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

[Cheman]

Radiation Pressure...

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 occurred?
Also, why is the momentum of light equal to E/c?

Thanks.

 

[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.

 

[pmb_phy]

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 occurred?
Also, why is the momentum of light equal to E/c?

Thanks.

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

 

[pervect]

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.

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.

 

[kurious]

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

 

[Cheman]

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

 

[Gonzolo]

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.

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.

 

[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...
$$E^2 = p^2c^2$$
$$E = pc$$
$$p = E/c$$

 

[Cheman]

Thanks. Are there any other ways to get p= E/c without using E=mc2 or something similar first? Its just that I've 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.

 

[pmb_phy]

Thanks. Are there any other ways to get p= E/c without using E=mc2 or something similar first? Its just that I've 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.

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

 

[ArmoSkater87]

Thanks. Are there any other ways to get p= E/c without using E=mc2 or something similar first? Its just that I've 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.

I think you can derive it by simply using Newtonian physics, but I am 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...

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

 

[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.

 

[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 possesses 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.