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Is cheating going on in Relativity?

  1. Sep 13, 2005 #1
    Is cheating going on in Einstein’s theory of relativity? For instance, the textbook “Fundamentals of Physics” (6th edition, published by John Wiley and Sons, jointly authored by David Halliday, Robert Resnick, and Earl Walker, on pages 925, 926 and 927, where the relativity theory was briefly introduced) said specifically and repeatedly that clocks would be used to measure time intervals. But I did not find any time interval that was read from the clocks. Instead all the time intervals used in the instruction were computed by means of Newtonian physics.
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  3. Sep 13, 2005 #2


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    Computing time intervals simply tells you what to expect when measuring the time. Indeed, time is measured by clocks; computations do not measure time. Not to worry, relativity is alive ,well and very robust, and has been so for roughly a century.

    Reilly Atkinson
  4. Sep 13, 2005 #3


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    Welcome to PhysicsForums!

    Not sure exactly what you mean about cheating. Relativity (both SR and GR) has been very well tested through the years. John Baez has some good stuff at his physics site, including this:

    What is the experimental basis of Special Relativity?
  5. Sep 14, 2005 #4


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    The most important concept in SR is that all clocks are local. In other words, the only one that matters is strapped to your wrist. Clocks strapped to other observers wrists will not, and cannot be ever be expected to agree. There is no universal clock. Special relativity is basically a four dimensional solution to Pythagorem's theorem. Wrap your head around that and you've got it... at least until you woozily stumble into the GR minefield and realized why you always hated geometry deep down inside.
  6. Sep 18, 2005 #5
    Thanks to reilly, DrChinese, and Chronos for your replies to my question. I however think the problem was not so simple.

    I wish all of you have access to the said textbook and then let us look into the thought experiment therein. The textbook authors claimed that the clocks will produce 2 different time intervals. But I believe otherwise and let me show you how.

    If we simplify the graphics and ignore the light, the thought experiment can be shown as follows:


    where S, R, P, and Q are points on two reference frames. A clock will move from S to R. And at P and Q each there is another clock. When the clock at S departs, it must register a reading of time. Let me call it the departure
    time "b". I believe the same departure time "b" will be recorded on another clock located at P. Don't you agree? (The textbook said, it was a light beam that departed. But it was all the same departure time, and that was why it was perfect all right not to mention the departure of light.)

    Now the clock will record an arrival time when it arrives at R. Let us call it "e". And the other clock at Q certainly will record the same arrival time "e" according to the design of the thought experiment. To obtain the time interval t, it is to deduct the departure time from the arrival time:
    t = e - b. This means that all 3 clocks would measure one identical time interval t, not two different time intervals claimed by the authors of the textbook.

    Consequently the relativistic nature of time cannot be proven unless relativistic advocates resort to cheating.
    Last edited: Sep 18, 2005
  7. Sep 18, 2005 #6

    Doc Al

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    Why don't you scan it in?

    To say "A clock moves from S to R" is very different from saying "A beam of light is emitted at S and arrives at R". Which is it? (I'll assume the latter.)

    You have two frames in relative motion, I presume. I assume S & R are in the "moving" frame. At the moment that the light is emitted at S, the point P passes by point S. Let's define the time read by the clocks at S and P to read the same at the moment they pass each other, which is the moment that the light is emitted.

    Not true at all! I assume the experiment is that the light arrives at point R just at the moment that point Q passes R. Each frame has its own clocks, presumably synchronized in their own frame: the moving frame has clocks at S and R; the other frame has clocks at P and Q. When the light arrives at R and Q, those clocks will not read the same times.

    You seem to just assume that the time of flight will be the same in both frames. But that ignores what is known about how time and distance measurements depend on the frame doing the measuring.

