Mordred

Might help if you look at the time aspect imsteas of writing \[a=\frac{dv}{dt}\] Write \[a(t)=\frac{dv}{dt}\] Secondly momentum includes mass via \[p=m*v\] So where is the issue ? Others have tried explaining this to you in the older thread

No worries hopefully you resolve your PC issues.

I like that article nice simple approach

This is more for fun but it also demonstrates another aid. Lets say you have a 2D vector field. We can graph this vector field as a function for simplicity lets do a 2D center of mass. I happen to know the function for this \[F(x,y)=(-x,-y)\] now here is a neat trick there is a handy tool Wolframalpha.com its entries can be tricky but the above is https://www.wolframalpha.com/input?i2d=true&i=plot+F\(40)x\(44)y\(41)%3D<-x\(44)-y> now lets say we want the curl of a magnetic field in 2D. \[F(x,y)=(-y,x)\] https://www.wolframalpha.com/input?i2d=true&i=plot+F\(40)x\(44)y\(41)%3D<y\(44)-x> this is just to demonstrate the usefulness of functions Here is what Function of sin(x) looks like see link https://www.wolframalpha.com/input?i2d=true&i=plot+sin\(40)x\(41) here is cos(x) https://www.wolframalpha.com/input?i2d=true&i=plot+cos\(40)x\(41) Now you can do the same with tan(x) and it will look completely different. However the main goal is to point out the WolframAlpha has some handy plotting capabilities as well as numerous others to help check your math or get a better feel for some of the more complex equations. The first graph is a converging field example gravity converging to a center of mass If I switch the sign of x I will type this in as WolframAlpha wants the format Plot F(x,y)=<( -y,x)> the field will be diverging from the center outward. Just copy and paste that line and it should work. Graph 2 is an example of a non diverging curl (rotationally symmetric) good example application is the magnetic field as opposed to the electric field which will be this example. I will let you enter that. However that depends on charge the opposite charge will be example 1. Plot F(x,y)=<( -y,x)> Hope that helps better visualize what functions do in terms of fields One particular handy use will come into play to better understand matrix mathematic for example \[\begin{pmatrix}a & b\\c&d \end{pmatrix}*\begin{pmatrix}e&f\\g&h \end{pmatrix}\] https://www.wolframalpha.com/input?i=[[a%2Cb]%2C[c%2Cd]]*[[e%2Cf]%2C[g%2Ch]] this will solve that operation for you and give a step by step (though you get far greater detail if you pay for membership) the details it does supply to non members is often as useful. Just an additional training aid to help you along

won't particularly matter when you consider absorption distance section. The quoted section makes no sense ALL frequencies are time dependent. The time element is determined via the frequencies in question. So that quoted section will need more than just some handwave statement specifically some form of mathematical proof. You cannot have the same time for all frequencies that is impossible by the very definition of frequency I also hate that terminology " inside time" sorry but that terminology doesn't work and gives the visualization of little matter balls for particles instead of field excitations which is something modern physics now teaches example under QFT I would assume you have some mathematics to your conjecture as its been on your books the past 6 years if not longer and there is absolutely no mathematics for your 2018 thread nothing more than blanket statements. Have you not done any mathematical development in the past 6 years ???? Without the related mathematics you literally have nothing of any substantial worth. You can't even prove to yourself mathematical viability let alone anyone else.

