Guest cephiyr Posted April 2, 2005 Share Posted April 2, 2005 I have a question on a part of the relativity theory that seems paradox to me. As far as I understand, a space craft travelling away from Planet Earth at high speed (i.e. close to c) will increase in mass and move slower through time. Thus the Astronaut onboard will age slower and when he gets back, everyone around him will have aged quicker. What I dont understand is, how we can tell that Planet Earth isnt actually the one travelling close to c. What I mean is, since speed is relative to other objects in space, how do we know that it is the astronaut that will age quicker ? Wouldn't it be possible that we see Earth moving away from the space ship at almost c, thus letting time move quicker on the space craft than on Earth. In my understanding there is no absolute reference frame in the theory of relativity, which makes me wonder how we can actually tell who is moving faster than somebody else at all ? Isnt it all relative ? How can our point of view determine whose time goes slower and who exerts more gravity ? I am sorry if this sounds quite like a laymans issue. I would much appreciate an explanation on this. Regards, Mark-Alexander Link to comment Share on other sites More sharing options...

5614 Posted April 2, 2005 Share Posted April 2, 2005 As you say, speed is relative. If there was only me and you in the universe and the distance between us was chaning, how could we tell if: (1) I'm moving (2) You're moving (3) Both of us are moving??? You can't for certain. If the gap increases I may be moving away from you, you may be moving away from me, we both might be moving away from each other... we cannot tell. So we need a 3rd person to look at the both of us and see who is really moving.... but what if that 3rd person is moving at exactly my speed.... to the 3rd person I'm not moving. Put your hand in front of your face and move your hand and your face in the same direction at the same speed... your hand stays right infront of your face... you can only tell that you are moving by looking at the thing (for me a computer screen) in front of you and you assume that that object is at rest, ie. it is not moving... if you do that again you dont assume that the screen is moving and you are still, you assume that the screen is at rest. Similarly scientists have to take what can be assumed as a rest frame (ie. an object which is "still".) A car going down the road... the road is assumed to be still, you don't say I was driving at 10,000mph in my Mini because you are saying your speed relative to the road surface, not the asteroid a few light years away travelling at 10,000mph! It is all relative. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted April 2, 2005 Share Posted April 2, 2005 Keep in mind the Universe cannot be defined as a frame in this sense. It includes all frames. Link to comment Share on other sites More sharing options...

swansont Posted April 2, 2005 Share Posted April 2, 2005 In my understanding there is no absolute reference frame in the theory of relativity' date=' which makes me wonder how we can actually tell who is moving faster than somebody else at all ? Isnt it all relative ? How can our point of view determine whose time goes slower and who exerts more gravity ?[/quote'] They both can say they are stationary and the other one is moving. Both see the other's clock slow down. And that's valid - until one of them accelerates. As far as the mass goes, we've had discussion on this (you can use the search function). But does it make sense that gravity for something else should increase because you are moving? The relevant value for gravity is the rest mass. Link to comment Share on other sites More sharing options...

