ABCD1234pop Posted January 20, 2012 Posted January 20, 2012 300,000Km/s ? Why isn't it some other number?
Tres Juicy Posted January 20, 2012 Posted January 20, 2012 300,000Km/s ? Why isn't it some other number? It is what it is mate
ajb Posted January 20, 2012 Posted January 20, 2012 (edited) We are free to use other systems of units. It is rather a historical accident as opposed to design that the speed of light is exactly 299,792,458 meters per second. The question you really want to ask is why is the speed of light the same in any inertial frame of reference? That is independent of the motion of the source. Edited January 20, 2012 by ajb
Tres Juicy Posted January 20, 2012 Posted January 20, 2012 We are free to use other systems of units. It is rather a historical accident as opposed to design that the speed of light is exactly 299,792,458 meters per second. The question you really want to ask is why is the speed of light the same in any inertial frame of reference? That is independent of the motion of the source. This is a much better question which deserves its own thread really
michel123456 Posted January 20, 2012 Posted January 20, 2012 (...) exactly 299,792,458 meters per second.(...) That is not what I expect from the meaning of the word "exactly". I thought the OP question was: why this number and not another? Why not 537,163,921 m/s or 56,842,489,667,535,578,975,422 m/s?
Appolinaria Posted January 20, 2012 Posted January 20, 2012 It is rather a historical accident as opposed to design that the speed of light is exactly 299,792,458 meters per second.
ydoaPs Posted January 20, 2012 Posted January 20, 2012 300,000Km/s ? Why isn't it some other number? We start with what we call Maxwell equations. These are the equations that describe how electric fields and magnetic fields interact. 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]\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. Edited for punctuation? I accidentally hit "Add Reply" instead of "Preview Post". 1
mississippichem Posted January 20, 2012 Posted January 20, 2012 (edited) Ah. I love the good ole curl of the curl identity. All the dels brighten my day. It has a nice ring to it. "the curl of the curl is the gradient of the divergence less the laplacian" Nice derivation ydoaps. Well said. Edited January 20, 2012 by mississippichem 1
michel123456 Posted January 20, 2012 Posted January 20, 2012 We start with what we call Maxwell equations. These are the equations that describe how electric fields and magnetic fields interact. 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]\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. Since , it is circular reasoning.
ydoaPs Posted January 20, 2012 Posted January 20, 2012 Since , it is circular reasoning. IIRC, this derivation was first done before we knew light was an EM wave.
D H Posted January 21, 2012 Posted January 21, 2012 Since , it is circular reasoning. That is an after the fact definition, just as defining the speed of light to be exactly 299,792,458 meters per second was an after the fact definition. The last meeting of the General Conference on Weights and Measures (Conférence Générale des Poids et Mesures, or CGPM) proposed to take this to a new level. It proposed making each of the Planck constant h, the elementary charge e, the Boltzmann constant k, the Avogadro constant NA, and the luminous efficacy Kcd a defined constant. There's still a lot of work to be done on this. It won't happen for at least four years (the next CGPM meeting). The proposed resolution just set the framework. There are still a lot of TBDs ("To be determined") in the resolution. 1
michel123456 Posted January 21, 2012 Posted January 21, 2012 I think the clear answer to the OP is "no one knows". However, one can imagine how weird world would that be a universe with Speed Of Light 1m/s for example, or even less. IOW there must be some relation between our dimension (and all dimensions from electrons to galaxies) and the value of SOL. Some expect this relation to be geometrical, but no one really knows yet.
granpa Posted January 22, 2012 Posted January 22, 2012 the speed of light is [math] \frac{1}{\sqrt{{\mu_0}{\epsilon_0}}} [/math] [math] \mu_0 [/math] and [math] \epsilon_0 [/math] represent the 'springiness' of space. the wavelength of light needs to be about the size of a cell or eyes wont work. the freqency of light is determined by teh frequency of the atoms which is determined by h (plancks constant) wavelength of light = speed of light / frequency of light so if the speed of light were different then the wavelenth of light would be the wrong size of our eyes.
