  # Pete

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## Everything posted by Pete

1. Yes. Most certainly. The correct expression for the kinetic energy, K, of a particle with proper mass m0 is $K =(\gamma - 1)m_0 c^2$ Its not just you. As I mentioned above, the increase in energy (regardless of the type of energy) will result in an increase in the strength of the gravitational field. I.e. increase in energy - > increase in active gravitational mass. Hence the widely used term mass-energy. Pete
2. By "physics university" I mean one of the universities which is quite well known for physics. E.g. MIT, Cal Tech, etc. When people say that physicists don't use inertial mass anymore they claim that they use energy instead since "its the same thing". I don't think I gave an example of what that's not true in general. If it was always true that E = m2 then one might want to say that this is a definition of inertial mass. But since the correct definition of inertial mass is m p/v then one can defined m as m = E/c2 if and only if this expression is true in all possible/concievable cases. If one can give 1 single counter example then such an assertion must be abandoned. A simple example is that of a rod at rest in the inertial frame S'. Let the rod lay on the x' axis and have a force exerted on each end such that the total force is zero. Then the rod does not accelerate in any frame. In this case the realtion E = mc2 is invalid. For those of you who understand the math/physics see a related example at http://www.geocities.com/physics_world/sr/rd_paradox.htm I'll make a more explicit example soon and post it. Pete
3. Since mass and energy are related then yes. An increase in the energy of a body will increase the mass of the body and hence increase the strength of the gravitational field. To be precise - The Principle of Equivalence states that a uniform gravitational field is equivalent to a uniformly accelerating frame of reference. This does not mean that all gravitational fields can be replaced by an accelerating frame. When the gravitational field is not uniform (or in the language of general relativity - when the spacetime is curved) then the principle of equivalence applies locally (restrict observations to regions of spacetime where the spacetime curvature, aka tidal forces can be neglected). If one is at rest in a rotating frame of reference then according to Eintein there is a gravitational field present. As Einstein wrote in The Foundation of the General Theory of Relativity, Albert Einstein (1916), Annalen der Physik 49 Newton would say no and Einstein would say yes. According to Newton it is an inertial force. According to Einstein its a gravitational force. Einstein postulates of GR imply that inertial forces are identical in nature to gravitational forces. He then concluded that since gravitational forces are "real" then so too must coriolis forces and centrifugal forces be real. If by this you mean that an increase in energy produces an increase in active gravitational mass then yes, that is correct. If you're in a rotating frame of reference and there is a particle moving in your frame then it will accelerate. The force related to this acceleration is called centrifugal force. It points radially away from the axis of rotation. There is another force acting on the particle as well - the Coriolis force which is a velocity dependant force (much like the magnetic force). Each of these are inertial forces. If the particles are also acted on by the wall then the it experiences other kinds of forces. But the gravitational forces here are the centrifugal force and the Coriolis force. Mind you - If you're at rest in an inertial frame and someone started spinning a space station around you then no gravitational forces will appear. The reason being that spinning the space station does not mean you've changed your own frame of reference. For gravitational forces to be produced one must change their frame of reference too. When you start to rotate with the space station then, according to general relativity, a gravitational field will be produced. Pmb
4. I see. I this particular case I found it hard not to due to the nature of the subject. Pete
5. ## What does GR say about time dialation when one is falling vs. on the ground?

