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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?

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    $\begingroup$ 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. $\endgroup$ – AtmosphericPrisonEscape Oct 11 '18 at 15:01
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    $\begingroup$ I'm voting to close this question as off-topic because this is a Physics question, not astronomy or cosmology $\endgroup$ – Carl Witthoft Oct 11 '18 at 17:10
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    $\begingroup$ 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. $\endgroup$ – Chappo Hasn't Forgotten Monica Oct 12 '18 at 0:10
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    $\begingroup$ FWIW, the Bekenstein-Hawking entropy of a black hole is A/4, where A is the area of the event horizon expressed in square Planck lengths. $\endgroup$ – PM 2Ring Oct 12 '18 at 12:50
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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, 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.

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  • $\begingroup$ Also, the formula for the Bekenstein-Hawking entropy of a black hole is simple in Planck units. $\endgroup$ – PM 2Ring Oct 12 '18 at 12:52
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    $\begingroup$ What do you mean that the unit systems are " inconsistent" ? $\endgroup$ – N. Steinle Oct 17 '18 at 0:49
  • $\begingroup$ @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 $E=mc^2$. $\endgroup$ – David Hammen Oct 17 '18 at 11:21
  • $\begingroup$ 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. $\endgroup$ – N. Steinle Oct 17 '18 at 14:41
  • $\begingroup$ 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. $\endgroup$ – N. Steinle Oct 17 '18 at 14:41
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The Planck length also still has meaning. This source 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.

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