Time dilation and Minkowski diagram

In summary: Lines parallel to the x' axis are the surfaces of constant space in the unprimed frame - they have the same x coordinate and the unprimed observer will say that all events on one of those lines happen... at the same place.
  • #1
71GA
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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 [itex]x,ct[/itex] is my rest frame as edges of a time interval [itex]\Delta t[/itex] happen on a same vertical line. This means that time interval [itex]\Delta t[/itex] should be the shortest time interval compared to time intervals in any other frames. Including frame [itex]x',ct'[/itex].

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

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

http://shrani.si/f/v/F4/2Fo9hrOu/avezo-1.png [Broken]
 
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  • #2
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
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
ghwellsjr said:
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?

So is my picture correct?
 
  • #5
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
Muphrid said:
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.

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

Please correct me if i am wrong.
 
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  • #7
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
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
Muphrid said:
From each point, you draw a horizontal line over to the ##ct## axis, and those are the times those intervals span.

One more thing. How come i have to draw a horizontal lines (lines parallel to [itex]x[/itex] axis) and why not lines parallel to [itex]x'[/itex] 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]
 
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  • #10
71GA said:
One more thing. How come i have to draw a horizontal lines (lines parallel to [itex]x[/itex] axis) and why not lines parallel to [itex]x'[/itex] axis? How do i know when to use each of these?

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
Nugatory said:
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.

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 parallel to [itex]ct'[/itex] axe.
 
  • #12
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
71GA said:
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 parallel to [itex]ct'[/itex] axe.

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.

And how do you measure time dilation? Well, you look at your clock at time T in your frame, and you see what the clock that's at rest in the other frame reads at the same moment (in your frame). Then you wait until your clock reads T+delta, and then see what the clock in the other frame reads at the same moment (in your frame). Compare the two readings on the other clock with the two readings on your clock to find the time dilation. To do all of this, you need a clock at rest in your frame (moving on a line parallel to your t axis), at rest in the other frame (moving on a line parallel to its t axis), and two "at the same time" lines in your frame, parallel to your x axis.
 
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  • #14
Muphrid said:
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.

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 [itex]x,ct[/itex] and then by using hyperbola i draw [itex]x'ct'[/itex]. I allso draw points [itex]1, 2...[/itex], [itex]1', 2'...[/itex]. Now i make a conclusion and i say: "Because i constructed frame [itex]x',ct'[/itex] by using frame [itex]x,ct[/itex] all the distances in a frame [itex]x',ct'[/itex] are already represented as observer in [itex]x,ct[/itex] 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 [itex]\Delta t[/itex] which is an interval between 2 clock ticks for a stationary observer in frame [itex]x,ct[/itex]. [itex]\Delta t[/itex] must lie on a vertical lines as 2 clock ticks aren't moving in respect to observer in [itex]x,ct[/itex].

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

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

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

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

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

STEP 5: I conclude, that [itex]\Delta t' > \Delta t[/itex] and so i have a proof for time dilation.
 
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  • #15
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
Muphrid said:
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##.

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.

lol.png
 
  • #17
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.
 
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  • #18
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
Fredrik said:
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?

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

[itex]
\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}
[/itex]

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
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.
 
  • #21
Muphrid said:
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.

I hope so too, BUT there is still one thing that confuses me. In my other thread on another forum (topic is located here, but you don't need to look at it) user Alfred Centauri states that i should construct my image like shown below.

Question 1: Why is he using tilted lines (parallel to ##x'##) and not hyperbola in combination with horizontal lines (paralel to ##x##) like we did?
Question 2: Is he wrong?
Question 3: Are there maybe more ways to construct this diagram?

qtLV4.png
 
  • #22
He's describing a passive transformation, where instead of actually boosting the pink dot along a hyperbola, you simply draw new axes and evaluate the coordinates with respect to the new axes. These are equivalent (though obviously not exactly the same). Let me explain. You can do one of two things:

1) You could boost the pink dot along a hyperbola and evaluate its coordinates with respect to the original ##x,ct## axes. This is what you attempted to do.

