imatfaal Posted May 10, 2012 Posted May 10, 2012 ! Moderator Note Wucko. Your latest post (#26) is off topic - please stop introducing divergent ideas into threads; if you wish to discuss this matter please do open a new thread. Please do not further derail the thread by responding to this modnote. 2
pmb Posted May 22, 2012 Posted May 22, 2012 (edited) &lt;br /&gt;Well, typically [math]p_\mu p^\mu \equiv m^2[/math]. Since [math]p^\mu =(E,p^i)[/math] where [math]p^i[/math] is the three-momentum, it's easy to see for a particle with m=0 that [math]E^2=p_i p^i[/math]. So massless particles have momentum by virtue of their energy.&lt;br /&gt;<br /><br />Those equivalences which Klaynos employed originated with electrodynamics. E = pc is derived from an EM calculation. Those calulations are the ones from which the 4-momentum are constructed and the invariance estabished. The expression E^2 - (pc)^2 = (mc^2)^2 is then used with E = pc to deduce the expression m = 0. I believe what you did here was to take that and work backwards to make your point. <br /><br />It's been a while since I've worked these equations. This will be pleasing to derive from the basics again. Please give me time though. I'm studying other material for the moment and will start on this tonight. Ill make these into web pages which will go in my website. Any help in PM would be appreciated. I choose PM so I won't have to get into side debates along the way.<br /><br />Let me start by deriving E = pc. Let b = beta = v/c, g = gamma = 1/sqrt( 1 - b<sup>2</sup> ), m = proper mass of particle.<br /><br />By definition of P, P = M(v)v. I show here <a href='http://home.comcast.net/~peter.m.brown/sr/inertial_mass.htm' class='bbc_url' title='External link' rel='nofollow external'>http://home.comcast....ertial_mass.htm</a><br /><br />that M(v) = gm. Therefore P = gmv. It seem that I never created a web page where I derive E = gE<sub>0</sub> so let's take on assumption. (it may be in <a href='http://home.comcast.net/~peter.m.brown/sr/work_energy.htm' class='bbc_url' title='External link' rel='nofollow external'>http://home.comcast....work_energy.htm</a>)<br /><br />Derivation of E = pc for a photon<br /><br />P = gmv, E = gE<sub>0</sub> ===> E/E<sub>0</sub> = g<br /><br />P = mv(E/E<sub>0</sub>) = mv(E/mc<sup>2</sup>) = v(E/c<sup>2</sup>) <br /><br />P = Ev/c<sup>2</sup> ===> Pc/E = v/c<br /><br />If the particle is a luxon, which is a particle for which v = c, e.g. a photon, the v = c.<br /><br />Pc/E = v/c = 1 ===> E = pc<br /><br />Energy-momentum relatinship for a photon: E = pc<br /><br />------------------------------------------<br />Next derivation: Derive E<sup>2</sup> - (pc)<sup>2</sup> = (mc<sup>2</sup>)<sup>2</sup><br /><br />P = E/c<sup>2</sup> ===> Pc = Eb ===> E<sup>2</sup>b<sup>2</sup> = (pc)<sup>2</sup><br /><br />E<sup>2</sup> - E<sup>2</sup>b<sup>2</sup> = E<sup>2</sup> - (pc)<sup>2</sup><br /><br />E<sup>2</sup>(1 - b<sup>2</sup>) = E<sup>2</sup> - (pc)<sup>2</sup><br /><br />E/gE<sup>2</sup> = E<sup>2</sup> - (pc) <sup>2</sup> = (mc<sup>2</sup>)<sup>2</sup><br /><br />Therefore E<sup>2</sup> - (pc) <sup>2</sup> = (mc<sup>2</sup>)<sup>2</sup><br /><br />At this point I can now define a 4-vector, the 4-momentum <b>P</b> as <br /><br /><b>P</b> = (E/c, <b>p</b>)<br /><br />This could not be done unless we hadn't derived the equations which allows us to derive the 4-momentum 4-vector as well as its properties. This is too long. All the equations are making mush out of my mind. LOL! I'll argue why one can't logically argue the other way around. Edited May 22, 2012 by pmb
