Understand Spacetime Diagrams: Calculating Proper Time

[Arman777]

I am trying to understand the basic conceptual ideas about the space-time diagrams. In spacetime diagrams we have events which are labeled as points on the diagram.

Let us call we have an event on point $A(0,0)$and another event on $B(4,0)$ measured by an inertial frame $S$. This inertial frame sees another inertial frame $S'$ which moves with a speed of $V = 0.6c$ in the $+x$ direction. Let us draw this.

I learned that in the SR diagram if there are two timelike geodesics the longest length belongs to the one which is the straigthest. So If $S$ is at rest, the proper time of the $S'$ must be shorter. But in this picture its higher. How is it possible ?

How are we labeling these events or what they are measuring ? I am so confused that I cannot even explain what I am really confused about. I know lorentz transformations but I am more confused about who measures which quantity, how is it labeled in the spacetime diagram. Since we are drawing the graph that shows that $S$ is at rest, is that mean all the measurements done by $S$? If so what does it mean $t'=5$ ? Does it says $S'$ measures its proper time as $5$ or does it says $S$ measures $S'$ proper time as $5$.

So $S$ measures an event at $B(4,0)$he calculates the proper time between events, $A$and $B$,

$$\tau(A,B) = \int_A^B \sqrt{-g(d\vec{x}, \vec{x})} = \int_A^B\sqrt{dt - V^2dt^2} = (t_B - t_A)\sqrt{1-V^2} = 4 $$

Now $S$ wants to calculate the proper time of the $S'$. What is the proper time equation that we can write for this case and why ?

 

[Orodruin]

I learned that in the SR diagram if there are two timelike geodesics the longest length belongs to the one which is the straigthest.

You mean world-lines, not geodesics. Geodesics are straight lines by definition.

But in this picture its higher. How is it possible ?

It is not, you are imposing the Euclidean metric of the plane for what you mean by "length" instead of using the actual Minkowski metric of the spacetime.

You cannot measure "lengths" in a spacetime diagram with a typical ruler intended to measure lengths in a Euclidean space.

 

[Pencilvester]

if there are two timelike worldlines the longest length belongs to the one which is the straigthest

This statement also refers to two worldlines connected at two points. Then it is the straight line between the two points that will have the longest length. Your two worldlines only meet at one point.

 

[Arman777]

It is not, you are imposing the Euclidean metric of the plane for what you mean by "length" instead of using the actual Minkowski metric of the spacetime.

You cannot measure "lengths" in a spacetime diagram with a typical ruler intended to measure lengths in a Euclidean space.

Yes I am aware of that. Thats what's make me confused. If its smaller then that then what does $5$ represents ?

This statement also refers to two worldlines connected at two points. Then it is the straight line between the two points that will have the longest length. Your two worldlines only meet at one point.

 

[robphy]

This might help you see what those tickmarks represent.

The ticks are modeled on the ticks of a light-clock.
The "light-clock diamonds" are traced out by the light-signals in a light-clock carried by the observer.
(The worldlines of the light-clock mirrors are implied by the spacelike diagonals of the diamonds.)
By Lorentz invariance, the area of all light-clock diamonds are equal.

In your example, the relative velocity is (3/5)c is arithmetically nice
because it leads to a pythagorean triple 3,4,5
with 4 along the hypotenuse, and 3 and 5 forming the legs of a Minkowski-right triangle.
This is one way to visualize the Minkowskian spacetime geometry, which as you see is different from Euclidean geometry.

edit: Can you see the TWO 3-4-5 triangles? The timelike hypotenuse has size 4, and the timelike leg (size 5) and the spacelike leg (size 3) are Minkowski-perpendicular (they are parallel to the diagonals of that observer's light-clock diamond).
(There are hyperbolas that are implicit in the diagram and the construction.)

For my construction, the more important feature is that the Doppler factor k=2.
(When k is rational, your Minkowski-right-triangle legs form a Pythagorean triple.)
This k and its reciprocal 1/k are the eigenvalues of the Lorentz boost transformation, with eigenvectors along the light-cone. So the red diamond is stretched by a factor of k along the frontward light-signal
and shrunk by a factor of k along the rearward light-signal.
This is made easier by drawing on "rotated graph paper".For more info:
check out my PF Insights.
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
See:
https://www.geogebra.org/m/HYD7hB9v# (visualizations)
https://arxiv.org/abs/1111.7254 (early draft)
https://aapt.scitation.org/doi/10.1119/1.4943251 (my article: Am.J.Phy. 84, 344 (2016))

 

[Pencilvester]

Yes that's also kind of confusing for me. I tried to calculate the proper time of the points between #A(0,0)andand and B'(5,0)##. How can we do that ?

Come up with an equation describing a timelike curve that passes through (0,0) and (5,0), and integrate. If you want to compare lengths, you’ll need to come up with at least two curves that pass through those points. Note that there will only be one curve that can describe a particle/observer that remains inertial throughout the trip from A to B (barring curves with discontinuous derivatives). So trying to set it up with two inertial observers is futile.

 

[robphy]

Yes that's also kind of confusing for me. I tried to calculate the proper time of the points between #A(0,0)$and$B'(5,0)##. How can we do that ?

"Proper time" is a property of a segment of a worldline... not of a pair of events.

"Proper time" is the "wristwatch time along a given worldline"... an arc-length, akin to distance along a path.
When Minkowski defined "proper time",
"proper" means ownership (like property)...
it's got nothing to do with proper vs improper.

Given two events, there is no [unique] "proper time between [timelike-related] events O and Z".

There is, however, a "proper time between events O and Z along an inertial worldline"
(often called the interval [the square-root of the square-interval])
as well as a "proper time between events O and Z along a given noninertial worldline",
and along a different noninertial worldline, and so on.Below are 5 worldlines from O to Z, each worldline has a proper time elapsed from O to Z.
The inertial worldline has the most proper time elapsed.
I'll leave it to you to count the ticks along each worldline.

(Note how the rotated graph paper easily displays 5 inertial frames of reference
through event O. Five-observer graph paper isn't needed.
Rectangular graph paper would not easily display the ticks, without calculation.
Hyperbolic graph paper might be okay from event O, but can't easily handle any turns by a worldline.)

 

[pervect]

I am trying to understand the basic conceptual ideas about the space-time diagrams. In spacetime diagrams we have events which are labeled as points on the diagram.

Let us call we have an event on point $A(0,0)$and another event on $B(4,0)$ measured by an inertial frame $S$. This inertial frame sees another inertial frame $S'$ which moves with a speed of $V = 0.6c$ in the $+x$ direction. Let us draw this.

View attachment 261242

So far, this is excellent.

I learned that in the SR diagram if there are two timelike geodesics the longest length belongs to the one which is the straigthest. So If $S$ is at rest, the proper time of the $S'$ must be shorter. But in this picture its higher. How is it possible ?

The question is a bit confused. I think it is likely that you are confused about what proper time is.

Proper time is something that's associated with a path or segment of a worldline on the space-time diagram. So, in your diagram, for instance, the straight line segment joining event A to event B has some proper time associated with it. It is the same for all observers. In this example, the value of this proper time for the worldline segment AB is 4 seconds. The value of the proper time of this segment is independent of the choice of frame, i.e. it is the same in frame S' as it is in frame S, it has a value of 4 seconds.

Proper times are always timelike, so the worldline must be timelike. Hmm, I suppose you need a definition of timelike, I'm not sure I have a precise one. Hopefully it's somewhat intuitive. If not, ask away, maybe I or someone can explain further.

 

[robphy]

Proper times are always timelike, so the worldline must be timelike. Hmm, I suppose you need a definition of timelike, I'm not sure I have a precise one. Hopefully it's somewhat intuitive. If not, ask away, maybe I or someone can explain further.

One can say that "proper times [wristwatch times] are only defined for observer worldlines",
which are curves whose [future] tangent vectors point into the interior of the [future] light cone.

 

[Arman777]

All of the answers helps a bit but not exactly. Suppose again we have 3 points, $O(0,0)$, $A(4,0)$ and $B'(5,0)$ such that,

Where $V = 0.6c$.

We all now that $|OA|= 4$ and everyone is okay with that. I asked about how to measure $|OB|=\tau(0,B)$. Now we know that $|OA|$ and $|OB|$ are frame invariant values because they are measured in terms of $\tau$ and $\tau$ is an invariant quantity.