[Tycho?] Posted April 2, 2005 Share Posted April 2, 2005 But does it make sense that gravity for something else should increase because you are moving? The relevant value for gravity is the rest mass. To be fair, in the realm where time and length can be different when viewed from different frames, it is rather difficult to discern what "makes sense" and what doesn't. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 2, 2005 Share Posted April 2, 2005 I have a question on a part of the relativity theory that seems paradox to me.As far as I understand' date=' a space craft travelling away from Planet Earth at high speed (i.e. close to c) will increase in mass and move slower through time. Thus the Astronaut onboard will age slower and when he gets back, everyone around him will have aged quicker. What I dont understand is, how we can tell that Planet Earth isnt actually the one travelling close to c. What I mean is, since speed is relative to other objects in space, how do we know that it is the astronaut that will age quicker ? Wouldn't it be possible that we see Earth moving away from the space ship at almost c, thus letting time move quicker on the space craft than on Earth. In my understanding there is no absolute reference frame in the theory of relativity, which makes me wonder how we can actually tell who is moving faster than somebody else at all ? Isnt it all relative ? How can our point of view determine whose time goes slower and who exerts more gravity ? I am sorry if this sounds quite like a laymans issue. I would much appreciate an explanation on this. Regards, Mark-Alexander[/quote'] Mark, suppose that two twins are in a space station which is in orbit about the earth. Now, each is floating weightless, and they are in the same inertial reference frame, and they are aging equally rapidly. Let each of them have an identical wristwatch on. Now, let one of the twins accelerate in some direction. Thus, one of the twins is still at rest in the original inertial reference frame S. Let the direction of the acceleration in this frame be in the i^ direction. That is the direction of increasing x coordinate in the frame. Now, don't worry about the direction of the accleration anymore, focus on the magnitude of the acceleration. Denote the magnitude of the acceleration by a. The magnitude of the acceleration is the time rate change of speed in the frame. Speed in a frame is distance traveled/ corresponding time of travel Let v denote the instantaneous speed of the twin who is accelerating in the original frame, frame S. Therefore: [math] a = \frac{dv}{dt} [/math] The equation above is a scalar equation, we aren't worrying about direction of relative motion at all right now. Now, here is the time dilation formula: [math] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/math] Let S` denote a reference frame which is permanently attached to the twin who accelerates in frame S. Thus, the speed of frame S` in S is the v in the formula above. The symbol c denotes the speed 299792458 meters per second. Now, consider an event which begins when the twins are together, and ends when the twins are 50 light years apart. According to the time dilation formula: Dt is the amount of time that passes according to the watch at rest in S. Dt` is the amount of time that passes according to the watch at rest in S`. As you can see, they are not equal according to SR. Specifically more time passes in S, than passes in S` for the same event. Now, the formula above does not involve the acceleration a of the ship, it only makes reference to the instantaneous relative speed v. So if you look at things from the point of view of the twin in frame S`, he sees his brother accelerate away from him at speed v as well. So yes there is a paradox created. The logical conclusion is this... If the time dilation effect can happen, it can happen in at most one of these two frames, not both. In other words, the formula cannot be true in both reference frames. Regards Link to comment Share on other sites More sharing options...

Janus Posted April 3, 2005 Share Posted April 3, 2005 Mark' date=' suppose that two twins are in a space station which is in orbit about the earth. Now, each is floating weightless, and they are in the same inertial reference frame, and they are aging equally rapidly. Let each of them have an identical wristwatch on. Now, let one of the twins accelerate in some direction. Thus, one of the twins is still at rest in the original inertial reference frame S. Let the direction of the acceleration in this frame be in the i^ direction. That is the direction of increasing x coordinate in the frame. Now, don't worry about the direction of the accleration anymore, focus on the magnitude of the acceleration. Denote the magnitude of the acceleration by a. The magnitude of the acceleration is the time rate change of speed in the frame. Speed in a frame is distance traveled/ corresponding time of travel Let v denote the instantaneous speed of the twin who is accelerating in the original frame, frame S. Therefore: [math'] a = \frac{dv}{dt} [/math] The equation above is a scalar equation, we aren't worrying about direction of relative motion at all right now. Now, here is the time dilation formula: [math] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/math] Let S` denote a reference frame which is permanently attached to the twin who accelerates in frame S. If the twin accelerates then S' is not an inertial reference frame and remain permanently attached to the twin. If it is an inertial refernce frame it cannot remained attached to twin. Thus, the speed of frame S` in S is the v in the formula above. The symbol c denotes the speed 299792458 meters per second. Now, consider an event which begins when the twins are together, and ends when the twins are 50 light years apart. According to the time dilation formula: Dt is the amount of time that passes according to the watch at rest in S. Dt` is the amount of time that passes according to the watch at rest in S`. As you can see, they are not equal according to SR. Specifically more time passes in S, than passes in S` for the same event. only as measured by S' Now, the formula above does not involve the acceleration a of the ship, it only makes reference to the instantaneous relative speed v. So if you look at things from the point of view of the twin in frame S`, he sees his brother accelerate away from him at speed v as well. No, he can easily tell that he was the one who actually accelerated. He can feel the acceleration, he can fire a laser perpendicular to his path and notes that it curves, etc. And while the time dilation formula holds when used by either twin while they are at constant velocity, and holds for the twin that does not accelerate (remains in the same reference frame the whole time.), It is incomplete for the twin who undergoes acceleration while he is accelerating. For him he has to add an additional transformation that takes into account his acceleration and the distance between himself and his brother. So yes there is a paradox created. No paradox as long as you take all the consequences of Relativity into account, including length contraction and Relativity of simulataneity. The logical conclusion is this... If the time dilation effect can happen, it can happen in at most one of these two frames, not both. In other words, the formula cannot be true in both reference frames. Regards No, it is not. Again you must take all the effects involved into account to properly address this. In fact, when you do take them into account, the time dilation formula must be applied in [/i]all[/i] frames in order for there not to be a paradox. I would strongly urge you not to try and explain Relativity to others, you do not grasp it properly yourself. Link to comment Share on other sites More sharing options...

