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Entanglement Makes Quantum Particles Measurably Heavier, Says Quantum Theorist
KentuckyFC writes: Physicists have long hoped to unify the two great theories of the 20th century: general relativity and quantum mechanics. And yet a workable theory of quantum gravity is as far away as ever. Now one theorist has discovered that the uniquely quantum property of entanglement does indeed influence a gravitational field and this could pave the way for the first experimental observation of a quantum gravity phenomenon. The discovery is based on the long-known quantum phenomenon in which a single particle can be in two places at the same time. These locations then become entangled — in other words they share the same quantum existence. While formulating this phenomenon within the framework of general relativity, the physicist showed that if the entanglement is tuned in a precise way, it should influence the local gravitational field. In other words, the particle should seem heavier. The effect for a single electron-sized particle is tiny — about one part in 10^37. But it may be possible to magnify the effect using heavier particles, ultrarelativistic particles or even several particles that are already entangled.
We only need to measure the mass of a 9.10938291 × 10^-31 kilogram particle accurate to 1 part in 10^-37. Alternatively, we can speed the electron up to 0.999c so it weighs more, then entangle it, and then measure it's mass to 1 part in 10^-37, with less than 5 sigma of measurement error.
Either way, I should have it done by lunch time.
No, really. We cannot measure anything to 37 decimal places. Not even close. If you measure the speed of light there will be uncertainty. If you insist the speed of light is exact by definition, then the uncertainty is in the length of a meter or the duration of a second or both.
Yo mama's so fat her wave function collapses into multiple eigenstates.
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The size of the visible universe is only on the order of 10^9 light years, so that won't do it. But combine it with the range of the weak force, which has been measured and our direct measurement capability spans a range of about 10^39 (weak force ~= 10^-18 meters, lightyear ~= 10^12 meters, visible universe ~= 10^9 lightyears) so we can at least comprehend this number in a concrete way. Measuring it even indirectly... not going to happen with your basic bathroom scale. But we are talking about finding a way to relate effects on the Planck scale to the cosmogical scale so big exponents should be unsurprising. Unlike the outspoken AC above I will wait for peer review before adjusting my bullshit meter. Naturally, we all want to believe there is some big breakthrough here after the endless low calorie diet of contrived mathematical attempts to unify the big theories for the last too many decades. Is this one it? Seems unlikely just based on the long run of failed attempts. But I will just sit back with popcorn and enjoy the show. At worst, a refreshing break from the usual multidimensional mathematical salad parade. I'm particularly interested in more eplanation of how two entangled particles became one, at least according to the press.
When all you have is a hammer, every problem starts to look like a thumb.
That's a theoretical analysis, not an experimental measurement, and is likely to be particularly dubious since we don't have a working theory for quantized general relativity yet. Interesting, but the phrase "does indeed" in the summary is a significant overstatement.
not to spoil it for you but it's time travel. i'll be making my announcement in 2044 and personally demonstrate that you can travel 5 minutes back in time. needless to say, i forgot to carry the one.
Anons need not reply. Questions end with a question mark.
We can't measure anything using any instrument anywhere to a precision of 1/10^37th. Bullshit meter is off the charts
We can't make any single measurement which contains 37 digits and have each of those digits accurate, that's true.
Just out of curiosity, how do radios work? I'm told that the measurement units for an antenna nanovolts per meter. Does the receiver make a 12-volt measurement to 8 digits of accuracy in order to recover the signal?
Or does the receiver amplify the signal so that it's large enough to be readily detected?
And is there no way to make multiple measurements so that the effect adds up? Can we do a million measurements added together to make the signal a million times stronger?
Given that two particles can emitted by a single source entangled, sent a long distance apart, and remain entangled,
And that if one particle becomes disentangled the other particle instantaneously becomes disentangled,
If we can measure the entanglement of a particle by its mass,
Then we can communicate faster than light.
But the no-communication theorem states that, during measurement of an entangled quantum state, it is not possible for one observer, by making a measurement of a subsystem of the total state, to communicate information to another observer.
So I think this means that either the no-communication theorem is wrong, or the change in mass of an entangled particle cannot be measured.
(T>t && O(n)--) == sqrt(666)
FWIW, it appears from the paper that this extra "mass" is an artifact of analyzing entangled particles in a linearized gravity framework and observing a stress-energy tensor term that seems to appear higher for entangled particles and radiated away as particles move to decoherence. This perhaps might be considered the mass of the entanglement.
On the other hand, wouldn't it be cool if the reason for the observed equivalency of gravitational mass and inertial mass was somehow related to quantum entanglement? (yes I know this is unrelated to this phenomena, but still)...
One part in 10^37 is not measurably heavier. No measurement in science has anything like 37 significant figures*.
*No, the cosmological constant does not count, as it was not measured from quantum principles, but from cosmological ones.
The discovery is based on the long-known quantum phenomenon in which a single particle can be in two places at the same time.
Wrooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooong.
Your question doesn't have a simple answer, but if it did, it would involve signal-to-noise ratio within a given bandwidth. A radio receiver with a bandwidth in the audio range (~10 kHz) can amplify a signal by about ten trillion times its original power or a few million times its original voltage, before hitting the thermal noise floor of -174 dBm/Hz. These figures aren't exact (for one thing, they neglect the impedance change from a 50-ohm antenna input to an 8-ohm speaker) but the basic idea is correct: the noise floor at 25C in a 50-ohm system is -174 dBm/Hz + 10*log(bandwidth) dBm.
You can improve SNR by making your measurement near absolute zero, but you can't get rid of the noise entirely because some of it isn't strictly thermal in nature. Synchronous demodulation can let you recover information from below the noise floor, given a carrier of known phase. There are other tricks and hacks, but the bottom line is that you are still going to be at least ten or fifteen orders of magnitude away from being able to work with 37 significant figures in any real-world physical measurement. Integration times for such a measurement would have to approach heat-death-of-the-Universe durations.
Leon's getting larger.
Excuse me? E^2 = p^2c^2 + m^2c^4 is the correct statement (or to be pickier, the four vector P = E/c - \vec{P} has conserved length equal to mc). Photons have zero mass, so for them E^2 = p^2c^2. You are thinking of E = \gamma m_0 c^2, which works fine for massive particles where m_0 \ne 0, not so well for light where \gamma = \infty because light travels at the speed of light.
BTW, does /. grok latex if one wraps it, that is, does $$E = \gamma m_0 c^2$$ work? Might as well try it...
No, apparently not. I suppose I'll have to look at actual documentation to see if there is any way to make it work.
Hey /. Dudes! You keep changing the site, improving it and so on! A 21st century website that cannot speak latex is so, not-even-20th century, and when that site is devoted to technology, it is vaguely insulting. Even wordpress can often understand $latex E = \gamma m_0 c^2$.
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Even when the experts all agree, they may well be mistaken. --- Bertrand Russell.
The photon has zero rest mass, yes.
E = mc**2 is a nice popularization; it's also wrong. It's actually E**2=(mc**2)**2 + (pc)**2, where p is the momentum. When momentum is zero, you can usually simplify this to E=mc**2, but a photon's existence is defined mostly by its momentum. Since m is zero for a photon, this means the energy of a photon is given by entirely by E=pc.
Hope this helps!