doppler's debacle --------------- In 1842 Doppler tested a new theory about sound. From casual observations, the pitch of a sound source approaching the listener increased, while that of a receding source decreased. To experiment, Doppler placed a trumpet player on a railroad car, which he hired to run back and forth. Doppler measured the frequency by finding a matching key on a piano at trackside. How Doppler heard the music over the hideous noise of the train is lost in history. The discovery of this, the Doppler effect, worked for sound, and other mechanical waves. Given the slow speed of the train compared to that of sound in air, the formula Doppler derived was a simple one v[m/s] / c = (lambda'[m/s] - lambda[m/m]) / lambda[m/m] +---------------------------------------------------------+ | DOPPLER EFFECT FOR SOUND AT LOW SPEEDS | | | | v[m/s] / c = (lambda'[m/s] - lambda[m/m]) / lambda[m/m] | +---------------------------------------------------------+ c c is the speed of sound, not of light!, thru air. Note that we write lambda'[m/s] and not lambda[m/s]. The wavelength here is that as heard, observed, by the stational frame; it includes the travel time for the sound waves to reach us. The formula is arranged so that a recession of the source has a positive velocity; accession, negative. This is merely a sign convention. Doppler recognized three distinct scenarios for the observer and source: Observer still in air, trumpet moves thru air Trumpet still in air, observer moves thru air Both trumpet and observer move thru air Each yielded a different formula for the frequency, or wavelength, shift. The three cases are not equivalent or symmetrical. The one stated above is for speeds very small compared to that of sound, v[m/s] << c. All three equations relax to this simple form when the speeds are small enough. Absurd Doppler speeds ------------------- Doppler did his work with sound and extrapolated the results to light. There was in his day the contention about light being a particle or a wave. Doppler took the wave side of the argument. He hit on the idea that his wavelength shift could be applied to the stars. Stars, he thought, have colors because their emitted wavelengths are shifted by their motion to or from us. In fact his book about the shift was 'On the colored light of double stars and other stars in the heavens'! To determine the motion of stars, he needed a reference wavelength of a star with no radial motion. With Earth in a nearly circular orbit around the Sun, the Sun has essentially no radial motion. Its color represented the 'rest' wavelength for light emitted by stars. The nature of stars was totally unknown, altho it was realized that stars had to be sun-like luminous globes. Doppler worked up the wavelengths of starlight from the colors. Then he applied his sound-shift model on them. Using modern values, Doppler came up with the following speeds. +--------+-------+--------+----------+-------------+ | color | lambda|v[m/s]/c| km/s vel | comments | +--------+-------+--------+----------+-------------+ | violet | 350nm | -0.367 | -109.091 | accession | | indigo | 400nm | -0.273 | - 81,878 | | | | blue | 450nm | -0.182 | - 54,545 | | | | green | 500nm | -0.091 | - 27,273 | \|/ | | yellow | 550nm | 0.0 | 0 | Sun at rest | | orange | 600nm | +0.091 | + 27.273 | recession | | red | 650nm | +0.182 | + 54,545 | | | |far red | 700nm | +0.273 | + 81,878 | \|/ | +--------+-------+--------+----------+-------------+ These are crazy speeds! In physics of the early 19th century, there was nothing wrong with such speeds. What's more, they added to the fascination of astronomy! The figures here are worked up with the small-speed formula. The more elaborate ones, for any of the three cases, give similarly ridiculous results. If we include infrared and ultraviolet radiation, which were just discovered in the early 1800s, we can come up with speeds approaching or exceding light. For example if the wavelength is 2,000 nanometers, we the star is receding at, gulp, 2.636 times the speed of light! Doppler and astronomers in general had no dealings with radiation from the stars except visible light. Fizeau's fixup ------------ Fizeau in 1848 also worked on the wavelength shift between source and observer. He had the same early 19th century ideas about light. He recognized that the wavelengths to compare were not those of the star's overall color. From spectrometry, crude as it was