    Where exactly is the cheating? You seem to have given your version of the thought experiment; why don't you describe the version in the book.
  8. Sep 18, 2005 #7


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    What you are missing is the fact that in the rest frame of the clock which moved from S to R, the clocks that are resting at P and Q are out-of-sync, so the time interval between them is not the correct one in the first clock's rest frame (if you walked through a hallway and the clock on one end said 2:00 while the clock on the other end said 3:00, then you would not conclude it took you an hour to cross the hallway, you'd just conclude the second clock was ahead of the first one). This is known as "the relativity of simultaneity", and it's a pretty simple consequence of the idea that each frame should assume light moves at c. Just imagine a rocket going by you where a crewmember sets off a flash at the rocket's midpoint and synchronizes clocks on either end of the rocket by setting them to the same time at the moment the light from the flash hits them; since the rocket is moving forward in your frame, you will see that the clock at the front of the rocket is moving away from the point where the flash was set off while the clock at the back is moving towards it, so if you also assume light moves at c in all directions in your frame, you will conclude the light hit the back clock before the front clock, and thus that the two clocks are out-of-sync.

    You might want to take a look at a detailed example I worked out (with diagrams) on this thread, involving two rulers with clocks mounted at regular intervals sliding alongside each other, showing how in each ruler's rest frame it is the clocks on the other ruler which are running slow (and which are out-of-sync).
  9. Sep 19, 2005 #8
    Thank you, Al Doc. I am sorry that I did not give you enough information because I don't have the scanning device. But in my opening post I have made it clear between [clocks] and [equation]. We used clocks to measure time intervals, or we used equations to compute time intervals. These are two distinctly different operations, clocks and equations. If someone told me he would use clocks to do something and he switched to equations, of course he cheated me. This was exactly what had happened in the said textbook.

    In your reply to me you said: "When the light arrives at R and Q, those clocks will not read the same times."

    Why the clock at Q will NOT read the same arrival time as the clock at R did? You did not explain why. I believe otherwise, because this was a thought experiment. If all the clocks at S and P could be thought to read the same
    departure time, the same "thought" logic must apply to the arrival point. Hence when the light beam arrived at R and Q, the clocks there would logically read the same arrival time.

    Further, there could not be any synchronization problem either, simply because it was all a thought. In a thought an infinite number of clocks can be thought as synchronized perfectly. Imperfection can only happen in real experiments in the laboratory. If the clocks in different frames were synchronized in different ways, our thought can make compensations to them, adding or deducting a few nano-seconds to make difference disappear.

    As a last argument of mine, even if the clocks at R and Q would not read the same arrival time as you alleged, the cheating was still here, simply because the textbook authors produced the different time intervals by means
    of Newtonian physics. My logic was so glaringly clear that nobody could miss. Say I bought an orange and an apple from Mr. Newton's store, and I told my parents I bought them from Mr. Smith's store. There was no change
    to the fact that there were an orange and an apple. But I cheated my parents about the source.

    If my parents got sick after consuming the fruits, the blame would fall on Mr. Smith, not on Mr. Newton, as a result of my cheating.

    As to JesseM's reply, I believe he had the same argument since he mentioned synchronization. My point was how to identify the cheat, not to tell what the light would do in different reference frames. As to your relativity of simultaneity, I shall respond separately after I have studied your details.
  10. Sep 19, 2005 #9


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    But whose clocks do you add or deduct time from? Each observer has a set of clocks which are at rest with respect to himself and which he defines as synchronized, while seeing the clocks of moving observers being out-of-sync. And each observer uses his own set of clocks to mark the time of events--so if I see an event happen next to clock A in my system when it reads 2:00, and another event happen next to clock B in my system when it reads 3:00, I'll say that the time-interval between the two events was 1 hour, even if other observers using their own set of clocks get a different answer. So if I say that my clocks are synchronized and your clocks are out-of-sync, while you say that your clocks are synchronized and my clocks are out-of-sync, the situation is perfectly symmetrical, there's no objective way to decide which set of clocks is measuring time-intervals "correctly".

    Did you look at the diagrams of the two ruler/clock systems I provided on the other thread I linked to? They make the symmetry of the situation pretty clear.
    Where do textbook authors use Newtonian physics when justifying the ideas of relativity? I've never seen that before, can you quote the section you're talking about?
  11. Sep 19, 2005 #10

    Doc Al

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    You do realize that this is a thought experiment, don't you? We can't actually read what the clocks say, we have to deduce what they say according to the accepted laws of physics. One principle that was probably applied was that light travels at the same speed ("c") with respect to any frame.

