vuquta Posted February 13, 2010 Posted February 13, 2010 (edited) Here is an interesting problem. Please break it. The Burn Mark Problem The below experiment will avoid clocks and use light distance travel and frame distance travel. Since the SR clock synchronization method uses distance and light travel for its implementation, then distance and light travel are more fundamental than clocks. Assume a stationary light source O and target T at a distance d and a moving frame O’ moving at v in the direction of the negative x-axis. When O and O’ are co-located, O emits a light pulse. O concludes when light moves a distance d to hit the target T, the observer O’ moves to the x coordinate –vd/c. O decides to place a burn mark at that x location. Then O tells O’, when O’ is co-located with the burn mark, light moved a distance necessary to strike the target T. Since it is supposed to be one light ray, this is legitimate logic. BM |----O-----------------------------------T v<-O' On the other hand, when O’ is co-located with the burn mark, O’ concludes its distance to T is (vd/c + d)/λ. Also, when O’ sees the burn mark, O’ concludes O is located a distance (vd/c)/λ. Further, in O’, light travels a distance d/λ when O travels a distance (vd/c)/λ. Now, it is supposed to be a fact when O’ sees the burn mark, light has traveled a sufficient distance to hit the target T. Obviously, in the real external world, either light hit the target or it did not. However, when O’ sees the burn mark, it concludes light traveled a distance d/λ since the O frame traveled (vd/c)/λ. But the target is a distance (vd/c + d)/λ from O’ which is further away than the distance light traveled. Therefore, in the frame of O, when O’ is co-located with the burn mark, O contends T has been hit with the light. However, in the frame of O’, when O’ is co-located with the burn mark, O’ contends T has not been hit with the light. This is a contradiction. [Change the above] [Edit based on comments from Iggy: Therefore, in the frame of O, when O’ is co-located with the burn mark, O contends T light has traveled a sufficient distance to strike T. However, in the frame of O’, when O’ is co-located with the burn mark, O’ contends light has not traveled a sufficient distance to strike T. This is a contradiction.] Edited February 13, 2010 by vuquta
Iggy Posted February 13, 2010 Posted February 13, 2010 Therefore, in the frame of O, when O’ is co-located with the burn mark, O contends T has been hit with the light. However, in the frame of O’, when O’ is co-located with the burn mark, O’ contends T has not been hit with the light. This is a contradiction. Two events that are simultaneous in O may not be simultaneous in O'. http://en.wikipedia.org/wiki/Relativity_of_simultaneity
vuquta Posted February 13, 2010 Author Posted February 13, 2010 Originally Posted by vuquta Therefore, in the frame of O, when O’ is co-located with the burn mark, O contends T has been hit with the light. However, in the frame of O’, when O’ is co-located with the burn mark, O’ contends T has not been hit with the light. This is a contradiction. Two events that are simultaneous in O may not be simultaneous in O'. http://en.wikipedia.org/wiki/Relativity_of_simultaneity Yes, but this is not about clocks.and simultaneity. It is about light distance travel. The two frames without any clock measurements contradict each other on the physical distance travel of one light ray. Light moves the same way regardless of clocks.
Iggy Posted February 13, 2010 Posted February 13, 2010 Yes, but this is not about clocks.and simultaneity. It is about light distance travel. The two frames without any clock measurements contradict each other on the physical distance travel of one light ray. Light moves the same way regardless of clocks. You're talking about "when" things happen. In SR if, BM |----O-----------------------------------T v<-O' BM and T are simultaneous for O and O' is moving toward BM then O' must expect BM to happen before T. That is consistent with relativity. It is actually demanded by it. The order of events is relative in relativity, so I don't think there is any contradiction.
