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Light's intensity increases by the square?

  1. Aug 23, 2013 #1
    A reading from e=mc2 by David Bodanis,

    The big question though is why? Why is squaring the velocity of what you measure such an accurate way to describe what happens in nature?

    One reason is that the very geometry of our world often produces squared numbers. When you move twice as close toward a reading lamp, the light on the page you're reading doesn't simply get twice as strong. The light's intensity increases four times.

    When you're at the outer distance, the light from the lamp is spread over a large area. When you go closer, that same amount of light gets concentrated on a much smaller area.


    That's a wonderful explanation David, but we don't live in a 2D world. How would this work in our 3D world?

    Well, consider a cube. Now imagine a light source at the center. If you're at 4, what will be the volume?

    4 x 4 x 4 = 64 cube

    Now move twice as close towards the center. What will be the volume at 2?

    2 x 2 x 2 = 8 cube

    When you moved from 4 to 2, the volume didn't cut in half. It decreased by a square value. Now you can understand why light's intensity increases by squares.

    But my logic can't be correct. Let me show you why.

    The sole purpose of the above example was to show you that when you double the distance, you don't double the volume, you square it!

    Now move back to 4. Let's double the distance to 8. What will be the volume at 8?

    8 x 8 x 8 = 512

    8 is double the distance from 4. Is 64 squared equal to 512? No... So do you think that light will increase by a square value? No... So what gives?

    I need help:smile:
     
  2. jcsd
  3. Aug 23, 2013 #2

    Nugatory

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    Just fine, because in a 3D world surfaces are two-dimensional, and the intensity of light is a measure of how much light is hitting a surface. If we lived in a 2D world, the intensity of light would go as the first power of the distance.
     
  4. Aug 23, 2013 #3
    I'm picturing this as how much light is lost over space.

    There must be a proportionality of the intensity of light over space traveled.

    If you're standing on the outer edges of the 4 x 4 x 4 cube, you'll get X intensity of light. Now double your distance to the 2nd integer & you'll be standing on the edges of the 2 x 2 x 2 cube. Now will the intensity of light increase by half or by the square? The square. Why? Because you have not cut the area in half, you've square rooted it, so you square the light to get your intensity.

    It's pretty straight forward. You went from 64 cube down to 8 cube. You've square rooted your volume so you must square the light to get your intensity.

    BUT

    Get on the 8th integer & you'll be standing on the edges of the 8 x 8 x 8 cube, & you'll get X intensity of light. Now double your distance to the 4th integer & you'll be standing on the edges of the 4 x 4 x 4 cube. Now will the intensity of light increase by half or by the square? The square. Why? Because you have not cut the area in half, you've square rooted it, so you square the light to get your intensity.

    WRONG!

    8 x 8 x 8 = 512
    4 x 4 x 4 = 64

    When you go from 512 cube down to 64 cube, you're not square rooting your volume. So why would you keep squaring the light to get the intensity???

    Maybe I should just ask this,

    2 x 2 x 2 = 8 cube
    4 x 4 x 4 = 64 cube

    See when you double your distance from edges of 2 x 2 x 2 to the edges of 4 x 4 x 4 cube, you don't double your volume, you square it. So lets hold onto that & continue. Never mind, that's just a bunch of junk. If you square 64, that isn't equal to 8 x 8 x 8.

    8 x 8 x 8 is 4096. See it didn't square like it did when you went from 8 to 64. This is why I don't understand why light STILL continues to square or square root it's value
     
  5. Aug 23, 2013 #4

    jbriggs444

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    Don't square root the intensity of light. Square root the ratio of intensities. That'll give you the corresponding ratio of distances.

    Don't square the distance. Square the ratio of distances. That'll give you the corresponding ratio of intensities.

    None of this has anything to do with e=mc2
     
  6. Aug 23, 2013 #5
    Your way of looking at it is absolutely wrong. Assuming a center point, then you can assume equal amounts of light will hit each wall, or the 6 sides of the cube. Instead of assessing the volume which deals with cubes, you deal with surface areas instead, which deal with squares.

    Even the 3D world deals with surfaces, which are typically applied in 2 dimensions also.

    If you delve into the properties of photons you'll also see that light is not "lost" as it travels through space. Photons will simply travel through space forever unless it strikes matter and is absorbed, hence all the light emitted by the source will strike the surfaces inside the cube.
     
    Last edited: Aug 23, 2013
  7. Aug 23, 2013 #6
    If you're talking about the luminous surface density, it decreases by the inverse square rule (http://en.wikipedia.org/wiki/Inverse-square_law) because we live in a 3D universe

    If you're talking about e=mc^2 and why the square is in there it's simply because that's what SR spits out
     
  8. Aug 24, 2013 #7

    Nugatory

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    Nope. There's no loss at all as light travels through space, it just spreads out.

    You might try thinking in terms of concentric spheres with the light source at the center instead of the cubes you're working with now; the cubes confuse things because the points on the surface of a cube are not at all at the same distance from the center.

    Say you have a light source at the center emitting a given amount L of light per second. In one second, how much light passes through the entire surface of a sphere centered on that light source? Also L (the amount of light leaving the sphere by crossing its surface outwards has to be equal to the amount of light being added by the source). So if the total amount of light passing through the entire surface of the sphere in one second is L, how much light is passing through one square meter of the surface in one second? That's the intensity, and you'll calculate it as ##L/A##, where A is the area of the sphere in square meters. And because ##A=4\pi{r}^2##... There's your inverse square law.
     
  9. Aug 24, 2013 #8
    Another way to think about light traveling through space is the laser. The reason lasers have such a far range as opposed to, say, a flashlight is for the very reason that the light does not disperse outwards as fast because of its focused nature.

    There's obviously still a minuscule amount of photons that disperse over time due to a varying small angles at which they leave the laser/light source. Otherwise, you would not be able to see the laser "beam".
     
  10. Aug 25, 2013 #9

    jbriggs444

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    The reason you can "see a laser beam" is because it is shining through smoke, haze or dust so that a small portion of the beam is scattered widely. That is, of course, the same as the way you "see a flashlight beam".

    If you were seeing the beam because of photons that had emerged from the source at divergent angles then you would have to be looking directly at the source. That is a very bad idea. And you'd be seeing the source, not the beam.
     
  11. Aug 25, 2013 #10
    I shouldn't have used cubes in this demonstration. Light decreases in intensity over space traveled. So depending upon how large your cube is, the light will probably be more intense on one side than the other. :redface:

    I was under the impression that if you 2x the distance, you've ^2 the area. Not always true.

    I understand the ratio now. Each time you 2x the distance, the area increases 4x, therefore the light will be 1/4 in intensity. No ^2 here.

    Spray paint is a good example,

    vV56eQX.jpg

    Further you pull back the can, less intense the color gets ...because the area is increasing. That seems to be the whole logic. But what happens if you pull back the can & the area doesn't increase?

    Like,

    WtV8NGY.jpg

    Imagine the yellow box a source of radiation. Distance X is farther away, but the same area. So light should have the same intensity? If distance X was 100 kilometers away, I wouldn't think so.
     
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