If you use natural units such that c = 1, then space and time are measured in effectively the same units. We can get the same effect by using human units and multiplying and measurement of time by c. And so the Lorentz transform is: \(\begin{pmatrix} \Delta x' \\ c \Delta t' \end{pmatrix} \begin{pmatrix} \cosh ( \tanh ^{-1} ( \frac{v}{c} ) ) & \sinh ( \tanh ^{-1} ( \frac{v}{c} ) ) \\ \sinh ( \tanh ^ {-1} ( \frac{v}{c} ) ) & \cosh ( \tanh ^{-1} ( \frac{v}{c} ) ) \end{pmatrix} \begin{pmatrix} \Delta x \\ c \Delta t \end{pmatrix}\) which is a hyperbolic rotation of space-time. This is the 1908 revelation of Minkowski -- that the Lorentz transforms are a geometric transformation of the coordinates of a unified space-time just like ordinary rotation is a transformation of of the coordinates of the Euclidean plane. Space and time are so much like each other, that just a change of perspective mixes them. The ordinary rotation looks like this: \(\begin{pmatrix} \Delta x' \\ \Delta y' \end{pmatrix} \begin{pmatrix} \cos ( \tan ^{-1} ( m ) ) & -\sin ( \tan ^{-1} ( m ) ) \\ \sin ( \tan ^ {-1} ( m ) ) & \cos ( \tan ^{-1} ( m ) ) \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \end{pmatrix}\) ( m is the slope of a line ).
Bowser: Then you're best of ignoring everything posted by or in relation to Motor Daddy, who isn't doing relativity at all. Yes. Both need to change. To put it in the simplest way, the speed of something is the distance it travels divided by the time taken. If you change the time taken, then the only way to hold the speed constant is to also change the distance. Not exactly. The thing is, length contraction and time dilation work both ways. If I watch you flying past me, then I see your clocks running slower than mine and I also see you as shorter than normal. But when you look back at me, you see my clocks running slower than yours and me as being shorter than usual. In other words, we don't agree on whose time changed and which lengths shrank.
I've always heard that if a traveler left earth at the speed of light and came back, time will have passed slower for him than that on earth? So, if I left the earth at the speed of light and traveled the stars, the clocks on earth would appear to move slower than mine? What you're telling me is a contradiction from what I've read elsewhere, or maybe I've misunderstood what I've read elsewhere. The plot thickens.
That's the so-called Twin Paradox, which isn't really a paradox at all. But that's more complicated than simple time dilation, because the travelling twin must go away, turn around, then come back. During the constant-velocity parts of the trip, the travelling twin says the Earth twin is aging slower than him, and the Earth twin says the travelling twin is aging slower. What makes the final difference is the turning around part of the trip, during which BOTH twins agree that the travelling twin ages slower.
ooh...oh....I think I have identified the difference of opinion we have. First I have to thank James R It's the best explanation that I ever heard. The difference comes in how we understand the property of light. Wiki:Special relativity,Postulates To me this means, in this regard, as sound. -Relative to the ground, speed of a bullet, fired from a plane is obtained by adding the velocity vector between plane and bullet and the velocity vector between plane and ground. This does not apply to sound or light. -Relative speed between bullet and plane not dependent on the plane speed. This does not apply to sound or light. What is your opinion?
Speed is just someones description of something moving a certain distance in a certain amount of time. For physics to be consistent, the same formula for "If I see it moving at speed u and I see someone moving at speed v, what speed would that someone see it moving at?" has to apply to light, sound and bullets. w = f(u,v) Clearly f(v,v) = 0 since a bullet fired at the same speed as the train would seem to stand still, and f(u,0) = u since a bullet fired at speed u should still look like speed u if seen from the train when the train is stopped. Moreover, f(0,v) = -v means that if I see the train moving at speed v, the train sees me moving at speed -v. Clearly f(u,v) = u - v will satisfy these conditions, but will not account for the consistency of light speed by various observers. Another alternative is that the family of functions \(f_K (u,v) = \frac{u - v}{1 - K u v}\) will work. And since the consistency of the speed of light means \(c = f_K (c,v) = \frac{c - v}{1 - K c v} = c \frac{1 - \frac{v}{c}}{1 - K c v}\) which means \(1 - \frac{v}{c} = 1 - K c v\) which means \(K = \frac{1}{c^2}\).
