A proper time interval $\Delta\tau$ for a given observer is a relativistic invariant. However, the calculation of $\Delta\tau$ requires reference to some arbitrary coordinate time t:

$$\Delta\tau = \int_{t_1}^{t_2} \! \frac{\mathrm{d}t}{\gamma}$$

Doesn't the invariance of $\Delta\tau$ require a "preferred" reference coordinate time?

According to Wikipedia:

Proper time is the pseudo-Riemannian arc length of world lines in four-dimensional spacetime.

Doesn't a unique arc-length require a fixed (ie preferred) metric?

For example, when discussing the twin "paradox", solutions typically involve drawing world lines against a fixed metric, where the different arc lengths are obvious. Does this not violate the spirit of SR? Is there a better way to look at it that does not involve a preferred reference frame against which arc lengths are defined?


2 Answers 2


No, the invariance of proper time $\Delta \tau$ doesn't require any preferred reference frame. It's the very definition of inertial systems that the proper time remains equal to $$ \Delta \tau = \sqrt{dt^2-c^{-2}(dx^2+dy^2+dz^2)}.$$ The set of coordinate systems in which this formula remains valid is continuously infinite - it's the set of all inertial frames.

The proper time has the simple form $\Delta\tau = dt$ in the rest frame of the moving object, but

  1. this rest frame is changing as the object accelerates;
  2. this rest frame is different for different objects in the Universe;
  3. the proper time may be calculated in all other inertial systems, via the formula above.

Indeed, the formula above only takes the simple form in coordinate systems in which the metric has the fixed form $$ ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 $$ i.e. $\eta_{\mu\nu}=\mbox{diag} (+1,-1,-1,-1)$. Up to the squaring and a multiplication by a power of $c$, this formula is totally equivalent to the first one.

But it's the whole point of the Lorentz symmetry that this form of the metric holds not only in one reference frame but in all (infinitely many) reference frames related by an arbitrary $SO(3,1)$ transformation. So the spirit of STR is that the metric is completely fixed, but it is fixed and equal in all inertial systems.

In general relativity, the metric itself can't be fixed - it is allowed to be curved in any way - but it's still true that locally, physics of general relativity reduces to physics of special relativity where the metric may be fixed to the form above, and it is true in infinitely many inertial frames.

This statement is totally analogous to the "proper length" in an ordinary Euclidean space. The Pythagorean theorem $ds^2 = dx^2+dy^2(+dz^2)$ for the length of a vector also doesn't require any particular coordinate system; it has the very same form in all coordinate systems related by an arbitrary rotation (and/or translation). The case of the Lorentz symmetry in STR is totally analogous, with an extra coordinate (time) added (with the opposite sign in the Pythagorean theorem).


Relativity is not that hard if in classical mechanics some typical examples of "relativity" are considered first. For example, a tree from a distance looks smaller and needs calculations to get the right (proper) tree size. In relativity such calculations involve time intervals too but the principle is the same: one obtains different "raw" experimental data in different RFs and needs recalculations to get the "proper" data. The same is valid for time intervals. They are "seen" differently from different RFs and to get the "proper" periods one needs a simple calculation involving the relative velocity.

  • 1
    $\begingroup$ A very good analogy. $\endgroup$
    – Marek
    Feb 16, 2011 at 17:26
  • $\begingroup$ I agree it's a good analogy, but it is non sequitur with regard to the question. $\endgroup$
    – user1247
    Feb 16, 2011 at 18:07
  • $\begingroup$ Another equally good analogy is that the value of Vladimir's musings seems high from Marek's viewpoint with the origin somewhere in Prague. But from my reference frame associated with Pilsen, the value and relevance for the question seems very low. ;-) $\endgroup$ Feb 16, 2011 at 18:56
  • $\begingroup$ I saw the Lubosh's answer marked as the best one so I did not try to really answer but to facilitate understanding. $\endgroup$ Feb 16, 2011 at 19:38

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