Proper (and coordinate) times re the Twin paradox

[Orodruin]

The difference is in the topology.

There is no difference in the topology of Minkowski space and Euclidean space, they are homeomorphic. The difference is geometrical, not topological. I agree with the rest. The main difference between Minkowski space and Euclidean space is the inner product. Among other things, the inner product on Minkowski space results in a Pythagorean theorem that is similar to, but yet very different from, the one we are used to from Euclidean space, namely
$$
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2.
$$
With $s = c\tau$ and $\tau$ being proper time along a world-line, computing the proper time along a curve is completely analogous to computing the length of a curve in Euclidean space - the only difference being in the different Pythagorean theorem.

As to why Nature is better described by Minkowski geometry than by Euclidean geometry, it is just how Nature seems to work.

 

[SiennaTheGr8]

@Grimble

As for the "contradiction" between what @Mister T and @Dale said, it comes down to a semantic issue that I think I can help resolve.

1. "Proper time" is what a traveler's wristwatch logs.
2. Sometimes we loosely speak of the "proper time between events"—this is shorthand for "the proper time that would elapse during an inertial journey between a pair of timelike-separated events" (equivalent to the "spacetime interval" between them).
3. For every inertial journey there's an inertial frame for which the traveler is at rest (all events on the traveler's world line occur at the same spatial location in this frame). The "coordinate-time clocks" in this frame tick at the same rate as the traveler's wristwatch, so in practice we don't always make a distinction between an inertial traveler's proper time and the coordinate time of the traveler's rest frame (but strictly speaking they are different concepts).

When @Mister T says "If the two events occur in the same place (in some frame) then the time that elapses in that frame is a proper time," he means that there's an equality between the coordinate-time that elapses in that frame and the proper time that would elapse for a traveler during an inertial journey between the events in question.

 

[Mister T]

Can you help resolve this apparent conflict between what you say and what Dale is saying?

But I don't see an apparent conflict. Perhaps if you could explain what you find conflicting?

 

[Dale]

I'm feeling a bit stupid here...
but how is part of the Pythagorean theorem is equivalent to proper time?

Look at the formulas:
$ds^2=dx^2+dy^2+dz^2$
$ds^2=-dt^2+dx^2+dy^2+dz^2=-d\tau^2$
They both represent an arc length, one in a 3D space with signature (+++) and the other in a 4D space with signature (-+++) which is equivalent to (+---).

When we say a rotation, what exactly are we rotating and how?

We are rotating our coordinate system. How is with a Lorentz transform, that is a rotation (boost) in spacetime.

I think that if we take two events with different time coordinates and increase the x coordinate of one of them then the connecting line is rotated,

No. That is not a rotation, that is a different line between different points. We leave the events the same and rotate the coordinate system. It is important to make a distinction between the geometric objects and their coordinates.

Think of, for example, the lightning strikes on the train thought experiment. The two flashes of light are two separate events. Those events have coordinates in the train frame, and those same two events have different coordinates in the embankment frame. The coordinate system is rotated (boosted), but the events stay the same.

Which part of my analysis are you referring to?

The part where you said “In motion it also travels between locations”. You are neglecting the change in time coordinates.

Do this exercise. Use units where c=1. Start with the origin as your first point since it just trivially transforms to the origin. Now, for your second point start with $(t,x)=(1,0)$. Calculate the proper time. Then do a Lorentz transform to a frame with $v=0.6$ relative to the first. Then calculate the proper time again.

How does it differ from Euclidean geometry and what makes it wrong to use Euclidean geometry?

Euclidean geometry can’t work because it only has one type of length. Spacetime needs two, one for the kind of “distance” we measure with rulers and the other for the kind of “distances” we measure with clocks. If $ds^2>0$ we call the interval spacelike and we measure it with a ruler. If $ds^2<0$ we call the interval timelike and we measure it with a clock.

 

[Orodruin]

Look at the formulas:
$ds^2=dx^2+dy^2+dz^2$
$ds^2=-dt^2+dx^2+dy^2+dz^2=-d\tau^2$
They both represent an arc length, one in a 3D space with signature (+++) and the other in a 4D space with signature (-+++) which is equivalent to (+---).
We are rotating our coordinate system. How is with a Lorentz transform, that is a rotation (boost) in spacetime.

