abskebabs Posted May 15, 2006 Share Posted May 15, 2006 When recently thinking about this, I suddenly realized why I am not more bothered about the statement in relativity that the speed of light is the same in all frames of reference. Naturally it seems quite counter intuitive to me now, just like a lot of quantum theory. I admit I do not know either to a great degree, but I know more about quantum theory having done some A level coursework at school on early quantum theory(e.g. work done by Planck, Einstein, de Broglie, Bohr's model of atom etc). To get to the point, I imagined I was in a vehicle moving at a very high speed, let's say about a 1% of the speed of light(if we ignore relativistic effects on mass). If I shined a torch while on the vehicle, from my frame of reference it would seem like the light would be travelling at (3*10^8)m/s but the light would also seem to be travelling at the same speed to an observer who was outside the vehicle and watching both me and the vehicle. Naturally, this would not make sense intuitively, for example; if I threw a tennis ball out of a car while travelling in it, it would sem to me to be travelling at a certain speed, but to an external observer, its immediate speed would be the sum of my speed as well as the speed I observe it to be travelling. The fact that light does not follow this rule shows it is something that I cannot apply such simple analogies to and should be considered very differently, and I guess this is defined well in Relativity. In fact to be honest I was quite shocked and amazed when I finally realised this, as well as being a little annoyed that it took me so long to realise the implications of such a profound truth. Have I made any mistakes when looking at this, or is all this correct scientifically? Also would anyone else like to add other things that can be observed from relativity that are also quite counter intuitive when you think about them, and why? Thanks a lot. Link to comment Share on other sites More sharing options...
swansont Posted May 15, 2006 Share Posted May 15, 2006 Velocities in general do not add linearly. Two objects travelling toward each other at 0.5c (according to some other observer) do not see each other as travelling at c. They will see the other as travelling at 0.8c. The transformation between reference frames is Lorentzian (nonlinear), not Galilean (linear), though it looks Galilean at low speed. Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 15, 2006 Share Posted May 15, 2006 its immediate speed would be the sum of my speed as well as the speed I observe it to be travelling. The fact that light does not follow this rule shows it is something that I cannot apply such simple analogies to FYI' date=' in relativity light and matter do follow the same velocity addition rule. But this rule is not the one you cite, as has been noted already. The rule is given by the following equation. [math']u^{\prime}=\frac{u+v}{1+\frac{uv}{c^2}}[/math] Note that when [imath]uv[/imath] is much less than [imath]c^2[/imath] the term [imath]\frac{uv}{c^2}[/imath] is much less than 1, and the equation is approximately given by the following. [math]u^{\prime}=u+v[/math] This is the simple rule you stated. Link to comment Share on other sites More sharing options...
abskebabs Posted May 15, 2006 Author Share Posted May 15, 2006 FYI' date=' in relativity light and matter do follow the same velocity addition rule. But this rule is not the one you cite, as has been noted already. The rule is given by the following equation. [math']u^{\prime}=\frac{u+v}{1+\frac{uv}{c^2}}[/math] Note that when [imath]uv[/imath] is much less than [imath]c^2[/imath] the term [imath]\frac{uv}{c^2}[/imath] is much less than 1, and the equation is approximately given by the following. [math]u^{\prime}=u+v[/math] This is the simple rule you stated. Ah.. I see. Does this mean the observed speed of light to me if I am in a vehicle travelling at let's say 0.1c and shine a torch, and the speed of the light observed by someone else from a static frame of reference(relative to me) will be different? If so how? Thanks for your help, I appreciate it. Link to comment Share on other sites More sharing options...
Janus Posted May 15, 2006 Share Posted May 15, 2006 Ah.. I see. Does this mean the observed speed of light to me if I am in a vehicle travelling at let's say 0.1c and shine a torch, and the speed of the light observed by someone else from a static frame of reference(relative to me) will be different? If so how? Thanks for your help, I appreciate it. To answer that, just make u = .1c and v = c in the equation given, like thus: [math]u'= \frac{.1c+c}{1+\frac{c(.1c)}{c^2}}[/math] [math]u'= \frac{1.1c}{1+\frac{.1c^2}{c^2}}[/math] [math]u'= \frac{1.1c}{1+.1}[/math] [math]u'= \frac{1.1c}{1.1}[/math] [math]u'= c[/math] Thus for you, the light travels at c, and for the "static" observer, the light travels at c. Link to comment Share on other sites More sharing options...
abskebabs Posted May 15, 2006 Author Share Posted May 15, 2006 Thus for you, the light travels at c, and for the "static" observer, the light travels at c. This is exactly what I have found amazing about this(perhaps I'm just stupid!)! It would seem to travel at a certain speed away from me, and even though I would already be m,oving; it would travel at this same speed to someone else who wasn't "moving." I think I just need to get used to this idea and read about relativity... Link to comment Share on other sites More sharing options...