pmb Posted May 22, 2012 Posted May 22, 2012 <br />One of the pseudoscientific instances is the confusion of correlation with causality. Physics seems to be biased in the dirrection of confusing causality with correlation. Its an ideology as much as a science. Thats the problem also.<br /><br /><br /><br />This thread i so fagmented along the train of thoughts that I can't follow it. Splitting a thread can make it very disturbing for me to read. My mind is messed up. Perhaps from the pain meds I'm on. Bad juju! In any case, if you want to see why the speed of light travels at the speed of light then let's start off with what we're really looking for. I want to determine what the speed of an EM Wave is where light is an EM wave. The speed is derived here http://www.scienceforums.net/topic/66476-supernovae-and-time/ in the first post. Best wishes, Pete
ydoaPs Posted May 22, 2012 Posted May 22, 2012 You tell me, i have my own theory. Since no one has shown the derivation, I'll post the derivation and explanation from my post in this thread. Maxwell equations: [math]\bigtriangledown\cdot{E}=\frac{\rho}{\epsilon_0}[/math] [math]\bigtriangledown\times{E}=-\frac{\partial{B}}{\partial{t}}[/math] [math]\bigtriangledown\cdot{B}=0[/math] [math]\bigtriangledown\times{B}={\mu_0}{\epsilon_0}\frac{\partial{E}}{\partial{t}}[/math] These are the equations that describe how electric fields and magnetic fields interact. Think of [math]\bigtriangledown\cdot[/math] as describing whether or not a vector field is pointing inward or outward, think of [math]\bigtriangledown\times[/math] as describing which in which direction and how tightly a vector field is curled, and think of [math]\frac{\partial}{\partial{t}}[/math] as being the rate of change of the vector field. A vector field is a space where there is a vector at every point. A vector is a mathematical object with both a number and a direction. Having no charges to worry about with light, we can set the charge density equal to zero which makes the equations: [math]\bigtriangledown\cdot{E}=0[/math] [math]\bigtriangledown\times{E}=-\frac{\partial{B}}{\partial{t}}[/math] [math]\bigtriangledown\cdot{B}=0[/math] [math]\bigtriangledown\times{B}={\mu_0}{\epsilon_0}\frac{\partial{E}}{\partial{t}}[/math] Now, let's take the curl of the curl equations and see what happens. [math]\bigtriangledown\times\bigtriangledown\times{E}=-\frac{\partial}{\partial{t}}\bigtriangledown\times{B}=-{\mu_0}{\epsilon_0}\frac{\partial^2{E}}{\partial{t^2}}[/math] [math]\bigtriangledown\times\bigtriangledown\times{B}={\mu_0}{\epsilon_0}\frac{\partial}{\partial{t}}\bigtriangledown\times{E}=-{\mu_0}{\epsilon_0}\frac{\partial^2{B}}{\partial{t^2}}[/math] Since [math]\bigtriangledown\times(\bigtriangledown\times{V})=\bigtriangledown(\bigtriangledown\cdot{V})-\bigtriangledown^2{V}[/math] for any vector field V, we can write: [math]-{\mu_0}{\epsilon_0}\frac{\partial^2{E}}{\partial{t^2}}=-\bigtriangledown^2{E}[/math] [math]-{\mu_0}{\epsilon_0}\frac{\partial^2{B}}{\partial{t^2}}=-\bigtriangledown^2{B}[/math] which we rearrange to get: [math]\frac{\partial^2{E}}{\partial{t^2}}-\frac{1}{{\mu_0}{\epsilon_0}}\cdot\bigtriangledown^2{E}=0[/math] [math]\frac{\partial^2{B}}{\partial{t^2}}-\frac{1}{{\mu_0}{\epsilon_0}}\cdot\bigtriangledown^2{B}=0[/math] which are the electromagnetic wave equations. The speed term is [math]\frac{1}{\sqrt{{\mu_0}{\epsilon_0}}}[/math] where [math]\mu_0[/math] is the permeability of free space and [math]\epsilon_0[/math] is the permattivity of free space. Plug in the numbers and that's how we get c. <br /><br /><br />This thread i so fagmented along the train of thoughts that I can't follow it. Splitting a thread can make it very disturbing for me to read. My mind is messed up. Perhaps from the pain meds I'm on. Bad juju! In any case, if you want to see why the speed of light travels at the speed of light then let's start off with what we're really looking for. I want to determine what the speed of an EM Wave is where light is an EM wave. The speed is derived here http://www.scienceforums.net/topic/66476-supernovae-and-time/ in the first post. Best wishes, Pete While it does not allow html, this board does use a similar formatting system.
