Black holes

[Vorpal]

And why is this? I mean, there's nothing particularly unusual in what Vorpal writes, is there? Nothing that you won't find in a suitable, contemporary physics textbook, is there?

As far as I'm aware, the notable deviation from many GTR textbooks is that on the subject of what the "gravitational field" actually is, the stance most of them seem to take is roughly "we don't talk about that." Sometimes very literally: MTW effectively spend an entire section explaining their reasons why they leave the term "gravitational field" undefined and hence why they're not going to use it, ever. (That's certainly a very consistent view: they still have all the physically relevant quantities in their mathematics anyway; they just refuse to give any of them that label.)

Under that (non-)interpretation, Synge's comment doesn't mean anything at all, so we need to define what "gravitational field" means. It so happens that Einstein did define it to be the connection coefficients/Christoffel symbols Γ^α_μν and the gravitational potential to be the metric components. There are good reasons why this is sensible, and also a reason to reject it--Γ's are not tensorial as a whole. But it remains the case that there isn't a better candidate, and its coordinate-dependent nature actually embraces the equivalence principle: gravity would be explicitly an inertial ('fictitious') force.

And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there.

P.S. I really should have made the last bit more explicit in my previous post.

 

[Farsight]

Nope, not in that frame, I wouldn't. "It's not the fall that kills you--it's the sudden stop at the end." My terminal ends is because the ground puts a tremendous force on me.

When you're in a box falling to earth your inertial reference frame is continually changing but you can't tell. When that change stops occuring all of a sudden, then you can.

I don't think you understand what local means.

Of course I do.

Of course--what is or isn't rest is frame-dependent. But the major difference is that in Newtonian physics, making a frame that comoves with a gravitationally freefalling particle gives an accelerated frame--it doesn't follow a geodesic. In GTR, it does: the acceleration four-vector along the trajectory identically vanishes. That's the point of EP.

And again, it only works for a region of infinitesimal size.

Synge's argument is about the relationship between the gravitational field and the gravitational tidal forces. You're simply claiming some relevance without even a word as to why it is so.

I've said it already - you cannot transform away the Riemann curvature. All you can do is take a small region of a curve, a region so small that it looks flat. But we know it isn't flat.

Since you yourself specified the case where the charge is not moving in our local inertial frame, the electric field is what you get to measure. Remember Lorentz force and it'll be obvious.

The Lorentz force isn't relevant. If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple.

For the purposes of this discussion, it doesn't matter whether there are or aren't any point charges. Read what I wrote again: I'm saying that even if that particular statement of yours was 100% absolute truth, nothing about my point changes. Because what you said literally adds nothing more than a side comment of "... and the field comes from the potential." Yes, everyone knows that. I don't mind if you consider it the mostest fundamentalest thing there ever was. What I said was:
(*) the gravitational field and gravitational tidal forces are different things
and the electromagnetic field was brought in to illustrate why Synge's argument is mistaken: because we can define a tensor that describes electromagnetic tidal forces, in the sense of deviation of equally-charged test particles, and directly confirm that its vanishing does not imply that the electromagnetic field vanishes.

Forget your equally-charged test particles. Use a single test particle. When you set it down and find it doesn't move, then you say there's no electromagnetic field present.

Or, more bluntly: Synge's implication is very clearly false for other fields.

No it isn't, you've misunderstood the scenario.

Why should we believe him about the gravitational field?

Because he's right. You can't make the Riemann curvature vanish. All you do when you take an infinitesimal region is sweep it under the carpet. Take that too literally and you end up with a flat hill and a misunderstanding of the gravitational field.

Evidently, you don't agree with it, because if you accepted it as valid, you'd see that Synge is wrong immediately. Hence my question about your reasons against it.

No I don't see that Synge is wrong immediately. Instead what I see immediately is that you got the scenario wrong.

It's just poking fun at the fact that for someone who incessantly insistent that he was "with Einstein", you spend extraordinary amounts of time arguing against him.

No I don't. I spend far more time saying Einstein said x. And I've said already that he wasn't perfect.

And I'll put this last:

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

What's wrong with that?

You drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen.

If you believe the analogy to be valid, then the gravitational field is the connection, and it's trivial that all you need for a homogeneous but nonvanishing gravitational field is to take flat spacetime in a uniformly accelerated frame. So I ask yet again: do you have any reasons to consider the analogy wrong?

Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime. The principle of equivalence is a principle that shows you why the two are similar, but it doesn't tell you that they're identical.

 

[Farsight]

...And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there.

