Examples and Exercises for Minkowski Space/Time Diagrams

Go to: http://www.trell.org/div/minkowski.html#Tutorial

And work through these:

1) Slope and stretch of worldline for the moving observer.

Inputt = d = 0, i.e., both events are at the origin at time 0. Input relative velocity0.3 c. Notice the slope of worldlinet' and the approximate distance between the time and distance markers. Change the velocity to0.5 cand recalculate. Notice that the slope relative to thetaxis has increased and that the distance between the markers has increased.

Exercise 1a: Try other velocities like0.7 cand0.8 c.

Exercise 1b: Now study thed'axis when different velocities are chosen. What is the relationship between the angle betweentandt'and the angle betweendandd'?

2) Change point of view

If observert'moves at velocityvin relation to observert, one may change the point of view and say that observert moves at velocity-vin relation to observert'. 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 wast anddfor event B is nowtanddfor 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.

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

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

Exercise 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 thet'andd'fields are blue. Input v = 0.6 c, t' = 4 and d'=0. This indicates that both events are on the worldline of themovingobserver. Suppose both events represent 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 thetanddfields 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 nonmovingclock is slow.
Conclusion:Bothobservers will conclude thatthe other clockis slow (by a factor of 4/5=0.8). /

Exercise 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 theotherclock?

Exercise 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?

Exercise 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 secondsbeforeevent A, they arenotsimultaneous. 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 secondsafterevent 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 "Playtworldline" and follow the animation of the events as measured by observer with worldlinet". Also click the button "Playt'world line" and follow the events as measured by observer with world linet'. 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 themovingobserver, butnotin the nonmoving reference frame. Here event B happens 2.25 secondsafterevent A. Notice that event B is placed on the axisd'.Allpoints on this axis are simultaneous to the origin in the moving reference frame. Likewise: All points on any line that is parallel tod'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!

Exercise 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?

Exercise 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 thatevent A happens before event B. Change the velocity to v = 0.5 c. Now the two events aresimultaneousaccording to the moving observer. Finally set v = 0.8 c. This time the moving observer will measureevent B to happen before event A. All three possibilities of temporal order are present, only depending on the velocity of the observer. Use thePlay worldline-buttons for all three scenarios to confirm the numerical and graphical data.

Exercise 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 byi2= (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.

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

[In three spatial dimensions the Lorentz interval, i, is given by the equationi2= (ct)2- (x2+ y2+ z2).
(In some texts the sign is switched 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 withspacelike intervals. Notice also that for all these instances the spacetime interval squared isnegative. In all such cases the spatial relation between events A and B cannot be reversed. With spacelike intervals the spatial separation dominates -d2> (ct)2- and there can be no causal 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 withtimelike intervals. Notice also that for all these instances the spacetime interval squared ispositive. In all such cases thesequence in timebetween 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 observer.

There is a third type of interval to consider. Let(ct)2= d2which 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.

Exercise 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 willseeevent A simultaneously, at t = d = 0. When will theysee(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 theShow/hidebutton. Notice that the nonmoving observermeasuresevent B to happen at t = 5 but heseesthe event at t = 9. The moving observermeasuresevent B to happen at t' = 3.7097 and shesees the event at t' = 5.8919. Notice the angle of the light path. Notice that the same light path passes through both observers.

Exercise 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 A1be the separation of the twins and event A2the reunion. These events are both on thetworldline of the twin that stays at home (i.e., at rest). Event B is the turn around point for the travelling twin. This event must therefore be on thet'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 traveler. 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 A2which 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.

Exercise 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!)

Exercise 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 traveler?

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 parallel 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.

Inputv = 0.3 candd = 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 findt? This equation from the inverse Lorentz transformation will help: