What does "invariance of proper time" mean?

[ghwellsjr]

I think that what he meant by $d\tau_1$ and $d\tau_2$ is $\sqrt{dt^2-dx^2}$ and $\sqrt{dt'^2-dx'^2}$.

No, that's not what he meant, at least that not what he said. He said:

$d \tau_1^2 = dt^2$

and:

$d \tau_2^2 = dt'^2 - dx'^2 = \frac{dt'^2}{\gamma^2}$

And I'm not just referring to the trivial squared/square root difference. Look also at his post number #20.

 

[ChrisVer]

It's the same thing...
$$d \tau_1^2 = dt^2 - dx^2= dt^2 (dx=0)$$ or if you like write it $$d \tau_1 = \sqrt{dt^2 - dx^2}=dt$$
for the same reason...
On the other hand:

$$d \tau_2^2 = dt'^2 - dx'^2= dt'^2 (1 + \frac{dx'^2}{dt'^2}) = \frac{dt'^2}{\gamma^2}$$ or else:

$$d \tau_2 = \sqrt{dt'^2 - dx'^2}= \frac{dt'}{\gamma}$$ ...

I don't understand what you are trying to say...

 

[ghwellsjr]

Just wanted to say that I like Matterwave's approach. Of course all other mathematical / geometrical concepts that have been mentioned are good and necessary, but if we want to be didactic and pedagogical, one has to unveil the phsyical background which those concepts "represent". Thus in terms of measurement, I think it is clear in the literature that proper time between events A and B is what a watch being present at both events measures, ie its ticks between A and B. Hence we are talking, yes, about timelike events, since a material and hence infra-luminal object (the clock in question) has "moved" in spacetime from one to the other. In the frame where such clock is at rest, the same has only moved in time. In other frames, it has moved both along space and time. And when all frames calculate the spacetime interval between the two events, by plugging into an identical formula [sqr (∆t2 - ∆x2)] their respective time and distance measurements, they all get the same value, that is why it is invariant.

But I would go one step further. This measurement is invariant because it reflects a physical happening, which is often the key to solving a problem. (You know, reality is usually invariant...) The measurement of the watch present at both events, its chronological ticks are the direct image or reflection of a phsyical issue that may be the object of a practical problem. For example, you may want to know if Mr X, who has witnessed event A, will survive until event B, knowing that his health will only allow 20 beats more of his heart (20 biological ticks). Or you may want to know if his clock itself will survive to event B, in view of its mechanical state.

If Mr X. wants to get to event B after being at event A, as long as they are time-like separated, he can get there before his heart gives out, no matter how soon that is. But not if he thinks that he has to be limited to the spacetime interval between those two events.

For example, here is a spacetime diagram showing Mr X reaching event A and wanting to get to event B before 20 more beats of his heart:

He can calculate the spacetime interval to event B as 24 beats so if he makes a bee-line to it, he will succumb before getting there as shown here:

However, if he just continues for a while longer without accelerating until sometime later, he can get to event B before his 20th beat:

Now the point of this exercise is to address the OP's question which is about the invariance of Proper Time. The spacetime interval is one example of an invariant Proper Time where an inertial clock goes between the two events in question. Note that these two events are not at the same Coordinate Distance in this RF. But another example is the second non-inertial path which has a different Proper Time interval between the two events but is also invariant. My point is that we don't want to be mislead that it is only an inertial clock that measures an invariant Proper Time interval, the same is true for any clock that passes through both events.

