Effect of gravitational time dilation on chemical reaction kinetics Suppose we have two identical barrels filled with water in which a specific amount of a chemical is dissolved. We add an enzyme (that slowly degrades that chemical with a constant rate (i.e. first order)) at the exact same time to the two barrels. Now, one of the two barrels is transported away from earth (i.e. away from massive objects).

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*As soon as half of all molecules of the chemical in this barrel are degraded, a photon is sent to earth. When the photon is received on earth, how much later/earlier is that (in earth time) after half of the molecules in barrel on earth have been degraded.


*As soon as half of all molecules of the chemical in this barrel are degraded, the barrel is transported back to earth. Once back on earth, is the percetage of degraded chemicals in both barrels the same?
 A: According to gravitational time dilation the passage of proper time is faster with increasing gravitational potential. So not only clocks will tick faster, all processes will happen faster including chemical reactions.

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*The speed of light has to be taken into account. The photon has been emitted before "half of the molecules in barrel on earth have been degraded".


*This question can not be answered because because it depends on details like the gravitational potential at the upper barrel and its transportation back to earth.
A: There is a general reasoning that takes into account your chemical reaction and any other time dependent process. The reason for this is that no matter the reaction, it is always going to take the same amount of time in its own proper time. Gravitational effects would be negligible for the reaction itself, as it is when one considers atoms and atomic scaled objects.
First notice that if one were to construct your setup even for inertial observers in Minkowski spacetime at rest with respect to each other and separated by a distance $L$ with a reaction that takes a time $T$, the setup would have to be changed a bit so the computation can be done. For example, one could assume there are two synchronized clocks in each of the reference frames that would perform the reaction and that they start the reaction at $t=0$. Then, the time so that the light ray emitted at $T/2$ from wither source to reach the other would be simply $T/2+L$. We would see a delay for the communication, which would be associated to the time light takes to travel between them.
In general spacetimes we would have to keep track of two things: The time the event takes to occur and the time the light rays take to travel between the observers.
To solve the problem assume the proper time so that the reaction takes place is $T$. If one observer is at the radial coordinate $r_1$ and the other one at $r_2$ and both are static in Schwarzschild spacetime, then the coordinate time  $t$ that will represent the times the reaction takes to occur will be given by their proper times,
$$T_1=\frac{T}{f(r_1)},$$
$$T_2=\frac{T}{f(r_2)},$$
where
$$f(r)=1-\frac{2M}{r}.$$
To solve the problem I will assume the reactions start at the same coordinate time, which could be achieved by sending a light ray to both from a point in between (synchronizing their clocks).
Now we need to compute the proper time it takes for one of the observers to receive a light ray emitted by the other. The coordinate time parametrization of the light ray can be obtained via
$$0 = ds^2 = - f(r) dt^2 + \frac{dr^2}{f(r)}\Rightarrow \frac{dt}{dr} = \frac{1}{f(r)} \Rightarrow \Delta t = r_2 - r_1 + 2M log\left(\frac{r_2-2M}{r_1 - 2M}\right).$$
Thus, the total $coordinate$ time it takes for a light ray originating from $r_1$ to reach $r_2$ is given above. In terms of coordinate time, we then have the total time to reach $r_2$ given by $\Delta t + T_1/2 $. Subtracting half the time it takes for the reaction at $2$ to take place we obtain the final answer in coordinate time:
$$\Delta t + \frac{T_1}{2}-\frac{T_2}{2} = r_2 - r_1 + 2M log\left(\frac{r_2-2M}{r_1 - 2M} \right)+ \frac{T}{2}\left(\frac{1}{1-\frac{2M}{r_1}} - \frac{1}{1-\frac{2M}{r_2}}\right).$$
To obtain the time in the reference frame of $2$, we simply multiply the result above by the redshift factor $$\frac{1}{\sqrt{1-\frac{2M}{r_2}}}$$.
I really enjoyed doing this computation :)
