# A clock devoid of motion

My understanding is that every clock mechanism we have depends on motion of something w.r.t the observer. From atomic transitions to clockwork gears. So, does this property/constraint makes every clock available to use inherently susceptible to effects of relativistic motion?

For example, if you travel at 99% of speed of light with a clock, then there'd be some particle in the clock whose motion, w.r.t you, is also in the same direction, and when added up, it'll surpass 100%.

For above situation to be avoided, every particle that makes up the clock will have to be:

1. Stationary w.r.t observer, but if this happens, the (frozen) clock can't tell time to the observer.
2. Move in a direction other than that of observer's motion, If this happens, clock can't travel with observer.

Please help me out of this confusion.

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Please restate, since every simple digital display clock answers your question. It is only the clock with hands that move with respect to the face of the clock ( if you call that "observer"). The internal motions are self consistent and have nothing to do with the observer. You must mean something else – anna v Oct 10 '12 at 6:19
Yes I do mean something else, I mean no motion of any kind, think electrons in the circuit. – user12926 Oct 10 '12 at 6:21
Time itself is defined by change. If there were no change of anything with respect to time no time could be defined. The internal motions that are used to measure time do not involve an observer. Unless you call observer electronic interactions. – anna v Oct 10 '12 at 6:25
Question updated. – user12926 Oct 10 '12 at 6:31
Yous still talk about the "w.r t the observer" in the body of the talk. see my answer to a similar question physics.stackexchange.com/questions/39141/… . – anna v Oct 10 '12 at 6:34

## 1 Answer

For example, if you travel at 99% of speed of light with a clock, then there'd be some particle in the clock whose motion, w.r.t you, is also in the same direction, and when added up, it'll surpass 100%.

No, that is not the case. Velocities in relativity add according to the relation

$$u' = \frac{u + v}{1 + uv/c^2}$$

instead of $u' = u + v$ which is what your intuition would tell you. The result of this is that if you are traveling at 99% of the speed of light relative to Observer Bob and some piece of a clock is traveling at any (subluminal) velocity relative to you, Observer Bob will still see the piece of the clock traveling at some speed less than the speed of light, because of time dilation and length contraction.

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