[lyner]

a gravitational field acts as a point source outside the mass causing the field

Only at infinity. The gravitational potential for a dumbell shape, for instance  is not  spherical. If you happen to be on the Moon, you are attracted to the Moon, not to the centre of  mass of Earth and Moon. The total system doesn't behave like a point if you are within / near it.

[another_someone]

But ,just batting along in a straight line, it seems reasonable that they should be in the same place.

But does the electron actually travel in an absolutely straight line?  It is a probability wave, so it could be anywhere that is approximately along that line, and so could not its centre of mass and charge be in different places within that probability field?

[lyner]

That's,surely, what duality is about. If it behaves, in our experiment, as if it is a particle then it is a particle. You surely don't want it to 'wiggle' just because it can also behave as a wave. The probability field could / would be more or less uniform with no 'constraining' fields.
btw, is the wave transverse or longitudinal?

[another_someone]

That's,surely, what duality is about. If it behaves, in our experiment, as if it is a particle then it is a particle. You surely don't want it to 'wiggle' just because it can also behave as a wave. The probability field could / would be more or less uniform with no 'constraining' fields.
btw, is the wave transverse or longitudinal?

In the case of a photon, an electromagnetic field is transverse, is it not?

As for a particle wiggling its way along - would that not be akin to brownian motion - what is so unbelievable about that?

In any case, we are not really discussing whether a particle as such is doing the wiggling, since we are hypothesising about different components of the particle (the different forces associated with the particle) acting in loose association, so it is really a question as to whether one or other component can wiggle in relation to the other components (a little like thinking of an electron as a molecule containing atoms of mass, electric force, weak force, etc.; and the various atoms being subject to brownian motion within the molecule, even as the molecule as a whole may appear to move in a straight line).

I realise that there will be a point where this molecular notion of a particle will break down, but in general terms we have been talking about a coupling between forces that hold them together within the same particle, and that coupling must have some energy, so one can draw parallels with the energies holding the atoms in a molecule.

[lyner]

we all know about photons; I was talking of De Broglie waves.
But in any case, where is this photon wiggling? Is it wiggling all over the Universe?

[another_someone]

we all know about photons; I was talking of De Broglie waves.

Maybe I am naive in my understanding of quantum theory, but in the case of a photon, are they actually two different things?  Is the De Broglie wave of a photon something other than the electromagnetic wave we are accustomed to?

But in any case, where is this photon wiggling? Is it wiggling all over the Universe?

Valid question, but the first question is "is it wiggling", then we can ask "how much is it wiggling"?  It may be a very small wiggle indeed, although common sense would indicate that if it wiggles, it could potentially wiggle all over the universe (to the edges of its probability wave), but spending the vast majority of its time near the centre line of its wave function (where the the probability of its being found are highest).

But, again, we come back to the question about whether we are even talking about where we find the particle itself, or just about where we find the focus of its various fields, with different fields focusing on different locations within the probability wave at one time.  In a very particular sense, we do see this with an electromagnetic wave, where the focus of the electric and magnetic fields are indeed in different locations (the magnetic field is at its greatest where the electric field is at its lowest).  This is a particular situation because of the close association between electric and magnetic fields (although we now regard the electric and weak fields as having a similar close association, but no such association is understood to exist between the electric field and mass, which is why the question of colocation between those particular attributes would probably be the most interesting).

[shmengie]

In over my head.  But I'm trying to follow this conversation, and my tiny brain is having difficulty.

Would it be correct in stating:

Given two particles are on a similar trajectory; Will quantum theory be the best prediction of how the particles will interact, or will relativity be the best prediction of how the particles will interact?

Or rather, how can it be determined what exactly will happen?

Or is my tiny brain peering in the wrong direction again?

[JP]

It depends how big they are and how fast they're moving:

Tiny particles (subatomic) moving slowly = nonrelativistic quantum mechanics
Large particles moving quickly (near light-speed)= special relativity
Tiny particles moving quickly = relativistic quantum mechanics

Also, only special relativity gives an answer that tells you "exactly what happens."  The two types of quantum mechanics will give you an exact answer, but it will only be able to predict the probabilities of the various outcomes (because that's how the world works for tiny particles).

[LeeE]

a gravitational field acts as a point source outside the mass causing the field

Only at infinity. The gravitational potential for a dumbell shape, for instance  is not  spherical. If you happen to be on the Moon, you are attracted to the Moon, not to the centre of  mass of Earth and Moon. The total system doesn't behave like a point if you are within / near it.

The key word there was 'acts' - with any shaped mass, from any position, you end up with a single vector towards that mass, implying a point source (or axis if directly aligned along the axis).

If you are standing on the moon, and therefore have the same velocity as the moon, you need to be attracted to the common center of mass of the Earth-Moon system to follow the same path as the Moon in that system, do you not?  Certainly, you would be aware of the local attraction between you and the moon, but as well as only being a subset of the total system under consideration, you'd also be within the system and not outside it, which was the other condition.

[lyner]

f you are standing on the moon, and therefore have the same velocity as the moon, you need to be attracted to the common center of mass of the Earth-Moon system to follow the same path as the Moon in that system, do you not?

A conker on a string would point near the centre of the Moon if you were near the surface of the moon. That would be the 'point mass' which would be at work on your conker (even if you and the conker happened to be zooming past the Moon). The centre of mass is an over-simplification of the situation under some / many circumstances. As I said, it works fine at infinity / great enough distances.
Prof Herman Bondi pointed this out to me in a lecture in 1965 and it makes sense if you do the sums.
As a direct equivalent situation, a distributed set of charges, although having a total charge of zero, can still be attracted to a charged rod - that's how pieces of paper are picked up on your charged comb. They do not become charged but become polarised by induction. If the centre of charge were what counted, the paper wouldn't be attracted.
You have to avoid falling into common conceptual traps.

[JP]

I agree with sophiecentaur.  I nice conceptual experiment would be to put your mass in a black box (you'd actually need a sphere) and test it for point-charge behavior by measuring the gravitational force at a bunch of points all around this box. Unless the mass is spherically symmetric, or you're at infinity, you'll end up with different answers for the location of the point-mass depending on where you're testing the field. 

Now a neat effect is that due to gravity and electric fields both having sources (monopoles), spherically symmetric source distributions for both fields will act as point sources located at the center of the mass.  (This is termed Gauss's Law.) 

[lyner]

Yes - but if your particle (say an alpha particle) is not spherically symmetrical it would not be expected to behave, nearby, like either a point charge or a point mass; indeed, the charge arrangement would consist of two positive charges and the mass arrangement would correspond to four masses; an entirely different configuration.
btw Gauss' law doesn't actually specify the distribution on the surface - just the total integral over the surface. If the source is not a true point the distribution over a surface near the centre need not be symmetrical - although it approaches uniform as the enclosing surface gets bigger.
Gauss' law is a neat method to get around mere details like the above in order to give a useful result.

[LeeE]

You have to avoid falling into common conceptual traps.

I constantly try but haven't yet found a way to guarantee it

[lyner]

"Do as I say, not as I do."