Also as we can notice we can calibrate the coordinates by using hyperbolas such that,

$$-t^2 + x^2 = |OA| = |OB| = -\overline{t}^2 + \overline{x}^2$$

For instance, if we say $A(1,0)$ then since $x=0$ this means that $t = |OA| = 1$ and $\overline{t} = |OB| = 1$. Now at this point we can try to calculate the $|OB|$ such that

$$\tau(0,B) = \int_0^B dt\sqrt{1-V^2}$$ in the reference frame of $\overline{S}$ the $\overline{t} = |OB| = \tau(0,B) = 1$ and we want to know what this coordinate is in terms of $t$. By using above equation we can say that $$ \Delta t_{\text{|OB| measured in } \overline{S}} = \Delta t_{\text{|OB| measured in S}}{\sqrt{1-V^2}}$$

Going back to the original question.
Let us do this for to calculate the $B(t,0)$ measured by $S$. By using the property of the invariance of the $\tau$ and we know that $x = \overline{x} = 0$, $t = \overline{t} = 4$ now by using above equation.

$$\Delta t_{\text{|OB| measured in } \overline{S}} = 4$$
$$\Delta t_{\text{|OB| measured in S}} = \text{What we are looking for}$$

So we have

$$\Delta t_{\text{|OB| measured in S}} = \frac{4}{\sqrt{1-V^2}} = 5$$

The inital diagram confused my mind because it says $\overline{t} = 5$ however that's not true at all.

I am sorry but its hard for me to understand the rotated paper without understanding the basics of the SR.

 

[Ibix]

All of the answers helps a bit but not exactly. Suppose again we have 3 points, O(0,0)O(0,0), A(4,0)A(4,0) and B′(5,0)B'(5,0) such that,

These coordinates aren't consistent with your diagram, assuming by the prime on $B'$ you mean that the coordinates are in the yellow frame and that the hyperbola you've drawn is focussed at $O$. If that's the case, the coordinates of $B$ in the yellow frame must be (4,0), since it clearly lies on the time axis and it lies on the hyperbola of constant straight-line interval from the origin.

What you've done is the Minkowski equivalent of drawing a circle centred on the origin of a Euclidean plane and insisting that two points on the circle are at different distances from the origin.

 

[Arman777]

These coordinates aren't consistent with your diagram, assuming by the prime on $B'$ you mean that the coordinates are in the yellow frame and that the hyperbola you've drawn is focussed at $O$. If that's the case, the coordinates of $B$ in the yellow frame must be (4,0), since it clearly lies on the time axis and it lies on the hyperbola of constant straight-line interval from the origin.

What you've done is the Minkowski equivalent of drawing a circle centred on the origin of a Euclidean plane and insisting that two points on the circle are at different distances from the origin.

With respect to $\overline{S}$ B has (4,0), with respect to $O$, $B(5,0)$, that's what I mean. Is it wrong ?

 

[robphy]

The rotated graph paper answers (from your first post)

...So If S is at rest, the proper time of the S′ must be shorter. But in this picture its higher. How is it possible ?

How are we labeling these events or what they are measuring ? I am so confused that I cannot even explain what I am really confused about. I know lorentz transformations but I am more confused about who measures which quantity, how is it labeled in the spacetime diagram.

The numbers on each t-axis are tickmarks according the wristwatch (clock) carried by the observer traveling along that worldline. So, the diamonds visualize those clock-ticks from a light-clock. Each set of diamonds determines a coordinate grid that that observer uses to measure [that is, assign coordinates to] an event.
So, the diamonds answer the question
Q: "what they are measuring ?"
A: The ticks of the clock carried along that worldline.
Why it is like this... that's the relativity part.
But, at this stage, the diamonds are a replacement of the tickmarks given on your diagram [from whatever software you are using]... a replacement with more physical interpretation (even if you can't explain the details).From your latest post,

Suppose again we have 3 points, $O(0,0)$, $A(4,0)$ and $B′(5,0)$such that,

Also as we can notice we can calibrate the coordinates by using hyperbolas such that,

$−t^2+x^2=|OA|=|OB|=−\bar t^2+\bar x^2$​

there's either a typo or some misunderstanding.

Yes, $|OA|$ and $|OB|$ are frame invariant quantities.
If A and B lie on the hyperbola, as you have shown, then $|OA| = |OB|$

But when you say "$A(4,0)$ and $B′(5,0)$" ,

• does the "prime on the B" mean that this is "5 tickmarks along the axis OB, and 0 tickmarks along the other orange line"? If so, then that event B cannot be on that hyperbola. It must be $B′(4,0)$.
• is the "prime on the B" a typo that shouldn't be there, and that you mean (5,0) as tickmarks along the black lines (the usual rectangular grid)? If so, then you must mean $B(5,3)$

 

[Arman777]

The numbers on each t-axis are tickmarks according the wristwatch (clock) carried by the observer traveling along that worldline.

