5614 Posted March 28, 2005 Posted March 28, 2005 OK: A knows B is moving relative to A. B knows A is moving relative to B. As far as each of them are concerned one is moving. The fact that one is moving relative to the other surely means that one's clock is moving at a different speed relative to the other??? (I am not familiar with the light clock derivation of the time dilation formula)
Johnny5 Posted March 28, 2005 Posted March 28, 2005 The fact that one is moving relative to the other surely means that one's clock is moving at a different speed relative to the other??? There is nothing sure about this. This is perhaps the most complex problem in the history of theoretical physics. (I am not familiar with the light clock derivation of the time dilation formula) Well I am. So either go off and familiarize yourself with it, or have me explain it to you. Either or. (Somehow I think you're pulling my leg though, but no matter...) You will use the Pythagorean theorem in the derivation.
5614 Posted March 28, 2005 Posted March 28, 2005 OK, I got this: http://www.glenbrook.k12.il.us/gbssci/phys/Class/relativity/relderiv.html So go on with the light clock thing. And no I was not pulling your leg.... I am learning this and time dilation as we speak, hence I'm now asking the questions, since Swansont said "But they can only be running a different speeds to a third observer." it has confused me... the clocks are moving at different speeds relative to each other, therefore (time dilation) the clocks must be moving at different speeds too... Maybe you could answer them with the assumption that SR is correct?
Johnny5 Posted March 28, 2005 Posted March 28, 2005 Here is what I am going to do. I am going to derive the time dilation formula, using the light clock example. Then we can all see what assumptions go into the conclusion. Measuring the speed of light In order to measure the speed of something, we need to be able to measure the distance it travels, and the time of travel. If the object travels non-uniformly (meaning it speeds up and slows down during the measurement) then all we can compute is the average speed it has. But there is one special case of motion, called by Galileo uniform motion, and in this case the average speed is equal to the instantaneous speed. What Galileo meant by "uniform motion" is that at each successive moment in time, the object has the same speed it previously had. In other words, the speed of the object is constant, for the duration of the measurement process. Let the measurement take place in deep space, far from any significant gravitational fields, and let the measurement process take place in an inertial frame of reference. The measurement is going to be of the speed of a photon, emitted by a laser. Let there be a mirror located a distance D away from the laser. The distance D is measured by a ruler which is at rest in the measurement frame. For the sake of definiteness, let the unit of length be the meter. Here is how the measurement is going to be made. A photon is going to be fired at the mirror, hit the mirror, and be reflected back. The time of travel of this photon is going to be measured by a single clock at rest in the measurement frame, located next to the laser. So we can know two measurements with almost complete certainty, or if you prefer with almost complete accuracy. The total distance travelled by the photon is 2D (since the photon goes back and forth), and the total time of travel is measured by a clock at rest in the measurement frame. Since the experiment is taking place in the vacuum of space, the photon does not encounter any impedance as it moves through the vacuum. Postulate 1: The speed of the photon during this experiment is constant. After the experiment is over, the experimenter will have two numbers. One will correspond to the time of travel, lets say in units of seconds. Let Dt denote the amount of time in this frame (frame where laser is at rest). Now, the other number will correspond to the distance travelled by the photon in this frame, in the amount of time as measured by this clock. The instantaneous speed of the photon is: v1 = 2D / Dt Let the experiment have been done. Let the computation yield the following value: v1 = 299792458 meters/second So an event took place in that frame, which started at the moment the photon exited the laser, and ended the moment the photon returned to the starting point. The time of this event, in this frame