Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativity of Simultaneity

  1. Sep 18, 2006 #1
    How can two events happening at the same time and place in one frame of reference happen at a different time and place in another frame of reference?

    For example:

    If you are standing at the side of a road and you see a car going at 0.6c and 8 light seconds down the road in both directions someone is shining a flashlight toward the car.

    From the car's point of view the two beams of light should reach his car at the same place and at the same time. But from the observer's point of view the two beams of light reach the car at two different times and at two different places.

    How can two events happen at the same place and time in one frame of reference and not in another?
     
  2. jcsd
  3. Sep 18, 2006 #2

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  4. Sep 18, 2006 #3
    Yes,
    How can two events happen at the same place and time in one frame of reference and not in another?
    If two events have the exact same co-ordinates, how can you use the Lorentz transformations to find two different sets of co-ordinates in another frame of reference?
     
    Last edited: Sep 18, 2006
  5. Sep 18, 2006 #4

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The coordinate system of one observer is necessarily different from the coordinate system of another; there is no "absolute coordinate system." The Lorentz transform allows you to convert coordinates for one observer into coordinates for the other. A simple derivation (shown on the Wikipedia link) demonstrates how the Lorentz transform does not preserve the simultaneity of events as seen by different observers.

    - Warren
     
  6. Sep 18, 2006 #5
    Yeah, but...
    If two events have the exact same co-ordinates in one frame of reference, how can you use the Lorentz transformations to find two different sets of co-ordinates in another frame of reference?
     
  7. Sep 18, 2006 #6

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are we going around in circles, or what? I just responded to that.

    - Warren
     
  8. Sep 18, 2006 #7

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If, in one inertial frame, the coordinates of event A are the same as the coordinates of event B, then, in all inertial frames, the coordinates of event A are the same as the coordinates of event B.
     
  9. Sep 18, 2006 #8
    if two events happen in exactly the same spot they appear to be simultaneous from all reference frames. Simultaneity is only variant when the events are in different locations.
     
  10. Sep 18, 2006 #9
    Ok then how do you explain the original situation "If you are standing at the side of a road and you see a car going at 0.6c and 8 light seconds down the road in both directions someone is shining a flashlight toward the car.

    From the car's point of view the two beams of light should reach his car at the same place and at the same time. But from the observer's point of view the two beams of light reach the car at two different times and at two different places."
     
  11. Sep 18, 2006 #10
    Everything became clear when i read this page.

    http://www.mth.uct.ac.za/omei/gr/chap1/node4.html

    I was just thinking of it as space and time contracting and expanding as opposed 2 different observers taking different space and time slices of spacetime.
     
  12. Sep 18, 2006 #11
    Its not a specific answer to your question, but draw the situation in a spacetime diagram and im sure itll work out. Do the 2 people with the flashlights turn on their flashlights at the same time in their frame or the cars?
     
    Last edited: Sep 18, 2006
  13. Sep 18, 2006 #12
    ]Have a critical look please at


    http://arxiv.org/abs/physics/0511062
    The best things a physicist can offer to another one are information and critics
     
  14. Sep 18, 2006 #13
    yes, the observer would view the car as viewing the two beams as not simultaneous lol. The observer however will view them as simultaneous as will the car. There is no way to view the two events as not simultaneous. Saying the car looks like it sees them happen at different times from some reference frame is like saying relativity is wrong because a moving car looks like it sees light travel at c-v from some reference frame.
     
  15. Sep 18, 2006 #14

    Ok,

    Let's try some math. In relativity, one interesting consequence of the Lorentz transforms is the relation:

    dt'=gamma* (dt-V*dx/c^2)

    Now, you can quickly see that if in frame S you have dt=0 (the temporal distance between two events is 0), this implies:

    dt'=-gamma*V*dx/c^2 , gamma=1/sqrt(1-(V/c)^2)

    That is dt' , the time separation in S' is 0 if and only if dx=0 !!!

    What does this mean? It means that in order for us to judge the simultaneity (or lack of thereof) of the two events in S we need to have light signals sent from S to S'. If the light signals are being sent from the same point in S (dx=0) , then the simultaneity in S translates into a simultaneity in S' . Mathematically:

    dt'=0 if and only if dt=0 & dx=0

    If, on the other hand, the two events in S are separated by a nonzero distance dx, then, light will not arrive simultaneously to an observer in motion with the speed V wrt S. Hence, the event separation : -gamma*V*dx/c^2

    In Galilean kinematics, light speed is infinite so simultaneity is absolute. (unconditional). This is what throws a lot of people off.
     