No. That is not a rotation, that is a different line between different points. We leave the events the same and rotate the coordinate system. It is important to make a distinction between the geometric objects and their coordinates.

Think of, for example, the lightning strikes on the train thought experiment. The two flashes of light are two separate events. Those events have coordinates in the train frame, and those same two events have different coordinates in the embankment frame. The coordinate system is rotated (boosted), but the events stay the same.

The part where you said “In motion it also travels between locations”. You are neglecting the change in time coordinates.

Do this exercise. Use units where c=1. Start with the origin as your first point since it just trivially transforms to the origin. Now, for your second point start with $(t,x)=(1,0)$. Calculate the proper time. Then do a Lorentz transform to a frame with $v=0.6$ relative to the first. Then calculate the proper time again.

Euclidean geometry can’t work because it only has one type of length. Spacetime needs two, one for the kind of “distance” we measure with rulers and the other for the kind of “distances” we measure with clocks. If $ds^2>0$ we call the interval spacelike and we measure it with a ruler. If $ds^2<0$ we call the interval timelike and we measure it with a clock.

Arguably the most important type is the third, ie, null intervals with $ds^2=0$ although the points are distinct.

 

[Grimble]

Well, now I begin to see...
I seem to be a little hazy with exactly what some terms mean. (duh!)
rotation is not a line being rotated and stretched, as for example, if the later event is moved in space but keeps the same time value; but it is the line rotating and keeping the same length.
I am going to review and check where else I am getting confused; particularly:
worldlines, proper time, coordinate time, intervals and how I see them reflected in Minkowski diagrams.

 

[vanhees71]

Have a look here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Grimble]

Have a look here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Thank you for the reference but I'm sorry that I do not understand that level of maths, not even the symbols or techniques used.

 

[Grimble]

I am going to surprise you now, Dale, by doing just what you say:

Do this exercise. Use units where c=1. Start with the origin as your first point since it just trivially transforms to the origin. Now, for your second point start with (t,x)=(1,0). Calculate the proper time.

The proper time is the time difference in an inertial frame. In this case it will be 1.

Then do a Lorentz transform to a frame with v=0.6v=0.6 relative to the first. Then calculate the proper time again.

With a Lorentz factor γ = 1.25, the transformed coordinates of the second point of the world line will be: x = 0.6, t = 0.8. So the rotated length of the worldline remains at 1. The distance, measured along the x-axis is 0.6 and the time difference is 0.8.

In relativity, proper time along a timelike worldline is defined as the time as measured by a clock following that line.

Now in this case it seems to me that the time measured by a clock would be the time difference = 0.8.
So that should be the proper time, τ = 0.8

 

[jbriggs444]

I am going to surprise you now, Dale, by doing just what you say:
The proper time is the time difference in an inertial frame. In this case it will be 1.

The proper time is only going to match the coordinate time difference when using an inertial frame within which the object is motionless.

In any other frame, you can calculate the proper time using the formula $\sqrt{(\Delta t)^2 - (\Delta x)^2}$ = $\sqrt{1^2-0^2}$ = 1.

With a Lorentz factor γ = 1.25, the transformed coordinates of the second point of the world line will be: x = 0.6, t = 0.8.

No.

Rather than just play pretend Lorentz transformation by throwing time dilation and length contraction factors willy nilly at numbers and pretending that they stick, you need to be doing real Lorentz transformations.

As it stands, you have performed a Euclidean rotation, taking (1,0) to (0.8,0.6). The Lorentz transformation is a hyperbolic rotation. So the result you obtained cannot be correct.

You also need to show the work, not just the results.

So the rotated length of the worldline remains at 1. The distance, measured along the x-axis is 0.6 and the time difference is 0.8.

No, no, no. Wrong length formula.

Now in this case it seems to me that the time measured by a clock would be the time difference = 0.8.
So that should be the proper time, τ = 0.8

The [squared] invariant length of the interval is given by $(\Delta t)^2-(\Delta x)^2$ = $0.8^2 - 0.6^2$ = 0.64 - 0.36 = 0.18. It should have come out to be 1.0. So it is clear that you have messed up somewhere.

