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Everything posted by juanrga

  1. The term [math]T^{\mu\nu}h_{\mu\nu}[/math] is not "unspecified" but a well-known term with precise properties. Next picture is from Feynman textbook on gravitation Note that Feynman adds an extra lambda coupling constant, because this helps him in a perturbation expansion in orders of lambda, but usually lambda is absorbed into h and the expansion made in terms of orders of h.
  2. I participated in that Contest and my answer was "both": http://fqxi.org/community/forum/topic/853
  3. What I wrote is not a number but a physical quantity (with units of energy). "h" is the potential of the gravitational field. Agree but you wrote "anything with either non-zero momentum or non-zero energy" and In English, the construct "either ... or" is usually used to indicate exclusive or. The angle does not depend on the momentum p in a first approximation, when you add higher order corrections to the bending formula the angle depends on the momentum as well. In any case, the variation of the momentum of the light signal depends on its initial momentum p at any order, as shown above.
  4. Yes the acceleration of a test body depends on the mass of the source body. This is also true for light. Light bending around the Sun depends on the mass of the Sun: more mass more bending. The gravitational interaction is [math]T^{\mu\nu}h_{\mu\nu}[/math], where [math]T^{\mu\nu}[/math] is the energy-momentum tensor. Therefore anything with a nonzero energy-momentum tensor will feel gravity. What I am saying is not exactly identical to what you did mean "anything with either non-zero momentum or non-zero energy", because I am also including systems for which both momentum and energy are nonzero. The effect of gravity on a light pulse depends on its momentum. The variation of momentum of a light signal moving near a gravitational field (source mass M at rest) is [math]\frac{4GMp}{c^2r}[/math] and this gives the measured deflection angle for a ligth signal bending around the Sun [math]\alpha = \frac{\Delta p}{p} = \frac{4GM}{c^2r}[/math]
  5. The usual Newtonian expression [math]F = -\frac{GMm}{r^2}[/math] is a non-relativistic expression valid for massive particles, but not for massless particles as photons. In reality gravity acts between anything with a nonzero energy-momentum. Light (i.e., photons) is affected by gravity because has a nonzero energy-momentum. No, light does not break "its own top speed".
  6. Why would remain the same? In the first place the spatial coordinates are not Galilean, therefore one would wait a change in the form of the nabla operator. In the second place, there is not conservation of energy in an Friedmann universe. Concretely the variation of energy in each cell cannot be accounted by flows to/from adjacent cells, because all the cells are losing the same amount of energy at each instant due to cosmological expansion. Using the mass-rest-energy relationship we find the same conclusion for m inside each cell. I think that all of this is accounted in the link that you give. For instance pay attention to "Terms such as cosmological expansion [...] may be included as source terms".
  7. What the article does not mention is that the main problem is not that (i) dark matter has not been found where it was supposed to exist, but that it cannot explain lots of data. Some related news: More Evidence Against Dark Matter? Dark matter theory challenged by gassy galaxies result Dark matter challenge Dwarf galaxies suggest dark matter theory may be wrong Dark matter results spark debate Dark matter effect might be explained by modified way to calculate inertial mass In any case I agree with you, dark matter will be not found in the next four years; althought, I am rather sure that we will read news claiming that dark matter has been finally found. We already read exaggerated news as that in the past (although they are now forgotten).
  8. Scientific notation is a type of exponential notation where non-zero numbers are expressed as multiples of powers of ten... http://juanrga.com/en/knowledge/scientific-notation.html

  9. Any textbook gives the interpretation of the density matrix in a SINGLE continuous basis [math]|\alpha\rangle[/math]: The diagonal elements [math]\rho(\alpha,\alpha) = \langle\alpha| \hat{\rho} |\alpha\rangle[/math] give the populations. The off-diagonal elements [math]\rho(\alpha,\alpha') = \langle\alpha| \hat{\rho} |\alpha'\rangle[/math] give the coherences. But what is the physical interpretation (if any) of the density matrix [math]\rho(\alpha,\beta) = \langle\alpha| \hat{\rho} |\beta\rangle[/math] for a DOUBLE continuous basis [math]|\alpha\rangle[/math], [math]|\beta\rangle[/math]? I already know that when the double basis are momentum [math]|p\rangle[/math] and position [math]|x\rangle[/math], then [math]\rho(p,x)[/math] (the well-known Wigner function) is interpreted as a pseudo-probability. I may confess that I have never completely understood the concept of pseudo-probability [*], but I would like to know if this physical interpretation as pseudo-probability can be extended to arbitrary continuous basis [math]|\alpha\rangle[/math], [math]|\beta\rangle[/math] for non-commuting operators [math]\hat{\alpha}[/math], [math]\hat{\beta}[/math] and if a probability interpretation holds for commuting operators. I.e. can [math]\rho(\alpha,\beta)[/math] be interpreted as a pseudo-probability for arbitrary non-commuting operators beyond x and p? Can [math]\rho(\alpha,\beta)[/math] be interpreted as a probability for arbitrary commuting operators? [*] Specially because [math]\rho(p,x)[/math] is bounded and cannot be 'spike'.