GutZ Posted May 16, 2006 Share Posted May 16, 2006 This confuses the hell out of me. It's just the observed portion that changes right? The speed of light is the same for everyone right? Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 16, 2006 Share Posted May 16, 2006 This is exactly what I have found amazing about this(perhaps I'm just stupid!)! Not necessarily. Very smart physicists thought that the old velocity addition formula was correct for centuries. It's just the observed portion that changes right? What is "the observed portion"? The speed of light is the same for everyone right? It's the same for every inertial frame. Link to comment Share on other sites More sharing options...
Ragib Posted May 19, 2006 Share Posted May 19, 2006 Yes i understand your difficulties, it is quite profound and counter intuitive. Note though, 2 people will see the light at the same speed, but different wavelengths, due to the doppler shift. And the person still will see the moving one going through time extremely slowly. Link to comment Share on other sites More sharing options...
macrocosmos Posted June 7, 2006 Share Posted June 7, 2006 If I say that at C2 mass exhibits the same characteristics as energy, what would lead you to assume that FTL travel is unlikely? Link to comment Share on other sites More sharing options...
Ragib Posted June 8, 2006 Share Posted June 8, 2006 C2? I think you mean C squared. Well, having the same characteristics as energy, I would assume FTL travel is unlikely because Pure energy travels at the speed of light. And to accelerate to a faster speed would require you to be able to accelerate energy. I dont think theres any way to do that. Link to comment Share on other sites More sharing options...
JaKiri Posted June 8, 2006 Share Posted June 8, 2006 If I say that at C2 mass exhibits the same characteristics as energy You'd be wrong. You're misinterpreting E=mc^2. c^2 is a constant of proportionality that arises from the mathematical axiom that the speed of light is constant for all observers. All it says is that if you turned a given mass m into energy, you'd have m * 299792458^2 joules. (The more accurate version is E^2 = m^2c^4 + p^2c^2, which also leads to a fundamental property of matter which is E-P invariance (rest mass is constant, therefore E^2 - p^2c^2 = constant).) what would lead you to assume that FTL travel is unlikely? E=mc^2 has little to do with FTL, or just plain lightspeed travel, being impossible. It's the gamma factor, 1/SQRT(1-v^2/c^2). For example, the relativistic definition of momentum is gamma mv. As v -> c, gamma -> infinity, therefore a massive object moving at the speed of light would have infinite momentum, and would therefore require an infinite impulse to reach said speed. Beyond the speed of light, the gamma factor becomes imaginary, but that's another issue entirely. Link to comment Share on other sites More sharing options...
swansont Posted June 8, 2006 Share Posted June 8, 2006 Pure energy travels at the speed of light. What is "pure energy" ? (Or "impure energy" for that matter) Link to comment Share on other sites More sharing options...
Ragib Posted June 8, 2006 Share Posted June 8, 2006 O sorry, i guess i ddint use the correct words. When i say pure energy, i mean, electromagnetic radiation..What i ment for impure would be say..kinetic energy or heat energy. Link to comment Share on other sites More sharing options...
abskebabs Posted June 8, 2006 Author Share Posted June 8, 2006 It seems while revising, I have come up with another question related to this topic. Why is the derivation of the doppler effect for EM waves different to that of sound and water waves as an example, and consequently, why is the relationship [math]\frac{\bigtriangleup{f}}{f}= \frac{v}{c}[/math] only approximate when dealing with EM waves(when v<<<c), when it is exact(as far as I know) for sound and water waves. Also on a side note, what is the sign for an approximate equals in latex? I would have put it in the above equation, but as you can see I don't know how. Link to comment Share on other sites More sharing options...
abskebabs Posted June 10, 2006 Author Share Posted June 10, 2006 Does anyone have an answer to my question? I'm sorry if it seemed like a statement as I made a grammatical mistake and left no question mark. Link to comment Share on other sites More sharing options...
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