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Everything posted by Mordred

  1. charge isn't carried by only electrons with flow of charge you are describing the effective average of each individual atom in a conductor. The photons are the mediator of charge in interparticle interactions. In this case specifically atom to atom. This is effectively the same as your EM field descriptive. With flow of charge you are not specifying the flow of a specific particle but the flow of the effective charge at a given locale
  2. I'm prepared to hear your argument why do you feel the flow of charge doesn't have a rate of c ?
  3. yes you are correct I specified for of charge as that's the lanquage used in a lot of high school lessons. A lot of the high school physics textbooks also apply the flow of charge terminology and subsequently some of their exam questions will trip a student up if they don't realize the question applies flow of charge vs flow of electrons. ( there is one specific lightning question on flow direction for a specific charge that students will get wrong if they don't pay attention to the flow of charge specifically). lol at least in Canada flow of charge in high school is still taught today, as I often help volunteer as an assistant instructor to high school and undergraduate students
  4. I agree the earth spins slowly enough that it wouldn't make any significant differences. For the first part if your applying the line elements already derived such as he Schwartzchild metric etc the coordinates vs proper time are already incorporated. You still have to be aware that those same line elements will return the proper time. Same goes for expressions for proper time already accounting for coordinate time The commonly used proper time to coordinate time expression is as follows \[\Delta \tau =\int \sqrt{1- \frac{1}{c^2} \left ( \left (\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left (\frac{dz}{dt}\right)^2\right )}dt\] which works fine until you have other factors such as frame dragging, rotation or acceleration. For rotation as above \[d\tau=\sqrt{1-\frac{\omega^2 r^2}{c^2}}dt\]
  5. as the others mentioned there is a distinction between flow of electrons and flow of charge. As you noticed flow of electrons is extremely slow however flow of charge is near c.
  6. The history of Eather theory is quite interesting. I have a copy of a 1918 physics textbook that described it. It didn't include anything involving relativity and the entire particle model only comprised of protons and electrons. Neutrons were discovered roughly 1935 if I recall.
  7. Your work sheet looks good I don't see any problems or mistakes in that analysis well done.
  8. It simply means there is always something new to learn and if your dedicated in your studies you never stop trying to learn new aspects of a given theory. from a preliminary quick research it seems to me applying the Carter Constant for the elliptical orbit may give me a methodology. I wonder if Markus or Jadus agrees with that approach https://en.wikipedia.org/wiki/Carter_constant there we go I think I have an applicable method https://en.wikipedia.org/wiki/Boyer–Lindquist_coordinates#Spin_connection edit speaking of challenges I wonder at the steps to get get proper time for an observer at infinity to a clock orbiting Earth with the subsequent transverse blue/redshifts.
  9. lol its how I test myself for improvement. If I can help others also learn then bonus. One thing about any physics theory you never know enough
  10. I I like a good challenge lol and as the Earth also orbits the observer at sea level has numerous interesting effects though most are negligible in the case of the earth I am curious how to handle situations not so trivial
  11. hopefully the proper acceleration example I provided aids you both in this GL I'm looking into finding a method to handle the elliptical orbit to have an observer at sea level while an observer is in orbit. I have found a couple of Kerr metric examples in elliptical orbit still examining them to see if they will be useful. currently examining this one https://arxiv.org/pdf/0903.3684.pdf
