# Nature of Time Dilation and Length Contraction

Discussion in 'Physics & Math' started by Prosoothus, Apr 4, 2006.

1. ### DaleSpamTANSTAAFLRegistered Senior Member

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Yes, it does. E=F/Q. If F=0 then E=0.

-Dale

3. ### ProsoothusRegistered Senior Member

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Dale,

So if you have an electric field pushing at all sides of a particle equally, then you don't have an electric field, right?

5. ### DaleSpamTANSTAAFLRegistered Senior Member

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Exactly. The electric field is a vector field, so if the individual contributions are "equal and opposite" vectors then the vector sum is zero so the field is zero.

-Dale

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi Prosoothus,
You're not speaking Dale's language. The thing you are calling "Electric Field" is what Dale and most educated in physics call Electrostatic Potential.

The thing that Dale (and most educated others) means by Electric Field describes the way that the electric potential changes in space.

8. ### DaleSpamTANSTAAFLRegistered Senior Member

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Thanks Pete, I have not been following the discussion at all so I probably should not have inserted my comments.

Prosoothus, the electrostatic potential is constant inside a charged sphere, but it can be non-zero. The potential is related to potential energy, which can be non-zero inside a charged sphere, while the E-field is related to force, which is zero inside a charged sphere. The potential is a scalar field (since energy is a scalar quantity) and the E-field is a vector field (since force is a vector quantity). The two fields are related to each other through the gradient operator. Sorry if my comments caused any confusion.

-Dale

9. ### przyksquishyValued Senior Member

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Electric fields are vector fields. The only way a vector field can have no direction is if there's no field.
What you're essentially claiming here is that a dipole can move without the help of an external electric field.
What's a true electrostatic dipole?
Depends on the charge of the particle. If it's positively charged, you'd need an electric field acting in the direction you want the particle to go. If it's negatively charged, you need a field acting in the opposite direction.
You're probably right here. I was actually quite surprised to get no net result. I expect I made an error somewhere (I'm a sloppy mathematician at best), so I'll have another look at this when I have time.

Have you tried developing the mathematics for your theory yet?

10. ### ProsoothusRegistered Senior Member

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Pete and Dale,

OK. I thought the electric field was primary, while the force it creates was secondary.

One more thing, even though an electron in a negatively charged hollow sphere is not being pushed in any specific direction, isn't it being compressed by the sphere's electric potential anyway? Wouldn't that compression force, even though it may have no effect on the electron, imply that there is an electric field present in the sphere?

11. ### ProsoothusRegistered Senior Member

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przyk,

I'm saying that the electric potential inside a sphere has no effect on a monopole charge inside the sphere because the charge is being pushed, or pulled, equally at all sides by the fields in the sphere. If, on the other hand, you had a particle in the sphere that had an un-uniform electric field, like a dipole, the sphere would push against the side of the particle with the "more" negative field, forcing the particle to move.

If it existed, it would be the equivalent to the electric field what a magnet is to a magnetic field. It might be possible to create a true electrostatic dipole by converting an electric field to a magnetic field, then back again, but that would be messy.

Let me rephrase my question:

Let's say you had a negatively charged hollow sphere and a particle in the center of that sphere. You want to move the particle, but you couldn't change the charge or shape of the sphere. You could, however, change the strength, shape, or direction, of the electric field of the particle any way you wish. How would you modify the electric field of the particle to make it move in a specific direction?

I don't have the mathematical knowledge to do that. I'm not a college educated physicist or engineer like some of the other members on this forum. I never even went to college.

12. ### przyksquishyValued Senior Member

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The force on the particle is F = qE, where F and E are respectively the force and electric field vectors, and q is the particle's charge. The particle is not affected by it's own field, and fields dont push or pull each other in the way you seem to be suggesting.

Anyway, why are you still asking about charged spherical shells? Your theory is that non-zero gravitational fields propel dipolar photons in empty space, not that they'll move around inside a spherical shell of mass where there's no gravity.

13. ### ProsoothusRegistered Senior Member

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przyk,

The electric field, or electric potential if you wish to call it, is uniform and omnidirectional in the center of a charged hollow sphere. This field is the closest example to a non-uniform gravitational field that I can imagine.

Let me also make clear that I'm not saying that light travels at c because it is propelled by gravity, I'm saying that it travels at c because it is propelled by gravitational fields. By gravity it is assumed a gravitational field that is causing a standard mass to move in a specific direction. In other words, that type of gravitational field is unidirectional, or non-uniform. A photon would accelerate in a gravitational field even if that field is omnidirectional or uniform because the photon's own field is unidirectional and nonuniform so it doesn't need an external field that is unidirectional or non-uniform to accelerate.

