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What will be the events relative to the ground observer , a timelike or space like? 
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#1
Feb2512, 10:48 AM

P: 163

Please refer to the attached diagram
The train moves in the direction BA If the events seen in the ground observer simultaneously, it should be seen at A before B in the train In special case, the difference in times between A and B according to the train observer could be equal to the time a light signal takes to travel from A to B only if v/c=0.618 However, if the light travel from A to B relative to the train observer, it should do so relative to the ground one as well. But because of the experiment setup, still the ground observer sees the 2 events happen at the same time So what will be the events relative to the ground observer, a timelike or space like? Thanks,,, http://www.physicsforums.com/attachm...1&d=1330188385 


#2
Feb2912, 12:27 PM

P: 163

any answer to my question!



#3
Feb2912, 02:33 PM

P: 1,101

seperated spatialy and can be observed as happening simultaniously, spacelike I think.



#4
Feb2912, 03:15 PM

P: 132

What will be the events relative to the ground observer , a timelike or space like?
If two separate events are simultaneous in a particular frame, they have a spacelike separation in all frames.



#5
Feb2912, 04:48 PM

P: 1,101

Good point the distinction between spacelike & timelike is invariant.



#6
Mar112, 01:32 AM

P: 163




#7
Mar112, 03:30 AM

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In the first sentence (in the attachment), you're talking about A and B as if they are events. Later you're talking about them as if they are locations in space (specifically in the slice of spacetime that the ground considers space). This makes your argument quite confusing.
If A and B are events, it only makes sense to say that a light signal is sent from A to B if their separation is lightlike, but this would of course immediately contradict the earlier claim that there's an inertial frame in which A and B are simultaneous. Then you start talking about calculating the time it takes the signal to go from A to B. By then you are clearly thinking of A and B as locations, not events. The signal would be sent from an event E on A's world line and received at an event F on B's world line. The separation between E and F is obviously lightlike. 


#8
Mar112, 06:34 AM

P: 249

The ground observer will see the two "simultanious" events happen at two different times, this is because it will take longer for light to reach the ground observer, but this would only be an illusion since the photon would only be traveling a larger distance from one of the points. Similiar to how we look back in time when we look at the stars, the stars are not younger just because we see them that way. It just takes a certain amount of time for the light to reach the observer since it travels at a limited speed. So then, I would say the seperation of the events is just space like.



#9
Mar112, 07:15 AM

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#10
Mar112, 09:14 AM

P: 1,101

That being said, the star is actually "younger" here compared to over "there". not an illusion, a reality. Said differently, at the location here the "distant star" is not old yet, because of the spatial separation more time needs to pass until the star is "old" over here. It would be a safe bet the star is even older over there but that proof does not exist over here, it is not a "reality" here, more an "illusion" of the "mind's eye", specifically an assumption. ( of course physics can predict this future accurately, calculating the age of the star over there, and intuitively is reffered to as "it's age", but comparatively thats relative) (mumbo jumbo here, but I think it could even be physicaly possible for there to be a "closed timelike curve" (or spacelike? idk) that "transports" the light to the observer location faster than light years) 


#11
Mar112, 01:28 PM

P: 163

If A and B are simultaneous for the ground observer, they must have a spacelike separation for the train observer. This should not permit Δt` to be long enough to allow B receiving any light from A. But that may happen if v/c=0.618 ! 


#12
Mar112, 07:29 PM

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#13
Mar212, 12:25 AM

P: 163

The derivation of Δt` according to LT is also based on that assumption and therefore multiplying AB by ##1/\gamma## to yield AB`. So multiplication by ##1/\gamma## occurs on both sides of the equations i & ii and it will cancel each other out And even if you don`t want to consider ##1/\gamma##, no problem, as the quadratic equation still remains valid because AB` will cancel each other out again 


#14
Mar212, 02:18 AM

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I think you need to think about what you're actually calculating using the length contraction formula. You can't plug in the distance between two events and expect to get the distance between those same events in another coordinate system. That's not just wrong, it doesn't make sense. Think about it this way instead: In a spacetime diagram where the axes are the t and x axes of the ground's frame, the world lines of the two points on the ground are vertical lines. A and B are two points on those two lines. Since they are simultaneous, there's a horizontal line that goes through both of them. The length of that line segment from B to A is what you called AB. But ##(AB)/\gamma## is the length that an observer on the train would assign to a line with slope 1/v between two events where such a line intersects the two vertical lines. 


#15
Mar212, 05:58 AM

P: 132

If you assign coordinates to the events A and B in the ground frame, you can use the Lorentz transformation to work out the coordinates in the train frame: THE GROUND FRAME: Let's say that the distance between the two events in the ground frame is d. To make things easy, we'll put event A at the origin: Event A: t = 0, x = 0 Event B: t = 0, x = d. THE TRAIN FRAME: We define the train frame to have the same origin as the ground frame. It is moving relative to the ground frame along the x axis at a velocity of v. The Lorentz transformation gives us the coordinates of the two events in this frame: Event A: t' = 0, x' = 0 Event B: t' = [itex]\gamma[/itex]vd/c[itex]^{2}[/itex], x' = [itex]\gamma[/itex]d. Increasing the velocity difference between the two frames will not only increase the time separation between A and B, it will also increase the spatial separation: you can see from the coordinates that the distance in space between the two events has been multiplied by [itex]\gamma[/itex]. The spacetime interval between A and B stays the same. 


#16
Mar212, 08:34 AM

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#17
Mar212, 09:09 AM

P: 132

I'm not sure quite what you mean by saying that "A to B could be "seen" as a single object in motion relative to the train". Are you imagining that a single object could be present at both event A and event B? That won't work, since neither event falls within the light cone of the other. 


#18
Mar212, 09:16 AM

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