I am having a bit of confusion, let us say there are two observers observing two simultaneous lightning strikes with a finite distance between them. Now one observer is located such that he is in between the two lightning strikes, while the other is near to one strike and away from the other. In this scenario, the observer situated in between will perceive the lightning strikes to be simultaneous while the other will perceive them differently, since he is close to one and far from the other. Does this mean the Principle of Relativity holds even without actual motion?
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$\begingroup$ $E=mc^2$ is a result of SR that is devoid of reference to motion. $\endgroup$– Lewis MillerCommented Aug 24, 2017 at 15:34
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2$\begingroup$ @LewisMiller not true since m depends on velocity in this formula $\endgroup$– anna vCommented Aug 24, 2017 at 16:49
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1$\begingroup$ special relativity is necessary to explain the nuclear models, a nucleus at rest still needs special relativity to be modeled. $\endgroup$– anna vCommented Aug 24, 2017 at 16:51
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$\begingroup$ @annav hello sir, i think we were about to discuss something in the chat section not very long time ago if you remember $\endgroup$– Ajinkya NaikCommented Aug 24, 2017 at 16:56
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$\begingroup$ @annav if you still have time now, i am more than willing to discuss them now. $\endgroup$– Ajinkya NaikCommented Aug 24, 2017 at 16:56
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3 Answers
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You are mixing up two different things.
The observers know that light takes time to travel the distance from the strike to their eyes. So when they are calculating the time the strike occurs they allow for the travel time. Since both your observers make allowances for the light travel time they will both agree on the time the strikes occurred.
What special relativity tells us is that if one of the observers is moving relative to the other then even after both observers have allowed for light travel time they will still disagree about the timing of the strikes e.g. one observer might observe that the strikes happened simultaneously while another might observe that they are not simultaneous.
Physicists have a bad habit of talking about what an observer sees, and this does sound as though what we mean is what appears on the observer's eyeballs. We should really talk about what the observer observes, where the verb observe means to record the time and position of an event after allowing for things like the travel time of signals from the event to the observer.
answered Aug 24, 2017 at 15:44
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3$\begingroup$ both of the explanations yours and ajinkya's are perfect, but his explanation satisfies me more. $\endgroup$– user167295Commented Aug 24, 2017 at 16:48
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As already stated by John Rennie, the observers are already aware of the property that light travels at a constant speed, and they account for the given fact, so even if the lightning strikes are not simultaneous for them, they know that because of the fact that light travels at a constant speed.
But for any two observers moving relative to each other, even after accepting the fact that light travels at constant speed, they will have different agreements on the lightning strikes. In short, if two observers, observe those lightning strikes while both of them are at rest but different locations, they can subtract the time taken (t) for the lightning to reach both of them, and so both of them would get same remaining value, whereas in case of relative motion, even after subtracting (t) from their time measures, they would not get equal remaining values.
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Note that this scenario basically serves to relate space and time through velocity, which unifies them into the single concept 'spacetime.' Spacetime, plus the soured of light being constant in all inertial frames, are the basis for the formulation of special relativity as a theory.
So yes, special relativity holds in this case because it is a particular case of the conditions which special relativity was formulated to address.
