Jump to content

Mordred

Resident Experts
  • Posts

    9089
  • Joined

  • Last visited

  • Days Won

    32

Everything posted by Mordred

  1. the argument against an explosion is valid when you look at the angles and not just distance change between multiple points. That is literally the lesson the balloon analogy was originally designed to convey. Though a more accurate analogy for 3d is the raisin bread analogy
  2. I wouldn't say equivalent when you further consider photons scatterings such as with say Thompson scatterings. You can see the issue via this link. https://en.wikipedia.org/wiki/Thomson_scattering however that's not an issue with phonons defined as Phonon (lattice vibration, cause of thermal conduction in electric insulators, and participant in energy conversion involving kinetic energy of solids). as opposed to photons Photon (has the largest range of energy, carriers its energy even through vacuum, and interacts with electric and magnetic entities in energy conversion). Now consider a solid is held together by an EM field.
  3. Do you recall me mentioning phonons under the conduction case ? I do recall providing a gauge invariance requirement with a link detailing where this requirement becomes an issue with a lattice. Now I'm not saying its incorrect to use photons in this case however you must also factor in the conductivity of the material. If you look at the article the same equation for the quanta of phonons are identical to the equation for photons. We use the phonons simply to make the distinction between the conductive case to the radiative case. This also frees up certain key conservation laws with particle to particle scatterings. Those include conservation of charge, color, flavor, energy momentum, spin etc. Using phonons as a quasi particle we can avoid these issues.
  4. The angles would be needed between any 2 points. In an explosion scenario the angles would vary between any two or more points. This would include a metric that is expanding outward from an origin point. (specifically if one wanted to have commoving coordinates radiating outward from an origin for an inhomogeneous and anistropic expansion ) A homogeneous and isotropic expansion no angles change including those of the metric points. That also includes curvature terms once k is set it doesn't change over time. If K=0 in the past it will remain k=o to the present. With k=0 there is no time dilation with its uniform mass distribution.
  5. You do not show a metric for an explosion. An explosion radiates from starting point outward. So it is anisotropic. The metric equations you have are essentially homogeneous and isotropic. You might want to consider the inner product of your vectors in terms of the symmetry relations of the Minkowski metric. \[\mu\cdot\nu=\nu\cdot\mu\] describing an explosion of an ensemble of particles from an origin point would require angles. for example take 3 particles on a sheet of graph paper. Set an origin point a 0,0. Now place a particle at 1.1 another at 3,5 the last at say 5,3. Now expansion the graph lines increase in scale. (using a draftman's scaling ruler would be useful simply set the new scale at 1:2 ratio and redraw your grid lines). None of the angles change between 1.1, 3,5 and 5,3. Those angles would change if those particles were radiating outward from the origin point.
  6. Glad to see you understand where your errors were coming about. GL in your studies on the 3 scenarios.
  7. Those links written by Dean is literally garbage, for his supposed degrees he certainly didn't learn the the first thing about how to write a decent paper. Though quite frankly there is nothing in those links that deals with physics. Seems more appropriate for philosophy but I honestly don't even see anything worth discussing even under philosophy.
  8. heat transfer through a vacuum isn't conduction. Why do I get the impression You don't understand why and when you have conduction, convection and radiation ? Stefan Boltzmann is a radiation law not a conduction law. Just in case lets be absolutely clear. Both conduction and convection involve the medium containing atoms and molecules. In the former the atoms are stationary, in convection the atoms are moving (a fluid). In radiation there is zero atoms or molecules you simply have the EM field via photons for transference as the transfer occurs in a VACUUM not a medium.
  9. That's why I'm adding links for reference. Thermal conductivity is governed by very different relations that it is in thermal radiation. I am describing thermal conduction specifically for the case of solids. None of the equations for thermal radiation such as Plancks radiation law, Bose-Einstein statistics, Stefan Boltzmann, etc apply as thermal conduction is distinct from radiation. For solids the thermal conductivity of a material follows the Fourier Law of Heat conduction which for a 3d solid is \[\vec{\hat{q}}=-k(\frac{\partial T }{\partial x}\vec{i}+\frac{\partial T}{\partial y}\vec{j}+\frac{\partial T}{\partial z}\vec{k})\] most times you will see it as \[\vec{q}=-k\vec{\nabla} T\] such as this wiki link https://en.wikipedia.org/wiki/Thermal_conduction q is the heat capacity while k is a constant of proportionality for a materials conductivity. One can use the above law to mathematically describe the scenario Swansont just gave. (Though that's typically one of the earliest lessons articles will give on Fouriers Law.) Lattice networks are typically the homogenous and isotropic scenario using the Fourier law in that link. Quantum mechanically though the Fourier is represented by the quantum mechanical master equation for heat transfer lol just so you know I'm not blowing fluff on the name lol here is an example article on arxiv https://arxiv.org/abs/0711.4599 very few ppl will understand the mathematics of this article
  10. Swansont beat me to it. Once again there is a difference in thermal conduction of a material and thermal radiation. In solids you also have to factor in the conductivity of the material which will also involve density of states. little side not there is a reason why photons aren't used in a crystal lattice. The reason has to do with momentum invariance within the lattice. There is only 1 vector in a lattice where the momentum is invariant and this doesn't represent the overall momentum of the entire lattice network. Instead they use a quasi-momentum and phonons as a quasi particle for the distinction. example details here https://www.ucl.ac.uk/~ucapahh/teaching/3C25/Lecture12p.pdf phonons are used to represent lattice waves, while its common to use magnons for magnetic waves and keeping photons for EM waves. All three directly apply to thermodynamics of a solid or more accurately its conductivity
  11. In the case of radiative heat transfer using photons is common as a mediator. However heat transfer is not the same thing as an objects temperature. In order to understand how photons can be used in thermal radiation one has to also understand the interaction with the molecules and atoms of a substance/state. So one also has to understand how this alters its temperature
