Mordred

No you were never wrong to question it not in the slightest. It's also why I chose to work with you to straighten it out. It's all good glad we could work it out. After all its how we all learn I will try to be more accurate and clear in further mentions of rapidity.

Why do you keep declaring Einstein theory does this or does that even after everyone has told you most your misconceptions are wrong. The Earth does not suddenly age different simply because one observer looks at it. There is literally billions of observers on Earth they all do not have any effect on the rate the Earth ages. That is pure nonsense. The Minkowskii metric doesn't even state that. By the way thanks for providing the math I asked for. Unfortunately so far as pointed out all the evidence with regards to c being invariant is something you shouldn't ignore.

Excellent precisely what you should be of it in terms of

Your fairly close to the right idea. Without going into the quantum regime too intensely. In essence the overall electron spin up/ spin down alignments contained in each domain gets altered. Some electrons will switch from spin up to spin down or the overall orientation changes by some angle. So the fields of the magnet is already present even when it's not interacting with another object. So the charge currents are essentially zero (it's never truly zero as there is always some electron exchanges). So one can equate this to the PE term (potential energy) When the nail interacts with the magnet. The interaction of the magnet including the B field provide directivity of the charge current that results from the interaction between the magnet and the nail. We see this directivity in the magnetic field lines. The tighter the field lines the greater the amount of force. So further away the field lines diverge and gets weaker. (1/r^2). So in essence the electrostatic field does the work. The B field interaction in essence provides directivity of the charge current. A charge current is a kinetic energy term.

I like your example +1 in point of detail Amperes law teaches us that all magnetic phenomena is the result of electric charges in motion. Faraday discovered moving magnets generates an electric current. Maxwell and Lorentz in essence put together the final touch that E and B are not separate entities but are inexplicitly intertwined. So even a point charge has E and B fields. Now it takes a charge to produce an electromagnetic field, but just as importantly is that it takes another charge to detect an electromagnetic field. Now when you have an ensemble of charges you use the principle of superposition which tells us the interaction of two charges is unaffected by the presence of others. So you can compute the force resulting from each charge to the test charge and sum up to the total vector sum for total force on the test charge. Now you probably recognize I just described the electrostatic field. However with that field you now have to think in terms of charge density and charge currents. (By the way this applies to QFT as well) including the Feymann path integrals, just an FYI). So in point of detail the force on the test charge results from the sum of force of the individual point charges mediated by the EM field. Now we can further break down this Electrostatic field into surface charge, line charge, continuous distribution and volume charge. Each has has its own integral combined with Coulombs law. for example charge distribution \[E_r=\frac{1}{4\pi \epsilon_0}\int\frac{1}{r^2}\hat{r}dq\] line distribution \[E_r=\frac{1}{4 \pi\epsilon_0}\int \frac{\lambda(\acute{r})}{r^2}\hat{r}d\acute{l}\] surface charge \[E_r=\frac{1}{4 \pi\epsilon_0}\int\frac{\sigma(\acute{r})}{r^2}\hat{r}d \acute{a}\] and volume charge which we use most often. as being the one most referred to with Coulombs law \[E_r=\frac{1}{4 \pi\epsilon_0}\int\frac{\rho(\acute{r})}{r^2}\hat{r}d\acute{\tau}\] So knowing that according to Amperes law magnetism is the result of electric charges in motion. One has to ask well how does a permanent magnet work. What materials are more likely to make a magnet which materials would make a stronger magnet? To better understand that one has to understand how readily a material accepts domain realignment via a process called hysteresis. However it should be more clear that the charge distributions described by the formulas above directly relates to the sum of coulomb force to the test charge "d" is domain while the identifier after it is the domain type. the "r" with the hat is the distance from the domain to the test charge. So ferromagnets has domains with domain walls the walls are potential difference separations each domain has its own hysteresis. Histeresis describes a phenomena that when you pass a magnet near a ferrous material the alignments of the point charges do not return to the original configuration. (ever have a screw driver that you often use to work on an electric circuit eventually become a permanent magnet ? ) its due to hysteresis. hope that helps better understand the electrostatic field and ferromagnetism So now you should be able to answer the question :" Where does the energy come from" in the permanent magnet case...think domain charge densities and hysteresis due to the magnet interacting with the nail. This will also help when you look at things like Currie temperature and how it effectively it can be used to realign domains The domain alignments has potential energy there is no outside interaction so no current flow but you still have a charge density. When you place the nail near the magnet to interact the interaction exchange results in a charge current flow. This describes a kinetic energy term mediating the force. Now unfortunately a lot textbooks teach flow of electrons in a copper wire etc. It isn't the flow of electrons, its the flow of charge. Electrons could not flow through a medium fast enough for one thing. However the flow of charge can as charge is mediated by photons. It serves as the momentum carrier to alter the spin alignments of the electron ensemble edit forgot to add the primes (I tend to use acute ) are the source coordinates of the given domain for example \(d \acute{a}\). The symbols \(\lambda, \sigma, \rho\) is charge per unit (length, area, volume). The above also helps better understand induction. Your inducing charge current.

