Expressions for the energy density of electromagnetic waves

[99wattr89]

I hope this is the right forum for a first year undergraduate problem!
The problem I've been working on is here: http://i.imgur.com/IhTtL.png

I think that I have the correct answers, but I'm not sure. I think that the energy densities of the E and B field components will be (0.5)(ε0)(E^2) and (B^2)/(2μ0), that the ratio between the two is just 1/(c^2) and that the characteristic impedance of free space will be (Eμ0)/B.

Can anyone confirm or deny those results? Thank you.

 

[tiny-tim]

welcome to pf!

hi 99wattr89! welcome to pf!

(try using the X² button just above the Reply box )

I think that I have the correct answers, but I'm not sure. I think that the energy densities of the E and B field components will be (0.5)(ε0)(E^2) and (B^2)/(2μ0), that the ratio between the two is just 1/(c^2)

yes

and that the characteristic impedance of free space will be (Eμ0)/B.

which is … ?

 

[99wattr89]

Thank you! And thanks for the tip!

I've been playing about with the result, and tried subbing in the energy equations form the first part of the question - that gives me an expression with μ0, ε0, and the E and B energies. I think I see how to finish it now - the E and B field energies at the same, right? Because if they are then the two cancel, and I get √(μ0/ε0), which I found in a couple of books as an expression for vacuum impedance.

The only trouble is justifying why the E and B waves have equal energy. I thought that maybe it was because in a loss-less system where E is generating B and B is generating E they would have to have the same energy - but in a wave they could both have been created by some unidentified source, which wouldn't necessarily have to give the same energy to both.

 

[tiny-tim]

… the E and B field energies at the same, right? Because if they are then the two cancel, and I get √(μ0/ε0), which I found in a couple of books as an expression for vacuum impedance.

i suspect the question wants an answer in terms of µ_o and c

The only trouble is justifying why the E and B waves have equal energy.

no, the electric field and magnetic field (in an EM wave) have equal strength, not energy

EDIT: actually, that's not correct, i should say that the strengths are related by E_max = cB_max

(btw, in a more convenient system, both µ_o and c would be 1, and both the strengths and the energies would be equal )

 

[99wattr89]

i suspect the question wants an answer in terms of µ_o and c no, the electric field and magnetic field (in an EM wave) have equal strength, not energy

(btw, in a more convenient system, both µ_o and c would be 1, and the energies would be equal )

I see, so the force created by each is the same, but not the energy carried?

Double checking, I don't actually get this result though, I get μ0 for the impedance. Could my ratio for the two energies be backwards? I was pretty hazy on how to find the ratio for them so I could easily have that wrong. When I did the E energy over the B energy I got μ0ε0E²/B², so I thought that meant a ratio of μ0ε0 - is that right? What confuses me is that there are E and B terms in the ratio and I don't know how to get rid of them initially I just ignored them, but I don't think that's right.

If the ratio of E energy over B energy is 1/μ0ε0, then I get a result of μ0c, which I think is correct, but I can't get that ratio.

 

[vela]

i suspect the question wants an answer in terms of µ_o and c

I think the problem is looking for a specific value.

no, the electric field and magnetic field (in an EM wave) have equal strength, not energy

(btw, in a more convenient system, both µ_o and c would be 1, and the energies would be equal )

That's backwards. If the waves have the same energy in one set of units, they'll have the same energy in all sets of units. Using SI, the E and B field amplitudes have different units, so you can't compare them directly, but just looking at the numbers and ignoring the units, you'd see "E">>"B". When you use units where c=1, the fields have the same units, and E=B.

 

[tiny-tim]

I see, so the force created by each is the same, but not the energy carried?

i got that slightly wrong … see the EDIT above

using E² = c²B² wil give you the correct result …

If the ratio of E energy over B energy is 1/μ0ε0, then I get a result of μ0c, which I think is correct, but I can't get that ratio.

in reply to your earlier question …

The only trouble is justifying why the E and B waves have equal energy. I thought that maybe it was because in a loss-less system where E is generating B and B is generating E they would have to have the same energy - but in a wave they could both have been created by some unidentified source, which wouldn't necessarily have to give the same energy to both.

yes, they are sort-of generating each other

from Maxwell's equations, we get ∇²E = µ^oε^o∂²E/∂t² and ∇²B = µ^oε^o∂²B/∂t²,

from which we deduce that there's a wave with speed 1/√(µ^oε^o), and E_max = cB_max, see http://en.wikipedia.org/wiki/Electromagnetic_radiation#Derivation

 

[99wattr89]

I think the problem is looking for a specific value.


That's backwards. If the waves have the same energy in one set of units, they'll have the same energy in all sets of units. Using SI, the E and B field amplitudes have different units, so you can't compare them directly, but just looking at the numbers and ignoring the units, you'd see "E">>"B". When you use units where c=1, the fields have the same units, and E=B.

i got that slightly wrong … see the EDIT above

using E² = c²B² wil give you the correct result …


in reply to your earlier question …


yes, they are sort-of generating each other

from Maxwell's equations, we get ∇²E = µ^oε^o∂²E/∂t² and ∇²B = µ^oε^o∂²B/∂t²,

from which we deduce that there's a wave with speed 1/√(µ^oε^o), and E_max = cB_max, see http://en.wikipedia.org/wiki/Electromagnetic_radiation#Derivation

Thank you both for all your help! I don't understand the equations listed on that wikipedia page, or how E/B=c comes out of them, but maybe that's more second year material?

I certainly have the answer at least, so thanks again.