# About extremal aging and SR proper time

1. Apr 15, 2014

### RiccardoVen

Hi,
I know there are already other posts about extremal aging, but all of them are actually closed
and none of them is actually answering to my doubt.
I've just started T&W "exploring black holes", and I just faced the "extremal aging" principle. Actually, this concept doesn't fit very well with what I have understood from SR ( of course I dont' really mean this principle is wrong, just I cannot fully understand it yet ).

1) if we take a time like trajectory in flat ST crossing two events, we know the proper time is the one measured by a clock that passes through both events. In this particular frame, looking at the metric, we can write:

(ΔTau)^2 = (Δt)^2 - (Δs)^2

From here we can see the time delation, i.e. ΔTau is always <= Δt, meaning the proper time is always MINIMUM related to all Δt measured by other observers pasting the rest frame ( sorry for my English, it's difficult to express properly so profund ideas in a language different than mine ).

2) The extremal aging principle, actually seems to state almost the opposite ( at least to me), i.e. the trajectory taken is always the one making the time extremal. For massive particles ( i.e. not photons ) this result actually in MAXIMIZING the time between the 2 events.
This seems logic to me, as depicted also from this site:

http://www1.kcn.ne.jp/~h-uchii/extrem.aging.html

so we can see the straight vertical line, which is actually a geodesic for flat ST, is the one which maximizes the ΔTau. Also the way in which Wheeler is using the twin paradox for introducing "natural" paths which makes time extremal is clear to me: the Earth is actually seen as an inertial frame which takes a "natural" path ( geodesic ) and hence makes the stay-at-rest twin getting older than the other twin, experiencing hence the maximum Tau.

The problem is when I try to mix these 2 principles ( i.e. proper time which is MINIMUM within the rest frame Vs extremal aging, which makes t MAXIMUM for "natural" paths ).

Hope my question and my doubt is clear. I know there must be a stupid assumption I'm making
which makes me taking the wrong conclusion, but I cannot see it.

thanks, regards

2. Apr 15, 2014

### Staff: Mentor

You are confusing two things:
1) Proper time,which is the interval between two points in spacetime on a given path, and which is the same in all frames. Different frames may use different values for the $x$ and $t$ coordinates, but the proper time will come out the same.

In flat spacetime the straight-line path between two events always has the longest proper time. For example, if I just sit still for two years while my my twin flies away at .5c for a year and then spends another year returning.... If we measure time in years and dstnce in light-years, we separate at (x=0,t=0) and reunite at (x=0,t=2), and the metric calculation tells us that the amount of proper time that elapses on my straight-line path is two years. But my twin travels from (x=0,t=0) to (x=.5,t=1) and from there to (x=0,t=2); calculate the proper time on that non-straight path and you'll get $\sqrt{3}$ which is less than two years - the straight-line path is longest.

2) Coordinate time, which is just the time values that we choose to assign to events. Different frames generally use different rules for assigning times to events, so different frames generally end up with different values for $\Delta{t}$, none of which necessarily have anything to do with the proper tme between the two events. Time dilation represents the ratio of $\Delta{t}$ calculated using one frame's definition of $t$ to $\Delta{t}$ calculated using another frame's definition of $t$.

One thing that contributes to the confusion is that the frame-dependent coordinate time between two events is equal to the frame-independent proper time between them if the two events also happen at the same location ($\Delta{x}=0$), which is to say a frame in which the clock is at rest.

Last edited: Apr 15, 2014
3. Apr 15, 2014

### ghwellsjr

This is only true for an inertial clock that passes through two events. This would apply to the "stay-at-home" twin. For the "traveling twin", you have to consider at least one other event, the event where he turns around. Then you calculate his Proper Time to get from the first event to the third event and add that to his Proper Time to get from the third event to the second event. The sum of the two Proper Time intervals for the traveling twin will be less than the one Proper Time interval for the stay-at-home twin.

Remember, in the rest frame of the stay-at-home twin, his Δs is zero so his ΔTau (his Proper Time) is equal to Δt, the Coordinate Time of the frame. In the same frame, Δs for the traveling twin is not zero so his ΔTau is less than his Δt which is presumably half of the stay-at-home twin's Δt.

Does that help? Ask again if not.

4. Apr 15, 2014

### Fredrik

Staff Emeritus
A maximum point of one function can be a minimum point of another. Consider f and g defined by $f(x)=x^2$ and $g(x)=-x^2$ for example.

The proper time of any timelike curve between the two events is equal to the smallest possible time coordinate difference in an inertial coordinate system. So I suppose you could say that the inertial coordinate system in which the geodesic through the two events coincides with the time axis, is the minimum of a function that takes each inertial coordinate system to the time coordinates difference it associates with this pair of events.

The proper time function takes timelike curves to non-negative real numbers, so its maximum is the curve with the greatest proper time. The proper time can be calculated by adding up contributions of the form $\sqrt{(\Delta t)^2-(\Delta x)^2}$, so we get the maximum value when all the $\Delta x$ contributions are zero.

So the maximum point in the domain of the proper time function is a curve that coincides with the time axis of a coordinate system that minimizes some other function.

Last edited: Apr 15, 2014
5. Apr 15, 2014

### RiccardoVen

This is probably the most enlighting answer I could expect. Many thanks for it.
And this is more or less what I was thinking before reading your answer ( thanks for all other answer as well, of course ).
What I realized by myself before reading it sounded like this:

my real source of confusion was applying MAXIMUM and MINIMUM relating them to proper time
without realizing the domain was more or less quite different:

1) when we say MAXIMUM, we are actually taking the "BEST" trajectory between many potential
time-like trajectories, i.e the one which actually maximixes the proper time. In flat ST, this trajectory is of course a straight line and coincides incidentally with the the time axis of one of the possible frame of reference.

2) when we say MINIMUM, we are actually talking about the SAME trajectory but seen from different coordinate system ( so not many potential trajectory, but just one of them ). All these
frames agree about the proper time as Nugatory was saying, so actually I really meant the Δt measured in each frame takes a MINIMUM in the rest frame, in which ΔTau = Δt, since in that frame Δs = 0.

Hope this resemble more or less your answer, comparing to the solution I gave to my doubt.
Let me know what you think about it.

Regards

6. Apr 15, 2014

### Fredrik

Staff Emeritus
Sounds good. Although, I wouldn't describe the "minimum" case as being about "trajectories". It's just about different time coordinate differences between the same pair of events.

7. Apr 15, 2014

### RiccardoVen

Yes I agree with you, Fredik. Of course the minimum doesn't concern about different trajectories at all, as I depicted.
Good. Thanks.

Regards