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Special Relativity and Reference Frames

  1. Oct 23, 2008 #1
    Hi,
    I am a bit confused about a certain aspect of Special Relativity and I am hoping someone can clear this up for me.

    Let's say we have an event and two observers. Observer A stands a few meters from the event, Observer B is hundreds of thousands of miles from it. Relative to the event and to one another, both observers are at rest. Since light travels at a finite speed, this event will take longer to reach Observer B than it will to reach Observer A. So, in an "absolute" reference frame, the event occurs earlier for Observer A than for Observer B. It seems the way to solve this difference is to require separate frames of reference for each observer. However, I am not certain that I am correct in my thinking here because I know that special relativity concerns moving observers. How do we account for this and am I correct in my thinking that we must assign separate reference frames to each observer even though each is at rest to the other and with respect to the event? Thank you for your help.
     
  2. jcsd
  3. Oct 23, 2008 #2

    Dale

    Staff: Mentor

    Hi pjhphysics, welcome to PF

    First, (a very minor mistake) an observer cannot be at rest relative to an event any more than a line can be parallel to a point. Observers are at rest relative to objects or other observers, which means that their worldlines are parallel.

    Now, the main source of your confusion seems to be the very common misunderstanding that relativity somehow deals with optical illusions or effects due to the finite speed of light. This is not the case. The key concepts of relativity (time dilation, length contraction, relativity of simultaneity) are all things that remain after accounting for the travel time of light. So for example, let's say in your example that observer A is 1 light-second away from the event and observer B is 1 light-hour from the event. If the event happens at 1:00:00 then observer A will recieve the light from it at 1:00:01 and observer B will recieve the light at 2:00:00. Observer A will account for the 1 second delay and observer B will account for the 1 hour delay and they will both agree that the event occured at 1:00:00. In other words, both observers are intelligent observers who will account for any delays caused by the finite travel time of light.

    The bottom line is that all observers who are at rest wrt each other will agree on the timing of events, regardless of where they are located.
     
  4. Oct 23, 2008 #3
    Thanks for your response.
    I see your point that this is not really an issue of special relativity. For clarity, may I pursue this question a bit further with the understanding that this question is no longer an issue of SR?
    As you say, "all observers who are at rest wrt each other will agree on the timing of events" AFTER compensating for the time delay caused by the speed of light.
    Without that compensation, however, this event would occur at different times for each observer. Therefore, we may say that in order for these events to occur simultaneously for each observer, we must take their position with respect to the event into account, if only for the purpose of compensating the disparity. That is, their position relative to the event (though static) must be considered...
    I've tried to reiterate what you said just to be certain my understanding is correct.
    Again, thanks for your help.
     
  5. Oct 23, 2008 #4

    Fredrik

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    You actually have to specify in what frame the distance is "a few meters" for this part of the sentence to make sense, but ok, let's just say that this is true in some frame, and that we're describing everything in that frame.

    An event is a point in spacetime, so "at rest relative to the event" doesn't make sense. But let's say that we're talking about the event where a light bulb is switched on, and interpret your assumption as: "Both observers and the light bulb are at rest in the frame we mentioned earlier".

    This is true in the frame mentioned earlier, but if A, B and the bulb aren't on a straight line in that frame, there might be other frames in which it isn't true.

    You're saying that "In frame C, the event has a lower time coordinate in frame A than in frame B". That doesn't make sense.

    If we disregard the first part of the sentence and just consider the statement to be that the event has a lower time coordinate in frame A than in frame B, it's just wrong. Both observers can compensate for the travel time of light, so why wouldn't they? If the don't, they would have to assign the same time coordinate to the event where the light was emitted and and the event where it was observed. That would make the speed of light = "spatial distance between the events" / zero = undefined.

    If I switch my cell phone on at 12:00:01 and at the same time see the light from an explosion on the moon, I'm not going to assign the same time coordinate to the two events. I would assign the time coordinate 12:00:00 to the explosion and 12:00:01 to the phone being switched on. The observation of the explosion is a third event, with time coordinate 12:00:01.

