In a Lorentzclock the light is bounced between the two mirrors with his speed of appr. 300.000km/s. Now when things speed up their time goes slower. But the light is always c, so in the Lorentsclock it also goes with c independent on the speed of the clock. But how is it possible that times slows down, but the light in your Lorentsclock doesn't/can't go slower while every bounce would be a second.

Perhaps this situation could be compared with light in a big gravity. But there the light is making a bigger distance due to curvature of spacetime. So the light still is c but looks going slower because the distance is bigger. Is that also the case with light if you travel with it?

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    $\begingroup$ I am down voting this question, as its not clear. $\endgroup$ – AMS Jul 15 '16 at 9:59
  • $\begingroup$ Ok, the question is why time could slow down when speeds increase while the light in your Lorentzclock can't slow down while the speed of this ligth gives you the time. So this clock shouldn't go slower as the clock on earth. $\endgroup$ – Marijn Jul 15 '16 at 10:02
  • $\begingroup$ Is there really nobody who understands my question, or could correct it?! $\endgroup$ – Marijn Jul 15 '16 at 10:23
  • $\begingroup$ @Marijn: Your question really is extremely unclear. Every observer sees light traveling at the same speed $c$. The observer traveling with the clock sees the light traveling a certain distance. The "stationary" observer sees the light traveling a greater distance. Therefore the "traveling" observer does not conclude the clock has slowed down, but the "stationary" observer does conclude that the clock has slowed down. Does this answer your question? If so, then others have answered it also. If not, then I don't understand what your question is. $\endgroup$ – WillO Jul 15 '16 at 13:06
  • $\begingroup$ It is true that the traveling observer do not conclude that his clock is slowed down unless he goes back to earth and sees that the clock of the stationary observer is ahead. Now the question is how this possible is as both use a Lorentsclock depending on the speed of light which has in all frames the same speed. $\endgroup$ – Marijn Jul 15 '16 at 13:11

What you are asking is just another version of the Twin Paradox. A twin leaves on a near lightspeed journey to space. When he comes back his twin is now much older than he is. In a Lorentz clock each bounce is a second. Since the speed of light remains the same, the distance goes larger between bounces.enter image description here enter image description here The distance of the light appear as longer to the observer but not to the person holding the clock. This is why the person holding he clock doesn't experience everything in slow-motion.

The distance between the clock's mirrors is what is changing. The observer sees the mirrors moving. Imagine that there are three meter markers . A 150000meter marker, a 300000 meter marker, a 450000 meter marker. If the ship is going at 99.999999999999999999% the speed of light, then, by the time the the first bounce happens. The mirrors have moved. In the eyes of the observer, for the light to keep the light bouncing, it has to take a longer, diagonal path.

  • $\begingroup$ That is true, but now the question is how does that occur with a lorentzclock as for both persons light travels wiht the same speed in their clocks $\endgroup$ – Marijn Jul 15 '16 at 10:52
  • $\begingroup$ @Marijn because in a Lorentz clock each bounce is a second. Since the speed of light remains the same, the distance goes larger between bounces. $\endgroup$ – Jimmy360 Jul 15 '16 at 10:55
  • $\begingroup$ So that is also true for the person who holds this clock in its own hands like they are in the same frame of reference? $\endgroup$ – Marijn Jul 15 '16 at 10:58
  • $\begingroup$ @Marijn No, they are two different distinct frames of reference, and therefore they observe two distinct things. $\endgroup$ – Jimmy360 Jul 15 '16 at 10:59
  • $\begingroup$ But how could one whole thing (a person with a clock in his hands) consists of two different frames of reference. Is the distance between them making this distinction, if so I don't understand why there should arise a picuture like you added. But if it is true than my problem is solved. $\endgroup$ – Marijn Jul 15 '16 at 11:09

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