In SR we've learned that the time dilation for an observer moving clock w.r.t one fixed in a frame at rest is

$$\tau = \gamma \tau_0 = \frac{\tau_0}{\left(1-v^2/c^2\right)^{1/2}}$$

ref: "Special Relativity - A.P. French" and many others

In this case being gamma > 1, it implies delta t < delta tau

No moving to GR, the basic starting expression for calculating the elapsed coordinate time from proper time for a observer clock located in a mass gravitational field and moving with velocity v w.r.t a frame at rest in the body mass center is (approximating square root at first order for v << c)

$$\Delta t = \int_A^B \left(1+\frac 1 {c^2} U + \frac{1}{2c^2} v^2\right)d\tau$$

ref: "Relativistic time transfer - ITU-R TF.2118-0" and many other

To note that all terms in the integral are positive, also excluding the presence of gravity (U=0), meaning that it would always result delta t > delta tau

This is an opposite result w.r.t. SR expression!

Can anyone clarify this (apparent) contradiction? Thanks in advance.

  • $\begingroup$ From the definition of relativistic time dilation effect it would be expected that an observer clock (proper time) would a time measurement slowed down with respect to a fixed clock (coordinate time), due the fact the it is moving with velocity v and it is subject to a gravitational field, both supposed to produce slow down effect with respect to a stationary clock not subject to gravity (coordinate time clock). The SR expression consequence would be then expected one.... but what about GR based expression then? $\endgroup$
    – Gianni
    Sep 5, 2020 at 8:52
  • $\begingroup$ Please use MathJax for equations instead of scanned images in the future. Doing so makes the question more accessible. I'll fix this question so you can see how to use it. $\endgroup$ Sep 5, 2020 at 10:43
  • $\begingroup$ Thanks, I'll try to do it next time. $\endgroup$
    – Gianni
    Sep 6, 2020 at 11:38

2 Answers 2


Can anyone clarify this (apparent) contradiction?

It's just different nomenclature. There is no contradiction.

French's equation 4-5 is $$\tau = \gamma \tau_0 = \frac {\tau_0} {\left(1-v^2/c^2\right)^{1/2}}$$ Note that French's equation 4-5 uses two taus, $\tau$ and $\tau_0$, to represent the time difference between two events as measured by two different observers. The latter ($\tau_0$) represent the time difference as measured by an observer at rest with respect to the two events. The former ($\tau$) represents the time difference as measured by an observer moving with respect to the stationary observer.

French's $\tau$ is coordinate time ($\Delta t$ in more modern nomenclature) while his $\tau_0$ is proper time ($\Delta \tau$ in more modern nomenclature). A more modern way to write French's equation 4-5 is thus

$$\Delta t = \gamma \Delta \tau = \frac {\Delta \tau} {\left(1-v^2/c^2\right)^{1/2}}$$

With this modernized rewrite it is obvious that there is no contradiction.

  • $\begingroup$ thanks for the clarification. I got the point of the terminology, and with this my issue is addressed from a mathematical point of view. Still not fully clear to me looking at the explanation French give to the tau_0 and tau elapsed times, because they are very close to reversed logic (which has triggered also my doubt). $\endgroup$
    – Gianni
    Sep 6, 2020 at 9:18
  • $\begingroup$ That's particularly relevant for Earth time dilation examples, where the tau_0 time difference should be applied to a "real" clock proper time measure, by definition moving w.r.t. Earth center, in spite from French definition tau_0 should be taken by a stationary observer(!). And if we apply full relativity considering the clock stationary, then we should consider the Earth center moving w.r.t to it, and this is never applied. It looks reference frames are not used consistently there. Maybe I'm still missing some bits and pieces. $\endgroup$
    – Gianni
    Sep 6, 2020 at 9:25

Using bionomial approximation, French's equation equals: $$\tau = \gamma \tau_0 = \frac {\tau_0} {\left(1-v^2/c^2\right)^{1/2}}\approx{\left(1+\frac{v^2}{2c^2}\right)\tau_0}$$

The second equation introduced by you, substituting $U=0$, also implies: $$\Delta t\approx{\left(1+\frac{v^2}{2c^2}\right)\Delta\tau}$$

There is no contradiction, I think!


... the elapsed coordinate time from proper time for an observer clock located in a mass gravitational field and moving with velocity $v$ w.r.t a frame at rest in the body mass center ...

However, remember that an observer located at the center of a planet measures that the clock located on the surface of the planet runs faster. I am doubtful that the boldfaced sentence may not indicate the Schwarzschild observer but rather the observer at the center of the planet. This can justify the difference between the two equations if you think $Δτ$ does not match $τ_0$.

  • $\begingroup$ Thanks for your feedback. My doubt was not in the formula per-se but in the French's explanation that seems to say that τ0 is the coordinate time (i.e. Δt in the second expression)and τ is proper time (Δτ in the second expression). That implies the first expression to be the opposite of the second one. But above explanation "recommend" to consider them the same, by matching Δτ with τ0 and Δt with τ (not intuitive). $\endgroup$
    – Gianni
    Sep 7, 2020 at 20:17
  • $\begingroup$ @Gianni My pleasure. I think you had better consider them the same, unless you make sure that in no way $\Delta\tau$ matches $\tau_0$. $\endgroup$ Sep 7, 2020 at 20:41
  • $\begingroup$ Sure I will, until someone reinvents relativity! But I still struggle to find a clear rationale for it. $\endgroup$
    – Gianni
    Sep 9, 2020 at 7:04

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