swansont Posted April 3, 2005 Share Posted April 3, 2005 The logical conclusion is this... If the time dilation effect can happen' date=' it can happen in at most one of these two frames, not both. In other words, the formula cannot be true in both reference frames. [/quote'] That's not the logical conclusion. You've assumed some things that you haven't stated. All you can logically conclude is that if the two observers witness some event, they will disagree on when that event occurred. I second the suggestion by Janus. You should not attempt to teach anyone physics that you do not understand nor accept. Link to comment Share on other sites More sharing options...

swansont Posted April 3, 2005 Share Posted April 3, 2005 '']To be fair, in the realm where time and length can be different when viewed from different frames, it is rather difficult to discern what "makes sense" and what doesn't. A valid objection. I should have phrased it in terms of mass being an intrinsic property, and can't depend on an external observer's motion. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 3, 2005 Share Posted April 3, 2005 If the twin accelerates then S' is not an inertial reference frame and remain permanently attached to the twin. If it is an inertial refernce frame it cannot remained attached to twin. It is obvious that S` is not an inertial reference frame' date=' since Newton's third law is untrue in frame S`. I never claimed that S` was an inertial reference frame. I know that it isn't so you misunderstood something i said. Moving on... S` is the rest frame of the spaceship. The origin of S` is the center of inertia of the ship. The center of inertia of the ship is permanently at rest in reference frame S`, so that it MUST be permanently attached to the ship. This was stipulated to be true. Moving on to your next objection... No, he can easily tell that he was the one who actually accelerated. He can feel the acceleration.... First of all, force and accleration are not equivalent; force and acceleration are not the same thing. The man in the ship feels force, not acceleration. It is incorrect to say that he is the one who can tell that he accelerated. Observers in both reference frames, see the other as accelerated. Instead, it is correct to say that he can tell that his spaceship is being subjected to an external force. One way for him to infer that he is in a non-inertial reference frame, is to see that his ships engines are on, and from that he can infer that Newton's third law is untrue in frame S`. Therefore, he can infer (from inside the ship) that his ship's rest frame is a non-inertial frame. So if he looks outside the window and sees a moon in free space, accelerating without an action/reaction pair, he will hardly be surprised. (Where I have assumed his ship is using a conventional propulsion system of course). he can fire a laser perpendicular to his path and notes that it curves, etc. This is a very complicated problem you know. Suppose that S` is an inertial reference frame. Then, it would be the case that the center of inertia of his ship is at rest in an inertial reference frame. Therefore, when he turns on the laser, we should expect the light beam to move in a straight line, away from the emitter. And by experiment, we know the speed of this is going to be 299792458 meters per second. Suppose there is a moon at rest in free space, relative to a point a distance D away from it. Now, let R denote the vector from the center of inertia of the moon, to this point. Let the direction of travel of the spaceship be in a straight line which is perpendicular to R. So this is a very simple situation to visualize. As the ship's center of inertia moves along this straight line, the center of inertia is accelerating in the rest frame of this moon. (There are no external forces acting upon the moon, therefore the rest frame of the moon is an inertial reference frame). Now, keep in mind that S` is really a non-inertial reference frame. Now, at the moment that the rocket is located at the tip of R, let the laser be turned on. The goal is to now find the path of the photon in reference frame S, as well as reference frame S`. Now, focus on the plane which contains the straight line which the rocket is moving along, and the center of inertia of the moon. Keep in mind that any straight line, and any point not on that line, lie in one and only one plane. Furthermore, R lies in this plane as well, and the length of R is D, as measured by a ruler at rest in frame S. Let R lie on the positive y axis of frame S. Thus, D is the perpendicular distance from the line of travel of the rocket, and a point not on that line (the center of inertia of the moon). Here is the Euclidean construction for R... Construction for the shortest path from a point to a line In fact, we can now write R in terms of the j^ vector of reference frame S. [math] \vec R = D \hat j [/math] Now, let the X axis of reference frame S be parallel to the line of rectilinear motion of the center of inertia of the rocket. Now, let the i^ vector of reference frame S point in the direction of motion of the center of inertia of the rocket. I wish I could draw this. Here is what I will do, I will use Euclid's drawing found here. Let the center of inertia of the moon be permanently located at the point C, in Euclid's figure. As I previously stipulated, the point H in Euclid's figure is a distance D away from the point C in Euclid's figure. When the laser on board the rocket is located at point H, the laser