vuquta Posted February 13, 2010 Author Posted February 13, 2010 (edited) You're talking about "when" things happen. In SR if, BM |----O-----------------------------------T v<-O' BM and T are simultaneous for O and O' is moving toward BM then O' must expect BM to happen before T. That is consistent with relativity. It is actually demanded by it. The order of events is relative in relativity, so I don't think there is any contradiction. You are making good points. Yes, but I am not using any clocks. It is temporal logic and nothing more. Further a light ray moves. You must conclude in O when the burn mark is hit by O', T is hit by the one light beam. So, when O' is co-located with ther burn mark, light is no where near T in the frame of O'. The connection between the frames is the one light beam. Frames cannot control it. Hence, if the frame of O' had a co-located observer at T, that observer would conclude T is not hit while a correspnding observer at T in O would conclude T is hit. BM and T are simultaneous for O and O' is moving toward BM then O' must expect BM to happen before T. That is consistent with relativity. It is actually demanded by it. The order of events is relative in relativity, so I don't think there is any contradiction. Yes, it is the case that BM is moving toward O'. But, the issue here is that it is one rod for O and length contracted in O'. So, BM moving toward O' has no impact on what light does or how it moves. There is no provision in SR where relative motion of the light emission points have any impact on the travel of light. O and O' are light emission points for each frame. It is the receiver T that would normally be at issue under SR. SR never discusses the divergence of the light emission points in each frame. Edited February 13, 2010 by vuquta
swansont Posted February 13, 2010 Posted February 13, 2010 (edited) Yes, but I am not using any clocks. It is temporal logic and nothing more. Same thing. Your "temporal logic" must still follow SR. You can't eliminate time dilation and simultaneity issues by omitting a physical clock from the problem. Merged post follows: Consecutive posts merged Therefore, in the frame of O, when O’ is co-located with the burn mark, O contends T has been hit with the light. However, in the frame of O’, when O’ is co-located with the burn mark, O’ contends T has not been hit with the light. This is a contradiction. They will agree of light hits the target, but they will not agree on when. Does O' contend that light never hits the target? That would be a contradiction. If O' contends that it hits later, then it isn't a contradiction and is probably the same error of simultaneity that you have made in every thread you have started here. Edited February 13, 2010 by swansont Consecutive posts merged.
vuquta Posted February 13, 2010 Author Posted February 13, 2010 Same thing. Your "temporal logic" must still follow SR. You can't eliminate time dilation and simultaneity issues by omitting a physical clock from the problem. Merged post follows: Consecutive posts merged They will agree of light hits the target, but they will not agree on when. Does O' contend that light never hits the target? That would be a contradiction. If O' contends that it hits later, then it isn't a contradiction and is probably the same error of simultaneity that you have made in every thread you have started here. Actually, let's remove the target. When O' and the burn mark are co-located, O contends the light beam is vd/c + d from the burn mark O' contends the light beam is d from the burn mark. Since this is one light beam, it cannot travel two different distances from a co-located point of measure between two frames. Length contraction/expansion does not resolve this problem. Oh, simultaneity does not resolve the twins contradition I proposed.
swansont Posted February 13, 2010 Posted February 13, 2010 Actually, let's remove the target. When O' and the burn mark are co-located, O contends the light beam is vd/c + d from the burn mark O' contends the light beam is d from the burn mark. Since this is one light beam, it cannot travel two different distances from a co-located point of measure between two frames. (emphasis added) Why? From what principle of relativity does this follow? It contradicts the existence of the phenomenon of length contraction. You can't just make up statements about the how the world behaves. Length contraction/expansion does not resolve this problem. Since you haven't applied length contraction here, this is a non-sequitur.
vuquta Posted February 13, 2010 Author Posted February 13, 2010 (emphasis added) Why? From what principle of relativity does this follow? It contradicts the existence of the phenomenon of length contraction. You can't just make up statements about the how the world behaves. Since you haven't applied length contraction here, this is a non-sequitur. OK, I will apply it. I thought it obvious it was insufficient to resolve it. When the burn mark and O’ are co-located. O stationary O concludes the one light beam is a distance vd/c + d from the burn mark. O concludes O’ will measure this distance as (vd/c + d)/λ. O’ stationary O' concludes the one light beam is a distance d/λ from the burn mark. O' concludes O will measure this distance as d/λ2. Obviously, sans length contraction, frames cannot disagree on the length, from a co-located point, of a single light beam. So, where is the flaw? Looking above, the term vd/c actually represents the distance between the two light emission points. Each frame believes its own emission point is the correct one.
