**Previous section: ****Spacetime Maps****.**

__Next section__**: ****Spacetime Trigonometry****.**

The **history **of a material particle is a timelike curve *X *= *X*(*τ*), as illustrated in the figure.

The path length *c**∆**τ* of an interval *∆**X *along the history is the ** proper time **interval

**. Hence,**

*registered by a clock traveling with the particle* *∆**X **∆**X̃ =* *|**∆**X**|*^{2} ≈ (*c *∆*τ*)^{2}

and, in the limit, |*dX**|** *= *c d**τ*

The *proper velocity* *V *= *V *(*τ*) of a particle is defined by

which must be distinguished from the

*relative velocity***v**=

*d*

**x**

*/dt*

**in a given inertial system.**

**Exercise: **Prove

* *(Click

__here__for a solution to this exercise.)

The proper velocity *V *of a **free particle **is constant, so its history is a straight line given by the equation

* **∆**X = **X*(*τ*) − *X*(0) = *Vc*τ = *Vc**∆*τ

The time axis of an inertial system is the history of a free particle at rest at the origin so that *proper time**τ* of another free particle can be associated with *observer time **t*, and the position vector **x **for an event *X *= *ct *+ **x **is the ** directed distance** from the particle history to the event. From the discussion above and this last diagram, we can see the relationship between the proper time

*τ*of particle 2 and the proper time

*t*of particle 1. We just integrate

*γ*

*d*

*τ*=

*dt*from 0 to

*τ*. This is easy because for free particles

*γ*is constant:

*γ**c**τ *= *ct*(*τ*) - *ct*(*τ*=0) = *ct* - *ct*_{0} (= *c**∆**t* in our diagram)

Similarly, *the time axis of **every **inertial system can be identified with the history of a free particle**.*

For a given observer (free particle) the position **x **and time *t *of a given event *X *can (in principle) be determined by *radar ranging** *with light signals, as illustrated in the figure below. Note that the event *X *is simultaneous with event *X*_{2} = *ct*. This geometric construction is Einstein’ s operational procedure for *synchronizing clocks **at distant events**.*

We (the observer) would send a light pulse at *ct*_{1} to be reflected at *X* and received at *ct*_{3}. The intervals of the blue light paths are zero (lightlike), so we get

*ct* = (*ct*_{3} + *ct*_{1}) / 2 and |**x**| = (*ct*_{3} - *ct*_{1}) / 2* *.

We can verify these formulas from the diagram or from the following expressions for the two light paths:

(*ct* - *ct*_{1})^{2} - **x**^{2} = 0 = (*ct*_{3} - *ct*)^{2} - **x**^{2} ⇒ |**x**| = *ct* - *ct*_{1} = *ct*_{3} - *ct* .

Below is an interactive diagram showing our previous diagram in a different light. It takes a while to get used to the features of spacetime maps, to construct and read them properly, and to avoid jumping to incorrect conclusions. It is interesting how a simple minus sign in the definition of the spacetime interval can wreak such havoc on our intuitive expectations, because our reflex is naturally to see lengths in Euclidean fashion. For example, in the previous diagram, we see right angles and equal sides on triangles and may feel comfortable with that. However, even in that simple example, with its 3D feel, we may tend to forget that the observer is simply sitting at a lab table and reflecting a light flash from a mirror.

The following diagram is a bit harder on the intuition and therefore worth the mental exercise. Now our lab table is sitting in a spaceship traveling in the positive x-direction with respect to an earth observer whose history is the numbered vertical time axis. The worldline of the lab table is the green line in the diagram. We send the flash of light towards the mirror in the nose of the ship. You can drag the event *X*3 (the end of the light pulse's round trip) with the mouse to see how the ship speed changes, as well as to check the *γ* factor (see the definition above) and the lab table's proper time τ, whose values are updated in the diagram's yellow boxes. Dragging *X*3 to the earth observer's history line takes the ship speed to zero and then this diagram corresponds to the previous diagram.

While using the interactive diagram, check out the following details:

1. The only meaningful ship speeds are for *v*/*c *< 1.

2. Earth time of event *X*3 is always equal to the *γ* factor times the spaceship's proper time τ from the origin to *X*3.

This ** time dilation** will be discussed in the following pages.

3. Move *X*3 to (ct,x) = (4,2), so the speed of the ship is *c*/2.

It may not look like it, but ** the red and green triangles are the same as in our previous diagram**!

The speed of light being the same to all observers means that a lightlike line will make a 45° angle with the

worldline of any observer, even with the tilted green line of the lab table's history. The smallest appearing red

triangle angle (near

*X1*where the light flash is emitted) and the largest appearing green triangle angle (near

*X*3

where the refelected flash returns to the ship's detector) are both 45° angles! In fact, the red and green

triangles are

**from the spaceship's perspective,**

*both right angled isosceles triangles sharing a common side*just as in the previous diagram, even though they look distorted in this diagram, because we are viewing things

from the perspective of earth coordinates and clocks.

