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Everything posted by DrRocket

  1. All force fields define the work expended to move an object between two points along a given path. But only fields that arise from a potential, conservative fields, are such that then points alone determine the work; i.e. the work is path-independent and is 0 for all closed paths. Not all fields are potential fields. The electrostatic field is a potential field.
  2. But it is rather difficult to have superconductivity (electrical) unless there are electrons around, so I think the reporter misinterpreted the significance of the Cooper pairs reported by the scientists.
  3. Sure. But in the meantime taking time and space as primitive is the best that we can do. Any more fundamental, and useful (not a bunch of philosphical mumbo jumbo), definitions would be a great step forward. My understanding is that lattice (discrete) theories have been attempted but have failed. If you can build better predictive theories than GR and QFT on some new concept of time and space that would be terrific. People are trying. Results thus far are underwhelming. Ver Linde has a pre-print in which he sees "space" as emergent from information -- but I see way too much vague hand waving in his "logic", and it has been awfully quiet since he gave his talk..
  4. I suspect that the problem goes way back to very early science classes where everyone was exposed to the littany differentiating between mass and weight (no arguement there) and being inculcated with the notion that mass is fundamental. Mass is not quite so fundamental as folks were told in those early days. It just is not that easy. What you measure on a labratory scale is relativistic mass, in the version of "invariant mass" taken in a reference frame of zero momentum. The mass that shows up in general relativity is also relativistic mass, but momemtum flux also shows up in the stress-energy tensor, and curvature is independent of any net linear velocity. So there is not a direct and easy correspondence between just rest mass or just relativistic mass either. GR is very subtle. In quantum field theories it is natural to consider mass as a charactristic of a particle and rest mass is the natural concept. What is really fundamental is the perspective that mass and energy are the same thing, and what is conserved is neither (rest) mass nor energy but mass/energy as a single entity. I suspect that the rigid "mass is rest mass" stance that is common these days comes from the particle physics influence. There are a lot more particle physicists than relativists. Perhaps you are right and we need a different word, or words. We are somewhat prisoners of our language, and "mass' is with us. It just is one of those words with several definitions. It is not a big problem as long as one is clear on the definition used in any given situation. Misner, Thorne and Wheeler attempt to break the language straight-jacket by using words like momeneregy, mass/energy, and of course spacetime. That does sometimes help, but their terminology is not widely used. Similar terminology issues exist in mathematics -- compact vs the French (compact + Hausdorf), Riemannian vs Pseudo-Riemannian, etc. -- but mathematicians seem more accepting of adopting local author-dependent conventions, perhaps because mathematicians are very careful about the definitions that are used in any specific theorem. It isn't any different. The terminology does reinforce that mass and energy are the same thing, two sides of a single coin.
  5. There is a rotational effect, the Lens-Thirring effect, aka "frame dragging" associated with a massive rotating body. I don't know much about it. ????????
  6. Start with the electric field from a static arrangement of xharges.. A point charge creates a spherically symmetric electric field that drops off like 1/r^2. The field due to an arrangement of point charges is the vector sum of the fields due to the individual chsarges. A electrostatic field is a conservative field. Therefore it is the gradient of some scalar field, called the potential. The voltage, aka potential difference, between two points in space is the difference between the values of the potential function at those two points. So, voltage depends on both the amount of charge ane the spatial arrangement of that charge.
  7. The Earth is not an inertial reference frame, though it is a fair approximation over short time intervals during its orbit around the sun.. The Michelson-Morley experiment was repeated at different times of the year and by various experimenters over a period of years -- different reference frames. http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/morley.html#c2
  8. You have the relative complement of a box of dimensions L+2, W+2, H+2 and a box of dimensions L, W, H (I assume that "around the box" implies "outside of the box) -- a solid rectangle with another solid rectangle removed from its center. This is not so much an exercise in logic as an exerecise in converting imprecise language into precise language. One has to figure out what the speaker actually means, which can be subject to interpretation. Edit: The comments below are correct. The outer box is actually a rectanglular solid in which the edges and corners are rounded off, with a radius of curvature of 1.
  9. Right, and you don't see your meter stick shrink either. Both you and the driver of the car will agree on the speed of the car relative to you (or of you relative to the car). The time dilation and length contraction effects compensate so that the ratio is the same in both reference frames. Were this not true the resulting asymmetry would allow identification of a preferred frame.