It isn't schizophrenic, it's really important. When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. And I was scathing of your "eerie" because you don't understand why that comparison can be made.

 

[DeiRenDopa]

... A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. ...

Funny how Farsight's 'absolute time and space' fundamental worldview keeps appearing, in his posts, isn't it?

 

[dafydd]

Funny how Farsight's 'absolute time and space' fundamental worldview keeps appearing, in his posts, isn't it?

If beams of light do not bend then how does gravitational lensing occur? I'm looking forward to Farsight's explanation.

 

[Brian-M]

That's wrong. If the clock at the ceiling stayed synchronised with the clock at the floor, things wouldn't fall down.

You're affirming the consequent.

The normal point of view is that time light passes at different speeds because the gravitational attraction is different. Your point of view reverses this, claiming that gravitational attraction is different because time light is passing at different speeds.

You can't know that if time light wasn't passing at different speeds at different heights then "things wouldn't fall down" until you first demonstrate that your point of view is correct.

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

Is it nonsense?

Increase the distance of your floating lab/city from the planet, and then increase the mass of the planet to compensate so that you're still experiencing the same amount of gravity. Suddenly, the tidal forces in your lab are smaller than they were before.

Repeat this process indefinitely. As distance approaches infinity, tidal force (difference in gravity) approaches at zero.

Sure, you can't actually get infinite distance. But at some finite distance you're going to reach a point where no practical test inside your lab can detect the presence of any gravitational variation with distance.

Now here's something to think about. If things fall down because light is moving slower at the lower elevation as you claim, shouldn't differences in the speed of light over the same elevation (as in my example above) cause things to fall with different force? But if gravity follows the inverse square law (as I assumed in my example above), then this can't possibly be true.

 

[W.D.Clinger]

...And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there.

It isn't schizophrenic, it's really important. When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. And I was scathing of your "eerie" because you don't understand why that comparison can be made.

It isn't schizophrenic so much as ignorant. Farsight doesn't realize that the coordinate-dependent Christoffel symbols (which, according to Einstein, describe the gravitational field) can be nonzero even in flat spacetime.

Had Farsight made an effort to understand the metric form posted by sol invictus, he might not have made that mistake. On the other hand, Farsight would have had to learn calculus first...

 

[Farsight]

If beams of light do not bend then how does gravitational lensing occur? I'm looking forward to Farsight's explanation.

Huh? I said this:

When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself.

Gravitational lensing occurs when beams of light bend. They bend because they move through a region of space where a gravitational field is present. When this happens we call it curved spacetime, when it doesn't and the beams of light go straight, we call it flat spacetime.

 

[Farsight]

You're affirming the consequent.

No, you are.

The normal point of view is that time light passes at different speeds because the gravitational attraction is different. Your point of view reverses this, claiming that gravitational attraction is different because time light is passing at different speeds.

My point of view is Einstein's point of view. He said a curvature of rays of light can only take place when "die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert", which translates into "the speed of light varies with the locality". When it doesn't, there is no gravitational attraction.

You can't know that if time light wasn't passing at different speeds at different heights then "things wouldn't fall down" until you first demonstrate that your point of view is correct.

Einstein was the author of general relativity, his point of view was correct. And you know it's correct from the parallel-mirror gif. And you know about the wave nature of matter don't you? You know we can make an electron out of light via pair production? And that the electron has spin angular momentum and a magnetic dipole moment? And that it can be diffracted? So think of the electron as a standing wave of light. A circle of light, as it were. Put it in a gravitational field and think it through. Divide the circle into four flat quadrants to make it even simpler:

..←
↓....↑
..→

Starting from the left and going anticlockwise, at a given instant we have light travelling down like this ↓. There’s a gradient in c from top to bottom, but all it does is make the light look blueshifted. A little while later the light is moving like this → and the lower portion of the wave-front is subject to a slightly lower c than the upper portion. So it bends down a little. Later it’s going this way ↑ and looks redshifted, and later still it’s going this way ← and bends down again. These bends translate into a different position for the electron. It falls down:

○
↓

The reducing c bleeds rotational motion out into linear motion. But only half the cycle got bent, so only half the reduced c goes into kinetic energy aka relativistic mass. That’s why light is deflected twice as much as matter, and gravity is not the sort of force that increases the relativistic mass. Simple.

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

Is it nonsense?

Yes. You've got a reducing speed of light that keeps on reducing, so much so that it and ends up being negative. There's no such thing as a negative speed. A negative speed is nonsense.

Increase the distance of your floating lab/city from the planet, and then increase the mass of the planet to compensate so that you're still experiencing the same amount of gravity. Suddenly, the tidal forces in your lab are smaller than they were before.