Let us Google up the Lorentz transformation.

$$t' = \gamma (t-\frac{vx}{c^2})$$ $$x' = \gamma (x-vt)$$

As @Dale has suggested, we will use v=0.6c which results in a gamma of 1.25

For the starting point (0,0) we have t=0, x=0
The transform yields: [YOU DO THE MATH]

For the ending point (1,0) we have t=1, x=0
The transform yields: [YOU DO THE MATH]

The length formula for the interval between the transformed starting point and the transformed ending point yields: [YOU DO THE MATH]

 

[Dale]

The proper time is the time difference in an inertial frame. In this case it will be 1.

Sure, but use the correct formula to calculate it.

With a Lorentz factor γ = 1.25, the transformed coordinates of the second point of the world line will be: x = 0.6, t = 0.8.

This is incorrect. Use the Lorentz transform formula.

Now in this case it seems to me that the time measured by a clock would be the time difference = 0.8.
So that should be the proper time, τ = 0.8

Please use the actual formula, not just guessing from Wikipedia. We have covered the formula for proper time several times in this thread. It is the formula that looks like the Pythagorean theorem almost.

 

[PeterDonis]

The proper time is the time difference in an inertial frame.

No, it isn't. It's the spacetime interval along the worldline.

in this case it seems to me that the time measured by a clock would be the time difference

No. Proper time is the spacetime interval along the worldline. It is not the coordinate time difference.

 

[Grimble]

The proper time is only going to match the coordinate time difference when using an inertial frame within which the object is motionless.

In any other frame, you can calculate the proper time using the formula $\sqrt{(\Delta t)^2 - (\Delta x)^2}$ = $\sqrt{1^2-0^2}$ = 1.

Rather than just play pretend Lorentz transformation by throwing time dilation and length contraction factors willy nilly at numbers and pretending that they stick, you need to be doing real Lorentz transformations.

Ah, yes! I can see I am getting confused by two frames in one diagram. I have been seeing the Primed frame in terms of the inertial frame, rather than using its own coordinates.

As it stands, you have performed a Euclidean rotation, taking (1,0) to (0.8,0.6). The Lorentz transformation is a hyperbolic rotation. So the result you obtained cannot be correct.

I am trying to grasp this terminology, but what is the difference? What exactly is a hyperbolic rotation?

You also need to show the work, not just the results.

Yes, I see that.

The [squared] invariant length of the interval is given by $(\Delta t)^2-(\Delta x)^2$ = $0.8^2 - 0.6^2$ = 0.64 - 0.36 = 0.18. It should have come out to be 1.0. So it is clear that you have messed up somewhere.

Let us Google up the Lorentz transformation.

$$t' = \gamma (t-\frac{vx}{c^2})$$ $$x' = \gamma (x-vt)$$

As @Dale has suggested, we will use v=0.6c which results in a gamma of 1.25

For the starting point (0,0) we have t=0, x=0
The transform yields: [YOU DO THE MATH]

t' = γ (t - ^vx/_c²), x' = γ (x - vt) using c = 1
t' = 1.25 (0 -0) = 0 x' = 1.25 (0 - 0) = 0

For the ending point (1,0) we have t=1, x=0
The transform yields: [YOU DO THE MATH]

t' = γ (t - ^vx/_c²), x' = γ (x - vt) using c = 1
t' = 1.25 (1 - (0.6 x 0)), x' = 1.25 (0 - (0.6))
t' = 1.25, x' = - 0.75

The length formula for the interval between the transformed starting point and the transformed ending point yields: [YOU DO THE MATH]

(Δs)² = ((Δt)² - (Δx)²)
(Δs)² = ((1.25)² - (-0.75)²)
(Δs)² = ((1.5625 - 0.5625) = 1

 

[Mister T]

I have been seeing the Primed frame in terms of the inertial frame, rather than using its own coordinates.

This seems to be a point of confusion. Both frames are inertial, that is, both the primed frame and the unprimed frame are inertial frames.

Moreover, proper time is the time that elapses, not in just any inertial frame, but in the inertial frame where the two events happen in the same place. In this case, it's the unprimed frame since the value of $x$ is the same for both events.

