# Electromagnetic waves of transmitters

1. Apr 15, 2009

### Niles

Hi all.

I have heard that cell phones heat up the brain, when you are talking in one because of the EM-waves at ~800-900 MHz.

Now the transmitter of a radio station at e.g. 100 MHz sends out EM-waves at a frequency of 100 MHz in all directions. Does this mean that all our brains are being heated up by the waves coming from the radio station?

Also, do these waves get weakend by scattering?

Last edited: Apr 15, 2009
2. Apr 15, 2009

### Born2bwire

Cell phones do not really heat up the brains too much (this is something that is regulated). They do not penetrate very deep, most of the heat is going to be on your ear or your skin. But all EM radiation is going to produce heat on your body, most of it is so weak you do not notice it though you are able to sense infrared fairly well. The lower wavelengths of the radio stations means that they will be less affected by scattering than your cell phone. It will take a longer distance and larger objects to obstruct them in comparison. For the most part, the waves weaken mostly due to the space loss factor, that is the inevitable attenuation by a factor of 1/r^2 due to the expansion of the wave as it propagates outward.

3. Apr 15, 2009

### Niles

Ahh, I see. I think I know the answer to this question, but I will go ahead and ask it anyway: Will I recieve more EM-radiation by putting my new radio clock on the table next to my bed? (Persoanlly I thinm not).

So the key point is that the intensity is so small that the heat cannot be felt? And in the case with the cell phones, it is because the transmitter is relatively close?

Last edited: Apr 15, 2009
4. Apr 16, 2009

### Born2bwire

Your radio is strictly a receiver. It will have some secondary radiation from the excited currents in the antenna but I doubt it is going to make any real impact.

Well, for the brain specifically the heat isn't felt in anyway because the brain doesn't have any nerves to sense heat or touch. But the main point is that the energy of the fields are low enough that they do not create much heat. The energy is absorbed as it penetrates through the body. Most of the energy is going to be absorbed on the skin and outer layers. This is especially true for higher frequencies. For example, the military's Active Denial System uses a wavelength that is mostly absorbed just under the skin where the nerves are. Because of this, the nerves feel an intense amount of heat and pain despite the fact that your body isn't really being heated at all (the worst injuries they have had are a few minor burns). With cell phones it isn't because the antenna is close by (if anything this makes things worse since the power drops of by the inverse square law), but the fact that the wave has to pass through parts of your ear, skin, skull, and fluids before it reaches the brain. All of these will dissipate some of the energy. In addition, cell phones are designed with this in mind, there are specific standards about the absorption of energy by the body that are allowed.

5. Apr 16, 2009

### Niles

Ok, that is reassuring. But since the transmitter sends out EM-radiation 24 hours each day, then doesn't that mean that we are all heated up by a certain amount always?

What I meant what that I have seen a documentary about cell phones, and they said that the network is made up of "cells", where there is a transmitter in each "cell". So I thought that the transmitter must be closer to the cell phone than in the case of e.g. the radio, and because of 1/r2, then the intensity is greater for the cell phone. When this is combined with the fact that the frequency is also higher for the cell phone, then this might explain why the heating is more "severe".

6. Apr 16, 2009

### Staff: Mentor

By the way, the wattage of a cell phone or a dissipated signal from a radio station is so low that if you tried to use it to heat up a cup of water, you couldn't even measure the temperature change. Don't worry about this - it is not an issue.

7. Apr 16, 2009

### Born2bwire

Yes, you are being bombarded by heat inducing EM rays 24 hours a day, 7 days a week, but, just to make you more comfortable, you are also being bombarded by ionizing radiation like gamma waves too, mostly via natural processes unless you spend more time at particle accelerators than most people (don't forget your film badge).

The higher frequency by itself does not affect the severity of the heating, classically the energy density of a wave is solely dependent upon its amplitude, not its frequency. It depends on several factors. The first is regarding the wavelength with respect to the absorber. A longer wavelength is going to give up less energy over the same distance as a higher frequency, but, if you have a material that is very conductive then this will not matter since even very long wavelengths will be fully absorbed before transmitting through. Another factor is that the dielectric constants and conductivities are often frequency dependent. The absorption spectrum of water is very frequency dependent for example.

The ubiquity of cell phone towers isn't going to be a problem for radiating the public at large. Again, because of the space factor loss the transmissions from cell phone towers will be rather benign. The difference from other sources say radio or TV is that a cell phone is a transmitter and receiver. Transmission from your cell phone is the main source of local heating on the body.

Oh hush, I'm trying to sell these EM shields that you attach on to the back of your cell phone that protect the brain from head cancer.

8. Apr 16, 2009

### Niles

I did not know that. Is the probability of the radiation exciting an atom also proportional to the amplitude?