No problem with that.

Repeat this process indefinitely. As distance approaches infinity, tidal force (difference in gravity) approaches at zero. Sure, you can't actually get infinite distance. But at some finite distance you're going to reach a point where no practical test inside your lab can detect the presence of any gravitational variation with distance.

But things in your lab still fall down, and the parallel-mirror light-clock at the ceiling still runs faster than the one at the floor.

Now here's something to think about. If things fall down because light is moving slower at the lower elevation as you claim, shouldn't differences in the speed of light over the same elevation (as in my example above) cause things to fall with different force?

Yes. If there's a big difference between the speed of light at the ceiling as compared to the speed of light at the floor, the force of gravity is large. If there's a lesser difference, the force of gravity is less. When there's no difference, things don't fall down.

But if gravity follows the inverse square law (as I assumed in my example above), then this can't possibly be true.

It is true. When you measure the difference between the speed of light at the ceiling as opposed to the floor in a very very tall building, you find that on floor 100,000 the difference is less than it is in the basement. If you could take an equatorial slice through the planet and measure the speed of light at various locations, when you plot them out what you end up with is the upturned hat. When you look at one small region of this there's still a gradient in gravitational potential and a gradient in c. There's no discernible curvature of the gradient, you can't detect any tidal force, but things still fall down, and light still curves when it moves through space, so we say spacetime is curved. Way out a zillion miles to the left or right where there's no discernible gradient, your measurements can't detect any curvature of light, or things falling down, and then we say spacetime is flat.

 

[Vorpal]

Farsight, you're simply repeating irrelevancies.
1) Absolutely no one is claiming that you can transform away Riemann curvature at events where it's nonzero. Repeating it incessantly is completely pointless, because no one is questioning this fact.
2) The real issue is whether the vanishing of Riemann curvature at some event implies that the non-presence of the gravitational field there. This is something you, and Synge, simply assert as true, and provide no reasoning for that claim.
You've simply made no connection between (1), which everybody knows, and (2), which is what we're trying to have a discussion about. And failing.

The Lorentz force isn't relevant. If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple.
...When you set it down and find it doesn't move, then you say there's no electromagnetic field present.

You're mistaken. If you have a charged test particle that starts at rest and it doesn't move, it in no way implies that there's no electromagnetic field present. It only implies that there is no electric field present.

You drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen.

Nonsense. This is a basic calculus exercise for STR. Suppose a particle starts from rest and undergoes constant force:
[latex]\[ F = \frac{dp}{dt} = m\frac{d(\gamma v)}{dt} = mc\left(\cosh\alpha\right)\frac{d\alpha}{dt}\] [/latex]
Where I've applied the substitution v/c = tanh α, so γv/c = sinh α and γ = cosh α. Just integrate: Ft = sinh α, and therefore:
[latex]\[ v = c\tanh\left(\sinh^{-1}\left(\frac{Ft}{mc}\right)\right) = \frac{Ft/m}{\sqrt{1 + (Ft/mc)^2}} \][/latex]
Note that the tanh form makes it obvious at the speed is bounded by c.

Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime.

I asked for reasons, not mere repetition of baseless assertion.

 

[dafydd]

Huh? I said this:

When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself.

Gravitational lensing occurs when beams of light bend. They bend because they move through a region of space where a gravitational field is present. When this happens we call it curved spacetime, when it doesn't and the beams of light go straight, we call it flat spacetime.

Thank you. You didn't mention flat spacetime in your previous post.

 

[Brian-M]

My point of view is Einstein's point of view. He said a curvature of rays of light can only take place when "die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert", which translates into "the speed of light varies with the locality". When it doesn't, there is no gravitational attraction.

Can you provide a link to the source of this quote? I'd like to see if the actual text supports your assertion that it's the change in the speed of light that causes objects to exhibit gravitation attraction.

Einstein was the author of general relativity, his point of view was correct.

That's the assumption we're operating from. Whether your point of view is an accurate representation of his point of view is a different subject altogether.

And you know it's correct from the parallel-mirror gif.

I don't know anything is correct from the parallel-mirror GIF. It's just an animation, and animations aren't necessarily an accurate representation of reality. Anything can happen in an animation.

But since it was created as a graphical representation of our pre-existing understanding that clocks lose synchronization at different elevations, we can assume that clocks (whether atomic clocks, or parallel-mirror light clocks) do lose synchronization at different elevations with or without the GIF.