  10. Wrong. Read the paper. Note the reference to variable mass, noted in my earlier post, which makes no sense if one is in the rest frame of the object. It was your claim that [math]E=mc^2[/math] applies only in the rest frame: It is quite clear that this is not true, since as you have asserted [math] E^2=m_0^2c^4 + p^2c^2[/math] is valid in any inertial frame and as shown previously and below it is eqiuivalent to [math]E=mc^2[/math] You do, of course have to worry about what you mean by mass, and [math]m[/math] is "relativistic mass". But you have to worry about what you mean by mass in any case. If you cling to "mass = rest mass" in all cases then what you measure on a laboratory scale is not mass. What you measure on a laboratory scale is in fact the relativistic mass in a reference frame in which the net momentum of the system of molecules is 0, what is called "invariant mass". Invariant mass is not the sum of the rest masses of the individual molecules. So, using a dogmatic "mass = rest mass" approach results in a macroscopic definition of mass that is not additive -- the mass of a system, as measured in the conventional manner, is not the sum of the masses of its constituent parts. You can adopt that stance, but it is, kindly, unconventional, and contrary to normal useage at the macroscopic scale. It requires some new definition of mass for the old reliable [math] F=ma[/math] and [math]F=\dfrac {E}{c^2}a[/math] just doesn't satisfy. Replacing the more rigorous [math]F=\dfrac {dmv}{dt}[/math] with [math]F= \dfrac {d (\frac{E}{c^2})v}{dt}[/math] is equally unappealing. I misinterpreted what you meant by "rest frame energy of the photon. But the statement that "The equation is not valid in any other frame" is incorrect, as shown below. Yes, it is a matter of what you can derive. And [math]E=mc^2[/math] is derivable from and equivalent to [math] E^2 = m_0^2c^4 + p^2c^2[/math] and I showed you the derivation. I'll repeat it here. [math]E = \sqrt{m_0^2 c^4 + p^2c^2}[/math] [math]= \sqrt{m_0^2 c^4 + \gamma ^2 m_0^2 v^2 c^2}[/math] [math] = \sqrt { m_0^2c^4(1 + \gamma^2 \frac {v^2}{c^2}})[/math] [math]= \sqrt{\gamma^2 m_0^2 c^4}[/math] [math]= mc^2[/math] so long as [math]m_0 \ne 0[/math] So, "mass=rest mass" in all situations is in fact dogmatic. Better to fit your definition to the problem at hand, being clear of the definition used. It has the advantage that you don't have to give up [math] F=\dfrac {dp}{dt}[/math] which is rather useful for macroscopic dynamics problems; you just have to accept [math]m[/math] as relativistic mass. I don't care about Lev Okun's view of history. The mathematics is right here. There is nothing radical in what I am telling you. This is in any number of texts on relativity. If you accept [math]E^2 = m_0^2 c^4 + p^2c^2[/math] you must also logically accept [math]E=mc^2[/math] for they are the same equation (for [math]m_0 \ne 0 [/math]). The bottom line is that "mass" has several different meanings, all of which are legitimate in mainstream physics.
  11. That is simply wrong. [math] E=mc^2[/math] is valid in any inertial frame if [math]m[/math] is relativistic mass. From the Einstein paper : "The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 1020, the energy being measured in ergs, and the mass in grammes." The reference to variable mass makes it clear that Einstein was considering what is now called "relativistic mass". BTWThere is no such thing as "the rest frame of the photon" in special relativity -- that would invoke singularities in Lorentz transformations that one cannot live with. You can do this either way, there is no need to be dogmatic. [math]E = \sqrt{m_0^2 c^4 + p^2c^2}[/math] [math]= \sqrt{m_0^2 c^4 + \gamma ^2 m_0^2 v^2 c^2}[/math] [math] = \sqrt { m_0^2c^4(1 + \gamma^2 \frac {v^2}{c^2}})[/math] [math]= \sqrt{\gamma^2 m_0^2 c^4}[/math] [math]= mc^2[/math] so long as [math]m_0 \ne 0[/math] The notion of relativistic mass goes back to Einstein, and was taught as a matter of course for many years. The convention to regard mass as "rest mass" is relatively recent and not universal. In either case we are talking about a convention, and not a question of "right" or "wrong" Regarding mass as rest mass is very convenient in quantum field theories (where m_0 can easily be 0), and less so for macroscopic problems in special relativity. In general relativity "mass' is even more murky, and nobody knows what to do about quantum gravity.
  12. Yes, you have to worry about what you mean by mass. That was the point. No, I am not making up a new equation, Einstein made up [math]E=mc^2[/math] (expressed a bit differently in the first paper, but clearly involving variable mass). http://www.fourmilab...tein/E_mc2/www/ If you use [math]E^2 = m_0^2c^4 +p^2c^2[/math] you have to come to grips with what you mean by momentum, and that will take you right back to "relativistic mass". Using total energy is fine. But you need to know what it is. Since what it is is [math]\gamma m_0 c^2[/math] for a body of positive rest mass you are right back to relativistic mass. Best to remain flexible and use the approach most convenient for the problem at hand. I find the recent tendancy to be dogmatic and demand that mass be rest mass rather amusing, and limiting.
