# Space-time diagrams

1. Nov 1, 2013

### rushikesh

I really find difficult understanding what spacetime diagram does relativity use while defining its notion of space and time and moving observers.

Can we place both a still and moving observer in the same space time diagram as in the figure (A), then the moving observer measures time going slow. Or does the moving observer also measure time with the same speed as in figure (B). In that case how does one see the other's time slowed down?

2. Nov 1, 2013

### Staff: Mentor

You can put them both in the same diagram, but you have to draw different x and t axes for the two observers. Any single event is represented by a single point in the diagram, but it will have different x and t coordinates according to which pair of axes you use.

3. Nov 1, 2013

### rushikesh

And what about the figure (B), is it valid?

4. Nov 1, 2013

### ghwellsjr

No, it is not valid. You are showing the passage of time for O2 contracted instead of dilated. If you want to see a lot of spacetime diagrams, including some that illustrate how one observer can determine the Time Dilation of a moving clock, do a search on my name for the word "diagram".

5. Nov 1, 2013

### rushikesh

Taking one of your above spacetime diagrams, why should the green lines (moving observers) have longer distance between two time coordinates? can't they be the same as the blue line (still observer) or if they had to be longer then why not as shown in this below figure

(and also tell me how do you post such an enlarged picture within the post, but mine appears small)

6. Nov 1, 2013

### ghwellsjr

Although in your diagram, the distance on paper shows the spacing between times as longer, that's strictly because of the motion. They need to have their time coordinates stretched out. Here's how to do it correctly. Start with a diagram showing a stationary observer like you already did. Then use the Lorentz Transformation process for both x and t to see what the coordinates are in a new frame moving with respect to the original one. Hint: use units where c=1. Draw a new diagram. Once you see how time is dilated for moving objects, you can just bypass this step and draw them in correctly based on the value of gamma.
Go to advanced editing mode. Scroll down and click on "Manage Attachments". Upload your pictures and close the window. Right-click on one of the file names and select "copy link location" or something similar. Click in the edit window where you want the image to appear. Select the "Insert Image" tool (I think that is what it is called) and paste it in the dialog box (overwriting what was already there). I like to center my drawings.

Last edited: Nov 2, 2013
7. Nov 2, 2013

### rushikesh

Ok, now let me put it different way. Lets forget about the space time diagrams and the coordinate system. Simply view time as a four dimensional space. Here it is a pure space and one dimension cannot have any distinctness with respect two any other dimension, just as in our three dimensional space, all three dimensions are not distinct from each other in any way.

Now we can consider various moving observers as going at different angles like the branches of a tree as shown below. (more angle means more relative speed) But two observers moving relative to each other will measure their time going at an equal speed i.e. the length/distance between their two points in time is equal for both observers as shown here (since we have accepted that their direction in a 4d space do not actually matter)

So considering this, my question is how do the two observers see each other's time as dilated? Can we explain it without using the general sp-time diag.

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8. Nov 2, 2013

### PAllen

The problem is that while SR relates time and space into spacetime, that does not mean there is no difference between time and space. The analog of the Pythagorean theorem for spacetime has plus signs for the spatial sides and a minus sign for the time side. An x-t plane has fundamentally different geometry than Euclidean. There is a trick to hide some of the difference using imaginary units for time, but that doesn't change the geometric difference.

9. Nov 2, 2013

### ghwellsjr

Time Dilation is a coordinate effect. Observers cannot see Time Dilation. It is different in each coordinate system moving with respect to each other. Different coordinate systems do not change what the observers in them can actually see. If you want to forget about coordinate systems, then you have to forget about Time Dilation.

You don't need a spacetime diagram to explain Time Dilation but you do need a coordinate system. So you can explain Time Dilation in purely mathematical terms but it is a lot easier to visualize using a diagram.

10. Nov 2, 2013

### WannabeNewton

Let $O$ be an observer with 4-velocity $\xi^a$ at some event $p$ on $O$'s worldline. Consider an event $q$ infinitesimally separated from $p$ and let $\eta^a$ be the vector connecting $p$ and $q$ i.e. the infinitesimal space-time displacement vector from $p$ to $q$. Furthermore, let $O'$ be another observer who passes by $O$ at $p$ with 4-velocity $\xi'^a$.

At $p$ itself, $O$ can decompose all vectors into a part parallel to $\xi^a$ and a part orthogonal to $\xi^a$. The set of all vectors at $p$ orthogonal to $\xi^a$ form the infinitesimal simultaneity slice of $O$ at $p$ so when we decompose a vector into a part parallel to $\xi^a$ and a part orthogonal to it, we are decomposing it into its time part relative to $O$ and its spatial parts relative to $O$ respectively. The parallel and orthogonal projection operators relative to $\xi^a$ are given respectively by $k^{a}{}{}_{b} = -\xi^a \xi_b$ and $h^{a}{}{}_{b} = g^{a}{}{}_{b} + \xi^a \xi_b$ where $g_{ab}$ is the metric.

Now the length of the time part of $\eta^a$ relative to $O$ is just the elapsed time $t$ between events $p$ and $q$ relative $O$ i.e. $t = \left \| k^{a}{}{}_{b}\eta^{b} \right \| = -\eta^{a}\xi_{a}$; similarly $s = \left \| h^{a}{}{}_{b}\eta^b \right \|$ is the spatial distance between $p$ and $q$ relative to $O$. Also, the length of the time part of $\xi'^a$ relative to $O$ is just the gamma factor $\left \| k^{a}{}{}_{b}\xi'^b \right \| = -\xi'^b\xi_b = \frac{1}{\sqrt{1 - v^2}} = \gamma$.

