Wireless routers vs. microwave ovens

[kevinisfrom]

TL;DR Summary

I believe most wireless router uses the same frequency as a microwave oven, yet microwave ovens heats up food while routers do not. If an energy of the light/radiation is only dependent on frequency, how is it that microwaves heats up food while wireless routers do not?

I believe most wireless router uses the same frequency as a microwave oven, yet microwave ovens heats up food while routers do not. If an energy of the light/radiation is only dependent on frequency, how is it that microwaves heats up food while wireless routers do not?

 

[PeterDonis]

If an energy of the light/radiation is only dependent on frequency

It isn't. The total energy carried by a beam of radiation depends on the beam's amplitude as well as its frequency. The amplitude of radiation being put out by a microwave oven is orders of magnitude higher than for a wireless router.

 

[kevinisfrom]

Can you share the formula? I always see E = hv

 

[PeterDonis]

I always see E = hv

This is a quantum formula for the energy of a single photon with frequency v. It has nothing to do with classical EM radiation, which is what we are dealing with for both the wireless router and the microwave oven.

 

[kevinisfrom]

So what happened to the amplitude?

 

[kevinisfrom]

This is a quantum formula for the energy of a single photon with frequency v. It has nothing to do with classical EM radiation, which is what we are dealing with for both the wireless router and the microwave oven.

Does this mean radio waves with high enough amplitude can have the same 'heating' effect as a microwave oven?

 

[PeterDonis]

So what happened to the amplitude?

What happened to it where? As I said, the formula you quoted has nothing to do with either wireless routers or microwave ovens; it's a quantum formula for the energy of a single photon. If you want formulas relevant for wireless routers or microwave ovens, you need to look for classical EM formulas for things like the energy flux or radiated power of a radiation source.

 

[PeterDonis]

Does this mean radio waves with high enough amplitude can have the same 'heating' effect as a microwave oven?

Radio waves in the same frequency range, yes. But wireless routers are physically incapable of putting out anything like that kind of amplitude of radiation.

 

[kevinisfrom]

Radio waves in the same frequency range, yes. But wireless routers are physically incapable of putting out anything like that kind of amplitude of radiation.

What about waves 1 m in wavelenght? With high enough amplitude, does it have the same heating effect as a microwave? According to what you said, with high enough amplitude, the power also increases. Hypothenically, does this mean any wavelenght of light can produce the same heating effect as a microwave oven if you just tune the amplitude of its wavelenght?

 

[kevinisfrom]

What happened to it where? As I said, the formula you quoted has nothing to do with either wireless routers or microwave ovens; it's a quantum formula for the energy of a single photon. If you want formulas relevant for wireless routers or microwave ovens, you need to look for classical EM formulas for things like the energy flux or radiated power of a radiation source.

How then does a photon or particle have a frequency without an amplitude?

 

[sysprog]

The amplitude of radiation being put out by a microwave oven is orders of magnitude higher than for a wireless router.

I wonder how fast the FCC would get me for running my wifi signal through a 1500W linear amplifier.

 

[f95toli]

What about waves 1 m in wavelenght? With high enough amplitude, does it have the same heating effect as a microwave? According to what you said, with high enough amplitude, the power also increases. Hypothenically, does this mean any wavelenght of light can produce the same heating effect as a microwave oven if you just tune the amplitude of its wavelenght?

yes, assuming that what you are trying to heat will absorb the radiation. It turns out that the things we typically try to heat in a microwave oven (food) are good absorbers of microwave. This is largely because the food contains water. You might have noticed that if you heat food in a plastic container the food gets hot, but the plastic does not (expect from contact with the food).

 

[f95toli]

How then does a photon or particle have a frequency without an amplitude?

Photons do not have amplitude. A somewhat correct way of thinking about this is that the amplitude is related to to average NUMBER of photons: the more photons you hit your target with, the hotter it will get.
This is of course because each photon carries an energy hf, more photons-> hotter food.

Note that I said somewhat correct, this is NOT actually correct in the general case. "Properly" describing light (or microwaves) using QM is actually very hard, and there are no simply explanations. For something like a microwave oven you are much better off using classical electromagnetics.

 

[PeterDonis]

does this mean any wavelenght of light can produce the same heating effect as a microwave oven

No. The heating effect depends on how well radiation of the wavelength being used is absorbed by the material being heated, as well as on the amplitude of the radiation.

 

[Vanadium 50]

I wonder how fast the FCC would get me for running my wifi signal through a 1500W linear amplifier.

Not as fast as it would take for your innards to cook!

