<p>The statement of the title is not correct, however I think just a few terms get incorrectly used.</p>
<p>In general it's simply geometry or simple physics: the <a href="https://en.wikipedia.org/wiki/Inverse-square_law" rel="nofollow noreferrer">inverse square law</a>:</p>
<p>You have a star with a certain <a href="https://en.wikipedia.org/wiki/Luminosity" rel="nofollow noreferrer">luminosity</a> (thus overall energy output), let's call it <span class="math-container">$L$</span>. At a distance <span class="math-container">$r$</span> this gives an illumination or <a href="https://en.wikipedia.org/wiki/Irradiance" rel="nofollow noreferrer">irradiance</a> of x W/m² as the total light created by the source is distributed over a sphere of radius <span class="math-container">$r$</span>. The sphere has an area of <span class="math-container">$A(r)=4\pi r^2$</span>. Now, if you double the distance <span class="math-container">$r$</span>, the area to illuminate the whole sphere is now <span class="math-container">$A(2r) = 4\pi (2r)^2 = 4\pi \cdot 4 \cdot r^2$</span>, thus four times the area, thus you get only 1/4 of the light per unit area. The distance doubled, the brightness decreased by a factor of four.</p>
<p>If you now talk about <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="nofollow noreferrer">(apparent) magnitude</a>, it's different as magnitude is the <em>negative logarithm</em> of brightness. So you double distance, you simply add a certain fixed amount. Due to how the apparent magnitude is defined (5 magnitudes represent a factor of 100 in brightness, <span class="math-container">$m - m_{ref} = -2.5\log_{10}(I/I_{ref})$</span>, a doubling of distance of a star increases its magnitude linearily by about 1.50 magnitudes.</p>
<p>It's very hard to define &quot;ease of visibility&quot;. So anything which has a magnitude less than 6 can be visible, if you have normal eye sight and a dark night and sky. As our eyes (and all our senses) also work logarithmically, if the apparent magnitude of an object decreases by 1 (or 1.5), it gets quite much better visible already. Whether that subjectively qualifies as four times as bright ... I cannot say and probably depends on subjective perception and time and circumstance as well: sensitivity of human eyes also depends on the surrounding brightness, adaption time, etc...</p>

CC BY-SA 4.0
2021-11-05T10:10:15.420Z
Thx for the "inverse square law" link. It was really interesting and I especially liked the chart. // Actually I don't bother too much about labels, what I want to know is if a star is four times easier to see in the night sky if it just half the distance away. Is it?
Constantthin
CC BY-SA 4.0
2021-11-05T10:20:02.477Z
That's quite a bit different question. Yet I amended a closing paragraph
planetmaker
CC BY-SA 4.0
2021-11-05T10:26:09.153Z
How do you know that the star "appears 72% closer than before"? What observation is that based on? I have no perception of distance of stars at all - or light sources. I can only judge them, when I see their motion with respect to anything I can relate to.
planetmaker
CC BY-SA 4.0
2021-11-05T10:31:11.360Z
I don't understand the hypothetical situation. Elaborate. What circumstance would put me in the situation that it would appear closer? Or are you *asking* what circumstance would make *me* assume it is 72% closer (thus at 1/4 distance)? Hint: I wouldn't. I would first assume a brightness change and not a distance change.
planetmaker
CC BY-SA 4.0
2021-11-05T15:26:08.823Z
Let's rephrase this: (Astronomical) distance is never visible. It is always an inferred property - from brightness, from knowledge about the type of object, from spectral analysis about its movement in line-of-sight, or from positional comparison over an extended period of time. But there is no-way I can *see* its distance change. I can only derive a distance change from analysing data I (or other people) gathered. And brightness alone NEVER does the trick - at least I need to know what I look at AND know that this type of object has a fixed absolute brightness. Few objects do.
planetmaker
CC BY-SA 4.0
2021-11-05T17:18:06.727Z
Given your comment, This seems to be a case of : ask more precisely. Maybe in a new question, explain your problem in detail and with context. Then explain your attempts to solve it in detail and what you got and why you think it is not good (enough). In particular also state your assumptions you make for your solution, don't generalise it unnecessarily. Then answers more likely help you along.
planetmaker
CC BY-SA 4.0
2021-11-06T12:09:55.503Z
@constantthin planets are more complicated with brightness as one possibly has to consider the phase function,especially for the inner planets
planetmaker
CC BY-SA 4.0
2021-11-05T10:05:38.740Z
2021-11-05T12:00:18.580Z
2021-11-05T12:00:18.580Z
planetmaker
<p><a href="http://curious.astro.cornell.edu/our-solar-system/82-the-universe/stars-and-star-clusters/measuring-the-stars/392-how-does-the-brightness-of-a-star-depend-on-its-distance-from-us-beginner" rel="nofollow noreferrer">This</a> website explains:</p>
<blockquote>
<p>The apparent brightness of a star is proportional to 1 divided by its distance squared. That is, if you took a star and moved it twice as far away, it would appear 1/4 as bright; if you moved it four times the distance, it would appear 1/16 as bright.</p>
</blockquote>
<p>It then goes about explaining how this works.</p>
<p>The problem I have is that I don't understand the reasoning.
Thus: in simpler terms than this website used, how does this work?</p>

CC BY-SA 4.0
2021-11-05T08:46:42.787Z
They don't. Your title is an incorrect statement. Perhaps you are talking about brightness?
ProfRob
CC BY-SA 4.0
2021-11-05T09:16:39.193Z
@ProfRob. Aha, there is a difference between apparent  magnitude and apparent brightness. But if so, what exactly is the difference between the two?
Constantthin
CC BY-SA 4.0
2021-11-05T09:20:11.323Z
Although this website implies that the two are the same:  https://astronomy.swin.edu.au/cosmos/a/Apparent+Magnitude
Constantthin
<star><distances><apparent-magnitude>
Why does apparent magnitudes get increased fourfold when distances get halved, instead of simply getting doubled?
Constantthin
# Why does apparent magnitudes get increased fourfold when distances get halved, instead of simply getting doubled?