is going to fire a photon in the direction of the vector which points from H to C (in Euclid's figure). Also, the vector R points from C in Euclid's figure, to H in Euclid's figure. Now, in Euclid's drawing, let the direction of motion of the rocket be from A to B. Thus, the motion of the center of inertia of the rocket lies entirely along the infinite straight line through points A,B. In fact, I will introduce a third frame, which is at rest in S, which I will refer to as the Euclidean frame. The origin of the Euclidean frame is the point H in Euclid's figure, and the point C lies on the positive y axis of the Euclidean frame, and the line through AB coincides with the x axis of the Euclidean frame, and the i^ direction of the Euclidean frame points from A to B. Since S is an inertial frame, and at least three noncollinear points of the Euclidean frame are at rest in S, it must be the case that the Euclidean frame is also an inertial reference frame. Let the i^ vector of S point in the same direction as the i^ vector of the Euclidean frame. Now, the origin of frame S` is permanently attached to the center of inertia of the rocket. Let the j^ vector of frame S` point in the same direction as the j^ vector of the Euclidean frame, and let the i^ vector of frame S point in the same direction as the i^ vector of the Euclidean frame. At this point, we should be able to determine the path of the photon in frame S, as well as frame S`. Since the object being fired is a photon, this complicates things quite a bit. For either one will hold that the speed of a photon is c in all inertial frames, or another will hold that the speed of a photon c is relative to the emitter. But in this problem here, the emitter is being subjected to an external force, prior to the laser being turned on. Hence the rest frame of the emitter is a non-inertial reference frame. To complicate matters, there is no general agreement on whether or not the mass of a photon is zero. The general concensus is that the "rest mass" of a photon is zero, but that the gravitational mass of a photon is non-zero. For now, let us suppose that the photon has mass. In fact, let the mass of the photon fired be quite substantial. In other words, think of the photon which is fired towards the center of the moon, as having been fired by a cannon. Thus, there will be an impulse on the ship at the moment in time at which the center of inertia of the ship is located at the point H in Euclid's figure. And that impulse will be in the direction of R, which is the same direction as the -j^ vector of the Euclidean frame. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 Yesterday, I was trying to find the trajectory of a photon fired at a right angle to the direction of the acceleration of a craft in the rest frame S` of the craft, as well as its path in the inertial rest frame of a moon, which the craft was flying by. As I recall, the problem isn't simple. The point I left off at, was right at the moment that the laser fired the photon at the center of mass of the moon. Here is the diagram that I was using: Euclid Book I, proposition 12 In the diagram C corresponded to the center of inertia of the moon, and the rest frame of the moon was considered to be an inertial reference frame, and I was referring to that frame as frame S. As I recall the rocket was moving along the line AB, and the direction of its acceleration a, in frame S was from A to B. The origin of S was the center of inertia of the moon. The point H was on the positive y axis of S. And the X axis of S was parallel to straight line AB. And the i^ vector of frame S pointed from left to right in Euclid's diagram. Then I defined another inertial frame which I was just referring as Euclid's frame. It was a frame in which at least three non-collinear points in it were at rest in frame S, from which we can also infer that Euclid's frame was also inertial. Keep in mind that frame S was stipulated to be inertial. The origin of Euclid's frame is the point H in Euclid's figure, the X axis of Euclid's frame is the line AB in Euclid's figure, the i^ direction of Euclid's frame was from A to B, and the point C lies on the positive y axis of Euclid's frame. And there was one other frame, S`, which is the rest frame of the rocket ship. Now, the speed of the rocket in frame S is changing in time, but the direction of motion of the rocket is along line AB. Now, there comes a moment in time when the rocket is at the point H in Euclid's figure. Right at that moment in time, the laser fires a photon right at the center of inertia of the moon, which is the point C in Euclid's figure. This is where I left off at yesterday. Now, if the photon is extremely massive, then there will be a substantial force on the rocket, which would cause frame S` to deviate off of it's straight line path AB. And that of course will affect the mathematical solution to the path of the photon in frame S`, after it is emitted from the laser. But, in reality, the inertial mass of the photon is negligible, in comparison to the mass of the rocket. In theory, the mass of the photon is literally zero, although there is no general consensus on this. But the reality is, that to a good approximation, the emission of a single photon by the laser, will hardly affect the direction of the acceleration of the rocket in frame S. So, for now, we can totally ignore this effect, and treat the motion of the center of mass of the rocket as constrained to move along line AB even after the photon is emitted