Is all the trajectory of a photon, since its emission, always accompanied by a gravitational wave? Whenever a photon is emitted, the atom/electron that emits it loses the mass equivalent to the energy $E$ of the photon.
This loss of mass automatically determines a corresponding loss of gravity or intensity of the spherical gravitational field distributed around the atom/electron.
Therefore, when the photon moves away from the electron that emitted it, it is always accompanied by a wave, or by a gravitational step, which means the correction of the intensity of the previous field by a new field now less intense, as the energy $E$ has lost corresponding to the photon that goes away.
Is all photon in its trajectory always accompanied by this hypothetical spherical correction of the gravitational  field intensity (as if it were a shadow) coming from the atom / electron responsible for the emission?
I'm assuming that the gravitational wave and the photon develop the same velocity c.
Does light only go where gravity also reaches? It's just a guess, no need to answer.
 A: One way to phrase your question is asking if the rate for the transition $A^*\rightarrow A+\gamma$ (where $A^*$ is some excited state of the atom $A$, taken to be neutral for simplicity to avoid similar issues in QED) should actually be zero since, actually, there always should be gravitons around, corresponding instead to the following processes $A^*\rightarrow A+\gamma+g$, $A^*\rightarrow A+\gamma+2g,\ldots$.
It turns out that the answer to this is affirmative,  because of a well-known property of gravity associated with the graviton being massless and hence producing infrared (IR) singularities. They require the inclusion of an IR regulator $\lambda$ (to be chosen smaller than the smallest physical mass scale in the problem) an that the rate for any exclusive process such as the one of your question $A^*\rightarrow A+\gamma$ actually vanishes when the IR regulator is removed, $\lambda\rightarrow 0$. Moreover, the rate that is  instead finite is the same one you wanted  but with gravitational bremsstrahlung (that is emission of infinitely many soft gravitons of total finite energy) added in the initial and final state. The classical limit of such a process corresponds to the emission of classical gravitational bremsstrahlung.
More explicitly, S. Weinberg (see e.g. vol.I chapter 13.2 and original references therein), generalizing earlier work for a similar phenomenon with photon-bremsstrahlung in QED by Yiennie, Fratsci and Suura, has shown that the rate $\Gamma^\lambda_{\alpha\rightarrow \beta}$ for any process $\alpha\rightarrow \beta$ (with or without gravitons) involving at most gravitons (real or virtual) with energy $E$ above the IR regulator , $\lambda<E$, vanishes as
$$
\Gamma^{\lambda}_{\alpha\rightarrow\beta}=\left(\frac{\lambda}{\Lambda}\right)^A\Gamma^{\Lambda}_{\alpha\rightarrow\beta}\rightarrow 0\,,\qquad A=\frac{G}{2\pi}\sum_{n\,, m}\eta_n \eta_{m} m_n m_m\frac{1+\beta_{nm}^2}{\beta_{nm}\sqrt{1-\beta_{nm}^2}}\log\frac{1+\beta_{nm}}{1-\beta_{nm}}>0
$$
for $\lambda\rightarrow 0$, where $m_n$ is the mass of the $n$-th particle line, and $\eta_{n}=\pm 1$ depending whether such a particle is ingoing or outgoing, and $\beta_{nm}=\sqrt{1-m_n^2 m_m^2/(p_n p_m)^2}$ is the relative velocity between the pairs $n$ and $m$.
What is instead finite when the IR cutoff is removed is the rate $\Gamma^{\lambda;\, E}_{\alpha\rightarrow\beta}$ for the same process but with arbitrarily many soft gravitons emitted with energy larger than $\lambda$ and below some $E$,  such that
$$
\Gamma^{\lambda;\, E}_{\alpha\rightarrow\beta} \rightarrow \left(\frac{E}{\lambda}\right)^{A}\Gamma^{\lambda}_{\alpha\rightarrow\beta}=\left(\frac{E}{\Lambda}\right)^A\Gamma^{\Lambda}_{\alpha\rightarrow\beta}
$$
up to a known overall o(1)-function that I am omitting for simplicity.
A: What you say is a correct first intuition, even though there are a few modifications. The first thing to consider is that the total four-momentum of the system is conserved in this process. That means that the atom/electron recoils with an equivalent spatial momentum in the opposite direction. As such, its gravitational field is also traveling at a non-zero speed in the center-of-mass of the system. (I have discussed this also in this answer.) This has to be taken into account to avoid argumentation traps with momentum/energy considerations. The photon, on the other hand, carries a gravitational field traveling at the speed of light with it. This is known as an impulsive gravitational wave and it can be modeled as the Aichelburg-Sexl ultraboosted particle. See the book of Griffiths & Podolský for a nice discussion.
This being said, it is true that when we consider the mass-energy $M$ of the emitting atom to be much higher than the mass-energy $\delta M$ of the photon, we can assume that by passing through a distance $r$ from the atom, the monopole moment is approximately changed as $M/r \to (M - \delta M)/r$. However, the point is that if you only consider this effect, you are vastly under-representing what is actually happening. The photon has its own fully anisotropic field with many strong multipoles. In particular, the dipole will typically be as strong as the monopole in an emission process such as this. As a result, the gravitational field of the photon can be considered to be approximately vanishing on "the other side" of the sphere $r=const.$ after a very short time after the emission. So the picture of a traveling spherical wave is not quite accurate. In fact, the wave will likely be best represented as a near-planar pulse.
(Caveat: all of the above is a classical discussion with the photon approximated as a point-like particle. At finite wave-lengths and considering quantum effects things will be "smoother" and "less sharp". However, most of the conclusions will still hold.)