 

[Ebeb]

Grimble, let me visualize this for you.
(I use a Loedel diagram. A Loedel diagram uses same unit lengths for both reference frames (Minkowski diagram doesn't), which makes it easier to understand that the 'time dilation' is actually not due to a stretching of reference units... or worse, a stretching of proper time units... )

---------------------------For clarity I show how green ref frame gives coordinates to the event[red clock diplays 1.25] :

 

[Dale]

t' = γ (t - ^vx/_c²), x' = γ (x - vt) using c = 1
t' = 1.25 (1 - (0.6 x 0)), x' = 1.25 (0 - (0.6))
t' = 1.25, x' = - 0.75(Δs)² = ((Δt)² - (Δx)²)
(Δs)² = ((1.25)² - (-0.75)²)
(Δs)² = ((1.5625 - 0.5625) = 1

Notice how both the x’ and the t’ coordinates differ from the x and t coordinates of the same event. And yet the spacetime interval was the same.

 

[Orodruin]

Notice how both the x’ and the t’ coordinates differ from the x and t coordinates of the same event. And yet the spacetime interval was the same.

Also, not only the spacetime interval - but the events themselves are the same! Even if they are described by different numbers.

 

[Grimble]

This is Minkowski geometry, not Euclidean geometry. Increasing the spatial displacement (the difference between starting and ending x coordinates) while holding the temporal displacement (the difference between starting and ending t coordinates) constant reduces the length of the (timelike) interval between the end points.

Yes, I can see that, but just what do those measurements represent?
It seems to me that:
the spatial displacement is the distance along the x axis;
the temporal displacement is the distance along the ct axis;
then what is the timelike interval between the endpoints? If both events have the same x coordinate, it is in a resting frame, and the timelike interval = the temporal displacement.
If they have different x coordinates then the frame is moving, the temporal displacement could be labelled ct', and the timelike interval must be the vertical distance to the x axis...
So which of them is the proper time?
if Δτ² = (Δct² - (Δx²)
then surely τ must be the vertical cathetus, x the other catheus and ct the hypotenuse

Yet this is confusing if τ is the proper time, it is the temporal displacement along the rotated worldline which must be ct - the temporal displacement...

Please help me I am confused again...

 

[jbriggs444]

vertical distance to the x axis

This is not Euclidean geometry. Getting out a ruler and measuring the distance on a piece of ruled paper with perpendicular grid lines will not work. I would suggest either learning to read a Minkowski diagram or dropping the visualization entirely and working with the abstract math. You have a distance metric and a formula for transforming coordinates.

 

[Nugatory]

If they have different x coordinates then the frame is moving,

No. The frame is not moving, and indeed the words "The frame is moving" are just some words strung together without meaning, making no more sense than something like "The quadratic formula is moving". A frame is a mathematical convention for assigning coordinates to events, and mathematical conventions aren't thing sthat can move... And until you are absolutely clear on what a frame is you will be wasting your time trying to understand anything more.

If two events have different x coordinates that alone tells us nothing about the two events.

If two events have different x coordinates and the same t coordinates, then the spacetime interval between them is the difference between the x coordinates and is the spatial distance between them in whatever frame we are using to assign the coordinates. There exists no frame in which that distance is zero - that is, we are necessarily working with events that happen at different points in space so there is no frame in which the two events happen at the same place (have the same x coordinate).

If two events have different t coordinates and the same x coordinates, then the spacetime interval is the difference between the the t coordinates and is the amount of time that a freefalling clock present at both events will measure between the two events. There will be no frame in which the two events happen at the same time (have the same t coordinate).

If both the x and the t coordinates are different, then the spacetime interval tells us whether there exists a frame in which both events happen at the same place or both events happen at the same place - at most only one of these is possible, but one or the other will exist as long as the spacetime interval is non-zero.

 

[Mister T]

So which of them is the proper time?
if Δτ2 = (Δct2 - (Δx2)
then surely τ must be the vertical cathetus, x the other catheus and ct the hypotenuse

$\Delta \tau$ is the proper time. Note that the hypotenuse is not the longest side of the triangle!

