# Passage of time in special relativity

Discusses how time passes for a non-inertial observer in special relativity [1,2].

## World lines

For any object moving through space, one can plot its spatial position against its temporal position with a spacetime diagram. For an object which stays in a single inertial frame, the world line of this object will be a straight line.

World lines do not have to be straight lines. We could also have objects which accelerate randomly and do not stay in a single inertial frame, and we would have no problem plotting a world line for the object as in the diagram below.

For objects which stay in a single inertial frame, we can use the equations for time dilation to figure out how time passes for these objects relative to another inertial frame. But what about objects which do not stay in a single inertial frame? What do they experience relative to a particular inertial frame, and how would we figure out how time passes for these objects?

## Spacetime intervals

Before thinking about how time passes for moving objects in general, let’s think about how it passes for an object moving at a constant velocity. In other words, an object that stays in a single inertial frame. The spacetime diagram on the left below shows the world line of such an object. To get the amount of time that passes for this object, we measure it in the inertial frame where the moving object is stationary, which is the spacetime diagram on the right below.

We can also find the time which passes for the moving object without transforming a world line into the intertial frame where it is stationary. First, we can define the spacetime interval as follows.1

$ds^2 = dt^2 - dx^2 - dy^2 - dz^z$

It can be shown that this remains the same no matter the inertial frame. In other words, the amount of time or length between two events will be different depending on the inertial frame, but the spacetime interval between two events will stay the same.

In the inertial frame where a world line is not moving, we have

$dx = dy = dz = 0$

so it must be the case that

$ds^2 = dt^2$

In other words, the time which passes for an object moving at constant velocity is the same as its spacetime interval. To distinguish the time which passes for the object from the time as observed from other inertial frames, we often call the former the proper time for the object, and $$\tau$$ is the symbol which usually denotes the proper time. So we can say that the proper time for an inertial world line is the same as its spacetime interval.

$d\tau = ds$

## Proper time in general

Now that we know how to figure out the time which passes for an object moving at constant velocity, how would we figure it out for an object which does not stay in a single intertial frame?

If a moving object switches between a finite number of inertial frames as in the spacetime diagram below, then it is a simple matter of adding up the time which passes for the object in each of these inertial frames.

We know that the time which passes for the object in each segment is equal to the spacetime interval of the segment, so the total proper time is the sum of these spacetime intervals.

$\tau = \sum \Delta\tau = \sum \Delta s$

What about an object which changes its velocity smoothly as in the spacetime diagram below?

We can think of this world line as an infinite number of infinitely small segments of straight world lines, each of them continuously switching over to different inertial frames. The total proper time would be the sum of these, and anyone who has even a passing familiarity with calculus would recognize that this is the same as the integral over the spacetime interval.

$\tau = \int d\tau = \int ds$

Given that $$ds = \sqrt{dt^2 - dx^2 - dy^2 - dz^2}$$, this gives us a way to figure out the proper time for any world line without having to transform infinitely tiny bits of spacetime diagrams an infinite number of times.

### With parameterized coordinates

If you are computing the proper time numerically by chopping up a world line into little pieces and adding up the spacetime intervals, this is enough. It’s a lot harder if you want to derive an analytical solution in the form of an equation. I don’t even know where to start with integrals of the form $$\int \sqrt{dt^2 -dx^2 - dy^2 - dz^2}$$.

It is easier if it is changed into an integral of a single variable. This can be done by parameterizing the coordinates $$t$$, $$x$$, $$y$$, $$z$$ with a single variable, let’s say, $$\theta$$. In other words, they would be functions of the form $$t(\theta)$$, $$x(\theta)$$, $$y(\theta)$$, $$z(\theta)$$, taking advantage of the fact that a world line is a line and not something like a surface.2 The integral for proper time would then be of a more tractable sort.

$\tau = \int \sqrt{\left(\frac{dt}{d\theta}\right)^2 - \left(\frac{dx}{d\theta}\right)^2 - \left(\frac{dx}{d\theta}\right)^2 - \left(\frac{dx}{d\theta}\right)^2} \, d\theta$

While it can be the case that $$\theta = \tau$$, it doesn’t have to be.3 The variable $$\theta$$ may not have any special physical meaning, since the most important factor in deciding how to parameterize the coordinates would be to make the calculations simpler, not that $$\theta$$ have any physical significance.

With all that said, I expect most such integrals would be beyond my ability to solve analytically. I’m satisfied with understanding the concept that proper time is equal to the sum of the spacetime intervals and being able to have a computer calculate the sum numerically to get the proper time.

### Staying still for the longest time

Spacetime diagrams are useful for visualizing a lot of things in relativity, but spacetime intervals are not quite intuitive with spacetime diagrams. It would be nice if the length of a world line is proportional to the spacetime interval, but it isn’t. The length is $$\sqrt{t^2+x^2}$$, while the spacetime interval is $$\sqrt{t^2-x^2}$$.

There are still some aspects of spacetime intervals and proper time that can be gleaned from a spacetime diagram at a glance. Namely, if you want to experience the most time between two events, then you need to stay in a single inertial frame. In the diagram below, the world line which stays still at $$x=0$$ will experience more time than the world line which meanders about.

We can see this by parameterizing the world lines by $$t$$.4 For the straight world line, $$ds = dt$$. However, for the meandering world line, it will be the case that $$ds = \sqrt{dt^2 - dx^2}$$, so $$ds$$ will be smaller than that for the straight world line at any point where $$dx \neq 0$$. This means the total spacetime interval for the straight world line is larger than that of the meandering world line. In other words, the stationary world line experiences a longer amount of time.

Since spacetime intervals remain the same in any inertial frame, we can conclude that the shortest world line in a spacetime diagram between two fixed events will have the largest proper time, and any deviation from a straight world line will experience a shorter amount of time between the two events.

## References

[1]
Sean Carroll. 2022. The biggest ideas in the universe: Space, time, and motion. Penguin Publishing Group.
[2]
Leonard Susskind and Art Friedman. 2017. Special relativity and classical field theory: The theoretical minimum. Basic Books.

1. Here we use natural units for convenience, where $$c = 1$$. when using actual measurement values, multiply the time by the speed of light.↩︎

2. A world “surface” can be parameterized, but it would need two variables.↩︎

3. In fact, if we are trying to compute the proper time, it will almost certainly be the case that $$\theta \neq \tau$$, since if they were equal, we would already know $$\tau$$ and wouldn’t be trying to do complicated calculations to obtain the proper time.↩︎

4. For the straight world line, $$t = \tau$$, but this is not the case for the meandering world line.↩︎