  10. As stated in my previous message the above expression also holds for massless particles. Substituting [math]m=0[/math] we obtain the well-known [math]E=|\mathbf{p}|c[/math]. The equations in your attached pictures are compatible with this. For instance using [math]E=h \nu[/math] we obtain [math]|\mathbf{p}|= E/c = h \nu/c = \hbar |\mathbf{k}|[/math]. Multiply by unit vector both sides and you obtain your second attached equation relating momentum and wave vectors. http://en.wikipedia....ical_properties
  11. Effectively [math]m_e[/math] is electron mass. This is an excellent point. Precisely I clarified the meaning and usage of different 'equivalent' units in my recent article about SI derived units. Look the part starting with (bold added here) Consider the half dozen of examples given in the next paragraphs of the article; for example, the quantity "torque" may be thought of as the cross product of force and distance, suggesting the unit "newton metre", or it may be thought of as energy per angle, suggesting the unit "joule per radian". Regarding momentum it seems natural to use [kg m/s] for massive particles such as the electron but [J s/m] for massless particles such as the photon. The unit [N s] makes more sense when dealing with forces at the classical level (e.g., when solving Newtonian equations using the observed forces) but I would not recommend its use at quantum level (e.g. when solving the Schrödinger equation where there are no forces involved at all). Formally it is correct, because it is a tautology [math]E = E[/math]; recall that relativistic mass is defined by [math]m_{rel} \equiv E/c^2[/math]. However there are a number of objections to the use of relativistic mass concept and it is deprecated in most of modern physical literature (in all of fundamental literature). Some objections to its use are given in the Am. J. Phys. paper Mass versus relativistic and rest masses. It is interesting that Einstein Never Approved of Relativistic Mass The modern equation replacing the above is [math]E= \sqrt{m^2 c^4 + p^2 c^2}[/math] For a photon [math]m=0[/math] and you obtain the well-known expression [math]E = |p| c[/math] found in the Wikipedia page that you cited above.
  12. That paragraph and the first equation [math]\mathrm{p}=m \mathrm{v}[/math] in the wikipedia are only valid for massive particles at low velocities. For an electron moving at about [math](1/10) c[/math] its momentum is not given by the Newtonian [math]m_e \mathrm{v}[/math] but by the relativistic [math]\mathrm{p}=m_e \gamma \mathrm{v}[/math] with [math]\gamma[/math] being the time-dilation factor. Below in the section 3.2 they give the correct momentum "for massless particles such as photons". The answer is [math]p=E/c[/math]. The mass of a bunch of photons is zero and [math]P=E_T/c[/math] holds but now with [math]P[/math] being the total momentum and [math]E_T[/math] the total energy.
  13. It was explained in #4 how massless particles such as photons generate gravity and are affected by gravity. In the first place the logical structure is hierarchical and we can surely answer "why?" at different levels (this is why Weinberg emphasizes that theoretical physics is about answering "why?"). In the second place if you consider a linear logic, then either you consider infinite levels of "why?" or an underlying fundamental level which cannot be answered. However, if we consider nonlinear logics then it is admissible to obtain a self-contained logical structure. In fact, structures of this type are under research in meta-mathematics with the goal of giving mathematics its own mathematical foundation. The goal is to define and give a foundation for mathematics within the scope of mathematics! This involves the use of recursive theory and other advanced topics.
