
When Gravity Breaks Down - dnetesn
http://nautil.us/blog/when-gravity-breaks-down
======
graycat
As we know, the path of a photon can be bent by a large mass with gravity.

Okay: Pass a photon, or electron, or other particle, through a beam splitter
and, thus, get its wave function in two parts.

Then the question seems to be, do the parts also get bent by gravity? Is there
any doubt they won't.

For a particle with mass, surely it has gravity. So, the two parts of its wave
function no longer have gravity -- is that the question she is asking? Maybe
depending on the beam splitter, the two parts of the wave function have 60%
and 40% of the gravity of the whole particle?

Get a lot of particles, say, neutrons, that do feel and generate gravity and
run them through a beam splitter. Bunch #1 goes north and bunch #2 goes east.
So far they still generate gravity? Some miles away from the beam splitter we
put detectors, one for each bunch. With a 50%-50% beam splitter, with the law
of large numbers, we detect 50% of the particles at each detector. That is,
each detector gets a part of the wave function for ALL the particles but
detects only half of the particles.

Suppose the detector for bunch #1 is some miles farther from the beam splitter
than the detector for bunch #2. So, the wave functions for bunch #2 hit their
detector before the wave functions for bunch #1 hits their detector.

Now look at the wave functions for bunch #1: About half of those wave
functions have already collapsed due to detections at the detector for bunch
#2. So, the gravity generated by the remaining wave functions of bunch #1 are,
from the law of large numbers, half what they would be without a detector for
bunch #2. So, for bunch #1, we have had faster than speed of light
communications from the detector for bunch #2 to a measurement of the gravity
of bunch #1.

Yes, we can't get a gravity detector sensitive enough, but for that theory
that's not important. Instead, for the bunches to generate gravity gets to be
an issue anyway.

~~~
comicjk
If you have a gravity detector that can distinguish where a photon's energy is
localized, it also works as a photon detector in the QM measurement sense, and
will collapse the wavefunction in the same way.

Also, remember that gravity waves move at the speed of light; you can't detect
a photon's gravity from a distance without waiting for the same light speed
delay that you would get from the photon itself.

~~~
graycat
> If you have a gravity detector that can distinguish where a photon's energy
> is localized, it also works as a photon detector in the QM measurement
> sense, and will collapse the wavefunction in the same way.

I've wondered about that: The moving particle, photon, electron, neutron, is
throwing off a gravity wave as it is moving if we don't have a gravity wave
detector detecting that wave. So, just having the detector, maybe a second
away, collapses the wave function?

My guess is that as a wave function is split into multiple parts going off in
different directions and getting far apart, the particle is not yet
"localized" in any of the parts. The idea that there is a particle,
"localized" but we just don't yet know where it is I have a tough time
accepting. Instead, until there is an _interaction_ that transfers the energy,
although maybe not the gravitational energy, there is no localization. Or, all
the parts of the wave function both feel and generate gravity until the
collapse from an interaction with, say, a detector.

Or, if only one part of the wave function has the particle, then when two
parts of the wave function are combined, as in just Young's double slit, we
should not get the interference we do get. Or, it seems we get the
interference of the wave function parts, then get the detection; there never
was anything localized until the detection at which time all the parts of the
wave function have to collapse everywhere, instantaneously, even across 1
billion light years -- no I don't like to believe that, but I'm just looking
at the interference of two parts of the wave function in Young's double slit:
If the particle was really in just one of the two parts of the wave function,
then tough to believe in the interference we do see.

~~~
j1vms
> So, just having the detector, maybe a second away, collapses the wave
> function? My guess is that (...)

This gets to to crux of the issue the author (and others in contemporary
physics, Hawking included for that matter) was raising. We can conjecture, via
theoretical physics, a whole slew of possible models of the universe that
combine general relativity and quantum mechanics. But, there is a lack of
experimental evidence (or interpretation of existing results) to resolve the
ongoing "debates". And there hasn't been any real progress, in that respect,
in the past 80 years.

We need to test the edge cases of both theories as best we know how. For
example, tunnel to the center of the earth (or another planet?) and verify our
existing models in the gravity well, maybe repeat the double-slit experiment
remotely while we're there. I'd say a task much more difficult, engineering-
wise, then flying a human being to another planet and back.

~~~
garmaine
Gravity decreases as you move inside the Earth.

------
_bxg1
Nautilus always does such a great job of putting concepts in layman's terms
and making them flavorful.

------
gpsx
There appears to be a very basic error in this writeup. I say there appears to
be because it seems to be so easy to misunderstand language used in casual
descriptions of quantum mechanics. Neither I nor the writer is a lawyer. My
apologies if I misunderstood what was written. My experience here in general
is when something seems very clearly wrong, it is my misunderstanding. None
the less...

Quantum mechanics does not say a particle can be in two places at the same
time. A wave function does not say a particle is at point A AND point b. The
wave function says the particele is at point A OR point B.

To take this a step further, if the particle has an electric field, then the
wave function would look like the following:

(particle at A AND electric field around point A) OR (particle at B AND
electric field around point B)

The wave function would NOT be:

(particle at A OR particle at B) AND (electric field around A OR electric
field around B)

And also not:

(particle at A AND particle at B) AND (electric field around A and B)

Presumably the same applies to a gravitational field.

~~~
drdeca
I like this description of it :

[https://www.smbc-comics.com/comic/the-talk-3](https://www.smbc-
comics.com/comic/the-talk-3)

key quote:

'Superposition doesn't mean "and", but it also doesn't mean "or".' 'It means a
complex linear combination of a 0 state and a 1 state. You should think of it
as a new ontological category: a way of combining things that doesn't really
map onto any classical concept.'

~~~
SuoDuanDao
I like that quote myself, but one thing I don't understand is - are the
amplitudes always of a magnitude 1? Because if qbits are really always the
same magnitude, why represent them as two-dimensional? Wouldn't the math be
easier if we converted it to polar form and just had one measurement per
superposition?

Sorry if this is a dumb question, I'm not formally trained in any of this
stuff...

~~~
millstone
It's a good question. The cartoon says that qubits are "unit vectors in 2D
Hilbert space." However because the length is 1, you can think of it as a
vector in 2D projective Hilbert space. Projective space is what you get when
you throw away a vector's length and only talk about its angles.

Some intuition: if you have a particle and you look everywhere for it, your
chance of finding it must be 100%. It can't be 50% (where could it be?) and it
can't be 200% (duh).

QM in one dimension is formulated as "a 1d complex unit vector", i.e. a
complex number of magnitude 1, which is routinely represented in polar form (a
complex exponential).

QM in two dimensions is a "2d complex unit vector" which is really 3d: you get
three linearly independent components, and then the fourth one is decided for
you. You can think of this as spherical coordinates, but complex numbers are
much easier to manipulate, so that's what we do.

------
skybrian
If existing theory covers all practical experiments, maybe we don't need a new
theory? Sometimes you just gotta declare victory and move on to some other
field.

~~~
garmaine
You could have said the same thing in 1880, and indeed many people famously
did declare “the end of physics” at that time.