then, he saw that a star emits light over most colors but with certain wavelengths missing. These are the dark lines in the star's spectrum, indicative of various chemical elements in the star. Fizeau compared the wavelengths of lines of a recognized element in the star with those from a source at rest. His motion in the stars, was of orders more modest extent. Speeds of recession or accession were in the ones to tens of kilometers per second. Fizeau and astronomers following him for most of the 19th century had one hell of a time getting the measurements of wavelength from stellar spectra. The image was examined by visual micrometry. Due to the thick prisms in the early spectroscopes, the image was dim and the lines were tough to discern. Spectra of only the brightest stars could be confidently examined. Due to this proper application of the wavelength shift, the effect is sometimes called the Doppler-Fizeau effect (Fizeau-Doppler effect, in France), There was a broad disparity of speeds quoted for a given star. Each astronomer had to independently squeeze out measurements of wavelength shifts, each surely different from those of others. Despite heroic valiant efforts, there was not much improvement in the measurements. As example, a star is receding from ys at 600KPS, within range of star motions in the Milky Way. The Doppler-Fizeau shift of a spectral line of wavelength 500nm would ne V / c = (lambda' - lambda) / lambda where c is the lightspeed of300,000KPS. lambda * V / c = lambda' - lambda (lambda * V / c) + lambda = lambda' lambda' = (lambda * V / c) + lambda = (500nm * 60KPS / 300,000KPS) + 500nm = 0.10nm + 500nm = 500.10nm An astronomer would have to discern that the line was displaced toward longer wavelength by 1/10 nm from its home position at 500nm. If he records that shift was 0.20nm, his speed for the star is 120KPS, twice as great. Or if he measures the shift to be 0.05nm, the speed is 30KPS. It is tough to inspect a spectrum by eye thru the telescope and errors like these were common. Photography eventually could capture a permanent image of the spectra by the 1880s. By the end of the 19th century, major projects to photograph star spectra were underway. From the photos, spectrograms, catalogs of stellar speeds were issued. One of the 19th century catalogs of spectra, the Henry Draper catalog, is still in use today (2005) for selecting candidates for extrasolar planets. Frequency versus wavelength ------------------------- Doppler really worked with his music waves by their frequency rather than by wavelength. This is still done today for sound and many parts of the electromagnetic spectrum. In the optical band we typicly deal with wavelength, not frequency. The choice is a function of legacy and the devices we use to explore the waves. It is no more correct to use the one or the other, altho in many situations, the opposite method may look peculiar at first. Wavelength and frequency are related thru the speed of the wave. The wave speed is the product of the frequency times wavelength. Knowing any two of these parameters fixes the third. Wavelength is almost always symbolized by lambda. Frequency is symbolized by either 'f' or 'nu'. +----------------------+ | FREQUENCY-WAVELENGTH | | | | lambda * f = c | | | | lambda = c / f | | | | f = c / lambda | +----------------------+ This is a general relation which applies to any flow of items of a given physical length. For example, a freight train with cars of 15m length passes us at a crossing gate. 12 cars roll by per minute. How fast is the train going? The 'wavelength' is 15 meters; frequency is 12 'cycles' per minute. The train speed is c = lambda * f = (15m) * (12/min) = 180m/min -> 10,800m/hr = 10.8KPH. By the physiology of human vision, wavelength is associated with color. The spectrum has a sequence of color across its wavelength range. The longer wavelengths are at the red end; shorter, blue (indigo and violet) end. Astronomers speak of a shift of wavelength toward the longer, red, end of the spectrum as a 'redshift'. A shift toward shorter, blue, end is a 'blueshift'. Historical blunder ---------------- Since Doppler and Fizeau, the shift of spectral lines was accepted as the hallmark for motion of approach and reproach. Only the component of motion along the line of sight, the radial component, shows up by this technique. Hence, the speed is commonly called 'radial velocity'. By convention positive speeds are