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Let's say that an observer is moving with the speed of light relatively to an atom that he wants to look into. He has equipment that precise that he can observe the atom and what is inside.

From Einstein's theory we know that for light particles, everything else that moves with velocity smaller than the speed of light, 'looks like frozen, no move'. How would the elements inside the atom look?

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Length contraction: $L'=L_0 \times \sqrt{1-\frac{v^2}{c^2}} = 0 \text{ in your example}$ In physical expression, he will measure nothing and will see nothing since everything has a 0 length relative to him in addition to the fact that no material body can reach the speed of light in vacuum and that you can't observe electrons. –  user29727 Oct 20 '13 at 11:12
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The FAQ should say: please, no speed of light observer questions –  jinawee Oct 20 '13 at 11:32
    
@jinawee I thought we should ask dumb questions. –  user29727 Oct 20 '13 at 11:49
    
@Adobe Ironic mode on? And this question has been asked again and again. –  jinawee Oct 20 '13 at 11:57
    
Those users who are familiar with the basics of relativity and quantum field theories might also be interested in "physics on the light-cone" of or "light-cone lQCD" in this context. –  dmckee Oct 20 '13 at 16:02

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First of all, an observed, which is a material body, cannot reach the speed of light in vacuum since it requires an infinite amount of energy.

Second, you can't observe electrons, protons or quarks. There is something called the Uncertainty Principle, it states that you can't be sure about both the location of a particle and it's momentum at the same time. If you were sure where an electron, for instance, is located then you can't be sure about it's momentum that is where it will go and so you can't observe electrons or other particles. Further information about the uncertainty principle: [1]

Third, let's suppose the observer goes at $c$, and that he can observe electrons, then we know from special relativity that length is contracted relative to the observer, and we can calculate that contraction using this formula: $$L'=L_0 \times \sqrt{1-\frac{v^2}{c^2}}$$ where $v$ is the speed of the observer. In your example the observer is moving at $c$, so: $$\begin{align} L'&=L_0 \times \sqrt{1-\frac{c^2}{c^2}}\\ &=L_0 \times 0\\ &=0 \end{align}$$ So the observer will see everything being contracted to a length of $0$, this means that he will see nothing. Can you see something that has $0$ length, no volume? No. So he will not be able to observe what's inside the atom.

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Thank you, I didn't know about Length contraction due to c-speed. –  user31396 Oct 20 '13 at 12:35

protected by Qmechanic Dec 6 '13 at 21:48

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