There are two time dilation effects in relativity. One is gravitational time dilation and there is time dilation due to a motion. The specifics of the problem and question will determine the exact response that you're looking for. I.e. who is measuring what etc. Let consider a specific example ("time orthogonal spacetime"/no frame dragging). In such a field time intervals dt (time measured by 'coordinate' observer - precise meaning depends on exact situation) are related to proper time intervals, dT (time read by clock which is in g-field and moving), by dt/dT = 1/sqrt(1 + 2*Phi/c2 - v2/ c2) where v = speed of particle and Phi = gravitational potential. Consider, for example, a person who is in free-fall in a uniform g-field. Then to that free-fall observer no gravitational field exists anywhere (Phi = 0 everywhere in region where field is uniform) so there is no gravitational time dilation to speak of. There is only time dilation due to motion. If, however, the gravitational field is not uniform then the gravitational field does not vanish everywhere for a free-fall observer and in general there will be gravitational time dilation. In case I made an error please see the Global Positioning System (GPS) example in Taylor and Wheeler's book at http://www.eftaylor.com/pub/projecta.pdf Pete
6. But then again Steve Carlip (GRist at UC Davis) also claims 'nobody' uses it. sigh! I'll have to wait until I can get to a decent research library and see for myself. Pete
7. When it would apply, sure. I see no reason for someone would take offense at someone saying that something was incorrect. Why would anyone find that offensive??? Pete
8. And the basic answer I gave was The second basic question I responded to was The rest was in response to swan's comments. His responses were about invariants, scalars etc. so II responded to that. Its my experience that people who make the assertions swan did seem to be particle physicists/nuclear/atomic scientists. But such groups have a limited use of relativity, since they rarely do anything which requires speaking of anything other than proper mass. For that reason such groups use only the term "mass" rather than "proper mass" or rest mass and are unaware of what GRists and cosmologists use. I've looked into this one particular subject point in more detail over the years and that research brought me to applications not found in either particle physics, general relativity or cosmology, the main users of relativity. One might have otherwise thought it strange to hear that E = mc2 is not always correct. I know that most people would find that fact strange to hear. Yet its true. Its all described in the relativity literature if one knows where to look and it can't be found in particle/GR/cosmology texts etc. I see no point in leaving out interesting points I know of. One can simply feel free to ingore them if they wish. Let me give you an example of certain points I'm refering to; The expression F = ma was not an expression given by Newton. It was given by Euler. For this reason F = ma has been come to be know as Euler's definition of mass. This definition can be used to define what can be called Mach's definition of mass. However Newton's was F = dp/dt. This is the definition used in relativity (the topic of this forum). This yields a definition of mass which has come to be called Weyl's definition of mass. This is the definition I stated above and the one people in relativity use. If a basic question is asked then I'll give a basic response. If that response is objected to or contradicted and I believe that objection/contradiction is incorrect then I'll state so. I see that as the purpose of a discussion forum. Thanks Pmb
9. By the way, some of what I've relayed has come from speaking to research physicists. For example, on this point I once asked a very well-known research physicist if "light has mass" to see his response. He told me that sometimes its useful/helpful to think in those terms. He thinks this stongly enough to teach it to his students in upper level courses at a major "physics university." So let us not neglect how a research physicists thinks and only focus by what appears in his published papers only (not that he leaves it out but I haven't scanned Physical Review D for example for such usage). Pete