As stated, we can write $\xi'^a$ as $\xi'^a = k^{a}{}{}_{b}\xi'^b + h^{a}{}{}_{b}\xi'^b$ hence $t' = -\xi'^a \eta_{a} = \gamma t - (h^{a}{}{}_{b}\eta_{a})\xi'^b$. If we take the events $p$ and $q$ such that they have zero spatial distance between them relative to $O$ then $\left \| h^{a}{}{}_{b}\eta_{a} \right \| = 0$ but $h^{a}{}{}_{b}\eta_{a}$ is a purely space-like vector so this means $h^{a}{}{}_{b}\eta_{a} = 0$ thus $t' = \gamma t$ which is the usual kinematical time dilation factor.

Obviously this is more long-winded than the usual derivations that require the use of a frame/coordinate system but this was just to show that the kinematical time dilation can be derived using only space-time geometry i.e. without working in some frame/coordinate system. Time dilation is still an observer dependent effect of course but with the above it's easy to see that time dilation arises from the 4-velocities of different observers not being parallel at a given event (i.e. being at an angle to one another at that event) in space-time.

Last edited: Nov 2, 2013
11. Nov 2, 2013

### robphy

Since I am fan of trigonometry, here is a trigonometric interpretation of the time-dilation factor:
it is the hyperbolic cosine of the spacetime-angle (rapidity) between the observers' 4-velocities.
(The hyperbolic tangent of that angle is the relative-velocity between those observers.)

$$\gamma=\cosh(\theta)$$
$$\beta=\tanh(\theta)$$

12. Nov 2, 2013

### TheBC

@Rushikesh.
Reciprocal time dilation is a result of different 3D space worlds of simultaneous events cross-cutting through 4D spacetime.
Einstein told us to look at it 4-dimensionally.
<< From a "happening" in three-dimensional space, physics becomes, as it were, an "existence" in the four-dimensional "world". >> Albert Einstein. "Relativity: The Special and the General Theory." 1916. Appendix II Minkowski's Four-Dimensional Space ("World") (supplementary to section 17 - last section of part 1 - Minkowski's Four-Dimensional Space).
<< Since there exists in this four dimensional structure [space-time] no longer any sections which represent "now" objectively, the concepts of happening and becoming are indeed not completely suspended, but yet complicated. It appears therefore more natural to think of physical reality as a four dimensional existence, instead of, as hitherto, the evolution of a three dimensional existence. >> Albert Einstein, "Relativity", 1952.

13. Nov 7, 2013

### rushikesh

Ok, I found out that I was talking about the Loedel's space-time diagrams and it is possible to explain time dilation in the tree example above. You can read more on it here: http://spacetime-blockuniverse.weebly.com/reciprocal-time-dilation.html

14. Nov 7, 2013

### ghwellsjr

A Loedel space-time diagram is nothing more than another Inertial Reference Frame (IRF) transformed to a speed such that two inertial observers are moving at the same speed in opposite directions. As such, they have the same Time Dilation. But they still cannot directly see that Time Dilation. They have to make exactly the same measurements, assumptions and calculations that they would in any other frame to determine each others Time Dilation. All IRF's are equally valid. None is preferred. Not one where an observer is at rest. Not one where both observers are traveling at the same speed. The observers can't know about the existence of a frame unless they make a whole lot of measurements, assumptions and calculations.

15. Nov 7, 2013

### PAllen

Yes, these diagrams can be helpful, but they're really only convenient for two observers at a time.

Last edited: Nov 7, 2013
16. Nov 10, 2013

### rushikesh

I regret not being from a physics background to understand the above.

Can you clarify it more?

17. Nov 10, 2013

### WannabeNewton

See robphy's post (post #11).

18. Nov 10, 2013

### rushikesh

what is observers' 4 velocity you are talking about

19. Nov 11, 2013

### TheBC

But the big advantage of loedel diagram is that all time and space units have equal length. This makes reading reciprocal time dilation and length contraction as a result of relativity of simultaneity in 4D spacetime a lot easier.

20. Nov 11, 2013

### ghwellsjr

Maybe you could show us how it is a lot easier.

Here is a spacetime diagram showing two spacecraft departing from each other at the same opposite speed of 0.6c in this Inertial Reference Frame (IRF). I'm defining the speed of light to be 1 foot per nanosecond. The dots represent 1 microsecond intervals of Proper Time along each worldline:

Note that the spaceships are both 4000 feet long in this IRF.

However, if we look at the IRF corresponding to the rest frame of the Black&Blue spaceship, we see that its Proper Length is 5000 feet:

We can also easily see the Time Dilation of the Red&Green spaceship by dividing the Coordinate Time of 17 usec by the Proper Time of 8 usec for the top red dot yielding a Time Dilation factor of 2.125 which also the value of gamma for a speed of 0.882353 (the relativistic sum of 0.6c and 0.6c).

We can take the inverse of gamma, 1/2.125 = 0.471 to confirm that the Length Contraction of the Red&Green spaceship is 5000 times 0.471 or 2353 feet. At the Coordinate Time of 17 usecs, we can see that the green line is 2353 feet further to the right than the top red dot.

Here is the IRF in which the Red&Green spaceship is at rest:

And we can confirm that the Time Dilation and Length Contraction factors are reciprocal, that is, in the rest frame of each spaceship, the other spaceship is Time Dilated and Length Contracted to the same degree.

Can you please show us how the first diagram shows these effects any easier than the other two diagrams?

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Last edited: Nov 11, 2013