Concept of simultaneity when moving at the speed of light

  • #1
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Hi,

I have been pondering a special relativity question about the concept of simultaneity when moving at speed of light.

Say for instance I took a round trip to Alpha Centauri 4 light years away, travelling at the speed of light both out and home (using negligible time to turn around). Then no time would have passed for me, while 8 years have gone by on Earth, but does that imply that everything that happened on Earth in the meantime happened simultaneously as seen from my perspective?

From what I know the answer to the question above is 'no' because time like events can never appear to be simultaneous to any observer. Yet, it still seems to me that if 8 years passed on Earth, and no time passed for me, then everything that passed in-between should be simultaneous.

I realise that this is just an example of the twin paradox, and I feel I can make sense of it when the round trip is made below the speed of light, but not when travelling at the speed of light.

Thanks in advance for your help :)
 

Answers and Replies

  • #2
phinds
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Your problem is that you have based everything on an impossible premise. There is no such thing as traveling at the speed of light (for things, like you, with mass), and there IS no frame of reference for light itself, so it's no good asking "how would light see it?" because that is a meaningless question.
 
  • #3
Ibix
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It's impossible for a massive body such as yourself (no offence) to travel at the speed of light. That is a feature of special relativity that protects us from having to resolve the various divide-by-zero errors that you get if you try to define a frame co-moving with light.

In other words, I'm afraid that the question you're asking doesn't really make sense.
 
  • #4
UltrafastPED
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You cannot travel at c; light can do that, but it always travels at c wrt all inertial reference systems. And light is not an observer.

Instead you can actually calculate everything for the case v=0.8c (the numbers work out nice). Then work it again with v=.99c, and so on to get your limit.
 
  • #5
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Alright, so in the end the answer was relatively (bad pun, I know) simple. Thank you all for your answers.
 

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