OR

2) You can draw new ##x', ct'## axes and evaluate the coordinates of the pink dot with respect to these new axes, the ##S'## frame. This is what Alfred Centauri did.

(1) is called an active transformation. (2) is called a passive transformation. They are closely related. I even suggested using a passive transformation earlier because I felt it would be clearer.
 
  • #23
Muphrid said:
He's describing a passive transformation, where instead of actually boosting the pink dot along a hyperbola, you simply draw new axes and evaluate the coordinates with respect to the new axes. These are equivalent (though obviously not exactly the same). Let me explain. You can do one of two things:

1) You could boost the pink dot along a hyperbola and evaluate its coordinates with respect to the original ##x,ct## axes. This is what you attempted to do.

OR

2) You can draw new ##x', ct'## axes and evaluate the coordinates of the pink dot with respect to these new axes, the ##S'## frame. This is what Alfred Centauri did.

(1) is called an active transformation. (2) is called a passive transformation. They are closely related. I even suggested using a passive transformation earlier because I felt it would be clearer.

If i add passive transformation along the active one i get the distance which i wrote as ##\Delta t''##. What is this? It is not the same value as my ##\Delta t'## which i got using the active transformation.

Here is the picture. I only added a line parallel to ##x'## axis and marked a distance ##\Delta t''##.

http://shrani.si/f/N/h8/1dEH9KWY/lol.png [Broken]
 
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  • #24
It may not look that way, but I assure you the numerical value of ##\Delta t''## is indeed ##1.15##. I think the drawing looks a little worse than it should (is that hyperbola really correct?), but otherwise, the gist of it gets everything across that it should.
 
  • #25
Muphrid said:
It may not look that way, but I assure you the numerical value of ##\Delta t''## is indeed ##1.15##.

"Numerical value of ##\Delta t''##" as measured in frame ##x'ct'## or as measured in ##x,ct##?

Muphrid said:
I think the drawing looks a little worse than it should (is that hyperbola really correct?), but otherwise, the gist of it gets everything across that it should.

It is not a real hyperbola, so maybe this is the case. Could anyone provide a picture which proves this?
 
  • #26
71GA said:
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.

Hmm - there's a problem here already, as the inertial observer experiences the longest proper time between events,

You are missing at least two lines on your diagram, and I don't see them in in any subsequent posts, so I'm guessing that you're still missing something. I'd suggest adding three more lines, and two more points:

One important pre-requisite, which I'll call PRE Make sure that the Lorentz interval, also called the proper time, between the origin and point 1 is the same as the Lorentz interval between the origin and point 1'. I think you addressed this by saying that points 1 and 1' should both be on the same hyperbola of (ct)^2 - x^2 = constant.

The first line that you are missing is the set of points that are simultaneous with event 1 in the (t,x) frame. Hint: this set of points is parallel to the x axis.

The second line is the set of points that are simultaneous with event 1 in the (t', x') frame. Hint: this set of points is parallel to the x' axis. The important thing to note is that this set of points is DIFFERENT that the set of points of our first line. Otherwise, we won't be using this line.

The third line that you are missing is the set of points that are simultaneous with event 1' in the (t', x') frame. Hint: this is parallel to the x' axis, similar to line 2, but not similar to line 1.

So those are the three lines I suggest you add - now come the points, and the meaning of time dilation on your graph.

In the (t,x) frame, line 1 represents the set of points simultaneous with the event you labelled as "1". Find point 2, where this line intersects Ranja's worldline. Time dilation says that point if point 1' occurs after point 2, because using the simultaneity conventions of the (t,x) frame, the clock in the (t', x') frame is ticking slowly. (Be sure that the precondition PRE is met!)

In the (t', x') frame, line 3 represents the set of points simultaneous with the event you labelled as 1'. Find point 3, where this line intersects Ziga's worldline. Time dilation says that point 1 comes after point 3. Again, be sure the precondition PRE is met.
 
  • #27
71GA said:
"Numerical value of ##\Delta t''##" as measured in frame ##x'ct'## or as measured in ##x,ct##?

##\Delta t''## is the time coordinate of the point ##1## with respect to the ##S'## frame.

It is not a real hyperbola, so maybe this is the case. Could anyone provide a picture which proves this?

I personally cannot, and while the diagrams are handy for seeing things, at some point I think one must rely on the equations to get something precise. This is why I'm not really a fan of all the delta's you've been using. It's easier to use coordinates because then you can put those coordinates into equations.

Let point 1 lie at ##(c t_{(1)}, x_{(1)})##. This is in the ##S## frame. Equivalently, it also sits at ##(ct_{(1)}', x_{(1)}')## in the ##S'## frame.

Your ##\Delta t = t_{(1)}##.

Your ##\Delta t' = t_{(1')}##, the time coordinate of ##1'## in the ##S## frame.

Your ##\Delta t'' = t_{(1)}'##, the time coordinate of point ##1## in the ##S'## frame.
 
  • #28
pervect said:
Hmm - there's a problem here already, as the inertial observer experiences the longest proper time between events,
Doesn't observer in frame ##x,ct## experiences the shortest time in my case? I would allso like to clear something up before i start writing. Are you writing this for the "passive transformation" or the "active transformation"?

pervect said:
The first line that you are missing is the set of points that are simultaneous with event 1 in the (t,x) frame. Hint: this set of points is parallel to the x axis.

The second line is the set of points that are simultaneous with event 1 in the (t', x') frame. Hint: this set of points is parallel to the x' axis. The important thing to note is that this set of points is DIFFERENT that the set of points of our first line. Otherwise, we won't be using this line.

The third line that you are missing is the set of points that are simultaneous with event 1' in the (t', x') frame. Hint: this is parallel to the x' axis, similar to line 2, but not similar to line 1.

So those are the three lines I suggest you add - now come the points, and the meaning of time dilation on your graph.

Are those lines really needed? If not i would prefer not to include them as they make my picture unreadable. If i want to explain it to someone i have to use multiple pictures and not only one. Otherwise it gets messy.

pervect said:
One important pre-requisite, which I'll call PRE. Make sure that the Lorentz interval, also called the proper time, between the origin and point 1 is the same as the Lorentz interval between the origin and point 1'. I think you addressed this by saying that points 1 and 1' should both be on the same hyperbola of (ct)^2 - x^2 = constant.

How do you mean the same? Explain please.

pervect said:
In the (t,x) frame, line 1 represents the set of points simultaneous with the event you labelled as "1". Find point 2, where this line intersects Ranja's worldline.

I think that my "point 2" is in the origin of frames.

pervect said:
Time dilation says that point if point 1' occurs after point 2, because using the simultaneity conventions of the (t,x) frame, the clock in the (t', x') frame is ticking slowly. (Be sure that the precondition PRE is met!)

"Point 2" is the origin and it is an event which happens before event labeled ##1'## does.

pervect said:
In the (t', x') frame, line 3 represents the set of points simultaneous with the event you labelled as 1'. Find point 3, where this line intersects Ziga's worldline. Time dilation says that point 1 comes after point 3. Again, be sure the precondition PRE is met.

Why are these lines and points so important for me to include in the picture?
 
  • #29
71GA said:
How do you mean the same? Explain please.

Prevect is saying that you should ensure the spacetime interval between point 1 and the origin has the same numerical value as the spacetime interval between point 1' and the origin. Saying that they both lie on the same hyperbola (which is a curve of constant spacetime interval) is enough to accomplish this, and I believe that's what you did.

A hyperbola is a curve of constant spacetime interval just as a circle is a curve of constant Euclidean interval (distance).
 