pmb Posted May 23, 2012 Posted May 23, 2012 (edited) <br />Sure is. The dispersion relation (the relation between energy E and momentum p) for the free electromagnetic field is [math]E = pc[/math], which is a simplification of the more general dispersion relation for free fields ([math]E^2 = m^2 c^4 + p^2 c^2[/math]) resulting from a mass-like term that is zero (m=0). From that, you directly get a group and phase velocities [math] \frac{\partial E}{\partial p} = \frac Ep = c[/math] for wave packets of the free electromagnetic field.<br /><br /><br /><br />That makes no sense to me. Especially The relation you gave is<br /><br />[math] \frac{\partial E}{\partial p} = \frac Ep = 1/c[/math] <br /><br />Dispersion relations apply only to waves and a particle moving with velocity <b>v</b> is not a wave. A photon is not a wave. This is a classical photon, not a quantum one. You can't even think of a classical photon as you could quantum one. Here, in classical mechanics, which relatiivity is of, photons move on classical trajectories.<br /><br /><br />Since no one has shown the derivation,...<br /><br />That's not true. I just posted a link to the derivation you posted a while back in another thread. Just clicking on the link I gave would have taken the reader to your derivation. That post was praisworthy for doing the workl. Since I posted a link to your post aqbove there really is no need to redo what I've aleady done. Soyou went from praiseworthy to redundant. But I suppose you had a need. I might have done the same thing in your position - and I'm be redundant too. lol! Also imatfal posted the following in post #7<br /><br />Maxwell's equations will also pop out a speed for Electromagnetic Radiation through vacuum<br />Relation between electricity, magnetism, and the speed of light<br /><br /><br /><br /><br /><br />Nobody asked him for clarity so its assumed that nobody wanted any or were to embarassed to ask. Hence the reason to link to your post elsewhere. But repeating it here is work you didn't need to do. And its not because we're lazy that it hasn't been provided. I've seen this question many times and decided yesterday to create a new web page to do the work out once and for all. And that will still be gone because in days to come the derivation on my website will be easy to find. This one, not so much.]<br /><br /><br /> Edited May 23, 2012 by pmb
juanrga Posted May 23, 2012 Posted May 23, 2012 (edited) I agree, (I've always struggled with massless particles having momentum) but are these equasions not back engineered? Momentum p is not given by mv for massless particles. There is not problem with a particle having m=0 and nonzero p Edited May 23, 2012 by juanrga
pmb Posted May 23, 2012 Posted May 23, 2012 (edited) he Physis FAQ it says that light has inertial mass. I read this years ago and forgot the details but I like it forhe most part. The new FAQ is here<br /><a href='http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html' class='bbc_url' title='External link' rel='nofollow external'>http://math.ucr.edu/...hoton_mass.html</a><br /><br />He left me out for credit which upset me a bit but no more. The author uses the symbol [math]p = m_{rel}v[/math]. The [math]m_{rel}[/math] here is a particles relativistic mass. The value is found by deviding the momentume buy its mass. It can also be found otherways. Seems that most GR texts noways define the photons inertial mass using its wavelength/frequency/energy.<br /><br / My appologies for the messyness of my posts. For some reason there are these darn HTML break line symbols floating all over the place and its ruining my posts. I keep asking the moderators why this happens but nobody is telling me. Perhaps the PM messages they might be sending me are not getting to them just as mine may not be getting to them. Very confusing. Note: I made a terrible error above. [math] \frac{\partial E}{\partial p} = \frac Ep = 1/c[/math] is so incorrect I just slapped myself. LOL!! <br />I agree, (I've always struggled with massless particles having momentum) but are these equasions not back engineered? They certianly explain the relationships but not the reason, that is, they do not explain why the photons leaving your display are so repulsed by the atoms in the LED and so attracted to the atoms in your retina.<br /><br />Go easy on me!<br /><br /><br /><br /> There are famous physicists who teach that light has mass. E.g. Richard Feynman and Alan Guth come to mind. Guth I know personally and he told me that he sometimes finds it useful to think of light has having mass. Others are the exact opposite. I was wondering something. Some of the physics I learned came from well-known physicists. When I say something like says something that Guth I know personally ... I fell like I'm unnecessarily name-dropping, which I hate. Is this an unwise way to argue opinion in physics? Edited May 24, 2012 by pmb