 

[ghwellsjr]

It's the same thing...
$$d \tau_1^2 = dt^2 - dx^2= dt^2 (dx=0)$$ or if you like write it $$d \tau_1 = \sqrt{dt^2 - dx^2}=dt$$
for the same reason...
On the other hand:

$$d \tau_2^2 = dt'^2 - dx'^2= dt'^2 (1 + \frac{dx'^2}{dt'^2}) = \frac{dt'^2}{\gamma^2}$$ or else:

$$d \tau_2 = \sqrt{dt'^2 - dx'^2}= \frac{dt'}{\gamma}$$ ...
Sure, if you're going to drop the subscripts and merely point out that we use the same formula to calculate the Time dilation of each observer, of course there's no problem with that. That's what I programmed into my application that draws my diagrams. But you started with this from post #20:

But you have two reference frames in each diagram...one moving and one stationary...
In any diagram the times are not invariant because the one [red or blue] line is a Lorentz transformed RF of the other [blue or red].
This is reflected in writing [stationary=1, moving=2]:
$d \tau^2_1 = dt^2_1 = dt^2_2 - dx^2_2 = \frac{dt^2_2}{\gamma^2}= d \tau^2_2$

That's what was confusing, especially when you said there were two reference frames in each of my diagrams and that the times are not invariant. I still don't know why you would write something like that.

And then you accused me of making false statements in post #22:

I am saying you have an observer on the blue line, that doesn't move wrt to it, so as the blue line progresses in spacetime, he is measuring the time $dt_1$ alone with the ticking of the clock (his $dx=0$),,, So your RF is that which moves together with the blue one...
Now your red observer is moving with respect to the blue. His measurements are thus not only $d \tau_1=dt_1$ (as you tried to note with the simultaneous point of acceleration) but he is also measuring $dx$ . So the proper time he measures it
$d \tau_2^2 = dt_2^2 - dx^2_2 = dt_2^2 (1 - \frac{dx_2^2}{dt_2^2}) = \frac{dt_2^2}{\gamma^2}$

What you meant by the non matching simultaneous times is that you are trying to say that $dt_1 =dt_2$ which is obviously wrong since they are not at rest wrt each other.
What is though the same for both of them in any diagram, is that $d \tau_1 = d \tau_2$

I have asked for clarification. Or a retraction would do fine.

 

[Saw]

My point is that we don't want to be mislead that it is only an inertial clock that measures an invariant Proper Time interval, the same is true for any clock that passes through both events.

Ok, noted. If I understand it well, what several of you were pointing out is that what is technically called "Proper time" and is invariant is not only the time measured by a clock traveling inertially between the two events (which is the ST interval) but also the time measured by a clock present at both events after moving through some non-inertial path.

If so, ok, I guess that my attempt at "making the explanation more physical" should bear in mind that complication. That would require some rephrasing. But the basic idea is still valid, isn't it? The proper time is a direct and simple image of what happens in the example. You could learn it directly and simply by putting a watch around the wrist of Mr X, while he jumps from one RF to another (which also makes the watch change RFs). Or, if you are located in another RF which is less lucky, you will have to use the combination of several time and distance measurements so as to indirectly, with a little more complication, reach the same conclusion about the proper time between the two events and hence about what actually happened, which is obviously the same as for the first observer (ie, since proper time and the reality that it mirrors are invariant).

 

[ChrisVer]

From a prespective there are two RFs in each of your diagram... In the one we are talking about, the blue line observer is the rest RF and the red line observer is on a Lorentz boosted RF moving with $u$ relative to the blue. Think about a muon and an observer on earth... if you are on the muon's rest RF then the observer is on a boosted frame wrt you..
by times I meant coordinate times. Obviously $dt \ne dt'$... the observer at rest will see that time to reach the acceleration point to be $dt=10 ~yrs$ but the time for the moving one will be $dt'=10 \gamma ~yrs$ (but the elapsed proper time will be the same for both)...

