# Why is the momentum of light equal to E/c?

1. Aug 24, 2004

### Cheman

I read on a website about relativity that "Before Einstein, it was known that a beam of light pushes against matter; this is known as radiation pressure. This means the light has momentum. A beam of light of energy E has momentum E/c." My question is how do we know there is this radiation pressure? Are there any examples where it has occured?
Also, why is the momentum of light equal to E/c?

Thanks.

2. Aug 24, 2004

### Gonzolo

People have lifted small samples of glass with lasers, (a tiny sphere). A very energetic laser pulse hitting a piece of steel or paper can also make a surprisingly loud noise, easily above 70 dB. Enough to give you a headache after a few minutes if you keep it going at 10 Hz. Such a pulse makes a piece of paper move suffiently that I am sure it would crumble a castle of playing cards.

3. Aug 24, 2004

### pmb_phy

This result was obtained by requiring momentum to be conserved. When particles interact with electromagnetic radiation there are forces exerted on the particles giving them momentum. In order for momentum to be conserved there must be momentum in the electromagnetic radiation. When the radiation, which has momentum, is absorbed or reflected from a surface then the radiation exchanges momentum with the wall. The time rate of change of this momentum results in a force. Radiation pressure is this force per unit area. The relationship E = pc is derived with this in mind and by requiring energy to also be conserved.

Pete

4. Aug 24, 2004

### pervect

Staff Emeritus
Levitation of glass spheres is pretty cool, but I believe the mechanism involved isn't direct light pressure, it's due to the dipoles in the glass interacting with the fields in the beam. The total energy is lowest when the glass is in the most intense part of the electric field.

5. Aug 24, 2004

### kurious

A paddle wheel in a vacuum will turn when light strikes it from one side.

6. Aug 24, 2004

### Cheman

Ok, cool, so we have some sweet examples. Could any one please elaborate on where the p= E/c equation come from?

7. Aug 24, 2004

### Gonzolo

From what I understand, what you're saying explains why the sphere stays in the center of the beam. But photon momentum, which is equivalent to radiation pressure, explains why the sphere doesn't fall on the ground.

As for p = E/c, well there are 3 ways to see it :

1. Special Relativity says that for a particle, $$E^2 = m^2c^4 + p^2c^2$$

2. Quantum mechanics says that for a photon :

$$E = hf = hc/\lambda = hkc = pc$$

3. Classical EM (Maxwell) found it earlier for electrical charges. The E-field from the light makes the charge move laterally, and the B-field makes the laterally moving charge move forward. Calculating this gives p = E/c. The relation B = E/c gives slight insight. You need the Poyting vector, energy density, Lorentz force and integrals to make this convincing and I don't feel like typing these right now.

Last edited by a moderator: Aug 24, 2004
8. Aug 24, 2004

### ArmoSkater87

I'll show how the equation Gonzolo showed tell you that p=E/c.

$$E^2 = m_0^2c^4 + p^2v^2$$ <---Total energy of particle with rest mass, $$m_0$$, going at velocity $$v$$.
Since light has zero rest mass and travels at velocity c...
$$E^2 = p^2c^2$$
$$E = pc$$
$$p = E/c$$

9. Aug 25, 2004

### Cheman

Thanks. Are there any other ways to get p= E/c without using E=mc2 or something similar first? Its just that ive seen p=E/c used in many "proofs" for E=mc2, and would like to know where it come from without having to reley on that equation. Gonzolo, could you please explain the second way you mentioned - to do with photon energy; I don't quite understand your method.

10. Aug 25, 2004

### pmb_phy

Yes. It can be done with the principles of electrodynamics by requiring that energy and momentum be conserved. E = pc then falls out of the derivation.

Pete

11. Aug 26, 2004

### ArmoSkater87

I think you can derive it by simply using newtonian physics, but im not sure...

$$F = \frac{p}{t}$$

$$E = Fd = \frac{pd}{t} = \frac{mvd}{t}$$

$$m = \frac{Et}{vd}$$

$$p = mv$$

$$p = (\frac{Et}{vd})v$$

$$p = \frac{Et}{d} = \frac{E}{d/t}$$

$$p = \frac{E}{v}$$

Since light travels at c...

$$p = \frac{E}{c}$$

12. Aug 27, 2004

### Gonzolo

Actually, I should have used h bar and lambda = 2Pi/k instead of h and 1/k above. Same result. Einstein could have used this when working with E = mc2, as E = hf came before.

ArmoSkater87, that brings you there, and some of those relations are needed. But I think it assumes that light is a particle.

It possible to rigorously show from Maxwell's equations that an EM wave can give momentum E/c to a charged particle. You need to be comfortable with vector analysis to go from Maxwell's equations (curls, divergence, complex plane wave) to what a plane lightwave looks like. Then if you accept this (that a plane lightwave has E and B perpendiular to each other and to the propagation direction), you can show how these E and B fields push a charge forward, giving it E/c momentum. A complete demonstration may be long for a post, but I will see what I can do.

13. Aug 28, 2004

### Gonzolo

Here it is. There are more than one way to go through it. I tried to make as simple as possible, yet rigourous enough. I assume it is understood that a plane EM wave has sinusoidal E and B perpendicular to each other as well as to their propagation direction.

An EM plane wave is incident on a charge q at rest in free space (B = E/c). The total force (Lorentz Force) on the charge is :

$$F = qE + qvB$$

The total work done on the charge over time $$\Delta t$$ is only due to the first term :

$$\Delta W = F \Delta d = qE\Delta d$$

because the magnetic force never does work. So:

$$\frac{\Delta W}{\Delta t} = qE\frac{\Delta d}{\Delta t}$$

$$\frac{\Delta W}{\Delta t} = qEv$$

$$v = \frac{\Delta W}{\Delta t}\frac{1}{qE}$$

The change in the charge's momentum, according to Newton's 2nd law is:

$$\frac{dp}{dt} = qE + qvB$$

But since the E oscillates in opposing directions, over a few cycles, it doesn’t give any net momentum to the charge in either direction, so that after a sufficient time $$\Delta t$$, the first term cancels itself out.

$$\frac{\Delta p}{\Delta t} = qvB$$

In vector form, this shows that the momentum is perpendicular to both v (and so E) as well as B because of the cross product in vector form. The right-hand rule shows it is parallel to the plane wave's propagation direction. Inserting the v from above, as well as B = E/c :

$$\frac{\Delta p}{\Delta t} = q\frac{\Delta W}{\Delta t}\frac{1}{qE}\frac{E}{c}$$

$$\Delta p = \frac{\Delta W}{c}$$

Assuming all the energy is absorbed, this is how much momentum the particle receives. The radiation can thus be said to possess momentum.

Interestingly, it seems that the E field gives the energy while the B field gives the momentum. Also, this demonstration shows that radiation will only give momentum and energy to a charged particle, (what about a neutron?). An atom could be considered as many charged particles.

Last edited by a moderator: Aug 28, 2004