Why does heat lose its energy dramatically as I move back?

Say I have a fire around 0.5 meters in front of me, I can clearly feel the heat, however, as I move even very slightly back, say 1 meter back, I will notice a dramatic fall in the heat I measure from the fire.

Similarly, the Sun radiates very hot radiation that finally reaches us and heats our planet for living things to exist. Now I wonder why heat is lost as it travels slightly.

My first hypothesis is that as light/heat (radiation) travels, the light loses its energy through interactions with particles and air molecules, gives them energy and therefore loses its energy. This may be a valid (or credible) hypothesis locally (on Earth), however, in space, this loses its credibility because there are clearly no masses of particles that normally reduce the energy of the sun's rays.

What actually causes this dramatic fall of heat as I move away, even slightly?

If dispersion happens, what about this heat? Next, if radiation or this heat is carried in photons and photons are particles and waves, how can these waves or particles disperse heat as if photons act as a fluid in large numbers?

  • $\begingroup$ You might want to think about Diffusion perhaps. $\endgroup$
    – mick
    Dec 21, 2014 at 22:53
  • $\begingroup$ You might want to think about your question, the answers bellow are correct $\endgroup$
    – Daz Hawley
    Dec 21, 2014 at 22:56
  • 2
    $\begingroup$ @DazHawley Yes, I have not down-voted any of them, but I don't think I get a full-description about how this "dispersion" happens. $\endgroup$
    – user43495
    Dec 21, 2014 at 22:57
  • $\begingroup$ well the further away you move from the heat source the more thinlly spread it becomes because the area it has to fill is larger $\endgroup$
    – Daz Hawley
    Dec 21, 2014 at 23:01
  • $\begingroup$ Light (and heat) decrease in intensity with the square of the distance. $\endgroup$
    – CoilKid
    Dec 21, 2014 at 23:13

5 Answers 5


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 Stefan–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).


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.

  • $\begingroup$ Part of what I said imho $\endgroup$
    – mick
    Dec 21, 2014 at 22:51
  • $\begingroup$ @mick I don't believe momentum and density of moving particles is the best answer to this question. The geomitry you mention is correct but it is not made very clear how. (Also note, I didn't see your answer before posting... No hard feelings, bro) $\endgroup$
    – Steeven
    Dec 21, 2014 at 23:03
  • 2
    $\begingroup$ This is mostly correct, but I think that it would be more complete if you also addressed the convection currents around the campfire since the hot gases and not the radiation is what does most of the heating. As any Boy Scout knows, it's way hotter 2 feet above the fire than it is 2 feet to the side. $\endgroup$
    – Geoffrey
    Dec 22, 2014 at 1:52
  • 1
    $\begingroup$ @Geoffrey the OP's example says that the fire is in front of him, so I think radiation will be the dominant factor. But it is insightful knowing where other factors come into play, for example when heating marshmallows. PS: please keep it metric :P $\endgroup$
    – fibonatic
    Dec 22, 2014 at 13:50

The heat radiates away from the source equally in every direction (until it hits something). Imagine the sun. It is roughly spherical and radiates energy (some as heat) in all directions. The energy is radiated in a sphere. As the energy travels away from the source (the sun) the sphere expands but the amount of energy remains the same. When the sphere is twice as large the surface area is four times as large. This means that only one quarter of the energy is felt for the same area. This is called the inverse square law because for every $n$ increase in size the amount of energy in the same area is $1\over n^2$.

Exactly the same rule applies to your camp-fire. So when you're a metre away the fire feels 1/4 as hot as when you're half a metre away.


The simple answer is the inverse square law which states that the intensity (power per unit area) of a heat source drops off with $1/r^2$ where $r$ is the distance (assuming a point source).

Looking at your camp fire: imagine people standing side by side around the fire. Each gets a share of the heat. When they make the circle wider, more people can get some of the heat - but since no additional heat is generated each of them will get a bit less. In a 2D situation (no heat traveling up or down, only sideways) this would mean heat drops off with the inverse of the distance (twice as far = twice as many people in the circle). However for the sun etc the heat can go up, down and sideways - so if you imagine a "three dimensional camp fire" then when you increase the radius by 2x you can get 4x more people (area is sphere scales with radius squared).


The heat simply disperses and the further away from the heat source you get the more space the heat has to disperse into

  • $\begingroup$ If radiation is carried through photons and photons are particles then how can this energy be spread out more, surely this photon must interact with something or am I wrong in thinking this if so can you please expand about it in more detail? $\endgroup$
    – user43495
    Dec 21, 2014 at 22:58
  • $\begingroup$ imagine a football team coming from the changing room in the tunnel near the source they ar close together, intense, and as they run out on to the pitch they spread out, so if you was stood in the tunnel you would definately bump into one of the players but on the pich you are less likely too $\endgroup$
    – Daz Hawley
    Dec 21, 2014 at 23:06

Informally speaking heat is determined by the (average) momentum and density of moving particles.

So if particles go slower, you get less heat. Also if there are less particles in a fixed volume its less hot.

Now from simple geometry you can easily show that going away from the heat source means cooling down.

Hope that helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.