, 2009 #12
    It's good that that doesn't seem to make sense to you. It shouldn't, because it's WEIRD. Sadly, most people initially don't understand what's so strange about light moving at the same speed for everyone. So you're on the right path!

    Here's my explanation.

    Think about water waves. We can do experiments easily on them. We see that water waves always move at a constant speed relative to their medium (the body of water). Imagine a pool. Better than that, imagine a pool on a bus. Start with the bus at rest. Drop a stone in the pool, and water ripples out in all directions. These waves move at some speed, let's say v_water. Let's restrict our bus to one dimension, so the waves travel towards the front of the bus at +v_water and to the rear of the bus at -v_water. You are standing outside the bus. You see the water move at the same speeds, +v_water and -v_water.

    Suddenly, the bus starts moving with a velocity of v_bus. A passenger on the bus drops another stone. The water waves travel at the speeds +v_water and -v_water, as they did before. But to you standing on the street, you have to add the velocity of the bus into the speed of the waves, so the waves are moving at v_bus - v_water and v_bus - v_water.

    Light is a wave. It reflects, refracts, and diffracts just like water. The difference between water waves and light, though, is light can move through a vacuum. Water waves cannot. If waves move at speeds relative to their medium, how can we reconcile that water moves without an apparent medium? One solution was an all-permeating gas called the aether in space. This theory was experimentally falsified. No matter what way they measured it, the speed of light always seemed to be the same. This would be as if no matter how fast our bus was moving, the waves always traveled the same velocity. Insanity!

    Einstein's special relativity solved that contradiction by taking the constancy of the speed of light as a law and deducing the logical conclusions. Like non-Euclidean geometry, the results were not inherently contradictory. Rather, they were just bizarre and non-intuitive.

    The best way to visualize time dilation is with a "light clock". Take two mirrors, one above the other, and a light pulse bouncing between them. Let's say that a person moving with the click sees it "tick" (make a complete cycle) every x seconds (x is actually a very, very small number because light is very, very fast!) Now, relative to you, set that clock in uniform motion to the right. The light, instead of traveling straight up, is traveling up and to the right. When it finally reaches the mirror on the top, it will have traveled more distance than when it was at rest. But the speed of light, by hypothesis, is constant, so it must have slowed its vertical velocity to compensate for the horizontal velocity. The frequency of the clock decreases as a consequence, as the light takes longer to travel between the mirrors. Time has slowed down for the clock in the most literal way possible!

    But the situation is symmetric. If you and a friend each had a light clock, and your friend is moving at a uniform speed relative to you, then YOU are moving relative to THEM. To you, your friend is slow. And to them, you are the one who is slow. And they are both right. There is no way to distinguish between motion and rest, and that is the beautiful part of relativity.
     
  14. Jan 26, 2009 #13

    tiny-tim

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    books on relativity

    Hi cam875! :smile:
    That's Brian Greene's book, isn't it?

    I don't have a copy now, though I did read it some time ago, and I remember it as a book that covers the whole subject, right up to string theory.

    You really need to fully understand the laws of Special Relativity first …

    and for that, I'm sorry, but you do need to understand the basic equations.

    First, get any short book on Special Relativity out of the Public Library (they're all about as good as each other), and become familiar with it, and then read Greene. :smile:
     
  15. Jan 26, 2009 #14
    Or you can get a treatment of special and general relativity right from Einstein for free:

    http://www.bartleby.com/173/

    And for once in physics, the ideas/concepts are more difficult to understand than the math!!! Or at least less intuitive.

    I'm just now rereading SIX EASY PIECES ,a series of lectures by Richard Feynmann ,and he does an interesting job with vectors, relativity, etc. But you can't really beat Greene's explanations...it's a great introduction.
     
  16. Jan 26, 2009 #15
    yeah the book is very good so far but how much harder is the math compared to classical mechanics.
     
  17. Jan 27, 2009 #16
    alright well if someone asked you a simple classical mechanics question such as if a man travels at a speed of 100 m/s and travels for 1 second how far does he travel, does that kind of question need more details in relativity such as who was the observer because of time dilation or does it not matter.
     
  18. Jan 27, 2009 #17

    tiny-tim

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    No, so long as the same observer does all the measurements. :wink:
     
  19. Jan 27, 2009 #18
    but what if one person measured the time it traveled for and another different person measured the distance it traveled and than they came together to come up with a speed for the object that was moving. Than would the circumstances change since there was multiple observers recording the information to be used.
     
  20. Jan 27, 2009 #19
    It depends what you mean by "came together." It would be possible for two observers moving at different speeds to make their measurements and then have the two observers to return to Earth or something to compare results. The problem then is that one of the observers must have undergone an acceleration. When things accelerate, the symmetry of the situation is broken and we get a situation commonly explained as the Twin Paradox.
     
  21. Jan 28, 2009 #20

    tiny-tim

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    Yes, that would be, not a whole new ball-game, but two ball-games. :smile:

    and with two ball-games, there are no winners! :rolleyes:
     
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