  13. I don't think the answer is settled, but something similar was hypothesized by Wheelere, the geon. http://en.wikipedia.org/wiki/Geon_(physics)
  14. http://depts.washington.edu/ssnet/Weinberg_SSN_1_14.pdf
  15. he subject is called heat transfer. There are three mechanisms, convection, conduction and radiation. Material properties are central to conduction and radiation. The problem of keeping coffee hot in a cup involves all three mechanisms. Heat transfer is definitely physics. However, many specialized texts are written for and by mechanical engineers.
  16. Right http://en.wikipedia.org/wiki/Peculiar_velocity http://en.wikipedia.org/wiki/Proper_motion
  17. In any theory there are some things that are taken as "primitive" and understood without definition. You cannot write a dictionary starting from the initial premise that no words are understood. "Space" (or "distance") and "time" are such terms. The best that we have are the operational definitions "distance is what rulers measure" and "time is what clocks measure". No better definition currently exists. Philosophers tend to obfuscate rather than illuminate when they address the issue. More fundamental definitions could be very valuable. but I am not going to hold my breath in anticipation.
  18. That depends on what one means by "mass", in special relativity. The current fashion, apparently inspired by elementary particle physics, is to equate "mass" ([math]m[/math]) with "rest mass" ([math]m_0[/math]), wich indeed is invariant, essentially by definition. But there is also the very useful concept of "relativistic mass", [math]m = \gamma m_0 [/math]. [math] E=mc^2[/math] is correct (admitting that the zero-rest-mass case must be handled separately) and easily remembered if one interprets [math]m[/math] as relativistic mass. If one works only with rest mass one is stuck with [math]E^2 = m_0^2c^4 +p^2c^2[/math]. This latter equation applies to the zero-rest-mass case, but also raises the question of what one means by momentum, [math]p[/math]. In the case [math]m_o \ne 0[/math], [math]p=mv[/math] where [math]m = \gamma m_0[/math] so one still has to deal with relativistic mass, at least implicitly. There is also the concept of "invariant mass", which is the relativistic mass of a system of particles, measured in center-of-mass coordinates so that the total momentum is zero, so that the mass is [math] E/c^2[/math]. This corresponds to the mass that would be measured by a labratory balance for a macroscopic object -- a hot bucket of water in principle weighs more than a cold one. If instead one were to cling to "mass" as being "rest mass" in the macroscopic setting, then the mass measurement of a laboratory balance would not equal the sum of the masses of the molecules of which an object is composed -- creating all sorts of confusion. My position is to consciously adopt whatever convention is appropriate for the problem at hand, keeping in mind that different authors use different conventions in different situations. "Mass" is not author-invariant. http://en.wikipedia....cial_relativity The situation in general relativity is even more murky: http://en.wikipedia....eral_relativity http://matheuscmss.w...s-applications/
  19. Yes. Flip the polarity swithch on your voltmeter, Or just interchange the leads. The range (in absolute value) is independent of polarity.
  20. One thing that should be abundantly clear is that you are ill-equipped to determine what I think. My mind is most certainly not made up, nor in my experience is that of any other professional mathematician or scientist. There is no such thing as dogma in science and questioning current theory is the essence of research. You are correct in only one regard; there is indeed no point in wasting time with a fool.
  21. There was an experiment being planned about a year ago, I think it was to be the first direct measurement of length contraction. The trick is to make two sinultaneous position measurements of a rapidly moving object. I have not heard anything further.
  22. Let k be a commutative ring and let[math]E_1,E_2,...,E_n[/math] be k-modules. Consider the class L of all multilinear functions [math]f:E_1 \times E_2 \times ... \times E_n \rightarrow F[/math] where [math]F[/math] varies with [math]f[/math]. The elements may be viewed as the objects of a category. If [math]f:E_1 \times E_2 \times ... \times E_n \rightarrow F[/math] and [math]g:E_1 \times E_2 \times ... \times E_n \rightarrow G[/math] are two objects in L, a morphism [math]f \rightarrow g[/math] is a homomorphism [math]h:F \rightarrow G[/math] such that [math] h \circ f = g [/math] The tensor product of [math]E_1,E_2,...,E_n[/math] denoted [math]E_1 \otimes E_2 \otimes ... \otimes E_n[/math] is a universal repelling object in this category.
  23. "To summarize , I would use the words of Jeans, who said that ‘the Great Architect seems to be a mathematician’. To those who do not know mathematics it is difficult to get across a real feeling as the beauty, the deepest beauty, of nature. C.P. Snow talked about two cultures. I really think that those two cultures separate people who have and people who have not had this experience of understanding mathematics well enough to appreciate nature once." – Richard P. Feynman in The Character of Physical Law
  24. One can always revert to the physicist's definition: A tensor is a symbol having four corners, the two on the right being particularly appealing, to which one may attach an unlimited number of indices.
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