9. Apr 17, 2009

### Born2bwire

Long answer: Yes. However, this question mixes classical and quantum mechanics. The idea of the wave's amplitude being proportional to the energy density is classical EM. In Quantum Mechanics, the EM wave is propagated in packets of energy/momuntem called photons. The energy of a photon is only dependent upon its frequency. The classical macroscopic EM field can be related to the Quantum Electrodynamic photons using Quantum Electrodynamic theory. However, there are many situations where you can mix classical EM with Quantum Mechanics. The situation you have asked is one situation that we can do so. We can assume that an EM plane wave of amplitude $$E_0$$ and angular frequency of $$\omega$$ is incident upon an atom. We can treat the EM wave in the classical sense and the atom in the Quantum Mechanical sense. In this way, we can see that the probability of absorption is directly dependent upon the square of the amplitude, or, equivalently, the energy density of the wave. However, it is also inversely proportional to the square of the difference in energy between the desired transition and the incident wave's photon.

So, the probability to transition to a specific state is proportional to the classical wave's energy density (amplitude squared) and inversely proportional to the square of the difference in energy between the desired excited state (some frequency $$\omega_0$$) and the energy of the exciting wave's photon (frequency $$\omega$$).

$$P_{a \rightarrow b} \propto \left( \frac{E_0}{\hbar \left( \omega_0 - \omega \right)} \right) ^2$$

It is interesting to note that if we want to find the probability of stimulated emission (an EM wave striking an excited atom resulting in the atom emitting a photon of the same frequency as the incident wve) we find that the probabilty is exactly the same. So the probability for an atom to absorb an EM wave is the same as the probability for stimulated emission.

Anyway, all this is mostly irrelevant to your original question. Classically, the heat from the nonionizing radiation is due to the fact that the EM waves induce currents in conductive medium. These currents give up heat through ohmic loss. Another classical method is the way that microwaves work. Polar molecules, like water, have a slight dipole moment. An EM wave will force these polar molecules to oscillate as they try to line up their dipole moments with the alternating electric field in the EM wave. This kinetic energy gives rise to heat.

10. Apr 17, 2009

### Niles

Thank you for that very interesting and good answer. I have one last question, which is related to your post:

You emphasize that it is in classical electrodynamics that the energy density is proportional to the square of the amplitude. Does is mean that it is less accurate, because it is the classical interpretation?

11. Apr 17, 2009

### Born2bwire

No, it isn't any less accurate from the macroscopic perspective. Classical EM is still a very relevant and very accurate method of modeling electromagnetics. However, I have found that people, after getting a brief spattering of quantum mechanics, often confuse and mix the classical and quantum models. I'm not sure why this is, I think it may be due to the fact that most people only learn quantum mechanics to any large degree while quantum electrodynamics is usually left to the realm of graduate students. Either way, I have found it best to try and be clear about the differences between the classical and quantum models.

The relationship of energy between the classical and quantum models is a common mix-up. Most people are aware that a photon's energy is solely dependent upon its frequency. However, some people will then apply that to the classical case, thinking that if I increase the frequency of electromagnetic radiation impinging upon a scatterer then I am increasing the energy being absorbed. But in terms of macroscopics, where the classical model is valid, this is not necessarily true.

12. Apr 17, 2009

### Niles

Ahh, I see. But can you be more specific of what you mean by "macroscopic"? I mean, we now know that the energy density for EM-radiation (classically) depends on the amplitude, and thus the frequency of the EM-radiation of cell phones isn't relevant here.

But in a microwave oven (atleast this is how I have understood it), the water molecules are given kinetic energy because of the high frequency of the EM-radiation.

I wouldn't say that one case is more "macroscopic" than the other (but perhaps I do not know the correct interpretation of the word "macroscopic").

13. Apr 17, 2009

### Born2bwire

The exact point between the quantum world and the "macroscopic" world is very fuzzy. In some ways, the macroscopic measurement is a statistical measurement of the quantum world. We can observe quantum behavior in EM waves easily enough. We can run a laser through layers of absorbers between the laser and a CCD. With enough absorbers, we can see to it that we observe individual photons on the CCD. But if we were to make a general measurement of the laser, for the most part it would follow a macroscopic measurement, a measurement of the mean value of the quantum states.

So for most real world measurements, we are doing macroscopic measurements where we are taking an average of the quantum states. Most quantum effects do not occur until we are working on very small length scales or with waves so weak that we are working with small groups of photons.

For the microwave oven, it works on the oscillating dipoles that I mentioned earlier. The water molecules are polar and so they have a weak dipole moment. 2.45 GHz is not a very high frequency in the large scope of things, and if you wanted to increase the temperature faster, you would use a higher frequency. 2.45 GHz was chosen because we need a balance between heating quickly and being able to permeate the heat through a food item evenly (so you do not want to cook to quickly, you'll get the frozen burrito, in addition, you want the waves to penetrate into to the food to cook the inside as well). Temperature is the average kinetic energy of a medium's molecules. So this is a classical effect.

14. Apr 17, 2009

### Niles

But isn't this the same that happens inside the brain (or the ear, skin, etc.), when the cell phone radiation hits you?