is Dt Now, consider the exact same event, as viewed in a frame in which the laser gun is moving from left to right, at a constant speed V. Since the laser gun (and mirror) are moving to the right at constant speed V, the photon traverses a triangular path in this frame, specifically an isosceles triangle is traced out. The altitude of the triangle is D. However, the total distance traveled by the photon is the sum of the lengths of the sides of the isosceles triangle. In this frame, there is a ruler which is at rest, which also has units of meters, and two clocks which are at rest. Let us stipulate that the clocks are synchronized, they are of identical construction to the clock which is at rest in the frame in which the laser is at rest. Let us call these two clocks Clock A, and Clock B, and let us refer to the clock that is attached to the laser as Clock L. Now, at the moment the photon is fired out of the laser, let Clock A and Clock L coincide. At the moment the photon has returned round trip, let Clock B and clock L coincide. Let the reading on clock A at the moment the photon was fired, be denoted by T1. Let the reading on clock B at the moment the photon went round trip be denoted by T2. Since Clock A and Clock B are synchronized and tick at the same rate, the total time of travel in this frame is T2-T1 Definition: Dt` = T2-T1 We can break the isosceles triangle down into two right triangles. One of the legs of which has length D. The other leg has the following length: V Dt` / 2 Convince yourself of this, ask questions if you don't understand. Now, using the Pythagorean theorem, the hypotenuse H of this right triangle is related to the lengths of the legs as follows: H2 = D2 + (V Dt` / 2 )2 So the speed of the photon in this frame is given by: V2 = 2H/ Dt` Since H2 = D2 + (V Dt` / 2 )2 it follows that: 4H2 = 4 [ D2 + (V Dt` / 2 )2 ] Therefore: (2H) 2 = 4 [ D2 + (V Dt` )2/4 ] Therefore: (2H) 2 = (2D)2 + (V Dt` )2 Taking the square root of both sides we have: 2H = [ (2D)2 + (V Dt` )2 ]1/2 Therefore we have: V2 = [ (2D)2 + (V Dt` )2 ]1/2 / Dt` V2 is the speed of light in this frame, and V1 is the speed of light in the other frame. Since the other frame is an inertial reference frame, and this frame is moving uniformly relative to that, this too is an inertial reference frame. Recall: v1 = 2D / Dt Therefore: (V1)2 = (2D / Dt) 2 Squaring both sides of V2, we obtain: (V2)2 = [ (2D)2 + (V Dt` )2 ] / (Dt`)2 (V1Dt) 2 = (2D) 2 (V2)2 = [ (V1Dt) 2 + (V Dt` )2 ] / (Dt`)2 So we reach the following unconditionally true statement: (V2Dt`)2 = (V1Dt) 2 + (V Dt` )2 It involves the speed of light V1 in the rest frame of the laser. It involves the speed of light V2 in a reference frame in which the laser is moving at a constant relative speed V. It involves the time measurement Dt of this event E in the rest frame of the laser. It involves the time measurement Dt`of the exact same event, in a frame in which the laser is moving at constant speed V. At this point in the argument, we can either choose to invoke the fundamental postulate of the theory of special relativity, or not. Fundamental postulate SR: The speed of a photon in any inertial reference frame is 299792458 m/s. If we invoke the postulate, then V1=V2. Let us invoke the postulate, so that we can have shown how the postulate of SR leads to the time dilation formula. Assumption 1: V1=V2 Therefore: (V1Dt`)2 = (V1Dt) 2 + (V Dt` )2 Since we have assumed that there is a unique value for the speed of light in any inertial reference frame, let us set aside the symbol c, for this speed. Therefore we can write: (cDt`)2 = (cDt) 2 + (V Dt` )2 Since c^2 is nonzero, we can divide both sides of the statement above by c^2 to obtain: (Dt`)2 = (Dt) 2 + (V/c Dt` )2 From which it follows that: (Dt`)2 - (V/c Dt` )2 = (Dt) 2 From which it follows that: (Dt`) 2 [1 - (V/c)2 ] = (Dt) 2 Presuming that the quantity [1 - (V/c)2 ] is nonzero, we can divide both sides of the equation above by it, to obtain: (Dt`) 2 = (Dt) 2/ [1 - (V/c)2 ] Now, take the square root of both sides of the statement above, to obtain the following statement which will have the same truth value as the previous one: Dt` = Dt/ [1 - (V/c)2 ]1/2 The previous formula is the well known time dilation formula of SR. We can now remove the assumption, so that we have proven that the following statement is true: Theorem: If the speed of a photon is the same in any inertial reference frame then Dt` = Dt/ [1 - (V/c)2 ]1/2 Where the symbols have the meaning used in the derivation. QED Notice that if we assume that v1=v2 in the derivation above, we do not arrive at the result that: Dt` = Dt Delta t is the time of the event E in the rest frame of the laser, and Delta t` is the time of the same event, in a frame in which the laser is moving at a constant relative speed V. End of Derivation 1 So we have finally derived the time dilation formula from the fundamental postulate of the theory of special relativity. The next goal, is to show that if the fundamental postulate of the theory of SR is true, then the Lorentz contraction formula must be true. Let the length of the ruler in the frame in which the ruler isn't moving be denoted by L0. Regardless of the postulate of SR, the laser traversed the entire length of the ruler in the following amount of time in the rest frame of the ruler: Dt` And the laser was traveling at speed V in this frame. So, the total distance traveled by the laser in this frame is: V Dt` Which is just the length of the ruler in this frame, therefore: L0 = V Dt` Now, let us re-assume that the fundamental postulate of SR is true. Therefore: Dt` = Dt/ [1 - (V/c)2 ]1/2 Therefore: L0 = V Dt/ [1 - (V/c)2 ]1/2 Therefore: L0 [1 - (V/c)2 ]1/2 = V Dt Delta t is the time of event E in the rest frame of the laser, and V is the speed of any point on the ruler, in the rest frame of the laser. Let L denote the total distance traveled by any point on the ruler in the rest frame of the laser, during event E. Therefore we have: V Dt = L From which it follows that: L = L0 [1 - (V/c)2 ]1/2 The previous statement is called the Lorentz-Fitzgerald length contraction formula. We can now remove the assumption, so that we have proven that the following statement is true: Theorem: If the speed of a photon is the same in any inertial reference frame then L = L0 [1 - (V/c)2 ]1/2 Where the symbols get their meaning from the experiment. QED End of derivation 2 We can combine derivation one and two, as follows: Theorem: If the speed of a photon is the same in any inertial reference frame then (Dt` = Dt/ [1 - (V/c)2 ]1/2 AND L = L0 [1 - (V/c)2 ]1/2) QED Now think about how to interpret the length-contraction formula. At the beginning of the event, the point 0 on the ruler coincided with the laser. At the end of the event, the point L0 coincided with the laser. So, in the rest frame of the laser, the total distance traveled by the point 0 on the ruler was L, which is less than L0 Therefore, if the postulate of SR is true then the ruler is shorter in the rest frame of the laser. Keep in mind, that the length of the ruler L in this frame, is to be measured by a second ruler, which is at rest in the laser frame, and has the same units as the first ruler. Note that nothing in this derivation proved that the postulate of SR is true. However, we did succeed in proving the following fact: If either the length contraction formula or time dilation formula are false then the postulate of SR is false. Therefore, we can logically falsify SR, if either formula (LCF or TDF) leads to a contradiction. Let us make just one assumption. Assumption: The Length contraction formula is true. We will now encounter problems with simultaneity. Under the given assumption, it is provable that: X before Y and Y before X Where X,Y are different moments in time. Keep in mind that a moment in time doesn't last for any amount of time; consequently, the time dilation formula does not apply to moments in time, rather the time dilation formula applies to amounts of time. I will not prove this here. Regards
swansont Posted March 28, 2005 Posted March 28, 2005 Why not? A and B are the only things in the universe. Lets make a virtual frame just for this example... relative to this virtual frame which is about to disapear A is moving at 10mph and B at 20mph.... virtual frame just died... A and B still retain that speed. So to A it looks like B is going at 10mph And to B is looks like A is going at 10mph There's a speed difference in that one is moving relative to the other. Therefore isn't one of the clocks going to be going slower? A and B see each other moving at 10 mph. One value. There can be no other. Each one sees the other's clock run slow.
swansont Posted March 28, 2005 Posted March 28, 2005 Are you familiar with the light clock derivation of the time dilation formula? Are you? Do you agree that the light clock runs slow, according to the other observer?