    Last edited: Sep 18, 2006
  16. Sep 19, 2006 #15
    in this reply i have quoted A. Einstein in The Principle of Relativity, pp 38 - 42.

    all our judgments in which time plays a part are always judgments of simultaneous events. for example, if I say that at 11:00 am, 9/19/2006, in Burbank, California, a train arrived, I am saying that the pointing of the small hand of my watch to 11:00 am and the arrival of the train are simultaneous events.

    let A = a point of space where there is a clock and an observer who determines the time value of an event in the immediate proximity of A by finding the position of the hands (of the clock) which are simultaneous with this event.

    let B = a point of space where there is another clock (in all respects resembling the one at A) and another observer who determines the time value of an event in the immediate proximity of B by finding the position of the hands (of the clock) which are simultaneous with this event.

    let A-time = the time value of the event in the immediate proximity of A (determined by the observer with the clock at A)

    let B-time = the time value of the event in the immediate proximity of B (determined by the observer with the clock at B)

    let the common A-and-B-time = the time value of the event in the immediate proximity of either A or B (either one determined by either the obsever with the clock at A or the observer with the clock at B). a common A-and-B-time cannot be defined unless we establish that the time required by light to travel from A to B equals the time required by light to travel from B to A. thus to say that there is now a common A-and-B-time is to say that the clock at A and the clock at B synchronize.

    let the stationary system (K) = a system of coordinates in which the equations of Newtonian mechanics hold good.

    let the stationary rigid rod (K') = a stationary rigid rod lying along the axis of x of K. the length of K' is L as measured by a measuring-rod which is also stationary. let a uniform motion of parallel translation with velocity v along the axis of x of K be imparted to K'.

    let the length of K' in the moving system (the moving system in this case is K' itself) = L = the length of K' ascertain by the following operation: an observer moving together with the given measuring-rod and K' measures the length of K' directly by superposing the measuring-rod, in just the same way as if all three were at rest.

    let the length of the moving system K' in the stationary system K = rAB = the length of K' ascertain by the following operation: by means of stationary clocks set up in the stationary system K and synchronizing, an observer ascertains at what points (x1 and x2) of the stationary system K the two ends (A and B) of the moving system K' are located respectively at a definite time. the observer then measures the distance between these two points by the measuring-rod already employed, which in this case is at rest.

    at the end A of the moving system K' there is a clock, and at the end B there is another clock. these two clocks and the stationary clocks in the stationary system K synchronize. with the clock at A there is an observer, and with the clock at B there is another observer.

    at tA (the time value determined by the observer with the clock at A) a ray of light departs from A towards B. at tB (the time value determined by the observer with the clock at B) the ray of light is reflected from B. and at t'A (the time value determined again by the observer with the clock at A) the ray of light reaches A again.

    observers moving with K' would find that in K

    c = [rAB / (tB - tA)] + v = [rAB / (t'A - tB)] - v

    but observers in the stationary system K' declare that in K

    c = L / (tB - tA) = L / (t'A - tB)

    for example, let's say that v = 0.9*c, and that at the definte time tA (11:00 am, 9/19/2006), a ray of light departs from A towards B, and at the definite time tB (11:30 am, 9/19/2006), the ray of light is reflected back to A, and at the definite time t'A (12:00 pm, 9/19/2006), the ray of light reaches A again.

    observers moving with K' would find that in K

    299792458 m/s = (rAB / 1800 s) + 0.9*299792458 m/s
    rAB = 5.396264*10^10 m
    and
    299792458 m/s = [5.396264*10^10 m / (t'A - tB)] - 0.9*299792458 m/s
    (t'A - tB) = 94.736842 s

    but observers in the stationary sytem K' declare that in K

    299792458 m/s = L / 1800 s (tB - tA = t'A - tB = 1800 s)
    L = 5.396264*10^11 m

    so in the interval of time from 11:00 am thru 11:30 am, the observers moving with K' see that in K, rAB = 5.396264*10^10 m, and the observers in the stationary K' see that in K, L = 5.396264*10^11 m.

    and in the interval of time from 11:30 am thru 12:00 pm, the observers moving with K' see that in K, the interval of time is 11:30 am thru 11:31:34.7 am, and the observers in the stationary K' see that in K, the interval of time is 11:30 am thru 12:00 pm.

    so we see that we cannot attach any absolute signification to the concept of simultaniety. for when two events are viewed as simultaneous events from a stationary system of coordinates, the same two events can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

    in other words, when the position of the hands of the clock at A and the position of the hands of the clock at B are viewed as 12:00 pm in K from the stationary system K', the position of the hands of the clock at A and the position of the hands of the clock at B are looked upon as 11:31:34.7 in K when they are envisaged from the moving K'.
     
    Last edited: Sep 19, 2006
  17. Sep 19, 2006 #16

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Do you have a point that you are trying to make here? It would help if you quote the post that you are replying to. Otherwise, I don't think there is a need to provide a brief lesson on a part of introductory special relativity that one can get out of any modern physics text. If you wish to do this as a "lesson", then post it in the Tutorial section.

    Zz.
     
  18. Sep 19, 2006 #17
    Ok, I figured it out myself. The answer is that the car does not see the two beams of light reach his car and the same place and time. He sees them reach his car at t=4 and x=0 and t=16 and x=0 but the key to the answer which I overlooked was that unlike the observer the car does not see the two beams of light start towards him at the same time but rather 10 light seconds away and at t=-6 and t=6 instead of 0 and 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativity of Simultaneity
Loading...