4 ms·

Doesn't radio have inverse cube falloff? That's the thing that somewhat worried me about mobile phones, it has inverse cube falloff and has to speak to a tower

Doesn't radio have inverse cube falloff? That's the thing that somewhat worried me about mobile phones, it has inverse cube falloff and has to speak to a tower miles away, so putting it against your brain for hours on end (I don't like calling, so that's not an issue to me) seems like a bad idea.

lobf 8y ago
Why?
pdonis 8y ago
> Doesn't radio have inverse cube falloff? No, inverse square. > has to speak to a tower miles away But the tower has a very sensitive antenna, so your phone doesn't need to put much power into transmitting. The effect of the phone on your brain and body is small compared to ordinary thermal noise at your body's temperature. > putting it against your brain for hours on end If you're going to be talking that much and are worried about this (which I don't think you need to be, but everyone has to make their own judgments), you can use a bluetooth headset. Bluetooth transmit power is an order of magnitude smaller, at least, than cell phone transmit power.
- freyr 8y ago
  Inverse square, in free space.
deleted 8y ago
[deleted]
rdl 8y ago
It's inverse square -- think of it as the surface area of the sphere a specific distance from the source at any given instant. Not the volume of the space between the frontier and the source.
- hyeonwho4 8y ago
  Although the free space attenuation is inverse square, interference effects can increase power attenuation for radio waves in the far field limit. If there is reflection off the ground, the power attenuation is closer to inverse quartic in the far field. https://en.m.wikipedia.org/wiki/Two-ray_ground-reflection_model https://en.m.wikipedia.org/wiki/Two-ray_ground-reflection_mo...
noobermin 8y ago
Geometry in three dimensions requires inverse square. You probably have incident on your body more energy sitting under a light bulb than from a cell phone, not to mention the nuclear fireball in the fucking sky.
- mindslight 8y ago
  Lumping different wavelengths into the general of "radiation" isn't really scientific. And adding more technical-sounding hyperbole to the conversation doesn't actually help.
etimberg 8y ago
Power drops off with the square of distance in free space. To get power of an RF wave you S = E x H [1]. Both the E & H fields decrease with distance. If you double the distance and both E&H decrease by half then |S| decreases by a factor of 4. Interestingly though, in some environments you can actually get better than free space loss. I remember an off-hand comment from a professor in university that in concrete tunnels you can get R^1.6-1.8 because the tunnel acts as a waveguide 1. https://en.wikipedia.org/wiki/Poynting_vector https://en.wikipedia.org/wiki/Poynting_vector
dreamcompiler 8y ago
Doesn't matter whether it's radio or light or sound; if it's a point source (zero-dimensional) in three-dimensional space, it's inverse square falloff. If it's a line source (one dimensional), it's inverse (one over first power of distance) falloff. This is why fluorescent lights are so effective at illuminating large warehouses: At near distances, they are approximately line sources.

lobf 8y ago

Why?

pdonis 8y ago

> Doesn't radio have inverse cube falloff? No, inverse square. > has to speak to a tower miles away But the tower has a very sensitive antenna, so your phone doesn't need to put much power into transmitting. The effect of the phone on your brain and body is small compared to ordinary thermal noise at your body's temperature. > putting it against your brain for hours on end If you're going to be talking that much and are worried about this (which I don't think you need to be, but everyone has to make their own judgments), you can use a bluetooth headset. Bluetooth transmit power is an order of magnitude smaller, at least, than cell phone transmit power.

freyr 8y ago
Inverse square, in free space.

deleted 8y ago

[deleted]

rdl 8y ago

It's inverse square -- think of it as the surface area of the sphere a specific distance from the source at any given instant. Not the volume of the space between the frontier and the source.

hyeonwho4 8y ago
Although the free space attenuation is inverse square, interference effects can increase power attenuation for radio waves in the far field limit. If there is reflection off the ground, the power attenuation is closer to inverse quartic in the far field. https://en.m.wikipedia.org/wiki/Two-ray_ground-reflection_model https://en.m.wikipedia.org/wiki/Two-ray_ground-reflection_mo...

noobermin 8y ago

Geometry in three dimensions requires inverse square. You probably have incident on your body more energy sitting under a light bulb than from a cell phone, not to mention the nuclear fireball in the fucking sky.

mindslight 8y ago
Lumping different wavelengths into the general of "radiation" isn't really scientific. And adding more technical-sounding hyperbole to the conversation doesn't actually help.

etimberg 8y ago

Power drops off with the square of distance in free space. To get power of an RF wave you S = E x H [1]. Both the E & H fields decrease with distance. If you double the distance and both E&H decrease by half then |S| decreases by a factor of 4. Interestingly though, in some environments you can actually get better than free space loss. I remember an off-hand comment from a professor in university that in concrete tunnels you can get R^1.6-1.8 because the tunnel acts as a waveguide 1. https://en.wikipedia.org/wiki/Poynting_vector https://en.wikipedia.org/wiki/Poynting_vector

dreamcompiler 8y ago

Doesn't matter whether it's radio or light or sound; if it's a point source (zero-dimensional) in three-dimensional space, it's inverse square falloff. If it's a line source (one dimensional), it's inverse (one over first power of distance) falloff. This is why fluorescent lights are so effective at illuminating large warehouses: At near distances, they are approximately line sources.