 

[etotheipi]

If you are interested in how a microwave works classically then you can look into dielectric heating. I think this paper gives a good overview, they state that the microwave power absorbed by a dielectric per unit volume goes as $$P/V = \omega \epsilon_0 \epsilon_2 E_{eff}^2$$(N.B. that the $\epsilon_2$ they use here is the imaginary part of the complex permittivity, $\epsilon = \epsilon_1 + \epsilon_2 i$, I think Wikipedia uses a different notation)

You can see that the power absorbed varies with the amplitude of the electric field (as well as the absorption properties of the material - you can check up some values for water). Then it is maybe less mysterious why you are not cooked by your router

 

[kevinisfrom]

absorption properties of the material

Is absorption of a material related to attenuation of the wave?

 

[etotheipi]

Is absorption of a material related to attenuation of the wave?

AFAIK the attenuation pertains to the conductivity of the dielectric. The complex part of the permittivity is the one that depends on the conductivity, so only in a perfect dielectric will there be no attenuation.

 

[kevinisfrom]

AFAIK the attenuation pertains to the conductivity of the dielectric. The complex part of the permittivity is the one that depends on the conductivity, so only in a perfect dielectric will there be no attenuation.

How would the microwave change after it is absorbed in a material? I was thinking that the amplitude would be attenuated and that's where the transfer of energy occurs, but assuming a perfect dielectric, can you have absorption but no attenuation?

 

[hutchphd]

The microwave energy is initially absorbed by making water molecules rotate. This energy rapidly spreads to other other motions in the sample (which is what we call heating). The microwave amplitude lessens typically exponentially with penetration depth.
There are physical systems that exhibit attenuation without absorption but that is not relevant to this discussion.

 

[DaveE]

How then does a photon or particle have a frequency without an amplitude?

If you want to stick with the photon model, then you can think of amplitude (like power) as the number of photons per second that interact with the target. If I may be crude, it's either waves or bullets of light, in the high school physics paradigms.

 

[kevinisfrom]

The microwave energy is initially absorbed by making water molecules rotate. This energy rapidly spreads to other other motions in the sample (which is what we call heating). The microwave amplitude lessens typically exponentially with penetration depth.
There are physical systems that exhibit attenuation without absorption but that is not relevant to this discussion.

Is absorption of a material related to attenuation of the wave?

Sounds like adsorption is another way of saying energy transfer to increase molecular vibration through attenuation of the wave. Is this the right idea? Attenuation is a direct result of adsorption?

You mentioned attenuation without absorption. Just out of curiosity, what type of material would this be?

 

[hutchphd]

Absorbtion... Adsorption is something else.

Sounds like adsorption is another way of saying energy transfer to increase molecular vibration through attenuation of the wave. Is this the right idea? Attenuation is a direct result of adsorption?

Yes it can be.

You mentioned attenuation without absorption. Just out of curiosity, what type of material would this be?

I will leave it to you and google and your curiousity.

 

[davenn]

Summary:: I believe most wireless router uses the same frequency as a microwave oven, yet microwave ovens heats up food while routers do not. If an energy of the light/radiation is only dependent on frequency, how is it that microwaves heats up food while wireless routers do not?

I believe most wireless router uses the same frequency as a microwave oven, yet microwave ovens heats up food while routers do not. If an energy of the light/radiation is only dependent on frequency, how is it that microwaves heats up food while wireless routers do not?

ohhh
don't you realize the difference in power levels ?

microwave oven ~ 600 to 1000W
router ~ 10 - 25mW (milliWatt) a tiny fraction of 1W

 

[sophiecentaur]

You mentioned attenuation without absorption. Just out of curiosity, what type of material would this be?

It's not a material (energy has to be conserved and has to go somewhere) . You will notice that @hutchphd used the word 'systems' which involves a structure and not just a material. A partly silvered mirror will let a fraction of the light through and reflect the rest with no transfer to heat. A polariser will pass one polarisation of light or radio waves (reflecting the rest) and attenuating an unpolarised signal to 50% (ideal). Two polarisers, set at a chosen angle to one another can produce any value of attenuation, from 50% up to total attenuation with no heat transfer.

 

[vanhees71]

If you want to stick with the photon model, then you can think of amplitude (like power) as the number of photons per second that interact with the target. If I may be crude, it's either waves or bullets of light, in the high school physics paradigms.

One should not use the photon picture at this level at all. It's very misleading. There's more harm done with this utterly wrong picture of photons being something like massless classical point particles. There's no such thing in nature!

A photon is a very specific state of the free quantized electromagnetic field, a socalled "one-particle Fock state". It's not so easy to create such a state with certainty. Only for some decades the quantum opticians have an efficient way to produce true single-photon states using certain birefringent crystals (usually $\beta$-barium-borate (BBO) crystals):

https://en.wikipedia.org/wiki/Barium_borate

A classical electromagnetic wave is much better described as such within classical electrodynamics. Quantum mechanically it is a coherent state, which has not a well-defined photon number but is a certain kind of superposition of all Fock states with arbitrary photon numbers.