But even given that, we can't say that Einstein's theory is correct from this alone. (But we do have other reasons separate from this to assume Einstein's theory is correct.)

And you know about the wave nature of matter don't you? You know we can make an electron out of light via pair production? And that the electron has spin angular momentum and a magnetic dipole moment? And that it can be diffracted?

I don't see what this has to do with relativity.

So think of the electron as a standing wave of light.

WTF? ​

A circle of light, as it were. Put it in a gravitational field and think it through. Divide the circle into four flat quadrants to make it even simpler:

..←
↓....↑
..→

Starting from the left and going anticlockwise, at a given instant we have light travelling down like this ↓. There’s a gradient in c from top to bottom, but all it does is make the light look blueshifted. A little while later the light is moving like this → and the lower portion of the wave-front is subject to a slightly lower c than the upper portion. So it bends down a little. Later it’s going this way ↑ and looks redshifted, and later still it’s going this way ← and bends down again. These bends translate into a different position for the electron. It falls down:

○
↓

The reducing c bleeds rotational motion out into linear motion. But only half the cycle got bent, so only half the reduced c goes into kinetic energy aka relativistic mass. That’s why light is deflected twice as much as matter, and gravity is not the sort of force that increases the relativistic mass. Simple.

If there's a big difference between the speed of light at the ceiling as compared to the speed of light at the floor, the force of gravity is large. If there's a lesser difference, the force of gravity is less. When there's no difference, things don't fall down.

I'm going to take some time to think of an appropriate response to this.

 

[Farsight]

Can you provide a link to the source of this quote? I'd like to see if the actual text supports your assertion that it's the change in the speed of light that causes objects to exhibit gravitation attraction.

Sure. See Über die spezielle und allgemeine Relativitätstheorie and look at section 22. It's on page 51, about three-quarters of the way down.

That's the assumption we're operating from. Whether your point of view is an accurate representation of his point of view is a different subject altogether.

He said what he said.

I don't know anything is correct from the parallel-mirror GIF. It's just an animation, and animations aren't necessarily an accurate representation of reality. Anything can happen in an animation. But since it was created as a graphical representation of our pre-existing understanding that clocks lose synchronization at different elevations, we can assume that clocks (whether atomic clocks, or parallel-mirror light clocks) do lose synchronization at different elevations with or without the GIF.

OK. The gif just brings it home I guess.

But even given that, we can't say that Einstein's theory is correct from this alone. (But we do have other reasons separate from this to assume Einstein's theory is correct.)

Yes, it's a well-tested theory.

I don't see what this has to do with relativity.

It isn't relativity, but it is physics, and it is relevant.

WTF?

Yes, think of the electron as a standing wave of light. You can make it from light along with a positron in pair production. You can diffract an electron. Electrons interfere with one another. You can annihilate an electron with a positron and what you get is light. Read up on atomic orbitals and see the bit that says "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves".

I'm going to take some time to think of an appropriate response to this.

OK noted.

 

[Farsight]

Farsight, you're simply repeating irrelevancies.
1) Absolutely no one is claiming that you can transform away Riemann curvature at events where it's nonzero. Repeating it incessantly is completely pointless, because no one is questioning this fact.
2) The real issue is whether the vanishing of Riemann curvature at some event implies that the non-presence of the gravitational field there. This is something you, and Synge, simply assert as true, and provide no reasoning for that claim.
You've simply made no connection between (1), which everybody knows, and (2), which is what we're trying to have a discussion about. And failing.

How many more times do I have to tell you about flat hills? Find some plot of gravitational potential like the upturned hat, and just look at it. Even this picture will do. If you have no Reimann curvature then when you start from a zillion miles to the right you've got a totally flat line. Start from the left at the surface of the sun and without Riemann curvature you've got a constant slope that goes on forever. The former is where there is no gradient in gravitational potential, so light doesn't curve and nothing falls down. The latter is where the coordinate speed of light goes on increasing forever, and the force of gravity doesn't reduce with distance. Neither is a description of a gravitational field. You need that Riemann curvature. When you take an infinitesimal region you're ignoring the change in the gradient but you aren't ignoring the gradient! How much simpler can I make it?

You're mistaken. If you have a charged test particle that starts at rest and it doesn't move, it in no way implies that there's no electromagnetic field present. It only implies that there is no electric field present.

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back:

"Then in the description of the field produced by the electron we see that the separation of the field into electric and magnetic force is a relative one with regard to the underlying time axis; the most perspicious way of describing the two forces together is on a certain analogy with the wrench in mechanics, though the analogy is not complete".