recession; negative, accession. In the 1910s, Einstein figured out that one other means of shifting wavelengths was an expanding of space. This expansion can be cited as the fractional increase in scalefactor per unit time. No such enlargement of the universe was known in the 1910s, so this cause was largely ignored by astronomers. Hubble in the 1920s with Humason captured spectra of galaxies. galaxies were realized in 1923 to be other Milky Ways, but were still commonly called 'spiral nebulae'. Hubble & Humason correlated the distances and wavelength shifts for dozens of galaxies. Almost all shifts were toward longer wavelength, toward the red end of the spectrum. In 1929 Hubble announced his results. The galaxies were rushing away from us at high radial speeds in a direct function of distance. Hubble assigned the spectral shifts of the galaxies to real radial motion, speaking of radial velocity and distance as evidence for an expanding universe. Hubble expressed the relation as an increase of speed for unit increase in distance. This is the Hubble factor H0. It is cited as either (km/s)/Mly or (km/s)/Mpc. This was quite normal in his day because, in neglect of Einstein physics, the only cause of spectral line shift was true radial velocity of the source. Ever since then, right into the 21st century, many explanations of the expanding universe speak of real motion of galaxies thru space. The Hubble factor is actually the fractional increase of scalefactor per unit time. It gives the expansion rate directly with no need to invoke physical motion of the galaxies. Looking at the Hubble factor, we see, for a typical value of (75km/s)/Mpc H0 = (75km/s)/Mpc = (75) * (km/(s.Mpc)) The units are [length]/[time*length]. A parsec is 3.08e13 kilometers. Hence, we have H0 = (75) * (km/(s.((1e6pc) * (3.08e13km/pc))) = (75) * (km/(s.(3.08e19km))) = (2.435e-18) * (km/(Km.s)) = (2.435e-18) * (1/s)) = (2.435e-18)/s = 1/(4.107e17s) -> 1/(~13 billion years) This states directly that the universe expands at 2.435e-18 part every second. If the universe started from zero size, at this rate of growth, it would take about 13 billion years to reach its present size. With this simple and elegant explanation of the Hubble factor, textbooks today -- in the 21st century! -- commonly speak of galaxies as actually streaming thru space at hideously insane velocities. This is on top of the erroneous use of the acoustic model for the Doppler shift. Reality versus Perception ----------------------- In all the foregoing we never explained how the one platform knows what's going on with the other. In the derivations so far we made no explicit mention of any conversations passing between the stational and motional frames. In the universe there is really only one way to convey information from place to place: electromagnetic radiation. The books on relativity relentlessly refer to 'light' as the medium of conversation, but we may use any wavelength of EMR. All radiation have the one and same speed in vacuo, c. We do have nonEMR knowledge of other places in the universe. We enjoy rocks and dirt from the Moon, meteorites (including a bunch from Mars!), cosmic heavy ion influx, and solar wind particles. Early in the 21st century we could have retrieved surface samples from Mars, gas from a comet's tail, and dust from around an asteroid. There may be others, yet all combined the information content is minuscule compared to the content of the EMR we receive from the universe. So, for all practicality's sake we must communicate across the depths of space only by EMR. We treat time in this paper as a 'clock reading' or 'unit of time measure'. Time actually applies to ALL processes that have duration. The growth of flowers, hair, rust; length of rainfall, erosion, combustion, mechanical movement; aging of people, animals, structures are all dimensioned in time. Everything in the spaceship -- not only the ticks of clocks -- is time-dilated relative to Earth. Observed Time ----------- We saw that the time unit on the spaceship is experienced on Earth to be longer than experienced on the spaceship delT[m/s] = delT[m/m] / beta with beta ranging between zero and one. We now let the message of each second's tick come to Earth on a light pulse. A one pulse is sent from the spaceship, then the next, then the next, and so on, at one second intervals on the spaceship's clock. The pulse travels a certain distance from the spaceship to