as I always prefer to include the mathematics for my statements where applicable or I have handy here are the corrections for peculiar velocity (recessive velocity beyond Hubble Horizon Recessive Velocity corrections past Hubble Horizon approx z=1.46 \[E_Z=[\Omega_R(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_\Lambda]^{1/2}\] \[v_{r}=\frac{\dot{a}}{a_0}D\] \[\frac{\dot{a}(t_0)}{a_o}=\frac{H(z_0)}{1+z_o)}=\frac{H_0E(z_o)}{1+z_O}\] \[v_r=\frac{cE(z_o)}{1+z_o}\int^{z^{obs}}_0\frac{dz}{1+z_o}\frac{D_c(Z_o,Z_s)}{D_H}\] \(Z_{os}\) is the reduced redshift \[1+z_{os}=\frac{1+z_s}{1+z_o}\] for observerd source redshift z_s present epoch Observer \(z_0=0 ,E(Z_o)/1+(z_o)=1\) \[v_r=(o,z)=c\int^z_o\frac{dz}{E(z)}=c\frac{D_c(z)}{D_H}\] gives redshift as a multiple of speed of light The above is time dilation effects due to expansion and commoving volume to a commoving observer. Below is how to determine the ae of the Universe as a function of redshift the Hubble parameter can be written as \[H=\frac{d}{dt}ln(\frac{a(t)}{a_0}=\frac{d}{dt}ln(\frac{1}{1+z})=\frac{-1}{1+z}\frac{dz}{dt}\] Notice you have a natural logarithmic function in the last statement for scaling look back time given as Ie how to calculate Universe age \[t=\int^{t(a)}_0\frac{d\acute{a}}{\acute{\dot{a}}}\] \[\frac{dt}{dz}=H_0^{-1}\frac{-1}{1+z}\frac{1}{[\Omega_{rad}(1+z^4)+\Omega^0_m(1=z0^3+\Omega^0_k(1+z)^2+\Omega_\Lambda^0]^{1/2}}\] \[t_0-t=h_1\int^z_0\frac{\acute{dz}}{(1+\acute{z})[\Omega^0_{rad}(1+\acute{z})^4+\Omega^0_m(1+\acute{z})^3=\Omega^0_k(1+\acute{z})^2+\Omega^0_\Lambda]^{1/2}}\]

This is more a related FYI and its not something one finds in textbooks. I can pretty much guarantee very few if any members will be aware of this little side detail. As we are all aware our universe is expanding. What is often overlooked when it comes to light propagation in this expanding volume is something called Absorption distance. I will try to keep this as simple as possible however the essence of absorption distance is that the higher the density of a particle ensemble the greater the likely hood of mean free path interference resulting in a reduction in number density of received photons. In order to factor this is will involve the absorption or scattering cross section of the particle population as well as the cross section of photons. Were going to assign X(z) to the absorption distance. \(n_0\) as the number density, \(\sigma_0\) as the cross section. Latter two being present epoch density. The evolution of number density as a function of redshift is \[n(z)=n_o(1+z)^3\] I won't get into the mathematical proof of the above it will invariably involve the density evolution of matter and radiation via the FLRW metric. We will be using the proper distance increment \( dl\) (take a ruler and keep the number of divisions on that ruler the same but as you go further back in time the spacing between each division decreases) \[dl=\frac{a(t)}{a_0}D_c=\frac{D_c}{1+z}\] where D_c is the commoving distance unfortunately the cross section of a photon beam is not consistent with redshift however we can choose to ignore geometric effects of orientation and orientation of gas structures for simplicity. In a commoving cosmic volume along the proper distance increment centered on the Z beam the probable number of targets (absorption/ scatterings)that can potentially interact with the beam is \[DN_t)=n_z\sigma_b(z)dl=n_o\sigma_b(z)(1+z)^3dl\] where \[dl=cdt=\frac{c}{H_o}\frac{d_z}{(1+z)}{(E(z)}=D_H\frac{dz}{(1+z)E(z)}\] \[DN=dn_t(z)[\frac{\sigma_o}{\sigma_b(z)}]\] where \[\frac{\sigma_o}{\sigma_b(z)}\] is the fractional area per structure the redshift dependence of DN/dz can be written as a dimensionless quantity \[\frac{DX}{Dz}=\frac{1+z^2}{E_z}\] upon integration we get the Absorption distance \[X_z=\int^z_o\frac{DX}{d_z}dz\] Now consider the above if for example DM was interacting with the photon beam the entire time of flight travel. Would we ever get the signal at say z=1100 ? conditions affecting the mean free path of photons after surface of last scattering is relatively clear and we are all aware of the opacity previous to last scattering on the mean free path of light. Just for consideration of a likely side effect should photons interact with DM along the mean free path via absorption/emission @DanMP this is the side effect I am referring to should DM interact with photons along the mean free flight path though the only cross section I have atm for DM would be under assumption of sterile neutrinos being DM. As that is part of my current research I have been doing in my Nucleosynthesis thread.

