Mordred

Well as this is in philosophy how about complexity of choice? Though mathematically this can also be broken down to degrees of freedom. ( Boolean logic per example,) not all forms of complexity is physics based. Though mathematics is universal lol. Choices and decision making is added complexity without being based on the amount of mass nor energy (mass being part of your degrees of freedom) energy the ability to perform any degree of freedom. Energy is a given requirement. So mass /energy cannot define complexity these are just factors. Same on degrees of freedom, other factors include chemical reaction, (energy is usually replaced with temperature on global average), Mathematically its feasible to reduce chemical reactions to degrees of freedom but I'm no chemist. For choices ie decisions it also can be reduced to degrees of freedom. ie Yes no being two degrees. The other consideration is degrees of freedom doesn't describe structure. Matter tends to be more orderly than a gas. etc. Complexity covers a wide range of factors we can mathematically reduce and describe it though. The degrees of freedom covering a large portion. Recall the above on mathematics modelling complexity. Here is a 865 page article covering just this "Dynamics of complex systems" https://www.google.ca/url?sa=t&source=web&rct=j&url=https://fernandonogueiracosta.files.wordpress.com/2015/08/yaneer-bar-yam-dynamics-of-complex-systems.pdf&ved=0ahUKEwjKh6zsv_PTAhUOS2MKHdvUAikQFggxMAQ&usg=AFQjCNG1tdJqxohINLSqEoM88orAfOJt8A&sig2=08TMx3bZ9J1Q07GNT-ZgoA

Well you definitely have done a lot of mathematical research. The math details itself will take a considerable time to go through. I should note a peer review done by a forum regardless of how good that forum is. Would not carry any real authority. About the best we can help with is spotting errors or possibly finding ways to improve the article. On that note a better math notation in many places to improve readability might be a suggestion. For example definition 3 can readily be simplified by simply using tensors. On this particular case the Euclidean Minkowskii tensor would be suitable. Particularly since your introduction mentions both QM and relativity. Which both metrics are primarily 4d. ( once you apply the Schrodinger and Dirac equations) under the QM side. Reduce definition 3 to the Minkowskii tensor and the show the vector inner dot product is a far more elegant method to show the vector displacement elements. [latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex] [latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex] [latex] u*v=v*u[/latex] Though your paper did not include the time coordinate component under definition 3. (adding the full 4d geometry is my other recommendation) Particularly in the infinity issues your paper is addressing. You are specifying R^3 as opposed to R^4 not sure why this choice at this stage. Particularly since QFT already uses displacement in its metrics via action and the Hamiltons. I am curious why you have not included the time coordinate in your observable displacements? Within the full paper I do not see the time component which is primarily needed to detail rate of displacement. you won't be able to property define mass and energy as per your conclusion section without the time component.

Or due to being in a more orderly arrangement the crystalized structure can be shown as less complex

There is several steps to handle uniform acceleration. This involves rapidity. This paper details the required steps. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.physik.uni-leipzig.de/~schiller/ed10/Uniform%2520relativistic%2520acceleration.pdf&ved=0ahUKEwj4__7oyO7TAhVE5mMKHb0HAT8QFggnMAM&usg=AFQjCNE8A2T74ijQXQJHYZLeaKuSLK5roQ&sig2=Cj-Fu2-U5nxBCYf5dDQjaw In particular the section on hyperbolic spacetime paths. equations 2.8 to 2.10 are of particular note. As you noticed the basic Minkowskii matrixes and Lorentz boost involve constant velocity. When you undergo acceleretion you need to calculate the rapidity under 4 momentum 4 velocity. The groups handle this under rotations. (in essence you need to use instantaneous velicities) These examples will help I'll dig around for better examples the ones I have are in my textbooks. Equation 18 here is your uniform acceleration (without change in direction) transformations and how they derive it. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://aether.lbl.gov/www/classes/p139/homework/eight.pdf&ved=0ahUKEwj4__7oyO7TAhVE5mMKHb0HAT8QFggqMAQ&usg=AFQjCNEp_TbZhvDInabh8on8EEslpMnQOg&sig2=_yNYf6cM_Vvva4V_bbG5_A Here is a workup I've posted in the past as another example this is the twin paradox turnaround. (change of direction =acceleration change but not constant). The above articles detail uniform acceleration. The transforms with rapidity are as follows. [math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math] The formula you posted are Lorentz but only under constant velocity. The above matrix provide the changes due to rapidity. Some articles may refer to it as your hyperbolic rotation. first define the four velocity. [latex]u^\mu[/latex] [latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex] this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex] the four velocity has constant length. [latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex] the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex] [latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex] so the acceleration and velocity four vectors are [latex]c \frac{dt}{d\tau}=u^0[/latex] [latex]\frac{dx^1}{d\tau}=u^1[/latex] [latex]\frac{du^0}{d\tau}=a^0[/latex] [latex]\frac{du^1}{d\tau}=a^1[/latex] both vectors has vanishing 2 and 3 components. using the equations above [latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex] in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here? the last equation defines constant acceleration g. with solutions [latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex] from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex] hence [latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex] similarly [latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex] so the solution to the last equation is [latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex] hence [latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex] with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex] so [latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex] hence [latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex] and finally [latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex], [latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex] the space and time coordinates then fall onto the Hyperbola during rotation [latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex]