    The reason why the clocks at S and P read the same at the instant that they pass each other is because we agreed to set them to read the same time at that single instant. For two systems of coordinates, you can always set the origin to be the same. But what other clocks read at other times must be deduced from physics, not merely assumed. (If you just assume that all clocks read the same time, then you will get results that contradict known laws of physics.)

    You seem to have the wrong idea as to what a "thought experiment" means. It doesn't mean "just think whatever you want". It means that you set up a situation (one that may be impractical to do in real life) and apply real laws of physics to deduce what happens. You don't just assume whatever you want.

    A clear application of the assumptions of relativity (that the speed of light is invariant, for one) will show that the two frames must disagree as to how the other frame's clocks are synchronized.

    What "Newtonian Physics" was used? I suspect that what was used was something like: Distance = speed X time. That's just as true for Einstein as for Newton.

    Once again: Where's the cheating? To show "cheating" you should lay out, step by step, the actual argument made in the textbook (not just your version) and point to the exact statements that you question.
  12. Sep 20, 2005 #11
    Doc Al and jesseM, and everybody else who have responded to my charge, I am sorry I could not copy and transmit the textbook's full text over here due to my computer skills. I shall drop the matter for good. If my charge had hurt any body, I apologize to them all. Sorry again.
  13. Sep 20, 2005 #12


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    No one was asking you to transmit the full text (which would be illegal anyway), just to quote a paragraph or two where they mentioned Newtonian physics in connection to relativity, or perhaps scan a page if you don't feel like typing it.
  14. Sep 21, 2005 #13


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    Not at all Sam, you asked an honest question... and got gang banged!! The best, and most honest part, was when you asked sincerely. I respect that, Sam. I rule in your favor.
  15. Sep 21, 2005 #14


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    Footnote: I think you are profoundly wrong, Sam, but I like the way you think.
  16. Sep 21, 2005 #15


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    You may simply be one of those people who likes to learn things "by debate", so to speak, using an argumentative skeptical style. In many fields this is common, and you'll even find this sort of thing in physics, especially in true "cutting edge" areas.

    The problem is that on the Net there are many anti-relativity cranks and crackpots who are convinced that they are right and mainstream scientists are all wrong, and who argue their views very aggressively and persistently. Anyone who comes onto an Internet discussion group and tries to discuss relativity in a similar style is likely to be treated like one of those people, regardless of their actual motives.
  17. Sep 23, 2005 #16
    Thanks to everybody. As JseesM suggested, I shall quote some from the said textbook. On page 925 of the textbook, under the title The Relativity of Time, there were these words: “If observers who move relative to each other measure the time interval between two events, they generally will find different results.
    Why? Because the spatial separation of the events can affect the time intervals measured by the observers."

    The textbook then gives a thought experiment involving an observer Sally on a train moving at v relative to another observer Sam on the ground, represented in two figures. I do not know how to draw these figures. Let me describe the first figure and see whether you can visualize.

    Sally will send a light pulse from her clock toward the ceiling of the cabin where she is in. To her, the light travels along the height (D) of her cabin, and comes down along the same height (D) and arrives at her clock. The textbook says: "Sally measures a certain time interval t between the two events, related to the distance 2D from source to mirror by t = 2D/c."

    The quote immediately above has two parts. The first part means Sally would obtain information from her clock, reading a departure time and an arrival time; the second part means she needs not read, because the time interval is related to the distance traveled by light.

    Don't you think these two actions, saying one thing and doing a different thing, constituted a lie?
  18. Sep 23, 2005 #17


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    Your passage described a light-clock.

    The duration between emission and reception, t, represents "1 tick" of that clock... a perfectly good unit of time. One could regard the first emission as the time reading of "0 ticks".

    Upon the first reception, if the source reflects back or re-emits a light signal, waiting for a second reception yields a second tick.

