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Simultaneity - Albert Einstein and the Theory of Relativity

  1. Apr 2, 2009 #1
    The following is the Einstein’s train but instead of train consider two airships of equal in lengths.

    http://www.youtube.com/watch?v=wteiuxyqtoM&feature=related

    Both are moving with same speed but opposite in direction. After certain time they cross each other such that one is exactly above the other (front of upper ship is exactly above the rare of lower and rare of upper is exactly above the front of lower). The stationary observer would see that light strike the rare and front of each ship at the same time. While both the moving observer would notice that light strikes the front before the rare end of their respective ship. So who is right?

    Upper moving observer
    Lower moving observer

    If both are right then both are wrong also. It means they noticed the same thing as stationary observer.
     
  2. jcsd
  3. Apr 2, 2009 #2

    Janus

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    I'm not sure what you are trying to get at here. You do realize that, due to length contraction, the observers in each ship will measure the other as being shorter and that from their perspective, the fronts and rears of the ships can not line up all at the same time. For instance the front of your ship may line up with the rear of the other ship, but the front other ship will not have reached the rear of your ship yet.
     
  4. Apr 3, 2009 #3
    The point with the relativity of simultaneity is that all observers are correct...it's just that they might not agree...as already implied by Janus most hypothetical constructions contain false assumptions or premises which lead to erroneous results....setting up a good example is not easy....

    Wikipedia does a decent job explaining with diagrams at

    http://en.wikipedia.org/wiki/Relativity_of_simultaneity
     
  5. Apr 4, 2009 #4
    Yes, I agree, but my intention was about the length contraction which can be corrected if both the observers communicate each other at the time of strikes.

    Here is the detail of length contraction

    The upper moving observer would notice that

    “The front of upper ship may line up with the rear of the lower ship, but the front lower ship will not have reached the rear of upper ship yet.”

    while at the same time it would happen to the lower moving observer that

    “The front of the lower ship line up with the rear of upper but the front of the upper ship will not have reached the rear of lower ship yet.”

    Thus we can get the actual length if either of strike is timely communicated.

    I thank you both for responding
     
  6. Apr 4, 2009 #5

    Janus

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    You really have to be careful when you use the term "at the same time" when dealing with Relativity.

    In your example, the proper observations for observers located at the front of the upper ship and rear of the lower ship would be:

    Upper ship:
    The front of my ship is even with the rear of the lower ship, but the front of the lower ship hasn't reached the the rear of my ship yet.

    Lower ship:
    The rear of my ship is even with the front of the upper ship, but the rear of the upper ship has already passed the front of my ship.

    Only when occupants in different ships are even with each can the observers agree that the observations were made "at the same time".
     
  7. Apr 4, 2009 #6
    Let’s do the other way.

    Both upper and lower moving observers have synchronized clocks. They don't have any connection.

    Both observers pauses their clocks when the lights strike their respective FRONT end.

    Later on, would the clocks show same reading, if compare at any station? If yes, then there is no physical length contraction.

    Similarly we can repeat the same rear ends.
     
  8. Apr 5, 2009 #7

    Janus

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    Synchronized according to which frame?
     
  9. Apr 5, 2009 #8
    At stationary observer’s station and from where both ships started their same computerized journey in a loop but in opposite direction. In order to facilitate crossing, There are gradients in the loop (understood) before and after crossing . They came back to their original position after completing their mission.
     
  10. Apr 22, 2009 #9
    For Other Users, I want to simplify the problem by defining few symbols.

    Call the upper ship A and lower B. Call the front and end of each ship Af, Bf and Ae and Be respectively. The observers Oa and Ob sit at the middle of each ship. The stationary observer is Os.

    As the ships pass Os and all 3 observers are instantaneously in line they may all observe each other's clocks and will find that both clocks on the ships are at the same time but slower than the Os's clock. Both ship look equally long to Os. He will therefore see both new flashes from the f and e of both ships at the same time. He might even conceivably "prove" that both ends of the ships were exactly together when the flashes struck by taking a single photo of Af, Bf, Ae and Be illuminated by the flashes.

    Now we consider events as seen from ship B. Ship A appears shorter as seen from B. Therefore first Bf and Ae meet and a flash strikes Bf/Ae. A moment later all 3 observers are aligned and another moment later Be/Af meet and a second flash strikes. Ob sees flash the Bf/Ae flash first followed by the Be/Af flash.

    Oa has an exactly symmetrical experience but sees flash Be/Af first followed by Bf/Ae.