Johnny5 Posted March 28, 2005 Posted March 28, 2005 Are you? Do you agree that the light clock runs slow, according to the other observer? I am aware of this... Lorentz contraction appears. In other words, you cannot have one without the other. One more thing, in my approach, I am going to reach a statement which is true, regardless of whether or not time dilates. So I am going to isolate the point in the derivation at which the postulate of relativity is invoked. It will be clear, its been a few years since i've ran through the derivation.
Guest Kceum Posted March 29, 2005 Posted March 29, 2005 Space and time are the same thing. It's not a 3 dimensional world plus time, it's a 4 dimensional world. An inertial frame is rotated by it's velocity which causes time and space to interchange but still encompass the same thing. The Lorentz transformations simply describe the relationship. By the way, time and age are not the same thing at all. Time and age are as alike as weight and mass.
Johnny5 Posted March 29, 2005 Posted March 29, 2005 Space and time are the same thing. This is false. It's not a 3 dimensional world plus time, it's a 4 dimensional world. I don't understand your statement. Space is three dimensional, in the sense that at most three infinite straight lines can meet at a point, and be mutually perpendicular. Not more than three. Time has no spatial extent, and hence it is not a dimension in any geometrical sense. An inertial frame is rotated by it's velocity which causes time and space to interchange but still encompass the same thing. The Lorentz transformations simply describe the relationship. This I do not understand. By the way' date=' time and age are not the same thing at all. Time and age are as alike as weight and mass.[/quote'] Weight and mass are not the same thing. Weight is a force, mass is not. You should try to think about how mass is measured. Go into space, and let there be a bowling ball there, which is at rest relative to you. Put an ordinary scale between your hand, and the bowling ball. Let there be a ruler aligned with the bowling ball, in the direction you are about to push. Now, push the ball in such a way, so that the reading on your scale is constant for amount of time Dt. Explain to me the acceleration of the bowling ball in the rest frame of the ruler please.
Johnny5 Posted March 29, 2005 Posted March 29, 2005 Do you agree that the light clock runs slow, according to the other observer? No.
Johnny5 Posted March 29, 2005 Posted March 29, 2005 Where does your analysis disagree? I concluded that the length contraction formula is false. Let two rulers have identical lengths when at rest with respect to one another. Let them be situated in a common inertial reference frame, and let the ends currently coincide with one another. Now, symmetrically push the rulers apart, then bring them to rest, so that they are still parallel, but separated in space by a trillion miles, and each had the same force exterted on it and so on. They are now both at rest in the original frame again, but are separated by a trillion miles. Now, make them head for each other by applying identical forces. Thus, they are now in relative motion, with relative speed V, in the original frame. Each is moving in a direction parallel to the length of the other, and they are eventually going to pass by one another. Think about things in the center of mass frame of the two rulers if you want. The speed of ruler 1 is V/2 towards the center of mass and its moving from left to right. The speed of ruler 2 is V/2 towards the center of mass and its moving from right to left. Ruler 1:A-----------B Ruler 2:___________A`----------B` There will come a moment in time, at which B will coincide with A`. That moment in time marks the beginning of an event, which ends when A coincides with A`. My question is, when A` coincides with A, does B` simultaneously coincide with B??? I say yes. Check the truth value of your statements in multiple frames. If simultaneity is absolute, then A` coincides with A simultaneously to B` coinciding with B in all frames of reference. Here swansont... Consider things from the rest frame of ruler two. Suppose the length contraction formula is true. Therefore, in this frame, Ruler one is length contracted, so that we can conclude that it's true that A coincides with A` before B coincides with B`. Now, consider things from the rest frame of ruler one. In this frame, Ruler two is length contracted, so that we can conclude that it's true that B` coincides with B before A` coincides with A. So we have concluded this: A` coincides with A before B` coincides with B, AND B` coincides with B before A` coincides with A. This conclusion came from assuming the length contraction formula is true. There's my problem. Regards
swansont Posted March 29, 2005 Posted March 29, 2005 I concluded that the length contraction formula is false. Non sequitur. The presence of disagreement in the light clock demonstration does not depend on length contraction. Length contraction never enters into the derivation.
swansont Posted March 29, 2005 Posted March 29, 2005 If simultaneity is absolute' date=' then A` coincides with A simultaneously to B` coinciding with B in all frames of reference.[/quote'] You can't assume simultaneity is absolute. That's circular reasoning. One of the consequences of relativity is that simultaneity is not absolute.
