Cell Phone Energy
So, lets start with cell phones. I looked at my cell phone battery, and it looks like it is rated for 1 A, at 3.5 V. So when it is running at its peak it should put out about 3.5 W of power in electromagnetic waves (assuming it reaches its rating and all of that energy is fully converted into radiation). But what form does this energy take? Well, its electromagnetic radiation, so its in the form of a bunch of photons. In order to determine the energy of each photon, we need to know the frequency of the radiation. Surfing around a bit on wikipedia, I discovered that most cell phones operate in the 33 cm radio band, or somewhere between about 800 - 900 Mhz.How much energy does each ~ 1 Ghz photon have? We know that the energy of a photon is:
\[ E = h \nu \sim 7 \times 10^{-25} \text{ J} \sim 4 \times 10^{-6} \text{ eV} \]
it will be convenient to know the photon energy in "eV's". 1 eV is the energy of a single electron accelerated through a potential of 1 volt, or
\[ 1 eV = (1 \text{ electron charge} ) * ( 1 \text{ Volt} ) = 1.6 \times 10^{-19} \text{ J} \]
So my cell phone is sending out signals using a bunch of photons, each of which has an energy of about 4 micro eVs. Lets consider the energy scales involved in most molecular processes and compare those scales with this energy.
Molecules
Great. We have a number. What what does it mean? A number in physics means little without some context. Lets try and consider what photons can do to molecules. I can think of three different processes: first, a photon could knock out an electron (i.e. ionize the molecule), second the photon could make the molecule vibrate or wiggle, or third the photon could make the molecule rotate.Lets see if we can estimate the energies for these three different types of processes. Lets first collect some of the information we know about atoms and molecules so that we can continue our estimations. I know that most atoms are about an angstrom big, or 10^(-10} meters. I know the charge of an electron and proton.
Ionization
What are typical molecular ionization energies? Well we could try and estimate it. Whats the energy stored in an electron and proton being about an angstrom apart? Well, remembering some of our electrostatics we have\[ E = \frac{ k q_1 q_2 }{ r} \sim (9 \times 10^9) \frac{ (1.6 \times 10^{-19} \text{ C} )^2 }{ (1 \ \AA)} \sim 14 \text{ eV} \]
which is pretty darn close to the ionization energy of hydrogen at 13.6 eV. So I will claim that since all atoms are about the same size, typical ionization energies across the board are about 10 eV, in order of magnitude.
Vibration
What about making our molecules vibrate? Well what are the energies of molecular bonds? They ought to be quite similar to the ionization energies of molecules, but as we know they are a tad weaker. Bond energies for most molecules are on the order of a few eV. The oxygen-hydrogen bond in water for example has a binding energy of 5.2 eV. What does that have to do with vibration? Well, if we consider a material as made up by a bunch of atoms all stuck together with springs, we can estimate the spring constant. Assuming that a typical binding energy of 3 eV or so, and a typical atomic separation of 1 angstrom or so, we can estimate the spring constant for atoms, knowing\[ U = \frac{1}{2} k x^2 \]
\[ 3 \text{ eV } \approx \frac{1}{2} k ( 1 \A )^2 \]
\[ k \approx 100 \text{ N/m} \]
And now having estimated the spring constant, we can estimate how much energy there is in a quanta of atomic vibration, i.e. figure out the corresponding frequency from
\[ \omega = \sqrt{ k / m } \]
and quantize it in units of hbar. We discover that a quanta of atomic vibration typically has energies on the order of
\[ U = \hbar \omega = \hbar \sqrt{ \frac{ k }{ m } } \]
\[ = 6.6 \times 10^{-16} \text{ eV s } \sqrt{ \frac{ 100 \text{ N/m} }{ 2 \times \text{ mass of a proton } } } \sim 0.1 \text{ eV} \]
So molecular vibration energies are about a tenth of an electron volt.
Rotation
I can also make molecules rotate. What is the energy of the lowest rotational mode of a molecule. Well, bohr taught us that angular momentum is quantized in units of h, planck's constant. Imagine a two atom molecule, with two atoms separated by an angstrom. The energy of a rotating object can be written\[ E = \frac{ L^2 }{2 I } \]
in analogy to the energy of a moving object
\[ E = \frac{ p^2 }{2m} \]
where I is the moment of inertia for a molecule. We will estimate
\[ I = 2 m r^2 \sim 2 (1 * \text{ mass of proton} * ( 1 \text{ \AA} )^2 \sim 3 \times 10^{-47} \text{ kg m}^2 \]
So we can estimate the rotational energy for a small molecule
\[ E = \frac{ h^2 }{ 2 m r^2 } \sim 80 \text{ meV} \]
and this is for a small molecule, and will only go down for larger molecules, as I will increase. So I will call typical rotational energies 1 meV for medium sized molecules.
Heat
Another relevant energy scale to discuss when we are talking about brains is the energy due to the fact that our brain is rather warm. Body temperature is about 98 degrees Fahrenheit, or 37 degrees Celsius, or 310 Kelvin. Statistical Mechanics tells us that temperature is an average energy for a system, and in fact the Equipartition theorem tells us that when a body is in thermal equilibrium, every mode of it has
\[ \frac{1}{2} k_B T \]
amount of energy in it. For our brain that means
\[ E = \frac{1}{2} k_B T = 2 \times 10^{-21} \text{ J } = 13 \text{ meV} \]
i.e. just the fact that our brains are hot means that every degree of freedom in our brain already has 13 millielectron volts associated with it. Comparing to our results above, this is comparable to the rotational energies of molecules, but a tad less than their vibrational energies, which means that we should expect most of the molecules in our head to already be rotating.