Hence, the conclusions from our previous diagram still follow:

From the ship's perspective, the green triangle gives:

*c*τ = *c*τ_{2} = (*c*τ_{3} + *c*τ_{1}) / 2 = (3.46 + 0) / 2 = 1.73 = √3

|**x**| = (*c*τ_{3} - *c*τ_{1}) / 2 = *c*τ_{3} - *c*τ = *c*τ - *c*τ_{1} = √3

From the earth observer's perspective:

*c*t = (*c*t_{3} + *c*t_{1}) / 2 = (4 + 0) / 2 = 2

|**x**| = 3 - 1 = 2 = (*c*t_{3} - *c*t_{1}) / 2 = (4 - 0) / 2

4. Again, move *X*3 to (ct,x) = (4,2), so the speed of the ship is *c*/2.

From the earth observer's perspective, the light flash travels 3 space units of distance, then bounces back

for one space unit of distance before getting detected at the emitter. Thus, the light flash travels a total of

4 units of space distance. Meanwhile, the spaceship has traveled only 2 units of space distance, which checks,

since the ship is traveling half as fast as the light flash.

5. The spaceship's green history is a line of constant spacial coordinate (x = 0) in the ship's frame.

We can lay out grid lines parallel to this line to show ** the ship's spacial coordinate lines** as seen on the

earth observer's spacetime map. From the diagram, it is easy to see if we call the angle on the map between

the ship's green history line and the earth's numbered vertical history line "a", then tan

*a*=

*v*/

*c*.

The *X*..*X*2 red line, from the way we constructed it (described under the previous diagram), is a line of

constant time in the ship's coordinates, often called ** a line of simultaneity** for ship observers. By laying

out lines parallel to it, we draw

**as seen on the earth spacetime map.**

*lines of the ship's time coordinates*The symmetry of these red and green ship's coordinate lines in the diagram tells us the red lines make

the same angle "a" with earth's x-axis as the green lines make with earth's ct-axis.

To complete the grid construction, we just need to know how far apart to draw the lines. We found a hint

about that in detail 2 above. Because we can see the slope of the ship's green history line on the earth's map

is always *c*/*v*, we can write the equation (in earth coordinates) of that green history line which passes through

the origin as *ct* = (*c*/*v*)*x*. We'd like to find where to draw the red line corresponding to cτ = 1. If we correctly

draw that one, all other lines will be easy to duplicate. We know (from the discussion in the previous

paragraph) that if the slope of all the green ship coordinate lines is *c*/*v*, then the slope of all the red ship

coordinate lines is *v*/*c* on the earth map. To find the earth coordinates corresponding the ship coordinates

of (cτ=1, *x*=0), calculate the spacetime interval between the origin and this point:

(cτ)^{2} - 0^{2} = 1 = (*ct*)^{2} - *x*^{2} = (*ct*)^{2} - (*vt*)^{2} = (*ct*)^{2 }(1 - *v*^{2}/*c*^{2}) = (*ct*)^{2} / *γ*^{2}

*ct* = *γ* and *x* = *vt* = (*v*/*c*) *ct *= (*v*/*c*) *γ*

Notice we've verified the observation from detail 2 above. Now we have the slope and a point on the ship's

green worldline corresponding to (cτ=1, *x*=0), but in earth coordinates, so we can calculate the *ct* intercept of that

line. Now plug the earth point (*ct*=*γ*, *x*=*v**γ*/*c*) into the red line equation *ct *= (*v*/*c*) *x + b* to get its *ct*-axis intercept:

*b* = *ct* - (*v*/*c*) *x* = *γ* - (*v*/*c*)*v**γ*/*c* = *γ* (1 - *v*^{2}/*c*^{2}) = *γ* / *γ*^{2} = 1 / *γ*

You might want to check out these numbers in the interactive diagram again. We follow the ship's green

worldline from the origin to a proper time of cτ=1. This corresponds to the earth coordinates of (*ct*=1.15, *x*=0.58)

when the spaceship has a speed of *c*/2. We verify again that the earth time *ct* is the *γ* factor times the spaceship's

proper time *c*τ. And the red line of simultaneity for cτ=1 hits the earth *ct*-axis at 1/*γ*. The interactive diagram

draws the grid lines, based on these conclusions, using:

*a* = angle between green line and *ct*-axis = angle between red line and *x*-axis = Arctan(*v*/*c*).

*i* = integer ranging from - 25 to + 25 for 50 extra grid lines of each color.

Red lines connect earth map points from (0, i/γ) to (cos a, i/γ + sin a).

Green lines connect earth map points from (i/γ, 0) to (i/γ + sin a, cos a).

The interactive diagram has buttons to turn ON & OFF the red and/or green spaceship coordinate grid lines

as seen on the earth observer's spacetime map. You could practice by reading off spaceship and earth

coordinates for events *X*1, *X*2, *X*3 and *X*. You could make a table of both sets of coordinates and compare

various intervals to check that they are independent of which coordinates are used (invariant interval).

__Previous section__**: ****Spacetime Maps****.**

**Next section: ****Spacetime Trigonometry****.**