from the laser (which is mounted to the rocket). But what is certainly true is this, at the moment that the laser emits the photon, that direction is at a right angle to the initial direction of the rocket through frame S. And the photon is emitted at a moment in time when the center of inertia of the rocket is at point H, in relation to the point C (the point C is permanently the center of inertia of a moon). So now the goal is to find the path of the photon in reference frame S, and reference frame S`. We need to keep in mind that S` is a non-inertial frame. Well think of things before the photon is emitted. Imagine that you are holding a baseball in your hand, as your rocket ship is accelerating along line AB. There is a force upon your hand. Then right when you pass the point H, release the baseball. The mass of the rocket + fuel just decreased by one baseball mass. But here is the thing to note. The moment that you release the baseball, the baseball goes from being in a non-inertial reference frame to being in an inertial reference frame. Can anyone else see that? The baseball will now be moving in a straight line at a constant speed, in reference frame S, in the direction from A to B. And had it been thrown at the center of inertia of the moon, instead of merely released, it would still be moving in a straight line at constant speed in frame S, only this time the direction would have been a diagonal line, in the first quadrant of the Euclidean frame. In other words, let v denote the velocity of the baseball in Euclid's frame at the moment it was thrown from the rocket straight (at point H) right at the center of inertia of the moon. [math] \vec V = v_1 \hat i + v_2 \hat j [/math] So now, the ship was accelerating at the moment when the baseball is thrown, and therefore the baseball had some initial speed v1 in the positive i^ direction, and the baseball was thrown right at the moon, so that at that same moment, the baseball was given some speed v2 in the positive j^ direction of Euclid's frame. So the path of the baseball in Euclid's frame, is going to be a diagonal line, in the first quadrant, and it's slope will be: [math] m = \frac {v2}{v1} [/math] Rise/run In other words, the dt's cancel out, and because of the way I have set up the problem we have: [math] m = \frac {v2}{v1} = \frac{dx}{dy} [/math] Now, what started me thinking about this, was the fact that swansont said the path of the emitted photon in the rest frame of the rocketship was curved. What I really want to do is find the formula for that curve, in the rocket frame, which I know is a non-inertial frame. Somewhere in this argument, I am going to fix the speed of a photon relative to its emitter as c in the center of mass frame. Since the ship is so much more massive then the photon, we are considering that the CM of the ship remains traveling on line AB, even after the photon is emitted. Ok swansont, I think I've worked out the basic idea of the path of the photon in the non-inertial reference frame of the ship. I know that once the photon is emitted, it's going to move in a diagonal line in frame S, and the Euclidean frame. But, the rocket is continually ejecting mass, and therefore accelerating along the line AB, and I've not yet specified the details of it. Consider the simplest case non-zero constant acceleration. Let the the rate of acceleration be g = 9.8 meters per second. So we can think of the rocket in free-fall, in the direction from left to right. Thus, in equal amounts of time, it's jumping larger and large distances, and the sum of those distances is the sum of squares... like this... X-X----X---------X----------------X-------------------------X And this change in distance is obeyed in the S frame, and Euclid's frame. So that means approximately this... In the rest frame of the ship, frame S`, the photon appears to have an acceleration g in the -i^ direction, and a constant speed v, in the j^ direction. So, if I'm not mistaken, the path of the photon after emission, in the rest frame of the rocket, frame S`, is a parabola, in the second quadrant of frame S`. Now I could have easily made a mistake, but I just wanted to check what Janus said, which was, let me go find the quote, so I get it right... he can fire a laser perpendicular to his path and notes that it curves, etc. Yes ok I agree with that, the path is a parabola. The next thing Janus says, is this: And while the time dilation formula holds when used by either twin while they are at constant velocity, and holds for the twin that does not accelerate (remains in the same reference frame the whole time.) This I didn't understand Janus. The next thing Janus says is this: It is incomplete for the twin who undergoes acceleration while he is accelerating. For him he has to add an additional transformation that takes into account his acceleration and the distance between himself and his brother. Here, it actually seems to me, as though you are arguing my point, which is that the time dilation formula is a false statement in reference frame S`. I do believe that I said this..., "If the time dilation effect occurs, then it is true in at most one of the two frames, not both." Janus then says this: No paradox as long as you take all the consequences of Relativity into account, including length contraction and Relativity of simulataneity. Well yes there is a paradox created if you believe that the formula is true in both frames, which is what I told this person. The time dilation formula cannot be a true statement in both