Solving the Twin Paradox with Lorentz Transformation

[Ken Natton]

Well, all I can say, Austin0 is that I do not understand where my interpretation changes. It seems perfectly consistent to me. Again, I suppose you could be pedantic and say that they would only ever disagree about spatial distances and time durations if they were both taking a Newtonian Mechanical view. If they had both studied relativity, they would understand that distances and times would not appear the same for each other and if they were comfortable with the mathematics, they could even work out exactly what the distances and times would appear to be for each other. But the point is that they don’t really need to. If they do their Minkowski calculations with their own distance and time measurements and arrive at a consistent spacetime distance, then there is no disagreement.

As I have said, I accept that the diagrams that appear further up this thread show best how the two viewpoints change over the duration of traveling twin’s journey. It just seemed to me that there was a much simpler way of understanding why traveling twin is younger when the journey is over, and why such a notion is consistent with relativity theory.

 

[Ich]

And by net journey, I meant the line that joins the start of the first component vector on the space time diagram with the end of the last component vector, which is, I think, the sum of all the component vectors.

Ok, so that's distance "bee-line". Yes, that's the (length of the) sum of the vectors.

So traveling twin’s history involves a complicated sequence of component vectors, but the understanding of the age difference with her twin depends only on this summary vector.

I don't agree. The age differnce comes from the different path lengths. The traveling twin's path is shorter than the Earth path. The reason is the http://en.wikipedia.org/wiki/Triangle_inequality#Reversal_in_Minkowski_space" in Minkowski space. The sum of the lenghts is shorter than the length of the sum.
Conceptually, it's nothing else than the explanation why (in euclidean space) two sides of a triangle are longer than the third. I think you would not explain it with "motion along x adds to motion along y", its easier if you don't refer to the (anyway arbitrary) coordinates and stick to the geometry.

 

[Ken Natton]

Well there you go, I asked for engagement, and now it comes so thick and fast that I can’t keep up with it. Be careful what you wish for, I suppose.

I have read enough of this and other threads to know that you are someone whose comments I must attend to, mathienste, but your comment does not align with the understanding I had taken, that whatever the variance in viewpoints of time and spatial distance, spacetime distances are consistent for all observers. All I can do is return to the text from which I thought I had taken this understanding, and see if I can find where I misunderstood it, or return to you if I conclude that I had not misunderstood it.

 

[yossell]

edit: beaten by ich

The vectors I referred to I had meant to be the spacetime vectors, which are, as I understand it, the world lines.

The spacetime vectors are not the world lines. Assuming that by vector you mean a straight directed segment joining two points, we can identify vectors with straight line journeys. World lines can be curved, and can be segmented - but a vector can't.

And by net journey, I meant the line that joins the start of the first component vector on the space time diagram with the end of the last component vector

I don't know what this means. What's the first component vector? What's the last component vector? Are you breaking the journeys up into a series of smaller journeys and treating the overall journey as a sum of these smaller journeys?

So traveling twin’s history involves a complicated sequence of component vectors, but the understanding of the age difference with her twin depends only on this summary vector.

You need to know the journey to compute the age - the starting point and ending point aren't enough. The path is essential. It's true that minkowski separation between two points is independent of coordinate system - so you're right about that - but you can only read the time off this figure for a clock that travels inertially - i.e. in a straight spatio-temporal line - between the two points.

To work out the time for the twin who travels, if he does it in two distinct journeys, you can just sum the times of the two journey. But, as it's a different path, this sum won't be the same, any more than the sum of two sides of a triangle equals the third.

Hope this helps

 

[Austin0]

Well, all I can say, Austin0 is that I do not understand where my interpretation changes. It seems perfectly consistent to me. Again, I suppose you could be pedantic and say that they would only ever disagree about spatial distances and time durations if they were both taking a Newtonian Mechanical view. If they had both studied relativity, they would understand that distances and times would not appear the same for each other and if they were comfortable with the mathematics, they could even work out exactly what the distances and times would appear to be for each other. But the point is that they don’t really need to. If they do their Minkowski calculations with their own distance and time measurements and arrive at a consistent spacetime distance, then there is no disagreement.

As I have said, I accept that the diagrams that appear further up this thread show best how the two viewpoints change over the duration of traveling twin’s journey. It just seemed to me that there was a much simpler way of understanding why traveling twin is younger when the journey is over, and why such a notion is consistent with relativity theory.

Hi Hen Natton
I think there is a miscommunication here. I doubt that atyy disagreed with your final relativistic interpretation. I know I don;t in any way.

What I was referring to was the contrast between : 1)the Newtonian view and also the view based on the assumption of reciprocal dilation with2)the full relativistic analysis and conclusion.

That the disagreement between the expected results in 1)[i.e same age] and the reality [age differenence]
compared to 2) where the expections agree with the end result,, is a large part of the cause for the popularity and perreneial reoccurrence of the "paradox"

The Newtonian view is the problem for people with no real understanding of SR
The assumprion of reciprocal dilation is the problem for people with some limited understanding of SR. For some people with a broader understanding of the principles involved it is not a problem with regard to the Twin;s "P" itself which is a non=issue.
But it raises questions regarding the various valid and consistent methods of resolution and the role of recirpocal dilation in those methods.

I hope this may clarify things.

 

[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.