 

[Dale]

Yet this is confusing if τ is the proper time, it is the temporal displacement along the rotated worldline which must be ct - the temporal displacement...

Please help me I am confused again...

You would probably be a bit less confused if you used the standard terminology. “Temporal displacement” is ambiguous and non standard, discard it. The standard terms are “coordinate time” and “proper time”.

“Coordinate time” is just the value of the time coordinate in a given coordinate chart. In an inertial frame it represents a system of clocks that are Einstein-synchronized. In other frames it may have little or no physical meaning.

“Proper time” is the time given by a clock following a given worldline. It is only defined on the worldline so there is no synchronization involved. It always has a clear physical meaning.

 

[Grimble]

You would probably be a bit less confused if you used the standard terminology. “Temporal displacement” is ambiguous and non standard, discard it. The standard terms are “coordinate time” and “proper time”.

Yes, sorry about that but I was responding to this quote:

This is Minkowski geometry, not Euclidean geometry. Increasing the spatial displacement (the difference between starting and ending x coordinates) while holding the temporal displacement (the difference between starting and ending t coordinates) constant reduces the length of the (timelike) interval between the end points.

so I was trying to use the same terminology; it is part of what makes things confusing when different terms are used as it becomes impossible to be sure what is being referred to...

I am going to try and find a way of understanding Minkowski diagrams. It seems they are more different than I have understood from Wiki - I know it isn't the best place but as a pensioner in the highlands of Scotland I am limited to what I can find on the internet.
Has anyone any better suggestions for my level of learning?

 

[Dale]

I have found the relativity Wikipedia entries to be pretty reasonable. Not 100%, but better than 90%.

My recommendation is to work problems. The only way to understand this stuff is to apply it and practice it. This material is not intuitive, so you have to rely on math. Come back here frequently and we can check your math and offer recommendations.

Regarding $d\tau^2=dt^2-dx^2$. The quantity that is invariant is $d\tau$. Both $dt$ and $dx$ are frame variant.

In Euclidean geometry you have $ds^2=dx^2+dy^2$. Here $ds$ is invariant and both $dx$ and $dy$ are frame variant under rotations. If you do a whole series of rotations you will find that after each rotation you have a different $dx$ and a different $dy$ and the same $ds$. You will also find that the set of all points with the same $ds$ traces out a circle, meaning that in Euclidean geometry distance is defined by circles and circles are unchanged under rotations.

In spacetime, if you do a whole series of boosts you will find that after each boost you have a different $dx$ and $dt$ and the same $d\tau$. You will also find that the set of all points with the same $d\tau$ traces out a hyperbola, meaning that in Minkowski geometry timelike distance (proper time) is defined by hyperbolas and hyperbolas are unchanged under boosts.

 

[Dale]

I am limited to what I can find on the internet.
Has anyone any better suggestions for my level of learning?

I also liked the Susskind lectures on SR:

 

[Grimble]

Regarding dτ²=dt²−dx². The quantity that is invariant is dτ. Both dt and dx are frame variant.
Minkowski
In spacetime, if you do a whole series of boosts you will find that after each boost you have a different dx and dt and the same dτ. You will also find that the set of all points with the same dτ traces out a hyperbola, meaning that in Minkowski geometry timelike distance (proper time) is defined by hyperbolas and hyperbolas are unchanged under boosts.

OK. Tyring to look at the maths rather than picturing it, I have a difficulty seeing why a hyperbola
(Please excuse - and correct - if my terminology is incorrect, if so I am sure you can see what I am trying to say...)
Yes, taking a proper time, τ=1=dt² - dx² on a cartesian diagram with axes t and x, one will indeed have a hyperbola. I can see that that seems to be what Minkowski was doing in his Space and Time lecture.
That is what happens when we measure the t coordinate along the vertical t axis. Yet even in Fig. 1. he is measuring t' from the origin (null point) of his diagram.

The distance light traveling from the null point in a spacetime diagram is ct and ct²=a² + b² + c² for any point in space that light reaches.

ct² - a² - b² - c²=0 is a lightlike or null interval.

So isn't ct a measurement from the origin rather than a measurement along the t axis? And if so how does that form a hyperbola.

