I Proper (and coordinate) times re the Twin paradox

[Herman Trivilino]

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.

 

[Dale]

Yes, that is the length of the “hypotenuse”

 

[vanhees71]

I'd not use "Euclidean jargon" in connection with the Minkowski space. It's hard, but one must hammer ourselves the fact into the brain that a Minkowski diagram must not be read in Euclidean terms! The lengths defined by the fundamental form of Minkowski space are not what we are used to in the Euclidean plane. The Minkowski plane is rather a hyperbolic space. The indefiniteness of the fundamental form is the key element making it a spacetime manifold, allowing for a causality structure. This cannot be achieved with a proper scalar product of a Euclidean (affine) space.

 

[Dale]

Fair enough, but there is no Minkowski term for the “hypotenuse” so the scare quotes is the best I can do.

 

[Herman Trivilino]

Calculate the length of what?

The spacetime interval between the two events. One event is the emission of the light flash and the other is the reception. Note you correctly calculated the value to be zero. The events have a lightlike separation. You seem to be confusing the spacetime interval with the proper length. The value of the spacetime interval equals the proper length only for events with a spacelike separation.

Edit: And the length of a worldline is the proper time. The spacetime interval for events with a timelike separation.

 

[Grimble]

I am finding this intriguing, you are opening up a whole new world to me.
I have never understood, nor seen any indication of, hyperbolic geometry in relation to Minkowski Diagrams.
Yes, I have seen the hyperbola in Minkowski's diagrams in his Space and Time Lecture but had no idea that that led to a completely different geometry...
So perhaps you can understand why I have been so reluctant to let go of the Euclidean perspective.
But are we not still dealing with the same equations?
Isn't the Minkowski hyperbolic treatment just one way of depicting them, because the hyperbola seems to be associated with losing the uniformity of scale on the diagonal lines, which in turn gives rise to what I see as anomalies: planes of simultaneity and jumps on the time axis of twin paradox diagrams.

Don't get me wrong I am not saying there is anything wrong with what you are saying, only that it is intriguing how far I have gone down the wrong road because I have failed to appreciate that hyperbolic geometry is integral with Minkowski diagrams.

For example this from the introduction to Minkowski Diagrams

Minkowski diagrams are two-dimensional graphs that depict events as happening in a universe consisting of one space dimension and one time dimension. Unlike a regular distance-time graph, the distance is displayed on the horizontal axis and time on the vertical axis. Additionally, the time and space units of measurement are chosen in such a way that an object moving at the speed of light is depicted as following a 45° angle to the diagram's axes.

Nowhere can I find any suggestion that it involves a different metric (if that is the correct term?)

 

[Dale]

But are we not still dealing with the same equations?

Clearly not:
$ds^2=dx^2+dy^2+dz^2$
is not the same equation as
$ds^2=-dt^2+dx^2+dy^2+dz^2$

 

[Grimble]

YEs, but one is the aggregate length in 3 dimensions and the other in 4, so does ds represent the same quantity in each case?

 

[Dale]

YEs, but one is the aggregate length in 3 dimensions and the other in 4, so does ds represent the same quantity in each case?

It is not the same quantity, but there are some similarities. In both cases it is an invariant measure of distance in the space. The differences are that in space it is called distance and there is only one kind of distance, while in spacetime it is called the spacetime interval and there are three different kinds of spacetime intervals (space like, time like, and null).

 

[Grimble]

Well if it is not the same quantity then both equations are true... but cannot be equated - or even compared if ds represents different quantities? Or am I missing something here?

The terms in these equations represent specific properties; I agree they are different things, but choosing a type of diagram changes how they are represented on that diagram, not what they represent.
It is still the same Spacetime, consisting of three Cartesian dimensions of space plus another dimension for time
Surely a²+b²+c² is still the aggregate length in Minkowski Spacetime; while -ct²+a²+b²+c² is the Spacetime interval and both are invariant intervals.
The Mathematics doesn't depend on how they are drawn.
It just seems difficult to be sure what the terms mean when the same term ds² means two different things...
It can be very confusing

 

[jbriggs444]

Surely a²+b²+c² is still the aggregate length in Minkowski Spacetime; while -ct²+a²+b²+c² is the Spacetime interval and both are invariant intervals.