    If they both "stop their watches" on the first flash they see then it will be a flash coming from the fronts of their respective ships but it will not be the same flash! In other words the first flash for Oa is the second flash for Ob and vice versa.
    Thus if both see their respective first front flashes at the same time on their watches. Then there is no length contraction by mathematical theorem;

    when A=B. B=C then A=C.

    Thus both clocks which went through the same journey can show real result while figment by observers (Oa,Ob)

    Here is Yahoo Team answer
    “So at least the above observations show that a contraction is not excluded. If there were no length contraction then all flashes would occur together for Oa, Ob and Os and all of them would experience the sequence of flashes in a similar way to that described above. However it would prove impossible then to meet the condition that the light from each event travels at the same speed relative to each observer.”
     
  11. Apr 22, 2009 #10

    JesseM

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    What about the velocities of ships A and B from the perspective of the "stationary" observer Os? Are the two ships moving in opposite directions in his rest frame, or are they moving in the same direction at different speeds? It would be helpful if you'd add some numbers to your example, like "both ships are 10 light-seconds long in the frame of Os, and in this frame ship A is moving to the right at 0.6c and ship B is moving to the right at 0.8c", something like that.
     
  12. Apr 22, 2009 #11
    For simplicity please click on the following link

    http://www.youtube.com/watch?v=wteiu...eature=related

    Now please consider ship instead of train in above link. There is other similar ship B. Both moving observers synchronized their clocks at stationary observer and started same computerized journey in same one loop, with a symmetrical gradient facility at crossing, but in opposite direction. After certain time they cross each other such that one is exactly above the other (front of upper ship is exactly above the rare of lower and rare of upper is exactly above the front of lower). Although in relativity it’s impossible. But anyway due to the length contraction

    The moving observer of A would notice that

    “The front of A may line up with the rear of the B, but the front B will not have reached the rear of A yet.”

    And the same would happen to the moving observer B that

    “The front of the B line up with the rear of A but the front of the A will not have reached the rear of B yet”

    The stationary observer would see that light strike the rare and front of each ship at the same time.

    While both the moving observers pauses their clocks as soon as they notice that light strikes the front before the rare end of their respective ship.

    Thus there should be no length contraction if both clocks show same reading if compared at stationary observer’s station. If not, how?
     
  13. Apr 23, 2009 #12

    JesseM

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    By "symmetrical gradient facility" do you mean equal speeds? (is English not your first language?) If so, I assume they are equal in the frame of the stationary observer?
    I guess by "rare" you mean "back"? Also, why do you say it's impossible in relativity? It's quite possible that in the frame of the stationary observer the front and back ends of the two ships will line up at the same time.
    Yes, both of these are correct.
    What do you mean by "both clocks"? The front end of A when lightning strikes it reads the same time as the front end of B when lightning strikes it, but these are two different strikes. For ship A, the time of the clock at the front end of A when lightning strikes it is earlier than the time of the clock at the back end of A when lightning strikes it. Likewise, for ship B the time of the clock at the front end of B when lightning strikes it is earlier than the time of the clock at the back end of B when lightning strikes it.

    Here's an example. Suppose both ships are 20 light-seconds long in their own rest frame. In the frame of the stationary observer, suppose they are moving at 0.5c in opposite directions, A moving to the right and B moving to the left. This means if the ships have clocks at either end that are synchronized in their own frame, in the frame of the stationary observer the clock at the back of each ship shows a later reading than the clock at the front of the ship, by 10 seconds (in general, if two clocks are synchronized and a distance x apart in their own rest frame, then in a frame where they're moving at speed v, one clock will be ahead of the other by vx/c^2). So if lightning strikes both ends simultaneously in the frame of the stationary observer, the clock at the front end of A might read -5 seconds when the lightning strikes it while the clock at the back end of A might read +5 seconds when the lightning strikes it, and likewise the clock at the front end of B (which is next to the back end of A) might read -5 seconds while the clock at the back end of B (which is next to the front end of A) might read +5 seconds. So, all of these events occur next to each other:

    -Lightning strike #1
    -clock at front end of A hit by lightning, reads -5 seconds
    -clock at back end of B hit by lightning, reads +5 seconds

    And these three events also occur next to each other:

    -Lightning strike #2
    -clock at back end of A hit by lightning, reads +5 seconds
    -clock at front end of B hit by lightning, reads -5 seconds

    Now we can consider how things look in the rest frame of one of the ships, say ship A. In A's frame, A is at rest while we can use relativistic velocity addition to show that B must be moving at (0.5c + 0.5c)/(1 + 0.5*0.5) = 1c/1.25 = 0.8c to the left in A's frame. Because of this, B is Lorentz-contracted by a factor of sqrt(1 - 0.8^2) = 0.6, so B is only 20*0.6 = 12 light-seconds long in A's frame. B's two clocks are also running slow by a factor of 0.6 in A's frame, and the clock at the back of B is ahead of the clock at the front by vx/c^2 = (0.8c)(20 light-seconds)/c^2 = 16 seconds.