  14. When one explains something (explanandum) in terms of something more fundamental, one take the fundamental as granted (explanan). In particular I have explained why the gravitational force is attractive using the spin of the gravitons (the force carriers). Of course, this open the door to a new question: why is the spin of gravitons just that? And again we will use something more fundamental to explain the spin of the gravitons. This is as both logic and physics work: we explain stuff using elementary stuff and next we explain elementary stuff using more elementary stuff. Regarding Weinberg, he is not referring to the world at the macroscopic level, nothing more far from reality! He is referring to fundamental aspects of the world, what he calls "principles". In fact he writes near the quote given above: Emphasis in the original. Weinberg is completely right; descriptive science is just a part of science. Answering "why" is another part of science.
  15. Yes/no Yes because many predictions made by string theorists about what we would observe at LHC have been now falsified. We did not see any gluino with mass below the TeV. No because string/brane theory is maintained in a safe state by changing its parameters. E.g. Kane 'predicted' a gluino mass of about 600 eV, when this was ruled out, he changed his 'prediction' to 1 TeV. Now that gluinos have been ruled out up to 1.24 TeVe, he is going to change his 'prediction' again...
  16. As Steven Weinberg writes in the preface of the volume I of his textbook on quantum fields: Bold face from mine.
  17. Not a surprise for me. SUSY is very difficult to accept for me. I also said time ago that LHC would not detect extra-dimensions or strings.
  18. The source of gravity is the stress-energy-momentum tensor [math]T^{\mu\nu}[/math] whose components are energy, momentum, and stress. In the non-relativistic approximation it reduces to mass. The tensor explains why massless particles as photons are affected (and generate) gravity. Thus, in very broad terms, anything with energy generates gravity. The gravity generated by the Earth is the result of the gravity generated by each particle. The concept of force is not fundamental, the relevant quantity is the gravitational potential. The force of gravity can be derived from it. The reasons for which the force of gravity is attractive can be understood by studying the details of the gravitational interaction. It is found that this consists of spin-0 (repulsive) and spin-2 (attractive) terms, but their 'weights' are different (1/3 and 2/3 from memory) and do not cancel. Summing both contributions one finds a net attractive force.
  19. Is is likely that dark matter is nothing real, but not in the sense that you seem to imagine "the quantum particle of nothing itself". In alternatives to the dark matter 'paradigm', the astrophysical/cosmological observations are explained using the observed mass. In those models dark matter does not exist.
  20. Important derived scientific quantities with special names: areic, flux, massic, moment, rate, wavelength... http://juanrga.com/en/knowledge/derived-scientific-quantities-with-special-names.html

  21. I agree on the existence of problems with current education system, from school to the post-doc level. I disagree on the origins. As it is said often "there is nothing more practical than a good theory". Precisely I started the project knowledge to provide a series of up to date educative resources for free. Several of the articles are devoted to the scientific methods (yes plural) and they illustrate very well what I am trying to say. We agree with William R. Robinson, Donald J. Wink, & William S. Harwood on that the "scientific method", as presented in many textbooks and encyclopedias, is not how "science is done" by anthropologists, biologists, chemists, geologists, medical scientists, physicists, and others. http://juanrga.com/en/knowledge/scientific-methods.html http://juanrga.com/en/knowledge/heuristic-methods.html http://juanrga.com/en/knowledge/didactic-methods.html
  22. The conclusion is that the expansion velocity is proportional to distance (more far more faster). Of course, when more far is the object of us more time needs the light to reach us, and this is why we speak about seeing them "in the past". To check if the universe is slowing down in expansion or not, we must measure the expansion at different times (e.g. today, two years ago, five years ago...) and see what happens. The rate of expansion is not slowing down but increasing http://news.nationalgeographic.com/news/2011/10/111004-nobel-prize-physics-universe-expansion-what-is-dark-energy-science/ http://map.gsfc.nasa.gov/universe/uni_expansion.html
  23. (i) A statement is not an idiom and (ii) both statements were made in the same idiom: English.
  24. Starting from [math]E = \sqrt{m^2c^4 + p^2c^2 }[/math] for a massless particle such as the photon set [math]m = 0[/math] and you obtain [math]E = |p|c[/math] for a non-relativistic particle expand the square root in a power series and ignore the higher order terms because [math]v \ll c[/math] [math]E = \sqrt{m^2c^4 + p^2c^2 } = mc^2 + \frac{p^2}{2m} + \cdots [/math] you obtain the 1/2 factor characteristic of the non-relativistic theory.
  25. Effectively, you are completely right. In relativity [math]v_{A+B} \neq v_A + v_B[/math] in the general case.
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