Relativistic transformation of the wave packet length

Let us suppose we have an excited atom at rest. It has a certain mean lifetime $\tau_0$. If we wait sufficiently long time $t>>\tau_0$, we will find a deactivated atom and a (spherical) electromagnetic wave function of photon with about $\tau_0\cdot c$ long layer with non zero probability to find a photon within. Something like a fast expanding probability "ring" with a $\tau_0\cdot c$ width of the ring.

Now, let us consider this system in a moving reference frame. It seems to me that this width $\tau_0\cdot c$ is relativistic invariant: it is a difference between two "fronts" of electromagnetic wave rather than a length of a material body subjected to the Lorentz contraction. Is it correct? In other words, whether this picture relativistic invariant?

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What is the difference between an ""...electromagnetic wave function of photon..."" and an electromagnetic wave? –  Georg Nov 16 '11 at 17:36
@Georg: An electromagnetic wave implies a flux of photons distributed according to the wave packet. So in a fog atmosphere you see the whole wave packet propagating, whereas one photon can only be found at one place within the wave packet. –  Vladimir Kalitvianski Nov 16 '11 at 17:44
An electromagnetic wave implies an electromagnetic wave. Either use photons to describe the light, or do it as electromagnetic wave. This mix is superfluous. In general, as long as propagation is concerned, one should stay with the wave description, it is much better (simpler). –  Georg Nov 16 '11 at 20:51
@Georg: One can obtain an EMW just with placing N excited atoms, where N >> 1. –  Vladimir Kalitvianski Nov 16 '11 at 22:03

No, the probabilistic density isn't Lorentz-invariant. First of all, the probabilistic density isn't a Lorentz scalar: it is a time component of a 4-vector whose spatial components encode the probability current.

Second, it is obviously not true that the ring will be equally thick in all directions. While it's true that the photons are moving by the speed of light in any inertial system, they're moving from a different center because the atom itself is moving (from the moving frame's viewpoint).

So the ring from a moving reference frame's viewpoint will look like one of the black or purple squeezed annuli on the picture below:

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Yes, I reasoned in the following way: the atom is moving at $v=c-\epsilon$, and it takes longer time to radiate a photon: $\Delta t = \tau_0 /\sqrt{1-(c-\epsilon)^2 /c^2}\approx \tau_0\cdot\sqrt{\frac{c}{2\epsilon}}$. This time dilation is not, however, sufficient to compensate the closeness of $v$ to $c$. As a result, the wave packet length is shorter in front of the atom: $L=\epsilon\Delta t < c\tau_0$. –  Vladimir Kalitvianski Nov 16 '11 at 18:32
It is funny, but $L$ is shorter than the "Lorentz contracted $l$ ", if $l$ were applicable here: $L=l/2$ . –  Vladimir Kalitvianski Nov 17 '11 at 13:47