    The duration of a tick is proportional to the separation, D, between the "mirror" on the ceiling and the source. That's what t=2D/c means... a relationship relating t and D.. not a definition of t. (If you want a clock with a finer resolution, reduce the separation D.)
  19. Sep 23, 2005 #18


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    Why is it a lie? It just says that the time interval she actually measures using her clock will in fact match up with what she would predict if she assumed light travels at c in both directions in her own frame. They're not saying that Sally "needs not read", they're just telling you what relativity predicts she will read on her clock. Predictions should match observations in any successful theory, no?
    Last edited: Sep 23, 2005
  20. Sep 23, 2005 #19


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    I'm sorry, but I don't see anything wrong with this quote. All it says is that the time that Sally reads on the Clock and 2D/c will be the same amount of time. It's like saying that if I start a stopwatch as someone starts to drive away from me at 60mph, then that person turns around after 10 miles and comes back, my stop watch will read 20 min upon his return. If he turns around after 20 miles, my stopwatch will read 40 min, and if he turns around after 5 miles it will read 10 min. There is a direct relation to the time I read on my stopwatch and the distance he drives before turning around.
  21. Sep 25, 2005 #20
    It seemed that JesseM and Janus and most people believed that Sally's clock would produce a time interval equal to the one produced by Newton's equation.

    First of all my allegation had nothing to do with the equality between time intervals. My allegation was, if the textbook authors eventually used Newton's equation to compute the time interval, they should not in the beginning have said they would use clocks to do the job. They need not set up any clocks at all. Too much ado about nothing. They should not use the word "measure". They should just say they will compute. As a result the first paragraph I quoted in the former post of mine should read like this:

    "If observers who move relative to each other COMPUTE the time interval between two events, they generally will find different results. Why? Because the spatial separation of the events can affect the time intervals COMPUTED by the observers."

    The change of one action to another (from measure to compute) has illogical consequences. When computation has produced a shorter time interval, the explanation we can give is that we have put in (into Newton's equation) a shorter distance. I believe we cannot say, the reason for the equation to have produced a shorter time interval is because the clock ticked too slow. Can we accept such a logic?

    But in my view this was exactly what was happening in teaching the relativity theory. When a shorter time interval was obtained from Newton's equation, we were told the clock was malfunctioning, or time dilated.

    Such a practice (a shorter distance producing a shorter time interval) means time dilation has been used to prove the existence of time dilation. Is this logic sound?

    Secondly, if the textbook authors believed that the clocks would produce a time interval equal to the one computed from Newton's equation, they were obliged to demonstrate HOW. Or at least they should have said a few words
    about it.

    Let us represent your belief like this:

    (e - b) = t = 2D/c.....wherein e is the arrival time, b the departure time, shown on clocks; t is the resulted time interval, which equals the computed time interval 2D/c.

    As is customary that two reference frames are moving at v relative to each other. As a result when the light pulse departs and arrives at the clocks, the two frames have created a distance. Let us call this distance g. This
    means, to create the distance g the frames have taken the same time interval t as the light pulse has done. Don't you agree? If you do, the situation can be represented with this: t = g/v. This equation must be true especially if we
    refer to Janus example of stopwatches and cars.

    Since you believed that the measured time interval (e - b) is equal to the computed time interval 2D/c, it means
    (e - b) = t = 2D/c = g/v. This equation means that
    both frames have obtained the one and only time interval t, whether the observers used the measuring technique or the computation one.

    But the textbook told us that there would be two "different results." So, there must be something wrong? Don't you agree?
    Last edited: Sep 26, 2005
  22. Sep 25, 2005 #21


