Basic input The user can input 3 variables, relative velocity and the time value and distance value, initially for event B. Values are entered in the three fields shown by the top red arrows. Note that input fields have a heavier border and are coloured different shades of light blue. The relative velocity is entered as a fraction of c, the speed of light (299 792 458 m/s). Event A is always at t = t' = 0 and d = d' = 0. To the bottom right are buttons labelled Calculate, Toggle input mode and Reset. When the button Calculate is pressed the animation will calculate the time and distance values for event B as measured by the observer in the worldline t'. These values are shown in the pink output fields. The animation will also calculate the invariant spacetime interval (the Lorentz interval) squared. These calculations are done separately for the two worldlines t and t'. The two calculations of the invariant interval are shown in the light yellow output fields. The animation calculates and displays the slope and scale of the worldline t' depending on the relative velocity. The distance line d' associated with the worldline t' is also displayed. The button Toggle input mode toggles the calculation mode of the animation and the respective input variables. In the initial setup the animation is using t and d as input and calculates with the inverse Lorentz transformations. Using the toggle button you may instead use t' and d' as input and calculate with the original Lorentz transformations. The activated input fields are coloured light blue and have a heavier border. The button Reset clears the display area, the user input and the calculated variables. The graph Event A is shown with a green ball. Event B is shown with a red ball. To find the distance of event B you follow the grey line from the red ball and parallel to the t (ct) axis and read off the d axis. To find the time of event B you follow the grey line from the red ball and parallel to the d axis and read off the t (ct) axis. When the t' (ct') axis and the d' axis are shown you read off the spacetime coordinates for event B in a similar way. To find the distance of event B in the t' reference frame you follow the red line parallel to the t' (ct') axis and read off the d' axis. To find the time of event B in the t' reference frameyou follow the red line parallel to the d' axis and read off the t' (ct') axis. Notice that the t' (ct) and d' axes are scaled according to the inverse Lorentz transformation in order to show the correct time and distance values. Real-time playback

When the data for the t' worldline are calculated and displayed, two new buttons appear: Play t worldline and Play t' worldline. The time will pass as shown by the clocks and the two events will appear as predicted by the respective measurements according to the to worldlines. This display is most effective with sound activated on your computer as the two events are accompanied by a couple of chords. Event A appears at t = 0 and d = 0.

 Observation times The Minkowski diagram shows the events as measured according to the reference frames. This should not be confused with observations of the events. Unless event B lies directly on a worldline of one of the observers, observation time will be different from measured time. To show the observation times it is necessary to calculate the travelling time for the light from the event to the observer. The animation can show the light path and the observation times. When an event has been calculated, the button Show/hide light path to observers appears in the bottom right of the animation. When this button is toggled on, the graph shows the light path from event B to the two observers at t' and t as a weak yellow line. Since light moves at v = c = ±1 in this diagram, this line is always at ±45o to the t and d axis. The observation times are marked with eyes on the two worldlines. The observation times are calculated and shown in a nearby text field. An example is shown in the illustration to the right. The observation times may be removed from the graphic by toggling the Show/hide button. 1) Slope and stretch of worldline for the moving observer.

Input t = d = 0, ie., both events are at the origin at time 0. Input relative velocity 0.3 c. Notice the slope of worldline t' and the approximate distance between the time and distance markers. Change the velocity to 0.5 c and recalculate. Notice that the slope relative to the t axis has increased and that the distance between the markers has increased.

Excercise 1a: Try other velocities like 0.7 c and 0.8 c.

Excercise 1b: Now study the d' axis when different velocities are chosen. What is the relationship between the angle between t and t' and the angle between d and d'?

2) Change point of view

If observer t' moves at velocity v in relation to observer t, one may change the point of view and say that observer t moves at velocity -v in relation to observer t'. The two velocities are equal in magnitude and opposite in direction. If we change the point of view of the motion, the locations of events A and B will be exchanged as well. What was t and d for event B is now t and d for event A and vice versa.

An example: Input v = 0.6 c, t = 6, d = 5. Notice that t' = 3.75 and d' = 1.75. Now switch point of view by inputting v = -0.6 c, t = 3.75 and d = 1.75. Notice that now t' = 6, d' = 5. In other words, the change in point of view keeps the same coordinates for the two events.

Excersise 2a: Input v = 0.25 c, t = 4, d = 6. Switch point of view. What are the new input values?

Excercise 2b: Input v = 0.8 c, t = 2.5, d = -1. Switch point of view. What are the new input values?

Excercise 2c: Input v = -0.6 c, t = 5, d = -5.5. Switch point of view. What are the new input values?