RPenner, This is good Stuff!!Please Register or Log in to view the hidden image! Now let's see if you have any understanding of that equation above? An Aether experiment was carried out by Hippolyte Fizeau in 1851, to measure the relative speeds of light in moving water. Albert Einstein later pointed out the importance of the experiment for special relativity. Since the time, that Einstein pointed out to us the importance of this new velocity equation; we comfortably use the above equation, that you so eloquently described. http://en.wikipedia.org/wiki/Fizeau_experiment Einstein got rid of the "Aether" Frame? You seem to be using the above equation as a frame of reference. Can you explain that frame?:shrug:
Um, who are you quoting? I have only used relative speed to actual objects (bullets, sound, light), not hypothetical yet unevidenced objects without even a physical model. Aether models were seriously plagued by a multiplication of contradictory hypothetical properties before Michelson-Morley, FitzGerald, Lorentz, Einstein and Minkowski skewered the whole idea as unnecessary nonsense predicated on dodgy mechanical analogies. For just one example relevant to the Wikipedia page you introduced, the 1914 dispersion correction demands a different aether to be dragged a different amount for every frequency of light. Fooking insane that is. The modern view of light (accurately tested in the parts-per-trillion range) is that all matter and energy is transmitted by particles with space-like and time-like periodicities much along Newton's conception of the corpuscles of light. Introducing any sort of non-preferred preferred frame (as any Aether frame must be to be compatible with experiment) is ludicrous at this point and anti-scientific. The only need to introduce the topic is to render 19th century electromagnetism papers comprehensible. Aether is a relic of interest to science historians, not physicists. So since no one on this entire thread introduced the concept or evidence for a luminiferous aether, it seems you are catering to your own pseudo-scientific persuations rather than participating in the thread. This will just cause your reputation with people who actually do physics on this forum to sink even lower than set by your contemptible performance self-publishing and debating AlphaNumeric. http://sciforums.com/showthread.php?t=107547
Also, a year ago, on the same page as my first link from post #97 on this thread, I linked to an even older post from 2008 where I demonstrated that I was well-acquainted with H. Fizeau's experiment. http://www.sciforums.com/showthread.php?p=2039656#post2039656 Even Motor Daddy should be familiar with H. Fizeau's experiment since Einstein devotes chapter 13 of his 1920 pop physics book to it. http://www.bartleby.com/173/13.html
This was my point! You are just writing down an equation that fits situation or question; without understanding!Please Register or Log in to view the hidden image! This is correct. There are lots of different Aether Models out there. Which one is the correct one??? The concept of the "Aether" being dragged by objects is a failed model! That is one (1) concept of a particular aether model that is not a correct; nor or good model. We don't drag the Aether. The Aether drags us! However, the dispersion model listed on that Wikipedia page that you call "Fooking Insane" is an actual measurable effect; and was confirmed by the Dutch physicist Peter Zeeman in 1914. And this brings me back to my original question to you! In your equation that you wrote down, you accept that it works for relative motion and velocity. That equation was originally derived by considering light moving through a medium. Yet, you claim that you wrote the equation down "without even a physical model" for consideration. Why?? You wrote \(f_K (u,v) = \frac{u - v}{1 - K u v}\) Index of Refraction Please Register or Log in to view the hidden image! Thus, "Light" propagating through different materials changes speed based on the density of the material given by the Index of Refraction (n). For example, the refractive index of a water medium is (n = 1.33), meaning that in a vacuum, light travels 1.33 times as fast as it does in water. Velocity of Light in Medium \(u = \frac{c_{Light}}{n}\) "Relative Velocity" of Light relative to fixed or stationary External Observer \(W = \frac{\frac{c_{Light}}{n} - v}{1 - \frac{\frac{c_{Light}}{n} {v}}{c^2_{Light}}} = \frac{\frac{c_{Light}}{n} - v}{1 - \frac{v}{{n} c_{Light}}\) 1) Velocity of Light in Vacuum is Independent of Medium or Observer -> \(c_{Light}\) 2) Velocity of Light in Stationary or moving Medium -> \(u = \frac{c_{Light}}{n}\) in Vacuum (n = 1) 3) Velocity of Moving Medium -> \(v_{Fluid Velocity} = -v\) 4) "Relative Velocity" of Light Relative to Stationary Embankment or External Observer -> \(W = f_K (u,v) = \frac{u - v}{1 - \frac{u v}{c^2_{Light}}\) This is the understanding that I was looking for. 1) The Relative Velocity is a Unique Frame of Reference. 2) In the above equation the index of refraction is equal to one (n = 1) in a medium where the average speed of the medium is equal to the speed of light. This means that when the index of refraction is equal to one (n =1) there is no relative velocity between the speed of light and the average speed of the medium. However there is relative velocity between the motion of the fluid (v = -v) and the speed of light isotropy that an external observer and an observer in the proper center of mass frames measures; using the following equation. \(W = \frac{c_{Light} - v}{1 - \frac{v}{c_{Light}} \) I agree mostly; and I would only slightly disagree, and claim that the modern view of light is that "Light" has both a particle and wave nature. There exist a "Wave–particle duality" to the nature of light. http://en.wikipedia.org/wiki/Light#Wave–particle duality The modern theory that explains the nature of light includes the notion of wave–particle duality, described by Albert Einstein in the early 1900s, based on his study of the photoelectric effect and Planck's results. Einstein asserted that the energy of a photon is proportional to its frequency. More generally, the theory states that everything has both a particle nature and a wave nature, and various experiments can be done to bring out one or the other. In my humble opinion, the Aether Frame is not a preferred or a "special" frame, it is a frame just like any other frame of reference. The Aether Frame if it exists should behave like a gas, and would exert a force on matter just like a force is exerted on matter as it moves through air, when moving at speeds very close to the average speed of the air molecules of the gas. Also many physicists are returning to the Aether just under the disguise of different names: Non-Baryonic Matter, Quintessence, Vacuum Expectation Value Energy, Ground State Energy, Zero Point Energy, Weakly Interacting Massive Particles (WIMPs), and Dark Matter.