Then what I said about the simultaneous point of acceleration is that the proper time to reach acceleration, both of them measure, is the same. The coordinate time interval is not. Simultaneity is related to the coordinate time [but the clocks are not synchronized anymore] and not the proper time. In analogy to the muon example I mentioned above, the acceleration point can be replaced by some event- let's say the muon's decay- so the elapsed time the muon will measure is $dt=2.2 ~\mu s$ whereas for the observer on Earth it can be much larger depending on their relative velocity by the gamma factor. Obviously in both the reference frames [muon, observer] the event of the decay doesn't happen simultaneously, but they both agree on some value- the muon's lifetime[ at rest]=proper time

 

[ghwellsjr]

My point is that we don't want to be mislead that it is only an inertial clock that measures an invariant Proper Time interval, the same is true for any clock that passes through both events.

Ok, noted. If I understand it well, what several of you were pointing out is that what is technically called "Proper time" and is invariant is not only the time measured by a clock traveling inertially between the two events (which is the ST interval) but also the time measured by a clock present at both events after moving through some non-inertial path.

Proper Time is the time measured by a clock no matter what it does. You don't have to stipulate anything about it being inertial or non-inertial. You don't have to limit it to any particular events. Every tick of the clock is an event. Every fraction of a tick of the clock is an event. Every interval between any of these events is invariant.

If so, ok, I guess that my attempt at "making the explanation more physical" should bear in mind that complication. That would require some rephrasing. But the basic idea is still valid, isn't it?

Yes, I can't think of anything more physical than the ticking of a clock. You don't have to work at making it more physical though. It's already there.

The proper time is a direct and simple image of what happens in the example. You could learn it directly and simply by putting a watch around the wrist of Mr X, while he jumps from one RF to another (which also makes the watch change RFs).

I cringe whenever I hear someone using this phraseology. Mr X doesn't jump frames just because he accelerates. Note that in my example in post #38, Mr X is never at rest. I'm sure what you mean is that he starts out at rest in one Inertial Reference Frame (IRF) and after he accelerates, he is at rest in another IRF. Although that is true, it is also true that he may start out at rest in one IRF and then he accelerates and is no longer at rest in that same IRF. But in my example, Mr X starts out at one speed according to the IRF in which I defined the scenario and after acceleration, ends up at a second speed according to the same IRF. In what sense do you consider him to jump or change frames?

Or, if you are located in another RF which is less lucky, you will have to use the combination of several time and distance measurements so as to indirectly, with a little more complication, reach the same conclusion about the proper time between the two events and hence about what actually happened, which is obviously the same as for the first observer (ie, since proper time and the reality that it mirrors are invariant).

Now you've got me worried. Are you talking about your Mr X scenario? If so, who is the first observer?

 

[ghwellsjr]

From a prespective there are two RFs in each of your diagram... In the one we are talking about,

That would be this diagram:

the blue line observer is the rest RF

I have remade a diagram showing just the portion of interest for the rest RF of the blue line observer:

and the red line observer is on a Lorentz boosted RF moving with $u$ relative to the blue.

To show this, we start with the red line observer in his rest RF:

And then we boost him using the Lorentz Transformation so that he is traveling at a speed of u=0.6c:

And then we combine the boosted red line observer's RF with the blue line observer's rest RF to get this:

Think about a muon and an observer on earth... if you are on the muon's rest RF then the observer is on a boosted frame wrt you..
by times I meant coordinate times. Obviously $dt \ne dt'$... the observer at rest will see that time to reach the acceleration point to be $dt=10 ~yrs$

Yes, the blue line observer at rest has a Proper Time equal to the Coordinate Time of 10 years.

but the time for the moving one will be $dt'=10 \gamma ~yrs$

Since gamma at u=0.5c is 1.25, this evaluates to 12.5 years but I don't see anything in any of the diagrams that corresponds to that coordinate time. It seems to me that you should have used $dt'=8 \gamma ~yrs$ from the red observer's rest RF to his Lorentz boosted RF which evaluates to a Coordinate Time of 10 years. Please explain.

(but the elapsed proper time will be the same for both)...

This doesn't seem correct to me. The blue line observer's final Proper Time is 10 years and the red line observer's Proper Time is 8 years which are different, not the same. Please explain.