Earth, but the next one must travel a bit longer path because the spaceship is in motion, say, away from us. From the stational frame the increment of the spaceship's distance between pulses is delX[m/s] = V[m/s]*delT[m/s]. The next pulse in travelling this extra distance takes an increment of time longer to reach Earth. This is the distance run divided by the speed of the pulse delX[m/s] = V[m/s] * delT[m/s] delT[m/s] = delX[m/s] / c = V[m/s] * delT[m/s] / c The perceived length of the spaceship second from the Earth's platform is delT'[m/s] = delT[m/s] + delT[m/s] = delT[m/s] + V[m/s] * delT[m/s ]/ c = delT[m/s] * (1 + V[m/s] / c) where delT'[m/s] is the length of the time unit as made known to us by the medium of light, as distinct from delT[m/s] which is the real length of that unit. delT'[m/s] = delT[m/s] * (1 + V[m/s] / c) = delT[m/m] * (1 + V[m/s] / c) / beta There is an alternate form of this equation that appears in other works, but it is entirely equivalent to this here one delT'[m/s] = delT[m/m] * (1 + V[m/s] / c) / beta = delT[m/m] * (1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2)) = delT[m/m] * (1 + V[m/s] / c) / ((1 + V[m/s] / c) * (1 - V[m/s] / c))^(1/2)) = delT[m/m] * (1 + V[m/s] / c)^(1/2) / (1 - V[m/s] / c)^(1/2) = delT[m/m] * (( 1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2) +----------------------------------------------------------+ | OBSERVED TIME DILATION BETWEEN INERTIAL FRAMES | | | | delT'[m/s] = delT[m/m] * (1 + V[m/s] / c) ) / beta | | = delT[m/m] * ((1 + V[m/s] / c) / (1 -V[m/s] / c))^(1/2) | +----------------------------------------------------------+ Observed Time Dilation ---------------------- Consider the case of the spaceship moving away from Earth with positive V[m/s]. The delT'[m/s] equation yields delT'[m/s] = delT[m/m] * ((1 + (>0)) / (1 - (>0))^(1/2) = delT[m/m] * ((>1) / (<1))^(1/2) = delT[m/m] * (>1)^(1/2) = delT[m/m] * (>1) > delT[m/m] We observe -- by the message carried by light -- the spaceship second on Earth is longer than on the spaceship itself. Except for the size of the difference, the sense is the same as we found for the real divergence of time between the two platforms. When the spaceship accedes toward us, V[m/s]<0 and delT'[m/s] = delT[m/m] * ((1 + (<0)) / (1 - (<0))^(1/2) = delT[m/m] * ((<1) / (>1))^(1/2) = delT[m/m] * (<1)^(1/2) = delT[m/m] * (<1) < delT[m/m] We see the observed second of the spaceship to be less, smaller, than that on the spaceship. Time for this motional platform speeded up. We have that by using light (EMR) we can not observe the universe as it really is. The medium of light distorts or corrupts our perception of the world. In reality time on the motional frame always runs slower relative to Earth. However, we actually observe that time may be either faster or slower according as the motional frame is acceding or receding relative to us. There is no getting around this distortion. It is not in any way caused by some imperfection in our arts and science of observation. No manner of improvement in our observation powers -- so long as those powers live off of electromagnetic radiation as the vector of the information -- can correct this corrupted view of the universe. EInstein and Doppler-Fizeau ------------------------- We saw that the observed time dilation between the stational and motional platforms is delT'[m/s] = delT[m/m] * (1 + V[m/s] / c)) / beta = delT[m/m] * ((1 + V[m/s] / c) / ((1 - V[m/s] / c))^(1/2) The time unit does not have to be an external signal placed on the EMR wave, like a radio time broadcast or television view of a clock. It can be the time unit embedded in the wave itself. The time unit is merely the reciprocal of the frequency of the wave, with the factor c missed out. delT'[m/s] = 1 / f'[m/s] 1/f'[m/s] = (1 / f[m/m]) * (1 + V[m/s] / c)) / beta = (1 / f[m/m]) * ((1 + V[m/s] / c) / ((1 - V[m/s] / c))^(1/2) The wavelength of the wave, which is what astronomers are comfortable with, is c/f, from lambda*f = c c/f'[m/s] = (c/f[m/m]) * (1 + V[m/s] / c)) / beta = (c/f[m/m]) * ((1 + V[m/s] / c) / ((1 - V[m/s] / c))^(1/2) lambda'[m/s] = (lambda[m/m]) * (1 + V[m/s] / c)) / beta = (lambda[m/m]) * (1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2) lambda'[m/s] / lambda[m/m] = (1 + V[m/s] / c) / beta = (1+V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2)) lambda'[m/s]/lambda[m/m] - 1 = ((1 + V[m/s] / c) / beta)-1 (lambda'[m/s] - lambda[m/m]) / lambda[m/m] = (1 + V[m/s] / c) / beta) -1 = (1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2)) - 1 An other quirk of astronomy is that the lambda difference expression is symbolized by Z and the plain lambda'/lambda, being 1 greater than Z, is called 'Z+1'. So Z = ((1 + V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2)) - 1 Z+1 = (1 + V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2) Many astronomy books have the alternative form Z = ((1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2)) - 1 = ((1 + V[m/s] / c) / ((1 + V[m/s] / c) * (1 - (V[m/s] / c})^(1/2)) - 1 = ((1 + V[m/s] / c)^(1/2)) / (1 - V[m/s] / c}^(1/2)) - 1 = (((1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2)) - 1 Z+1 = ((1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2) +-------------------------------------------------------+ | EINSTEIN'S DOPPLER-FIZEAU EFFECT | | | | Z = ((1 + V[m/s] / c) / beta) - 1 | | = (((1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2)) - 1 | +-------------------------------------------------------+ The Low-Velocity Approximation ---------------------------- Altho the Einstein form of the Doppler-Fizeau effect is the correct one, based on the true cause of the effect, it is easy to see how the classical form can work so well. It turns out the speeds of stars and other bodies in our Milky Way galaxy are small fractions of c. So we have for V[m/s] << c Z = ((1 + V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2)) - 1 = ((1 + (<<c) / c) / (1 - ((<<c) / c)^2)^(1/2)) - 1 = (1 + (<<c) / c) / (1 - (<<<<c))^(1/2)) - 1 = (1 + (<<c) / c) / (1 - 0)^(1/2)) - 1 = (1 + ((<<c) / c) / 1^(1/2)) - 1 = (1 + ((<<c) / c) / 1) - 1 = 1 + (<<c) / c - 1 = (<<c) / c = V[m/s] / c +-----------------------+ | LOW VELOCITY | | DOPPLER-FIZEAU EFFECT | | | | Z = V[m/s] / c | +-----------------------+ The Missing Step -------------- As elegant as the Einstein-Doppler effect is, it presumes we know the velocity between the source and observer and we solve for Z, or Z+1. In astronomy we normally assess Z and we must work out V[m/s]. We need an expression V = func(Z). Yet few textbooks ever present such a function! We simplify V[m/s] to just V here Z+1 = ((1 + V/c) / (1 - V/c))^(1/2) (Z+1)^2 = (1 + V/c) / (1 - V/c) (Z+1)^2 * (1 - V/c) = 1 + V/c (Z+1)^2 - (Z+1)^2 * (V/c) = 1 + V/c (Z+1)^2 = (1 + V/c) + ((Z+1)^2 * V/c) = 1 + V/c + ((Z+1)^2 * V/c) (Z+1)^2 - 1 = V/c + ((Z+1)^2 * V/c) = (1 + (Z+1)^2) * V/c ((Z+1)^2 - 1) / (1 + (Z+1)^2)) = V/c V/c = ((Z+1)^2 - 1) / ((Z+1)^ 2 +1) +---------------------------------------+ | VELOCITY FROM EINSTEIN-DOPPLER EFFECT | | | | V/c = ((Z+1)^2 - 1) / ((Z+1)^2 + 1) | +---------------------------------------+ Space Travel Near c ----------------- One of the grandest fallacies in popular treatises on relativity is time-dilated space travel. The spiel goes like this: A spaceship flies to a star 100ly away at 0.8c. It takes by Earth time 125y for the spaceship to complete the journey. But the spaceship riders feel it took only 75y. beta is (1-(0.8)^2)^(1/2) = 0.6. delT[m/s] = delT[m/m]/0.6 = delT[m/m]*(1.66). A second on the spaceship is 1.66 Earth seconds. So Earth's 125y is completed in (125y)/(1.66) = (125y)*(0.6) = 75y. By increasing V[m/s] to 0.9c the time dilation is greater. The trip takes only 54.5y. This means that space travel is feasible however far the destination is merely by travelling fast enough to dilate the travel time to some comfortable fraction of a person's lifespan. Not. What's wrong? Simply put, it's a ruse. Altho the time cited is delT[m/s], the distance involved is delX[s/s]. That 100ly is the Earth's distance experienced by Earth. It is not merely unfair but conniving to commingle spacetime units from different places! The proper method of determining the spaceship's experienced travel time is to use both the delX[m/s] and delT[m/s]. We did not explore how distance units vary with relative motion. It turns out that the meter on the spaceship as felt by Earth delX[m/s] is also dilated. delX[m/s] = delX[m/m]/beta. Yes, the spaceship completes the trip in 75y, 0.6 that experienced on Earth. But the spaceship only travelled (100ly)*(0.6) = 60ly -- not 100! -- in that 75y. Earth feels it finished a 100ly trip. The spaceship experiences that it has 40ly more to go. What's more, it'll take (40y)*(1.25) = 50y to run out that last 40ly. So!, the spaceship eventually does travel the full 100ly, the 60ly+40ly, but it takes 75y+50y = 125y to do