# Time dilation and Minkowski diagram

1. Oct 24, 2012

### 71GA

After i figured out how to show length contraction in this topic. I tried to use a similar way to show time dilation in Minkowski diagram. Time dilation means that time interval between two events is the shortest in the frame in which those two events happen in same place. We call this frame its rest frame.

According to just said I set up my problem so that frame $x,ct$ is my rest frame as edges of a time interval $\Delta t$ happen on a same vertical line. This means that time interval $\Delta t$ should be the shortest time interval compared to time intervals in any other frames. Including frame $x',ct'$.

But when i draw time interval $\Delta t'$ it is obvious that it is shorter than interval $\Delta t$. So why do i get time contraction instead time dilation?

Marks $1'$ and $2'$ on $ct'$ axis are further appart than marks $1$ and $2$ on $ct$ axis (i even draw a parabola to show this). This is why $\Delta t = 1$ and $\Delta t' < 1'$.

http://shrani.si/f/v/F4/2Fo9hrOu/avezo-1.png [Broken]

Last edited by a moderator: May 6, 2017
2. Oct 24, 2012

### ghwellsjr

Are you getting mixed up by the term "time dilation" which mean seconds on a moving clock according to an Inertial Reference Frame take longer than the coordinate seconds of the IRF which means the moving clock advances by a smaller amount of time than the coordinate time?

3. Oct 24, 2012

### Muphrid

Your $\Delta t'$ should go all the way up to point $1'$. That is the time we're talking about when we talk about time dilation: it captures how we, in the unprimed frame, measure the time elapsed between an event at the origin and point $1'$.

4. Oct 24, 2012

### 71GA

So is my picture correct?

5. Oct 24, 2012

### Muphrid

It's not. $\Delta t'$ needs to extend all the way up to $1'$. For a given invariant interval, the time portion of that interval is shortest in the rest frame. But the way you've drawn it, the black portions do not have the same invariant length, so you get an erroneous result.

6. Oct 24, 2012

### 71GA

If i understand this then first i have to extend $\Delta t'$ all the way up to $1'$. Secondly if i want to know how observer in rest frame sees this interval i have to project it to $ct$ axis? There i will see that projection is actually longer than $\Delta t$ .

Please correct me if i am wrong.

Last edited: Oct 24, 2012
7. Oct 24, 2012

### Fredrik

Staff Emeritus
You seem to have overlooked that the scale on the t' axis is different from the scale on the t axis. Check out section 1.7 in Schutz.

Edit: I wrote this before I saw the last two posts above.

8. Oct 24, 2012

### Muphrid

No. The two invariant intervals must be the same length, and you measure the times from those intervals (you don't adjust the times to be the same, which is what you just suggested). The time we measure for the two intervals in the unprimed frame is just the projection of those intervals on the $ct$ axis.

You correctly drew the hyperbola and saw that $1'$ must lie on that hyperbola. $\Delta t'$ goes up to $1'$. $\Delta t$ goes up to $1$. From each point, you draw a horizontal line over to the $ct$ axis, and those are the times those intervals span.

Perhaps you could draw a 2d Euclidean plane with a point on the y-axis that is then rotated. How would you measure the y-components of the unrotated and rotated point?

9. Oct 24, 2012

### 71GA

One more thing. How come i have to draw a horizontal lines (lines parallel to $x$ axis) and why not lines parallel to $x'$ axis? How do i know when to use each of these?

I draw a picture one more time. Can you tell me if it is correct?

http://shrani.si/f/2G/AQ/3uvVSUHW/minkowsky-diagram10.png [Broken]

Last edited by a moderator: May 6, 2017
10. Oct 24, 2012

### Staff: Mentor

Lines parallel to the x axis are the surfaces of constant time in the unprimed frame - they have the same t coordinate and the unprimed observer will say that all events on one of those lines happen "at the same time".

Lines parallel to the x' axis are the surfaces of constant time in the primed frame - they have the same t' coordinate and the primed observer will say that all events on one of those lines happen "at the same time".

And how do you measure the length of something? You notice where the two ends are "at the same time" and then you measure the distance between them. So you use the lines parallel to the axis that belongs to the frame you're working with.

11. Oct 24, 2012

### 71GA

But here i am not dealing with length contraction, i am dealing with time dilation. I measure time on a same vertical line or line paralel to $ct'$ axe.

12. Oct 24, 2012

### Muphrid

The interval you've labeled as $\Delta t'$ is the the time between the origin and $1'$ according to the $S$ frame, not the $S'$ frame. With the situation you have--an active transformation--this is the correct time to look at to see the effect of time dilation, but you've drawn the wrong conclusion about what time it represents. The set of axes you use to determine the coordinates determine which frame you're measuring with respect to.

To be honest, I think this active transformation has really done too much to confuse you. I would consider ignoring point $1'$ and simply looking at point $1$. Look at this point's coordinates with respect to the $S$ frame, and then compare them with the coordinates in the $S'$ frame. This should make the effects of time dilation apparent. You can find the coordinates of this point in the $S'$ frame by using lines parallel to the $x', ct'$ axes.

13. Oct 24, 2012

### Staff: Mentor

Same general principle.... The lines parallel to the ct axis are the curves of constant position (that is, object at rest) in the unprimed frame, and the lines parallel to the ct' prime axis are the curves of constant position in the primed frame. If you want to consider the time passing on a clock that's at rest in a given frame, you look at the the readings at two points on one of the constant position curves.