 

[Dale]

A null interval forms a cone (called a light cone), which is a degenerate hyperboloid.

So isn't ct a measurement from the origin rather than a measurement along the t axis?

No. The “hypotenuse” is 0. In fact, the coordinate time axis is usually labeled ct explicitly just to make it clear that coordinate time is being treated geometrically and considered as a coordinate distance.

 

[Grimble]

Hi Dale; I have been trying to work through what you say and most of it fits together very well, however, when you say

No. The “hypotenuse” is 0.

I am unsure what you mean. As I understand it:

In geometry, a hypotenuse is the longest side of a right-angled triangle, the side opposite the right angle.

So how does that make any sense?

Unless you mean that x and ct are not the sides of a right angle triangle... in which case there can be no hypotenuse?

 

[Orodruin]

As I understand it:

That is true in Euclidean geometry, not in Lorentzian geometry.

 

[jbriggs444]

Unless you mean that x and ct are not the sides of a right angle triangle... in which case there can be no hypotenuse?

Note the scare quotes on @Dale's use of the word "hypotenuse". This is hyperbolic trigonometry we are talking about. The length of the "hypotenuse" is the square root of the difference of the squares of the sides, not the sum.

 

[Grimble]

Thank you, I have never come across hyperbolic geometry nor Lorentzian geometry for that matter.

Is there any explanation of the differences between classical spacetime diagrams and Minkowski diagrams?

 

[Dale]

I am unsure what you mean. As I understand it:

The formula for calculating the spacetime interval in units where c=1 is $ds^2=-dt^2+dx^2+dy^2+dz^2$. Notice the - sign in front of the $dt^2$ term. That makes it so that the “hypotenuse” is not the longest side.

As others have mentioned, the scare quotes were intended to remind you that this is Minkowski’s geometry rather than Euclid’s geometry. Not all of the axioms from Euclidean geometry apply here.

So, specifically for light, if we are plotting the world line for a pulse of light that was released at the origin and traveled for a coordinate time of 1 year then it will have traveled a coordinate distance of 1 light year. This will form a 45 degree slope from the origin, that is the “hypotenuse”. Now, as an exercise for you, calculate the length using the Minkowski formula.

 

[SiennaTheGr8]

@Grimble

You might check out Chapters 4 and 5 on this site from Tevian Dray.

 

[pervect]

It's not irrelevant to me. I have to look at the Euclidian representation of the graph. Knowing that longer Euclidian length = shorter worldline length is useful to me. In fact, what I was trying to point out is how the Euclidian length on a minkowski diagram has, in effect, the opposite meaning to what one's intuition would suggest.

For visualizations, I really like the approach Rob uses in post #23. To be able to interpret the diagram, one needs to be familiar with the concept of light clocks.

Then the dark red squares on the diagram are representations of stationary light clocks, and then 10 vertical red squares, laid corner to corner, represent the trip time as measured by the stationary observer with his stationary light clocks. The lighter red squares are also light clocks, but they represent the distance travelled.

The green and blue rectangles represent non-stationary (moving) light clocks. They diagram the elapsed trip time for a moving observer. One can see from the diagram that there are only 8 blue rectangles, and 6 green ones.

The paper referenced in Rob's post, "Relativity on Rotated Graph Paper", gives some additonal interesting insights into the diagrams, such as the fact that all the squares and rectangles on the diagram have the same area when drawn with the correct scale (where light travels at 45 degree angles). There's a preprint of the paper on arxiv, though I gather the published version (which I don't have access to) is slightly different. Ther'es also an insight article here on PF.

 

[Grimble]

So, specifically for light, if we are plotting the world line for a pulse of light that was released at the origin and traveled for a coordinate time of 1 year then it will have traveled a coordinate distance of 1 light year. This will form a 45 degree slope from the origin, that is the “hypotenuse”. Now, as an exercise for you, calculate the length using the Minkowski formula.

Calculate the length of what?
As you have given the coordinate time - 1yr and thecoordinate distance - 1 ltyr, are you referring to the proper length or the length of the worldline?

In any case ds²=−dt²+dx²+dy²+dz² = -1 + 1 = 0.