Iggy Posted February 13, 2010 Posted February 13, 2010 (edited) Since this is one light beam, it cannot travel two different distances from a co-located point of measure between two frames. It can. Look at a space time diagram: Does a ray of light need to be measured the same spatial distance in both frames? I get the feeling you are either not very familiar with relativity or not too fond of it. Either way, this tutorial might be helpful: http://casa.colorado.edu/~ajsh/sr/paradox.html because it starts with a 'paradox' like your example and works through the solution. Edited February 13, 2010 by Iggy typo
vuquta Posted February 13, 2010 Author Posted February 13, 2010 (edited) It can. Look at a space time diagram: Does a ray of light need to be measured the same spatial distance in both frames? I get the feeling you are either not very familiar with relativity or not too fond of it. Either way, this tutorial might be helpful: http://casa.colorado.edu/~ajsh/sr/paradox.html because it starts with a 'paradox' like your example and works through the solution. Hey thanks for the lesson. Your world diagram needs clock differentials to resolve and explain the light distances. We are not using clocks only distances. Note your t-axis. This does not resolve the below. when the burn mark and O’ are co-located. O stationary O concludes the one light beam is a distance vd/c + d from the burn mark. O concludes O’ will measure this distance as (vd/c + d)/λ. O’ stationary O’ concludes the one light beam is a distance d/λ from the burn mark. O’ concludes O will measure this distance as d/λ2. Hence, an observer at the burn mark from O' claims the light stretches down the x-axis by a distance d/λ in its measurements. A co-located observer from the O' frame at the burn mark claims light stretches down the x-axis a distance of vd/c + d in its measurements. Since we are not using clocks and all we have is length contraction as a frame to frame differential, there is no explanation for this immense light path length differential for co-located observers. Merged post follows: Consecutive posts mergedIt can. Look at a space time diagram: Does a ray of light need to be measured the same spatial distance in both frames? I get the feeling you are either not very familiar with relativity or not too fond of it. Either way, this tutorial might be helpful: http://casa.colorado.edu/~ajsh/sr/paradox.html because it starts with a 'paradox' like your example and works through the solution. Also, just a comment about this "paradox". No solution resolves the fact that the origin of the two light spheres are separated in space by vt after any time t. I can do sphere to sphere mapping using LT and convert the points of the light sphere from one to the other and back. Note, this translation must be done point by point to work correctly. That is all fine and good except the two light spheres have two distinct origins in the space of either frame. That is a physical impossibility. There is no resolution to this problem anywhere in the literature for the origins and LT dioes not address diverging origins. However, here is a point I developed for you to play with. x = vtλ /(1+λ) You will find this is on both light spheres at any time t and further, t' = t using LT. Edited February 13, 2010 by vuquta Consecutive posts merged.
swansont Posted February 13, 2010 Posted February 13, 2010 OK, I will apply it. I thought it obvious it was insufficient to resolve it. When the burn mark and O’ are co-located. O stationary O concludes the one light beam is a distance vd/c + d from the burn mark. O concludes O’ will measure this distance as (vd/c + d)/λ. O’ stationary O' concludes the one light beam is a distance d/λ from the burn mark. O' concludes O will measure this distance as d/λ2. Obviously, sans length contraction, frames cannot disagree on the length, from a co-located point, of a single light beam. So, where is the flaw? Looking above, the term vd/c actually represents the distance between the two light emission points. Each frame believes its own emission point is the correct one. If there was a target there, you would be demanding that the light beams hit the target simultaneously, in both frames. This is an artificial constraint that you cannot put on the problem. Your "contradiction" is a confirmation that you have applied an improper constraint on the problem. Just like you have done in all of the other scenarios you have presented. Lorentz transforms are just math. Any application of them that leads to a contradiction is contained in the boundary conditions you artificially apply to the situation. All you are doing is demonstrating that the constraints are contradictory — SR is inconsistent with absolute simultaneity. This is not surprising to people who understand relativity. Merged post follows: Consecutive posts mergedWe are not using clocks only distances. Can't separate the two. They are related; the velocity four-vector is invariant.