    Also suppose that in A's frame, the center of the ship is at position x=0, while the back is at position x=-10 and the front is at position x=+10. Then we can talk about the position of both ships at different times in A's frame, and also the readings on clocks at different times in A's frame:

    At t=-5 in A's frame

    -back of A at x=-10, front of A at x=+10

    -clock at back of A reads -5, clock at front of A reads -5

    -back of B at x=+10, front of B at x= (10 - 12) = -2 (you can see that B is 12 l.s. long in this frame, and that the back end of B lines up with the front of A at this moment)

    -clock at back of B reads +5, clock at front of B reads (5 - 16) = -11 (you can see that the clock at the back of B is 16 seconds ahead of the clock at the front of B in this frame)

    Now, since B is moving at 0.8c in this frame, 10 seconds later at t=+5 in this frame, both ends of B will have moved 8 light-seconds to the left. Also, since both of B's clocks are ticking at 0.6 the normal rate in this frame, in 10 seconds the times on each clock will have increased by 6 seconds. So:

    At t=+5 in A's frame

    -back of A at x=-10, front of A at x=+10

    -clock at back of A reads +5, clock at front of A reads +5

    -back of B at x=+2, front of B at x=-10 (you can see both ends have moved 8 l.s. from their previous positions, and the front of B lines up with the back of A at this moment)

    -clock at back of B reads (5 + 6) = 11, clock at front of B reads (-11 + 6) = -5 (you can see both clocks have moved ahead by 6 seconds from their previous times)

    So, from all this you can see that in A's frame B is shorter than A, but it's still true in this frame that when the front of A lines up with the back of B, A's front clock reads -5 and B's back clock reads +5, and that when the back of A lines up with the front of B, A's back clock reads +5 and B's front clock reads -5.

    Please look over this example and tell me if there's anything you don't understand.
     
  14. Apr 25, 2009 #13
    English is my first language for the last eight year. I think its not too bad/ gibberish to be considered with disdain and also I am a little gay with editing so leniency please.

    Same computerized journey means both are moving with same speed but opposite in direction while the crossing detail is as follow which explain symmetrical gradient facility

    (For simplicity imagine a parallel circuit diagram for crossing)

    “A” leave the normal loop and move up grade, straight, down grade and then enter the normal loop from where“ B” left

    “B” leave the normal loop from opposite direction but move down grade, straight, up grade and then enter the normal loop from where “A” left.

    A is well above B and both even with each other for Os when they are at straight portion in crossing facility.

    why do you say it's impossible in relativity?
    What do you mean by “both clocks”?

    Let Oac and Obc are clocks used by observer Oa and Ob respectively. Thus

    First Strike:
    Oac is paused by Oa when Af even with Be and
    Obc is paused by Ob when Bf even with Ae

    Although in relativity both A and B can not line up for Oa and Ob to each other in their respective frame of references. Thus

    For Oa the first front strike of Bf would be second and
    For Ob the first front strike of Af would be second

    Thus both Oa and Ob started kook fight for their claim of first strike.

    Oa claimed that he had paused his Oac before Obc while
    Ob also claimed that he had paused his Obc before Oba

    At last both Oa and Ob agreed for simultaneous strike when they compared their clocks READING for their respective front strikes at Stationary Observer’s station. Because both clocks went through the same jouney.

    Here is point which I want to draw one attention.

    Now since Af was even with Be and Bf with Ae in the first strike which occurred simultaneously as confirmed by the clocks Oac and Obc therefore no real length reduction. Although both Oa and Ob are also right for their apparent/ figment reduction from their perspective frame of reference.

    I think aforementioned scenario was not futile or hogwash and I can also understand your answer too but I do not want to bother you further therefore I want to withdraw myself from this discussion.

