# Video: Photon Momentum

In this video, we will learn how to calculate the momentum of a photon given its frequency or wavelength.

16:26

### Video Transcript

In this video, our topic is photon momentum. Photons, particles of light, do indeed have momentum, but so little that it’s easy not to notice. If we start out by considering an individual photon, we can recall that this object has no mass to it. From that perspective then, we might think that it also has no momentum. But experimentally, if we run this photon into a sheet of material, then whether the photon is absorbed by the material or reflected back in the direction it came from. In both cases, the surface that the photon runs into recoils a bit from this interaction. And interestingly, the amount of recoil depends on whether the photon is absorbed or reflected.

In the case of photon absorption, the recoil is less. But in the case of reflection, it’s relatively larger. We can explain this observation in terms of momentum transfer. If we assume that our incoming photons does have some amount of momentum, then when that photon is absorbed, it transfers that momentum to the absorbing surface. But when the photon is reflected back in the direction it came from, then the photon undergoes two times as much momentum change as it did when it was simply absorbed. And that greater difference in the photon’s momentum accounts for the relatively greater recoil that the surface experiences. This is in keeping with the law of momentum conservation.

So photons indeed do possess momentum. And mathematically, we can write that momentum this way. We can say that the momentum of a single photon, we’ll call it 𝑃 sub p, is equal to Planck’s constant ℎ divided by the wavelength of that photon 𝜆.

Now we mentioned earlier that the momentum of an individual photon is very small. To see just how small it is, let’s say that this incoming photon has a wavelength of 450 nanometers. This is a wavelength corresponding to the indigo, violet, and blue portion of the visible spectrum. So as we calculate the momentum of this incoming photon, we know 𝜆. It’s 450 nanometers or, equivalently, 450 times 10 to the negative ninth meters. That leaves us with ℎ, Planck’s constant, which we can treat as having a value equal to 6.63 times 10 to the negative 34th joule seconds. So that value goes in our numerator.

Now, before we calculate this fraction, let’s consider the units involved in our expression. In our numerator, we have joules times seconds. And we can recall that a joule is equal to a newton times a meter. And then we can recall further that a newton is equal to a kilogram meter per second squared, which means we can replace this unit of joules with kilograms meter squared per second squared.

With that unit substitution done, we now see that some cancelation takes place. First, one power of seconds cancels out from the numerator of our expression. And then we can also cancel one factor of meters in numerator and denominator. We’re left then with kilograms meters per second. These are the standard units in which we report momenta.

Now that that’s done, let’s go ahead and calculate this fraction. To three significant figures, we find a result of 1.47 times 10 to the negative 27th kilograms meters per second. The extreme smallness of this value is one reason why photon momentum is a fairly recent discovery. To see that it’s there at all, extremely precise measurement equipment is needed.

Now, by itself, a transfer of momentum of this amount to a surface won’t have very much effect on the motion of that surface. But what if instead of single photons or tens or hundreds or even millions of photons, we had many more than that incident on the surface? The cumulative effect of many photons being absorbed or reflected by a surface can have a significant impact on the motion of that surface. This is the idea behind a device that’s called a solar sail.

A solar sail is a surface that’s attached to lightweight unmanned spacecraft. The sail is designed to absorb or reflect incoming photons, perhaps from a star nearby. And in doing so, the momenta of these photons is transferred to the sail. And this provides a force thrusting the sail and therefore the spacecraft it’s attached to along.

There are three important factors which affect just how much thrust force is generated on a solar sail. The first is whether the sail absorbs incoming photons or whether it reflects them back. Reflecting versus absorbing leads to a greater transfer of momentum and therefore a greater thrust force. Another factor is the number of photons incident on the sail. The more photons, the more thrust force. And lastly, if we look back at our equation for photon momentum, we can see the third important factor. It’s the wavelength, 𝜆, of the photons incident on the surface. Photons having a shorter wavelength and therefore a smaller value of 𝜆 will have a relatively larger momentum.

Now, there’s another mathematically equivalent way of writing this equation. It’s related to the fact that light not only acts like a particle, like a photon as we’ve been describing it, but it also behaves like a wave. And for that reason, it’s described by this wave speed equation, which says that the speed of a wave 𝑣 is equal to the frequency of the wave times its wavelength.

Now, in the case of light, we know what the speed of our wave is. It’s the speed of light in vacuum 𝑐. So we can say that the speed of light is equal to the frequency of that light times its wavelength. And then if we divide both sides of this equation by 𝑐 times 𝜆, what we find is that on the left-hand side the speed of light cancels out. And on the right, wavelength cancels out. We’re left with an equation that tells us that one over wavelength is equal to 𝑓 divided by 𝑐. This is the wave speed equation for photons.