so. This is exactly how long it takes on Earth! There plain is NO saving of time by travelling near c. No matter what inertial platform you ride, it'll take, for a give fraction of c, the same (100)/(fraction of c) years to traverse 100ly. Twin Paradox ---------- With the time dilation effect in Einstein's relativity system there arises the question of getting clocks out of synch by the relative motion between them. That is, if one inertial observer is moving across an other inertial observer and compare their clocks, they differ. The motional observer's clock somehow runs slower and that person ages less when compared to the clock of the stational observer. But the motional platform can be taken as the home observer and now it's the stational fellow that loses time and ages slower. To take a specific example, consider a spaceship in the motional frame m travelling at V[m/s] = 0.8c. For this speed beta is 0.6 beta = (1 - (V[m/s] / c)^2)^(1/2) = (1 - (0.8)^2)^(1/2) = (1 - (0.64))^(1/2) = (0.36)^(1/2) = 0.6 Recall that delT[m/s] = delT[m/m] / beta meaning that Earth, the stational s frame, experiences the second of time on the spaceship to be 1/beta times the second experienced on the spaceship itself. Earth measures the spaceship time as slowed down or dilated; its seconds are bigger. Hence, for a given journey in Earth years the spaceship takes fewer years and so it ages less than Earth. By picking V[m/s] close to c we can leverage the time dilation effect so a voyage of some long time on Earth takes on the spaceship only a few days! In this way we can set off to the other stars how so ever far off they stand within the lifespan of a human rider on the spaceship. In the meantime generations upon generations of humans on Earth came and went. This does not in fact actually happen. The spaceship in its own time takes just as long to finish its journey as by the time kept on Earth. There is no paradox but a severe misinterpretation. The m platform goes out 12 lightyears, to tau Ceti. At 0.8c the trip should take 15 years, as measured on Earth. Due to the time dilation we experience that the spaceship took only (15y)*(0.6) = 9y. (The dilation was derived for the unit of time; the interval, the count of units, is the reciprocal of this dilation.) So!, what would take 15y on Earth is accomplished in only 9y on the spaceship, a saving of 6y. Or the motional observer ages 6y less than the stational observer. But we committed -- along with armies of inattentive writers on relativity -- a colossal booboo. First off, we assumed that after our 15 years the spaceship in fact reached tau Ceti. Did it? Well, the spaceship sees the relative speed between it and Earth as that same 0.8c, by the symmetry of the inertial frames. So in 9 of its own years it travelled (9y)*(0.8c) = 7.2ly. It didn't finish its trip yet! What went wrong is that the common treatments fail to account for the space dilation. We mixed together the time on the spaceship with the space on the Earth. The m frame travelled 12ly in Earth space over 9y in spaceship time. Apart from the seemingly faster-than-light speed, (12ly)/(9y) = 1.333c, we simply can not meld the spacetime measures from two different inertial platforms. It's like posting sales paid in French francs and labeling the entries 'US dollars'. Because the speed is measured the same (but in opposite directions) in the two platforms, yet the time is dilated, the space must also be dilated by the same factor. V[m/s] = delX[m/s] / delT[m/s] = delX[m/m] / delT[m/m] delX[m/s] / delT[m/s] = delX[m/m] / delT[m/m] = delX[m/m] / (delT[m/s] * beta) = delX[m/m] / (delT[m/s] * beta) delX[m/s] = delX[m/m] / beta The spaceship's lightyear which Earth experiences is longer than the lightyear the spaceship itself experiences. So in a given span of Earth time the spaceship crossed fewer lightyears than on Earth. The time is 6y less simply because the spaceship never got to tau Ceti; it has 4.8ly yet to go. At 0.8c it'll take (4.8ly)/(0.8c) = 6y to cover the remaining distance and arrive at tau Ceti. This is added to the 9y already run and the whole trip takes (9y)+(6y) = 15y. There is no 'paradox' at all. We took the clock reading on the m frame before it finished the trip we set it out to fulfill. It is the pure and simple fact that from Earth the spaceship really did traverse only 7.2ly in 9y when our clock ticked off 15y. If we wait until the m frame completes the 12ly trip, we experience