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Time dilation analysis help please

  1. Apr 6, 2008 #1
    Can someone please have a look at my analysis here and tell me if there are any mistakes
    the rest of the pistures are in my other posts
    ___________________________________________________________________________
    Einstein’s explanation of time dilation is given here http://en.wikipedia.org/wiki/Time_dilation

    Referring to fig 1
    In Einstein’s explanation you have 2 identical clocks. Clock 1 is at rest wrt the observer. Clock 2 is in a space ship zooming along wrt the observer.

    At the red end of the clock is a light source, at the blue end a detector.

    If L is the length of the clock. The observer, observing the clock at rest wrt him would see that the light pulse would take L/c time to traverse the length of the clock.

    Einstein postulated that as the observer would see the photons in clock 2 travel a longer path to move from the red end to the blue end when compared with length of the path the photons in the clock 1. Therefore clock 2 must tick slower (time passes slower) when compared to clock1.


    Referring to fig 2
    As in Einstein’s explanation again you have 2 identical clocks. Clock 1 and clock 2 are at rest wrt the observer. This is the trivial case. Again at the red end of each clock is a light source, at the blue end a detector.

    If L is the length of the clock. The observer, observing both clocks would see that the light pulse would take L/c time to traverse the length of the clock.

    Referring to fig 3
    As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
    Clock 1 is at rest wrt the observer, clock 2 is in a space ship that is moving wrt the observer.

    As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.

    If L is the length of the clock.

    The observer, observing the clock 1 would see that the light pulse would take L/c time to traverse the length of the clock.

    T = L/c

    As can be seen the path travelled by the photons in the clock2 is shorter than the path of the photons in the clock1. This is also what the observer would see.

    The time the photons in clock2 would take would be at most

    T = L/(c + V)

    As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c + V).

    I note the limit of the function t = L/(c + V) as V tends to infinity so t (time) goes to 0


    Referring to fig 4
    As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
    Clock 1 is at rest wrt the observer, clock 2 is in a space ship that is moving wrt the observer.
    You should now note both clocks have been rotated through 180 degrees compared to fig 3. In Fig 3 the light source is at the right the detector at the left. In fig 4 the reverse.

    As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.

    If L is the length of the clock. The time taken in Clock 1 would be L/c time to traverse the length of the clock.

    T = L/c

    As can be seen the path travelled by the photons in the clock2 is longer than the path of the photons in the clock1. This is also what the observer would see.

    The time the photons in clock2 would take would be at least

    T = L/(c - V)

    As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c - V).

    I note the limit of the function t = L/(c - V) as V tends to infinity so t (time) goes to infinity

    I have three questions

    Is this analysis right?

    How is it that the observer can get two different results simply by rotating the clock?

    The theory is a moving clock ticks slower then a stationary clock. In fig 3 clock 2 is ticking faster than clock1 which refutes the accepted theory?


    WHAT AN OBSERVER IN THE SPACE SHIP SEES


    Referring to fig 5
    I have a clock in the space ship.
    The space ship is flying at constant velocity V.
    At the red end of the clock is a light source, at the blue end a detector.
    As the speed is constant the contraction is constant and the time rate is constant.
    The observer, point A and point B form an isosceles triangle.

    The length of the clock is L which equals the length from B to the observer and also A to the observer.

    I NOTE THE DRAWING IS NOT TO SCALE.

    At some initial time the ends of the clock will be located at points A and B. At some time in the future as the space ship is moving the ends of the clock will be located at points A hat and B hat.

    At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.

    By the time the pulse from the red end of the clock reaches the observer he will be located at point x.

    When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.

    By the time the pulse from the blue end reaches the observer the observer will be located at point z.

    Analysis

    Referring to fig 7
    Using the cosine rule L3 can be calculated.
    Times will be proportional to lengths as time is simply length/c

    Angle g = 60 degrees
    Angle e = 120 degrees
    Cos 60 = 0.5
    Cos 120 = -0.5
    L3^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
    L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )

    t1 is the time it takes for the pulse of light to travel from B hat to point z

    Referring to fig 8
    Using the cosine rule L1 can be calculated.