So then returning to our equation for photon momentum, we know that ℎ divided by 𝜆 can also be written as ℎ times one over 𝜆. And one over 𝜆, based on the equation we’ve just developed, can be replaced by 𝑓 over 𝑐. Making that substitution, we arrive at a second equivalent way of writing photon momentum. It’s equal to Planck’s constant times the frequency of the photon divided by the speed of light.

Knowing all this about photon momentum, let’s get some practice with these ideas through an example.

A laser produces 4.00 times 10 to the 27th photons, each with a frequency of 4.25 times 10 to the 14th hertz. What magnitude of momentum does producing these photons impart on the laser? Use a value of 6.63 times 10 to the negative 34th joule seconds for the Planck constant. Give your answer to three significant figures.

Okay, so in this example, we have a laser. Let’s say that this is our laser. And this Laser is producing photons, which have a frequency — we’ll call it 𝑓 — of 4.25 times 10 to the 14th hertz. Now, each one of our 4.00 times 10 to the 27th photons possesses some amount of momentum.

We can recall that that amount — we’ll call it 𝑃 sub p, the momentum of a single photon — is equal to Planck’s constant ℎ times the frequency of the photon divided by the speed of light in vacuum. So here’s the idea. Each one of these photons as it moves along has some momentum. And by the law of momentum conservation, this same amount of momentum must be imparted in the opposite direction on the laser. That is, as a photon is given this amount of momentum to the right, it imparts this same amount of momentum in the opposite direction on the laser.

If we add up the momentum of all the photons, then we’ll get some much larger momentum amount. And once again, by momentum conservation, this will be equal to the total amount of momentum imparted on the laser.

Notice that our question asks us to solve for the magnitude of momentum imparted. Which means that if we solve for this amount, the total momentum possessed by all the photons, then we’ll have answered our question. Because that’s equal to the magnitude of momentum imparted on the laser. So how will we solve for this total amount of momentum possessed by the photons emitted?

We can start by calculating the momentum of each individual photon. We’ll call that 𝑃 sub p. And then realizing that all of the photons that our laser emits have the same frequency. We can say that if capital 𝑁 is the total number of photons emitted, then capital 𝑁 multiplied by 𝑃 sub p will be equal to the total momentum of all the photons emitted. We’ll call that simply 𝑃.

This momentum, as we said, is equal to the total momentum of the photons. But it’s also equal to the magnitude of the momentum imparted to the laser. So in solving for 𝑃, we’ll find the answer we want. We can start doing this by replacing 𝑃 sub p, the momentum of an individual photon, with ℎ times 𝑓 divided by 𝑐, according to this equation. So the total momentum 𝑃 that we want to calculate is equal to Planck’s constant. We’re told to treat this value as equal to 6.63 times 10 to the negative 34th joule seconds. Multiplied by the frequency of each photon 𝑓. This value is given as 4.25 times 10 to the 14th hertz. Divided by the speed of light in vacuum 𝑐. We can approximate that value as 3.00 times 10 to the eighth meters per second. All multiplied by the total number of photons emitted by our laser. That’s 4.00 times 10 to the 27th.

When we substitute all these values in, before we go and calculate the total momentum 𝑃, let’s consider the units in this expression. First, in our numerator, we see that we have units of joules, which can be written equivalently as a newton times a meter. And a newton, we can recall, can also be written as a kilogram meter per second squared, which means that we can replace a joule with a kilogram meter squared per second squared.

With that substitution made, let’s now consider the units of hertz. These indicate number of cycles per second and can be replaced with the units inverse seconds. With these replacements, we see that some of our units can cancel out. In particular, in our units for the Planck constant, we see that one factor of seconds can cancel out. And then along with this, one factor of meters in our numerator and our denominator can cancel. And finally, one factor of inverse seconds can cancel as well. This leaves us with the overall units for this expression of kilograms meters per second. These are the units we expect for momentum. So it looks like we’re on the right track.

When we compute this fraction on the left and keep three significant figures in our answer, we find a result of 3.76 kilograms meters per second. This is the total momentum of all the photons emitted. And therefore, it’s the magnitude of momentum the photons impart on the laser.

Let’s look now at a second example of photon momentum.

A solar sail is a proposed method of spacecraft propulsion that uses the momentum of photons as a source of thrust. When photons hit the sail, they are absorbed and their momentum is transferred to the sail. If laser light with a wavelength of 200 nanometers is used to propel the sail, how many photons must hit the sail for it to gain one kilogram meter per second of momentum? Use a value of 6.63 times 10 to the negative 34th joule seconds for the Planck constant. Give your answer to three significant figures.

All right, so in this example, we have this structure called a solar sail. A solar sail is a surface, we could sketch it like this, that is designed to absorb or reflect incoming photons. The idea is that every incoming photon has some amount of momentum. And when those photons are, say, absorbed by the sail, that momentum is transferred to the sail. The effect of this is to create an overall force on the sail and the spacecraft it’s attached to. This force can move spacecraft through space. And so, indeed, these incoming photons are a source of thrust.