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    What are you talking about when you say "Newton's equation"? Do you just mean speed = distance/time? This equation is not specifically Newtonian, it's just as true in relativity as in Newtonian physics.
    Even if you feel the clocks and rulers are unnecessary, this hardly qualifies as "cheating" as long as the computed result matches what you'd actually get using clocks and rulers. In any case, the clocks and rulers are necessary, because if you have two observers in different reference frames, each observer uses his own clocks and rulers to determine speeds, and this explains how the two observers can both say the same light beam moved at c even if they disagree on the length of the path that the beam took; otherwise this wouldn't make any sense.
    But what they mean here by "compute" is "subtract the measured time of the first event according to that observer's clocks from the measured time of the second event", so clocks are still involved. If Sally is on the train and Sam is on the ground, and Sam uses his own clocks to mark the time that the light beam leaves the floor and the time it returns to the floor (he will need two separate synchronized clocks for this since the train is moving while the clocks need to be at rest relative to him, and each clock needs to be next to the event it's measuring so that you don't have to worry about delays due to the finite speed of light), then if he subtracts the measured time on his own clocks that the light beam left the floor from the measured time on his own clocks that it returned to the floor, he'll get a larger time interval between these two events than Sally got using her onboard clock (which means from his point of view, Sally's clock must have been ticking slow). But he also measures the light beam to have taken a longer path between the floor and the ceiling, since the train was moving horizontally and thus the light beam was moving diagonally in his frame rather than purely vertically; if he divides the distance he measured by the time he measured, he will also find that the light beam travelled at c in his frame.

    edit: I should add that when Sam "synchronizes" his two clocks at different locations, he does so using the assumption that light always travels at c in his frame; if he made a different assumption then he wouldn't measure Sally's light beam to travel at c. So it is true that the idea that light always travels at c is a starting assumption of relativity rather than something found empirically; however, if each observer creates a coordinate system using a network of rulers and clocks synchronized using this assumption, then the laws of physics will have exactly the same equations in each observer's coordinate system, so there's no physical experiment that will distinguish one of these coordinate systems from another, and that is an empirical fact that would not be true if we lived in a universe with different laws of physics.
    Last edited: Sep 25, 2005
  23. Sep 26, 2005 #22
    Yes, I did mean that. I used to called t=d/v Newton's equation. I now realized I should not. I also apologize for the incorrect use of the word "compute". Anyway, reading clocks is an act very different from using the equation.

    Let us make our discussion short. JesseM's language implied that he has access to the said textbook. If so, I believe he would be able to judge these 3 cases. (1) The two reference frames have created a distance; (2) A time interval will be obtained when this distance is put into the equation (t = d/v). (3) And the time interval so obtained is the time interval needed by the frames for creating that distance, and is equal to the one taken by light for its round trip. Do you think these 3 cases justified?
    Last edited by a moderator: Sep 27, 2005
  24. Sep 27, 2005 #23


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    I didn't mean to imply that, I just was inferring the textbook authors' meaning from the quotes you provided.
    What do you mean they "created a distance"? Each reference frame measures the distance between events using their own set of rulers, I dunno if that has any relation to what you're talking about here.
    If you know the object's velocity in your frame, I guess you could do this, but usually it's the other way around--you have clocks at two locations along the object's path, and you note the times on each clock at the moment the object passes them and subtract one from the other to get a time interval, then you divide by the distance between the two clocks.
    I still don't understand what it means for a time interval "create" a distance, a distance is just something you measure on a ruler. And the time interval for an object to cross a given distance will only be equal to the one for light to cross that distance if the object is travelling at light speed, if it's going slower then the time interval will be larger. But I'm probably misunderstanding what you're trying to say--maybe you could give me a numerical example of what you're talking about?
  25. Sep 27, 2005 #24


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    Sam, if I may jump in, as others have already argued, you may have a difficulty understanding what is mean to be a "thought experiment". In fact, a thought experiment is not an experiment at all, but a way of reasoning, which COULD eventually correspond to an experimental situation. It is, in a certain way, "armchair philosophy" :smile: But what's the purpose ? The purpose is to analyse the consequences of a certain set of (imagined or real) laws of nature. Now, in special relativity, we do the following:
    1) we still assume that, for an observer, space is 3-dim Euclidean space (meaning, you can use all highschool geometry), we call it E^3_observer
    2) we still assume that LOCALLY an observer has a clock that gives him "time". Whether this is a real clock, or a mathematical variable you call t_observer, doesn't matter of course.

