Total Posts:16Showing Posts:116
Why is light not subject to relativity?
Posts: 2,633
Add as Friend Challenge to a Debate Send a Message 
4/23/2014 8:17:22 PM Posted: 2 years ago This is something I can't seem to grasp my head around. I understand how all motion is relative, yet it seems so odd that light is not subject to the relativity of motion.
Are there any explanations behind this that anyone could offer to me? Thanks. 
Posts: 3,740
Add as Friend Challenge to a Debate Send a Message 
4/23/2014 9:30:38 PM Posted: 2 years ago At 4/23/2014 8:17:22 PM, PeacefulChaos wrote: What do you mean, why do you think light isn't subject to the relativity of motion? Do you mean because it's speed is constant, no matter what your reference frame is? If so, that doesn't mean light isn't subject to relativity, light has zero rest mass, and according to relativity theory something with zero rest mass will always travel at the speed of light, and don't experience time. It was the constant speed of light from which Einstein developed the Special Theory, you could say the theory of relativity is subject to light. "It is one of the commonest of mistakes to consider that the limit of our power of perception is also the limit of all there is to perceive." " C. W. Leadbeater 
Posts: 2,633
Add as Friend Challenge to a Debate Send a Message 
4/23/2014 9:37:06 PM Posted: 2 years ago At 4/23/2014 9:30:38 PM, Sidewalker wrote:At 4/23/2014 8:17:22 PM, PeacefulChaos wrote: Our physics teacher explained to us in class the following ideal scenarios. I don't have much time to spare, so I'll just quickly summarize what he told us: Suppose person "A" throws a ball to person "B" at 20 mph while person "B" is traveling at 40 mph away from the ball. From the view of person "A," the ball is traveling at 40 mph; however, from the view of person "B," the ball is traveling at 20 mph. In another scenario, however, we replace the ball with light. If person "A" directs a light source to person "B," and person "B" travels away from the light source at half the speed of light (c/2), the light will appear to still move at the speed of light instead of at the speed of c/2. 
Posts: 1,022
Add as Friend Challenge to a Debate Send a Message 
4/23/2014 10:09:27 PM Posted: 2 years ago At 4/23/2014 9:37:06 PM, PeacefulChaos wrote:At 4/23/2014 9:30:38 PM, Sidewalker wrote:At 4/23/2014 8:17:22 PM, PeacefulChaos wrote: This is where length contraction and time dilation come in; observer A and B will both see light travelling at the same speed, but at relativistic speeds depending on their frame of reference, they will observe time and distance differently. At slower relative velocities, time dilation and length contraction are negligible. * Fixed some numbers for you 
Posts: 23
Add as Friend Challenge to a Debate Send a Message 
4/23/2014 10:12:17 PM Posted: 2 years ago This will answer your question thoroughly: http://math.ucr.edu...

