Solar radiation pressure: Visual walkthrough — Physics deep dive

Let me name the characters before the play begins.

Step 1 — A photon is a tiny bullet that carries momentum

WHAT. We start with the smallest unit of light: a single photon. It has energy $E$ but zero mass. We claim it still carries momentum $p = E / c$ .

WHY this, and why now. Force is "momentum delivered per second." So before we can talk about a push from light, we must first prove light has momentum to deliver. Why the formula $p = E / c$ and not the school formula $p = m v$ ? Because $p = m v$ is the non-relativistic rule for objects with mass. A photon has $m = 0$ , which would give $p = 0$ — clearly wrong, since light does push. The correct rule comes from the relativity energy relation.

PICTURE. Look at the energy–momentum triangle below. The full relativistic law is $E^{2} = (p c)^{2} + (m c^{2})^{2}$ — a right triangle with legs $p c$ (momentum side) and $m c^{2}$ (mass side) and hypotenuse $E$ .

$E$ is the hypotenuse — total energy.
$m c^{2}$ is the "rest mass" leg. For a photon this leg shrinks to zero.
$p c$ is the momentum leg. When the mass leg vanishes, the triangle collapses to a straight line and $E = p c$ , i.e. $p = E / c$ .

See Photon momentum and relativity for why the triangle is the truth and $p = m v$ is only the low-speed corner.

Step 2 — The Sun sprays power over an expanding sphere

WHAT. The Sun pumps out a fixed total power $L_{⊙}$ . At distance $r$ , that power is smeared over the surface of a sphere of radius $r$ , whose area is $4 π r^{2}$ . The power per square metre is the intensity $S$ .

WHY. We can't push a spacecraft with the whole Sun's output — only the sliver that reaches its location matters. So we need "power per area at distance $r$ ." Why divide by $4 π r^{2}$ ? Because energy is conserved: the same total power crosses every sphere centred on the Sun. A bigger sphere has more area, so the same power is thinner. This is the Inverse-square law.

PICTURE. Two nested spheres. The same number of arrows (power) pierce both, but the outer sphere's arrows are spread out — fewer per tile.

Step 3 — Turn energy-per-second into momentum-per-second

WHAT. Each square metre catches $S$ joules of light every second. Step 1 says each joule carries $1/ c$ of momentum. So momentum landing per area per second is $S / c$ .

WHY. "Momentum per area per second" is force per area, which is exactly pressure. This is the conversion that changes an energy statement into a push statement. We use the $p = E / c$ tool from Step 1 precisely because it is the bridge from joules to kilogram-metres-per-second.

PICTURE. A one-second slab of sunlight hitting a 1 m² tile. Count the joules in the slab, multiply each by $1/ c$ , and you get the momentum stamped onto the tile that second.

This is the pressure on a surface that swallows every photon. Next we fix what happens if it bounces them.

Step 4 — Absorb vs reflect: why a mirror is pushed twice

WHAT. A black surface absorbs a photon and keeps its momentum $p$ — change of $p$ . A mirror reflects it: the photon arrives with $+ p$ and leaves with $- p$ , so the surface's momentum change is $p - (- p) = 2 p$ . A mirror feels double.

WHY. Newton's third law bookkeeping: whatever momentum the photon loses, the surface gains. Absorbing = photon loses $p$ (from $p$ to $0$ ). Reflecting = photon loses $2 p$ (from $+ p$ to $- p$ ). We track the change, because force is change of momentum per second.

PICTURE. Two lanes. Top lane: a ball splats and sticks (absorb) — one arrow's worth of kick. Bottom lane: a ball bounces back (reflect) — two arrows' worth.

This is why Solar sails are built as mirrors: they extract the full factor of two.

Step 5 — Tilt the plate: the $cos θ$ for projected area

WHAT. If the plate faces the Sun square-on, it catches sunlight over its full area $A$ . Tilt it by angle $θ$ (measured from the surface normal — the arrow sticking straight out of the plate) and it catches light only over its shadow, the projected area $A cos θ$ .

WHY. The sunbeam is a fixed river of photons. A tilted plate presents a narrower silhouette to that river, so fewer photons hit it. Why $cos θ$ specifically? Because the projected width of a tilted length onto the beam is the length times the cosine of the tilt — the same geometry as a shadow shrinking as you rotate a card toward edge-on.

PICTURE. A plate rotating in a parallel beam. At $θ = 0$ its shadow is full width $A$ ; as $θ$ grows the shadow narrows to $A cos θ$ ; at $θ = 90°$ the plate is edge-on and catches nothing.

Step 6 — Degenerate cases: check the edges

WHAT & WHY. A formula you trust must survive its extremes. Let us pin each corner to a picture so the reader never meets an untested case.

$θ = 0$ (face-on): $cos θ = 1$ , full push. The plate is a wall to the beam.
$θ = 90°$ (edge-on): $cos θ = 0$ , zero force. The plate is a knife-edge; the beam slides past.
$η = 0$ (jet black): factor $1$ , minimum push, all photons absorbed.
$η = 1$ (perfect mirror): factor $2$ , maximum push, every photon bounced.
$r \to \infty$ (far away): $S \propto 1/ r^{2} \to 0$ , force fades but never flips sign — SRP always pushes away from the Sun.
$A / m$ tiny (dense probe): acceleration $a = F / m$ is negligible; SRP hardly matters. $A / m$ large (sail, balloon): SRP dominates. Contrast with Atmospheric drag, which opposes motion instead of pointing radially outward.

The one-picture summary

Everything above, in a single chain: photon momentum → sphere spreading → divide by $c$ → reflectivity dial → tilt cosine → acceleration.

Recall Feynman retelling — the whole walkthrough in plain words

Sunlight is a downpour of tiny energy packets called photons. Even though a photon weighs nothing, relativity's little right-triangle rule says it still carries momentum $p = E / c$ — so it can nudge things (Step 1). The Sun blasts out a fixed amount of power, but by the time it reaches you, that power is smeared thin over a giant sphere, and the sphere grows with the square of the distance, so the light gets weaker as $1/ r^{2}$ (Step 2). Count the joules hitting one square metre each second and multiply each by its little $1/ c$ of momentum — that momentum-per-second-per-area is a pressure, $P = S / c$ (Step 3). If the surface is black it just catches the kick; if it's a mirror the photon bounces back and gives twice the kick, which is the $(1 + η)$ dial (Step 4). Tilt the surface and it presents a smaller shadow to the beam, catching less light — that's the $cos θ$ (Step 5). Check the corners: edge-on gives zero, mirror gives double, far away fades to nothing but always pushes outward (Step 6). Divide the force by the spacecraft's mass and you get the acceleration that actually bends orbits — and the number that decides whether SRP matters is how much area you carry per kilogram, $A / m$ .

Quick self-check

Where does

cos θ

come from in the simple model?

the projected (shadow) area

A cos θ

that the tilted plate shows to the beam.

Why does a mirror feel double the push of a black surface?

its momentum change is

2 p

(photon reverses from

+ p

- p

) versus

p

for absorption.

Why is

p = E / c

instead of

p = m v

photons are massless, so the relativity relation

E^{2} = (p c)^{2} + (m c^{2})^{2}

collapses to

E = p c

What makes intensity fall as

1/ r^{2}

fixed power spread over a sphere whose area

4 π r^{2}

grows as

r^{2}

What happens to SRP force when

θ = 90°

it is zero — the edge-on plate catches no light.

Which single number governs whether SRP matters for a craft?

the area-to-mass ratio

A / m

Visual walkthrough — Solar radiation pressure