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I thought a scenario like; lets say I am looking an object and there is nothing except this object. Is there a way to understand that if this object is stay on its position or if object moving with a constant speed and also I am moving as same constant speed with this object ? (consider there is not any friction etc.)

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    $\begingroup$ There is no way to tell the difference. This is what Galilean relativity is about. $\endgroup$ – Stéphane Rollandin Jul 5 at 19:20
  • $\begingroup$ If it is possible to compare clocks between constant velocity & zero velocity situations, can we say constant velocity time will lower than zero velocity time ? @StéphaneRollandin $\endgroup$ – Ozn Ozi Jul 5 at 19:34
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    $\begingroup$ It’s impossible. This is the principle that gave the theory of relativity its name. First noticed by Galileo in 1632 for mechanics, it was generalized by Einstein in 1905 to all physics laws. $\endgroup$ – J. Manuel Jul 5 at 20:13
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You cannot tell the difference. In fact many would say there is no difference. If you're really in a universe where there's nothing except the object like you say, there would be no difference between just sitting still and moving at a constant speed. There are no stars to see whizzing by, no atmosphere to make you feel the wind, nothing to make the light so you can measure redshift with special relativity.

If you're curious, this is one of the fundamental assumptions of special relativity.

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Constant velocity and speed have no meaning unless you specify the frame of reference with respect to which it is measured or observed. (The only exception is the speed of light).

If you see an object “moving”, then it is moving with respect to your frame of reference. If you see it as “still” it is still with respect to your frame of reference.

But in the frame of reference of somebody who is with the object in a windowless compartment the object is always at rest and the person has no no way of knowing what you are seeing.

All this assumes constant velocity.

Hope this helps

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  • $\begingroup$ Thank you for this answer too $\endgroup$ – Ozn Ozi Jul 6 at 7:55

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