elfmotat

Photons do not have a perspective because there is no valid reference frame in which they are at rest. Why do you always make me repeat myself? Am I going to have to say this 4-5 more times? I'll do it in advance, just to get it out of the way: they don't have perspectives. They don't have rest frames. You can't talk about the experience of a photon. Time does not pass for anything moving at c. There is no photon perspective. End of story.

But it's not a frame of reference. Like I said, photons do not have rest frames.

If you want to be a crank, fine, that's your decision. There is no "photon's perspective." Photons do not have rest frames.

How would you quantify this? How would you test it?

I don't particularly care why you received warnings. You're still only responding to the "gravity vs. gravitation" question with deflection. Are you going to explain it or not?

Look, it's not our fault that your nonsense is nonsense. You're getting honest feedback here. I know you'd prefer validation, but you're not going to get it because nothing you're saying makes any sense. That's the truth. I think it's safe to say that nobody has a clue what you mean when you say that gravity and gravitation are "different." People have asked for clarification several times now, but all you ever say is "I already explained the difference in a previous post." It's not our fault that you can't follow the rules either. You've been hijacking threads left and right.

So basically your question is, "what is the definition of 'alien?'"

It depends on what you mean by "relative." If you mean "frame-dependent," then it is not relative. If you mean "measured relative to some other object," then it is relative. This is all just semantics.

A better answer for why you can't possibly get any torque on the wheel, regardless of the distribution of matter around it, is because gravity is a conservative force. This means there's no "circulation" in the field, so no matter how clever you are in positioning the wheel it simply will not rotate.

Oh no, what are you on about? Neutrinos are massive particles. That means they don't move at c, and they have rest frames. Nothing moving at the speed of light has a rest frame.

Don't forget that φ increases in the counter-clockwise direction. That means eφ is oriented in the counter-clockwise direction. An easier way to see it might be to take points where we do know B. For example, on the positive x-axis (where y=0) we have x=r, we know the magnitude of B is Bφ (because it's the only nonzero component), and B is oriented in the +y direction. In other words, we have: [math]\mathbf{B} (y=0) = \frac{\mu_0}{2} J_z x \, \mathbf{e}_y[/math] You can do the same thing on the positive y-axis as well, which is where the minus sign should come from. The should be easy to see, because B will be oriented in the -x direction.

No, I'm saying that the net force on the wheel provides no torque. I.e. you can't get any rotation out of it. Yes, I can tell you for sure that there will be no torque on the wheel.

Say you're given the position vector r=(r,φ,z) and you want to find the components of the vector in Cartesian coordinates, (x,y,z). Try working that out first. Now by analogy, how would you find the components of some general vector A in Cartesian coordinates, given the same vector in cylindrical coordinates?

First, you've ignored some other criticism such as why there appear to be three lumps inside nucleons if quarks don't exist, as you claim. Second, several of us took the time to give several explanations for why gravity can't be electromagnetism. You're ignoring some of them (gravity is spin-2 while EM is spin-1, gravity is highly nonlinear while EM is linear, not everything interacts electromagnetically while everything interacts gravitationally), and giving half-baked nonsense explanations for the others. Positive charges certainly DON'T fly into space, and there's no such thing as aether. You're ignoring real science in favor of some seemingly-clever idea that is trivially false. You're not the first person to have this idea, and it's already been falsified in a thousand different ways. It simply does not work. Your time and effort would be much better spent learning actual physics. You're heading down the road to crackpotville. Turn back while you still can!

If all you know about physics is the inverse-square laws for gravity and electrostatics, I can see how it would be tempting to think that they would be related in some way. Unfortunately, things are not that simple. First: if the gravitational field is really just a small positive electrostatic field, what do you think would happen to positive charges on Earth? They would be shot into space! But we can accumulate positive charge in the lab very easily without it having any effect on the weight of anything, so this idea clearly has to be wrong. A more sophisticated answer is the one I was giving before: the EM field does not have enough degrees of freedom to explain gravity. An even more convincing answer is the one Mordred gave: some particles do not interact electromagnetically, for example neutrinos, while all particles interact gravitationally. And one final killing blow: if gravitational fields were really EM fields, then they would have no effect on light! One of the key aspects of the EM field is that it is linear. This means that the field will not interact with itself. Light itself is an EM field, and so it should not interact with external EM fields. But one of the famous experiments to test general relativity was the famous light-bending experiment, where the deflection of light due to gravity as it passed near the sun was measured. So we know that gravity does effect the path of light, meaning it cannot possibly be explained by EM fields!

The notation in that equation is a bit funny. Did you mean: [math]B_{\phi} \left( r \right ) = \frac{\mu_{0}}{2} J_z \, r[/math] ? Because if so that looks correct. Now the question is, how can you relate B in cylindrical coordinates to B in Cartesian coordinates? Plugging in [math]r=\sqrt{x^2+y^2}[/math] will give you [math]B_{\phi}[/math] in Cartesian coordinate, but that's not what you're looking for. You're looking for [math]B_x[/math] and [math]B_y[/math].

Maybe this will help you visualize it:

Yes. Any energy a free particle has at rest will be measured as inertia.

I just explained why not: there aren't enough degrees of freedom in the EM field. The only known way (AFAIK) to get both the EM field and the gravitational field together is by adding extra spacial dimensions in Kaluza-Klein type theories, for example string theories. So now we're getting into unified field theories. Anyway, this is probably well beyond the scope of what you're talking about.

The gravitational field couples to a rank-2 tensor (the stress-energy tensor) and the electromagnetic field couples to a rank-1 tensor (four-current). The EM field simply does not have enough degrees of freedom to explain gravity.

Looks like a scam. I would stay away.

Nope, not seeing it. Because it doesn't exist.

He's asking why there appear to be three distinct lumps inside nucleons. Momentum is a property, not a thing. That's not antigravity, that's magnetism. "Antigravity" has a specific meaning: repulsive gravity. Not magnetism, or something to counteract gravity, but repulsive gravity itself. Light does not need a medium to propagate. It's just an electromagnetic field.

Oh, the irony.

Using caps lock and big font does not imply I was angry. It implies I was trying to drill home a point that I've been repeating for two pages now, without much effect.