Does weak equivalence break down at the quantum level?

Dec 08, 2009 By Miranda Marquit feature

(PhysOrg.com) -- One of the givens in physics is the weak equivalence principle. This principle has been considered solid since Einstein proposed that it is not possible to detect the difference between uniform acceleration and a uniform static gravitational field. The uniqueness of freefall allows uniform acceleration, even between masses that are different, according to Einstein's postulate in the theory of General Relativity. The weak equivalence principle is well established amongst the science community, but it has yet to be demonstrated completely. This is where Phillippe Bouyer at Laboratoire Charles Fabry de l’Institut d’Optique, Campus Polytechnique in Palaiseau, France, and his colleagues are attempting to go.

“We are looking to see to which extent this principle holds,” Bouyer tells PhysOrg.com. “Does the principle break down at a certain level? And if it does, what is that level? If it breaks down, that opens up a whole range of fundamental questions, such as the existence of new interactions as predicted by many current quantum theories of gravity. We have created a test that we hope will open the field to see whether the weak equivalence principle holds at the quantum level.”

In order to test the weak equivalence principle, Bouyer and his peers Varoquaux, Nyman, Geiger and Cheinet at the Laboratoire Charles Fabry, and Landragin at SYRTE, suggest a technique using a parabolic flight plane outfitted with an atom interferometer. The device would be used to determine whether the acceleration is the same for two different atoms in free fall. Their proposed method is described in New Journal of Physics: “How to estimate the differential acceleration in a two-species atom interferometer to test the equivalence principle.”

“If you want to test the acceleration of atoms on the ground, you have very little time before the atoms hit the floor” Bouyer explains. “In parabolic flights, though, you have up to 20 seconds of free fall to see whether there is a difference in acceleration. When dealing with atoms, this is a rather long time indeed.” The atoms would be captured, laser cooled using well known techniques and then dropped in the “free falling” plane. It would then be possible to precisely measure the acceleration of the two different atoms using atom interferometry. “We plan to use rubidium and potassium,” he says, “since they are very different atoms with a difference in mass that is significant. This way, if they have no difference in acceleration, the equivalence principle is demonstrated. If there is a difference in acceleration, we know that it breaks down at the precision level of our measurement.”

One of the problems, though, is that the parabolic flight plane is very noisy. Bouyer and a team from various scientific institutes in France addressed this problem last year. They tested the operation of a specially designed atom interferometer in a free fall plane (results can be found in The European Physical Journal D), finding that the sensitivity of measurement was enhanced, allowing acceleration in rubidium to be measured. To cancel out the noise, Bouyer and his colleagues worked out a way to extract the acceleration using Bayesian statistical estimation. “Our dedicated statistical analogy allows us to extract the EP signal out of the noise of the environment,” he explains.

For now, the proposed experiment remains just that: proposed. But Bouyer is hopeful. “We are very close to getting two atoms at the same place, and using our interferometer in a state of free fall should enable us to take measurements that may help us demonstrate the viability of the equivalence principle at the quantum level.”

More information:
• Varoquaux, et al, “How to estimate the differential acceleration in a two-species atom interferometer to test the equivalence principle,” (2009). Available online: http://www.iop.org/EJ/abstract/1367-2630/11/11/113010/.

• Stern, et al, “Light-pulse atom interferometry in microgravity,” European Physical Journal D (2009). Available online.

Copyright 2009 PhysOrg.com.
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User comments : 19

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Alexa
1 / 5 (1) Dec 08, 2009
Weak equivalence is indeed violated by Casimir force, which is proportional to crossection of objects, instead of their mass, so that equivalence principle of relativity doesn't apply here - and no large speculations are required about it - question marks the less.

Basically it means, quantum scale begins at Casimir force scale, which roughly corresponds the wavelength of cosmic microwave background, which roughly corresponds the size of human brain waves.
Alexa
2 / 5 (2) Dec 08, 2009
Anyway, it's surprising, how deeply scientists are surprised by fact, pair of dual theories (relativity and quantum mechanics) are inconsistent mutually. Especially if they know already, these theories are giving quite different predictions, concerning energy density of vacuum or cosmological constants. I mean different in one hundred orders of magnitude..

http://en.wikiped...astrophe
KBK
1 / 5 (1) Dec 08, 2009
The whole deal is that it is tied up in the proper 20 equations in 20 unknowns that makes up the full and original Maxwell's equations. Not the shortened Heaviside version. In the original 20 equations, the whole deal was about the spiral motion of the situation, where an infinitesimal amount of time and an infinitesimal difference in final rest position was the whole point of the complex equations. This created the egress or origin point for the complex spiral vectoral origins of the Quanta of TIME. In the original math, the spiral motion created infinitesimally small differences and there was no symmetry. Thus our observed character of time as having a uni-directionality, a permanent past and an open future. Time as flow and a specific temporal quanta associated with the given specific energies in question. All there, plain as day, in the original mathematical works.