  • #30
Muphrid said:
##\Delta t''## is the time coordinate of the point ##1## with respect to the ##S'## frame.

I thought so.

Muphrid said:
Your ##\Delta t'' = t_{(1)}'##, the time coordinate of point ##1## in the ##S'## frame.

I could use this to prove that for observer in primed frame thinks time interval of an observer in unprimed frame is longer. I think.
 
  • #31
If 1' is supposed to be the event on the time axis of S' that has the same time coordinate in S' as 1 has in S, then the events 1 and 1' will be on the same hyperbola ##-(ct^2)+x^2=-(c\Delta t)^2##. I don't know what formula you used to draw the curve in your diagram, but it clearly isn't that hyperbola (since it intersects the x=ct line).

Note that for any point (ct,x) on that hyperbola with x,t>0, we have ##ct=\sqrt{x^2+(c\Delta t)^2}>x##, so the hyperbola must be drawn above the 45° line. Also note that $$\frac{ct}{x}=\sqrt{1+\frac{(c\Delta t)^2}{x^2}}\to 1$$ as ##x\to\infty##. So the hyperbola will get closer and closer to the 45° line as x grows, but never intersect it.
 
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  • #32
71GA said:
Why are these lines and points so important for me to include in the picture?

The lines are the graphical illustration of "Ziga thinks Rinja's clock is running slow" and "Rinja thinks Ziga's clock is running slow". I.e .the diagram represents the concept of time dilation.

Your diagram as is doesn't contain those concepts, alas. I'm not quite sure what you think your diagram IS proving.

I originally wrote a lot more, but I think it was too wordy, so I deleted most of it.
 
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  • #33
If i add passive transformation along the active one i get the distance which i wrote as Δt′′. What is this? It is not the same value as my Δt′ which i got using the active transformation.

Here is the picture. I only added a line parallel to x′ axis and marked a distance Δt′′.

Alright, after getting confused myself because i have not done such minkowski diagrams for a long time, here are my 2 cents.

I call the diagram crossing E0/E0' which is the synced reference point.

Accordingly, the time-interval measured by Ranja between 0' and 1' on your diagram, would be a clock Ranja carries and which measures 1 second between those two event point, E0' and let's say E1'.

I would label the time interval between E0'/E1' as Δt ( you did not label this one at all).

So you can directly apply the formula as wikipedia has it.

Δt' = Δt * γ (according to my labels)

As you can see in the diagram, the time interval between E0'/E1' is measured to be longer than 1 second seen from Ziga's point of view.

Now let's check the interval between 0/1 in your diagram. Let's call those events E0/E1. A clock Ziga carries, would measure 1 second between E0/E1.

I would label this time-interval as Δt2 (you labeled it Δt). The time interval you labeled as Δt'' i would label as Δt2'.
Again, as you can see, while Ziga measures this time interval between E0/E1 to be 1 second, Ranja measures this as Δt2' > 1 second. Again, the same formula applies

Δt2' = Δt2 * γ (according to MY labels)

The parallel line to x' you drew, which goes through 1 and crosses Ranja's time axis, is equivalent to the parallel line to the x-axis which goes through 1' and crosses Ziga's time axis.
The cross-points allow to directly see the time-intervals as measured by the observer at rest. (i am too rusty on such diagrams to understand how minkowski did this magic)

This is another great achievement of genius Minkowki, allowing us to draw two diagrams within one space overlaping in such a way that we can extract information about time dilation and length contraction without actually having to do the maths.
 
Last edited:
  • #34
Here is a variation of the spacetime drawing that eliminates the hyperbola and provides an observer perspective resulting from time dilation.
On the left, an event e occurs at U(t,x)=(1,1).
Observer A, moving at .6c relative to U, intercepts the reflected light from e at event 6, at U(1.25,.750).
An arc with r=1.25 intersects a vertical line from 6.
That point is projected to the ct axis, giving A(t,x)=(1.00,.600). A thinks event e occurred at e6.
The right figure uses the same method for .3c and .9c.
If you have any CAD experience, the software can do the calculations.

https://www.physicsforums.com/attachments/52325
 
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  • #35
Jeronimus said:
Alright, after getting confused myself because i have not done such minkowski diagrams for a long time, here are my 2 cents.

