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grav
2009-Jul-29, 03:41 PM
I have been attempting to rewrite the laws of physics using only invariants that work for any theory ascribed to it, whether Relativity, ballistic, that of an ether, or otherwise. In the course of doing so, when starting to change from one dimensional to two dimensional, I ran into a conceptual problem that I don't quite see how to apply yet in a normal fashion, that is, by using straight lines to determine how objects will travel inertially. It seems it could be a problem with ballistic and Relativity theory alike, or perhaps that of simultaneity, but probably just my conceptualization.

Let's say we have two observers that are passing each other, with some distance between their perspective paths, perpendicularly to their line of travel. At some point in space observer A emits electrons that travel directly toward where observer B will be such that they will be received in regular intervals. That is, observer B will receive one electron, then another one second later, say, then another one second after that, etc, as observer B continues to travel inertially along a stragiht line path.

Here's the dilemma. Observer A emits these electrons all at once, from a single point in space, so that both observers should agree that all of the electrons have been emitted from a single point in space. From observer A's perspective, if the electrons are emitted when observer B is approaching, then the electrons emitted more directly toward observer B will travel faster and those emitted that are to be received after observer B passes will travel more slowly. Observer B is to receive one photon per interval forever. So here's the thing. Let's say B is in a spherical ship, although that shape may flatten out somewhat according to Relativity from A's perspective but that doesn't matter much. According to A, as the electrons reach B, aimed directly at the center of the ship, they will strike first more toward the forward side of the ship on the hull and then as B passes, they will strike more toward the back. So that should be the case as B sees things also, although the angles of impact will be slightly different according to Relativity for each observer. According to B, however, B considers himself stationary, and all of the electrons were emitted at the same time from a single point in space, so should travel in a straight line from that point in space where A emitted them to B, but just at different speeds, and strike in the same place on the hull of the ship forever. The only way I can see that this would occur differently so far is if B is rotating while the rotation slows, or B does not see all of the electrons travel in a straight line from a single point in space, or does not see all of the electrons emitted at once from the same point in space somehow, or all of the above, or something else. So what am I missing?

grav
2009-Jul-29, 04:21 PM
Okay, wait. Sometimes it must help when I describe these things in a post or something, as I seem to get it soon after. Relativity reduces to ballistic in the low range, so since we're not using light, the paths of the electrons drawn back to their origin should travel with A. So B cannot see them emitted from a single stationary point in space from his own perspective in his own coordinate system, but travelling with A's coordinate system as they should always be drawn back to the origin at A, I would think, since that is what A sees. That is why gravitational systems and so forth travel with the mass when the mass is travelling inertially. So what about when using light? Would the paths of the photons still always be drawn back to A, so that the origin moves with A? It seems they should since Relativity says that gravity travels with the mass, and gravity travels at the speed of light. But then, wouldn't that be somewhat ballistic also?

grant hutchison
2009-Jul-29, 04:22 PM
Did you draw a diagram? :)
You have a faulty picture of what A will see when the electrons strike B.

After B has passed A, A will see that B is catching up with electrons that are moving more slowly than B is. So both A and B will see that these electrons strike B from the front, not the rear.
If B draws a sightline in the direction from which the first electron hits him, that sightline will cut across the fan of electron tracks emitted by A, and will intersect every electron as it progresses along its track. In order to encounter B as you describe, the electrons have to keep that alignment, with B and with each other, as they each move along their separate diverging tracks. From B's point of view, they all come rattling down the same linear track; from A's point of view, they spread out in a fan, but maintain a linear pattern that always points towards B.

Grant Hutchison

grav
2009-Jul-29, 05:02 PM
Did you draw a diagram? :)
You have a faulty picture of what A will see when the electrons strike B.

After B has passed A, A will see that B is catching up with electrons that are moving more slowly than B is. So both A and B will see that these electrons strike B from the front, not the rear.
If B draws a sightline in the direction from which the first electron hits him, that sightline will cut across the fan of electron tracks emitted by A, and will intersect every electron as it progresses along its track. In order to encounter B as you describe, the electrons have to keep that alignment, with B and with each other, as they each move along their separate diverging tracks. From B's point of view, they all come rattling down the same linear track; from A's point of view, they spread out in a fan, but maintain a linear pattern that always points towards B.