granpa Posted May 24, 2012 Posted May 24, 2012 The speed term is [math]\frac{1}{\sqrt{{\mu_0}{\epsilon_0}}}[/math] where [math]\mu_0[/math] is the permeability of free space and [math]\epsilon_0[/math] is the permattivity of free space. Plug in the numbers and that's how we get c. you can think of permeability and Permittivity as the springiness of space
pmb Posted May 24, 2012 Posted May 24, 2012 I was wondering something. Some of the physics I learned came from well-known physicists. When I say something like says something that Guth I know personally ... I fell like I'm unnecessarily name-dropping, which I hate. Is this an unwise way to argue opinion in physics? I just looked up the term "name-dropping". From http://www.thefreedictionary.com/name-dropping To mention casually the names of illustrious or famous people in order to imply that one is on familiar terms with them, intended as a means of self-promotion. Let me state this now and perhaps I'll come to terms with it. The purpose of my posting a quote from Guth had nothing to do with implying anything about my familiarity wih Dr. Guth. The purpose was physics related and not fame related. As if I'd ever become famous by name-dropping here. LOL!! Although now that I think back ... I did do a bit of shameful name-dropping here. It was more along the way of being proud to know and work with them. After all that part is in print so ...
imatfaal Posted May 24, 2012 Posted May 24, 2012 I just looked up the term "name-dropping". From http://www.thefreedi...m/name-dropping Let me state this now and perhaps I'll come to terms with it. The purpose of my posting a quote from Guth had nothing to do with implying anything about my familiarity wih Dr. Guth. The purpose was physics related and not fame related. As if I'd ever become famous by name-dropping here. LOL!! Although now that I think back ... I did do a bit of shameful name-dropping here. It was more along the way of being proud to know and work with them. After all that part is in print so ... To be honest Peter - I like to bask in the reflected glory; it's like a degrees of separation game: I have discussed physics with you, you have worked with Edwin Taylor, who co-authored one of the most important texts in physics with John Archibald Wheeler, who collaborated with Einstein
D H Posted May 28, 2012 Posted May 28, 2012 <br /><a href='http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html' class='bbc_url' title='External link' rel='nofollow external'>http://math.ucr.edu/...hoton_mass.html</a><br /><br /> Stop using HTML! Your posts routinely look like garbage. There's a "Preview Post" button right next to the "Add Reply" button. Use it!
mindless Posted May 28, 2012 Posted May 28, 2012 There are two issues here. The first is "why is the speed of light a constant?" and the second is "why do photons travel at this constant velocity?". The answer to the first question is given in plenty of detail at: http://en.wikibooks.org/wiki/Special_Relativity/Spacetime#The_modern_approach_to_special_relativity and is the result of the existence of Lorentz Invariance. The answer to the second question is more interesting nowadays. In the frame of reference of a photon there is no separation between its source and its destination and the transit takes no time. For a photon emitters and absorbers are in contact...
juanrga Posted May 28, 2012 Posted May 28, 2012 There are two issues here. The first is "why is the speed of light a constant?" and the second is "why do photons travel at this constant velocity?". The answer to the first question is given in plenty of detail at: http://en.wikibooks....cial_relativity and is the result of the existence of Lorentz Invariance. No. There systems/processes which do not verify Lorentz invariance and still the speed of light is a constant. General relativity was developed for those systems/processes. Therefore the constancy of light speed cannot be "the result of the existence of Lorentz invariance". The answer to the second question is more interesting nowadays. In the frame of reference of a photon there is no separation between its source and its destination and the transit takes no time. For a photon emitters and absorbers are in contact... There is not such thing as "the frame of reference of a photon".