It might help to consider even in atoms electrons never stay still. In permanent magnets those electrons are moving around within the atoms of the magnet as well as the environment. However magnetism isn't a force nor is it a form of energy. A common analogy is to think of it as a translator with the E field. It results from the E field current and can thus be used to affect the E field through induction. lol you also run into articles etc stating permanant magnets have no E field but that wouldn't be true. The atoms have electrons and is held together by the EM field. So when you move the magnet to the nail your really just inducing electric current in a field already present which does the work via electromagnetic induction. (keep in mind I'm keeping the mediator photons out of the equation for this discussion) ie keeping it classical rather than quantum lol

that would be useful perhaps you should start showing the related math to go with which Lorentz Eather variation your using as there are numerous changes and revisions over the course of its development. One of those variations violated conservation of energy/momentum due to symmetry loss with regards to the preferred frame

\[{\small\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline Field & \ell_L& \ell_R &v_L&U_L&d_L&U_R &D_R&\phi^+&\phi^0\\\hline T_3&- \frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}&0&0&\frac{1}{2}&-\frac{1}{2} \\\hline Y&-\frac{1}{2}&-1&-\frac{1}{2}&\frac{1}{6}&\frac{1}{6}& \frac{2}{3}&-\frac{1}{3}&\frac{1}{2}&\frac{1}{2}\\\hline Q&-1&-1&0&\frac{2}{3}&-\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}&1&0\\\hline\end{array}}\]

You know you really need to define what you consider Real. Ignoring math when it doesn't agree with your your conjecture and using the term Real doesn't help. You cannot tell me LET doesn't have the same symmetry relations so your Real excuse makes zero sense

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The better question is what gives rise to the B field. The E field current generates the B field. With permanent magnets the E field current is generated due to the electron charge alignments as per ferromagnetism so there is always an E field current allowing the B field If you take a rotor for example and law it on its side so the opening is facing you the E field current will be through the center of the rotor heading either towards you or away from you depending on the magnetic pole alignment. What allows the rotor to turn depends on varying the E currents with the different poles of the rotor. Typically 3 poles for 3 phase motors. The phase shifts provides the differences in current in each pole. It is still the E field performing the work. For DC motors it's much the same you send pulses at selected poles to generate the field variations to induce rotation. The number of poles is equal to the number of signal wires the device has and you send pulse patterns to the DC stepper motor.

The E field not the B field. Doesn't matter if it's a permanent magnet or an electromagnet it's still the same. To help understand the permanent magnet if you look at inductance it does have both the E and B fields . The reason why the B field has less energy and doesn't do the work involves how the field diverges as opposed to the E field coupled with the Lorentz force law via the right hand rule. That directly relates to Swnsonts previous statement with regards to the cross product for the B field as opposed to the inner product of the E field

Just a reminder I did post the related mathematics using Maxwell equations to describe what Swansont is stating in this thread previously

Here this applies regardless of model and applies to all those models. There's your symmetry \[\vec{A}\rightarrow B= A \leftarrow \vec{B}\] It doesn't matter what the vector represents that symmetry applies. The goal isn't to eliminate that symmetry but to preserve it. The Gamma factor is the correction to preserve that symmetry. To restore to Galilean relativity and preserve all the physics described by Newtonian physics. The same thing occurs with LET it must also do the same in order to have any validity with observational evidence. That's what the axiom "the laws of physics must be the same regardless of observer" literally describes.

Let's put this bluntly if you handed me Galilean relativity , SR relativity, GR relativity and LET. I can point out that all the above have the same symmetry relations involving signals sent between A to B and B to A regardless of model.

There is no difficulty understanding this symmetry. Neither Einstein nor Lorentz nor anyone that looks at the equations would have difficulty understanding that symmetry. This is why I've been telling you to look at the transformation equations. The symmetry relations is included in the very formulas. It is the Lorentz ether based model that has the difficulty in maintaining that symmetry. The Einstein models has no issue preserving that symmetry. I can mathematically show you quite easily those symmetry relations. They will have identically the same reasons as the symmetry relations of the signals sent between two cops using nothing but Doppler shift.

We need to get you over this misunderstanding. No faraway observer causes a distant object to suddenly change the rate it ages. That doesn't happen in any form of relativity. No matter who else is observing observer A on Earth Observer A clock will run the same rate. Observer B who examines his clock will see his clock running normally regardless of other observers. It is only when you compare clock A and Clock B that you notice the two do not run the same. This applies to light clocks as well. If an observer watches a series of pulses between A and B. The speed of light remains constant. The wavelength of each pulse (redshift/blueshift) will vary. It's no different than a cop using a speed radar. If you have two cops monitoring each other with a speed radar they will both get the same readings. The total redshift from A to B will be identical to the total redshift from B to A. That's due to directional symmetry. It doesn't matter if the redshift is caused by motion or gravitational. That directional symmetry still applies. If either A or B accelerates the resulting redshift change in the pulses sent during the time the acceleration occurred with have a variation in its redshift . This will be noticed by both observers monitoring the others emitter signal. So both observers monitors the signal sent to the. Will see the same difference on both signals. Thr redshifted signals are not sufficient alone to determine who is aging. As both observers will still see the same redshift variations at either end.

