I've suggested (& published in 21 journal papers) a new theory called quantised inertia (or MiHsC) that assumes that inertia is caused by horizons damping quantum fields. It predicts galaxy rotation & lab thrusts without any dark stuff or adjustment. My University webpage is here, I've written a book called Physics from the Edge and I'm on twitter as @memcculloch. Most of my content is at patreon now: here

Monday 4 July 2016

MiHsC and Gravity from Uncertainty?

I've always liked Heisenberg's uncertainty principle, and a few years back I managed to derive something that looks like gravity from it (see references). This approach is very appealing: it feels somehow like the open channel, and over the weekend I managed to derive something that looks like MiHsC inertia from it (with caveats, see below). I won't go into details before publication, but to explain vaguely: the uncertainty principle says that the uncertainty in momentum of an object (Dp=D(mv)) times the uncertainty in position of it (Dx) is equal to the reduced Planck's constant (hbar), see the equation in the diagram:

This is pure quantum mechanics, but what happens if now we apply it on the macroscale where it is not supposed to be valid, and add relativity? When any object (the blue arrows in the diagram) accelerates, say, to the left (black arrows), a relativistic Rindler horizon forms to the right (the black curves) and blocks a huge chunk of space since information can no longer get from beyond that horizon to the object. With greater acceleration the horizon comes closer (see the diagram). Why not say that this horizon reduces the uncertainty of position, Dx? (the red arrows). It obviously does since the 'knowable space' shrinks. If we assume that and apply the uncertainty principle, then the momentum (or energy/c) uncertainty goes up and this becomes available to produce the inertial force to oppose the original acceleration (blue arrows).

I did the maths over the weekend, and the inertial mass predicted this way looks like that predicted by MiHsC (which explains galaxy rotation without dark matter and cosmic acceleration.. etc) My previous derivations are actually equivalent, but this way is rather elegant. There is a 27% difference though, which can be explained by the crudity of the model I've used so far: I have assumed the horizon is spherical.

What is becoming clearer is that MiHsC is inevitable if you take seriously both relativity and quantum mechanics, and allow an interaction between them on large scales.

References

McCulloch, M.E., 2014. Gravity from the uncertainty principle. Astrophys. and Space Sci., 349, 957-959.

qraal said...

Aberration surely plays a role?

Bud Haven said...

The two boundaries become discrete. The transactional boundary moves. The boundary in the direction of movement is always Hubble divided by the speed of light . The infomational boundary where objects can signal to you but you cannot interact with them because of Hubble is Hubble horizon-acceleration or the Rindler boundary behind you. Two masses moving relative to each other create waves They can be electromagnetic or gravitational The maximum wave length you can perceive is limited by your Hubble horizon. Or your Hubble horizon- Rindler when experiencing acceleration.

Mike McCulloch said...

qraal: I assume you mean relativistic aberration, or relativistic beaming. It will, but only when v->c. This would be a non-acceleration way to cut down dx.

qraal said...

Hi Mike
It's fascinating that you get a value so close, yet not close enough, but suspiciously like the ratio between 'matter' and 'dark energy'. I feel that's telling us something, but can't put my finger on it.

Unknown said...

it makes sens since perturbations coming from horizon side diminish => particles become less nervous ! dx decreased dp must increase to reset jolty state !.

ricvil said...

The commutation relation will remains the same form in a frame with acceleration?
Or in other words, the Fourier representation of momentum operator still valid in a accelerated frame?
The UP is a direct result of these representations for quantitative answers.