mindless Posted May 28, 2012 Posted May 28, 2012 (edited) No. There systems/processes which do not verify Lorentz invariance and still the speed of light is a constant. General relativity was developed for those systems/processes. Therefore the constancy of light speed cannot be "the result of the existence of Lorentz invariance". There is not such thing as "the frame of reference of a photon". In General Relativity light travels along null geodesics (See http://en.wikipedia....l_relativity%29 ) which embodies a very closely related concept to the idea of light travelling along paths specified by a zero space-time interval. "Lorentz Invariance does not apply globally within the General Theory of Relativity because such invariance expresses certain relations between global orthogonal rectilinear coordinate systems, and within the General Theory of Relativity (and in most non-Euclidean spaces) there do not exist such coordinate systems (because under General Relativity mass distorts the flatness of space). Local Lorentz Invariance does however apply." Invariances: the structure of the objective world. By Robert Nozick This local Lorentz Invariance occurs at the point where the photon is located and represents the same constraint that occurs in Special Relativity and gives rise to the existence of a particular constant velocity for all local observers no matter how fast they are moving ©. General Relativity does not undermine the insights available from Lorentz Invariance, it builds on them. Your point about there being no such thing as the frame of reference of a photon is well taken and might be replaced by "in the limit as an inertial reference frame (IRF) approaches the velocity of a photon" there is no separation between an emitter and an absorber of electromagnetic radiation in the direction of travel of the IRF in flat space-time. Still a very interesting point given that almost all EM interactions that we observe occur in flat spacetime. Consider also that we can never observe a photon, what we observe is an interaction that can be explained by the concept of the transfer of a quantum of energy. What we consider to be a host of little particles in flight could also be regarded as a resonance between an emitter and an absorber. Imagine the IRF discussed above, the one moving at .9999... c in the direction taken by a photon, it might observe the interaction between the photon emitter and absorber as the direct resonant contact between emitting and absorbing outer shell electrons. (The space between these electrons having been contracted to very, very nearly zero). You have made me amend the overly broad position that I took earlier but the substance of my points still stands: that there is a constant velocity 'c' (amended to a constant velocity for all observers in flat spacetime and for local observers in curved spacetime) and that EM interactions can be regarded, in flat spacetime at least, as direct interactions between emitters and absorbers with no room left for greater or lesser velocities. Edited May 28, 2012 by mindless
juanrga Posted May 28, 2012 Posted May 28, 2012 (edited) In General Relativity light travels along null geodesics (See http://en.wikipedia....l_relativity%29 ) which embodies a very closely related concept to the idea of light travelling along paths specified by a zero space-time interval. "Lorentz Invariance does not apply globally within the General Theory of Relativity because such invariance expresses certain relations between global orthogonal rectilinear coordinate systems, and within the General Theory of Relativity (and in most non-Euclidean spaces) there do not exist such coordinate systems (because under General Relativity mass distorts the flatness of space). Local Lorentz Invariance does however apply." Invariances: the structure of the objective world. By Robert Nozick This local Lorentz Invariance occurs at the point where the photon is located and represents the same constraint that occurs in Special Relativity and gives rise to the existence of a particular constant velocity for all local observers no matter how fast they are moving ©. General Relativity does not undermine the insights available from Lorentz Invariance, it builds on them. Which confirms my point of that the constancy of light speed cannot be "the result of the existence of Lorentz invariance". We can select two observers whose frames are not related by a Lorentz transformation and still the speed of a light signal sent from one to the other is constant and equal to c in general relativity. Your point about there being no such thing as the frame of reference of a photon is well taken and might be replaced by "in the limit as an inertial reference frame (IRF) approaches the velocity of a photon" there is no separation between an emitter and an absorber of electromagnetic radiation in the direction of travel of the IRF in flat space-time. Still a very interesting point given that almost all EM interactions that we observe occur in flat