Now I am left wondering why does the heat become lost as if travels slightly.

It is *not* lost. It is spread more out.

If you stand so close to the heat source that you are hit by, say 1/10 of it's radiation (1/10 of all photons sent out hit you), then when standing further away you are maybe only hit by 1/100.

The heat radiation sent from the source like the sun spreads out with distance. The same amount of energy every second is spread out at a larger and larger sphere as it travels away from the source.

Surface area of such a sphere is $A=4 \pi r^2$. Doubling your distance to the campfire means:

$$A_2=4 \pi r_2^2=4 \pi (2 r_1)^2=4 *4 \pi r_1^2=4 A_1$$

The area that the radiation is spread over is four times as large for just the double distance.

This of course regards the sun with only space surrounding it. Since the campfire is place on ground, downwards radiation in that case will be smaller.

**Solar constant**

For your interest, the Suns' intensity on our planet Earth is called the *solar constant* (or solar coefficient) $S$. In units $\mathrm{W/m^2}$. This tells us how much radiation that reaches a square meter on *Earth* (or any other planet at the same distance from the Sun) every second. Mercury which is closer will have another solar constant.

To find the Suns' intensity at any distance, you will need to know how much energy is generated per second within the Sun, that is the *power* of the Sun $P$. This energy will be spread out while travelling:

$$P=S*A=4 S \pi r^2$$

where both intensity $S$ and area $A$ are for a *specific sphere* at a *specific distance*. The Suns' own intensity at its surface is then found by insertings the Suns' own radius and isolate S.

Also, our Sun can be viewed as a socalled *blackbody*. That is, it emits radiation very efficiently. The Stefanâ€“Boltzmann law of blackbody radiation then gives us the emissive power from the Suns' surface:

$$P=\sigma T^4 A= \sigma T^4 4 \pi r^2 $$

with $T$ being the surface temperature of the Sun (around $5800^\mathrm{o C}$ if I remember correctly).

**Campfire**

Regarding the campfire as pointed out in the comments, *convection* might be considerable if you are standing very close to the campfire or maybe even reaching over it.

By natural convection, heated air will flow upwards, and this will carry a lot of energy that way. The radiation itself is negligible at that position.

Walking slightly further away might remove the convection effect entirely. This will feel like a huge decrease in heating. Adding a little windy weather, you might not feel any wind while being in "equilibrium" in the convection zone near the campfire. But walking two steps away can have a large cooling effect now that forced convective *cooling* from the wind acts as well.

Each photon will be at rest relative to the other, since they both travel in the same direction at the same speed. A photon at rest has zero frequency, hence zero energy. So, at least to first order, there should be no gravitational interaction between the two photons *within their mutual rest frame*.

I'm not sure what an outside observer would see. It seems to be accepted that gravitational waves interact with each other very very weakly, which suggests that the same should be true of electromagnetic waves.

## Best Answer

Yes the effect is real, potentially at least, but no it's not measurable.

As an aside, the redshift of light by its gravitational interaction with (homogeneous and isotropic) dust is exactly what the FLRW metric predicts, but this clearly isn't what the question means.

The gravitational field of a beam of light is calculated in the paper On the Gravitational Field Produced by Light by Tolman, Ehrenfest and Podolsky (the link is a 1.5MB PDF). A test particle feels a force normal to the light ray (there is no force parallel to the light ray), so some of the light's energy would be transferred to the test particle. In principle, when a light ray passes a dust particle we'd get something like this:

Some of the light's energy would be transferred to the particle and it would be red shifted as a result. The change is too tiny ever to be measurable, but in principle it would be there.

However this isn't likely to happen outside of a physics lab. For starlight shining through a dust cloud the interaction would actually look like:

because the light is symmetric aound the dust particle the particle feels no net force and is not accelerated. Therefore no energy is transferred to it and the light is not red shifted. As a result there is no cosmological red shift due gravitational interactions with dust.