19
$\begingroup$

The planck length is defined as $l_P = \sqrt{\frac{\hbar G}{c^3}}$. So it is a combination of the constants $c, h, G$ which I believe are all Lorentz invariants. So I think the Planck length should also be a Lorentz invariant! But there seem to be some confusion about that, see e.g. the following paper Magueijo 2001: Lorentz invariance with an invariant energy scale:

The combination of gravity $G$, the quantum $h$ and relativity $c$ gives rise to the Planck length, $l_p$ or its inverse, the Planck energy $E_p$ . These scales mark thresholds beyond which the old description of spacetime breaks down and qualitatively new phenomena are expected to appear. ... This gives rise immediately to a simple question: in whose reference frame are $l_P$ and $E_P$ the thresholds for new phenomena?

But if $l_P$ is a Lorentz invariant their is no question about that. $l_P$ is the same in all reference frames! Another confusing issue is that the Planck mass (from which the Planck length is derived) is often derived by setting equal the Compton length $\lambda_C = \frac{h}{m_0 c}$ ( a Lorentz invariant 4-length) and the Schwarzschild length $r_{s} = \frac{2Gm}{c^2}$ (which I believe is not a Lorentz invariant, since in the derivation of the Schwarzschild metric it is assumed to be a 3-length, measuring a space distance). But since Compton wavelength and Schwarzschild radius are not lengths of the same kind I think such a derivation is not valid. So my question is:

Is the Planck length a Lorentz invariant and if so, how to derive it then without using the Compton wave length and the Schwarzschild radius ?

Improve this question
$\endgroup$
  • 6
    $\begingroup$ This question doesn't make sense because constants are trivially constant by definition. Invariants on the other hand are non-trivial objects that transform as scalars under the given group. E.g. length of a vector is invariant with respect to rotations, mass is Lorentz invariant because it is a norm of energy-momentum vector, etc. $\endgroup$ – Marek Jan 28 '11 at 20:26
  • 1
    $\begingroup$ But why are papers written about that question? The second author on the paper I'm citing is actually Lee Smolin, maybe you heard about him? Also it is clear that $c$ is Lorentz invariant because this was measured, but what about $h$ and $G$? Is there an experiment showing the Lorentz Invarianz of these two values? $\endgroup$ – asmaier Jan 28 '11 at 20:39
  • 5
    $\begingroup$ Dear Marek, +1 point for you, good for you, and it will be harder for me to jump ahead of you in the table. ;-) @Asmaier, you should perhaps avoid papers by that author unless your goal is to unlearn all of physics. Constants, like $\pi$ or $e$ or $\hbar,c,G$ and their powers and products, are invariant under everything. What is disputable is whether "the length of a particular object" is Lorentz-invariant. Well, as Vladimir wrote, the transverse distances are but the longitudinal are not: the latter are Lorentz-contracted. $\endgroup$ – Luboš Motl Jan 28 '11 at 20:48
  • 1
    $\begingroup$ When one says that "c is Lorentz invariant" one is actually making a statement about particular events: a pulse of light leaves a lighthouse at event A and hits a boat at event B. All inertial observers measure the same speed of light based on their observations of these events. What events are you talking about when you ask whether h or G are Lorentz invariant? $\endgroup$ – Greg P Jan 28 '11 at 20:50
  • 2
    $\begingroup$ It is neither, until it is measured. This is a subject of research and speculation. There are proposed theories that make $l_P$ lorentz invariant, and there are theories where $l_P$ is only significant as a proper length. Nobody can answer which is right. $\endgroup$ – Malabarba Jan 28 '11 at 20:50
8
$\begingroup$

A possible answer to the last part of the question: the article Six Easy Roads to the Planck Scale, Adler, Am. J. Phys., 78, 925 (2010) contains multiple "derivations" that you might (or might not) find more satisfactory than the one you mention.

As far as the rest of the question is concerned, others have made the most relevant points. I think a fair summary of what Magueijo is getting at is something like the following:

One frequently hears that "interesting new physics" happens when some length $l$ is less than the Planck length. The Planck length is manifestly Lorentz invariant. The other length $l$, if it is the physical length of some object, is manifestly not Lorentz invariant. What meaning, then, can one assign to such statements?

It seems to me that reasonable people can differ over whether this is an interesting question. I don't find it manifestly insane, myself.

Improve this answer
$\endgroup$
  • $\begingroup$ The Planck length must be talked about relative to some fixed reference frame. All observers in that reference frame agree what the planck length is. $\endgroup$ – Jerry Schirmer Jan 28 '11 at 23:56
  • $\begingroup$ So, should we rephrase that "in any reference frame, interesting new physics happens for any length smaller than Planck length"? Not that I buy it -to me, it has always been Planck area-, but it could fit the requisite of meaningfulness. $\endgroup$ – arivero Jan 29 '11 at 0:18
  • 4
    $\begingroup$ What I don’t get is: why would the length l be the physical length of some object? Forget quantum gravity and Planck scale physics and all those mysterious things. People also say that interesting things happen when you reach physical scales like the electron mass or the QCD scale. In all these cases it is completely obvious they refer to some Lorentz invariant scale. Similarly, any effect indicating something interesting happening at the Planck scale can be phrased in a Lorentz invariant way, even if people don’t always bother to make that explicit. $\endgroup$ – user566 Jan 29 '11 at 1:39
  • 2
    $\begingroup$ I know of no single indication that large quantum gravity effect happens when the coordinate length of some object becomes small, the same way that protons don’t suddenly fall apart when viewed from another inertial frame. $\endgroup$ – user566 Jan 29 '11 at 1:40
  • 1
    $\begingroup$ I had a look at the cited paper. All derivations for the Planck length are done using non-relativistic QM and Newtonian Gravity, this means in the limit v<<c. The author doesn't seem to care about that he often compares the Lorentz invariant Planck length with Lorentz non-invariant 3-lengths of some objects. So I think the statement, that new physics should be happening below some Planck scale is not a Lorentz invariant statement. And I doubt that it's a good idea to use Lorentz non-invariant statements like that as a guideline on the way to a theory of Quantum Gravity. $\endgroup$ – asmaier Feb 1 '11 at 23:10
5
$\begingroup$