that it took after all exactly 15y, the same as on Earth. Or on the spaceship the riders feel that time has not slowed at all and they age just as fast as the folk on Earth. At 0.8c it really takes 15y to get to tau Ceti. They are fully 15y older than when they left Earth. Simultaneous Events ----------------- As referred to a given inertial platform, two events are simultaneous if in that platform they occur at the same time. They occur at different places in that platform, so their X coordinates are not the same. But they share the same cT coordinate. In Newton physics two events which are simultaneous for any one observer must be simultaneous for all observers. This comes from the external and absolute nature of time in Newton physics. In Einstein physics we must set aside the earthly concept of absolute simultaneity. In the motional frame two events happen at two places in the spaceship at the same time. The X coordinates differ but the cT coordinates are the same. delX[m/m] <> 0 and delT[m/m] = 0. Spacetime Interval for Simultaneous Events ---------------------------------------- We start with the spacetime interval equation and set delT[m/m] to zero. delX[m/s]^2 - c^2 * delT[m/s]^2 = delX[m/m]^2 - c^2 * delT[m/m]^2 = delX[m/m]^2 - c^2 * 0 = delX[m/m]^2 -c^2 * delT[m/s]^2 = delX[m/m]^2 - delX[m/s]^2 We saw in the Twin Paradox that delX[m/s] is dilated relative to delX[m/m] by delX[m/s] = delX[m/m] / beta So -c^2 * delT[m/s]^2 = delX[m/m]^2 - delX[m/m]^2 / beta^2 = delX[m/m]^2 - delX[m/m]^2 / beta^2 = delX[m/m]^2 * beta^2 / beta^2 - delX[m/m]^2 / beta^2 = (delX[m/m]^2 * beta^2 - delX[m/m]^2) / beta^2 = (delX[m/m]^2 - delX^2 * (V[m/s]^ 2 /c^2) - delX[m/m]^2) / beta^2 = (-delX[m/m]^2 * V[m/s]^2 / c^2) / beta^2 delT[m/s]^2 = (-1 / c^2) * (-delX[m/m]^2 * V[m/s]^2 / c^2) / beta^2 = (delX[m/m]^2 * V[m/s]^2 / c^4) / beta^2 = delX[m/m]^ 2 *V[m/s]^2 / (c^4 * beta^2) delT[m/s] = delX[m/m] * V[m/s] / (c^2 * beta) +------------------------------------------------+ | EINSTEIN SIMULTANEITY | | | | delT[m/s] = (delX[m/m] * V[m/s] / (c^2 * beta) | +------------------------------------------------+ Platforms Can Not Agree --------------------- So long as V[m/s] is nonzero -- there are in fact two distinct platforms -- the right side of the above equation must be nonzero. Now in deriving this Einstein simultaneity relation we set delT[m/m] = 0, by the definition of simultaneity in the motional frame. Yet in the stational frame delT[m/s] is not zero. There is a nonzero interval of time between the stational frame's experience of the one event and the other. Or the events in the spaceship that are occurring at the same instant in the spaceship are experienced as occurring in sequence on the Earth. In general events which in one platform occur together can not occur together in any other platform. There is no absolute simultaneity under Einstein physics. This conclusion is even more violative against our common sense than the time and length dilation! It does seem so perfectly natural that all frames must agree that two simultaneous happenings for any one do occur at the same instant for all. Yet this just is not true. It is only the very minuscule speeds between observers we cope with that masks this relativity of time in our everyday life. Simultaneous in Space and Time ---------------------------- If we insist that the two events in the spaceship occur not only at the same time but also at the same place, we have delX[m/m] = 0 as well as delT[m/m] = 0. We then get the trivial result that in the Einstein simultaneity equation delT[m/s] = 0. The stational frame experiences the events to happen together. But in this case, there are not two separate happenings. We have merely duplicated the same set of coordinates, so there is really nothing for relativity to act on. Lorentz Transformation -------------------- We derived the relation between the time on the spaceship, the motional frame, as experienced by Earth, the stational frame, in two situations. First, we saw events in the motional platform occurring in the same location but at two different times; we zeroed the terms with delX[m/m]. Then we looked at two events occurring at the same time but in two different locations by zeroing the delT[m/m] term. Thus we obtained delT[m/s] = delT[m/m] / beta delT[m/s] = (V / c^2) * delX[m/m] / beta Similar relations join the space coordinates between the two platforms, which we did not explicitly derive