Wireless routers vs. microwave ovens

  • #1

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?

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
PeterDonis
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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.
 
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  • #3
Can you share the formula? I always see E = hv
 
  • #4
PeterDonis
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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.
 
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  • #5
So what happened to the amplitude?
 
  • #6
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?
 
  • #7
PeterDonis
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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.
 
  • #8
PeterDonis
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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.
 
  • #9
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?
 
  • #10
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?
 
  • #11
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PeterDonis said:
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. :oops:
 
  • #12
f95toli
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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).
 
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  • #13
f95toli
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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.
 
  • #14
PeterDonis
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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.
 
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  • #15
Vanadium 50
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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!
 
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  • #16
etotheipi
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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 :smile:
 
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  • #17
absorption properties of the material
Is absorption of a material related to attenuation of the wave?
 
  • #18
etotheipi
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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.
 
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  • #19
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?
 
  • #20
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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.
 
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  • #21
DaveE
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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.
 
  • #22
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?
 
  • #23
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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.
 
  • #24
davenn
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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 realise the difference in power levels ?

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

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