# Must all types of clocks agree on the time that has passed since the spaceship started accelerating?

The above would be true if the spaceship were not accelerating, because SR says that the physical laws are all same in inertial frames of references. If there were a gap in the time of the 2 clocks, a person on the spaceship could measure that gap and calculate his velocity, hence contradicting SR. So, the time on all clocks must be same, irrespective of mechanism, in an inertial frame. But does the same hold true for non-inertial frames?

Edit : The clocks are at the same height.

• Could you please expend a little bit on what you mean by an atomic clock, vs a light clock? It is not immediately clear what you mean. – Mikael Fremling Oct 8 '16 at 11:28
• @MikaelFremling my focus isn't on the type of clock, but on the reason why all clocks must be affected the same way. Apologies for the misunderstanding – ryannmahajan Oct 8 '16 at 16:04

This depends on exactly what you are asking. Clocks at different heights in an accelerating spaceship run at different rates. This is discussed in the question Which clock is the fastest inside an accelerating body?.

All clocks are affected by the time difference in the same way, so it doesn't matter whether the clock is atomic, a light clock, a mechanical clock or whatever, all clocks at the same height in the accelerating spaceship will run at the same rate. However a light clock is traditionally quite large i.e. the light travels some large distance then reflects back, and we measure the time by the duration of the light beam's round trip. If the light beam travels up the accelerating spaceship then it will pass through regions of measurably different time dilation and the time it measures will differ from the time measured by a more compact clock.

So if the light clock is small, or if it's arranged so the light beams stay at the same height as they travel, then the light clock will measure the same time as all the other clocks. If the light clock is large and the light beams travel vertically then there will be a difference.

• Why must all clocks be affected in the same way? – ryannmahajan Oct 8 '16 at 16:00
• @ryannmahajan: time dilation is a geometrical property of spacetime and not dependent on any details of how individual clocks work. For more on this see What is time dilation really?. – John Rennie Oct 8 '16 at 16:27
• You're absolutely right. Look at this way : We generally prove gravitational time dilation by showing that a clock is faster at a height or that photons lose energy and their frequency is lowered and so on. The point is, we can't possibly show it for all the mechanisms (can't show that thought processes and cancer development will be faster). We can, by argument, prove it when acceleration is not involved, as I have in the question's description. But is there such an argument for when acceleration is involved? – ryannmahajan Oct 10 '16 at 10:40
• @ryannmahajan: time dilation is not due to photons losing energy. It arises from the geometrical properties of spacetime. This is what I discuss in the article I linked in my comment above. This approach treats both accelerated and non-accelerated motion. – John Rennie Oct 10 '16 at 15:42
• Off topic. Just like to say thanks for the help. I'm unsubbing from physics. It's a waste of time for experimentalists and the moderation is starting to annoy me. More in meta, but it really doesn't matter. regards Dirk – user56903 Oct 12 '16 at 7:44

The answer depends on the geodesic of a photon in the accelerated frame. The accelerated frame is the Rindler wedge which is illustrated below. The metric in that frame is $$ds^2~=~dT^2~-~dX^2~-~dy^2~+~dz^2$$ with a transformation to the $t,~x$ coordinates of the flat Minkowski spacetime as $$T~=~x~sinh(at),~X~=~x~cosh(at),$$ for $a$ the acceleration. The line element for the path of an accelerated observer in Minkowski spacetime is then $$ds^2~=~a^2x^2dt^2~-~dx^2~-~dy^2~-~dz^2.$$ Let us look at geodesics in the Rindler wedge, which we can find by a variation on the above metric. This leads to $$\frac{d^2t}{ds^2}~+~2x^{-1}\frac{dx}{ds}\frac{dt}{ds}~=~0$$ $$\frac{d^2x}{ds^2}~+~x\left(\frac{dt}{ds}\right)^2~=~0,$$ and we have $d^2y/ds^2~=~d^2z/ds^2~=~0$ From these last two we have $y~=~y_0$ $z~=~z_0$ as constants. The first geodesic equation gives $dt/ds~=~kx^{-2}$, for $k$ a constant. The differential equation for the $x$ coordinate is then $$\frac{dx}{ds}~=~\sqrt{s~+~kx^2~-~y_0^2~-~z_0^2}$$ A solution for the $x$ coordinate position is $$x~=~\sqrt{s^2~+~2s\sqrt{k^2~-~x_0^2(y_0^2~+~z_0^2)}~-~s^2(y_0^2~+~z_0^2)},$$ which is a semicicular arc. This shows that geodesics and null geodesics are semicicular arcs that emerge and enter perpendicular to the Rindler horizon.

With some further geometry one can look at this arc and find its radius. If one is on an accelerated frame a photon clock that measures time should have the photon bounce between mirrors spaced a distance much smaller than the radius of this arc. In that way one can satisfy the criterion for a proper local frame. As the distance between the mirrors increases and approaches scale of these perpendicular arc the photon clock time deviates from the proper time. We may think of these semi-circular arcs as virtual photon loops, where the accelerated observer witnesses the vacuum as the production and destruction of photons across the Rindler horizon. The radius of these loops will be on the order to the distance $d~=~c^2/a$ the horizon is from the observer. For the mirrors in the photon clock a distance $d_{cl}~<<~d$ this means the observer is measuring time in intervals at the UV cutoff in photons produced in Unruh radiation.