    Angle e = 60 degrees
    Cos 60 = 0.5
    L1^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
    L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )

    t2 is the time it takes for the pulse of light to travel from point A to point x

    from fig 5
    L2 = L – V(t3)
    t3 is the time it takes for the pulse of light to travel from point A to point B hat.

    The time difference the observer will see between the first pulse of light he sees and the second is:

    Time_Diff1 = (L1 – (L2 + L3))/c
    L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
    L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
    L2 = L – V(t3)




    Referring to fig 6
    I have a clock in the space ship.
    The space ship is flying at constant velocity V.
    At the red end of the clock is a light source, at the blue end a detector.
    As the speed is constant the contraction is constant and the time rate is constant.
    The observer, point A and point B form an isosceles triangle.

    The length of the clock is L which equals the length from B to the observer and also A to the observer.

    I note the clock in fig 6 is rotated 180 degrees wrt the clock in fig 5 the red and blue ends have changed.

    I NOTE THE DRAWING IS NOT TO SCALE.

    At some initial time the ends of the clock will be located at points A and B. At some time in the future as the space ship is moving the ends of the clock will be located at points A hat and B hat.

    At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.

    By the time the pulse from the red end of the clock reaches the observer he will be located at point x.

    When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.

    By the time the pulse from the blue end reaches the observer the observer will be located at point z.

    Analysis

    Referring to fig 7
    Using the cosine rule L1 can be calculated.
    Times will be proportional to lengths as time is simply length/c

    Angle g = 60 degrees
    Angle e = 120 degrees
    Cos 60 = 0.5
    Cos 120 = -0.5
    L1^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
    L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )

    t1 is the time it takes for the pulse of light to travel from point A to point x

    Referring to fig 8
    Using the cosine rule L3 can be calculated.

    Angle e = 60 degrees
    Cos 60 = 0.5
    L3^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
    L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )

    t2 is the time it takes for the pulse of light to travel from point B hat to point z

    from fig 6
    L2 = L + V(t3)
    t3 is the time it takes for the pulse of light to travel from point A to point B hat.

    The time difference the observer will see between the first pulse of light he sees and the second is:

    Time_Diff2 = (L1 – (L2 + L3))/c
    L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
    L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
    L2 = L + V(t3)

    Observations

    Further observation:
    L3 in fig 6 = L1 in fig 5
    L1 in fig 6 = L3 in fig 5
    V(t3) in fig 5 = V(t3) in fig 6

    The time difference measured in fig 5 must equal the time difference measured in fig 6.

    Time_Diff1 = (L1 – (L2 + L3))/c
    Time_Diff2 = (L1 – (L2 + L3))/c
    Time_Diff1 = Time_Diff2

    Time_Diff2 = (L1 – (L2 + L3))/c
    L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
    L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
    L2 = L + V(t3)
    Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – (L + V(t3) + sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) ) )
    Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L - V(t3) - sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )


    Time_Diff1 = (L1 – (L2 + L3))/c
    L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
    L3 = sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
    L2 = L – V(t3)
    Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – (L – V(t3) + sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) ))
    Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3) - sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )

    Time_Diff1 = Time_Diff2

    sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3) - sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
    =

    sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L - V(t3) - sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )

    As can be seen Time_Diff1 does NOT equal Time_Diff2

    As all frames of ref are completely contained within a larger frame I note for any observer, unless they are in the absolute frame of ref, Time_Diff1 will not equal Time_Diff2 as they will have some residual velocity due to the velocity of the larger containing frame of ref.

    This analysis demonstrates that an experiment can be performed in a moving frame of reference to show the observer at rest with that frame of ref that they are in a moving frame of ref.

    As all frames of ref are completely contained within a larger frame of ref. By using this apparatus it may be possible to determine if there is an absolute frame of ref. By using this apparatus and changing the direction and velocity of a frame of ref until Time_diff1 = Time_diff2 then the direction and velocity of that frame of ref would be the same as the absolute (or preferred) frame of ref.
     