3) Time dilation and the slowing of clocks.

 Toggle the input mode so that the t' and d' fields are blue. Input v = 0.6 c, t' = 4 and d' = 0. This indicates that both events are on the worldline of the moving observer. Suppose both events represents a tick of a clock that ticks every second. From the moving observer's point of view, the clock shows that 4 seconds has passed. Notice that the nonmoving observer measures that 5 seconds has passed. The nonmoving observer will conclude that the moving clock is slow. Now turn the situation around. Toggle the input mode so that the t and d fields are blue. Input v = 0.6 c, t = 4 and d = 0. The clock is now on the worldline of the observer that is at rest. The nonmoving clock shows that 4 seconds has passed. But according to the moving observer, 5 seconds have passed. She will correctly conclude that the nonmoving clock is slow. Conclusion: Both observers will conclude that the other clock is slow (by a factor of 4/5 = 0.8). Excercise 3a: Repeat this experiment with v = 0.8 c. What are the t and d values for the nonmoving observer be when the moving observer's clock shows t' = 1.5? What is the rate of time dilation each observer will calculate for the other clock?

Excercise 3b: Repeat the experiment with v = 0.5 c. What are the t and d values for the nonmoving observer when the moving clock shows t' = 3? What is the rate of time dilation?

Excercise 3c: After observing the slowing of clocks for 3 different velocities, 0.5 c, 0.6 c and 0.8 c, what will you conclude regarding the relationship between the relative velocity and the time dilation?

4) Simultaneity

Let v = 0 and input t = 0, d = 4. The two events A and B are simultaneous in both reference frames. Now change the velocity to v = 0.4 c. Notice that t' changes to -1.7457. This means that in the reference frame of the moving observer event B is measured to happen almost 2 seconds before event A, they are not simultaneous. But in the nonmoving reference frame they are still happening at the same time!

Change the relative velocity to v = -0.4. Now event B is measured to happen almost 2 seconds after event A according to the moving observer. To conclude: Events that are simultaneous to one observer are generally not simultaneous to another observer moving relative to the first one.

Repeat this experiment. Each time, click the button "Play t worldline" and follow the animation of the events as measured by observer with worldline t". Also click the button "Play t' world line" and follow the events as measured by observer with world line t'. Notice when and where the two events appear in the two respective worldlines.

Another example. Input v = 0.6 c, t = 2.25 and d = 3.75. Notice that t' = 0. Event B is now simultaneous to event A in the reference frame of the moving observer, but not in the nonmoving reference frame. Here event B happens 2.25 seconds after event A. Notice that event B is placed on the axis d'. All points on this axis are simultaneous to the origin in the moving reference frame. Likewise: All points on any line that is parallell to d' are equidistant in time from the origin in this reference frame.

Use the "Play worldline"-buttons also for this experiment and for the two exercises that follow!

Excercise 4a: Two observers are moving at 0.8 c relative to each other. For one observer two events A and B are simultaneous and separated by 3 units of space. How will the two events be measured by the other observer?

Excercise 4b: At a relative velocity of 0.8 c, what will the nonmoving observer measure if the moving observer measures event B to be simultaneous with event A and at the distance of 1.5? (Hint: Use the input mode for the original Lorentz transformations.)

5) Temporal order

Under some conditions the temporal order of events may be any of these: Before, simultaneous or after. An example: Let event B be located at t = 2, d = 4. First, try v = 0.2 c. Both observers now agree that event A happens before event B. Change the velocity to v = 0.5 c. Now the two events are simultaneous according to the moving observer. Finally set v = 0.8 c. This time the moving observer will measure event B to happen before event A. All three possibilities of temporal order are present, only depending on the velocity of the observer. Use the Play worldline-buttons for all three scenarios to confirm the numerical and graphical data.

Excercise 5: Set v = 0.6, t = 1 and d = 5. Then find a velocity where the two events are simultaneous for the moving observer and a velocity where event A happens before event B for the moving observer.

Examples 4 & 5 demonstrate clearly that there is no such thing as absolute simultaneity.

6) The invariant interval

In special relativity there is a quantity that is invariant (not changing) for all inertial reference frames. In other words, given some event B one can calculate an interval which is the same regardless of the relative velocity of the two observers. The equation for this interval (also called the Lorentz interval or the invariant spacetime interval) is given by i2 = (ct)2 - d2. The animation calculates the square of the invariant interval - and it calculates it separately for the two reference frames in order to show that it really is an invariant!