One more thing, a standard mass inside a a spherical shell of mass will remain stationairy not because there is no gravitational field inside the sphere, but because the mass is attracted equally at all sides by the field of the sphere due to the uniform gravitational field of the mass. If the mass in the center of the sphere did not have a uniform field, but instead had a field pointing in one direction, or even dipolar, that mass would start to accelerate in a specific direction.

14. ### imaplanck.BannedBanned

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What measurements have you taken to justify this as anything more than mind pollution?

15. ### kevinalmRegistered Senior Member

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Someone needs to learn the difference between vector fields and scalar potential fields, and how the two are related.

16. ### przyksquishyValued Senior Member

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First of all, make sure you don't confuse the electric field and electric potential - they're two different things. The electric field at every point in space is the force that would be exerted on a hypothetical 1 C test charge at that point. It's a vector field, and has the units of newtons per coulomb. The electric potential is the potential energy this test charge would have (with respect to some reference point); it is measured in joules per coulomb, or volts, and is a scalar quantity.

Mathematically, the electric field is the gradient of the electric potential, or E = (<sup>&part;V</sup>/<sub>&part;x</sub>, <sup>&part;V</sup>/<sub>&part;y</sub>, <sup>&part;V</sup>/<sub>&part;z</sub>).
The electric field is uniformly zero everywhere inside the sphere. The electric potential is constant; it's exact value is arbitrary.
I'm not following this. What does it mean to be propelled by a field but not by a force? Does this have an electric field analogy?
The field is zero inside the shell because the fields of all the particles on the surface of the sphere cancel out everywhere inside the sphere.
Actually the opposite would happen: the non-symmetric field of your mass would result in it exerting a net pull on the shell. The shell would move, but your dipolar particle wouldn't (defying Newton's third law and the law of conservation of momentum).

17. ### ProsoothusRegistered Senior Member

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przyk,

It is known that a test charge inside a hollow charged sphere would not be effected by the charge of the sphere. As you and Dale pointed out, since no force would be exerted on the charge, excluding any compression or expansion forces, you can argue that no electric field exists within the sphere. However, the two of you didn't take into consideration that just because the sphere doesn't exert a force on a uniformly charged particle, that doesn't mean that it won't exert a force on a non-uniformly charged particle. If the charged sphere exerts a force on a non-uniformly charged particle, like for example a dipole, then you would have to accept the fact that the electric field inside a hollow charged sphere is not equal to zero, right?

The particle is propelled by a force that is the result of the interaction between the particle's field and the external field.

I see that we generally agree (except that I still believe in Newton's third law

).

18. ### ProsoothusRegistered Senior Member

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imaplanck,

It's just mind pollution.

Seriously, I'm trying to develop a theory that not only explains why photons travel at c, and why they accelerate to c, but also explains experimental data that physicists use as proof of relativity without having to use the physically impossible phenomena of length contraction and time dilation.

Oh, and welcome to sciforums.

19. ### przyksquishyValued Senior Member

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It's not just a measurement, it can be deduced (fairly easily) from Gauss's law (one of the laws of electromagnetism). If you think there's a field inside the sphere, You also think Gauss's law, and consequently Maxwell's theory, is wrong.
Electric fields don't interact with one another, they just superpose.
Newton's third law isn't just believed as a law on it's own - it's a property of the known fundamental forces of nature.

Last edited: Jun 27, 2006
20. ### DaleSpamTANSTAAFLRegistered Senior Member

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Actually, it does. It is pretty clear with fundamental calculus concepts. Take any given smoothly varying non-uniform distribution of charge. The charge is uniform across a differential element. The force on the differential element is therefore zero. The integral of zero over the entire non-uniform charge distribution is still zero.

The case of the e-field inside a charged spherical shell is actually somewhat stronger than the above presentation. The force is equal to zero not just for a small uniform charge but for a point charge. This means that the above analysis not only applies for smoothly varying charge distributions but for any arbitrary charge distribution, even those with non-differentiable boundaries.

-Dale

21. ### CANGASRegistered Senior Member

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Mathematics is an attempt to create an understandable model of reality.

Reality is reality. Mathematics is not reality.

22. ### DaleSpamTANSTAAFLRegistered Senior Member

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No, math is logic. Physics is an attempt to create an understandable model of reality. Using math in physics simply ensures that your model is logical.

-Dale

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