frames. If it is true in at least one of the frames, then it is true in at most one of the frames, in this twin paradox example. From which it follows that it is true in exactly one of the frames. But if SR is totally wrong, then it's true in neither of the frames. So the best thing to say is this... In the twin paradox problem, the time dilation formula is true in at most one of the two frames. This leaves open the possibility that it's false in both frames, but closes off the possibility that it's true in both frames. Lastly, you say this... No, it is not. Again you must take all the effects involved into account to properly address this. In fact, when you do take them into account, the time dilation formula must be applied in [/i]all[/i] frames in order for there not to be a paradox. I would strongly urge you not to try and explain Relativity to others, you do not grasp it properly yourself. You are going to confuse that person, because the time dilation formula cannot be true in both frames, that is impossible, because then this is what is happening... Since the formula is true in frame S in which twin one is at rest, Then after one year aboard the ship, twin one is ten years older than twin two. Suppose that when the trip begins each twin is age 30. Since the time dilation formula is true, there is a moment in time at which twin one is age 31, and twin two is 40. Now, suppose that the time dilation formula is true in frame S`, which is the non-inertial frame in which twin two is at rest. Since the time dilation formula is true, when twin two is age 31, it must be the case that twin 1 is aged forty. Therefore, twin one reaches his fortieth birthday before twin two reaches his fortieth birthday, and twin one does not reach his fortieth birthday before twin two reaches his fortieth birthday. So the time dilation formula cannot be true in both frames, which is all I told this person. Kind regards Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 That's not the logical conclusion. You've assumed some things that you haven't stated. All you can logically conclude is that if the two observers witness some event' date=' they will disagree on when that event occurred. I second the suggestion by Janus. You should not attempt to teach anyone physics that you do not understand nor accept.[/quote'] But I do understand it, I gave the person the right answer. Even by your way of thinking... so i fail to see the problem. What have I assumed, that I haven't stated? Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 But I do understand it' date=' I gave the person the right answer. Even by your way of thinking... so i fail to see the problem. What have I assumed, that I haven't stated?[/quote'] No, you didn't. Given the symmetry of the situation, how can the formula only work in one reference frame? I'm not sure what you assumed, but your conclusion does not follow. You stated a few things, and then the conclusion appeared. The proof is missing. Why can't each observer measure the other's clock as being slow? Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 You are going to confuse that person' date=' because the time dilation formula cannot be true in both frames, that is impossible, because then this is what is happening... Since the formula is true in frame S in which twin one is at rest, Then after one year aboard the ship, twin one is ten years older than twin two. Suppose that when the trip begins each twin is age 30. Since the time dilation formula is true, there is a moment in time at which twin one is age 31, and twin two is 40. Now, suppose that the time dilation formula is true in frame S`, which is the non-inertial frame in which twin two is at rest. Since the time dilation formula is true, when twin two is age 31, it must be the case that twin 1 is aged forty. Therefore, twin one reaches his fortieth birthday before twin two reaches his fortieth birthday, and twin one does not reach his fortieth birthday before twin two reaches his fortieth birthday. So the time dilation formula cannot be true in both frames, which is all I told this person. [/quote'] Why do both measurements have to agree? You seem to have assumed that thay have to. The whole upshot of relativity is that the measurements you make depend on your frame of reference. There are no absolute measurements of time and distance. If you assume absolute reference frames, you find that relativity doesn't hold. That's your contradiction. One of those two is false, but you haven't shown your assumption to be true. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 No' date=' you didn't. Given the symmetry of the situation, how can the formula only work in one reference frame?[/quote'] Swansont, you know better than this. The two frames are not symmetrical. They are kinematically equivalent, as regards relative speed, and relative acceleration. They are not dynamically equivalent though. In the one frame, the net external force upon the twin is zero. In the other frame, the net external force upon the twin is greater than zero. Hence, the two frames are not totally equivalent. There is an enormous difference between the frames. One of them is an inertial reference frame, the other is not. Regards Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 I'm not sure what you assumed' date=' but your conclusion does not follow. You stated a few things, and then the conclusion appeared. The proof is missing. Why can't each observer measure the other's clock as being slow?