PeterJ Posted January 23, 2012 Posted January 23, 2012 Much confused about this. Am I right to suppose that for a photon time stands still, and if it went any faster time would have to go backwards?
granpa Posted January 23, 2012 Posted January 23, 2012 time dilation doesnt apply to light. only to things that move slower than the speed of light.
ajb Posted January 23, 2012 Posted January 23, 2012 Am I right to suppose that for a photon time stands still, You cannot really think like this. There is no notion of an inertial reference from of a photon. Thus one cannot talk about "from the photon's point of view". and if it went any faster time would have to go backwards? If a particle travels faster than the speed of light as measured in some inertial frame then it is always possible to find an inertial frame in which the particle is travelling backwards in time.
PeterJ Posted January 23, 2012 Posted January 23, 2012 You cannot really think like this. There is no notion of an inertial reference from of a photon. Thus one cannot talk about "from the photon's point of view". Um. What does 'no notion' mean here? Do you mean the idea makes no sense? Didn't Einstein do quite well by giving the photon a pov? Not arguing, just clarifying. If I were riding on a photon then wouldn't time stand still for me? If a particle travels faster than the speed of light as measured in some inertial frame then it is always possible to find an inertial frame in which the particle is travelling backwards in time. Why would it not be the very same frame in which the particle is measured as travelling faster than c?
D H Posted January 23, 2012 Posted January 23, 2012 Um. What does 'no notion' mean here? Do you mean the idea makes no sense? It makes no sense. Suppose you were riding along a beam of light. From your perspective, the photons that comprise that beam of light are not moving. There's a big problem here. One of the major tenets of relativity theory is that the speed of light is the same to all observers. So which is it: Are the photons moving at rest (v=0) or are they moving moving at c (299,792,458 meters per second)? This is a contradiction. The concept makes no sense. Didn't Einstein do quite well by giving the photon a pov? Just the opposite. He tried imagining riding along a beam of light at age 16. He realized then that big problems arose from this thought experiment. It took him a a decade to resolve those issues. By 1905 he had it figured out. He realized that the concept of riding alongside a beam of light makes no sense. No matter how much faster you try to go, light will always appear to travel at the same speed. As far as superluminal particles are concerned, it's important to keep in mind that most physicists don't think they exist.
PeterJ Posted January 23, 2012 Posted January 23, 2012 It makes no sense. Suppose you were riding along a beam of light. From your perspective, the photons that comprise that beam of light are not moving. There's a big problem here. One of the major tenets of relativity theory is that the speed of light is the same to all observers. So which is it: Are the photons moving at rest (v=0) or are they moving moving at c (299,792,458 meters per second)? This is a contradiction. The concept makes no sense. I would say the photons are at rest in their frame. Is this a problem? There would be no change and no possible observation of other photons or indeed anything at all. Is this not a possible solution? Just the opposite. He tried imagining riding along a beam of light at age 16. He realized then that big problems arose from this thought experiment. It took him a a decade to resolve those issues. By 1905 he had it figured out. He realized that the concept of riding alongside a beam of light makes no sense. No matter how much faster you try to go, light will always appear to travel at the same speed. But couldn't we just say that while this is true for all observations, the case of something traveling at c is a special case where observation becomes impossible? As far as superluminal particles are concerned, it's important to keep in mind that most physicists don't think they exist. What about the absorber theory of time? Does this not imply ftl particles? Or is this not an equivalent interpretation?
ajb Posted January 23, 2012 Posted January 23, 2012 (edited) I would say the photons are at rest in their frame. Is this a problem? You simply cannot find an inertial frame such that this is true. Edited January 23, 2012 by ajb
swansont Posted January 23, 2012 Posted January 23, 2012 I would say the photons are at rest in their frame. Is this a problem? There would be no change and no possible observation of other photons or indeed anything at all. Is this not a possible solution? We don't have a model for a photon in a rest frame, and no real way to test one if we did. So it's a big problem.
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