Look it up ---if you don't believe me.
OregonWind
5 / 5 (1) Dec 08, 2009
KBK

I don't believe you and where then should I look that up?
barakn
3 / 5 (4) Dec 08, 2009
Why not design this experiment for the International Space Station, which is in perpetual free fall (ignoring occasional station-keeping accelerations) and much less noisy than a jet?
barakn
3.4 / 5 (5) Dec 08, 2009
@OregonWind KBK is right, originally there were 20 equations. See, for example, http://en.wikiped...ic_Field

KBK's issue is that (s)he doesn't understand that Maxwell didn't use vector notation, instead presenting the equations in Cartesian coordinates. In many cases this forced the use of separate equations for each of the three axes. Modern vector notation reduces the number of equations but is more or less fully mathematically equivalent to Maxwell's original formulations. Try telling that to KBK though.
sender
not rated yet Dec 08, 2009
I think the answer these experiments aim to achieve have been proven by subwavelength collimation and spectral shifting of light from extragallactic anomalies, e.g. redshifting
Donutz
5 / 5 (1) Dec 08, 2009
There's one thing I've always wondered about weak equivalence: for a point mass, it makes sense; but for a mass with measurable width, i.e. perpendicular to the direction of acceleration, there would be a measurable difference between acceleration and gravity. For instance, two objects separated by distance X would be accelerated in an exactly parallel manner in a spaceship, but in a gravity well they've be accelerated radially and would be slightly closer together at the bottom of any drop. So in the real world, given good enough equipment, you *would* be able to tell whether you're in a spaceship or a gravity well.
sender
not rated yet Dec 08, 2009
so the tether idea from something like this would have worked out after all:

http://projectrho...able.jpg
Donutz
not rated yet Dec 08, 2009
so the tether idea from something like this would have worked out after all:

http://projectrho...able.jpg


Website declines to show the webpage without a login. that's just great. Pique my curiosity then yank the rug out.
nkalanaga
not rated yet Dec 09, 2009
Donutz is right on the difference in the real world, although it's hard to test, sinc ethe difference in angle is very small over most normal distances.
DuncanI
not rated yet Dec 09, 2009
An accelerated charged particle emits radiation. The same charged particle stationary in the gravitational field of a neutron star does what? If it also emits radiation, where does the energy come from? If it doesn't emit radiation then it breaks equivalence.
rmuldavin
not rated yet Dec 09, 2009
Does weak equivalence break down at the quantum level?
{{Comparing the acceleration of these two different atomic species constitutes a meaningful test of the UFF, as they combine a large mass ratio (almost a factor of two), very different nuclear compositions (37 protons and 50 neutrons for 87 Rb and 20/19 for 39 K) and almost equal laser wavelength and thus interferometer scale factors.}}

[comments-rm: the "20/19" appears reversed. Trying to clear my confusion:

Z/A(K): 19/39
Z/A(Rb): 37/85; 37/87

I'm easily confused, perhaps so was the editor of the article, or the type setter, or a noisy electron circuit? To assure myself I am not doing my counting correctly:

Taking the most relevant abundance Rb(85) at 72.2% or even the half-life given at 4.9x10^10 yrs, Rb(87), then doing prime number separations, and remembering R at Z=86 is not listed on my notebook Atomic Masses
rmuldavin
not rated yet Dec 09, 2009
[cut off:-rm]:

A(K@39)=>(19x2)+(1);
A(RB@85)=>(17x5)

Where am I going here? Trying to get some polyhedron structures so the authors can draw their K and Rb free falling differences.

Back when I get more done.

Best, rm
vacuum-mechanics
not rated yet Dec 09, 2009
An accelerated charged particle emits radiation. The same charged particle stationary in the gravitational field of a neutron star does what? If it also emits radiation, where does the energy come from? If it doesn't emit radiation then it breaks equivalence.


May be it is easier way to visualize this matter than to do with a neutron star! Does anyone (on our earth) found stationary electron(s) radiate?

Instead, for one whom familiar with the process of radio communication system, such as one which was used in mobile telephone system, would know that radio radiation signal was created and then radiate from the acceleration (oscillation) process of stream of electrons in the equipment!
Alexa
not rated yet Dec 09, 2009
If it doesn't emit radiation then it breaks equivalence..

Sorry, I didn't get it. Why it should violate equivalence principle, if it does nothing?
lomed
not rated yet Dec 10, 2009
Does anyone (on our earth) found stationary electron(s) radiate?
Stationary electrons do not radiate. Having looked up the formula, I find that in order to produce radiation a charged particle must have non-parallel and non-zero velocity and acceleration. Therefore, stationary electrons on Earth or a faraway neutron star will not radiate (unless they are just instantaneously at rest and thus can still fulfill the above requirements).

In fact, an electron dropped from rest in a uniform (or even radial) gravitational field (with that being the only force on the particle) will not radiate either since its velocity and acceleration are parallel.
flaredone
not rated yet Dec 13, 2009
..In fact, an electron dropped from rest in a uniform (or even radial) gravitational field (with that being the only force on the particle) will not radiate either since its velocity and acceleration are parallel.
I presume, if it will be accelerated in this gravity field it would radiate. It would radiate long EM waves to outside, even if it would remain attached to the surface of planet due the planet rotation. If something charged is moving, it radiates EM wave. If this wave is absorbed by observer, the the charged body is decelerated with respect to this observer.
lomed
not rated yet Dec 31, 2009
I was assuming the electron (and the neutron star/planet) was initially at rest with respect to the observer. If the electron is affixed to the surface of a planet that is rotating with respect to the observer; I agree that the observer would detect radiation.