Both, the waves transmitted by a WLAN router as well as those in a microwave oven are classical electromagnetic waves and you don't need any quantum field theory to describe them. Before doing quantum field theory you should learn classical Maxwell theory first. Otherwise you have no chance to really understand what a photon really is. It's NOT a massless classical particle in any sense. It is not even possible to define a position observable for it!

 

[sophiecentaur]

One should not use the photon picture at this level at all. It's very misleading.

You have to blame the 'educators' for this. At School, people are taught that the photon explanation of everything is the 'real' explanation; it represents a 'higher level' of understanding. That's why the word constantly comes up from people with limited knowledge in their attempts to 'help' someone with less knowledge. I can only hope that the level of qualification to teach Physics in UK schools (also the level of those who set the Curriculum) is improving.
Of course, the Corpuscular Theory of Light, which was one of the early models, is very attractive and 'concrete'. That theory is confused with the photon theory. Waves and other continuous phenomena are actually harder to grasp so I am not holding my breath.

 

[Delta2]

I disagree with @vanhees71 on this topic. Not on what exactly a photon is, he is probably right that a photon is not a massless classical point particle ( I don't know enough quantum field theory myself in order to be able to tell what exactly a photon is, I am aware of the fact that we run into deep trouble if we try to define a wave function for a photon, like the same way we define one for say an electron).

But I think we can use the "classical" photon model to explain phenomena like in this thread. I don't see anything wrong with saying that the antenna of the wi-fi router emits less photon per second than the tube of the microwave oven hence though photons are of the same frequency and energy, the power emitted by the wi-fi router is less than the power of the microwave oven. I think that even if I knew the exact math involved -within the framework of advanced QFT-to give an answer on what exactly is the power emitted by the wi-fi router , the simplified intuitive answer would be that the wi-fi router doesn't emit so many photons per second and that's why the power is much less.

Of course the simplified intuitive answer doesn't tell us the exact accurate and whole truth, but I believe for phenomena like in this thread , it is good enough.

 

[sophiecentaur]

Of course the simplified intuitive answer doesn't tell us the exact accurate and whole truth, but I believe for phenomena like in this thread , it is good enough.

That totally makes my point, above. What on Earth is the justification for introducing 'little bullets' into a situation which involves wave behaviour? If you were discussing how much water was arriving in a hole that you were firing a hose at, would you ever decide to talk about the number of molecules of water involved? Where would be the advantage of going microscopic to help ones 'intuition' in that case?
How well did you read through the post from @vanhees71 ? Your intuitive picture has no hope at all of describing how the signal diffracts round and through the media in a building and why certain wavelengths work better than others. Do your 'photons' just bounce around between the walls between the router and the computer?
Pretty well anyone who reads PF will know about the Young's Slits experiment. That most basic of diffraction situations just cannot be described without waves so how will ignoring them help in any discussion of the much more complicated situation of propagation of a signal round a house?
Come to terms with waves and things will start to become much clearer for you.

PS It is very difficult to describe what happens with a wire antenna if you want a photon explanation. You would wait a long time for enough little bullets to hit a thin piece of wire. Radio just wouldn't work if it weren't for the waves.

 

[Delta2]

That totally makes my point, above. What on Earth is the justification for introducing 'little bullets' into a situation which involves wave behaviour? If you were discussing how much water was arriving in a hole that you were firing a hose at, would you ever decide to talk about the number of molecules of water involved? Where would be the advantage of going microscopic to help ones 'intuition' in that case?
How well did you read through the post from @vanhees71 ? Your intuitive picture has no hope at all of describing how the signal diffracts round and through the media in a building and why certain wavelengths work better than others. Do your 'photons' just bounce around between the walls between the router and the computer?
Pretty well anyone who reads PF will know about the Young's Slits experiment. That most basic of diffraction situations just cannot be described without waves so how will ignoring them help in any discussion of the much more complicated situation of propagation of a signal round a house?
Come to terms with waves and things will start to become much clearer for you.

PS It is very difficult to describe what happens with a wire antenna if you want a photon explanation. You would wait a long time for enough little bullets to hit a thin piece of wire. Radio just wouldn't work if it weren't for the waves.

I think the simplified "classical" particle model and the wave model give both good explanations when we dealing with phenomena like this where we try to study the power emitted by an antenna or a tube.

For phenomena like diffraction probably the wave model is much better. I didn't say that the "classical" photon model is good for all cases, just for some cases like in this thread.