  12. Continuing from here for the orbiting Observer to Observer at CoM for simplicity. Once again will be using the Lewis Ryder reference in the previous post above. were going to set the orbital rotation along the z axis in cylindrical coordinates \[ds^2=-c^2dt^2+dr^2+rd\phi^2+dz^2\] \[\acute{t}=t, \acute{r}=r, \acute{\phi}=\phi-\omega t, \acute{z}=z\] \[\acute{ds}^2=-ct^2d\acute{t}^2+\acute{r}^2(d\acute{\phi}+\omega d\acute{t})^2+d\acute{z}^2\] \[=-(c-\omega^2\acute{r}^2)d\acute{t}^2+2\omega\acute{r}^2d\acute{\phi}^2 d\acute{t}^2+d\acute{r}^2+\acute{r}^2d\acute{\phi}^2+d\acute{z}2\] dropping the primes we have the invariant spacetime in a rotating frame \[ds^2=-(c^2\omega^2 r^2)dt^2+2\omega r^2d\phi^2 dt+dr^2+r^2d\phi^2+dz^2\] this becomes \[ds^2=g_{\mu\nu}dx^\mu d^\nu=g_{00}(dx^0)^2+g_{0i}dx^0dx^j+g_{ik}dx^i dx^k\] ijk sums over 1-3 where \[x^\mu=(x^0,x^1,x^2,x^3)=(ct,r,\phi,z)g_{\mu\nu}\] \[g_{\mu\nu}=\begin{pmatrix}-(1-\frac{\omega^2 r^2}{c^2}&0&\frac{\omega r^2}{c}&0\\0&1&0&0\\\frac{\omega r^2}{c}&0&r^2&0\\0&0&0&1\end{pmatrix}\] time interval between events \[ds^2=-c^2d\tau^2\] as world time not proper time relation between world time and proper time is given by \[d\tau^2\sqrt{-g_{00}}dt\] giving in the above case \[d\tau=\sqrt{1-\frac{\omega^2 r^2}{c^2}}dt\] lengthy process but hope that helps better understand the 3 clock scenario described by the OP... all the material as mentioned is from Lewis Ryder "Introductory to Relativity" Sagnac effect as shown above further detail here https://en.wikipedia.org/wiki/Sagnac_effect I'm also considering applying the Kerr metric to the rotating case to place observer at sea level instead of CoM have to think about that though
  13. With the rotations and boosts above of the Lorentz tranformations we can now examine the different clocks under proper acceleration which I will detail below for each case. (I will need to set one observer at CoM instead of sea level for simplicity) four velocity \[u^\mu=\frac{dx^\mu}{d\tau}=(c\frac{dx}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})\] \[A=\frac{du}{d\tau}\] without going through all the relations a constant acceleration along the x direction with constant acceleration g gives the following \[c\frac{dt}{d\tau}=u^0,\frac{dx^1}{d\tau}=u^1,\frac{du^0}{d\tau}=a^0,\frac{du^1}{d\tau}=a^1\] \[a^\mu a_\nu=-(a^0)^2+(a^1)^2=g^2\] gives two solutions \[a^0=\frac{g}{c}u^1,,a^1=\frac{g}{c}u^0\] from which \[\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0\] gives solution between observer A and B (A set at CoM, B falling observer) \[u^1=Ae^{g\tau/c}+Be^{g\tau/c}\] hence \[\frac{du^1}{d\tau}=\frac{g}{c}(Ae^{g\tau/c}-Be^{g\tau/c}\] hence \[x=\frac{c^2}{g}cosh(\frac{g\tau}{c}),, ct=\frac{ct}{g}sinh(\frac{g\tau}{c})\] space and time coordinates fall onto the hyperbola \[x^2-c^2t^2=\frac{c^4}{g^2}\] \[\frac{dx}{d\tau}=c \sinh(\frac{a\tau}{c})\] \[\tau=\frac{c}{a}\sinh^{-1}\] there is your proper time under constant acceleration for the falling clock Lewis Ryder pages Introductory to General Relativity pages 23 to 28 even thoug its from the twin paradox solution the Hyperbolic rotation is identical in the linear acceleration case https://en.wikipedia.org/wiki/Lorentz_transformation https://en.wikipedia.org/wiki/Hyperbolic_functions
  14. Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\] \[(x^0,x^1,x^2,x^3)=(ct,r,\theta\\phi)\] \[(x_0,x_1,x_2,x_3)=(-ct,r,r^2,\theta,[r^2\sin^2\theta]\phi)\] \[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\] \[\Lambda^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&\cos\theta&\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1\end{pmatrix}\] generator along z axis \[k_z=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}\] generator of boost along x axis:: \[k_x=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}=-i\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\] boost along y axis\ \[k_y=-i\begin{pmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0 \end{pmatrix}\] generator of boost along z direction \[k_z=-i\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0 \end{pmatrix}\] the above is the generator of boosts below is the generator of rotations. \[J_z=\frac{1\partial\Lambda}{i\partial\theta}|_{\theta=0}\] \[J_x=-i\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0 \end{pmatrix}\] \[J_y=-i\begin{pmatrix}0&0&0&0\\0&0&0&-1\\0&0&1&0\\0&0&0&0 \end{pmatrix}\] \[J_z=-i\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0 \end{pmatrix}\] there is the boosts and rotations we will need and they obey commutations \[[A,B]=AB-BA\]
