WHEN A PHOTON OF 1Kg STOPS A METEORITE OF 3,000Tm

First of all, I calculate the frequency of the photon with an equivalent mass of 1 kg. Its rest mass is 0.

$$E=m{c}^2$$

$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{0}{0}$$

0/0 is an indeterminate form that can take any value, for example 0/0 = 1.

$$m{c}^2=hf;1\times c^2=hf$$

$$hf=c^2$$

$$f=\frac{c^2}{h}=\frac{{(3\times {10}^8)}^2}{6.63\times {10}^{\mathrm{-34}}}=1.36\times {10}^{50}\text{Hz}$$

This frequency is very, very high.

For example, the frequency of yellow light is of the order of 508–526 × 10¹² Hz, which is much lower. The peculiar thing about this photon is that although it has a mass of 1 kg, we cannot see it because its frequency is far beyond the ultraviolet range, in an extreme extrapolation of the gamma-ray spectrum, outside the visible spectrum.

Now we suppose that there is an elastic collision of the photon with a meteorite which travels, for example, at a speed of 100 m/s or 360 km/h. We calculate what the meteorite’s mass is so that, when the photon collides with the meteorite, the light suddenly stops the stone. Of course, the trajectories of both bodies, when they collide, must pass through their respective centers of gravity.

The photon could also be absorbed by the meteorite instead of undergoing an elastic collision. Here, however, we examine only the elastic-collision scenario.

The momentum of each body must match.

The momentum of a photon of 1 kg, which travels at the speed of light, is:

$$p=mc=1c=1·3\times {10}^8=3\times {10}^8\text{kg}\frac{\text{m}}{\text{s}}$$

It must match the momentum of the meteorite, which travels at a speed of 100 m/s:

$$p=m_{m}v;1c=m_{m}100$$

$$m_m=\frac{c}{100}=\frac{3\times {10}^8}{100}=3\times {10}^6$$

$$m_m=3\times {10}^6\text{kg}=3000\text{tonnes}$$

This is the mass of the meteorite.

Why about half the equivalent mass is enough in an elastic collision

The calculation above gives equal and opposite initial momenta. This alone does not give the stopping condition for an elastic collision: we also need to include the photon after the collision and conserve both momentum and energy.

Consider an ideal head-on collision in which the photon bounces straight back and the meteorite stays intact, without heating or breaking apart. Because the photon reverses direction, the momentum it gives to the meteorite is the sum of the magnitudes of its incoming and outgoing momenta. At the meteorite’s low speed of 100 m/s, the photon’s incoming and outgoing energies are almost equal, so the transferred momentum is approximately twice its incoming momentum.

Let p_m be the magnitude of the meteorite’s initial momentum, K_m its initial kinetic energy, and E_in and E_out the photon’s incoming and outgoing energies. For the meteorite to stop, conservation of momentum and energy gives:

$$p_m=\frac{E_{\text{in}}+E_{\text{out}}}{c}$$

$$E_{\text{out}}=E_{\text{in}}+K_m$$

The photon takes away the meteorite’s initial kinetic energy. Combining these two equations gives the required incoming photon energy:

$$E_{\text{in}}=\frac{p_{m}c-K_m}{2}$$

For a 3,000-tonne meteorite moving at 100 m/s, p_m is approximately 3 × 10⁸ kg·m/s and K_m is approximately 1.5 × 10¹⁰ J. Using the rounded value c = 3 × 10⁸ m/s used above, the required photon energy is approximately 4.5 × 10¹⁶ J:

$$M_{\text{in}}=\frac{E_{\text{in}}}{c^2}\approx 0.5\text{kg}$$

So approximately half the equivalent mass is enough to stop the meteorite in this ideal elastic backscattering case. It is only approximately half because the outgoing photon gains the meteorite’s kinetic energy. Here, “equivalent mass” means E/c²; the photon’s rest mass remains zero.

With exactly equal and opposite initial momenta, as in the original 1 kg example, an ideal elastic collision with the photon bouncing straight back would reverse the meteorite’s motion rather than leave it at rest. This is a thought experiment, not a claim that a real meteorite could remain intact and reflect such an energetic photon. The energy calculations below still refer to the original 1 kg photon.

Conversely, a photon with an equivalent mass of 1 kg could stop a meteorite of approximately 6,000 tonnes moving at 100 m/s in the same ideal head-on elastic backscattering case. At a fixed meteorite speed, doubling the photon’s energy doubles the meteorite mass that can be stopped: both the meteorite’s momentum and its kinetic energy are proportional to its mass. At this low speed, the approximate stopping condition is:

$$m_m\approx \frac{2M_{\text{in}}c}{v}\approx \frac{2\times 1\times 3\times {10}^8}{100}=6\times {10}^6\text{kg}=6000\text{tonnes}$$

This rounded result uses c ≈ 3 × 10⁸ m/s, as in the original calculation. Using the exact value c = 299,792,458 m/s and the full relativistic expressions for the meteorite’s momentum and kinetic energy gives approximately 5,995.85 tonnes. The result assumes the same ideal elastic collision described above.

On the other hand, the energy needed to create the photon is:

$$E=m{c}^2=1c^2={(3\times {10}^8)}^2=9\times {10}^{16}\text{J}$$

Starting from the 1 kg equivalent-mass photon calculated above, its energy is approximately 9 × 10¹⁶ J. Since E = Mc² and c is constant, halving the photon’s equivalent mass also halves its energy. For a photon with an equivalent mass of approximately 0.5 kg, we therefore get:

$$E_{\text{0.5 kg}}\approx \frac{9\times {10}^{16}}{2}=4.5\times {10}^{16}\text{J}$$

Creating the photon needed for the ideal elastic-stopping case therefore requires approximately half the energy of the original 1 kg photon, before allowing for any production losses. The meteorite still has a mass of 3,000 tonnes: it is the photon’s equivalent mass that is approximately half a kilogram. The ice-melting comparison below continues to use the original 1 kg photon’s energy.

And knowing that 3.35 × 10⁶ J are needed to melt 10 kg of ice, then:

$$\frac{9\times {10}^{16}}{3.35\times {10}^6}=2.69\times {10}^{10}$$

At least this many times the energy needed to melt 10 kg of ice is needed to create the photon (γ).

Now new questions arise, such as: how do we create this photon of 1 kg? Is nuclear energy needed to do it? Is it possible to create it with current technologies? Can an electron be so excited that it releases a photon of this mass? It doesn’t seem to be possible because nucleons and electrons are very tiny.

Read the original 2018 PDF.

More on photon mass

For an earlier video on photon mass, see Photon mass.

For more on the rest-mass and relativistic-mass calculations, see Mass at rest and relativistic mass of photon.