The above is described under Bose-Einsten, Fermi-Dirac statistics. There is considerable details chapter 3 and 4 here http://www.wiese.itp.unibe.ch/lectures/universe.pdfParticle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis

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yes except inertia means at rest or at a constant velocity. The more energy a wave has, including its monentum. (think of rest mass as the binding energy). inertia mass being the momentum term. The light beam has no binding energy as photons have no charge. It is charge neutral. The binding energy of charge is what establishes the rest mass of a particle. (for the strong force its color charge, weak force flavor charge.) this binding charge sets the coupling constant value for each force. Thus resulting in rest mass.(the total binding energy due to the sum of a particles (charges) Inertia mass is gained when you increase in velocity. Both forms of mass can perform work, they both have an energy value and as such can equally perform action. Edit charge means simply to attract or repel.

NO NO once again

Lets replace complexity with entropy shall we this is in essence what your describing above. Though I will now show that greater complexity arises with lower concentrations of mass/energy. Rather than the opposite that you posted. The second law as you mentioned previous states entropy never decreases, lets assume this is true So lets start with the Big Bang at [latex]10^{43}[/latex] seconds. This is as far back as our understanding reaches, we do not know the origin of energy etc previous to this point. As I'm showing a cosmology example we need a comoving volume (expanding) the entropy in a comoving volume is given by the equation [latex]S=\frac{\rho+p}{T}R^3[/latex] [latex]\rho[/latex] is the energy density, p is the pressure, T the temperature, R the radius. now at this time all particles are in thermal equilibrium, they become indistinguishable with one another, at high enough density they all look and behave like photons. The photon has spin of one which has 2 degrees of freedom. (degree of freedom corresponds to their interactions) loosely put. As the universe expands we get the other particles dropping out of thermal equilibrium. photon= 2 degrees of freedom, w boson=4, z boson= 2, gluons= 16, Higgs =4, For the leptons. electron=4, electron neutrino=2, muon =4,muon neutrino=2,tau=4, Each quark has 12. degrees of freedom. so greater complexity of our system arises as our universe expands, more particles drop out of thermodynamic equilibrium and our system gains a greater number of degrees of freedom ( added complexity).

Why restrict time to a property of matter when the term matter only applies to fermionic particles? Bosons isn't matter. ie photon. Time is simply a property of any change, regardless of what your measuring. That property being the rate of change. Something is always changing once you have a volume. It makes no sense to apply time to a volume of zero. Anyways just a sidenote on time,