So $S$ measures 4 ticks and $\overline{S}$ measures 5 ticks w.r.t $S$ ? or

So $S$ measures 4 ticks and $\overline{S}$ measures 5 ticks w.r.t itself?

But when you say "A(4,0)A(4,0)A(4,0) and B'(5,0)B′(5,0)B′(5,0)" ,

• does the "prime on the B" mean that this is "5 tickmarks along the axis OB, and 0 tickmarks along the other orange line"? If so, then that event B cannot be on that hyperbola. It must be B'(4,0)B′(4,0)B′(4,0).
• is the "prime on the B" a typo that shouldn't be there, and that you mean (5,0) as tickmarks along the black lines (the usual rectangular grid)? If so, then you must mean B(5,3)

Okay here are my last thoughts,

$A(4,0)$ w.r.t $S$ that's okay for everyone.

$B(5,0)$ w.r.t $S$ beacuse that's what I calculated (if it is correct)

$B(4,0)$ w.r.t $\overline{S}$

maybe I should have putten a hat on the top so $B$ meaured by $\overline{S} ≡ \overline{B}(4,0)$. I hope this is correct othwerise I'll be so lost again.

That is also what Schutz says, when a clock moving on the $\overline{t}$ axis reaches $B$ it has a reading of $\overline{t} = 1$, but the event has coordinate $t = \frac{1}{\sqrt{1-V^2}}$ in $S$. So to $S$ it appears to run slowly.

Then writes,

$$ \Delta t_{\text{ measured in S}} = \frac{\Delta t_{\text{measured in }\overline{S}}} {{\sqrt{1-V^2}}}$$

For the interval $|OB|$ it becomes what I calculated.

$\Delta t_{\text{measured in }\overline{S}} = 4$ So $\Delta t_{\text{ measured in S}}=5$

 

[Arman777]

Now the problem is the to explain where the $3$comes. I guess I should just forget all of these and just focus on the lorentz transformations.

 

[robphy]

This might help.

What the time-dilation formula actually measures
is the component of OB along OA,
using an event on OA that OA says is simultaneous with B... that's your (5,0),
...whereas B itself is (5,3).
Trigonometrically, it is $OB\cosh(\theta) =OB \cdot \widehat{OA}$, using the rapidity (the Minkowski angle between timelike vectors, where $v=\tanh\theta$).
(I'm using vector notation since I'm assuming that is familiar to you, based on your other post on "two timelike vectors".)

Essentially, vector OB has (t,x)-components according to OA :
$(OB\cosh\theta,OB\sinh\theta)$,
where
$\cosh\theta=\frac{1}{\sqrt{1-v^2}}=\gamma$
$\sinh\theta=\frac{v}{\sqrt{1-v^2}}=\gamma v$ .

Geometrically,
the event on OA that OA says is simultaneous with B
is gotten by the intersection of OA and the particular hyperbola centered at O
whose tangent line at the intersection event meets B.
That's the geometry of simultaneity: "space is Minkowski-perpendicular to time"
like the tangent-line is Minkowski-perpendicular to the radius (as Minkowski declared).
[As in your "two timelike vectors" thread, @George Jones 's definition of spacelike is $g(u,v)=0$.)

 

[Ibix]

With respect to $\overline{S}$ B has (4,0), with respect to $O$, $B(5,0)$, that's what I mean. Is it wrong ?

It can't be at zero x coordinate in both. It manifestly doesn't lie on the axis in the black frame.

 

[Arman777]

What the time-dilation formula actually measures
is the component of OB along OA,

That make sense.

the event on OA that OA says is simultaneous with B

You mean A and B are simultaneous ?

Okay I understand everyting now. The graph was really helpful, thanks. It made a lot of sense after you labele the points.

 

[Arman777]

After thinking again my real confusion was due to fact that we can label these points with 2 different set of numbers. And the other thing was about the proper time and its relation with the lorentz transfomation.

 

[robphy]

You mean A and B are simultaneous ?

Okay I understand everyting now. The graph was really helpful, thanks. It made a lot of sense after you labele the points.

A and B are equidistant from O (intersection of worldlines),
both have 4 seconds of elapsed time along OA and OB, respectively.

Using the the OA (t,x)-coordinates,
A is at (4,0) and the distant event B is (5,3).
On worldline OA, the event that OA says is simultaneous with the distant event B (5,3)
is the local event [on OA] at (5,0), which occurs after event A (at (4,0)).
That's time dilation.

 

[Arman777]

I understand it now, thanks