<p>The Planck length also still has meaning. <a href="http://www.physlink.com/Education/AskExperts/ae281.cfm" rel="nofollow noreferrer">This source</a> describes it as:</p>

<blockquote>
  <p>The Planck length is the scale at which classical ideas about gravity and space-time cease to be valid, and quantum effects dominate. </p>
</blockquote>

<p>So we may not be sure what happens on such small length scales, but we can be sure that it is <em>not</em> even approximately described by "classical" space-time. Conversely, for larger distances, and absent very large concentrations of energy of some kind, it is reasonable to expect that classical space-time might be a reasonable approximation to the universe.</p>

CC BY-SA 4.0
2018-10-11T14:48:59.000Z
2018-10-11T14:48:59.000Z
Steve Linton
<p>First off, you're much more likely to see parsecs, light years, or astronomical units than Planck lengths in astrophysics.</p>

<p>However, for those who like to work in a <a href="https://en.wikipedia.org/wiki/Natural_units" rel="nofollow noreferrer">system of natural units</a>, the Planck length is <strong>the</strong> obvious choice for the unit of length. This choice has nothing to do with quantizing spacetime. It instead results from a desire to make conversion factors such as the speed of light <span class="math-container">$c$</span>, the reduced Planck constant <span class="math-container">$\hbar$</span>, the Boltzmann constant <span class="math-container">$k_b$</span>, and the gravitational constant <span class="math-container">$G$</span> all have a numerical value of one or some other nice number such as <span class="math-container">$
\frac{1}{4\pi}$</span>.</p>

<p>One way to look at this: Compare the conventional units used in the US and a few other backward countries. These countries have to use <span class="math-container">$F=kma$</span> rather than the <span class="math-container">$F=ma$</span> used in the metric system. The <span class="math-container">$k$</span> is a conversion factor that results from having those conventional units being an inherently inconsistent system. From a modern perspective, force <em>is</em> the product of mass and acceleration; no conversion should be needed.</p>

<p>That said, that the mass-energy equivalence equation expressed in metric units is <span class="math-container">$E^2=\bigl(m c^2\bigr)^2 + \bigl(pc\bigr)^2$</span> is a sign that the metric system, too, is inconsistent. With natural units, this expression simplifies to <span class="math-container">$E^2=m^2+p^2$</span>. Energy, mass, and momentum all have the same units; no conversion is needed.</p>

CC BY-SA 4.0
2018-10-12T12:52:34.120Z
Also, the formula for the Bekenstein-Hawking entropy of a black hole is simple in Planck units.
PM 2Ring
CC BY-SA 4.0
2018-10-17T00:49:05.613Z
What do you mean that the unit systems are " inconsistent" ?
Daddy Kropotkin
CC BY-SA 4.0
2018-10-17T11:21:20.393Z
@N.Steinle - I thought I explained that in the answer. The customary units used in the US are inconsistent with physics, which says that force *is* the product of mass and acceleration. No conversion factor should be needed, but there it is, the $k$ in $F=kma$. Similarly, the metric system is inconsistent with physics, which says that mass *is* a form of energy. Once again, no conversion factor should be needed, but there it is, the $c$ in $E=mc^2$.
David Hammen
CC BY-SA 4.0
2018-10-17T14:41:02.340Z
I hear what you're saying, and I think I really disagree here. You did say exactly that in your answer, but clearly if I'm still disagreeing then you might want to find a different way to explain it.... I don't buy your argument: you seem to be saying that if there exists a conversion factor then the system is "inconsistent with physics." But you really mean that the system of units is inconsistent with our **model** of reality, which is different than what you're saying. You seem to suggest that our models of reality, i.e. $F = ma$ or $E = mc^{2}$, ARE the reality, which is just false.
Daddy Kropotkin
CC BY-SA 4.0
2018-10-17T14:41:38.070Z
The measurement IS the reality, and we convert it to match a model. So, I would argue instead, that the models are inconsistent with the reality of measurement, not the other way around like you suggest.
Daddy Kropotkin
CC BY-SA 4.0
2018-10-18T00:28:26.037Z
@N.Steinle - What measurement? The speed of light is no longer a measured quantity. It is instead a defined constant with an exact numerical value, 299792458 metres per second. The same will happen across the International System of Units (SI) very shortly. A number of constants that are one in a consistent set of units will soon have exact defined values that are not unity. That said, a consistent system of units would beyond useless with regard to commerce, which is the primary reason the SI exists.
David Hammen
CC BY-SA 4.0
2018-10-19T20:58:10.840Z
The speed of light is defined as such because we have measurements to justify such a definition. It is not an arbitrary definition. Without measurements to justify our claims, theory is no more than a religion. But some theorists forget this, and also forget that their models are not the reality. (i'm a theorist btw, i'm not digging on theorists).
Daddy Kropotkin
CC BY-SA 4.0
2018-10-11T14:18:08.373Z
2018-10-12T07:34:11.310Z
2018-10-12T07:34:11.310Z
David Hammen
<p>We are not able to quantize spacetime and the theory of  <strong>loop quantum gravity</strong> has been proved wrong but we still use the Planck length in astrophysics. Why?</p>

CC BY-SA 4.0
2018-10-11T15:01:38.780Z
Also it would be nice to provide a reference for a disproven theory. From your question it is not clear what paper/conference you're referring to. Have all iterations and instances of QLG been proven wrong? The general concept? I'd like to know that, if it really happened.
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-10-11T17:10:58.303Z
I'm voting to close this question as off-topic because this is a Physics question, not astronomy or cosmology
Carl Witthoft
CC BY-SA 4.0
2018-10-12T00:10:38.867Z
Dunno @CarlWitthoft, it looks like an astrophysics question to me, which is definitely *on-topic*. If you don't think the question is useful, *you can downvote it*. Alternatively, if "physics" questions continue to be an issue, it might be worth us revisiting the [meta discussion](https://astronomy.meta.stackexchange.com/questions/450).
Chappo Hasn't Forgotten
CC BY-SA 4.0
2018-10-12T12:50:04.963Z
FWIW, the [Bekenstein-Hawking entropy](https://en.wikipedia.org/wiki/Black_hole#Entropy_and_thermodynamics) of a black hole is A/4, where A is the area of the event horizon expressed in square Planck lengths.
PM 2Ring
<space-time><quantum-mechanics>
Why do we use the Planck length?
Poin
# Why do we use the Planck length?