spacetime. Consider also that we can never observe a photon, what we observe is an interaction that can be explained by the concept of the transfer of a quantum of energy. What we consider to be a host of little particles in flight could also be regarded as a resonance between an emitter and an absorber. Imagine the IRF discussed above, the one moving at .9999... c in the direction taken by a photon, it might observe the interaction between the photon emitter and absorber as the direct resonant contact between emitting and absorbing outer shell electrons. (The space between these electrons having been contracted to very, very nearly zero). You have made me amend the overly broad position that I took earlier but the substance of my points still stands: that there is a constant velocity 'c' (amended to a constant velocity for all observers in flat spacetime and for local observers in curved spacetime) and that EM interactions can be regarded, in flat spacetime at least, as direct interactions between emitters and absorbers with no room left for greater or lesser velocities. But "in the limit as an inertial reference frame (IRF) approaches the velocity of a photon" the frame ceases to be localizable and cannot be a reliable reference frame (you cannot define/measure distances in it). Edited May 28, 2012 by juanrga
jozef Posted May 29, 2012 Posted May 29, 2012 You tell me, i have my own theory. Light is energy, energy spread of the environment, does not speed the spread of energy, the environment determines the speed of light, so the speed of light is not constant
mindless Posted May 29, 2012 Posted May 29, 2012 (edited) Which confirms my point of that the constancy of light speed cannot be "the result of the existence of Lorentz invariance". We can select two observers whose frames are not related by a Lorentz transformation and still the speed of a light signal sent from one to the other is constant and equal to c in general relativity. But "in the limit as an inertial reference frame (IRF) approaches the velocity of a photon" the frame ceases to be localizable and cannot be a reliable reference frame (you cannot define/measure distances in it). The speed of light is constant in General relativity in the same way as the speed of a racing car can be constant on a circular race course and this is the result of the light following null geodesics (the path at right angles to null vectors in a 4D manifold). To quote from the "speed of light" FAQ on John Baez's website: "If general relativity is correct, then the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime. The causal structure of the universe is determined by the geometry of "null vectors". Travelling at the speed c means following world-lines tangent to these null vectors. The use of c as a conversion between units of metres and seconds, as in the SI definition of the metre, is fully justified on theoretical grounds as well as practical terms, because c is not merely the speed of light, it is a fundamental feature of the geometry of spacetime." http://math.ucr.edu/...d_of_light.html The velocity of light and the speed of light over an extended interval are not constant in General Relativity but the speed over an infinitessimal interval is constant and found to be so by all observers. The essential feature of General Relativity is that the tangent space to any point in spacetime is a 4 dimensional Minkowski Space ie: locally spacetime is Lorentz Invariant. This can be imagined as there being at any point a local spacetime that is flat and can be described by Special Relativity. As the quote says, the "causal structure of of the universe is determined by the geometry of 'null vectors'" and the null vectors are where, to a very close approximation: 0 = dx^2 + dy^2 + dz^2 - (cdt)^2 As it points out in the chapter on the book on Special Relativity referenced above (http://en.wikibooks....cial_relativity ) Lorentz Invariance means that every observer observes a null vector to be a null vector whatever their own velocity. As a result there is a universal constant 'c' (see the link to the book for the simple maths of the derivation of 'c').'c' is the velocity at which a null vector occurs and this velocity is the same for all observers - a tautology of dynamics and geometry. Whether this constant velocity is "the speed of light in a vacuum" is another problem. Your point that "But 'in the limit as an inertial reference frame (IRF) approaches the velocity of a photon' the frame ceases to be localizable and cannot be a reliable reference frame (you cannot define/measure distances in it)." is the same as my proposal. At very nearly the speed of light an observer might be able to measure a few Planck lengths in a "stationary" frame of reference but at higher velocities there is no meaningful separation. I agree that a photon is not an Inertial Reference Frame but we are discussing the geometry of spacetime, a four dimensional manifold with three dimensions of one sign and one dimension of another sign, this manifold has paths within it, tangents to null vectors, that can be considered as direct connections from one place and time to another. "Direct" in the sense that anything following the path of a photon takes no time along that path and the path has no length. Either this extrapolation of the geometrical properties of spacetime is correct or the photon is "outside" of spacetime in some unknown way. Edited May 29, 2012 by mindless