Last edited: Oct 24, 2012
14. Oct 24, 2012

### 71GA

OK i hope now i get it... But to be sure i am not doing it wrong again i will list steps which bring me to the conclusion. Please take a look at steps below and if anyone among those steps is wrong please tell me.

STEP 1: I draw both coordinate frames. First i draw frame $x,ct$ and then by using hyperbola i draw $x'ct'$. I allso draw points $1, 2...$, $1', 2'...$. Now i make a conclusion and i say: "Because i constructed frame $x',ct'$ by using frame $x,ct$ all the distances in a frame $x',ct'$ are already represented as observer in $x,ct$ sees them. I think this is what you ment with "active transformation".

http://shrani.si/f/1w/N0/ES721h2/minkowsky-diagram10.png [Broken]

STEP 2: I add (in the picuture) a time interval $\Delta t$ which is an interval between 2 clock ticks for a stationary observer in frame $x,ct$. $\Delta t$ must lie on a vertical lines as 2 clock ticks aren't moving in respect to observer in $x,ct$.

http://shrani.si/f/s/oW/1IzLR4kj/path3920-51-3.png [Broken]

STEP 3: I project $\Delta t$ to the $ct'$ axis using hyperbola and i get a length from origin to $1'$ which represents $\Delta t'$ (BUT as observer in a frame $x,ct$ sees it!).

http://shrani.si/f/R/Wl/3IgVyisj/minkowsky-diagram10.png [Broken]

STEP 4: Just for better comparrison i transfer $\Delta t'$ to the $ct$ axis using normal rotation.

http://shrani.si/f/2A/I1/4EjyDWjm/minkowsky-diagram10.png [Broken]

STEP 5: I conclude, that $\Delta t' > \Delta t$ and so i have a proof for time dilation.

Last edited by a moderator: May 6, 2017
15. Oct 24, 2012

### Muphrid

Step 1: there should be no 2' on the x' axis. The curve you're using is a hyperbola, not a parabola. By "active" transformation, I mean you're actually remapping the points $1,2$ to $1',2'$, and ultimately, you will resolve the components of these primed points with respect to the unprimed ($S$) frame. In fact, you need not draw the $x', ct'$ axes at all, but it does make things clearer.

Step 2: this is correct.

Step 3: I think this is the big issue. To me, it's not clear if you mean to measure the interval between the origin and $1'$ with $\Delta t'$ here or what. There is an interval here, but it doesn't make sense to call this black line $\Delta t'$ and then claim to measure this interval along the axis (which is what you end up doing in step 4).

Step 4: No, there should be no ordinary rotation here. I drew an analogy to ordinary rotation to try to get you to think about how that entirely different problem would be done. All you should do is draw the horizontal line from $1'$ to the $ct$ axis.

My advice to you, to try to make this as unambiguously clear as possible, is to do the following: forget trying to use deltas. Just look at the coordinates. Don't draw the $x', ct'$ axes. The hyperbola is enough to get a sense of where $1'$ is. Simply draw perpendiculars from the $x, ct$ axes to $1'$, and this tells you the coordinates of $1'$ in the $S$ frame. The time coordinate of $1'$ will be larger (15% larger, by the math you've done) than the time coordinate of $1$.

16. Oct 24, 2012

### 71GA

Take a look again at below picture. Can you confirm now that it is correct. $\Delta t'$ is the time component you are talking about.

17. Oct 24, 2012

### Fredrik

Staff Emeritus
I haven't followed the entire discussion, but the most important details to keep in mind are these: (I'm using units such that c=1).

• The t' axis is drawn with slope 1/v, where v is the speed of the primed coordinate system in the unprimed coordinate system.
• The x' axis is drawn with slope v.
• The scale on the t' and x' axes is determined by drawing a bunch of hyperbola $-t^2+x^2=\mathrm{constant}$. For example, the hyperbola that intersects the t axis at (t,x)=(1,0) intersects the t' axis at (t',x')=(1,0). Since every event on the t' axis has x=vt, the event with (t',x')=(1,0) has (t,x) coordinates that satisfy $x=vt$ and $-t^2+x^2=-1$. It's easy to see that this means t=1/γ and x=v/γ, so $(t,x)=(1/\gamma,v/\gamma)$.
• When you want to know the (t,x) coordinates of an event, you draw a horizontal and a vertical line through that event, and see where those lines intersect the t and x axes.
• When you want to know the (t',x') coordinates of an event, you draw a line that's parallel to the t' axis and a line that's parallel to the x' axis through the event, and see where those lines intersect the t' and x' axes.

Schutz's GR book is a very good place to learn about these things. I linked to the most relevant section at Google Books in post #7.

Last edited: Oct 24, 2012
18. Oct 24, 2012

### Fredrik

Staff Emeritus
71GA, can you explain one thing...what is $\Delta t'$ supposed to be? I mean, I can only calculate it if I know its definition, so how do you define it?

19. Oct 24, 2012

### 71GA

If you want to calculate it you can use equation ($m$ is a unit of meter):

$\begin{split} \Delta t' &=\gamma \Delta t \\ c\Delta t' &= \gamma \, c \Delta t \\ c \Delta t' &= \gamma \,1m \\ c \Delta t' &= 1,15\, 1m \\ c \Delta t' &= 1,15m \end{split}$

But this is not about equation at all. It is about a graphicall image which can help reader to see things and clear them up.

20. Oct 24, 2012

### Muphrid

He didn't want a formula, just an explanation about what it represents, I think. This has been some point of confusion throughout the thread. I tried to clear it up by saying that $\Delta t'$ should be the time coordinate of $1'$ in the $S$ frame.

I think the last picture is valid, so you should be able to see what's going on now.