There are two objects (x and y) with x travelling at 10km/h and y travelling at 11km/h, both with respect to the earth. After 1h, from the perspective of the earth, y travels 11km, but from the perspective of object x, y travels 1km.

What is the “true” distance that object y travels? Is it the 1km that object x observed, is it the 11km that is observed from the earth, is distance relative, or is it none of the above?


3 Answers 3


To specify the distance an object has travelled, you need to also specify its position relative to some initial reference point. In the context of your question, there is no "true distance" or "absolute distance" an object has travelled. Instead, all distance measurements are relative and the position of an object is described by referring to some coordinate system or a point in space.

In your example, you have two objects moving at different speeds. You then went to specify their positions after a certain time, relative to the same point on the earth. You then calculated the relative distance between each object and got another value. So far so good.

But then you asked "What is the “true” distance that object y travels?" The answer is relative to what? Relative to the original point on earth, or relative to the other object, the moon, or what?

So the distance an object travels is always measured relative to some reference point, usually where the object begins its motion, or any other point in the past.

  • $\begingroup$ What if the object Y of the question was traveling in some kind of ground and had a hodometer? what the hodometer would measure? $\endgroup$
    – raulmd13
    Jan 12, 2022 at 7:30
  • 3
    $\begingroup$ It would still measure the distance travelled relative to its start point. $\endgroup$
    – joseph h
    Jan 12, 2022 at 7:44
  • 5
    $\begingroup$ @raulmd13 For a hodometer, the ground defines a reference frame. $\endgroup$
    – John Doty
    Jan 12, 2022 at 16:53

There is a "true" distance in relativity, but it involves both space and time. The true distance an object travels in spacetime between two points is called the "proper time", and is equal to the elapsed time on a clock carried along with the object. An observer moving relative to the object will believe that the clock's time is slowed by time dilation, but will also believe that the object is moving in space, and the spacetime distance $ds^2 = dt^2 - \frac{dx^2}{c^2} - \frac{dy^2}{c^2} - \frac{dz^2}{c^2}$ will be calculated by both observers to be the same.


So, like I said on my comment the way we define distance in galilean relativity is entirely, well... relative. We say that if an observer has a position vector $x$ then the distance they travelled is $\int_t |\dot{x}(t)|dt$ (if you are unfamiliar with calculus you can think that it's defined as summing up its velocity multiplied by infinitesimally small intervals of time). You can see from the structure of the definition that if you perform a change of reference (in galilean relativity, $x'=x-vt$, and restricting ourselves to inertial frames of reference), the distance then becomes $\int_t|\dot{x}-v|dt$ which is clearly different.
In other words, galilean distance is not a geometric property. Let me elaborate. In geometry, we can refer to geometric properties as properties that are true no matter how you look at them. Things like the sum of internal angles of a polygon, the length of a vector, the angle of two lines or vectors; all these things are properties of the geometric objects themselves, not of the ways we use to describe them. Take a vector; for example, the vector $v=1\,e_1+1\,e_2$, where the $e_i$ are the cartesian basis vectors $(1,0),(0,1)$. The length of $v$ is $\sqrt{2}$. Now take a new basis that is twice as big as the cartesian basis. The length of $v$ is then still $\sqrt{(\frac{1}{2}2\,e_1)^2+(\frac{1}{2}2\,e_2)^2}=\sqrt{2}$ (note that the components transform into $\frac{1}{2}$ because you're using a "ruler" that is twice as big. This is better visualized if you actually draw this vector out). Galilean distance is not this kind of property.
If our universe didn't abide by the laws of General and Special Relativity, this might be the end of the story. However, this geometrical invariance of the lengths of vectors in geometry inspires us to look for a new definition. You see, measuring length and distance depends on the kind of geometry you're looking at: measuring the lentgh of a line in a plane is not done the same way as measuring the length of a line in a spherical surface. Without getting too much into it, just so you can look it up, this "difference in measuring lengths" is described by a mathematical object called the metric tensor, which is, then, a kind of geometrical invariant ruler (in the sense that you can use it no matter what type of geometry you're looking at) that allows you to measure distances. What is done formally in special and general relativity is then to define a new geometry, composed of space and time, in which this metric tensor thingy is such that the distance travelled by an observer is also a geometric property. We call it the proper time because it coincides with the time measured by a clock attached to the observer in its journey, and it is given by: $$d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$$ (many people would call this the line element, but as the difference is up to a factor of c(the speed of light) I won't worry about that as it is essentially the same up to a matter of units). This expression is intimately tied to this metric tensor ruler and is made so that the speed of light is observed to be the same for every observer. So imagine you are sitting still in your room for 1 time unit (it doesn't really matter which it is), and consider only one spacial dimesion, with respect to which you are at its origin. Your position vector in this new geometry is $S=(1,0)$ so its length is just $c$. It can be shown that if you perform the analogous of a change of reference frames (while enforcing that the speed of light is invariant), then no matter what inertial reference frame you choose, the length of the path that your position traced is always $c$, much like the vector we discussed earlier.
This lengthy and probably disproportionate answer is my way of transmitting to you that while a sort of, say, kinematic distance is relative, geometric distance is in fact not, and it is in this duality that relativity builds its philosophy, effectively trying to translate all kinetic problems into geometric statements. I can finally answer your question, then, that the true distance that the observer $y$ travels is whatever distance is given by the formula with $d\tau$; "true" in the sense that no matter how you would be moving, in relation to $y$, you would always measure that distance.


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