juanrga Posted May 29, 2012 Posted May 29, 2012 (edited) To quote from the "speed of light" FAQ on John Baez's website: "If general relativity is correct, then the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime. The causal structure of the universe is determined by the geometry of "null vectors". Travelling at the speed c means following world-lines tangent to these null vectors. The use of c as a conversion between units of metres and seconds, as in the SI definition of the metre, is fully justified on theoretical grounds as well as practical terms, because c is not merely the speed of light, it is a fundamental feature of the geometry of spacetime." http://math.ucr.edu/...d_of_light.html The velocity of light and the speed of light over an extended interval are not constant in General Relativity but the speed over an infinitessimal interval is constant and found to be so by all observers. First. In the above quote Baez is writing about inertial frames. This is related to what I said about observers not conected by Lorentzian transformations. Second, in another part of the same link Baez confirms what I said regarding the constancy of the speed of light : a more modern interpretation is that the speed of light is constant in general relativity. Edited May 29, 2012 by juanrga
mindless Posted May 29, 2012 Posted May 29, 2012 (edited) First, the quote from the link addresses both Inertial Frames and spacetime geometry: "the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime." This was why I introduced the quote, because it supports my statement that Lorentz Invariance underlies the constancy of the speed of light. Lorentz Invariance means that null vectors are seen as null by all observers and hence there is a constant speed 'c'. Lorentz invariance is a feature of four dimensional spacetime with a signature of three positive and one negative dimension (or vice versa). It is a geometrical invariance and not dependent upon the presence of an inertial frame of reference. In other words the geometry of spacetime specifies the constancy of the speed of light, as it says in the quote. As for whether an Inertial Frame is required for the speed of light to be constant, which seems to be your contention, I can only point out that the geometry is present whether or not an inertial frame is under consideration and invariant null vectors are a property of the geometry. Second, the full quote from the link is: "A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity." We both agree that the speed of light is constant in General Relativity. The quote above is consistent with what I said above, eg: racing cars etc, the speed of light is constant in General Relativity when taken over a short period. Curvature means that the average speed between two points will not be constant over longer intervals because the light will not go in a straight line. We both know this. I cannot see how this quote resolves your contention that Lorentz Invariance does not underlie the constancy of the speed of light. In each short interval events are governed by the tangent Minkowski Space. The net effect is that along a small element of the path of a light ray the speed is always 'c'. Edited May 29, 2012 by mindless
juanrga Posted May 29, 2012 Posted May 29, 2012 (edited) As for whether an Inertial Frame is required for the speed of light to be constant, which seems to be your contention No. I have just said the contrary. In general relativity, the speed of light is also constant for non-inertial frames. We both agree that the speed of light is constant in General Relativity. The quote above is consistent with what I said above, eg: racing cars etc, the speed of light is constant in General Relativity when taken over a short period. We do not agree because I am saying the contrary than you: the speed of light is always constant in general relativity. The link by Baez says the same that I am saying: the speed of light is constant in general relativity. Edited May 29, 2012 by juanrga
mindless Posted May 30, 2012 Posted May 30, 2012 It only makes sense to talk about the "speed of light in a vacuum" and as a ratio of small or infinitessimal intervals (delta X/delta Y). Given this definition of "speed" I have also been saying that the speed of light is always constant in General Relativity. I have also been saying that the speed of light is constant in general relativity and have described how this might be expected from general or local Lorentz Invariance. Well, its been nice talking with you, I do not really have any more to add.