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  2. jcsd
  3. Apr 6, 2008 #2
    more pics

    more pictures for my last post
     

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  4. Apr 6, 2008 #3
    more pics

    more pictures
     

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  5. Apr 6, 2008 #4
    the entire post in one file

    the entire post in one file
     

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  6. Apr 6, 2008 #5

    Mentz114

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    Gold Member

    Rab99:

    I tried to read your first post and stopped when I came to this

    This made me unhappy because c+V implies something moving faster than light-speed, which I can't accept.

    I skipped to the end and found

    What is the absolute frame within which all frames are contained ?

    I don't believe this. I am convinced that this "it is not possible to distinguish a state of rest from a state of uniform motion" is true.

    Take me for instance, sitting in front of my computer in a room on earth. Assuming there are observers around the cosmos on other planets or spaceships, they will each see me in a different state of motion. I don't have an 'absolute state of motion', only relative states of motion.
     
  7. Apr 6, 2008 #6

    Dale

    Staff: Mentor

    Hi rab, I am not even sure this apparatus is a clock. Usually a clock has some sort of recurring event at one location in its rest frame, this doesn't. How would you use this to time something?

    Usually a light clock has a mirror on one side and then the emitter and detector on the same side. That way you are measuring the time between two events (emission and detection) at the same location in it's rest frame.
     
  8. Apr 6, 2008 #7

    JesseM

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    Science Advisor

    No, it doesn't imply that. If you have a light signal emitted from the front end of a rod of length L in your frame which is moving at speed v in your frame, then in your frame it will take a time of L/(c+v) to reach the back end. For example, suppose the back end is at x=0 when the light is emitted at t=0, while the front end is at x=L; then the equation for the back end's position as a function of time will be x(t) = vt, while the equation for the light moving from front to back will be x(t) = L - ct. So, the light will meet the back end when vt = L - ct, which means t = L/(c+v).
     
  9. Apr 6, 2008 #8

    Mentz114

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    Gold Member

    JesseM,
    OK, I wasn't paying attention. But what the OP is trying to do is false and hardly worth close attention. What do you think of his 'proof' that motion is absolute ?
     
  10. Apr 6, 2008 #9

    JesseM

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    Science Advisor

    Remember that two different frames can only agree on the length of the clock if the clock is oriented at right angles to the two frames' relative motion. So in the setup of your first diagram they'll both agree on the length, but in the setup of your second diagram, if the clock's length is L in the ship's rest frame then it will be only [tex]L*\sqrt{1 - v^2/c^2}[/tex] in the frame where the ship is moving forward at speed c.
    Usually in this thought-experiment we talk about the time for the light to go from one end to the other and back, in which case the time would be 2L/c for the observer on the ship.
    Here is where what I said above about length contraction becomes important. If you take into account that the length is shrunk in the frame where the ship is moving, then you will find that the time for the light to go from one end to another and back is [tex](2L/c)/\sqrt{1 - v^2/c^2}[/tex], exactly as predicted by the time dilation formula.

    To calculate the one-way time for the clock to go from the left end to the right end is more complicated. It is different in different directions in the frame where the rocket is moving, but it will nevertheless be the same in both directions for the observer on the rocket if he places watches at each end; and notes the times the light hits each end; to understand this part you must take into account the relativity of simultaneity, which is based on the fact that each observer synchronizes watches in different locations using the "Einstein synchronization convention" which is based on the assumption that light moves at c in all directions in their own frame. So if I want to synchronize watches at the front and back of the rocket in the rocket's own frame, I can set off a flash at the midpoint of the two rockets, and set them to read the same time when the light from the flash reaches them. But this means that in the frame of an observer at rest in a different frame who sees the rocket moving forward, my watches will be out-of-sync, because the watch at the back of the rocket is moving towards the point where the flash was set off and the watch at the front is moving away from that point, so naturally if this observer believes light moves at c in both directions he'll say the light reached the back watch before the front watch.