[/quote'] I try to assume as little as possible swansont, always assume as little as possible. As a matter of fact, the best proofs occur when you assume nothing at all (these kind of proofs are what empirical science is based on... i believe they are called experiments. To prove to yourself that two things fall at the same rate... DROP THEM... no reasoning necessary ). The next best proof, has only one simple assumption in it, and ends in contradiction. Also, if what you assume happens to be true, though you don't know it's true, you wont reach a contradiction to it, presuming your argument is logical, but by the same token, you will never figure out that the assumption was true, since you assumed it. If you thought that you concluded your assumption was true, that's what everyone calls circular reasoning, and they are right, circular reasoning is no good. Yes the proof is missing. As to why each observer cannot measure the other guys clock as slow, you keep running into a temporal paradox of the following kind: X before Y and Y before X and not (X=Y) Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 Why do both measurements have to agree? You seem to have assumed that thay have to. Both measurements of what the amount of elapsed time? The whole upshot of relativity is that the measurements you make depend on your frame of reference. There are no absolute measurements of time and distance. Yes, that is what relativity says. I understand relativity to the point where I know that is what it says. If you assume absolute reference frames, you find that relativity doesn't hold. That's your contradiction. One of those two is false, but you haven't shown your assumption to be true. What assumption? Link to comment Share on other sites More sharing options...

5614 Posted April 4, 2005 Share Posted April 4, 2005 I try to assume as little as possible swansont, always assume as little as possible. As a matter of fact, the best proofs occur when you assume nothing at all (these kind of proofs are what empirical science is based on... i believe they are called experiments. To prove to yourself that two things fall at the same rate... DROP THEM... no reasoning necessary ) Johnny5, I like your idea of "don't stop until you learn all the physics in the world" and I like the idea of learning physics and understanding it and all.... I personally just don't like the way you go about it [e.g. in the above quote]. For you to understand physics you are putting aside all of physics itself in aim of you starting from the beggining yourself... This is just not possible. Einstein, Tesla, Newton and hundreds if not thousands of others worked in teams of physicists to arrive at a single part of what I would now call the humans knowledge of physics. I don't think it is sensible of you to reject all of known physics in the hope of learning it yourself. You should accept proven physics theories... and sure they are called "theories" but they are (in todays modern world) tested extensively. Sure, so Newton said "no" to what was 'known' at the time and he's famous and all.... but that was a long time ago, in todays modern world there's almost certainly no place to correctly say "no that's wrong". Maybe try accepting GR and SR etc and trust the most famous physicists in the world to correctly derrive a formula and spend that time learning something that you don't know! (As opposed to deriving, proving or arguing against already proven and extensively tested theories that have been around for years.) Still friends, just my opinion Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 Johnny5' date=' I like your idea of "don't stop until you learn all the physics in the world" and I like the idea of learning physics and understanding it and all.... I personally just don't like the way you go about it [e.g. in the above quote']. For you to understand physics you are putting aside all of physics itself in aim of you starting from the beggining yourself... This is just not possible. Einstein, Tesla, Newton and hundreds if not thousands of others worked in teams of physicists to arrive at a single part of what I would now call the humans knowledge of physics. I don't think it is sensible of you to reject all of known physics in the hope of learning it yourself. You should accept proven physics theories... and sure they are called "theories" but they are (in todays modern world) tested extensively. Sure, so Newton said "no" to what was 'known' at the time and he's famous and all.... but that was a long time ago, in todays modern world there's almost certainly no place to correctly say "no that's wrong". Maybe try accepting GR and SR etc and trust the most famous physicists in the world to correctly derrive a formula and spend that time learning something that you don't know! (As opposed to deriving, proving or arguing against already proven and extensively tested theories that have been around for years.) Still friends, just my opinion I am reminded of Galileo, who after being excommunicated from the catholic church for maintaining that the earth moves around the sun, was heard to utter "and yet it moves." I know you were joshing me about the empiricism remark, but, in the spirit of Galileo, I would like to add this to my original comments... "And yet it's false." But what am i talkin about? Kind regards PS: Oh and one more thing, I have not the aptitude to redo everything from scratch, nor the lifespan. That's why I work on the part that's wrong. Link to comment Share on other sites More sharing options...