Nice That ones much easier to insert into complex equations

Well once many thanks I had that problem over a year ago so it's nice to finally know where was

\[{\not}\tiny\,\normalsize\partial\] ah so its the use of displaystyle throwing it off many thanks \[ S=\int dx^4 x\sqrt{-g}[\frac{M_P^2}{2}R-\frac{1}{2}(\partial\phi)^2-\frac{\lambda}{4}F^4_\infty-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{g^2}{4}F^2_\infty B_\mu B^\mu-i\bar{\psi}{\not}\tiny\,\normalsize\partial \psi+\frac{y}{\sqrt{2}}F_\infty\bar{\phi}\phi ]\]

nicely covered +1, well described

\displaystyle{\not}\partial \[\displaystyle{\not}\partial\] if anyone knows a way to make the Feymann slash more visible I would appreciate it greatly

Palatini Higgs Induced gravity scenario \[S=\int d^4x\sqrt{-g}[\frac{\xi h^2}{2}R-\frac{1}{2}(\partial h)^2-\frac{1}{4}h^4-\frac{1}{4}f_{\mu\nu}F^{\mu\nu}-\frac{g^2}{4}h^2B_\mu B^\mu-i\bar{\psi}{\not}\tiny\,\normalsize\partial\psi-\frac{\gamma}{\sqrt{2}}h\bar{\psi}{\psi}]\] {\not}\tiny\,\normalsize\partial scalar field h, vector field \(B_{\mu\nu}\) fermion field \(\psi\) above Abelion with standard \(B_{\mu}B^{\nu}\) kinetic terms \(F_{\mu\nu}F^{\mu\nu}\) scalar field expectation value \[G_{n,eff}\equiv\frac{1}{8\pi\xi h^2}\] to keep \(G_{n,eff}\) well behave non-minimal coupling \(\xi\) is constrained to positive values for semi-positive definiteness of the scalar field kinetic term. shown by a field redefinition \(h^2\rightarrow h^2\xi\) Einstein-Hilbert frame redefinition of the metric terms. \(g_{\mu\nu}\rightarrow \Theta g_{\mu\nu},,\Theta \equiv \frac{F^2_\infty}{h^2},,,F_\infty\equiv\frac{m_P}{\sqrt{\xi}}\) with rescaling of the vector and fermion fields \(A_\mu \rightarrow \Theta^{-1/2},,,,\psi \rightarrow \Theta^{-3/4}\psi\) \[S=\int d^4 x[\frac{M_p^2}{2}R-\frac{1}{2}m^2_P K(\Theta)(\partial\Theta)^2-\frac{\lambda}{4}F_{\mu\nu}F^{\mu\nu}-\frac{g^2}{4}F^2_\infty B_\mu B^{\mu}-i\bar{\psi} {\not}\tiny\,\normalsize\partial \psi-\frac{\gamma}{\sqrt{2}}T_\infty \bar{\psi}\psi]\] contains non-canonical term for the \(\Theta\) with kinetic coefficient \[K(\Theta)\equiv\frac{1}{4|a|\Theta^2}\] quadrupole at \(\Theta=0)\) and a constant \[a\equiv\frac{\xi}{1+6\xi}<0\] canonical field correction via field redefinition \[\Theta^{-1}=exp(\frac{2\sqrt{|a|\phi}}{M_P})\] \[ S=\int dx^4 x\sqrt{-g}[\frac{M_P^2}{2}R-\frac{1}{2}(\partial\phi)^2-\frac{\lambda}{4}F^4_\infty-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{g^2}{4}F^2_\infty B_\mu B^\mu-i\bar{\psi}{\not}\tiny\,\normalsize\partial \psi+\frac{y}{\sqrt{2}}F_\infty\bar{\phi}\phi ]\] https://arxiv.org/abs/1807.02376

One of its useful relations is how it increases. Rapidity increases linearly when describing the four momentum and four force of the object. So if you have some object/craft/particle undergoing constant acceleration. You need some constant force as well. So as we all know the particle velocity will never reach it will always approach but reach c. However the rapidity of the the object under constant acceleration grows linearly so it can grow arbitrarily large. Velocity has c as a limiting factor but that isn't true for rapidity. That is very useful when you want to use the four force to maintain the constant acceleration of the spacecraft.

yes its different than proper velocity. This might help first rapidity is a dimensionless parameter that alone is a significant difference from velocity however rapidity is the hyperbolic function relating (v/c) \[TanH\alpha=(v/c)\] it adds as \(\alpha=\alpha_1+\alpha_2\) with the following \[tanh\alpha=\frac{tanh\alpha_1+tanh\alpha_2}{1+tanh\alpha_1tanh\alpha_2}=v=\frac{v_1+v_2}{1+v_1v_2/c^2}\]