(Asked by: username_0 on 2021-11-05)

[This](http://curious.astro.cornell.edu/our-solar-system/82-the-universe/stars-and-star-clusters/measuring-the-stars/392-how-does-the-brightness-of-a-star-depend-on-its-distance-from-us-beginner) website explains:

> The apparent brightness of a star is proportional to 1 divided by its distance squared. That is, if you took a star and moved it twice as far away, it would appear 1/4 as bright; if you moved it four times the distance, it would appear 1/16 as bright.

It then goes about explaining how this works.

The problem I have is that I don't understand the reasoning.
Thus: in simpler terms than this website used, how does this work?

**Tags:** star, distances, apparent-magnitude

> **Comments:**
> **username_1**: They don't. Your title is an incorrect statement. Perhaps you are talking about brightness?  
> **username_0**: @username_1. Aha, there is a difference between apparent  magnitude and apparent brightness. But if so, what exactly is the difference between the two?  
> **username_0**: Although this website implies that the two are the same:  https://astronomy.swin.edu.au/cosmos/a/Apparent+Magnitude  

---

## 1 answers

### Answer by username_2 on 2021-11-05. Score: 4

The statement of the title is not correct, however I think just a few terms get incorrectly used.

In general it's simply geometry or simple physics: the [inverse square law](https://en.wikipedia.org/wiki/Inverse-square_law):

You have a star with a certain [luminosity](https://en.wikipedia.org/wiki/Luminosity) (thus overall energy output), let's call it $L$. At a distance $r$ this gives an illumination or [irradiance](https://en.wikipedia.org/wiki/Irradiance) of x W/m² as the total light created by the source is distributed over a sphere of radius $r$. The sphere has an area of $A(r)=4\pi r^2$. Now, if you double the distance $r$, the area to illuminate the whole sphere is now $A(2r) = 4\pi (2r)^2 = 4\pi \cdot 4 \cdot r^2$, thus four times the area, thus you get only 1/4 of the light per unit area. The distance doubled, the brightness decreased by a factor of four.

If you now talk about [(apparent) magnitude](https://en.wikipedia.org/wiki/Apparent_magnitude), it's different as magnitude is the *negative logarithm* of brightness. So you double distance, you simply add a certain fixed amount. Due to how the apparent magnitude is defined (5 magnitudes represent a factor of 100 in brightness, $m - m\_{ref} = -2.5\log\_{10}(I/I\_{ref})$, a doubling of distance of a star increases its magnitude linearily by about 1.50 magnitudes.

It's very hard to define "ease of visibility". So anything which has a magnitude less than 6 can be visible, if you have normal eye sight and a dark night and sky. As our eyes (and all our senses) also work logarithmically, if the apparent magnitude of an object decreases by 1 (or 1.5), it gets quite much better visible already. Whether that subjectively qualifies as four times as bright ... I cannot say and probably depends on subjective perception and time and circumstance as well: sensitivity of human eyes also depends on the surrounding brightness, adaption time, etc...

> **Comments:**
> **username_0**: Thx for the "inverse square law" link. It was really interesting and I especially liked the chart. // Actually I don't bother too much about labels, what I want to know is if a star is four times easier to see in the night sky if it just half the distance away. Is it?  
> **username_2**: That's quite a bit different question. Yet I amended a closing paragraph  
> **username_2**: How do you know that the star "appears 72% closer than before"? What observation is that based on? I have no perception of distance of stars at all - or light sources. I can only judge them, when I see their motion with respect to anything I can relate to.  
> **username_2**: I don't understand the hypothetical situation. Elaborate. What circumstance would put me in the situation that it would appear closer? Or are you *asking* what circumstance would make *me* assume it is 72% closer (thus at 1/4 distance)? Hint: I wouldn't. I would first assume a brightness change and not a distance change.  
> **username_2**: Let's rephrase this: (Astronomical) distance is never visible. It is always an inferred property - from brightness, from knowledge about the type of object, from spectral analysis about its movement in line-of-sight, or from positional comparison over an extended period of time. But there is no-way I can *see* its distance change. I can only derive a distance change from analysing data I (or other people) gathered. And brightness alone NEVER does the trick - at least I need to know what I look at AND know that this type of object has a fixed absolute brightness. Few objects do.  
> **username_2**: Given your comment, This seems to be a case of : ask more precisely. Maybe in a new question, explain your problem in detail and with context. Then explain your attempts to solve it in detail and what you got and why you think it is not good (enough). In particular also state your assumptions you make for your solution, don't generalise it unnecessarily. Then answers more likely help you along.  
> **username_2**: @constantthin planets are more complicated with brightness as one possibly has to consider the phase function,especially for the inner planets  

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