Example: Set v = 0.3 c, t = 4, d = 3. Notice that i2 = i'2 = 7. Try changing the velocity, e.g set v = 0.5 c. Try any other value for v. Try a negative value for v.

Excercise 6: Find the invariant interval for t = 2.5, d = 5.5. Find the invariant interval for t = 6, d = 2.

[In three spacial dimensions the Lorentz interval is given by the equation i2 = (ct)2 - (x2 + y2 + z2).
(In some texts the sign is swithced between the time and space dimensions).]

7) Timelike and spacelike intervals

In the animation the spacetime diagram is divided into 4 equal sectors. Two of these are labelled "Timelike intervals", two "Spacelike intervals". Choose any value for v and plug in these values for t and d: t = 3 d = 5, t = -1 d = 6, t = 4 d = -5, t = -3 d = -6. Notice that all instances of event B are in the sectors with spacelike intervals. Notice also that for all these instances the spacetime interval squared is negative. In all such cases the spacial relation between events A and B cannot be reversed. With spacelike intervals the spacial separation dominates - d2 > (ct)2 - and there can be no causual relationship between the two events. Another way to view this: The interval is such that there is not enough time for light to pass from one event to the other.

Now choose any value for c and plug in these values for t and d: t = 2 d = 1, t = 5 d = 3, t = 6 d = -1, t = -5 d = 4, t = -6 d = -3. Notice that all instances of event B are in the sectors with timelike intervals. Notice also that for all these instances the spacetime interval squared is positive. In all such cases the sequence in time between events A and B cannot be reversed. With timelike intervals the separation in time dominates - (ct)2 > d2 - and there is no fixed directional relationship between the two events: left and right can be exchanged depending on the motion of the ovserver.

There is a third type of interval to consider. Let (ct)2 = d2 which is to say v = d/t = c. In other words, the velocity is c, the speed of light. This type of interval is therefore called lightlike and it is represented in the animation by the two yellow worldlines representing light passing through event A.

Excercise 7: Given that event A as usual is in the origin, decide which of the intervals are spacelike, timelike and lightlike for the following events B. Use both the numerical and the graphical method to decide. Also use the "Play worldline"-function to confirm your choice.
a - t = 5, d = 3. b - t = -6, d = -3. c - t = 1.5, d = 5. d - t = 3, d = -3.5. e - t = 2.1, d = 2.0. f - t = 4.5, d = 4.5.

8) Observations

The two observers will see event A simultaneously, at t = d = 0. When will they see (observe) event B? To find out, one must calculate the travel time for light from event B to the worldline of the two observers. When a calculation has been done in the animation, a new button appears: Show/hide light path to observers. Input v = 0.4 c, t = 5 and d = 4. Calculate and toggle the Show/hide button. Notice that the nonmoving observer measures event B to happen at t = 5 but he sees the event at t = 9. The moving observer measures event B to happen at t' = 3.7097 and she sees the event at t' = 5.8919. Notice the angle of the light path. Notice that the same light path passes through both observers.

Excercise 8a: What is the angle of the light path relative to the time axis t?

9 ) The twin paradox

The twin paradox is the well-known fact that someone travelling at a relativistic speed ages more slowly than someone at rest. This can be demonstrated by an experiment with a pair of twins where one twin travels an interstellar distance and back at high speed.

Let event A1 be the separation of the twins and event A2 the reunion. These events are both on the t worldline of the twin that stays at home (ie, at rest). Event B is the turn around point for the travelling twin. This event must therefore be on the t' worldline. To achieve this we must set the values such that d / t = relative velocity. Let us try v = 0.75, d = 3 and t = 4. Note that t' = 2.6458, this is the elapsed time for the traveller. Elapsed time for the stay-at-home twin is of course 4. Now the travelling twin turns around. The return trip therefore has v = -0.75 and t = -4 and d = 3. t must be negative to achieve the negative relative velocity. An alternate reasoning is that the event A2 which is the common event when the two twins are back together again has the values t = d = 0, and to achieve this t and t' for event B on the return trip must be negative. Note that t' = -2.6458 and t = -4. Time elapsed must be the negative of this result since we started the home lap with a negative time value. Total elapsed time for the travelling twin is thus 2.6458 * 2 = 5.2916. And total elapsed time for the stay-at-home twin is 4 * 2 = 8.

Conclusion: the travelling twin ages slower! This is of course not really a paradox: The stay-at-home twin is staying in the same inertial reference frame but the travelling twin is switching reference frame at the turn around point. This can also be seen from the fact that the worldline of the stay-at-home-twin is continuous (vertical both times) but the travelling twin has two different worldlines with - in this case - an angle of about 74o.