11. I don't believe he was "technically" incorrect per se. I only meant to say that he was incorrect as far as how everyone in the relativity community defines the term mass. There are two ways in which people use the term "mass" in relativity. One is as I've described and the other, i.e. proper mass, is proportional to the square of the magnitude of the particle's 4-momentum. However this is a vauge term since 4-momentum can't always be defined for an object. 4-momentum can only be defined for an isloated object or a pointlike object (no spatial dimensions). But some people use the term mass to refer to proper mass (which is what swan calls rest mass). However it is my opionion that the magnitude of some 4-vectors are best refered to as proper quantities. Thus if dX = (cdt, dx, dy, dz) is a spacelike spacetime displacement then c^2dT^2 = dX*dX where dT is referred to as the proper time interval between the two events. If dX is a spacelike spacetime displacement, then c^2ds^2 = dX*dX where ds is proper distance. Likewise P = (mc, p) is the 4-momentum of a particle whose mass is m and 3-momentum is p. m02c2 = P*P where m0 is the particle's proper mass. The terms proper mass and rest mass do not refer to the same thing. Rest mass denotes the value of proper mass when the velocity is zero. However m(0) does not equal m0 unless dt/dT = 1. For example; In a wide variety of circumstances (i.e. time-orthogonal spacetimes, where g0k = 0 -- no spacetime dragging) dt/dT is a function of both the particle's speed and the gravitational potential of the particle. dt/dT = 1 only when v = 0 and the gravitational potential also is zero (more on this can be found in Moller's text on general relativity). Pmb
12. That is incorrect. That is not the correct definition of inertial mass. That is a relationship between force and inertial mass in those cases when the mass is constant. Force is not defined as F = ma. Its defined as F = dp/dt. The qauntity F/a does not have the same value as the quantity that might be considered to be "gravitational charge" as is the quantity m in in F=GMm/r2. The expression for gravitational force in relativity is not F=GMm/r2. Its the m in $$G_k = m\Gamma^{\alpha}_{k\beta} v_{\alpha}v^{\beta}$$ That is incorrect. The m in that expression is an equality under certain circumstances. Not all circumstances however. There are cases when E = mc2 is invalid. Note: E = mc2 is only total energy when the potential energy V is zero. For a photon the inertial mass can be expressed in various ways. If the magnitude of the photon's momentum is p then its inertial mass is given by m = p/c. If the photon's frequency is f then since E = hf the inertial mass is m = E/c2 = hf/c2. That is incorrect. The inertial mass of a photon is a very well defined quantity. That is also incorrect. The concept is widely used. For example; Textual Examples --- Relativity: Special' date=' General and Cosmological,[/i'] Rindler, Oxford Univ., Press, (2001) Cosmological Physics, John A. Peacock, Cambridge University Press, (1999) Understanding Relativity: A Simplified Approach to Einstein's Theories, Leo Sartori, University of California Press, (1996) Basic Relativity, Richard A. Mould, Springer Verlag, (1994) Introducing Einstein's Relativity, Ray D'Inverno, Oxford Univ. Press, (1992) Gravitation from the Ground Up, Bernard F. Schutz, Cambridge University Press, (2003) Journal Examples Apparatus to measure relativistic mass increase, John W. Luetzelschwab, Am. J. Phys. 71(9), 878, Sept. (2003). Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995). In defense of relativistic mass, T. R. Sandin, Am. J. Phys. 59(11) 1032 (1991). A simple relativistic paradox about electrostatic energy, Wolfgang Rindler and Jack Denur, Am. J. Phys. 56(9), Sept. (1988). An elementary development of mass-energy equivalence, Daniel J. Steck, Frank Rioux, Am. J. Phys. 51(5), May (1983). A Short Course in General Relativity, Foster & Nightingale, Springer Verlag, (1994). Observed Relativistic Mass Increase for 0.3 eV Electron, G. Gabrielse and H. Dehmelt, Bull., Am. Phys. Soc. 25, 1149 (1980). Online Examples http://web.uniud.it/cird/girepseminar2003/abstracts/pdf/mulaj.pdf http://physics.syr.edu/courses/PHY106/Slides/PPT/Lec16-Special-Relativity_2.pdf http://physics.syr.edu/courses/PHY106/Slides/PPT/Lec17-Special-Relativity_2.pdf http://www.physics.mq.edu.au/~jcresser/Phys378/LectureNotes/SpecialRelativityNotes.pdf http://astro.wsu.edu/allen/courses/astr150/Einstein.pdf http://www.phys.virginia.edu/classes/109N/lectures/mass_increase.html http://www.eas.asu.edu/~holbert/eee460/Relativity.pdf http://www.ucls.uchicago.edu/Academics/depts/science/physics/Relativity.pdf http://www.dur.ac.uk/Physics/students/physics_specialrelativity.html http://www.ph.rhul.ac.uk/course_materials/PH154/Relativistic%2520mass%2520and%2520dynamics.pdf http://www.lima.ohio-state.edu/physics/113sp03/113Lectures/113rela2.pdf http://www.lima.ohio-state.edu/physics/113sp03/113Lectures/113rela2.pdf http://www.physics.fsu.edu/users/ng/courses/phy2054c/hw/Ch26/ch26.pdf http://www.astro.washington.edu/tmurphy/phys110/faqs/AB05.05.html Particle Accelerator Labs http://humanresources.web.cern.ch/humanresources/external/training/tech/special/AXEL2003/AXEL-2003_L02_24Feb03pm.pdf http://www.neutron.anl.gov/hyper-physics/inertia.html http://aether.lbl.gov/www/classes/p139/animation/sr.html http://www.fpm.wisc.edu/safety/Radiation/2000%2520Manual/chapter12.pdf That assertion is dependant