I don't know if anyone is still watching this thread, but anyway, the 2001 paper referred to by the OP describes an idea called doubly special relativity (DSR). There is a WP article on it, which provides a more current view. Basically my impression is that DSR didn't work out well, and nobody, including Magueijo and Smolin, is really working on it anymore.

For an answer to this question in the context of loop quantum gravity, see Rovelli and Speziale, "Reconcile Planck-scale discreteness and the Lorentz-Fitzgerald contraction," http://arxiv.org/abs/gr-qc/0205108 .

The OP asked the question in a somewhat naive way, but that doesn't mean that the whole issue is trivial. In SR, we have a constant called c. It's constant by definition. But that doesn't mean that it's a trivial statement that when an observer sees a particle as having a velocity equal to c, that fact can be Lorentz invariant.

Improve this answer
$\endgroup$
3
$\begingroup$

To answer this question one should first define the Planck scale operatively, i.e. to define how different Lorentz observers would measure it in (perhaps thought) experiments.

Usually one defines coupling constants in terms of local, (quasi) static experiments that each observer can perform in its rest frame. Thus G, for example, is measured in such experiment and therefore is, by definition, Lorentz invariant, in spite of being a dimensionful quantity.

c on the other hand clearly cannot be measured that way, but it is not a coupling constant, but a dimensionful observer-independent scale used to define what Lorentz transformations are (Lorentz transformations are such linear spacetime transformations that allow for the presence of an invariant velocity scale c).

Improve this answer
$\endgroup$
  • $\begingroup$ There's been a recent paper on this on hep-th by Roberto Percacci et al. $\endgroup$ – WIMP Jul 20 '11 at 9:01
1
$\begingroup$

When it is transverse, it is Lorentz invariant. When it is longitudinal - it is not ;-).

Improve this answer
$\endgroup$
  • $\begingroup$ Does this hold when the velocity difference between two frames of reference is non-linear, say, with angular components? $\endgroup$ – CommaToast Mar 31 '15 at 16:40
1
$\begingroup$

A scalar is invariant if it has the same value when measured from any reference frame--the speed of light c is the most famous invariant scalar.

A scalar is a universal physical constant when it serves as a proportionality constant in a physical formula---e.g. Gn,Newton's gravitational constant,is the well-known proportionality constant in Newton's gravitational attraction force equation between two masses M1 and M2 separated by a distance R.

If Gn is measured in the rest frame of reference, M1, M2 and R have certain "rest" values sometimes called proper values and we get a certain "rest" value for Gn However, if we measure Gn from a moving frame of reference, then M1 and M2 have to be multiplied by the relativistic "gamma" and R*2 should be divided by "gamma"*2, which means that a factor of "gamma"**4 (fourth power) should be added to the right hand side of the equation as compared to the "rest"equation.

Hence it is obvious that the new Gn as measured from a moving frame is not necessarily equal to the "rest" Gn value.

This means Gn (or in short G) is not necessarily invariant in the sense mentioned above The issue for Planck's constant (E=hv) is more interesting. Einstein, in his first famous paper on relativity "the electrodynamics of moving bodies..." proves that h IS INVARIANT!!!

If we put everything together, it is not obvious that lp (planck's lenth) is Lorenz invariant (or, in short, invariant).

There is no reason why lp should be the minimum length in physics.

It is easy to show that extrema values of any physical quantities (minimum or maximum) are invariant and vice-versa.

Improve this answer
$\endgroup$
0
$\begingroup$

This is a funny question, and Marek does point out that a constant is simply a constant.

The Planck length is a length because it is the square root of the gravitational constant, G times some units that are units in naturalized units. The speed of light is unity and the Planck constant is momentum-length, which are reciprocals of each other. The gravitational constant in naturalized units is [Area], which connects up with the entropy per unit area of an event horizon.

The Planck area squared $\ell_p^2~=~G\hbar/c^3$ is a unit of a black hole event horizon, and for $n$ of these unit areas $A~=~n\ell$ this is associated with units of action, $$ A~=~n\hbar(G/c^3). $$ We may also write $n$ = $M/m_p$ for $m_p~=~\sqrt{\hbar c/G}$ the Planck mass. The area is also proportional to the entropy $S~=~kA/4\ell^2$ and so the event horizon is also a measure of the number of degrees of freedom unavailable for observation.

These units are all expressed according to black hole horizons. The Planck area is computed from the area of a de Broglie wave that is equal to its horizon area. So these are expressed according to invariant (horizons), and are thus invarant.

Improve this answer
$\endgroup$
  • $\begingroup$ G is a coupling constant and can run w/ energy as all other coupling constants. $\endgroup$ – WIMP Jul 20 '11 at 8:58
  • $\begingroup$ I don´t know what you mean by coupling constant and that it can run with energy, as all other coupling constants. Universal constants are universal constants, we were comment at my degree about renormalization and not constant values, but they were apparent and useful for the calculations, not really variant $\endgroup$ – Raúl Aparicio Bustillo Mar 10 '20 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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