I read, I think, some time ago that the "weight" of photons from the Sun hitting an area the size of a football field at noon on a sunny day would be about the "weight" of a dime?

Would appreciate it someone could flesh that out, verify if correct or false?


3 Answers 3


Photons are massless so their weight is 0. However, photons do have momentum so they can exert force. This force is due to their momentum and would occur even in the absence of gravity, so it is not a weight.

The solar irradiance during peak hours is approximately $1000 \mathrm{ \ W \ m^{-2}}$ and the size of a football field is about $7200 \mathrm{ \ m^2}$ for a total radiant power of $7.2 \mathrm{ \ MW}$. Since $p=E/c$ and $F=\frac{dp}{dt}$ we get that the force from this energy is $(7.2 \mathrm{\ MW})/c = 0.024 \mathrm{\ N}$.

In comparison, a dime has a mass of $2.268 \mathrm{\ g}$ which on the earth turns into a gravitational force, or weight, of $0.022 \mathrm{\ N}$.

So the force of the sunlight on a football field during peak solar hours is close to the weight of a dime.

  • 10
    $\begingroup$ I believe you're assuming 100% absorptance? As the other answer's comment points out, if photons are reflected, you get twice the momentum transfer. $\endgroup$ Commented Jan 11, 2022 at 17:35
  • $\begingroup$ @JonCuster You must also consider the pressure of the outgoing infrared that results from the sunlight heating the field. So, the actual pressure will be intermediate between the perfect absorber and specular reflector cases. $\endgroup$
    – John Doty
    Commented Jan 11, 2022 at 23:02
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    $\begingroup$ I’ve purged a debate about introductory relativity. Please do that somewhere else. $\endgroup$
    – rob
    Commented Jan 12, 2022 at 3:54
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    $\begingroup$ Some additional comments removed. If anyone has concerns about “censorship,” please raise them on Physics Meta, not here. $\endgroup$
    – rob
    Commented Jan 12, 2022 at 14:52
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    $\begingroup$ @samerivertwice your suggested defect is incorrect. For any object $m^2 c^2 = E^2/c^2 - p^2$ which, for a massless object like a photon this reduces to $p=E/c$. Gravity is not relevant to the momentum. A photon has momentum because it is massless and it has energy. The momentum of light is well founded both in theory and in experiment. Here is a lecture that covers both the theory and some experiments: ocw.mit.edu/courses/electrical-engineering-and-computer-science/… $\endgroup$
    – Dale
    Commented Jan 14, 2022 at 13:54

"Weight" can be understood as a type of force - standing on the floor, you impart a force on the floor.

Light can impart force on a surface due to the transfer of momentum involved. In other words, if a photon with momentum $p$ strikes a surface and is reflected in the opposite direction, a total momentum of $2p$ is imparted on the surface. This is called radiation pressure. In Newtonian mechanics, you need mass to have momentum, but in relativistic mechanics, you only need energy to have momentum. And photons certainly carry energy.

So how do we compare the radiation pressure on a football field to the weight of a dime? Consider the units: Pressure is force per area, as measured in newton per square meter, N/m$^2$. You would typically think about weight as measured in kilograms, but that is actually mass. The weight is the force on the object due to gravity, so if the dime has mass $m$ and we have gravitational acceleration $g$, the weight is $F = m g$.

Force is also defined as change in momentum. So if we say $N$ photons with momentum $p$ are being reflected off a football field with area $A$ per time $t$, in total their momentum is changed by $2 N p / t$. The pressure on the football field is $2 N p/(t A)$. If we imagine a dime spread over the football field, the pressure from this dime would be $m g/A$. So, by saying that the weight of a football field of photons is the same as that of a dime, we are saying

$$\frac{2 N p}{t} = m g.$$

As for whether it is true or false, that is a simple question of estimating the parameters in this equation, which I leave as an exercise for the reader.

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    $\begingroup$ +1, but one nit. "Weight" is a force, not a pressure. As you point out, you get a pressure when you distribute that force over the area of a football field. $\endgroup$
    – mmesser314
    Commented Jan 11, 2022 at 14:52
  • $\begingroup$ True, corrected. $\endgroup$ Commented Jan 11, 2022 at 14:54
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    $\begingroup$ Football fields are generally not very reflective, so you should reduce that factor of 2. $\endgroup$
    – PM 2Ring
    Commented Jan 11, 2022 at 15:18
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    $\begingroup$ "Force is also defined as change in momentum" - is this correct? I would say change in momentum is impulse (= force x time) $\endgroup$
    – Martin CR
    Commented Jan 11, 2022 at 15:20
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    $\begingroup$ @PM2Ring The albedo of grass is about .25 so to be even more accurate the factor should be 1.25 rather than 1 $\endgroup$
    – Kevin
    Commented Jan 13, 2022 at 17:43

Adding to Dale's answer, the dimensions of a dime are $17.91$ mm diameter and $1.35$ mm thick. So the volume is $341.1$ mm$^3$.

If you spread that out over a football field with an area of $7200 \cdot 10^6$ mm$^2$, the thickness is $4.72 \cdot 10^{-8}$ mm. Or about $1/2$ an atom thick.

  • $\begingroup$ @JasonGoemaat - Arrgh! Thank you! Fixed! $\endgroup$
    – mmesser314
    Commented Jan 13, 2022 at 1:13
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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Commented Jan 13, 2022 at 2:18
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    $\begingroup$ @BioPhysicist - It actually does, in a little broader sense. He is asking about what weight is equivalent to the force from the Sun on an object. Spreading the dime out over the field makes it clearer what weight to expect for other objects. $\endgroup$
    – mmesser314
    Commented Jan 13, 2022 at 15:47
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    $\begingroup$ It's relevant information, I just don't think it fully answers the question. You even identify that it is more of an addition to an existing answer. $\endgroup$ Commented Jan 13, 2022 at 17:54
  • $\begingroup$ Why don't you write 47.2pm (picometer)? $\endgroup$
    – U. Windl
    Commented Jan 17, 2022 at 8:07

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