on how one defines "mass" in the first place. That is incorrect. One can always speak of frame dependant quantities in relativity because those are the only quantities which are measureable. Its not quite accurate to say that relativistic mass is not invariant unless you're quite specific about what you mean by that statement. It can be very tricky. For example: If the 4-momentum of a particle is P and the 4-velocity of an observer is U then the relativistic mass as measured by that observer is proportional to the scalar product of P and U[/b]obs, i.e. mobs = P*Uobs In case you think this sort of thing doesn't appear in the relativity literature then I recommend reading Energy Conservation as the Basis of Relativistic Mechanics II, J. Ehlers, W. Rindler, R. Penrose, Am. J. Phys. 33, 995-997 (1965). From page 996 It is good to keep in mind that the components of a vector or tensor are defined in terms of the scalar product of the tensor or vector with a basis vector and as such is a scalar. For people who do that they run the risk of making an error in those cases when p/v does not equal E/c2. There are cases where the relationship p/v = E/c2 does not hold and therefore E = mc2 does not hold. A simple example would be the mass of a gas in a box. Its both useful and as such still used. Since that expression is invalid this arguement is flawed. Mass is the proportionality quantity between velocity and momentum. It therefore represents the resistance to changes in momentum, not velocity. More on this subject can be found listed in the references here http://www.geocities.com/physics_world/relativistic_mass.htm and in the journal articles listed here http://www.geocities.com/physics_world/mass_articles.htm Pete
13. Why do you think that there are "special" coordinate systems in which light is "really" moving on a straight spatial trajectory? Why true in Newtonian mechanics it is false in GR. That's correct. And that's why I was explaining to you. I.e. that whether the spatial trajectory of a beam of light is curved or not depends on the frame of reference. Then the question "Is the spatial trajectory is curved" is quite useless. But I can still speak of it. Its not true that frame-dependant statements are useless. The observer lives in a particular frame and the observer himself is a geometric object. Therefore statements like "spatial path is curved" has a geometric meaning independant of a coordinate system and depends only on observer and phenomena. The whole point of tensor stuff is that relativity demands that the laws of physics not depend on a particular coordinate system. It does not mean that observations are coordinate independant. Pmb
14. People should keep examples of geodesics in mind so as not to be confused. A few examples go a long way. Here are a few which are illuminating (1) geodesics on an infinite flat plane (2) geodesics on a sphere (3) geodesics on a cylinder (4) geodesics on a cone For #1 the geodesics are straight lines. For #2 the geodesics are great circles. For #3 the geodesics are spirals. See Figure 4 at http://www.geocities.com/physics_world/euclid_vs_flat.htm For #4 the geodesics are hard to picture. See figure 2 at http://www.geocities.com/physics_world/euclid_vs_flat.htm Absolutely. Its quite important to distinguish the two. You measure "space" with a ruler and "time" with a clock. As Einstein himself said in Science in Feb 17, 1921 issue page 783 Consider the inside of an elevator in flat spacetime in an inertial frame of reference. Let there be a rectangular set of spatial axes, i.e. an xy coordinate system whose axes are mutually perpendicular. A beam of light will move on a spatial curve described by x = ay + b. This is a straightline in space and is thus also a geodesic in space. The worldline is a line in spacetime for which x = ct (when x and "ct" axes are perpendicular). This too is a straight line, a straightline in spacetime - a geodesic. However if the elevator is uniformly accelerating then the curve the light will move on a spatial curve which is not a straight line. The curve in spacetime is also not a staight line. However it is a trajectory in spacetime is still a geodesic. Pete
15. Light does have mass (aka inertial mass aka relativistic mass). It has zero proper mass (aka rest mass) and that's what you were thinking about. As such it is effected by gravity. One simply has to turn to The Feynman Lectures Vol -I page 7-11. Section entitled Gravitation and Relativity That's not quite right. While it is true that light moves on geodescics is not quite right to say that a geodesic is a "straight line". Its more appropriate to say that its the straightest possible line. In this case it is a geodesic, not in space, but in spacetime. Light does not travel on a straight line is space. It travels on a curved line in space. Pete
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