# Why is proper time $d\tau$ equated to spacetime length $ds$?

Follow-up to this question: Why proper time is a measure of space?.

The selected answer to me tells us why proper time is an invariant quantity, but I'm still wondering why we equate it to $$ds$$. Can there not be two independent invariant quantities?

• Jun 17, 2022 at 11:01

Generally speaking, the line element can be written as $$ds^2=g_{\mu\nu}(x)dx^{\mu}dx^{\nu},$$ where $$g_{\mu\nu}(x)$$ are the metric components at the point $$(x^0,x^1,x^2,x^3)$$. For example, working in cartesian coordinates $$(t,x,y,z)$$, for the Minkowski metric $$g_{\mu\nu}(x)=\text{diag}(-1,1,1,1)$$, this simply becomes $$ds^2=-dt^2+dx^2+dy^2+dz^2.$$

Proper time is the time experienced by an observer as it reads on their own clock. This means that an observer would describe their position in spacetime, in their coordinates, say $$(t',x',y',z')$$, to be $$(\tau,x_0,y_0,z_0)$$, where $$\tau$$ is the proper time measured by the observer and $$x_0,y_0,z_0$$ are constants (because in their coordinates the observer is always at rest and everything else moves around them).

In particular, this implies that $$dx'=dy'=dz'=0$$ in the observer's coordinates. So, in this set of coordinates, the line element reads $$ds^2=g'_{\mu\nu}(x')dx'^{\mu}dx'^{\nu}=g'_{00}(\tau,x_0,y_0,z_0) d\tau^2.$$

Now, for an inertial observer (i.e. an observer moving freely on a geodesic), their coordinates are given by the inertial coordinates at the point where the observer is. These coordinates have a couple of important properties: the vanishing of the Christoffel symbols and the fact that the metric is given by the Minkowski metric at the point where the observer is (this goes for both statements). In other words, $$g'_{\mu\nu}(\tau,x_0,y_0,z_0)=\text{diag}(-1,1,1,1)$$ and the above becomes $$ds^2=-d\tau^2$$.

For a non-inertial observer, I must admit I have never seen a good explanation of what is meant by "their coordinates" in this case. I believe even in this case there exists a set of coordinates such that the metric at the point where the observer is is still given by the Minkowski metric, so the above argument for the line element is unchanged (the difference in this case being that the Christoffel symbols do not vanish anymore). Maybe someone more knowledgeable may correct me on this one if I am wrong.

In general, the separation between two events contains a spatial component and a time component.

If the separation is time like, the time is large and the distance is small. You can travel between the two events without exceeding the speed of like. Taking flat space for simplicity, if you do this at a uniform velocity you have selected an inertial frame of reference where the spatial component is $$0$$. The separation is entirely time only in this frame. In this frame, the time is the interval between them. This is the proper time.

The interval is invariant in any frame. From another frame you can figure out the proper time. It is equivalent to figuring out the time in this special frame of reference.

For curved space, the same argument applies along the geodesic that passes through the two events. This is unaccelerated motion.

This answer considers the generalization of special relativity to general relativity:

1. The infinitesimal arclength $$ds$$ between 2 infinitesimally close spacetime events/points $$P$$ and $$Q$$ on a Lorentzian manifold is by definition given as $$(ds)^2~=~g_{\mu\nu}dx^{\mu}dx^{\nu}.$$ It is an invariant independent of coordinate system. We'll assume that the signature of the metric tensor is $$(1,3)$$ rather than $$(3,1)$$ to avoid an extra minus sign in what follows.

2. Assume that $$(ds)^2>0$$ is positive, i.e. timelike. Now pick a coordinate system such that the points $$P$$ and $$Q$$ have same spatial components $$dx^1=dx^2=dx^3=0$$, i.e. a rest frame for an observer going through $$P$$ and $$Q$$.

3. Then $$(ds)^2~=~g_{00}(dx^{0})^2, \qquad g_{00}~>~0.$$ We could still reparametrize the temporal coordinate $$x^{0}= f(\tilde{x}^{0})$$. Then $$(ds)^2~=~\tilde{g}_{00}(d\tilde{x}^{0})^2, \qquad \tilde{g}_{00}=g_{00}f^{\prime}(\tilde{x}^{0})^2.$$ However, $$ds$$ is an invariant notion of lapsed temporal coordinate for an observer at rest, so it's natural to define the lapse in proper time as $$d\tau~=~\frac{ds}{c}.$$

4. Note that if $$(ds)^2\leq 0$$, we can't choose a rest frame and make an identification between proper time and arclength.

• so, let me see if i understand correctly. is the argument that we can always frame an event as if it happened in a frame where there is no spatial motion, only temporal motion, which would be proper time for this frame. and since we can always look at it this way, we can say that $ds = d\tau$ because $ds$ is invariant, so the fact that we can frame it this way in one frame means it takes this value in all frames? Jun 18, 2022 at 19:07