lol greatest training aid graph paper

I can live with bad terminology so now that is addressed we can move on. DM assuming it is a particle is considered professionally as being very stable so no issue there. The issue however still remains that even via absorption and emission of a photon your still involving medium like properties so time of flight for the photon beam is being additionally delayed. The other problem that makes this interaction distinct from redshift is accordingly via Snell's law the refraction angles (emission of photons will be frequency dependent on an equivalent refraction index. This leads to diffusion or divergence of the waveforms which is not the case with redshift. The redshift relation has no refraction index redshift has no frequency dependence on refraction. As redshift equates to time dilation ie gravitational time dilation, relativistic Doppler shift or even cosmological shift (with corrections beyond Hubble horizon) this is relevant and shouldn't be ignored. Granted the other detail I hadn't mentioned is such an interaction would also affect the KE terms so there should be a temperature increase/decrease

night and its great your starting to understand scalars and vectors keep at it. Here is an assistant article just on units and vectors for physicists https://www2.tntech.edu/leap/murdock/books/v1chap1.pdf in particular \[A\times B=\begin{pmatrix}i&j&k\\a_x&a_y&a_z\\b_x&b_y&b_z\end{pmatrix}\] which in other notation equals \[A\times B=(a_yb_z-a_zb_y)i+(a_zb_x-a_x b_z)J+(a_xb_y-a_yb_x)k\] this will correspond to that right hand rule I mentioned quite a while back also those matrix you have been looking at for the 2*2 matrix drop the k terms. You can project a 2d plane in any orientation in 3d. The 3*3 case is two complex planes where as the first case is one complex plane (first case SU(2) second case SU(3). Those are two fundamental particle physics groups which is where this lesson becomes valuable. For spin you need the SU(2) gauge group those vectors above are applied under all unitary groups U(1), SU(2),SU(3). for higher dimensions we will need higher vector commutations. we can only fill a matrix entry with scalar values so we need to get scalar values from vectors. That is why this lesson is so important. By the way once you understand the above you will have the necessary tools to understand Special Relativity and the Minkowskii metric. We can step into SR quite readily once you are comfortable with the above. Were literally on the edge of stepping you into electromagnetism via Maxwell equations

lets start with this statement an observable is any measurable quantity that includes time so how can an observable which we can measure lies outside of time when most observables are time dependent via the time dependent Schrodinger equations or velocity and acceleration. That statement alone goes against all modern physics

that is understandable it gave me trouble as well when I first learned this but if you figured out the 2d case your on the right track and doing well. Just a side note that plane your moving around is a complex plane and you may have noticed you can orient it in 3d. The entries for i and j had previous math operations done to it for the linearization they are complex conjugates denoted by the \[\hat{i}\] hat on top when it comes to quantum spin we will have to calculate these as the vertical will be imaginary numbers and the horizontal will be real numbers. ( we will need those for charge conjugation ie CPT charge, parity and time symmetry) later on hint your boundaries is the edges of the parallelogram for spin its the edges of the possible values ie the edges of the shaded area of the spin graph https://en.wikipedia.org/wiki/Spin_quantum_number the shaded area on that graph are the range of possible values ie the quantization which we have to cover specifically the math in this link https://en.wikipedia.org/wiki/Azimuthal_quantum_number hopefully though this will take considerable time we can get you to understand this link https://en.wikipedia.org/wiki/Angular_momentum_diagrams_(quantum_mechanics)#Inner_product but we need to sharpen your classical physics before taking on the challenges of QM

Yes the 4 entry box is a matrix. That matrix has entries those entries don't worry about just yet but they are derived by the inner products of the i,j vectors. ( part of the linear algebra that the link doesn't show so don't fret of the entries itself as I have taught you how to fill those entries for the determinant you are performing a math exercise using the entries \[\begin{pmatrix} a&b\\c&d\end{pmatrix}\] so lets set the entry values a=2 b=3 c=4 d=5 the formula for determinant is \[ad-bc\] so the determinant is (2* 5)-(3*4)\] so 10-12=-2 which if you look at the descriptive's in the link if the sign changes from a positive to negative determinant the orientation of the parallelogram is flipped over but also the value (2) is greater than 1 so it is stretched out. that is the general purpose of the determinant its to provide a means of scaling For quantum spin the main focus is the sign reversal itself specifically spin up and spin down. The entries for that matrix in the third link will be using a specific set of linear equations that will be determinant from linearizing the spin angular momentum operators. Which we haven't covered yet.

Agreed the one way light can however be used to detect DM hasn't been mentioned yet. Though c is still invariant nor is light interacting with DM. Via gravitational lensing and its relation to mass luminosity.

Yes but there is still no such thing as a quasi absorption not to my knowledge. Absorptions I can accept quasi-absorptions not so much lol. However irrelevant as absorption still amounts to light slowing down and absorptions still requires DM to interact with photons.