# Is a volumetric rate frame-invariant in general relativity?

Imagine that I have a radioactive material with a long half life. The atoms in this material decay at a certain rate $R$. The rate is the decay constant times the number density $R = \lambda N$. It has dimensionality:

$$\left( \frac{ \text{decays} }{m^3 s} \right)$$

Imagine that the material is on board a spaceship traveling at some significant fraction of the speed of light. Length is contracted and time is dilated.

$$\Delta t' = \Delta t \gamma = \frac{\Delta t}{\sqrt{1-v^2/c^2}}$$

$$L'=\frac{L}{\gamma}=L\sqrt{1-v^{2}/c^{2}}$$

The volumetric decay rate according to the lab reference frame is found by correcting for both the increased density (due to length contraction) and the decreased decay constant (due to time dilation).

$$N' = L' A = \frac{N}{\gamma}$$

$$R' = \frac{N'}{N} \frac{\Delta t'}{\Delta t} R = R$$

It's the same volumetric decay rate! Amusingly, the $Q$ value of the decay would be greater, but that's aside the point.

Question:

What if the material was put in a large gravity well? If you use the coordinates from outside the gravity well, would you obtain this same result?

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## 1 Answer

I don't know whether it applies to all physically possible metrics, but the volumetric decay rate you define does stay constant in a Schwarzschild metric. Well, it does if the box is small compared to the curvature i.e. the time dilation etc is constant thoughout the box. I would need to think more about what happens if the box is very large.

Anyhow the Scharzschild metric is:

$$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2$$

The time dilation is easy, as we see time moving more slowly for the box by a factor of $(1 - 2M/r)^{1/2}$. I had to think a bit about length contraction, but I think this is a sensible way to define it:

The Schwarzschild radial co-ordinate $r$ is defined as the radius of a circle with circumference $2\pi r$. So we can take a shell with circumference $2\pi r$ and another with circumference $2\pi (r + dr)$ and that defines our ruler of length $dr$. But the observer standing alongside the box would measure a different radial distance between the shells. Specifically they would measure the distance to be $dr/(1 - 2M/r)^{1/2}$. This distance is bigger than the observer at infinity measures, and therefore this means the shell observer's ruler is shorter than ours by a factor of $(1 - 2M/r)^{1/2}$. This factor is exactly the same as the time dilation factor, which means the time dilation and length contraction balance out, and the volumetric decay rate stays the same.

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Actually, a previous answer of yours inspired this question.  I thought the equation $d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$ would be used to answer, not for a specific geometry, but possibly completely generally.  My thinking was that the universality of distance between two spacetime points would lead to agreement on a "4-D volume", which led me to volumetric decay rates.  One last point: it shouldn't matter the size or shape of the box, no?  Everything here could be a differential volume.  Anyway, your answer is very clear and easy to understand. – AlanSE Oct 4 '12 at 13:20
I'm fairly certain you could make up a metric where the time dilation and length contraction didn't balance, then work backwards to find the match stress-energy tensor. What I'm not sure about is whether that stress-energy tensor would be physical i.e. not require exotic matter. In any case, questions like yours need careful thought. In SR we can always choose frames with the origins at the point of interest (the box) so the box is at (0, 0) in both frames and can easily be compared. In GR this is not the case and you need to think carefully how you do the comparison. – John Rennie Oct 4 '12 at 14:58
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