vuquta Posted February 13, 2010 Author Posted February 13, 2010 If there was a target there, you would be demanding that the light beams hit the target simultaneously, in both frames. This is an artificial constraint that you cannot put on the problem. Your "contradiction" is a confirmation that you have applied an improper constraint on the problem. Just like you have done in all of the other scenarios you have presented. Lorentz transforms are just math. Any application of them that leads to a contradiction is contained in the boundary conditions you artificially apply to the situation. All you are doing is demonstrating that the constraints are contradictory — SR is inconsistent with absolute simultaneity. This is not surprising to people who understand relativity. Merged post follows: Consecutive posts merged Can't separate the two. They are related; the velocity four-vector is invariant. Let's just get to the point. The following is not disputable. O stationary O concludes the one light beam is a distance vd/c + d from the burn mark. O concludes O’ will measure this distance as (vd/c + d)/λ. O’ stationary O' concludes the one light beam is a distance d/λ from the burn mark. O' concludes O will measure this distance as d/λ2. Now you claimed I am hitting boundary conditions or some such thing. Specify these boundary conditions to support this claim. What is clear is is O concludes the one light beam is a distance vd/c + d from the burn mark. O' concludes the one light beam is a distance d/λ from the burn mark. These cannot be refuted or explained by length contraction. And no, I do not need a target. Math is sufficient to demonstrate the contradiction in the theory. If you have an alternative explanation, then list it. What is above is a logical thought experiment within the domain of SR. The conclusions of the thought experiment are derived based on the logic of SR. There is nothing wrong with anything above.
phyti Posted February 13, 2010 Posted February 13, 2010 Here is an interesting problem. On the other hand, when O’ is co-located with the burn mark, O’ concludes its distance to T is (vd/c + d)/λ. Therefore, in the frame of O, when O’ is co-located with the burn mark, O contends T has been hit with the light. However, in the frame of O’, when O’ is co-located with the burn mark, O’ contends T has not been hit with the light. This is a contradiction. refer to drawing O' can only speculate or rely on the info from O with this incomplete scenario. If O actually measures the distance to the target with a return signal, which could be relayed to O', then O' would assign event 2 to his clock event 2', and calculate the distance as D(1+v/c)/g, in agreement with your answer, but occurring after passing M. Using SR transform, x' = g(x-vt) = gx(1+v/c)
vuquta Posted February 13, 2010 Author Posted February 13, 2010 refer to drawing O' can only speculate or rely on the info from O with this incomplete scenario. If O actually measures the distance to the target with a return signal, which could be relayed to O', then O' would assign event 2 to his clock event 2', and calculate the distance as D(1+v/c)/g, in agreement with your answer, but occurring after passing M. Using SR transform, x' = g(x-vt) = gx(1+v/c) Insufficient. Your model declares LT results are not logically decidable and other methods must be used to decide the problem like light transfer. I can prove to you frame to frame clock synchronization is not logically decidable. This is all over the literature. As such, frame to frame light signal transfer cannot decide the logic of this problem. I should not need to come behind LT with light signals to proclaim the logical consistency of SR. If you propose this, you are proclaiming LT is not sufficient to solve this problem correctly. With that, you and I are in agreement.
Iggy Posted February 13, 2010 Posted February 13, 2010 Your world diagram needs clock differentials to resolve and explain the light distances. We are not using clocks only distances. Note your t-axis. Velocity involves distance and time. If you want to know how far something travels with a given velocity between two events then you need to consider the time between events. It is a contradiction to say "We are not using clocks" and to assign velocity to things and talk about "when" things happen.
vuquta Posted February 14, 2010 Author Posted February 14, 2010 Velocity involves distance and time. If you want to know how far something travels with a given velocity between two events then you need to consider the time between events. It is a contradiction to say "We are not using clocks" and to assign velocity to things and talk about "when" things happen. Interesting. Velocity is also a state as compared to light. Light travel is more primitive than clocks as distance and light travel are used to sync clocks under SR. Hence, while light moves d, an object moves d(v/c). Note, clocks are a human thing. Clocks do not control nor alter this relationship.
phyti Posted February 14, 2010 Posted February 14, 2010 Insufficient. Your model declares LT results are not logically decidable and other methods must be used to decide the problem like light transfer. I can prove to you frame to frame clock synchronization is not logically decidable. This is all over the literature. As such, frame to frame light signal transfer cannot decide the logic of this problem. I should not need to come behind LT with light signals to proclaim the logical consistency of SR. If you propose this, you are proclaiming LT is not sufficient to solve this problem correctly. With that, you and I are in agreement. Clock synchronization is relative to each frame using 2-way light signals. Clock frequencies are a function of speed. Light is the universal measuring rod. You are actually measuring distance using a clock!