Posts: 3,740
Add as Friend Challenge to a Debate Send a Message 
4/24/2014 7:34:31 AM Posted: 2 years ago At 4/23/2014 9:37:06 PM, PeacefulChaos wrote:At 4/23/2014 9:30:38 PM, Sidewalker wrote:At 4/23/2014 8:17:22 PM, PeacefulChaos wrote: OK, I thought that's what you were referring to, the constant speed of light is a physical fact and it was the basis of relativity theory in the first place. Einstein took the fact that it had always been measured to be constant independent of reference frame and made it a fact of nature, he worked out the classical physics implications of that and what resulted was the Special Theory of Relativity. Light is inexplicable, you just can't get your head around it, unless you are a couple orders of magnitude smarter than everyone else, your head will explode if you try to. Einstein was that smart and even then, it made his hair explode. The fact is, light is mysterious and not amenable to logical understanding. According to relativity theory, here are some more facts about light for consideration. Light exhibits mutually exclusive characteristics, it contradicts itself in manifesting as either a wave or a particle. Temporally speaking, a wave has a periodicity, and spatially, it is spread out in space, a particle cannot have a periodicity, nor can it be spread out in space. A particle has a discrete location in time and space, a wave does not, a wave is spread out over a larger region of space and time. A wave has amplitude and a frequency, it exhibits the phenomena of diffraction and diffusion, a particle does not. Light exhibits both of these mutually exclusive properties depending upon how you experimentally examine it, which is a direct contradiction. Light has no mass, which is why it can move at the speed of light, with no rest mass it must move at the speed of light.. It has no dimensional existence in space, according to relativity, an object gets shorter in the direction of travel until, at the speed of light, it's length equals zero, hence, it doesn't have a spatial coordinate. From it's own frame of reference, it is stationary, it doesn't move, it can't move because there is no time for it to move in. It doesn't exist in space or in time. From it's reference frame, it is everywhere all at once. Much about light is inconceivable, nothing else behaves like it, and consequently, it violates any normal definition of logic and rationality. You simply can't get your head around it. "It is one of the commonest of mistakes to consider that the limit of our power of perception is also the limit of all there is to perceive." " C. W. Leadbeater 
Posts: 2,633
Add as Friend Challenge to a Debate Send a Message 
4/24/2014 9:24:04 PM Posted: 2 years ago To everyone who responded:
Thank you very much. I don't have the time to check all the sources you linked to me or respond quite yet, but I will when I get the time. Once again, thanks. 
Posts: 2,243
Add as Friend Challenge to a Debate Send a Message 
4/24/2014 10:26:32 PM Posted: 2 years ago I'll be hanging out in the science forums for a while. I might as well learn a thing or two.
Leader of the DDO Revolution Party 
Posts: 587
Add as Friend Challenge to a Debate Send a Message 
4/26/2014 12:05:48 AM Posted: 2 years ago To say that the speed of light is constant is to say that it's a constant for everyone, whatever our states of motion, unlike the speed of a baseball, which is not a constant.
But, in one sense, photons are not so different from baseballs: a baseball flies through the air with a measured speed that is related to how the measuring observer is moving, and this means its energy will be different for different viewers; likewise, light waves will have different wavelengths for differently moving observers, and this means that the energy of the light wave will be measured differently by different viewers, just like the baseball. For a baseball, energy is related to speed, but for a light wave, energy is related to wavelength. That's the difference. "The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." Sidewalker 
Posts: 2,633
Add as Friend Challenge to a Debate Send a Message 
4/27/2014 8:54:47 PM Posted: 2 years ago Your certainly right that this is an odd concept to get my head around. The most I've understood is that it's not subject to our frame of reference due to the fact that:
1. It has no mass. 2. It can behave as both a particle and a wave. I have another question, though. Why can't anything with mass accelerate to the point that it surpasses the speed of light? I know the answer to this is that doing so would require infinite energy, but what is the reasoning behind this? More specifically, what mathematical equation proves this? 