Grant HutchisonAh, that is another possibility I didn't consider. I will have to draw out a diagram and work out the math more precisely as you say to find out. If that is the case, though, wouldn't that mean that gravity from a mass that is moving inertially according to an observer would not move with the mass so that it appears to come from the current position of the mass wherever it may be, but from the stationary point in space where it was initially emitted according to the observer? Also, in the scenario, A would consider herself stationary, so would see the electrons moving directly away from her at all times, so shouldn't B see the entire "field" of electrons also always pointing back to A while A continues to travel, so that the field itself, at least its origin, moves ballistically with A, although the motions of the individual electrons are relativistic?

grant hutchison
2009-Jul-29, 05:17 PM
Ah, that is another possibility I didn't account for. I will have to draw out a diagram and work out the math more precisely as you say to find out. If that is the case, though, wouldn't that mean that gravity from a mass that is moving inertially according to an observer would not move with the mass so that it appears to come from the current position of the mass wherever it may be, but from the stationary point in space where it was initially emitted according to the observer?I'm not sure how we got from a simple scenario of relative motion to a hypothesis about general relativity. Richard might be able to describe for you how GR will work in that setting: it turns out that GR doesn't play the same game as those electrons. :)

Also, in the scenario, A would consider herself stationary, so would see the electrons moving directly away from her at all times, so shouldn't B see the entire "field" of electrons also always pointing back to A while A continues to travel, so that the field itself, at least its origin, moves ballistically with A, although the motions of the individual electrons are relativistic?B can backtrack the electrons to their point of origin in B's reference frame. In B's frame, A passed through that point of origin just as it emitted the electrons, but it doesn't stay there.

Grant Hutchison

grav
2009-Jul-29, 05:27 PM
B can backtrack the electrons to their point of origin in B's reference frame. In B's frame, A passed through that point of origin just as it emitted the electrons, but it doesn't stay there.But it always stays there according to A, I would think. According to A, at any point in time, a line drawn from the furthest electrons that are currently striking B's hull, all the way back to the very slowest ones that have barely travelled away from A should always be drawn back to A forever, shouldn't they? If so, then B would always see the line ending at A also, no matter where A might be. Yup, I definitely need a diagram. :)

grant hutchison
2009-Jul-29, 05:28 PM
Yup, I definitely need a diagram. :)You definitely do. :)

Edit: Here (http://www.bautforum.com/1053536-post40.html) is Richard pointing out that gravity manages to compensate pretty well for the relative motion of sources.

Grant Hutchison

grav
2009-Jul-29, 06:24 PM
After B has passed A, A will see that B is catching up with electrons that are moving more slowly than B is. So both A and B will see that these electrons strike B from the front, not the rear.I haven't finished with the math yet, but I've come to realize that if A is sending electrons directly toward the center of B's ship at all times, then there is no way B can catch up to the electrons, because the electrons would have to be travelling faster in the direction of B in order to do that. In other words, if the electrons are travelling directly toward the center of B's ship, then they will strike the hull from behind before they get there after B has passed A. In order to strike the hull from in front, the electrons would have to pass the center of the ship first along the direction of travel of B before B then catches up to strike the hull in front according to A, but that is not where A is aiming them.

grav
2009-Jul-29, 06:28 PM
Edit: Here (http://www.bautforum.com/1053536-post40.html) is Richard pointing out that gravity manages to compensate pretty well for the relative motion of sources.But that would mean that the field does travel ballistically with A, so always lines up with A when A is travelling inertially, always pointing to A's current position. Or actually, since A is travelling relativistically to B, and the origin of the field is just travelling with A, then the origin is travelling relativistically as well from that perspective, but anyway, you get the point.

grant hutchison
2009-Jul-29, 06:31 PM
I haven't finished with the math yet, but I've come to realize that if A is sending electrons directly toward the center of B's ship at all times, then there is no way B can catch up to the electrons, because the electrons would have to be travelling faster in the direction of B in order to do that. In other words, if the electrons are travelling directly toward the center of B's ship, then they will strike the hull from behind before they get there after B has passed A. In order to strike the hull from in front, the electrons would have to pass the center of the ship first along the direction of travel of B before B then catches up to strike the hull in front according to A, but that is not where A is aiming them.Think again. :)
You told us A fired the electrons towards the point at which B's centre would be when the electrons arrived. The slow-moving electrons swept up by B as it catches them from behind are indeed aimed so that they would cross the trajectory of B's centre just as B's centre arrived.

Grant Hutchison

grant hutchison
2009-Jul-29, 06:34 PM
But that would mean that the field does travel ballistically with A, so always lines up with A when A is travelling inertially, always pointing to A's current position. Or actually, since A is travelling relativistically to B, and the origin of the field is just travelling with A, then the origin is travelling relativistically as well from that perspective, but anyway, you get the point.You're missing the point. Gravity doesn't work in the simple way your field of electrons works. You can't learn about gravity by looking at this model.

Grant Hutchison

grav
2009-Jul-29, 06:36 PM
Think again. :)
You told us A fired the electrons towards the point at which B's centre would be when the electrons arrived. The slow-moving electrons swept up by B as it catches them from behind are indeed aimed so that they would cross the trajectory of B's centre just as B's centre arrived.