Now, in our example, these incoming photons that land on the sail are provided by a laser. The wavelength of these photons, we can call it 𝜆, is equal to 200 nanometers. We want to know just how many photons at this wavelength, and we’ll call that number capital 𝑁, would be needed to deliver a gain in momentum to the solar sail of one kilogram meter per second. We can refer symbolically to that gain in momentum as Δ𝑃.

To figure out how many photons are needed to create this change in momentum in our sail, we’ll need to know how much momentum each one of our individual photons with a wavelength of 200 nanometers possesses. Written as an equation, this amount of momentum for a single photon is given as Planck’s constant divided by the wavelength of that photon. In our problem statement, we’re told to treat Planck’s constant as exactly 6.63 times 10 to the negative 34th joule seconds.

So knowing ℎ and knowing 𝜆, we’ll be able to calculate the momentum of a single photon emitted by our laser. But recall that what we want to do is figure out how many of these photons will be necessary to impart a gain in momentum of one kilogram meter per second to the sail. To see how to calculate that, let’s clear a bit of space on screen.

Okay, much better. Now, like we said, we want to solve for capital 𝑁, the number of photons emitted by our laser in order to impart this change in momentum to our sail. Since Δ𝑃 is the total momentum change of our sail, we can say that it’s equal to the momentum of one of the photons emitted by the laser multiplied by the total number of those photons. We can write this because each one of our individual photons all have the same wavelength, 200 nanometers. Therefore, each one provides the same amount of momentum to the sail.

Now, just as an interesting side note, notice that our photons are being absorbed by the sail. This is in contrast to being reflected. If the photons were reflected back, then that means they would impart twice as much momentum change to our sail. And that’s because rather than simply having their forward momentum being stopped, like it is when they’re absorbed, reflection would indicate that that momentum is reversed in the opposite direction. This would double the change in momentum experienced by the sail.

And therefore, if our sail reflected photons, then over here at our equation from the momentum of a single photon, we would multiply this by two. That’s because ℎ over 𝜆 of momentum would be transmitted to the sail when a photon is stopped. And then that same amount would be transmitted when the photon is sent back the way it came. This is why we have a factor of two whenever our sail is reflective material.

But since our sail in this example is not reflective, but it absorbs photons, we won’t use this factor of two in our equation. In this case, the momentum transmitted to this sail by an individual photon truly is equal to ℎ over 𝜆. Plugging that expression in for 𝑃 sub p, we now have this equation where it’s capital 𝑁 that we want to solve for.

To do that, we can multiply both sides of the equation by 𝜆 over ℎ. This cancels out both 𝜆 and ℎ on the right-hand side. And we find that the wavelength of our photons divided by Planck’s constant all multiplied by the total gain in momentum of our sail is equal to the number of photons incident on it.

Our next step is to substitute in the values in the left-hand side of this equation. But before we plug in 200 nanometers for our wavelength, let’s change the units here from nanometers into meters. We can do this by recalling that one nanometer is equal to 10 to the negative ninth meters. Which means that we can reexpress 𝜆 as 200 times 10 to the negative ninth meters. So we’ll substitute that value in for 𝜆 in our equation. And we’ll replace ℎ with 6.63 times 10 to the negative 34th joule seconds. And Δ𝑃, we know, is equal to one kilogram meter per second.

Looking at this expression, we can note that a joule is equal to a newton times a meter. And that a newton is equal to a kilogram meter per second squared. Which means we can replace the unit joule with a kilogram meter squared per second squared. The value in doing this is that now we see that the units on the left-hand side of this expression entirely cancel out. That is, kilograms cancel out with kilograms. Meters squared in the numerator cancels out with meters squared in the denominator. And after canceling one factor of seconds in the denominator, all that’s left is one over seconds in the numerator and denominator. So that cancels as well. We’re left with a unitless result, which is good because we’re looking for a pure number.

To three significant figures, 𝑁 is 3.02 times 10 to the 26th. That’s the number of photons that would need to land on the sail in order to impart one kilogram meter per second of momentum gain to it.

Let’s summarize now what we’ve learned about photon momentum. Starting off, in this lesson, we saw that individual particles of light, photons, do not have mass, but they do possess momentum. We saw that the momentum of an individual photon is equal to Planck’s constant divided by the photon wavelength, or equivalently Planck’s constant times the photon frequency divided by the speed of light 𝑐. An implication of this is that photon momentum is directly proportional to photon frequency, and it’s inversely proportional to wavelength. And lastly, we saw that for 𝑁 identical photons, each with momentum 𝑃 sub p, the total momentum of the photons 𝑃 is equal to 𝑁 times 𝑃 sub p. This is a summary of photon momentum.