# Wave comparison

1. Dec 25, 2016

### Lord33

1. The problem statement, all variables and given/known data
Flat harmonic electromagnetic wave propagates in the positive direction in vacuo axis y. Vector electromagnetic energy flux density is given by: S(y,t)=Sm *cos(wt-ky)2.Wave value: k=(2*π)/λ=0.41 m-1,Amplitude Sm=26 W/m2.Compare this wave with another wave.

2. Relevant equations
ƒ=V/λ
k=(2*π)/λ
E=h*ƒ

3. The attempt at a solution
I find wavelenght from wave value λ=15m ⇒ it is a radio short wave ⇒next i find frequency from ƒ=V/λ=2*107⇒and finally i find photon energy E=h*ƒ=1,3*10-26 J =81*10-9neV.

I compared the result with the table, and he enters the period of short radio waves, ie, this wave is the same radio signal from your phone or radio?Maybe I made a mistake somewhere?

2. Dec 25, 2016

### Delta²

Radio or Mobile phone waves are not flat (I suppose you mean plane waves by the term flat) but they are spherical waves.In order to be more accurate, they are not even spherical waves but they resemble spherical waves when we are far away from the source antenna that produces the waves, they resemble spherical waves in the far field region as we say.

In real world we cannot produce plane(flat) waves cause they would require infinite amount of energy. We can produce spatially restricted plane waves inside waveguides https://en.wikipedia.org/wiki/Waveguide_(electromagnetism).

Last edited: Dec 25, 2016
3. Dec 25, 2016

### Lord33

What about the amplitude, how do I compare with short waves? If ampiluda will be 1026?

4. Dec 26, 2016

### Delta²

It doesn't matter what the amplitude is, we look only at the frequency (or equivalently the wave length) if we want to check whether it is a short wave or VHF/UHF wave or a microwave etc. Your calculations seem correct to me.

In my previous post I just wanted to say that plane waves (regardless of their frequency) are not physically realizable, unless of course they are restricted inside waveguides.

5. Dec 26, 2016

### Cutter Ketch

At any significant distance from the transmitter the radiation is indistinguishable from a plane wave. For example at a 1 m antenna located just 1 km from the transmitter the curvature of the wave over the length of the antenna is 125 microns or less than 1e-8 of the wavelength. Why confuse the issue with this?

6. Dec 26, 2016

### Delta²

For two reasons
1) Its good to be accurate even though the difference in real world might be small in the case you display(which it isn't in this case if you take into account point 2) see below)
2) You are "hiding" the fact that in plane waves the amplitude is constant, while in spherical waves the amplitude is inversely proportional from the distance from the source. That's why spherical waves require finite amount of energy, while plane waves require infinite energy.