Grant HutchisonYup. Thinking about it again, I can still see that actually. It depends on how the speeds and angles are matched. I guess I'll just finish the math and find out. :)

grav
2009-Jul-29, 06:39 PM
You're missing the point. Gravity doesn't work in the simple way your field of electrons works. You can't learn about gravity by looking at this model.

Grant HutchisonIf the field of electrons is found to always precisely line up with A at low speeds and high speeds alike, then that becomes an invariant, and they will line up at relativistic speeds and light speed.

grant hutchison
2009-Jul-29, 06:43 PM
If the field of electrons is found to always precisely line up with A at low speeds and high speeds alike, then they will line up at relativistic speeds and light speed.Yeah, but they don't.
From B's point of view, A moves steadily towards him, suddenly fires a bunch of electrons towards him at an assortment of speeds, and them sweeps on by. The electrons arrive along a single line of sight that doesn't ever contain A, except at the moment of emission.

Grant Hutchison

grav
2009-Jul-29, 07:34 PM
Yeah, but they don't.
From B's point of view, A moves steadily towards him, suddenly fires a bunch of electrons towards him at an assortment of speeds, and them sweeps on by. The electrons arrive along a single line of sight that doesn't ever contain A, except at the moment of emission.

Grant HutchisonYup, you're absolutely right. I still haven't finished the mathematics yet, at least for other cases, but I didn't need much for this one, and it's easy to see. Let's say A emits the electrons just as B passes the perpendicular point between the paths of travel. This is all according to A's perspective. Let's give the electrons a velocity of vx along the line of travel of B in the same direction, and a velocity of vy directly perpendicular between the paths. Using those two velocities, we can easily see that once the electrons have been emitted, vx will always exactly match B's speed for all of the electrons and approach B along a straight line directly on the y axis while the entire line of electrons then moves away from A. They will always strike B's hull in the same place in that case also. So what does that say about a field pointing to the current position of an object if they are seen to point to the place at which they were emitted according to the observer?

grant hutchison
2009-Jul-29, 07:44 PM
So what does that say about a field pointing to the current position of an object if they are seen to point to the place at which they were emitted according to the observer?Very little, as it turns out. :) See the link I gave to Richard's post on another thread, explaining how gravity and EM don't work like your particle streams.

Grant Hutchison

grav
2009-Jul-29, 08:45 PM
Okay, if according to A, the distance between the paths is d and B is at a distance of -d from the perpendicular point when the electrons are emitted, then when the first electron is being received by B, another electron that is to be received at twice the time will be travelling at half of vx and vy of the first electron and will be at a distance of d/2 in each direction from B wherever B might be, and these same ratios will always follow wherever B is, so the line of electrons moves with B at all times, and while one electron is reaching B, the other end of the line of electrons is always at the same relative distance along the axis as when the electrons were emitted and still at y=0. In other words, the original angle between A and B was 45 degrees and while A sees the line of electrons move away, the angle of that line to B is always the same, so always remains 45 degrees.

grav
2009-Jul-29, 08:54 PM
Now that that's settled, let's replace electrons with light. We can only use a single pulse for this as emitted by A to reach B at some point, because light always travels at the same speed, so will not be seen to fan out or spread into a line. However, the angles will be the same from the perspective of A and B except for any corrections for aberration on B's part. In order for B to see the light coming from the point at which it was emitted and the current position of A according to B to also coincide with that point, simultaneity issues must be accounted for, so I will work on that next.

publius
2009-Jul-30, 02:04 AM
My brain is just sort of stuck in neutral for some reason, and I just don't want to think too much about all this. It's just one of those weeks, I guess. :lol:

The field of a moving charge is a well studied subject, and indeed, it doesn't work like one charge throwing particles at each other. Grav, you need to learn a bit of vector calculus and EM theory, where this stuff can be handled slick as the proverbial owl manure. The 4-vector formulation of EM (based on relativistic geometric concepts, of course) is the slickest, and the field of a moving charge can be derived in a very cute manner.

See Leinard-Wiechert potentials:

http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential

All this has been worked out and it's all consistent. The above gives it in the 3-vector vector/scalar form, but believe me, the 4-vector form wraps it up in one 4-potential (scalar potential is the time-like part, and the vector potential is the space-like part -- the field, however, is a rank-2 antisymmetric tensor -- in this formulation the EM field is a invariant geometric object. Only the coordinate components change with frame)

The "information" about a moving charge travels at the speed of light. Think of the charged particle radiated out little circles (spherical surfaces in
3D). Those little circles expand at light speed in all frames. The field at any point comes from a "retarded potential", which is when one of those circles hits a field point at any given time.

However, EM has a little trick up its sleeve. Although the information comes from the retarded field point (where the particle was in the past by the light travel mean time), the actual E field will not point at the retarded location, but ahead, to the point where the particle would be if it continued to move at constant velocity. It will only miss and cause radiation if the particle accelerates. The thing to appreciate is this correction to the E-field that makes it "anticipate" the present position is an effect on the Coulomb part of the field, due to the changing B field present in a frame where the charge is moving.

