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Regarding light as a particle, is there something like momentum from light when it shines on objects?

We can see light as a particle or a wave. Regarding it as a particle, is there some momentum given to objects which are struck by light?

It should not be, as light has no mass - or does it have mass due to movement?

safkan
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Brooklyn
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4 Answers4

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Photons do have momentum, and the equation for the momentum of a photon is given by:

$$p = \frac{E}{c}$$ where E is the energy of the photon.

These equations can be derived from Einstein's Equation:

$$E^2 = p^2c^2 + m^2c^4$$

For a photon with mass = 0, the equation is reduced to:

$$E = pc$$

This means that mass is not required in order to have momentum, so long the particle has energy (which a photon surely has - as shown through the photelectric effect)

Thus, it is a common misconception that mass is required for momentum.

Edit: momentum of a photon can also be written as $$p = \frac{h}{\lambda}$$

What this λ represents in the context of a particle is simply the EM wave associated with that particular photon since particles themselves do not have wavelengths.

Hope this helps!

john
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  • If P=E/c then the equation E=pc seems like a stretch. It means the the speed of light is used twice. In other words the equation is saying E=(energy\c) (c). – Bill Alsept Mar 11 '22 at 16:05
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    @BillAlsept All john did was multiply both sides of the equation $p=E/c$ by $c$ to get $pc=E$. Nothing deep or controversial. – Andrew Mar 11 '22 at 19:33
  • @Andrew I’m not talking about John‘s equation. I’m talking about pc. That’s momental times the speed of light. Momentum (p)already has speed and then you’re multiplying it again by speed (c)? – Bill Alsept Mar 11 '22 at 20:12
  • @BillAlsept Energy has units of mass times velocity squared. Momentum has units of mass times velocity, and the speed of light has units of velocity. So the units work out. – Andrew Mar 11 '22 at 20:12
  • @Andrew OK so p is momentum, and momentum is mass times velocity. So in the equation pc, what is the velocity used in p? And if it’s the speed of light AGAIN, why? Where did it come from? – Bill Alsept Mar 11 '22 at 20:53
  • @BillAlsept Momentum is only equal to mass times velocity in non-quantum, non-relativistic physics. In classical electromagnetism, the momentum density of the electromagnetic field is $\vec{S}/c$ where $\vec{S}=\vec{E} \times \vec{B} / \mu_0$, and $\vec{E}$ is the electric field, $\vec{B}$ is the magnetic field, and $\mu_0$ is the vacuum permeability constant. Quantum mechanically, the momentum of a photon is $p=\hbar k$ where $k$ is the wavenumber. – Andrew Mar 11 '22 at 20:57
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In addition to the other correct answers, it's worth noting that even classically (i.e., even in the old days when people thought light was just a wave), the electromagnetic field carries energy and momentum. The momentum of a classical electromagnetic wave leads to radiation pressure. Of course we have a deeper understanding of this now this now in terms of quantum mechanics and photons, but the effect is there even in classical physics.

Andrew
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Yes the electromagnetic field carries a momentum density proportional to $\mathbf{E}\times\mathbf{B}$, when a charged particle is accelerated by said field some of this momentum is transferred. The same is true for energy and angular momentum.

AfterShave
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Yes it does transfer momentum. Although it has no mass, from the energy momentum relation you get the following

$ E = \sqrt{p^2c^2+m^2c^4} = h\nu$

And hence,

$ |p| = \frac{h\nu}{c}$

As Kurt G. pointed out, that was the idea for Crooke's radiometer, but eventually not the correct explanantion. Other examples I think would be the trail of a coment pointing away from the sun. Or the project "Breakthrough Starshot", where they want to accelarate a nano-spacecraft by laserlight.

Martin
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