5614 Posted March 29, 2005 Posted March 29, 2005 A and B see each other moving at 10 mph. One value. There can be no other. Each one sees the other's clock run slow[/b']. EXACTLY!!!!! So why did you say that: "But they can only be running a different speeds to a third observer. To each other, the speeds must be the same." -- Swansont (post #54) Why must there be a third observer? The clocks travelling through space-time at different speeds are therefore ticking at different speeds.... one clock needs to look at the other to see that.... I understand this all, I did a while back, just when you said a third observer is needed it confused me! Did you really mean that 3rd observer is needed for one clock to see the other running slow?
swansont Posted March 29, 2005 Posted March 29, 2005 EXACTLY!!!!! So why did you say that: "But they can only be running a different speeds to a third observer. To each other' date=' the speeds must be the same." -- Swansont (post #54)[/i'] Why must there be a third observer? The clocks travelling through space-time at different speeds are therefore ticking at different speeds.... one clock needs to look at the other to see that.... I understand this all, I did a while back, just when you said a third observer is needed it confused me! Did you really mean that 3rd observer is needed for one clock to see the other running slow? No. You have observers A and B, moving with respect to each other. A sees B moving at v. B sees A moving at v. There can be no other speed observed by either of them. You cannot have A seeing B move at v, while B sees A move at v' - that's what I mean by saying you cannot observe two different speeds. The system is completely symmetric.
Johnny5 Posted March 29, 2005 Posted March 29, 2005 Originally Posted by Johnny5 I concluded that the length contraction formula is false. Non sequitur. The presence of disagreement in the light clock demonstration does not depend on length contraction. Length contraction never enters into the derivation. Before going any further, did you follow my derivation of the time dilation formula? If you had an objection to any step, I want to hear it.
5614 Posted March 29, 2005 Posted March 29, 2005 No. You have observers A and B' date=' moving with respect to each other. A sees B moving at v. B sees A moving at v. There can be no other speed observed by either of them. You cannot have A seeing B move at v, while B sees A move at v' - that's what I mean by saying you cannot observe two different speeds. The system is completely symmetric.[/quote'] Yeah, I know! It just clicked....... "But they can only be running a different speeds to a third observer. To each other, the speeds must be the same." You meant the clocks moving at different speeds as in moving in distance or different mph...... I thought you meant running at different speeds as in the clock was ticking at different speeds. So I thought you meant that you could only see the clocks ticking different speeds (time dilation) if you were a third observer... you meant that they only move in distance at different rates for a third observer which is of course correct! OK, that's sorted... good, I still get time dilation!.... and it's freaky, I can remember that time dilation formula off by heart and I just saw it yesterday for the first time!
