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?


3 Answers 3


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

If you think of the proper time interval as the amount of time that you would age along that path it might make it easier to reason about the physics.

Between any two points that happen to you, you've aged a certain amount. That age is a proper scalar and is invariant under Lorentz transformation and all observers in the universe will agree that you have aged that much. It would be very weird if one observer thought you were 5 years old at some point (on your personal timeline) and one observer thought you were 5 seconds old at that point.

Those observers may themselves disagree on the amount of time they personally measure between those two points in their inertial coordinates, but they all agree on the amount of time that you measure between those two points.

And it sounds like you have some confusion around the semantic meaning of "preferred". When physicists are talking about a "preferred" reference frame they mean a hypothetical agreed upon universal coordinate system which would be the "correct" one. That reference frame doesn't exist. So I can make a calculation in a highly boosted reference frame to you and find a very different $\Delta t$ and $\Delta x$, but then I compute $\Delta \tau^2 = \Delta t^2 - \Delta x^2$ and it always will come out to the same answer as everyone else gets.

If there were a "preferred reference frame" then there would be a $\Delta t$ and $\Delta x$ measurement that would be the "right" one. It is correct that proper time is invariant and "special" but it is just a scalar value and a number (with units). There's no reference to any coordinate axis. You can do the same calculation in Cartesian coordinates, spherical coordinates, or highly boosted coordinates and they'll all agree.

And the principle here is that if you do physics in one reference frame that it agrees with physics done in another reference frame. That frees you to pick a reference frame to solve any given problem which makes the problem easier to solve. The best frame to pick for a problem will often be the one where something important (like the center-of-mass) is at rest. But that just makes it easier to do the calculations, that doesn't mean that way of doing the calculations is any more correct than a horribly chosen reference frame and coordinate system where the answer will ultimately agree (about invariant quantities).

Physics may be simpler for humans to reason about in a certain reference frame which takes advantage of scalar invariant quantities, but physics works identically in all reference frames. That is what physicists are really talking about when they talk about there being no "preferred reference frame".


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