Your definition of matter is incorrect. You really need to get around these basic physics definitions. The Pauli exclusion principle shows that bosons can stack on the same spacetime coordinate. However fermions in the same quantum number state will not stack but will take up a spacetime point. You can literally stack an infinite number of photons onto the precise same spacetime coordinate. You cannot stack two fermions in the same state onto the same coordinate. This is how matter is defined. "Matter has many definitions, but the most common is that it is any substance which has mass and occupies space." The occupy space applies in terms of the Pauli exclusion principle. There are classes of models in cosmology of toy universes that literally has no matter yet still has gravity. The Einstien field equations itself tells us that all forms of energy (ability to perform work) contribute to spacetime curvature via the energy/momentum stress tensor. This has been well tested. Declaring matter is needed when you obviously don't understand how matter is defined in physics won't get you far. (Its been mentioned numerous times to you only fermions count as matter) Arbitrarily choosing to ignore this will not make you right. Neither will ignoring the basic definition of energy or mass. For example the Cosmological constant is performing work. (it is causing accelerated expansion) yet there is no particle for the cosmological constant.( nor can there be one). The ideal gas laws show that when you have a field of particles the number density of those particles will decrease as the volume increases. However the Cosmological constant stays constant. (so it cannot be due to some unknown particle) Yet there is a pressure influence due to the cosmological constant that drives (contributes to) expansion. Pressure is force per unit volume. So how can the cosmological constant exert force without even being a type of particle field let alone a matter field.? Keep in mind energy is the "ability to perform work." you cannot perform work unless you have the required energy. So you cannot generate force which is a measure of work.. Matter has many definitions, but the most common is that it is any substance which has mass and occupies space. Defined by the Pauli exclusion principle Ignoring the bold part will never make you right. It just means your making garbage assertions without studying rudimentary and basic physics. Lets put this bluntly. 1) Every particle in the Standard model of Particle physics has an Energy value. Regardless if that particle has a rest mass. 2) Every particle therefore has the ability to perform work or exert an effective force. (cause action) 3) matter is a classification that only, only, only ...includes fermions

Excellent definition it pretty much sums up a field

Your welcome glad to be of help. As the other Planck units derive from the same principles using Planck length and time is in essence the same thing as applying the Planck constant.

good catch screwed up the SI sequence. Also should have used the SI notation. [latex]kg* m^2*s^{-1}[/latex]

Your close but although the Planck length and time are indeed involved its easier to simply realize the following. What you posted isn't fundamentally incorrect its just easier and more accurate to understand with the following In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction. the Planck constant itself is a quantum of action. This is why the minimal observable action or otherwise is in units of quanta. The minimal action of a spinning body from the above will equal the relations of the Planck constant. [latex]6.62607004*10^{-34} m^2/kg / s[/latex] 1 action or 1 quantum of action is the lowest possible rate of spin in this case in accordance to the rotation operators I defined earlier. Another key note is under QFT or QM all units are discrete regardless if the system is quantum or macroscopic. (one example being spinfoam under LQC loop quantum cosmology) in this case length,width, volume, area, momentum, energy, etc are all discrete units. (key side note from video. Recall Sean mentions a wave is the probability distribution function) never forget this key note. It is the probability of finding the position and momentum of a particle.(for example. All quantum numbers have a wavefunction. This includes angular momentum.. sidenote spacetime itself is discrete under QM/QFT your welcome just noticed your cross post. See my last post in particular what a wavefunction is defined as. (probability distribution) One key difference to understand between QM and classical physics. QM is the statistical probability of a measurable quantity whereas classical attempts exact determination of a measurable quantity. Waves wavefunctions included. So putting the above together visualize a highly irregular siusoidal wave just like you would see under an oscilloscope. When you have a sharp spike of 1 quanta or greater (but always using discrete confined by the Compton wave length that is your most probalistic position and momemtum (in essence your most likely particle location) though its a bit more complex than this heuristic descriptive aka Heisenburg uncertainty as one example) Which is one of the key reasons for QM being probabilistic. Requiring statistical mathematics (including all terminology ie superposition). The last part on electron knotis an accurate descriptive but its not the entire story there are no smaller Leptons for the electron to decay into and the conservation of lepton number applies. However its a bit OT to get into too much detail. I recommend a new thread to discuss Electron knots in greater detail Art Hobson has an excellent write up including the grizzly mathematics. note it is on arxiv working from phone atm. " There are no Particles there is only fields" https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/1204.4616&ved=0ahUKEwitn8Lhq-nTAhVIZFAKHdNnC4wQFggcMAA&usg=AFQjCNEqAKaDGcbyMG2ax22sA9BakBSaTQ&sig2=DavQ6RLz8gqGAchmIr7C4g Example quote.. quantized fields have certain particle-like appearances: quanta are unified bundles of field that carry energy and momentum and thus "hit like particles;" quanta are discrete and thus countable. But quanta are not particles; they are excitations of spatially unbounded fields". There is considerable detail on how this applies to the double slit experiment.