Did you catch that? It's the field. It exerts force in two ways, resulting in linear and/or rotational motion. If you're a charged particle and I set you down somewhere in the electron's electromagnetic field, you move linearly towards it or away from it, and the electron similarly moves towards or away from you. If I throw you through it, you go round and round as well. You only experience the rotational force when you have relative motion.

Nonsense. This is a basic calculus exercise for STR. Suppose a particle starts from rest and undergoes constant force:
[latex]\[ F = \frac{dp}{dt} = m\frac{d(\gamma v)}{dt} = mc\left(\cosh\alpha\right)\frac{d\alpha}{dt}\] [/latex]
Where I've applied the substitution v/c = tanh α, so γv/c = sinh α and γ = cosh α. Just integrate: Ft = sinh α, and therefore:
[latex]\[ v = c\tanh\left(\sinh^{-1}\left(\frac{Ft}{mc}\right)\right) = \frac{Ft/m}{\sqrt{1 + (Ft/mc)^2}} \][/latex]
Note that the tanh form makes it obvious at the speed is bounded by c.

Gravity isn't a force in the usual sense. Forget your constant force, it's a constant acceleration, and escape velocity is the flip side of the speed of an infalling body dropped from "a great distance". When you drop your body into a black hole, it supposedly ends up going at the speed of light. You cannot contrive a gravitational field where it continues to accelerate at 9.8 m/s/s ad infinitum and gets to the speed of light when it's only halfway there. That's why the infinite wall falls down. That's why I said the problem was that you drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen. Because the reducing c bleeds rotational motion out into linear motion. That's why it's obvious that the speed is bounded by c.

Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime.

I asked for reasons, not mere repetition of baseless assertion.

It's not a base assertion. If we're all in space where light doesn't curve, Vorpal putting the pedal to the metal doesn't make it curve one iota. Learn to tell the difference between what you see and what's there.

 

[Farsight]

Thank you. You didn't mention flat spacetime in your previous post.

My pleasure.

 

[Hellbound]

Yes, think of the electron as a standing wave of light. You can make it from light along with a positron in pair production. You can diffract an electron. Electrons interfere with one another. You can annihilate an electron with a positron and what you get is light. Read up on atomic orbitals and see the bit that says "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves".

Just because I like to throw in wrenches...

I do hope you realize that you can annihilate a proton with an anti-proton, and also get light.

And you can annihilate a neutron with an anti-neutron, and get light.

So are you suggesting that all particles are standing light waves?

 

[edd]

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back:

I'm really not sure you got Vorpal's point. You take a charged particle and place it at rest and it doesn't move. You'd seem to be saying that that demonstrates the absence of any magnetic field, since you're saying that it shows there's no electromagnetic field.

 

[DeiRenDopa]

Just because I like to throw in wrenches...

I do hope you realize that you can annihilate a proton with an anti-proton, and also get light.

And you can annihilate a neutron with an anti-neutron, and get light.

So are you suggesting that all particles are standing light waves?

By Jove Democritus!

That's it! The TRUE SECRET TO THE UNIVERSE!!!!!

Atoms are made up of neutrons, protons, and electrons. Each of which, in turn, per Farsight, is a standing wave of light.

A crystal, of an element such as iron, is then also just a bunch of standing waves of light. Molecules are also just mixtures of standing waves of light.

A virus, which is made up of molecules, is, then a complex mixture of standing waves of light.

So is a tree, you, me, an asteroid, a planet, a star, ...

And if you put enough standing waves of light together, and scrunch them down, you get ... wait for it ... a black hole!

What a waste, 25 pages of largely irrelevant posts, when we could have just considered black holes as yet another manifestation of light ...

 

[sol invictus]

You need that Riemann curvature. When you take an infinitesimal region you're ignoring the change in the gradient but you aren't ignoring the gradient! How much simpler can I make it?

You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature.

Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is.

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time.

You asserted "If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple.
...When you set it down and find it doesn't move, then you say there's no electromagnetic field present."

That's false, as Vorpal just tried to explain to you. If the electromagnetic field strength tensor in the rest frame of the particle has zero time-space components, but non-zero space-space components (that's a long-winded way of saying "there's a magnetic field, but zero electric field") then the particle won't move when you set it down.

So you're wrong. Admit it, learn from it, and move on. Can you do that?

 

[ben m]

So are you suggesting that all particles are standing light waves?

Hellbound, Farsight will talk to you as though he knows that some sort of topological field theory is true, and as though you're behind-the-times for NOT knowing that all particles are differently-knotted configurations of some underlying ur-field.