I call the diagram crossing E0/E0' which is the synced reference point.

Accordingly, the time-interval measured by Ranja between 0' and 1' on your diagram, would be a clock Ranja carries and which measures 1 second between those two event point, E0' and let's say E1'.

I would label the time interval between E0'/E1' as Δt ( you did not label this one at all).

So you can directly apply the formula as wikipedia has it.

Δt' = Δt * γ (according to my labels)

As you can see in the diagram, the time interval between E0'/E1' is measured to be longer than 1 second seen from Ziga's point of view.

Now let's check the interval between 0/1 in your diagram. Let's call those events E0/E1. A clock Ziga carries, would measure 1 second between E0/E1.

I would label this time-interval as Δt2 (you labeled it Δt). The time interval you labeled as Δt'' i would label as Δt2'.
Again, as you can see, while Ziga measures this time interval between E0/E1 to be 1 second, Ranja measures this as Δt2' > 1 second. Again, the same formula applies

Δt2' = Δt2 * γ (according to MY labels)

The parallel line to x' you drew, which goes through 1 and crosses Ranja's time axis, is equivalent to the parallel line to the x-axis which goes through 1' and crosses Ziga's time axis.
The cross-points allow to directly see the time-intervals as measured by the observer at rest. (i am too rusty on such diagrams to understand how minkowski did this magic)

This is another great achievement of genius Minkowki, allowing us to draw two diagrams within one space overlaping in such a way that we can extract information about time dilation and length contraction without actually having to do the maths.

I understand this now.

phyti said:
Here is a variation of the spacetime drawing that eliminates the hyperbola and provides an observer perspective resulting from time dilation.
On the left, an event e occurs at U(t,x)=(1,1).
Observer A, moving at .6c relative to U, intercepts the reflected light from e at event 6, at U(1.25,.750).
An arc with r=1.25 intersects a vertical line from 6.
That point is projected to the ct axis, giving A(t,x)=(1.00,.600). A thinks event e occurred at e6.
The right figure uses the same method for .3c and .9c.
If you have any CAD experience, the software can do the calculations.

https://www.physicsforums.com/attachments/52325

Could you please mark axis in your picture? Picture would make more sense to me then. :)
 

1. What is time dilation?

Time dilation is a phenomenon in which time appears to pass at a slower rate for an object or person moving at high speeds relative to another object or person. This is a prediction of Einstein's theory of relativity and has been observed and confirmed through various experiments.

2. How is time dilation related to Minkowski diagrams?

Minkowski diagrams are graphical representations of spacetime, which is the combination of space and time. They show how time and space are affected by an object's motion through spacetime. Time dilation is represented on a Minkowski diagram by the slope of the object's worldline, showing how time appears to pass at different rates for different observers.

3. Can you explain the concept of "time slowing down" in time dilation?

When an object is moving at high speeds, time appears to slow down for that object relative to a stationary observer. This means that the moving object experiences time passing at a slower rate compared to the observer. This effect becomes more significant as the speed of the object approaches the speed of light.

4. How does time dilation affect GPS systems?

GPS satellites use atomic clocks to keep accurate time, but their orbits cause them to experience time dilation. This means that the clocks on the satellites run slightly faster than clocks on Earth. To compensate for this, the clocks on the satellites are programmed to run slower, allowing for accurate time measurements for GPS users on Earth.

5. Is time dilation only applicable to objects moving at high speeds?

No, time dilation can also occur in the presence of strong gravitational fields. This is known as gravitational time dilation and is a result of the warping of spacetime by massive objects. This effect has been observed and confirmed through experiments, such as the famous Hafele-Keating experiment.

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