Then your not using the correct methodology and therefore getting the wrong answers. Use the correct formulas as described by relativity and you might have a chance.. Start with listing the difference in the transformation equations of Galilean relativity and Lorentz transforms. Secondly understand that the Lorentz transforms are under constant velocity. This makes absolutely no sense. All observers will always measure the velocity of a photon at c. So why would you ever add two observer velocities if both observers measure the signal at c your essentially adding c to c to equal c... obviously incorrect

acceleration is inertia change

What Strange posted is correct he also posted relevant links to each equation

Capiert would you please look up the definition of mass under physics. Your basic problem is your not considering what mass is defined as. Mass is resistance to inertia change. If you can't even pay attention to the very definition of mass then your never going to understand the key differences between massive and massless particles.

Well no it wouldn't be infinite. The reason is we have already established the top must be rotating so eventually the top can undergo a single rotation. What I have done is apply an educated guess based on QFT on the logic behind Sean Carroll's spinning top analogy under QFT action treatment. Remember were also using field waves so the Compton wavelength is involved as well as eugenstates. The difference between one eugenstate the next eugenstate is ie if your at some arbitrary groundpoint if you add a creation operator or subtract an annihilation operator the plus or minus eugenstates will be + or - [latex]\Delta E=\hbar w [/latex] Under QM treatments energy is in discrete packets. Unlike GR or classical where you can have any arbitrary energy value.

excellent question and one that can be answered under QM. yes to the first and yes to the spin of a macroscopic objects. Mathematically particle spin is mathematically the equivalent of rotational angular momentum. So lets put this all together. For a Quantum system the angular momentum is an Observable. Now what does that mean under QM? well an observable requires action. Observable also means that with the hypothetical ideal instruments we can measure this action. (Action equates to kinematic motion). So from the above under QM we need to associate this observable a Hermitean operator. [latex]\hat{L}=\hat{X}*\hat{P}[/latex] [latex] L=r*p=\begin{pmatrix}i&j&k\\x&y&z\\p_x&p_y&p_z\end{pmatrix}[/latex] the three Cartesian components are [latex] L_x=yPz-zPy, L_y=zPx-xPz, L_z=zPy-yPx[/latex] https://en.wikipedia.org/wiki/Angular_momentum where [latex] \hat{P}=i\hbar\triangledown[/latex] the definitions of the position and momentum operators above along with being Observable. (All operators in QM and QFT are observable). we can see from the last equation we are indeed quantizing the spin of a macroscopic top under QM formalism We cannot measure any value that is not quantized. Any instrument requires an interaction and all interactions require action and action requires discrete units of quanta as defined by their QFT /QM operators. (other operator examples creation, annihilation operators) ​from the above details we can derive [latex] \hat{L}_x=i\hbar(y\frac{\partial}{\partial_z}-\frac{\partial}{\partial_y})[/latex] [latex]\hat{L}_y=i\hbar(z\frac{\partial}{\partial_x}-\frac{\partial}{\partial_z})[/latex] [latex]\hat{L}_z=i\hbar(x\frac{\partial}{\partial_y}-x\frac{\partial}{\partial_z})[/latex]

Umm if the neutron had momentum of c it wouldn't be a neutron to start with.

Judging from the feedback I think this would be a good thread to pin.

You can also apply the field aspects to how Relativity works including 1) time dilation 2) length contraction. 3) redshift/blueshift and their frequency relations. This is one of the reasons why I'm glad Carroll mentions mass from the standpoint of how it works with vibrations. The video will be extremely handy to explain these advanced concepts and others to numerous threads currently ongoing. With literally no background knowledge required by the viewer. I am considering pinning this thread for a time so our regular members can readily go. "here watch this video " However I'd like to see what the OP thinks of it.

Good lecture, definitely handy to keep a link to. I particularly liked how he broke down the descriptive of fields with local neighbor potential. +1 this video covers a lot of the related replies I have been trying to teach here. I wish it was available when I first started studying physics. It took me years to put the pieces together to understand what he described in the first 50 minutes. Some additional information to digest. the portion of every particle field communicates mentioned in the video is forms of constructive/destructive interactions. The constructive interferences can give rise to new particles when the two waves combine to a quanta of energy in confinement. The portion on the above interactions accurately describe spacetime. That being the sum of all field values at a given coordinate. As every particle has a corresponding field, whose interactions can generate additional vibrations both constructive and destructive. To model all the dynamics at each point you use embedded geometries such as Hilbert space etc. The embedded geometries are your overlapping vibrations. String theory is also a collection of embedded geometries. Notice Sean Carroll's reference to string vibrations. In particular his correlation to mass.