Posts: 587
Add as Friend Challenge to a Debate Send a Message 
4/29/2014 3:58:49 AM Posted: 2 years ago At 4/27/2014 8:54:47 PM, PeacefulChaos wrote: Here it is: E=L*v*m(c^2) E is the kinetic energy of an object with mass m and velocity v, while c^2 is good old speed of light squared, and L is a factor called the "Lorentz factor", and it is the important part. By important, I mean it answers your question. You see, L^2 = (c^2)/[(cv)*(c+v)], which means that as v approaches the speed of light, the denominator of L approaches zero with c^2 in the numerator. That is, when v>c, L>infinity, and therefore so does E. I'm sorry for the ugly typesetting. A little nonmathematical reasoning: For any object with inertia ("ordinary" stuff with mass), there is some reference frame in which it is at rest. There is no such reference frame for light, because it has no inertia, but light is still in everyone's reference frames. What this means is that no one can catch up to a light wave, even though it might be right in front of us, moving at a finite speed. When we write down the equations, this fact pops up as "L>inf as v>c ". "The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." Sidewalker 
Posts: 587
Add as Friend Challenge to a Debate Send a Message 
4/29/2014 4:02:46 AM Posted: 2 years ago At 4/27/2014 8:54:47 PM, PeacefulChaos wrote: Also, this fact does not come into play within special relativity. This is an extra fact that comes from the principles of quantum mechanics. "The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." Sidewalker 
Posts: 2,633
Add as Friend Challenge to a Debate Send a Message 
4/29/2014 5:17:30 PM Posted: 2 years ago At 4/29/2014 3:58:49 AM, Poetaster wrote: I'll admit, I'm rather confused. I certainly understand the equation, but the problem is that I've seen so many variations of the equation. E.g. E = mc^2 (this is for motionless particles) E^2 = (mc^2)^2 + (pc)^2 (this is for particles with motion ... which I just found today) E = (mc^2)/square root ((1  v^2)/(c^2)) (this seems to allow the velocity of a particle to reach the speed of light, which perplexes me) Here is a link that gives the last one in a better format: http://www.emc2explained.info... It's the last picture. I'm assuming that, mathematically, it all works out, but it's quite odd to have so many variations of the same equation.
Thank you. This makes sense to me. Also, using the other equation I found from a video explaining this to me (you may have already seen this before), it also has the same conclusion along with yet another equation: E^2 = (mc^2)^2 + (pc)^2 V = c(pc/E) If the particle is massless, then: E = pc So for a massless particle, we get: V = c But if the particle has mass, then E does not = pc, and the two values can only get to become very close to one another. So we're left with something like this: V = c(0.999999999) And the velocity cannot reach the speed of light if it has mass.
Thanks again for explaining everything to me. 
Posts: 587
Add as Friend Challenge to a Debate Send a Message 
4/29/2014 6:47:48 PM Posted: 2 years ago At 4/29/2014 5:17:30 PM, PeacefulChaos wrote:At 4/29/2014 3:58:49 AM, Poetaster wrote: In this last equation, letting v>c causes E to diverge to infinity. The term in the denominator approaches zero because the ratio v^2/c^2 approaches 1. Thus, a massive object cannot reach c. Also, L is meant to be a function of v (it looks like multiplication the way I wrote it, but that's a typing mistake of mine). I'm assuming that, mathematically, it all works out, but it's quite odd to have so many variations of the same equation. Why assume? I'm confident that you could work it out yourself; it's all just algebraic rearrangement. Nothing forbidding. Here, try looking at this derivation of your second equation from the third one: http://en.wikipedia.org... , in the section "Origins of the Equation" Just copy down the steps on paper, following the algebra through. It doesn't sound substantial, but it helps clear your mind of wondering how it "works out" and instead pay attention to what the equations mean (that's the hardest part!) Of course, the first equation is obtained from the second by setting p=0, and from the third by setting v=0 (which is equivalent). If the particle is massless, then: This ties in with what I mentioned earlier: light has momentum p=hk, where k is the number of wave crests in a meter, and h is a constant. Thus, the closer together the crests of the light wave are, the more energy it has. This means it's "kinetic energy" (E=pc) is related to its wavelength instead of its speed. Thanks again for explaining everything to me. Sure thing. "The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." Sidewalker 
Posts: 2,633
Add as Friend Challenge to a Debate Send a Message 
4/29/2014 7:19:01 PM Posted: 2 years ago I'll definitely try to work it out algebraically later on.
But I have another quick question. Can the Lorentz factor be equal to zero? 
Posts: 587
Add as Friend Challenge to a Debate Send a Message 
4/29/2014 9:34:47 PM Posted: 2 years ago At 4/29/2014 7:19:01 PM, PeacefulChaos wrote: The Lorentz factor, (squared, because it's easier to type) is: L^2 = c^2/[(cv)*(c+v)]. I mean, just by looking at it, we can see that it can't be zero for v<inf, much less for v<c. On the other hand, the reciprocal of L can have a limiting value of zero. "The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." Sidewalker 