Then what I said about the simultaneous point of acceleration is that the proper time to reach acceleration, both of them measure, is the same.

Are you saying that the blue observer measures the Proper Time of the red observer at the point of acceleration to be 8 years, just like the red line observer measures? If so, what has that got to do with simultaneity?

The coordinate time interval is not. Simultaneity is related to the coordinate time [but the clocks are not synchronized anymore] and not the proper time.

Simultaneity is related to two different events and whether or not they have the same Coordinate Time. What two events are you talking about?

 

[georgir]

In Euclidean geometry, the distance between any points and the length of any line (curved or not) are invariant under rotation transformations. What does that mean?
It is something you would consider quite obvious so you might even say the question is stupid.
It means that even though coordinate values and coordinate differences can both change from the transformation, the sum of the squares of coordinate differences (and therefore its square root as well, which we call "distance") remains the same.

In relativity, space-time intervals between events ("distance") and proper time along any world-line ("length") are invariant under Lorentz transformation.
It means the same as above - specific coordinates and the differences between them can change, but the sum of the squares of the coordinate differences (with the distinction that time's square is taken to be negative) remains the same.

"Time" between two events is a difference in coordinates, and is changed by Lorentz transformations.
But "proper time along a specific worldline" is analogous to "length of a specific curve" in Euclidean geometry, and is not changed.
"Proper time" between two events in general is a bad phrase, just as "length" between two points - because there can be many lines connecting them. If someone uses it, they most likely mean "space-time interval", the analog of "distance", but the ambiguity is more dangerous in context of "invariance" discussions. Consider how "Length between two points is invariant" may be misinterpreted as the incorrect "The length of all curves between two points is equal" instead of the probably intended "The distance (length of the shortest line) between two points is invariant under rotation".

 

[ChrisVer]

This doesn't seem correct to me. The blue line observer's final Proper Time is 10 years and the red line observer's Proper Time is 8 years which are different, not the same. Please explain.

Then you are saying that a Lorentz transformation [Lorentz Boost] is going to change the proper time that we call invariant under Lorentz transformations? I am not the one who needs to explain but you. The problem is that the proper time is not that $dt'$ but it will be the $dt'/\gamma$. So indeed you have to measure $dt'=12.5 ~yr$ in order for the proper time to be invariant for both.
In order to see the 12.5yr in diagram you have to redo it in the red's rest reference frame... Then they will be equivalent but the red line will have to travel from t=0 to t=12.5 yrs...
The blue will measure $d \tau = dt = 10~yr$, so I got $d \tau$. Using the invariance of it, you can determine $dt' = d \tau ~\gamma=10 \gamma$

 

[ghwellsjr]

Then you are saying that a Lorentz transformation [Lorentz Boost] is going to change the proper time that we call invariant under Lorentz transformations?

No, I said the Proper Time of the turn around event for the red observer is 8 years in both his rest frame and in the frame in which he is traveling at 0.6c and I have shown this in several diagrams.

I am not the one who needs to explain but you. The problem is that the proper time is not that $dt'$ but it will be the $dt'/\gamma$. So indeed you have to measure $dt'=12.5 ~yr$ in order for the proper time to be invariant for both.
In order to see the 12.5yr in diagram you have to redo it in the red's rest reference frame... Then they will be equivalent but the red line will have to travel from t=0 to t=12.5 yrs...

OK, if you insist:

And now after boosting so that the red observer is traveling at 0.6c:

Is that what you want? As always, the Proper Time of the last (uppermost) dot is 12.5 years in both frames but I don't see how that relates to what you are explaining.

The blue will measure $d \tau = dt = 10~yr$, so I got $d \tau$. Using the invariance of it, you can determine $dt' = d \tau ~\gamma=10 \gamma$

Why are you talking about the blue observer or his Proper Time? We aren't boosting his rest RF. What has the blue observer or his Proper Time got to do with the red observer's Proper Time?