    With points 1 and 2, we can now define of course things like "a trajectory" (for an observer), which is a mapping from t_observer into E^3_observer, and from such a trajectory, we can define a velocity which is de derivative of this trajectory to t_observer. All this makes sense WITH RESPECT TO AN OBSERVER.

    Up to here we are on par with Newtonian physics, except that:

    3-Newton) Newton assumes that for all observers, the t_observer are equal
    3-Relativity) Relativity assumes that c is a constant for all inertial observers.

    From the 3-Relativity point, you can now DEDUCE that the time seen on the watch of Sally, when her light pulse bounced up and down, MUST BE equal to 2 D/c
    Indeed, the light pulse is going to cover a distance 2 D (this we can deduce, from Sally's point of view, because of 1) and because of 3-Relativity, we know that the speed of light is c, so we have to have that c = 2 D / time-seen-by-Sally
  26. Sep 27, 2005 #25
    I am sorry it was all my fault due to lack of information, specifically lack of graphics about the thought experiment. Let us drop this very textbook and go to Einstein's writing available online.

    I believe people cannot succeed in their attempt to prove that time is relative unless they cheat. This belief of mine included Einstein himself as the cheater. To find out whether there is cheating in the theory of relativity,
    and whether Einstein cheated, we shall refer to Chapter 9 of Einstein's own book: Relativity, The Special and General Theory, available at http://www.ivorix.com/en/einstein/ [Broken].

    With easy access to Einstein's own words, we are in a better position to exchange opinions than when I could not provide you enough information from the textbook Fundamentals of Physics named in my first posting.

    In the said chapter, Einstein designed a thought experiment to prove that simultaneity is relative. He described it in these words:

    "Up to now our considerations have been referred to a particular body of reference, which we have styled a railway embankment. We suppose a very long train traveling along the rails with the constant velocity v and in
    the direction indicated in Fig 1. People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. " And the Figure looks like this:

    Fig. 1.
    A ____________________m ____________________B
    A ____________________M____________________ B

    Einstein continued:
    "Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises :
    "Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

    "When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length AB of the embankment. But the events A and B also correspond to positions A and B on the train. Let m be the mid-point of the distance AB on the traveling train.

    "Just when the flashes (as judged from the embankment) of lightning occur, this point m naturally coincides with the point M but it moves towards the right in the diagram with the velocity v of the train. "

    Here I drew the following figure to represent Einstein's description, Figure 2:

    In Einstein's book, there was no figure like this showing a total of 4 events A, A, B, B. But his description of the graphics implied explicitly that there were 2 events on the train and 2 events on the embankment. Yet the above graphics may or may not agree with Einstein's later ideas. We shall encounter this doubtful situation in the last paragraph of my post.

    Einstein continued:
    "If an observer sitting in the position m in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A."

    Here I believe a cheat has been committed: "he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A." To appreciate my charge, let me repeat Figure 2:


    When Einstein said m is "hastening towards the beam of light coming from B", which B was he talking about?

    If he was talking about the B on the train (m's reference frame), I think he lied against his light-speed principle. If he was talking about the B on the embankment (M's reference frame), I think he lied again, against his own regulation: " People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. "

    I believe my thinking above is correct meaning Einstein did cheat.

    Einstein continued his cheating by saying:

    "Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A."

    I believe otherwise. I believe observers taking the railway train as their reference-body cannot come to the conclusion as Einstein did, but rather they will conclude that lightning flashes A and B emitted simultaneously.
    My belief can be justified like this:
    Since Am = mB, so that Am/c = mB/c = t.

    But Einstein continued and said:
    "We thus arrive at the important result:
    "Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity)."

    To me this statement of Einstein's was a big lie. First of all the first word "events" was false or highly contradictory, because, according to Einstein's description of his thought experiment, there were a total of 4 events, two on the train frame and another two on the embankment frame. I believe we as students of his theory are entitled to be told which events he was talking about (again?). If he was talking about only two events, which two?

    When you try to identify them, I believe you shall find that the entire sentence representing Einstein's "important result" does not make sense at all. If you do not find so, I am glad to learn from you and retract all my charges.
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