Gravity does one better than that, taking into account the acceleration of the particle, and will only miss if there is a change in acceleration ("jerk") over the light travel mean time. (And in the full analysis, it's a complicated non-linear mess, because curvature changes the null geodesics as it propagates, and spatial curvature in the strong field changes the notion of where something must "point" -- the "line of sight" becomes a curved path).

-Richard

grav
2009-Jul-30, 09:41 PM
Okay, let me ask this. If a source of constant light that is propagated in every direction were travelling inertially relative to us with some distance between us where the path becomes perpendicular, or to the closest point of passing, the same as with our last scenario, would the source of the light be observed at the same place it currently resides at the moment of observation?

grant hutchison
2009-Jul-30, 09:59 PM
Okay, let me ask this. If a source of constant light that is propagated in every direction were travelling inertially relative to us with some distance between us where the path becomes perpendicular, or to the closest point of passing, the same as with our last scenario, would the source of the light be observed at the same place it currently resides at the moment of observation?No. There's an aberration effect, classical or relativistic.
That's a different matter from the apparent direction of the source of a moving electrical or gravitational field.

Grant Hutchison

grav
2009-Jul-30, 10:15 PM
No. There's an aberration effect, classical or relativistic.
That's a different matter from the apparent direction of the source of a moving electrical or gravitational field.

Grant HutchisonOkay, that's what I'm starting to gather. I can see no way of lining them up using only SR, in fact relativistic aberration moves them even further apart than classical would, so I figure GR must be introduced and adds something else which will move the perceived motion of the origin of the field back to moving ballistically with the source when it continues to travel inertially, which I find ironic. Is that right?

grant hutchison
2009-Jul-30, 10:18 PM
I can see no way of lining them up using only SR, in fact aberration moves them further apart, so I figure GR must be introduced and adds something else which will move the perceived motion of the origin of the field back to moving ballistically with the source when it continues to travel inertially, which I find ironic. Is that right?I don't know. You're the only person who can know if you find something ironic.

Grant Hutchison

grav
2009-Jul-30, 10:21 PM
I don't know. You're the only person who can know if you find something ironic.

Grant HutchisonSomehow I knew you'd say something like that when I stated it that way. :) I meant the GR part.

grant hutchison
2009-Jul-30, 11:20 PM
I meant the GR part.Yeah. Just trying to discourage those stream-of-consciousness dependent clauses. :)
There's a gravitomagnetic component to a moving gravitational field, just as there's a magnetic component to a moving electrical field. So these things shake down in a different way from a simple stream of particles.

Grant Hutchison

publius
2009-Jul-30, 11:36 PM
Consider a frame where a point charge is moving, but a test particle at a field point is not. That is, we're in the rest frame of the test particle. In that frame, while there is a magnetic field, since the test particle isn't moving,
v x B = 0. The correction that makes the field point ahead of the light delayed image of the charge is to the E-field itself. The B field is time-varying, and thus we have a component of E due to dB/dt (or dA/dt if you think in terms of the potentials).

And so it is with the gravitational field of a moving source. It's a "gravitoelectric" correction (and one that is more complex than EM, of course).

Now, when the test particle is moving in a frame, then there is a magnetic component to the total force on tha particle.

The best way to think of this is in terms of the 4-vector formulation. The EM field, rather than being two fields, E and B, is one tensor field. What we call E and B are just coordinate components of that tensor field. The force on a test particle is given by a covariant tensor expression that holds in all frames. It's just the components of that vary from frame to frame.

THe field of an inertial point charge is the same invariant tensor field no matter what frame you're in. In the rest frame of that charge, it's all electric, and the 4-potential resolves into a single time-like scalar component, the familiar electric scalar potential. In moving frames, that same potential and field resolves into difference components, having both vector potential and magnetic components. But it's the same field.

-Richard

grav
2009-Jul-31, 01:13 AM
Thanks Grant and Richard. I guess it's about time I learned this stuff better. Let's say we have the poles of two magnets placed near each other such that the magnetic field lines run parallel from one pole to the other. If we place a charged particle between them that is stationary to the magnets so that v = 0, there is no force that acts upon the particle, right? Now let's say the particle is moving at some speed perpendicularly to the field lines. Now the particle's path will begin to circle around into an orbit, right? I always thought that was just the magnetic force acting upon the particle, but now I'm wondering if since the centrifugal force acts perpendicularly to the line of travel and also perpendicularly to the magnetic field lines, maybe that force is the electric component. So just starting off with basics here, is the centrifugal force considered an electric component that has been introduced due to the particle's motion in a magnetic field, or is that still just considered the magnetic force?