5614 Posted April 4, 2005 Share Posted April 4, 2005 PS: Oh and one more thing' date=' I have not the aptitude to redo everything from scratch, nor the lifespan. That's why I work on the part that's wrong.[/quote'] But going by your method of "assume nothing" the only way you can work out if something is wrong is by proving it, the only way you can work out something right is by proving, and once you've done that you then need to sort out all the problems in the wrong stuff.... that is effectively reworking physics, which as you said you do not have the aptitude or the lifespan. Galileo/Newton were a long time ago, there is not the same circumstances (drastically wrong physics etc) around for you to be the next one... even on a smaller level! Why is it you chose not to accept physics (such as SR + GR) which most physicists in the world happily accept? And when nearly all physicists who are working in the field of physics prove them correct on a daily basis? Link to comment Share on other sites More sharing options...

Dave Posted April 4, 2005 Share Posted April 4, 2005 I wouldn't necessarily say that they're proving theories to be right. It's more like contributing evidence which is in accordance with what the theories predict - you can't really prove anything to be 100% right in Physics. Just nitpicking, don't mind me Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 Swansont' date=' you know better than this. The two frames are not symmetrical. They are kinematically equivalent, as regards relative speed, and relative acceleration. They are not dynamically equivalent though. In the one frame, the net external force upon the twin is zero. In the other frame, the net external force upon the twin is greater than zero. Hence, the two frames are not totally equivalent. There is an enormous difference between the frames. One of them is an inertial reference frame, the other is not. Regards[/quote'] No, you've changed the conditions of the experiment. The two are moving at some speed relative to each other. No acceleration. Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 I try to assume as little as possible swansont' date=' always assume as little as possible.Yes the proof is missing. ... As to why each observer cannot measure the other guys clock as slow, you keep running into a temporal paradox of the following kind: X before Y and Y before X and not (X=Y)[/quote'] And yet you've assumed X and Y are the same for both observers. It's X before Y and Y' before X' Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 I am reminded of Galileo' date=' who after being excommunicated from the catholic church for maintaining that the earth moves around the sun, was heard to utter "and yet it moves." I know you were joshing me about the empiricism remark, but, in the spirit of Galileo, I would like to add this to my original comments... "And yet it's false." But what am i talkin about? Kind regards PS: Oh and one more thing, I have not the aptitude to redo everything from scratch, nor the lifespan. That's why I work on the part that's wrong.[/quote'] Then you must show it to be wrong, not assume it to be wrong, and come up with a contradiction. The contradiction only means that one part is false - and you cannot assume that your assumption isn't it. You also have to be willing to follow your assumptions to all of their conclusions. You seem to disappear when that happens, only to pop up in a new thread. "Alas, to wear the mantle of Galileo it is not enough that you be persecuted by an unkind establishment; you must also be right." (Robert Park, of the American Physical Society) Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 If you assume absolute reference frames, you find that relativity doesn't hold. That's your contradiction. One of those two is false, but you haven't shown your assumption to be true. What assumption? Read it again, please... Link to comment Share on other sites More sharing options...

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