    The equation to remember for the relativity of simultaneity is that if you have two clocks at rest relative to each other which are synchronized in their own frame and a distance L apart in that frame, then in the frame where the clocks are moving at speed v along the axis between them, the time on the back clock will be ahead of the time on the front clock by the amoung vL/c^2. If you take this into account, and also take into account length contraction, you'll find that the observer on the ship still measures a time of L/c between the light leaving one end of the light clock and the light arriving at the other end, even if you analyze the situation from a frame where the ship is moving.

    If you want an example along these lines, take a look at the one I gave in post #6 from this thread.
     
    Last edited: Apr 6, 2008
  11. Apr 6, 2008 #10
    this does have two events... the time the photon left the light source and thr time that it arrives at the detector. The time interval between these two events could be timed and used to set a clock
     
  12. Apr 6, 2008 #11
    JesseM
    Remember that two different frames can only agree on the length of the clock if the clock is oriented at right angles to the two frames' relative motion. So in the setup of your first diagram they'll both agree on the length, but in the setup of your second diagram, if the clock's length is L in the ship's rest frame then it will be only in the frame where the ship is moving forward at speed c.


    The clock at rest wrt the observer is an identical clock to the one in the space ship. Why is the length of the clock even relevent they are identical. Just assume the clock in the space ship is shorter wrt the one at rest with the observer?
     
  13. Apr 6, 2008 #12

    Doc Al

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    Staff: Mentor

    These two events take place at different locations in the rest frame of the apparatus.
     
  14. Apr 6, 2008 #13

    Dale

    Staff: Mentor

    Yes, but the interval between the two events is lightlike, not timelike. This means that there does not exist any frame where the events measure two points in time at a single point in space. The interval between two clock tick events needs to be timelike.
     
  15. Apr 6, 2008 #14
    This was extraordinarily difficult to follow but I think I got the point.

    No this was not a postulate, this is one of the consequences of two postulates.
    1) The speed of light is the same for all observers.
    2) The laws of physics is the same regardless of motion.
    Below you have taken issue with the second postulate.

    I'm not sure what you mean by "contained within a larger frame" but I suspect it has something to do with your setup in "fig 6". You have the observers contained in the same frame as point A and B. You then proceed with an analysis showing length contraction. In what frame of reference does this time dilation take place? There is no motion within that frame to say that A or B changed position at all so there is no velocity V to use in your analysis.

    By assuming it has a velocity at all you implicitly defined a third frame of reference that you never actually specified. You just assumed from an intellectual perspective, your minds eye. This make you the one in motion wrt the spaceship observer, A, and B, and must be treated as such. It is you, which is in motion separately from the system you defined, that is the only defined frame for which "Time_Diff1 does NOT equal Time_Diff2". It cannot be said of the observers in the spaceship.
     
  16. Apr 6, 2008 #15
    The observer has to agree that the clock in the space ship is shorter than his clock...full stop.
     
  17. Apr 6, 2008 #16
    One question: Is this observer in the spaceship or moving with it?
     
  18. Apr 6, 2008 #17
    What I mean by two frames are always contained within a larger frame is this
    you are walking on the earth and a car passes you by. your walking is in a frame which can be descirbed by a vector, direction and velocity, say vector X. The car can be described by say vector y. The earth is spinning on its axis with vector w. The vector for the car is now the resultant of Vector W, x and y. The earth is also moving around the sun with vector B the car and your vectors are now the resultant of vectors b, x, y and W. The earth is within the solar system which is moving by vector D which adds the all of the aformentioned vectors. The solar system which is within the milky way, moving by vector U. The milky way is within the galaxy moving at vector Q. The galaxy is within within some larger struscture moving at vector K. And if you belive in big bang theorey this whole maras of vectors is within the universe probably descirbed by vector H. Now the vector decribing the car and you is a resultant of all the above vectors. I would call this vector that is the reultant of all the vectors up to the largest structure the preferred vector or frame
     
  19. Apr 6, 2008 #18
    the observer and the clock next to the observer is in a diff frame of ref to the space ship
     
  20. Apr 6, 2008 #19
    two identical clocks one in the space ship one at rest wrt the observer
     
  21. Apr 6, 2008 #20
    what do you think of the fact that time1 does not equal time 2 or is it too poorly communicated?
     
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