Results
So going through some very rough calculations, we discovered that for molecules, there are three obvious ways you can get them hot and bothered, you can ionize them, make them wiggle, or make them rotate. There are some typical energy scales for these things, ionization energies are about 10 eV, vibrational energies are about 0.1 eV, and rotational energies are about 0.001 eV.
In addition, our brain already has about 13 meV of energy in every one of its degrees of freedom, and cell phone photons have an energy some 1/3300 of this.
And what was the energy for each cell phone photon? 0.000004 eV. Notice that this energy is about a hundred times weaker than typical rotational energies, and some 3300 thousandth times less than the natural thermal energy in our brains. Now you can begin to understand why most physicists are not too worried about the effects of cell phones. The radiation from cell phones is just not on the kind of energy scales that affect molecules in was that could potentially harm us. So I'm not too worried.
Not that it matters since you don't actually use the 3.5W number, but...
ReplyDeleteBatteries are rated by a voltage and a given capacity. In the case of the one you looked at, it was likely 1 amp-hour (the battery in my phone is 3.7v, 1.15 Ah).
3.5W is a huge amount of power. References on the internets claim that current consumption while talking on a cell phone is actually in the few hundreds of mA range, rather than the 1A you use. And of course, since power goes as the square of the current, this is a huge change in the amount power consumed.
You're right. I never ended up using the number, however, I think I was originally intending to go for worst case, my-phone-is-trying-to-kill-me kind of powers.
ReplyDeleteObviously your phone emits more then one photon.
ReplyDeleteUsing your analysis you can also "prove" that microwaves cannot cook anything.
But each photon has the same energy, and the cross section for any potentially photon doubling interactions is exceedingly small.
ReplyDeleteWhile its true that microwave ovens and cell phones both emit 'microwaves', the microwave band is very wide. Cell phones as I discussed emit radiation near 900 Mhz. Most microwave ovens operate at 2.45 Ghz. The energy corresponding to one of those photons is 4 times higher in energy.
True, I ignored dielectric heating, which is the dominate way microwave ovens heat food, but keep in mind that the power there is much higher, usually operating at 1000 watts or so. At these high powers, dielectric heating becomes much more important.
Whereas I contend the dielectric heating due to your cell phone operating at 100 mW or so is negligible compared to your bodies own heat generation.
I'm a vigorous cell phone user and do not worry about its adversay effects (except for wasting time).
ReplyDeleteHowever, we'd better be a little bit careful.
Cell phone signal is coherent radiation, not like incoherent radiation such as thermal radiation and ordinary light. Cell phone signal is the microwave-band equivalent of laser in the visible band.
Simply put, incoherent radiations cancel their effects because they are out of step. The effects of coherent radiations, however small, add up.
You are right to note a difference between coherent and incoherent radiation. However, once we've gone to a quantum mechanical viewpoint, we can ignore that, to a certain extent. For the simple treatment that Alemi gave above, the idea is that a single photon will interact with the matter and can potentially excite one of the processes above. His argument is that since the processes are higher energy than the photon energy, we don't have to worry about a single photon doing much.
ReplyDeleteQuantum mechanically, the 'adding up' photons is two photons striking the same piece of matter simultaneously, which isn't very likely. What you're referring to is primarily a classical, rather than quantum mechanical, effect. Basically what Alemi claims is that, since photons don't have the energy to excite motion of our molecules, they will pass right through us without damage. Now, if they weren't passing through, considerations of coherent and incoherent would be a concern, because that would determine the intensity we experience.
Basically, it doesn't matter if these photons are coherent or incoherent, because whatever intensity of radiation we get (which is the number of photons), all of the photons still have the same energy and will pass right through us.
Many "cell phones" today operate in the 1.9GHz band. This would increase the energy per photon, though not enough to change the argument.
ReplyDeleteAnyone reading this please understand that in the first sentence the author exposes himself as knowing nothing about this subject. The battery is rated in Amp Hours (Ah). This is not the same as an Amp. This is important to understand because the battery is so small in a cell phone and if it really produced anything close to 3.5 Watts, it would last only a very short amount of time. Cell phones produce only mAmps of power. That means that 100mA cellphone is 1/10 the power of 1W stated above and power is decreased by the inverse square law so, it is a lot less power than just 1/10th the total.
ReplyDeleteCompare this amount of radiated power (from a cellphone) to the Walkie Talkie radios that police, fire and others use that do output power in the 1-7 Watt range. This behavior has been going on for decades and there is no reported increase in brain damage in these groups.
very well done i love your work thanks for sharing
ReplyDeleteIn today's business world the difference between a home office and the field are as present as ever. Fortunately, enterprise mobility solutions are available for the executive that realizes the need for connectivity. Employees whose job entails that they consistently be traveling, such as field managers, and salespersons, are what many in the free business calls a 'Mobile Workforce'. These employees spend little time in any main office, instead doing their work in different satellite locations.
ReplyDelete