(Asked by: username_0 on 2018-10-11)

We are not able to quantize spacetime and the theory of **loop quantum gravity** has been proved wrong but we still use the Planck length in astrophysics. Why?

**Tags:** space-time, quantum-mechanics
> **Comments:**
> **username_1**: Also it would be nice to provide a reference for a disproven theory. From your question it is not clear what paper/conference you're referring to. Have all iterations and instances of QLG been proven wrong? The general concept? I'd like to know that, if it really happened.  
> **username_2**: I'm voting to close this question as off-topic because this is a Physics question, not astronomy or cosmology  
> **username_3**: Dunno @CarlWitthoft, it looks like an astrophysics question to me, which is definitely *on-topic*. If you don't think the question is useful, *you can downvote it*. Alternatively, if "physics" questions continue to be an issue, it might be worth us revisiting the [meta discussion](https://astronomy.meta.stackexchange.com/questions/450).  
> **username_4**: FWIW, the [Bekenstein-Hawking entropy](https://en.wikipedia.org/wiki/Black_hole#Entropy_and_thermodynamics) of a black hole is A/4, where A is the area of the event horizon expressed in square Planck lengths.  

---

## 2 answers

### Answer by username_6 on 2018-10-11. Score: 5 | Accepted answer

First off, you're much more likely to see parsecs, light years, or astronomical units than Planck lengths in astrophysics.

However, for those who like to work in a [system of natural units](https://en.wikipedia.org/wiki/Natural_units), the Planck length is **the** obvious choice for the unit of length. This choice has nothing to do with quantizing spacetime. It instead results from a desire to make conversion factors such as the speed of light $c$, the reduced Planck constant $\hbar$, the Boltzmann constant $k\_b$, and the gravitational constant $G$ all have a numerical value of one or some other nice number such as $
\frac{1}{4\pi}$.

One way to look at this: Compare the conventional units used in the US and a few other backward countries. These countries have to use $F=kma$ rather than the $F=ma$ used in the metric system. The $k$ is a conversion factor that results from having those conventional units being an inherently inconsistent system. From a modern perspective, force *is* the product of mass and acceleration; no conversion should be needed.

That said, that the mass-energy equivalence equation expressed in metric units is $E^2=\bigl(m c^2\bigr)^2 + \bigl(pc\bigr)^2$ is a sign that the metric system, too, is inconsistent. With natural units, this expression simplifies to $E^2=m^2+p^2$. Energy, mass, and momentum all have the same units; no conversion is needed.

> **Comments:**
> **username_4**: Also, the formula for the Bekenstein-Hawking entropy of a black hole is simple in Planck units.  
> **username_7**: What do you mean that the unit systems are " inconsistent" ?  
> **username_6**: @N.Steinle - I thought I explained that in the answer. The customary units used in the US are inconsistent with physics, which says that force *is* the product of mass and acceleration. No conversion factor should be needed, but there it is, the $k$ in $F=kma$. Similarly, the metric system is inconsistent with physics, which says that mass *is* a form of energy. Once again, no conversion factor should be needed, but there it is, the $c$ in $E=mc^2$.  
> **username_7**: I hear what you're saying, and I think I really disagree here. You did say exactly that in your answer, but clearly if I'm still disagreeing then you might want to find a different way to explain it.... I don't buy your argument: you seem to be saying that if there exists a conversion factor then the system is "inconsistent with physics." But you really mean that the system of units is inconsistent with our **model** of reality, which is different than what you're saying. You seem to suggest that our models of reality, i.e. $F = ma$ or $E = mc^{2}$, ARE the reality, which is just false.  
> **username_7**: The measurement IS the reality, and we convert it to match a model. So, I would argue instead, that the models are inconsistent with the reality of measurement, not the other way around like you suggest.  
> **username_6**: @N.Steinle - What measurement? The speed of light is no longer a measured quantity. It is instead a defined constant with an exact numerical value, 299792458 metres per second. The same will happen across the International System of Units (SI) very shortly. A number of constants that are one in a consistent set of units will soon have exact defined values that are not unity. That said, a consistent system of units would beyond useless with regard to commerce, which is the primary reason the SI exists.  
> **username_7**: The speed of light is defined as such because we have measurements to justify such a definition. It is not an arbitrary definition. Without measurements to justify our claims, theory is no more than a religion. But some theorists forget this, and also forget that their models are not the reality. (i'm a theorist btw, i'm not digging on theorists).  

---

### Answer by username_5 on 2018-10-11. Score: 3

The Planck length also still has meaning. [This source](http://www.physlink.com/Education/AskExperts/ae281.cfm) describes it as:

> The Planck length is the scale at which classical ideas about gravity and space-time cease to be valid, and quantum effects dominate.

So we may not be sure what happens on such small length scales, but we can be sure that it is *not* even approximately described by "classical" space-time. Conversely, for larger distances, and absent very large concentrations of energy of some kind, it is reasonable to expect that classical space-time might be a reasonable approximation to the universe.


---