    It's never wrong to do that, but it's not always necessary.
     
  6. Oct 23, 2008 #5

    Fredrik

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    Simultaneity explained:

    (I'm writing the coordinates of an event as (t,x), i.e. with the time coordinate first).

    Suppose that a light signal that's emitted in the positive x direction at (t1,0) gets reflected at some point along the x axis and returns at (t2,0), then the reflection event must be simultaneous with (t1+(t2-t1)/2,0). That's how simultaneity works. Pick t1=-T and t2=T for simplicity. The emission event is (-T,0) and the return event is (T,0). The reflection event must be simultaneous with (0,0), and since the speed of light is 1, the spatial coordinate must be T, so the reflection event is (0,T).
     
  7. Oct 23, 2008 #6
    Thank you for response,
    So here is a revised version...
    Light bulb, Observer A and Observer B are in the same reference frame F.
    Light bulb is switched on at 11:00:00.
    Observer A sees light bulb go on at 11:00:01
    Observer B sees light bulb go on at 12:00:00
    If neither observer compensated their time coordinate,
    at 11:30:00, the light bulb would be on for Observer A, but off for Observer B.
    To reconcile this fact, each observer must compensate their time coordinate in accordance with their distance from the bulb.
    So Observer A compensates by 00:00:01
    and Observer B compensates by 01:00:00
    This is correct?
    If so, is it correct to say, that though this is not an issue of special relativity, the position of the observer with respect to the bulb must be taken into account in order for each observer's view to be consistent with that of each other observer in the same reference frame F?

    Thanks!
     
  8. Oct 23, 2008 #7

    Dale

    Staff: Mentor

    No, the event occurs at some specific coordinates (in a given reference frame), i.e. (t0,x0). A distant observer's reception of a signal from the event is a completely separate event with different coordinates (t1,x1) where c=(x1-x0)/(t1-t0). Obviously, t1 and x1 are known to the observer, but if they do not also know x0, then they don't know anything about t0.

    In other words, if you do not have sufficient information to make the compensation then the emission event occurs at an unknown time, not at the time of the reception event. If you do have enough information then all intelligent observers make the compensation and relativistic effects remain even after that. Either way, the time of the reception event is never considered to be the time of the emission event.
     
  9. Oct 23, 2008 #8
    Quoting DaleSpam, "Either way, the time of the reception event is never considered to be the time of the emission event."
    So, under what circumstances would the observer have an effect upon the time of the emission event?
    My point in this inquiry is to consider how the state of the observer factors into the state of the observed object...
    In what cases would the object have different time locations, dimensions, etc. for different observers?
     
  10. Oct 23, 2008 #9
    Let me give context for my question:
    In an essay that is only partially about physics, I am trying to show that we cannot think of an object's state without taking the role of the observer into consideration.
    For example, if we have a light bulb that is sometimes on and sometimes off, and we ask "is the lightbulb on at time X?" We need specify time X in order to ask this question in a meaningful way. But, let's say we have two observers, that are at drastically different distances from the light bulb. Observer A is extremely close to the bulb and Observer B is many thousands of miles from it. If each observer specifies the same value for time X, they may receive two different answers to the question. The light will take longer to reach Observer B, and so, at the moment that Observer A experiences the light go on, Observer B will still see the light as off. In this respect, we might say that the state of the object, as it appears to its observer is dependent upon when we inquire into that state AND where the observer is in relation to the light.
    Is this not correct?

    Please help!
     
  11. Oct 23, 2008 #10

    Dale

    Staff: Mentor

    No, this is wrong. I already explained it above. If A and B are observers at rest wrt each other then they will agree on the time of any event, regardless of the location of the event.

    This is a quantum mechanical concept, not a relativistic concept. The most famous example of this is the electron double-slit experiment where measuring which slit the electron goes through alters the interference pattern.
     
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