When a clock is transported here and there into space and then brought to the same place it differs with the other clock. When particles are accelerated with high speeds and then brought to rest their mass again gets back to its original rest mass. Why?

Answer allegorically please.

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    $\begingroup$ The short way to put it, is that the rate of time's passage does change back. $\endgroup$ – Russell Borogove Aug 27 '13 at 20:16
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    $\begingroup$ Note that the modern standard way to talk about it is that mass is invariant in relativity. Nobody talks about relativistic mass in relativity these days. The factor of $\gamma$ is absorbed in other places and is not counted as part of the mass. $\endgroup$ – Ben Crowell Aug 27 '13 at 21:22
  • $\begingroup$ Why the clock doesnot decelerate to get back to its original time? $\endgroup$ – user28737 Sep 5 '13 at 11:54
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    $\begingroup$ @WaqarAhmad Compare to the classic "racing balls" demo. There is nothing special or unusual about this expect that you (and most people) persist in thinking that time is absolute. $\endgroup$ – dmckee Sep 22 '13 at 16:28

Because time is accumulating, to calculate the time lapse, you integrate. The elementary time interval transforms like mass. The difference is that the total time lapse is done by "summing" over all elementary intervals. For the mass, you don't do this.

For mass:

$$m=\frac {m_0} {\sqrt{1-\frac{v^2}{c^2}}}$$

For time:

$$dt=\frac {dt_0} {\sqrt{1-\frac{v^2}{c^2}}}$$

$$t=\int{ \frac {d t_0} {\sqrt{1-\frac{v^2}{c^2}}}}$$

  • $\begingroup$ false, due to the entropic nature of time. Say, 1kg become 2kg then again back to 1kg, total =4kg. Now let say 5beads are counted here when you go to space you counted 8beads at the meanwhile then when you come back x+5beads already counted on here and not the same 5beads. Because time is also moving here and in space simultaneously. $\endgroup$ – user28737 Aug 27 '13 at 18:34
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    $\begingroup$ This answer is 100% true. $\endgroup$ – user10851 Aug 27 '13 at 19:09
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    $\begingroup$ @WaqarAhmad 1. you didn't mention entropy, nor does it seem relevant. 2. why did you add up the masses? 3. why would the number of beads change? I don't see the analogy: the mass doesn't "actually get heavier" because you're not "actually moving". You can only talk about how things appear to someone other particular reference. $\endgroup$ – Robert Mastragostino Aug 27 '13 at 19:29
  • $\begingroup$ Ok i m picking the concept. when you are accelerated to high speed in space you remain at that speed forever(permanent) and thus time is slowed down for you forever. But when you accelerate mass, its mass increase and will remain that forever, provided any reaction force doesnot stop it. $\endgroup$ – user28737 Aug 31 '13 at 10:02

Because a clock record time, while your mass is just a physical property of your body.

To be clearer : You have two clocks that tick every second. They indicate 10:00. You put one (clock A) into movement. As seen by clock B, the frequency of clock A will be different. It will tick every 1/2 second, for instance. Put clock A again at rest next to B. Clock A will eventually indicates 11:00 and clock B 12:00. But one hour later, clock A will indicate 12:00 while clock B shows 13:00. Got it ? So the frequency of the clock will depend on its relative speed. The frequency will change if the clock is in movement, but will be back to the same value as before if you bring it back at rest.

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    $\begingroup$ So mass and the rate-of-change for time both return to normal. but elapsed time remains the sum of the changes. $\endgroup$ – Mr.Mindor Aug 27 '13 at 18:30
  • $\begingroup$ The way I like to think of it is this: mass is an intrinsic property of an object, while time is not - the object is MADE of mass, but EXISTS in time. $\endgroup$ – MattS Aug 27 '13 at 21:31

I think you mix up time (ie what this or that clock is showing right now), and time's speed (ie, how fast that clock is "ticking" compared to how another clock (possibly moving differently) is "ticking", as seen from a reference point.).

Like the first answer says, you should compare m with dt (time's "speed"), not t (ie, current time, the integration of that dt)

What changes when you move a clock in regard to another is that it's time's "speed" and its mass are changing, then go back to "normal".

The "current time" of the clock you are moving has changed "less" if you compare it to another clock "at rest". The clock that was moving will have, while moving faster, a slower "time speed", or "ticking rate", and then that "time speed" goes back to normal when it comes back to "rest" (compared to the reference clock). The accumulated time is therefore different, and will surely not "go back" (Going from T1 to T2

Of course this is all kind of mind boggling as the very word I used (ticking, backwards) are defined "in time", so it's kind of hard to separate whatr's physically happening from false information derived from the words themselves...


I dont think that's true.

Particles are constantly emitting radiation, and when a particle is moving at high-speed it emits less radiation and it also has a longer period of decay, so a high-speed particle loses mass at an inferior rate than when it was stationary, but still loses mass. You cannot stop a particle from decaying, and its rest-energy will never get 'unchanged'.

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    $\begingroup$ I am quite doubtful but even if that was true, that would be a negligible. In a first approximation, we can say that the mass is the same if you accelerate it then put it at rest. However the difference between two clocks with some relative speed has already been measured. $\endgroup$ – AnSy Aug 27 '13 at 17:11
  • $\begingroup$ It is true, thats why photons theoretically dont decay, because only at light speed particles dont lose energy. $\endgroup$ – eJunior Aug 27 '13 at 21:23
  • $\begingroup$ All that I was saying is that the relativity of time has a impact on the net-mass of the object also. $\endgroup$ – eJunior Aug 27 '13 at 21:32

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