  15. hrrm seems to me we will need the rotating frame described by the Sagnac effect. This will take me a bit of time to get down so please be patient I'm going to apply the form given by Lewis Ryder from his textbook as its layout is one of the easiest to follow also going to have to break it down further for readers not familiar with the Lorentz transforms in the x, y and z direction. (its worth the additional effort as it is informative to all readers.) the dust solution I simply had handy as it describes an unaccelerated frame stress energy tensor treatment as mentioned it wont work in this case by itself but the details of how to fill the two tensors are included in differential form Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\] \[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\]
  16. putting these here as a visual aid to fill the GR and stress energy tensor \[\frac{dx^\alpha}{dy^{\mu}}=\frac{dx^\beta}{dy^{\nu}}=\begin{pmatrix}\frac{dx^0}{dy^0}&\frac{dx^1}{dy^0}&\frac{dx^2}{dy^0}&\frac{dx^3}{dy^0}\\\frac{dx^0}{dy^1}&\frac{dx^1}{dy^1}&\frac{dx^2}{dy^1}&\frac{dx^3}{dy^1}\\\frac{dx^0}{dy^2}&\frac{dx^1}{dy^2}&\frac{dx^2}{dy^2}&\frac{dx^3}{dy^2}\\\frac{dx^0}{dy^3}&\frac{dx^1}{dy^3}&\frac{dx^2}{dy^3}&\frac{dx^3}{dy^3}\end{pmatrix}\] Schwartzchild metric Vacuum solution [latex]T_{ab}=0[/latex] which corresponds to an unaccelerated freefall frame [latex]G_{ab}=dx^adx^b[/latex] if [latex]ds^2> 0[/latex] =spacelike propertime= [latex]\sqrt{ds^2}[/latex] [latex]ds^2<0[/latex] timelike =[latex]\sqrt{-ds^2}[/latex] [latex] ds^2=0[/latex] null=lightcone spherical polar coordinates [latex](x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)[/latex] [latex] G_{a,b} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}[/latex] line element [latex]ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)[/latex] Dust solution no force acting upon particle [latex] T^{\mu\nu}=\rho_0\mu^\mu\nu^\mu[/latex] [latex]T^{\mu\nu}x=\rho_0(x)\mu^\mu(x)\mu^\nu(x)[/latex] Rho is proper matter density Four velocity [latex]\mu^\mu=\frac{1}{c}\frac{dx^\mu}{d\tau}[/latex] Leads to [latex]ds^2=-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2=-c^2dt^2(1-\frac{v^2}{c^2})^\frac{1}{2}=\frac{1}{\gamma}[/latex] [latex]T^{00}=\rho_0(\frac{dt}{d\tau})^2=\gamma^2\rho_0=\rho[/latex] [latex]\rho[/latex] is mass density in moving frame. [latex]T^{0i}=\rho_0\mu^o\mu^i=\rho^o\frac{1}{c^2}\frac{dx^o}{d\tau}\frac{dx^2}{d\tau}=\gamma^2\rho_0\frac{\nu^i}{c}=\rho\frac{\nu^i}{c}[/latex] [latex]\nu^i=\frac{dx^i}{dt}[/latex] [latex]T^{ik}=\rho_0\frac{1}{c^2}\frac{dx^i}{d\tau}\frac{dx^k}{d\tau}=\gamma^2\rho\frac{\nu^i\nu^k}{c^2}=\rho\frac{\nu^i\nu^k}{c^2}[/latex] Thus [latex]T^{\mu\nu}=\begin{pmatrix}1 & \frac{\nu_x}{c}&\frac{\nu_y}{c} &\frac{\nu_z}{c} \\\frac{\nu_x}{c}& \frac{\nu_x^2}{c} & \frac{\nu_x\nu_y}{c^2}& \frac{\nu_x\nu_z}{c^2}\\ \frac{\nu_y}{c}& \frac{\nu_y\nu_z}{c^2} & \frac{\nu_y^2}{c^2}& \frac{\nu_y\nu_z}{c^2}\\ \frac{\nu_z}{c} &\frac{\nu_z\nu_x}{c^2}&\frac{\nu_z\nu_y}{c^2}&\frac{\nu_z}{c^2}\end{pmatrix}[/latex] Schwartzchild metric Vacuum solution [latex]T_{ab}=0[/latex] which corresponds to an unaccelerated freefall frame [latex]G_{ab}=dx^adx^b[/latex] if [latex]ds^2> 0[/latex] =spacelike propertime= [latex]\sqrt{ds^2}[/latex] [latex]ds^2<0[/latex] timelike =[latex]\sqrt{-ds^2}[/latex] [latex] ds^2=0[/latex] null=lightcone spherical polar coordinates [latex](x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)[/latex] [latex] G_{a,b} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}[/latex] line element [latex]ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)[/latex] the above will not work directly but it is informative and will be helpful in a solution development. We will require additional terms I do know of one example that I have yet to add (the above has no acceleration for starters so this will involve rotations likely via rapidity.)