It makes little sense to use the term "invariant" to refer to $a^2+b^2+c^2$ in the context of Minkowski Spacetime.

Presumably the formula is shorthand for "the square of the difference in the x coordinates plus the square of the difference in y coordinates plus the square of the difference in z coordinates in some coordinate system".

The obvious question is: The x, y and z coordinates of what?

Without a t coordinate, the only rational answer is "between two lines".

And without an agreed upon coordinate system, the next question is "between which two lines".

Which brings us to the stated conclusion: It makes little sense to use the term "invariant" for this.

 

[PeterDonis]

if it is not the same quantity then both equations are true

True in their respective geometries, yes. But they are different geometries. Euclidean 3-space is not the same geometry as 4-D Minkowski spacetime. Each one has its own formula for $ds^2$. It makes no sense to say the formula for $ds^2$ in Euclidean 3-space is "true" in Minkowski spacetime.

It is still the same Spacetime

Euclidean 3-space is not spacetime. It's Euclidean 3-space.

Surely a2+b2+c2 is still the aggregate length in Minkowski Spacetime; while -ct2+a2+b2+c2 is the Spacetime interval and both are invariant intervals.

No, this is not correct. $dx^2 + dy^2 + dz^2$ is the invariant length in Euclidean 3-space. $- c^2 dt^2 + dx^2 + dy^2 + dz^2$ is the invariant length in 4-D Minkowski spacetime. This length is called a "spacetime interval" in the Minkowski spacetime, but that's just nomenclature.

 

[PeterDonis]

The Mathematics doesn't depend on how they are drawn.

Yes, but it does depend on which geometry you are in.

It just seems difficult to be sure what the terms mean when the same term ds2 means two different things...
It can be very confusing

Welcome to math and physics. Terminology will sometimes be confusing; you just have to learn how to figure out what is intended from context, and ask questions when something is not clear. What you should not do is assume that the same symbol must mean the same thing in a different context.

 

[Grimble]

The main difference in the two geometries is the number of dimensions, is it not?
In Euclidean 3-space, a²+b²+c² is invariant because there is no time component.
So in Minkowski Spacetime the equivalent would be that a²+b²+c² would be invariant at any single specific time.
Similarly, in classical mechanics using euclidean geometry, ds²=(ct')²+a²+b²+c² where ct' is the time axis for a moving body

 

[PeterDonis]

The main difference in the two geometries is the number of dimensions, is it not?

No. That's one difference, but not the main one. The main difference is that Euclidean 3-space is Riemannian (the metric has +++ signature) while 4-D Minkowski spacetime is pseudo-Riemannian (the metric has -+++ signature). That means, as @Dale said, that while in Euclidean 3-space there is only one type of interval/length, in Minkowski spacetime there are three: spacelike, null, and timelike.

In Euclidean 3-space, a2+b2+c2 is invariant because there is no time component.

No, it's because there is no such thing as a "time component" in Euclidean 3-space. There are only three dimensions, and they're all spacelike because the metric has +++ signature.

in Minkowski Spacetime the equivalent would be that a2+b2+c2 would be invariant at any single specific time.

There is no such thing as "any single specific time" because there is no preferred inertial frame in Minkowski spacetime. An interval that has zero $dt$ in one frame will have nonzero $dt'$ in any other frame. And "invariant" means the equation has to hold in every frame, not just one, so your claim is wrong.

in classical mechanics using euclidean geometry, ds2=(ct')2+a2+b2+c2 where ct' is the time axis for a moving body

Wrong. There is no such spacetime interval in Newtonian mechanics.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

The thread topic has been sufficiently discussed. The thread will remain closed.