Johnny5 Posted March 29, 2005 Posted March 29, 2005 You can't assume simultaneity is absolute. That's circular reasoning. One of the consequences of relativity is that simultaneity is not absolute. I didn't assume that simultaneity is absolute, actually I know what I did, and I know I didn't circularly reason. It will be advantageous to just say it... Suppose that you are an observer permanently stationed in inertial reference frame F. Now, there is a meterstick which you have which is made out of wood. It is your favorite meterstick, and you had it engraved with your name and put a nice coat of varnish on it and what not. So you know how long it is when it isn't moving relative to you. And your friend has an almost identical meterstick, varnished and all, but he has his name engraved on his. When these two rulers are placed side by side they have the exact same length, as far as sensory perception can tell. Now, you are stuck in this inertial reference frame F forever. Here comes the reasoning to be analyzed: Assumption 1: The Lorentz contraction formula is true, in any inertial reference frame. Since you are in an inertial reference frame, the Lorentz-contraction formula applies to any and all wooden rulers in your frame. It even applies to steel rulers, and titanium rulers as well, but you don't own one of them. So since the LCF is true in this frame which you are in, any and all rulers which are at rest in this frame, will have their maximum possible length, which is called their proper length, or rest length. Now, if for any reason whatsoever, one of these rulers is moving through your inertial frame F, instead of being at rest in F, its length will be less than it's proper length, since we have assumed that the LCF is true. Let us presume that your friends wooden meterstick with his name on it, is moving from left to right, along the x axis of reference frame F, with speed V as defined using the coordinates of frame F, and clocks at rest in frame F. If V is sufficiently large, your friends meter stick is now .25 meters, instead of one meter. Your meterstick with your name on it is at rest in your hand, and you are using it to measure the length of his meterstick when it passes by you. So here is where simultaneity problems arise, I tried to show it somewhere else, but its the same argument over and over again, so here it comes: Ruler 1: A__B Ruler 2:......A`____________B` Then later Ruler 1:......A__B Ruler 2:......A`____________B` Then later Ruler 1:........................A__B Ruler 2:......A`____________B` So because your friends ruler was length contracted, A` coincided with A before B` coincided with B. If instead we use the Galilean transformations, there would have been a moment in time, in your reference frame F where things would have been this way: Ruler 1:......A_____________B Ruler 2:......A`____________B` In other words the ends would have coincided simultaneously. But because we assumed that your friends ruler contracts if it has a nonzero speed V in frame F, then as your friends ruler passes by yours, it is impossible for the ends to coincide simultaneously. So there is a reductio ad absurdum proof that can be used (not circular reasoning), which can allow you to conclude that the LCF is false. I haven't shown it all yet. But I've shown enough where someone else can figure out the problems caused by assuming LFC is true.
Johnny5 Posted March 29, 2005 Posted March 29, 2005 No. You have observers A and B' date=' moving with respect to each other. A sees B moving at v. B sees A moving at v. There can be no other speed observed by either of them. You cannot have A seeing B move at v, while B sees A move at v' - that's what I mean by saying you cannot observe two different speeds. The system is completely symmetric.[/quote'] This is correct.
5614 Posted March 29, 2005 Posted March 29, 2005 Yeah I know, I just thought he meant speed of the clock ticking, not the physical speed the clock was travelling through space at... confusion over!
Johnny5 Posted March 29, 2005 Posted March 29, 2005 Yeah I know, I just thought he meant speed of the clock ticking, not the physical speed the clock was travelling through space at... confusion over! With this problem, for some reason the confusion is never over... Kind regards PS: Someplace I called it the most complex problem in the history of theoretical physics... I wasn't kidding.
J.C.MacSwell Posted March 29, 2005 Posted March 29, 2005 ...................................................................... If instead we use the Galilean transformations' date=' there would have been a moment in time, in your reference frame F where things would have been this way: Ruler 1:......A_____________B Ruler 2:......A`____________B` In other words the ends would have coincided [i']simultaneously[/i]. But because we assumed that your friends ruler contracts if it has a nonzero speed V in frame F, then as your friends ruler passes by yours, it is impossible for the ends to coincide simultaneously. So there is a reductio ad absurdum proof that can be used (not circular reasoning), which can allow you to conclude that the LCF is false. I haven't shown it all yet. But I've shown enough where someone else can figure out the problems caused by assuming LFC is true. This proves that, assuming LCF and your other assumptions are correct, that Galilean transformatons (especially at those speeds) are incorrect.
Johnny5 Posted March 29, 2005 Posted March 29, 2005 This proves that, assuming LCF and your other assumptions are correct, that Galilean transformatons (especially at those speeds) are incorrect. How so? I haven't finished the argument yet.
J.C.MacSwell Posted March 29, 2005 Posted March 29, 2005 How so? I haven't finished the argument yet. LCF system:2 ends cannot coincide simultaneously Galilean: 2 ends can coincide simultaneously
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