Sorry for a bit of a basic question, but want to clarify things in my head. Is proper time quantified by the amount of physical process that an object, or physical system undergoes, for example the decay of an unstable element? And is the reason why proper time is independent of the frame of reference because such a physical process occurs without needing to introduce any coordinate system, thus regardless of the reference frame that you measure the process occurring in, it will occur at the same rate in the same manner?

For example, consider two observers equipped with identical clocks. One of the observers is considered at rest on Earth and the other is in a spaceship moving at a considerable fraction of the speed of light. Now is the notion of proper time that each observer, in their own reference frame will observe their clock to tick (cycle) at the same rate (however they will not observe each others clocks to tick at the same rate, as coordinate time is not frame-independent. Indeed from their own reference frame they will observe the clock in the other reference frame, moving relative to them, to tick at a different rate, as they are using their own coordinate time to measure the process). The point being is that if they both have identical clocks, then if the clock given to the observer on Earth ticks in a certain way for that observer in their rest frame on Earth, then the clock given to the observer in the fast moving spaceship will tick in exactly the same way in their frame of reference on the spacecraft (i.e. it won't suddenly change how it ticks to being aperiodic or non-linear, etc.)


I think you are mixing up two different concepts, which is muddying the waters.

Firstly, relativity (both special and general) is a geometrical theory and the proper time for an observer has a precise definition as the length of a world line along which the observer travels (give or take a factor of $c$). This length is calculated using the metric. As Titus says, the proper time is equal to the time experienced by the observer, and it's easy to see why this is. In the observer's rest frame they aren't moving ($dx = dy = dz = 0$) so the world line is a straight line up the time axis. The length of this line is obviously equal to the elapsed time experienced by the observer.

It is an assumption in relativity that the proper time is an invariant i.e. any observer in any frame, accelerating, in a gravity well or whatever, will calculate the same proper time. To see how this assumption produces all the weird effects like time dilation or length contraction see How do I derive the Lorentz contraction from the invariant interval?, and many other related questions on this site.

There is a second (but related) fundamental principle in relativity that a local experiment will always produce the same result i.e. that if we put you in a sealed box in free fall you cannot tell how the box is moving by a local experiment. So if your experiment is to observe some clock mechanism, whether it's a caesium atom or a clockwork watch, it will always tick in the same way. But this process of timing is not what determines the proper time.

  • $\begingroup$ So is proper time as an invariant quantity the notion that an observer in their rest frame will experience time passing the same way (in their own frame) as another observer would experience time passing in their own rest frame, i.e. each persons "rest frame time" passes in the same way? Sorry, I'm not wording this very well, but I think I get what you mean. $\endgroup$ – Will Jun 9 '15 at 15:59
  • $\begingroup$ @Will: My last paragraph points out that physics is locally invariant. Since your brain is a mechanism (albeit a rather squishy one) that obeys the laws of physics and is necessarily local to you, it is going to operate in the same way regardless of what your rest frame is doing. So whether you are floating in space, orbiting the Earth or falling into a black hole your brain will operate the same way and therefore presumably experience time in the same way. $\endgroup$ – John Rennie Jun 9 '15 at 16:25
  • $\begingroup$ So how one experiences time is independent of the frame that we are at rest in (whether it be on the earth, or in a spacecraft speeding along at 0.7c). I think I get it now, thanks for your help. $\endgroup$ – Will Jun 9 '15 at 16:37
  • $\begingroup$ @Will: Yes. As a side note, I would be cautious about trying to interpret the flow of time. In relativity there isn't any flow of time - time is just a coordinate like the other coordinates. $\endgroup$ – John Rennie Jun 9 '15 at 16:38
  • $\begingroup$ Yes, good point. I was using it really as an aid to try an give an intuitive explanation of my thoughts on proper time, but perhaps I should've been more cautious. $\endgroup$ – Will Jun 9 '15 at 16:42

Proper time of an observer is time as measured by the observer's own clocks. So it's obviously frame-independent because calculating proper time of a given observer requires to use his own frame of reference.

  • $\begingroup$ Yes, that's kind of what I thought. Generally then, is proper time measured in terms of physical processes that occur in a frame-independent manner? For example, is this why 1 second is defined by the number of periods of transition between two hyperfine levels of a Cesium atom, because it is a regular periodic process and thus defines what a second means in a frame independent manner? $\endgroup$ – Will Jun 9 '15 at 13:24
  • $\begingroup$ To be honest, I don't know why we have taken such a definition of 1 second. Searching on the web (physics.nist.gov/cuu/Units/second.html) made me think that's because of mainly technical reasons. But if I understand your point correctly, you are wrong. One second does not mean nothing if you don't specify in what set of clocks (so "1 sec for who?"). There are indeed formulas that relate the length of time intervals of different observers. $\endgroup$ – Titus Petronius Jun 9 '15 at 13:34
  • $\begingroup$ What I meant really was that in the rest frame of observer A, a cesium atom (at rest with respect to observer A) behaves in a certain way, e.g. the way it transitions between hyperfine levels. Now, if an observer B also has a Cesium atom in their rest frame (that is at rest with respect to them) then they will observe it behaving in exactly the same way, i.e. the same periodic transitions etc. So the propertie of the Cesium atom (the transitions that it undergoes) are frame-independent... $\endgroup$ – Will Jun 9 '15 at 14:52
  • $\begingroup$ ... but not observer independent, as if observer A is moving relative to observer B, then observer B will observe these processes to occur at a slower rate. $\endgroup$ – Will Jun 9 '15 at 14:52

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