    I can guess from your response that You seem to be a great, very cool and open minded person and I can not find a suitable words to thank you for your answer to my question.
    I will add your name in my best friend's list. Regards, Adieu
     
  15. Apr 26, 2009 #14

    JesseM

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    Based on this, I guess the clock Oac is located at the front of ship A (position Af) and the clock Obc is located at the front of ship B (position Bf)? But shouldn't you also place clocks at the back end of ship A (Ae) and the back end of ship B (Be), which also stop when the lightning strikes them? That is what I did in my example in the previous post, showing that when the lightning strikes next to Af and Be, A's clock at Af reads -5 while B's clock at Be reads +5, and when the other lightning bolt strikes next to Bf and Ae, B's clock at Bf reads -5 while A's clock at Ae reads +5.
    Yes, you can see this is true from my numbers above, Oa claims he paused his clock at A's front end (Af) at a time of -5, while according to his clock at A's back end (Ae) lightning did not strike there until +5, which is when it was next to the clock at the front end of B (Bf) as it was paused. But Ob claims he paused his clock at B's front end (Bf) at -5, while according to his clock at B's back end (Be) lightning did not strike there until +5, which is when it was next to the clock at the front end of A (Af) as it was paused.
    No, each observer defines simultaneity in terms of reading on his own clocks, not the other observer's clocks. A says that lightning struck near his clock at Af when that clock read -5, and lightning struck near his clock at Ae when that clock read +5, so the strikes were not simultaneous according to A. Likewise B says that lightning struck near his clock at Bf when that clock read -5, and lightning struck near his clock at Be when that clock read +5, so the strikes were not simultaneous according to B either. It's true that when lightning struck next to Af, A's clock at that location read -5, and when lightning struck next to Bf, B's clock there read -5, but since these are clocks from two different observers this does not mean the strikes were simultaneous in either frame--this is your crucial mistake, each observer only defines simultaneity using multiple clocks of his own at different locations.
     
  16. May 1, 2009 #15
    Ok, here is addendum to your previous annotation in order to review the same scenario with different angle. I hope this would be more interesting than previous.

    Let Oaf, Obf, Oae and Obe are observers at front “f” and end “e” of A and B. A is upper ship and B is lower ship.

    There are four light sources “Ls” mounted at the bottom of f and e of A and similarly at top of f and e of B. Just besides each “Ls” there is a mirror “M” which can ensnare vertical flash from its respective opposite light source “Ls” when f of one ship line up with the e of other, from his frame of reference. Here is the detail of the lay out of mirrors and light sources.


    Ship A End ----------------------------------- Front
    Lsae Mae Lsaf Maf
    Ae, Oae Af, Oaf

    Front ---------------------------------- End
    Ship B Mbf Lsbf Mbe Lsbe
    Bf, Obf Be, Obe

    For Os, in elevation or top view,

    Lsae is well above Mbf and Mae is well above Lsbf
    Lsaf is well above Mbe and Maf is well above Lsbe

    where Lsaf= light source of A front, Mbf= Mirror of B front and so on.

    As for Os both A and B are equally longer therefore he would see both new flashes from the f and e of both ships at the same time. And also at the same time

    Vertical flash of Lsae is detected by Mbf and that of Lsbf by Mae
    Vertical flash of Lsaf is detected by Mbe and that of Lsbe by Maf

    The distance between A and B is too small as compared to distance traveled by light in one second and therefore longer path of the light can be neglected.

    Consider the event from A frame of reference

    For Oaf, the front light strike the Af/Be followed by Ae/Bf. This means
    Mbe is struck by vertical flash of Lsaf that Maf by Lsbe
    Since Oae is still waiting for Bf to be even with Ae, therefore
    Vertical flash of Lsae hasn’t yet touched B and or Mbf while
    Vertical flash of Lsbf has touched A but not yet Mae and thus Oae is still waiting for the event to be occur at Ae.

    Obf and Obe have an exactly symmetrical experience but sees flash Bf/Ae first followed by Be/Af

    Here is the detail of event as seen from B frame of reference

    For Obf, the front light strike the Bf/Ae followed by Be/Af. This means
    Mae is struck by vertical flash of Lsbf and Mbf by Lsae
    Since Obe is still waiting for Be to be even with Af, therefore
    Vertical flash of Lsbe hasn’t yet touched A and or Maf while
    Vertical flash of Lsaf has touched B but not yet Mbe and thus Obe is still waiting for the event to occur at Be.

    So here are the perplexities that eating into me.

    If each front observer deny the lining up of the front of the other ship with the end of his ship then how comes abovementioned mirrors detects all those vertical flashes from their respective opposite light sources???

    OR

    Event from A frame of reference
    If Oae is still waiting for Bf to be even with Ae, then how come

    Mae was struck by Lsbf and Mbf by Lsaf from B frame of reference


    Event from B frame of reference

    If Obe is still waiting for Af to be even with Be then how come
    Mbe was struck by Lsaf and Maf by Lsbf from A frame of reference??

    Please ignore if answer is the same as in previous post.
     
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