Solving the Twin Paradox with Lorentz Transformation

[Ken Natton]

I’m not really looking to relight the blue touch paper on this. I did want to acknowledge the contributions from yossell and Austin0. I think, for the most part I have sorted out what you guys are telling me – I’m not claiming to fully understand it all, but I can see some of where I went wrong. I suppose it was always a little naïve to think that there was any simpler way of putting it than had already been presented on the thread. I suppose that I need to keep reading and keep thinking.

But of all the responses I got yesterday, the one that really bothers me is matheinste’s post. Pretty central to the understanding that I thought I had taken is the idea that a clock traveling with an observer measures the time in his or her reference frame only. I confess that I have not previously encountered this term ‘proper time’. This seems to carry connotations of some absolute measurement of time, which matheinste’s assertion that

A clock traveling with an observer measures the proper time along the worldline of the observer. It is a measure of the spacetime path length and is frame independent.

seems to confirm. But this is fundamentally against what I thought special relativity asserts.

The ... time measured by a clock with the stay at home twin is not the same as the ... time measured by the traveler ...

is more in line with my understanding, but I don't see why it follows that

...and so the spacetime path length is not the same.

Would someone care to expand on this for me?

 

[yossell]

Hi Ken,

remember that, while times and (spatial)distances are all frame-relative in relativity, the Minkowski separation between two points is *not* frame-relative, and so the Minkowski `length' of a path in space-time is not frame-relative either. That's what Matheinste is referring to by his clock traveling with the observer - it's measuring the Minkowski length of the path and so measures something frame invariant.

Now, for an inertial clock traveling from event A to event B, these two events have the same spatial coordinate in that clock's frame - since the formula for a Minkowski separation between two events calculated in a frame F is: -temporal distance^2 + spatial distance^2 (quantities calculated from within frame F), and since for this frame spatial distance^2 is zero, the proper time alone effectively measures the Minkowski separation.

It's still true that, in other frames, other clocks will disagree about the temporal separation between these two events.

Hopefully that resolves the tension you felt between the first two quoted statements.

As for `so the spacetime length is not the same', I think there is nothing more to this than the idea that two paths between two events do not have the same (Minkowski) length - it is effectively the triangle inequality - the length of a straight path from A to B is not the equal to the length of the straight path from A to C plus the length of the straight path from C to B, even though the VECTOR (AB) equals the VECTOR SUM of (AC) with (CB)

 

[Austin0]

IMO
Your initial feeling that a clock meaqsures the time in its own frame is correct.
This in fact is proper time

Matheniste is referring to the calculation of that time from other frames which is derived from spacetime path length. And will agree with other frames calculations regarding the same frame

Given that the proper elapsed time of the stay at home twin is greater infers that the spacetime path length must be shorter.

I am gald if I have helped.

 

[matheinste]

Hello Ken.

EDIT. Yossell beat ne to it.

The interval is central to the geometry of spacetime and is frame invariant, every observer agrees upon its value. It is analogous to the distance between two points in Euclidean space which can be calculated from the coordinate values of the points, but because of the minus sign in the metric it can take positive, zero or minus values unlike the Euclidean distance which is always positive. Any path in spacetime can be approximated as closely as we wish by a series of small intervals along the path. So the "length" of a spacetime path is invariant because it is the sum of invariant intervals. For a timelike curve, that is one in which all intervals making up the curve are timelike and which is the only type of path along which a body possessing mass can move, the proper time, that is the time interval measured by a clock traveling with the object, is a measure of the path length and is invariant.

Do not confuse the spacetime path length with the spacetme distance (interval) between two events. There are an infinite number of spacetime paths between two events but only one interval. But all the path lengths (proper times) are invariant.The stay at home twin's path length is the same as the spacetime interval but the traveling twin's is not.

An object traveling inertially experiences the greatest possible proper time between two events at which it is present and so the stay at home twin, being inertial, ages more than than his twin whatever path his twin takes between the two events of separation and reuniting.

Matheinste.

 

[Ken Natton]

My opinion, for what it is worth, is who beat whom to reply is not important. Differing perspectives are always useful and those last three replies add up to something very helpful for me. I hope yossell and Austin0 won’t mind if I say that matheinste’s post #184 delivers the most solid boot in the backside in the direction of a better understanding for me. I’m sure you understand that I need to go and chew the cud for a while. I hope to cross paths with you all again very soon.

 

[yossell]

I hope yossell and Austin0 won’t mind ...

On the contrary, I am now a broken man. Expect a stiff letter from my lawyer.

 

[phyti]

Would someone care to expand on this for me?

First you should understand this is an idealized, isolated, thought experiment.
The Earth is not in the 'chosen' fixed frame of reference, and the outbound and inbound legs will not be equal. When a clock moves,it runs slower, a result of the speed of
light in space being constant and independent of its source. The clock used by the Earth twin is moving (relative to the sun, at the least). The clock used by the a-naut is moving
(relative to the earth). Both clocks are thus affected by time dilation, but we can
only measure the difference in speed of both clocks, which is not enough information to decide if the clock rates are different.
SR is designed to produce symmetrical results when comparing one linear motion to another, and the method used to establish simultaneity for a frame of reference produces
distorted times and distances. The simple observations of diverging or converging paths will not determine if there is a difference in the clocks (aging of the twins).

When one twin rejoins the other, the symmetry and uncertainty is removed. For the returning twin, either the outbound or inbound leg, or both had to be at a speed faster
than the speed of the other twin, therefore that clock would have accumulated less time than the other clock, and the loss cannot be totally compensated by any gain in the other
leg of the trip.
The 2-part trip will lose more time than the 1-part trip.