 

[Vanadium 50]

Let me point out that the first use of the photon concept (but not the word) was an equation that was not relevant to the problem at hand. Why the devil do people want to reinforce this misunderstanding by dragging in photons?

 

[vanhees71]

The classical particle model almost always leads to qualitatively (!) wrong results. Why should one teach it? For me the most difficult thing in my study of physics concerning concepts was to unlearn the wrong pictures about photons provided by popular-science textbooks. The same is true for the qualitatively (!) wrong picture of the atom the Bohr-Sommerfeld model provides as if the atom were a minature planetary system with point electrons running around point atomic nuclei.

I think you can teach a qualitative photon picture that is at least not as mistaken as the "bullet picture" in the following way. As an example take a polarization filter and start with the model of light provided by classical electrodynamics. You have to explain that electromagnetic fields (including light) are transverse waves (it's sufficient to discuss the plane-wave solutions) and that the polarization filter is a material that absorbs all field components in one polarization direction and let's all field components in the perpendicular direction through. Then you can explain what happens to a general em. wave, which formally leads to a projection operator.

Then you tell the students that the classical picture is not entirely right but that quantum theory tells us that matter can not absorb em. fields of any energy but only in integers of $E=\hbar \omega$. Since you cannot detect any electromagnetic field if it is not absorbed by the detector (that's of course an oversimplification ;-)) this implies that there's a state of the electromagnetic field which corresponds to a plane em. wave with frequency $\omega$ of minimal possible energy $\hbar \omega$, and there cannot be an electromagnetic-field state with a lower energy. This is what can be done on a qualitative level to make photons plausible.

It's of course not fully accurate and now you have to also explain the much less intuitive concept of probabilities. This is also not too difficult at this level: if you have a polarization filter and electromagnetic waves linearly polarized at an angle $\varphi$ relative to the orientation of the filter. The intensity $I$ behind the filter is given by $I/I_0=\cos^2 \varphi$.

If you now use this setup with a single-photon source (note that it's now plausible that photons are wave-like entities with a polarization as transverse electromagnetic waves naturally have; there's no problem to explain these wave-like properties of photons as in the erratic bullet model!) you have the following phenomenon. For a single photon $I/I_0=\cos^2 \varphi$ cannot be true anymore! What happens for a true single-photon state is that this photon is either completely absorbed or it goes completely through. All that this model can then predict is the probability for such a photon coming through being $\cos^2 \varphi$.

You can drive this qualitative model quite far, providing a simple example of a two-dimensional quantum system for polarization states of photons and develop the complete concept of quantum theory (2D Hilbert space description, observables, probabilities and Born's rule, and all that).

 

[sophiecentaur]

I think the simplified "classical" particle model and the wave model give both good explanations when we dealing with phenomena like this where we try to study the power emitted by an antenna or a tube.

Hmm. How does the little bullet model (and it really is no more than that) deal with how an antenna interacts with space or, indeed, how a 'tube' interacts with a circuit and, then with an antenna? I have read a fair few books on EM and circuits but I've never found one which tried to explain things using photons. There must be a good reason for that. Can you cite any reasonable source that uses the photon model?

Why the devil do people want to reinforce this misunderstanding by dragging in photons?

I can't think of one - except that waves are 'just too hard'. Strangely, the cleverest brains around shy away from using photons when discussing EM theory - that rather implies it must be the photon model that's actually 'just too hard'. Now, which side of the fence should we choose?

 

[vanhees71]

But why are waves considered "just too hard"? You have plenty clearly comprehensible around you (like water waves on a pond). You can do experiments with a slinky which is fun etc. etc. I think it's not difficult at all to introduce waves and even tell about the wave equation in the final high-school years.

For a very good book about the history of the photon concept, see

K. Hentschel, Photons - the history and mental models of light quanta, Springer (2018)

 

[etotheipi]

But why are waves considered "just too hard"? You have plenty clearly comprehensible around you (like water waves on a pond). You can do experiments with a slinky which is fun etc. etc. I think it's not difficult at all to introduce waves and even tell about the wave equation in the final high-school years.

A very common high-school problem is to estimate the thrust on a solar sail, and it's usually solved at that level by using the intensity on the sail to determine the rate of photon collisions (with each photon being of a known frequency, or from a known distribution of frequencies) to find the rate of change of momentum.

That's arguably easier to introduce than working out the radiation pressure using the Poynting vector (which requires a bit more E&M background), since by that point the idea of momentum is quite familiar to students. It is made worse by the fact that nowadays, in the British A Level at least, basically no electromagnetism is covered. Every exam question can be answered by copying down the formula booklet, which itself only goes as far as Faraday's law of induction.

Maybe the British system has it backwards, they should instead do more E&M and leave photons to a proper QM or QFT course at university? I think I agree yes