derek w Posted May 30, 2012 Posted May 30, 2012 (edited) I am no sure I understand the problem,but if the vacuum of space is expanding between the galaxies or galaxy clusters and shrinking into black holes,then anything that travels as a wave e.g. light must be effected by the changing medium through which it propagates?Light is either red shift or blue shift depending on whether the space is expanding away from us or shrinking away from us. Edited May 30, 2012 by derek w
pmb Posted May 30, 2012 Posted May 30, 2012 The speed of light is constant in General relativity in the same way as the speed of a racing car can be constant on a circular race course and this is the result of the light following null geodesics (the path at right angles to null vectors in a 4D manifold). I disagree. The speed of light in a uniformgravitational field was calculated by Einstein back in 1911. He got the result in Eq. (5) where I calculate the speed of light in such a field http://home.comcast.net/~peter.m.brown/gr/c_in_gfield.htm A modern textbook does the same thing. E.g see Gravitation and Spacetime - Second Edition by Ohanian and Ruffini page 197 Eq (47). Baez doesn't make a lot of sense to me. The comment "If general relativity is correct, then the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime. Makes no sense. Why is it a tautology. He doesn't say as far as I can see. Of course he does bury it in math so I might have missed it. It seems to me that he says one thing, i.e. In such a frame, the speed of light can differ from c, basically because of the effect of gravity (spacetime curvature) on clocks and rulers. which is 100% right. He's only talking about the local speed of light then he's being confusing since the variation in the speed of light comes from gravty, which he omits when he says that the speed of light is constant. His most misleading statement is this Finally, we come to the conclusion that the speed of light is not only observed to be constant; in the light of well tested theories of physics, it does not even make any sense to say that it varies. Which is untrue. The Shapiro time delay demonstrates that the speed of light varies with gravitational potential. I see a lot of diversion in his page. That speed of light in a gravitational field iis near trivial to calculate. I do it for both the uniform gravitational field as well as the Schwarzchild spacetime. That can be made very difficult by adding in a lot of differential geometry terminology but there's no real reason for it. Make it simple I say. E.g. instead of saying The essential feature of General Relativity is that the tangent space to any point in spacetime is a 4 dimensional Minkowski Space instead say At every point in spacetime there is a locally inertial frame of reference. Let's recall that a beam of light moving towards the event horizon of a black hole slows down as it gets to the horizon. If it could get there in a finite amount of time then it would stop on the horizon.
juanrga Posted May 30, 2012 Posted May 30, 2012 (edited) I disagree. The speed of light in a uniformgravitational field was calculated by Einstein back in 1911. His computation does not even use general relativity! He assumes a flat background spacetime, which is not the physical spacetime of general relativity. Which is untrue. The Shapiro time delay demonstrates that the speed of light varies with gravitational potential. I see a lot of diversion in his page. That speed of light in a gravitational field iis near trivial to calculate. I do it for both the uniform gravitational field as well as the Schwarzchild spacetime. That can be made very difficult by adding in a lot of differential geometry terminology but there's no real reason for it. Make it simple I say. E.g. instead of saying The essential feature of General Relativity is that the tangent space to any point in spacetime is a 4 dimensional Minkowski Space instead say At every point in spacetime there is a locally inertial frame of reference. Let's recall that a beam of light moving towards the event horizon of a black hole slows down as it gets to the horizon. If it could get there in a finite amount of time then it would stop on the horizon. Baez is right. The speed of light in general relativity is a constant. The shapiro time delay is perfectly compatible with the speed of light being a constant in general relativity. You are confused about modern physics because you rely on historical sources of about 1911 (even before general relativity was formulated!). It is possible to compute the speed of light in general relativity and check that it is a constant that equals c. Edited May 30, 2012 by juanrga
granpa Posted May 31, 2012 Posted May 31, 2012 it is constant 'locally'. http://en.wikipedia.org/wiki/Gravitational_time_dilation The speed of light in a locale is always equal to c according to the observer who is there. The stationary observer's perspective corresponds to the local proper time. Every infinitesimal region of space time may have its own proper time that corresponds to the gravitational time dilation there, where electromagnetic radiation and matter may be equally affected, since they are made of the same essence (as shown in many tests involving the famous equation E=mc2). Such regions are significant whether or not they are occupied by an observer. A time delay is measured for signals that bend near the Sun, headed towards Venus, and bounce back to Earth along a more or less similar path. There is no violation of the speed of light in this sense, as long as an observer is forced to observe only the photons which intercept the observing faculties and not the ones that go passing by in the depths of more (or even less) gravitational time dilation.
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