vuquta Posted February 14, 2010 Author Posted February 14, 2010 Clock synchronization is relative to each frame using 2-way light signals.Clock frequencies are a function of speed. Light is the universal measuring rod. You are actually measuring distance using a clock! Clock synchronization is relative to each frame using 2-way light signals. hmmmmmm. Einstein’s proclaimed the clock synchronization method is “free from contradictions”. http://www.fourmilab.ch/etexts/einstein/specrel/www/ This means the clock synchronization method is a logical truth and not a "relative truth". Now, if there exists multiple light emission points under SR, this means SR cannot logically decide the origin of the light path. Yet, SR claims the light path distance and thus the light emission point to the receiver is "free from contradictions". As such, SR is a failed theory.
Iggy Posted February 14, 2010 Posted February 14, 2010 Hence, while light moves d, an object moves d(v/c). That is correct. In either frame while light moves dL an object with velocity vo will move vo(dL/vL). Distance is velocity times time and (dL/vL) is the time that light travels. Hence, do = vo(dL/vL) is equivalent to do = vo*to if tL = to which is exactly what you mean when you say "while light moves...". You mean that both the light and the object have moved for the same duration. For example, in your thought experiment: BM |----O-----------------------------------T v<-O' if the velocity of O' is .6c relative to O then O' may find that the light moves 2 light seconds while O moves 1.2 light seconds between the two events (the emission and co-location with the burn mark). Between the same two events O would find light moves 2.5 light seconds while O' moves 1.5 light seconds. As you say: do = vo(dL/vL) in the frame of O': 1.2 = 0.6(2/1) in the frame of O: 1.5 = 0.6(2.5/1) Here is a space time diagram of this example where v = .6c: O and O' are co-located at the black dot when the light is emitted and O' is co-located with the burn mark at the green dot when both frames calculate the distance the light has traveled. O calculates 2.5 light seconds and O' calculates 2 light seconds. I understand you would rather O and O' agree on the spatial distance the light travels between events, but special relativity does not work that way. I suggest that your disagreement with relativity does not present a contradiction in the theory, but instead a contradiction between your expectations and the results of the theory. Since the results of the theory are shown to be experimentally accurate, you may want to examine your expectations--I would humbly suggest.
swansont Posted February 14, 2010 Posted February 14, 2010 Clock synchronization is relative to each frame using 2-way light signals. hmmmmmm. Einstein’s proclaimed the clock synchronization method is “free from contradictions”. http://www.fourmilab.ch/etexts/einstein/specrel/www/ This means the clock synchronization method is a logical truth and not a "relative truth". Clock synchronization is defined WITHIN a reference frame. You cannot synchronize identical clocks in separate frames, since they do not run at the same rate. Now, if there exists multiple light emission points under SR, this means SR cannot logically decide the origin of the light path. Yet, SR claims the light path distance and thus the light emission point to the receiver is "free from contradictions". As such, SR is a failed theory. SR predicts that the time and distance to the receiver will depend on the frame in which you make the measurement. The contradiction you find is in a constraint you have placed on the problem. The contradiction means that the constraint is false. Merged post follows: Consecutive posts merged What is clear is is O concludes the one light beam is a distance vd/c + d from the burn mark. O' concludes the one light beam is a distance d/λ from the burn mark. When does the light beam arrive? It matters. It's the crux of the problem. You have assumed that it is simultaneous in both frames, and that's the issue. The contradiction is not with SR, it's with your incorrect and unsubstantiated assertion that the light beams will arrive simultaneously in both frames.
Sayonara Posted February 14, 2010 Posted February 14, 2010 So, in summary "Relativity: you're doing it wrong". Vuquta, being wrong is not against the rules. But re-posting closed topics is. SR is a failed theory. In reality, which is where most people live, SR is one of the most accurately and consistently proven theories ever committed to paper. I suggest that the failure will be somewhere in your reasoning, and I think that this thread can very safely end on that note. 1
swansont Posted February 14, 2010 Posted February 14, 2010 I suggest that your disagreement with relativity does not present a contradiction in the theory, but instead a contradiction between your expectations and the results of the theory. Yes, this is precisely the problem.
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