Photons travel at the speed of light but are expressed in inertial reference via what we call wavelength and frequency. Only the wavelength and frequency are inertial reference dependent. We see blue and red shift in the wavelength/frequency. But C stays the same. Although only a loose analogy, it would be like having a ball that is moving in a linear way at constant speed, while also spinning. Inertial reference will only notice changes in the spin, which will alter the frequency of rotation. But the linear will not change. There is a logic to this. If we apply special relativity to something traveling at C, what we should see is something that exhibits the properties of a point-instant. Yet photons, although traveling at C, can show a variety of outward appearances based on a wide range of wavelength-frequency. There is a dissociation. A photon that looks like a point-instant would be the most energetic photon possible, having zero wavelength. No one reference can define that much energy, making C independent of reference, but implicit of all the references combined at the same time.
There is a fundamental difference between sound and light. This is not an opinion, this is a fact that has been verified by measurements. Sound has a constant speed in a given medium (density, temp, etc). At standard temperature and pressure the speed of sound in air is 343 m/sec. If a plane is moving at 300 m/sec the sound wave will move away from the plane in the direction of travel at 43 m/sec. The speed of the plane will not change the speed of the sound wave, BUT the speed of the sound wave relative to the plane will be different. The situation with light is much different. The speed of light is 300,000 km/sec. If the plane were moving at 200,000 km/sec the relative speed of the light wave would not be 100,000 km/sec it would still be 300,000 km sec. The speed of light is constant AND independent of the speed of the observer and the source.
Moderator request: Please split off and move to Pseudoscience (or delete) Magneto_1's two recent posts and my three replies. That's also the post where I derive the "agnostic" velocity addition law from physical postulates. Perhaps you are the one commenting "without understanding" since the "agnostic" velocity addition law has to apply to bullets and sound as well as light. And so, I don't say anything specific about light by introducing it. I let experiment say something specific about light. There are a lots of different religions out there. Which one does physics tell us is the correct one? (Despite the name, the answer is not Scientology.) None, because all posit unevidenced metaphysics. Some are blatantly counter-factual in physical prediction. And so it is with all Aether Models. Quibbling about specific models is just evidence of one's descent into pseudoscience. So you wish to throw out Newton's third law? On the same Wikipedia page is the derivation of the same physical law from the aetherless velocity addition law of special relativity. Lorentz's derivation may have been from a dragged aether model, with a different aether for every frequency of light. Einstein's chapter 13 results, and the results at the bottom of the same wikipedia page were derived without any aether model at all. Likewise, I introduced the "agnostic" velocity addition law, relying secretly on my 2008 post, without even thinking about any model of luminiferous aether because I was thinking about velocity, not discredited 19th century metaphysics. Relative to the co-moving frame of any quantity of water with a well-defined state of motion, the speed of light in that water is c/n. I would write "The velocity of light in an isotropic ponderable medium is a constant in the co-moving frame fixed by that medium, u = c/n.[/tex] It's pointless to copy the K if you are going to fix the velocity addition law as the velocity addition law of special relativity. I'm not sure that works in English. Then you don't know what a frame is. This would cause the planets to spiral into to the sun. This is why you have no physics reputation to be damaged.