  17. @joigusUsing the Schwartzchild solution as your basis is a good methodology for the sea level observer however be aware that to observe the clock on the orbital you will require the energy momentum terms in the T^{oi} and the T^{oj} directions. This will be further complicated in the Schwartzchild treatment as the radius to either the sea level clock and the observer at infinity will vary. I would further suggest using proper acceleration along with rapidity for the orbiting clock. I may have an appropriate solution for the orbiting clock but will have to dig it up later today
  18. yes protons are composite particles so are neutrons I'm referring to the elementary particles that have no internal structure any composie particle would be incredibly short lived due to the high density and resulting scattering
  19. Think of the first state as a quark gluon plasma state. Electrons neutrinos etc are present including the entirety of the standard model of particles. The energy density equates to 10^{90} photons but that's simply a calculated equivalency. That quantity is conserved throughout the expansion history though as mentioned is simply an equivalency not the actual number of photons. As mentioned they are in a state of thermal equilibrium so they all become indistinct from photons. When the electroweak symmetry break occurs then the other particles start to become distinct into neutrinos, electrons etc. Atoms come much much later.
  20. The ASPIC library has narrowed down the number of viable inflationary models to 74 lol
  21. neither inflation nor expansion create particles, expansion and inflation occur due to particle interactions. With inflation the two leading hypothesis for inflation is the quasi particle inflaton or the Higgs field. Both are viable though I don't believe anyone has ever observed an inflaton so in my own opinion I tend to favor the Higgs field. However that's only my opinion both are equally mathematically viable. The particles already exist in a state called thermal equilibrium in that state they are extremely short lived and highly energetic making them indistinct from one another until they drop out of thermal equilibrium. This occurs during symmetry breaking a subject that I have two threads currently active examining the related mathematics. One I have in speculation simply because some of the formulas are not typically found in textbooks but are proposed formulas from other peer review material. By thee rules of the forum I could readily consider them sufficient for mainstream but chose the Speculation forum simply to help avid confusion to other readers. (assuming they can follow the math, as I didn't include much in the way of explanation) Anyways both the inflaton field and the Higgs field incorporate the same Mexican hat potential, the potential energy at the top of the Mexican hat potential is the initial energy density. Inflation starts when the energy density starts a slow roll to the lower energy density we have today. The higher potential is oft refered to as the false vacuum. While its believed the true vacuum state we have today. The Inflation thread I have is currently in this forum for further detail. https://www.scienceforums.net/topic/128412-musings-of-a-mad-scientist-inflation-as-cosmological-constant/ this thread here has some of the mathematics I've been examining for both electroweak symmetry breaking and inflation I've been examining the slow roll itself as each stage should result in variations of the slow roll due to phase transitions. I'm still gathering the related formulas to put it all together so its layout is as I find the needed formulas lol so don't expect a clear cut order
  22. if there is any measurable deviation in the quark sea for the valence quarks alignment it would be extremely subtle I would think likely following a probability distribution naturally. A rotating charged particle generates a magnetic moment so here is one study of precision tests of a protons magnetic moment for examination. https://arxiv.org/pdf/1201.3038.pdf edit I should reword that to any particle with an angular momentum and a charge distribution term to avoid visualization of a spinning ball oops almost forgot there is a difference in the types of magnetic moments the proton uses the nuclear magnetic moment where as the electron applies the orbital magnetic moment the proton magnetic moment is subsequently lower than the the electron magnetic moment.
  23. well as I stated the OP never specified the quantity of protons added so I didn't assume a quantity slight correction here as protons don't generate pressure. Replace it with the curvature term however in other cases one also applies the pressure term as is the case with radiation. little FYI for everyone at the 10-{-43 sec} the mass density far exceeded the critical density which with femionic matter would have caused an instant collapse. However the system was also extremely hot and in thermal equilibrium where all particle species was so energetic to be relativistic. This lead to the negative pressure term that caused the initial expansion to get to the symmetry breaking contributions leading to inflation. Inflation regarded as either due to inflaton or Higgs inflation the effective equations of state are identical in both cases. for further clarity as I lost count on the number of people I've seen apply e=mc^2 for mass or energy and get incorrect results the full equation that would be needed to apply in this situation is the energy momentum relation \[E^2=(pc)^2+(m_0 c^2)^2\]
  24. that depends on the effective equation of state w=0 and the critical density if you add enough protons in a small enough volume you can readily get a collapse the calculated value I got assuming no previous rate of expansion with H=0 is \[1.6-8*10^26 kg/meter\] that would be the critical density value without if you add precisely that amount the universe geometry would be perfectly Euclidean and hence static. However if you add less then the system expands and vise versa if the amount exceeds the critical density you will get a collapse. that is incorrect expansion is not purely gravitational if it were every system would collapse under self gravity you need to apply the equations of state for each particle species (radiation, matter Lambda to the fluid equations which entail the deceleration/acceleration equation of the FLRW metric which incorporate the thermodynamic contributions each particle species has
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