  12. Try not to confuse how thermal energy is transferred until you understand why thermal energy is described via its mean average kinetic energy. There is a difference between the two both however are required. I do have to ask why you would think the current understanding of temperature is incorrect in regards to frequencies/vibrations etc ? Particularly when we can definitively show our understanding via previous experiments and observations ? One example I can think of, offhand that clearly demonstrates our understanding is using lasers to supercool atoms and molecules to form a Bose-Einstein condensate. Using lasers to cool something is not something one would expect but in essence the lasers are used to reduce the internal motion of particles within the condensate. Its not just any laser in that example its specific laser frequencies that vary depending on the condensate. Just another thing to keep in mind there is also a distinction between conductive heat transfer and radiative heat transfer in regards to Stefan-Boltzmann...
  13. I already have if you looked at the two equations I posted. However I ask you the following if you cause vibrations an object to increase via any mechanism including sound do you agree that the average temperature of resonating nearby objects will increase ? or that you can heat up metal by hitting it with a hammer? Any time you increase an objects vibration you also increase the molecular and atomic vibrations. So ignoring molecular and atomic vibrations makes absolutely no sense whatsoever.
  14. There is also the Cosmic neutrino background however that's irrelevant. What is relevant is that there a temperature even with particles that does not interact with photons. As far as considering why your ignoring the kinetic energy terms of molecular vibrations your simply incorrect to do so. As such your model will fail Seems to me that you have chosen to completely ignore the fundamental definition of temperature or at the very least a good portion of the kinetic energy terms involved.
  15. just to add to this with another example. How would a cosmic neutrino background which has no photon interactions have a blackbody temperature of 1.95 K ? I concur with Swansont on the rest of the quote
  16. So let me get this correct Joigus provided you with links directly related to the vibration of molecules and you feel that atoms and molecules have no vibrational modes is that correct ? Do you for some unusual reason feel that the details provided by Joigus didn't apply ? what happens to your vibrational and rotational temperature ? \[\theta_{vib}=\frac{h\tilde{v}c}{K_{B}}\] and \[\theta_{R}=\frac{hc\bar{B}}{K_{B}}\] You know I honestly cannot think of a single static particle. Every particle has harmonic oscillations including those of atoms and molecules. It would certainly make absolute zero viable as under QM absolute zero temperature is impossible.
  17. Try to wrap your study on the difference between an invariant vs a variant measurement. The former all observers agree on literally all observers. That's is what's used to calculate the expansion and age of the universe etc. The commoving observer is used to establish that needed invariance
  18. The term for the observer your looking for is the commoving observer. This is an observer at the same influences of the global spacetime conditions. In essence the flat global metric. The flat spacetime of the LCDM parameters has no significant curvature term for any time dilation effects. Subsequently the age of the universe is determined from the commoving observer. The formulas applies the Hubble parameter in its formula. There is absolutely no point in using the observations of an observer in a traveling spacecraft, particularly using redshift. One simply has to realize how useless thar would become by recalling that as the spacecraft approaches a destination you blueshift while at the same time getting redshift from the crafts origin point. Makes using observers on a spacecraft essential useless. Particularly once you start applying transverse Doppler for every angle not parallel to the travel direction. In essence an observer in a spacecraft would get different redshift values at every single angle of measurement. Kind of pointless to use that to calculate expansion or age.
  19. No problem, hope your getting better. Hospitals are never fun. I would suggest instead of thinking of alternative universes. Simply apply the quantum vacuum positive and negative frequency modes for particle production. QFT already has all the applicable formulas for that scenario. Modelling a 2 component scalar, boson of even a fermionic field under QFT has plenty of textbook support at the introductory level. In QFT field and momentum are your primary operators. Though having familiarity with Fourier transforms and the Langrangian/Hamilton greatly helps to understand and apply QFT.
  20. Correct any finite portion will not extend to infinity. However we do not know if the entirety of universe beyond our Observational portion is infinite or finite. Our Observable universe portion will always remain finite. Had findings of our universe term been precisely flat at unity then we likely would have had an open infinite universe. However their is a slight curvature term that raises the possibility of a closed universe.
  21. Your logic is not the same as my logic or Bob's logic. Logic means nothing. It's simply an aid. There is no preferred frame this is well established and if you understand the mathematics of GR with its application of invariant quantities you would understand the truth of that statement. Let's use the following math statement on the Lorentz transforms \[\mu \cdot \nu= \nu \cdot \mu)\] This statement describes the symmetric relations of the Lorentz transform.
  22. Thats a poor excuse, one cam still stick to the math and model any form universe conditions by applying the correct math alterations provided one correctly performs the correct math operations. I can easily adapt the FLRW metric to a form that does not require a homogeneous and isotropic dstribution as per the Cosmological principle or even adapt the metric for a rotating universe. That's because I know the needed math steps. You don't learn those by simply studying pop media literature. Regardless if you study the arxiv and professional papers instead of paying attention to pop media you will learn that pop media literature makes mountains out of mole hills. Pop media will always lead a reader astray. The cosmological Principle is quite secure by observational evidence a good example is the uniformity of the CMB. The scale however one must apply is per 100 Mpc.
  23. Ds is the separation distance interval is designated by (ct) though that's usually shortened to simply t. The interval allows time to have the same dimensionality of length via the interval
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.