Excercise 9a: Use the animation to calculate how much younger the travelling twin is after travelling 24 light years and back at 0.96 c. (Hint: Start by finding the travel time at 0.96 c!)

Excercise 9b: Use the animation to calculate how far you can travel and return while 500 years pass at home and you travel at 0.9999 c. What is your elapsed time as a traveller?

10) Length contraction

To use the Minkowski diagram to calculate and show the Lorentz-Fitzgerald length contraction requires some special consideration. Suppose we have an object of a certain length, say 7 units. What would it entail for the two observers to measure the length of this object? The nonmoving observer has no problem, the object is not moving relative to him, so he can use his distance axis d. We know that all points on this axis are simultaneous in the nonmoving reference frame. The moving observer also needs to measure the length of the object by measurements simultaneous at both ends of the object. That is only possible along the d' axis or a line parallell to this axis. We know that all points on the d' axis are simultaneous for the moving observer. In the case of the d' axis we also know that the time t' at all points is 0.

Input v = 0.3 c and d = 7. The meaning of this is that the rest length of the object is 7 units, and this is the length the nonmoving observer measures. How can we find t? This equation from the inverse Lorentz transformation will help: We know v and d, and we also know that t' = 0 since we are only interested in points along the d'-axis. This fact simplifies the equation to t - vd = 0 and t = vd. With v = 0.3 and d = 7 we easily calculate that t = 2.1. Intput this result.

Notice that this gives d' = 6.6776. This demonstrates that the object is shorter by appr 0.3224 when the velocity is 0.3 c.

Now input some new values: v = 0.4 c, t = 0.4 * 7 = 2.8. The result is now d' = 6.4156.

Excercise 10a: Find the length contraction at v = 0.5, 0.6, 0.7, 0.8, 0.9, 0.95, 0.99 and 0.999 c.

Notice that the length contraction increases more rapidly with increasing velocity. Notice also that the length contraction becomes extreme when v approaches c.

Excercise 10b: Suppose the object is moving along with observer t'. Find out how to use the animation to calculate the length contraction when v = 0.3 c and d' = 7. (Hint: Use the original Lorentz transformation, not the inverse!)

Answers to excercises

1b: The two angles are equal in size and opposite in direction.

2a: v = -0.25 c, t = 2.582, d = 5.164.

2b: v = -0.8, t = 5.5, d = -5.

3a: Event B happens at t = 2.5 and d = 2 in the rest frame. The rate of time dilation is then 1.5 / 2.5 = 0.6.

3b: For the nonmoving observer event B happens be at t = 3.4641 and d = 1.7321. The rate of time dilation is 3 / 3.4641 = 0.8660.

3c: The time dilation increases with increased velocity.

4a: One event is 4 units of time before the other and they are separated by 5 units of space.

4b: The nonmoving observer measures t = 2 and d = 2.5.

5: At v = 0.2 c the events are simultaneous in the moving reference frame. For any velocity < 0.2 c event A happens before event B.

6: i2 = -24, i2 = 32.

7: a & b - timelike. c & d - spacelike. e - timelike. f - lightlike.

8a: The angle is 45o in this example, it may otherwise be -45o. In any and all cases it is parallel to one of the two worldlines of light passing through event A.

9a: The stay-at-home twin ages 50 y, the travelling twin ages 14 y, the difference is 36 y.

9b: 249.975 ly and back, and the total travel time is 7.07 y, only a little over 1/100 of the stay-at-home elapsed time!

10a: The following table shows the length contractions for several values of c when the length for the nonmoving observer is 7 units.

0.1 c d' = 6.9649
0.2 c d' = 6.8586
0.3 c d' = 6.6776
0.4 c d' = 6.4156
0.5 c d' = 6.0622
0.6 c d' = 5.6
0.7 c d' = 4.999
0.8 c d' = 4.2
0.9 c d' = 3.05125
0.95 c d' = 2.18575
0.99 c d' = 0.98747
0.999 c d' = 0.31297

10b: Use this equation from the original Lorentz transformation: . Now you want to measure along the d axis and therefore t = 0. v and d' are given and we only need to find t'. Since t = 0, the equation reduces to t' = -vd' and we easily calculate t' = -2.1. Input v, d' and t', and, voila!

Sources

• Mook & Vargish: Inside Relativity. Princeton Univ. Press 1987
• Taylor & Wheeler: Spacetime Physics. 2nd ed. Freeman & co 1992
• Bais: Very Special Relativity. Harvard Univ. Press 2007
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