"Pseudoscience" I used your equation; the one that you just "blindly" wrote down without any understanding!Please Register or Log in to view the hidden image! And your reputation in physics is ????:shrug: This is the only thing worth responding too; and only because others are also reading!! No, the Aether will not cause the planets to spiral into the sun. Theory: The "Aether Vortex" carries the planets in orbit around the sun. And, it is this reason that the Michaelson & Morely, and other past experiments to measure the Aether failed. They were all trying to measure the "Relative Velocity" between the motion of the earth and the aether vortex. However, when it is considered that the "Aether Vortex" carries matter and planets in their orbits their is no relative velocity between a planet and the aether that is carrying the planet. How is this so, that "Aether Vortex" drags matter? Imagine, a ping pong ball and a flowing river. Drop the ping pong ball into the flowing river. The instant that the ping pong ball hits the river its is at zero velocity (v(ball) = 0) relative to the river. However after some time the river bombarding and impinging upon the ping pong ball will eventually accelerate the ping pong ball to flow with the uniform velocity of the flowing river. Hence there is no relative motion between the ping pong ball and the flowing river. Your relative velocity equation that you blindly state will predict the above experiment. Finally, the "Aether Vortex" and the entrainment is currently being measured with the Gravity Probe Experiment. The second is “frame-dragging,” which Joseph Lense and Hans Thirring proposed in 1918 using Einstein’s theory. This is the amount by which Earth twists spacetime as it spins on its axis. “Imagine the Earth as if it were immersed in honey,” says Francis Everitt, GP-B’s principal investigator at Stanford University. “As the planet rotates, the honey around it would swirl, and it’s the same with space and time.” The "Aether Vortex" model is being resurrected; Look Here! Best
The Moon, at times, crosses the path of your hypothetical "Aether Vortex" and thus refutes it. Obviously, that is not for me to say. My physics ability occupies the vast middle ground between those that know nothing and those that have added to Man's knowledge of the universe and the models we use to study in. In my estimation, your ability lies below that of those that know nothing, and your every post seems filled with self-praise and untrue things.
The moon has its own vortex! And for dessert: The "Tangential Velocity" component of each mass body prevents this catastrope from happening.Please Register or Log in to view the hidden image! I interpret this to mean; Magneto, you help and challenge me to "think" and increase the awareness of my worldview!! Thanks. Anytime.Please Register or Log in to view the hidden image!
When I make such an affirmation, I give a link and a quote from that link. Why you not do the same? From the Earth towards the Sun, starts simultaneously , a plane at speeds of V1=150,000 km/s and a short beam of light at speed of V2=300,000 km/s In about eight minutes the light beam reaches the Sun. At this moment the plane is halfway between Earth and Sun. This is not correct?
That's a problem about "rates" not about relativity. It's not about relativity until you place yourself in the shoes of another. Both Earth and the plane agree that the time that the light and plane started the journey is simultaneous (since they started from the same place at the same time, all observers agree they started from the same place at the same time, even if they disagree on where and when that was). We call this event O (for origin) and write the Earth coordinates as \(\left( x_O , \quad t_O \right)\). The plane coordinates we write \(\left( x'_O, \quad t'_O \right)\) Earth sees the light travel distance Δx = L in time Δt = L/c, so this is event A \(\left( x_A = x_O + L, \quad t_A = t_O + \frac{L}{c} \right)\). For Earth, the plane "at this moment" has moved a distance of Δx = v Δt = (c/2)(L/c) = L/2. So the Earth coordinates are for this event B \(\left( x_B = x_O + \frac{L}{2} , \quad t_B = t_O + \frac{L}{c} \right)\). Also at the same time, Earth is at event C \(\left( x_C = x_O , \quad t_C = t_O + \frac{L}{c} \right)\). So as a simple rate problem, Earth does see B as halfway between A and C. The Plane sees events differently. \(\left( x'_A = x'_O + \frac{\sqrt{3} L}{3}, \quad t'_A = t'_O + \frac{\sqrt{3} L}{3 c} \right)\), \(\left( x'_B = x'_O, \quad t'_B = t'_O + \frac{\sqrt{3} L}{2 c} \right)\), \(\left( x'_C = x'_O - \frac{\sqrt{3} L}{3}, \quad t'_C = t'_O + \frac{2 \sqrt{3} L}{3 c} \right)\). So while B is halfway in position between A and B, no two of A, B or C happen at the same time for the plane. According to the plane, "at the same time" the light hits the sun, the plane is at event D \(\left( x'_D = x'_O, \quad t'_D = t'_O + \frac{\sqrt{3} L}{3 c} \right)\) and the Earth is at event E \(\left( x'_E = x'_O - \frac{\sqrt{3} L}{6}, \quad t'_E = t'_O + \frac{\sqrt{3} L}{3 c} \right)\) and D is not halfway between A and E. This is also related to the addition of velocities since the speed of light as seen as the difference between A and O for both observers is \(\frac{\Delta x}{\Delta t} =\frac{\Delta x'}{\Delta t'} = c\). Thus D cannot be halfway between an object moving at speed f(c,c/2) = c and an object moving at speed f(0,c/2) = -c/2 except at the time when all three objects start off in the same place, O.
