Mordred

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 \[tanh\alpha=\frac{tanh\alpha_1+tanh\alpha_2}{1+tanh\alpha_1tanh\alpha_2}=v=\frac{v_1+v_2}{1+v_1v_2/c^2}\]

yes I'm fully aware rapidity is not equal to proper velocity. why do you think I went through the equations for the hyperbolic relations ? If I didn't know that? did you miss my earlier comment in regards to this ?

pray tell how am I not using it correctly in your understanding Yes rapidity's are additive that's literally one of its conveniences. I provided the link showing the vector additions for rapidity.

Lets look at the reality of the SR/GR symmetry relations. The reality is that they describe in many cases our conservation laws. For SR this directly relates to measuring physical properties. Properties such as the momentum/velocity accelerations particle spin etc etc etc including temperature Any measurable quantity is physically real. If anything it could be argued that the Eather isn't real as its not measurable.

If you placed two regions of a static space described by Eather. Those regions would be symmetric, The Eather treated as a field would be described by scalar functions. So it too is symmetric. The freefall state or more accurately the state which has no force or interaction is symmetric as per SR. The constant velocity conditions. pray tell how is symmetry not applicable to the Eather. If you want I can readily provide the mathematics for a static scalar field. It will be symmetric The Cartesian coordinates used are symmetric. so pray tell why would you believe symmetry doesn't apply to Lorentz eather ? or is it some particular symmetry relation your referring to perhaps

do you agree the mathematics are required to calculate the age in order for the twin paradox to have any meaning Yes or no. The transforms in the Ether Link was prior to the SR version or did you forget Lorentz had to regularly fix his theory ? So what your telling me is that in order to calculate the age difference I would use the SR transforms in the same manner as done in SR and GR and they would be equivalent. However You also claimed that SR and GR cannot resolve the twin paradox yet you cannot calculate the age difference without using SR/GR is that correct ?

this is the SR Lorentz transformations \[\acute{x}=(x-vt)\] \[\acute{t}=\gamma(t-\frac{vx}{c^2})\] they are not the same as in that link now calculate the age using the link you just provided. You need the math to calculate age so obviously the math is a huge part of the solution to the twin paradox. Yet you choose to ignore the mathematics while claiming Einstein theory is incorrect. Yet in the same breath claim they are the same