10
$\begingroup$

I've been reading Physics SE answers on Planck units such as this one and this one.

The general picture I get is that much of what is said about the Planck length (and the associated Planck units) is either speculation or outright false. However, one claim that keeps popping up in various forms (including in answers from both of my links) is that we don't know how to describe physics at a scale smaller than the Planck length.

People typically say this in two different ways.

One way is that there's something inherent in quantum gravity theories that makes it impossible to talk about distances lower than the Planck length. Is this true? And if so, what makes us believe this is true? According to wikipedia, the length of strings in string theory are on the order of the Planck length. But I don't know anything about string theory so I don't know the implications of that.

A second way is that the Planck length is the scale at which gravitational effects and quantum effects start being comparable in which case our current theories (quantum physics and general relativity) clash and we don't know how to describe what is happening. Is this true? And if so, what makes us believe this is true?

One argument I've heard for this second interpretation is that the Planck length contains G, c, and h bar, constants from quantum physics and GR, and thus when this is 1 both quantum and relativistic effects are important. However, this argument is incredibly dubious because this exact same argument could be made for a length equal to any constant times the Planck length or for the Planck mass, which is clearly not in any sense a limit. Is there some better more rigorous way to make this argument? Perhaps by looking at some well-known system and showing that GR and quantum effects are comparable exactly at the Planck length?

In sum I'm trying to get a better handle on what the Planck length really means. Any help would be appreciated.

$\endgroup$
5
  • $\begingroup$ No formal theory predicts any of that. They are heuristics ideas when people think, for instance, of what a theory of quantum gravity would look like. But there is not yet a theory of quantum gravity; so it is not clear that such a theory will predict the same as the heuristics. $\endgroup$
    – user126422
    Commented Jan 7, 2017 at 1:46
  • $\begingroup$ Thanks for your comment. If there is no formal proof, then why do people believe the heuristics? Also as a side note, you don't necessarily need a theory of quantum gravity to know where our current theories fail us, right? For example if you're looking at a system where both quantum and GR effects are significant it can be reasonable to say we're not entirely sure we know what's going on. $\endgroup$
    – user35734
    Commented Jan 7, 2017 at 2:26
  • $\begingroup$ Yes, agreed, I believe I was very unclear. $\endgroup$
    – user126422
    Commented Jan 7, 2017 at 3:08
  • $\begingroup$ If I could locate sources that employed mathetmatical formalism, AND also provided enough verbiage for me to get the basics of how they'd describe the formalism aloud, I'd post this as an answer: We need energy to see things so that we can measure them, and there is only so much energy currently accessible to us for that purpose, so the Planck length would be distinguishable from objects currently visible to us only thru magnification energies greater than those that our civilization's chosen to dedicate to the confirmation of hypothetical pseudo-visualizations available more economically. $\endgroup$
    – Edouard
    Commented Feb 16, 2021 at 3:29
  • 1
    $\begingroup$ @Edouard The limiting factor isn't how much energy is accessible, it's how much a probe can have without gravitational capture of the target preventing the intended measurement. The relevance of the Planck scale comes from comparing the Schwarzschild radius to the de Broglie wavelength. $\endgroup$
    – J.G.
    Commented Feb 16, 2021 at 7:53

2 Answers 2

5
$\begingroup$

The Standard Model and General Relativity are both successful in appropriate limits, but they cannot be consistently combined for scales below $\sim\ell_P:=\sqrt{\dfrac{G\hbar}{c^3}}$ for various reasons. (By $\sim$, I mean "give or take a multiplicative constant that's besides the point here and may be hard to compute".) For example, what happens if you try to probe such length scales with a photon? How will its wavelength compare to its Schwarzschild radius?

When you ask about the physical meaning or significance of such short length scales, that's where it gets contentious. I'll try to summarise the range of views on this, but I'll probably fudge or simplify a few details:

  • String theory says spacetime is infinitely divisible, but particles have a size $\sim\ell_P$. They therefore have worldsheets instead of worldlines, which smears Feynman diagram vertices as thus. This smearing removes the troublesome infinities from the treatment of gravity.
  • Loop quantum gravity, in a sense, says the opposite: particles aren't posited to have size, but spacetime is quantised. Particles live at distinct lattice points. The area and volume of an object have operators in the Hilbert space, and these operators have discrete eigenvalues $\sim\ell_P^{2\,\mathrm{or}\,3}$.
  • There have been attempts to combine ST with LQG (motivated by their respective pros and cons and their obtaining similar results from very different precepts, e.g. logarithmic corrections to the Hawking-Bekenstein entropy of black holes), but these are in their infancy. For now, it suffices to say such a union would introduce both deviations from the "point particles in infinitely divisible spacetime" idea that causes SM+GR problems.
  • Another proposal is that $[\hat{x}_\mu,\,\hat{x}_\nu]=i\ell_P^2 \theta_{\mu\nu}$ is a non-vanishing antisymmetric tensor. This is far from developing into a full theory of quantum gravity, but it's an idea that's been explored in such attempts. Just as quantum mechanics says $[\hat{x}_j,\,\hat{p}_k]=i\hbar\delta_{jk}$ without the eigenvalues becoming discrete, the above use of noncommutative geometry requires only that $\sigma_{x_\mu}\sigma_{x_\nu}\ge\frac{1}{2}\ell_P^2|\theta_{\mu\nu}|$, not that eigenvalues of $x_\mu$ cannot differ by arbitrarily small fractions of $\ell_P$.
$\endgroup$
4
  • $\begingroup$ The last point is new to me. Could you guide me to somewhere or could you expand on the point so that I can see more about it? $\endgroup$ Commented Apr 7, 2018 at 14:22
  • $\begingroup$ @YuzurihaInori en.wikipedia.org/wiki/… $\endgroup$
    – J.G.
    Commented Apr 7, 2018 at 14:29
  • $\begingroup$ @J.G. -As you've used formal notation, I'm wondering if you'd mind validating or refuting the comment (under the OP's good question) that I made a few moments ago. (I could probably craft an answer of my own out of Brian Greene's pop-sci books, but I doubt if it would get much accomplished, especially now, when the subject's not too urgent and millions of people still don't have a light bulb in their homes.) $\endgroup$
    – Edouard
    Commented Feb 16, 2021 at 3:37
  • $\begingroup$ (I'm not saying "pseudo"-visualizations to denigrate the processes involved: The adjective in quotes is intended only to point out the fact that visualizations not observationally or experimentally confirmable may be visualizing nothing.) $\endgroup$
    – Edouard
    Commented Feb 16, 2021 at 3:45
0
$\begingroup$

The last point is, in fact, the correct one. Planck length is the natural length scale constructed out of all fundamental constants. The involvement of Newton's and Planck's constant along with speed of light ensure the existence of quantum gravity at that length scale. However, your last argument is not correct. This is because the Planck length itself is not a hard limit after which quantum gravity takes over. The right way to say is that Quantum gravitational effects take over at length scale of the order of Planck length. This could be 3 times the Planck length and so on or more precisely $\mathcal{O}(10^{-33})$. At this length scale, the curvature will be of $\mathcal{O}\left(\frac{1}{l_p^2}\right)$, which will be a singular state and can't be explained by classical general relativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.