I Relativity of Simultaneity: Analyzing Wavelengths

Nickelodeon
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Am i right in thinking that relativistic simultaneity explanations tend not to mention the fact that one of the ‘simultaneous’ events may well be red shifted and the other blue shifted and by analysing wavelengths you could presumably work out whether the event was in reality simultanous or not.
 
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No (or rather, you are correct that they don't mention it, but that's because it's wrong). There's no sense in which events are "really" simultaneous or not. And events can't be red-shifted or blue-shifted.

Light emitted at an event may be red-shifted or blue-shifted, but even two light pulses emitted from two sources at the same event may be Doppler shifted with respect to each other if the sources are in relative motion.
 
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Nickelodeon said:
Am i right in thinking that relativistic simultaneity explanations tend not to mention the fact that one of the ‘simultaneous’ events may well be red shifted and the other blue shifted and by analysing wavelengths you could presumably work out whether the event was in reality simultanous or not.
This wouldn't work. For example, in the usual train car thought experiment, the light from the front of the car will be blue shifted and the light from the rear of the car will be red shifted for an observer in the car. But for an observer in the next car ahead both will be red shifted and for an observer in the next car behind both will be blue shifted. These three observers will agree on simultaneity (since they are all at rest wrt the train), but they will disagree on the Doppler shift.
 
Nickelodeon said:
Am i right in thinking that relativistic simultaneity explanations tend not to mention the fact that one of the ‘simultaneous’ events may well be red shifted and the other blue shifted and by analysing wavelengths you could presumably work out whether the event was in reality simultanous or not.
Look at it this way. There are two light sources. One is 1 light sec from you and stationary with respect to you and the other is coming towards you at some high fraction of c. When the moving source passes the stationary source, both emit a light pulse. Both light pulses leave at the same moment, and they both travel the same distance to reach you and at the same speed. The fact that you will see the pulse coming from the moving source as blue-shifted has no bearing on this.
put other pair of sources 1 light sec away in the opposite direction with the moving one traveling in the same direction as the first moving source was. Thus when it passes the second stationary source it is moving away from you and you will see its flash as being red-shifted. If both of these sources flash the moment they pass each other you get a similar situation as above. If you, standing half way between the sources when they flash see all four flashes simultaneously, you can conclude that all four where emitted simultaneously. It will not matter that you saw 2 non-shifted flashes, 1 blue shifted flash, and 1 red shifted flash. They all arrived together and they all left from the same distance from you, traveling at the same speed.
 
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Thanks for your responses. I was thinking that two objects were stationary with respect to each other but a long way apart. Me, the observer, did a fly by leaving the first behind as i approached the second at nearly c. An event was scheduled simultaneous on both objects. In this scenario i thought if the red shift matched the blue shift then that would confirm simultaneity.
 
You can confirm simultaneity in whatever frame the events were scheduled to occur simultaneously (Edit: although simple arrival times ought to be enough - the Doppler shift just tells you that you were moving, with respect to the sources). But that doesn't mean anything other than that they were simultaneous in that frame. Certainly you'll see the light coming from in front of you blue shifted and from behind red shifted, but again, all this tells you is that you're moving relative to the sources. It doesn't make the shared rest frame of the sources "more real".
 
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