delX[m/s] = delX[m/m] / beta delX[m/s] = (V/c) * c * delT[m/m] / beta If we remove the constraint of holding either the location or time of the events fixed in the spaceship, we get the more general transformation of time and space between the two frames. These are merely the addition of the constant-location and constant-time formulae from above. delT[m/s] = (delT[m/m] / beta) + ((V/c^2) * delX[m/m] / beta) = (delT[m/m]+(V/c^2)*delX[m/m])/beta delX[m/s] = (delX[m/m] / beta) + ((V/c) * c * delT[m/m] / beta) = (delX[m/m] + (V/c) * c * delT[m/m]) / beta These are the Lorentz transformation equations relating the spacetime in the motional frame to that experienced in the stational one. They were discovered in the 1890s by Lorentz in the era leading up to Einstein's synthesis of relativity. +-------------------------------------------------------+ | LORENTZ TRANSFORMATION BETWEEN TWO FRAMES | | | | delT[m/s] = (delT[m/m] + (V/c^2) * delX[m/m]) / beta | | | | delX[m/s] = (delX[m/m] + (V/c) * c *delT[m/m]) / beta | +-------------------------------------------------------+ Addition of Velocities -------------------- V is the velocity of the motional frame relative to the stational one, and is a short form for V[m/s]. beta is the Lorentz factor (1- (V/c)^2)^1/2. We now change the subscripts to let us introduce a second motional frame. The m platform is now the m1 frame; the new one, m2. delX[m1/s] = (delX[m1/m1] + (V/c) * c * delT[m1/m1]) / beta delT[m1/s] = (delT[m1/m1] + (V/c^2) * delX[m1/m1]) / beta Divide the first by the second delX[m1/s] / delT[m1/s] = ((delX[m1/m1] + (V/c) * c * delT[m1/m1]) / beta) / ((delT[m1/m1] + (V/c^2) * delX[m1/m1]) / beta) = (delX[m1 / m1] + (V/c) * c * delT[m1/m1]) / (delT[m1/m1] + (V/c^2) * delX[m1/m1]) = (delX[m1/m1] / delT[m1/m1]) + ((V/c) * c * (delT[m1/m1] / delT[m1/m1]) / (delT[m1/m1] / delT[m1/m1]) + ((V/c^2) * delX[m1/m1] /delT[m1/m1]) = (delX[m1/m1] / delT[m1/m1]) + (V/c) * c *1) / (1 + (V/c^2) * delX[m1/m1] / delT[m1/m1]) = (delX[m1/m1] / delT[m1/m1]) + (V/c) * c) / (V/c^2) * delX[m1/m1] / delT[m1/m1]) delX[m1/m1]/delT[m1/m1] is the change of position of a point as experienced by the m1 frame, or the speed of a point in a new motional platform, the m2 platform, as experienced by m1. Call this V[m2/m1]. delX[m1/s]/delT[m1/s] is the speed of this same point, in the second m2 frame, as experienced by the stational frame. Call it V[m2/s]. V[m2/s] = (V[m2/m1] + (V/c) * c * 1) / (1 + ((V/c^2) * V[m2/m1])) = (V[m2/m1] + V) / (1 + (V * V[m2/m1] / c^2)) With more careful subscripts, we have V[m2/s] = (V[m2/m1] + V[m1/s]) / ( 1 +(V[m1/s] * V[m2/m1] / c^2)) +--------------------------------------------------------- ---------+ | VELOCITY RELATION IN EINSTEIN PHYSICS | | | | V[m2/s] = (V[m2/m1] + V[m1/s]) / (1 + (V[m1/s] * V[m2/m1] / c^2)) | | | | V[m2/m1] = (V[m2/s] - V[m1/s]) / (1 - (V[m2/s] * V[m1/s] / c^2)) | +-------------------------------------------------------------------+ The second expression is a rotation of the first, as explained below. It is applicable when the speeds of the two motional platforms are given relative to the stational frame. Comparison with Newton -------------------- At speeds near the speed of light, the deviation from Newton shows up. For V[m1/s] = 0.6*c and V[m2/m1] of 0.8*c, Newton figures the combined speed V[m2/s] to be 1.4*c. However, V[m2/s] = (V[m2/m1] + V[m1/s]) / (1 + (V[m1/s] * V[m2/m1] / c^2)) = ((0.8) + (0.6)) / (1 + (0.8) * (0.6) / (1^2))) = (1.4) / (1 + (0.48) / (1)) = (1.4) / (1.48) = (0.9459) The combined speed is still less than lightspeed. In fact, even in the limits where both of the ingredient speeds are c, the resultant combined speed is at most c, but never greater than c. Hence, once again we have the imposition of a real limit on the relative speed achievable, that of light. To see this, we try it, V[m2/s] = (V[m2/m1] + V[m1/s]) / (1 + (V[m1/s] * V[m2/m1] / c^2)) = (c + c) / (1 + (c * c / c^2)) = 2 * c / (1 + (c^2 / c^2)) = 2 *c / (1 + 1) = 2 * c / (2) = c Fizeau's Experiments ------------------ Fizeau schemed to test the aether theory by measuring the speed of light in flowing water. In 1859 he arranged a set of tubes with water pumped thru them and sent light beams into them. By interferometry, which he invented, he assayed the speed of light upstream and downstream in the water. He found that the speed was slightly less than the Newton addition of velocities, but derived a wrong formula for it. It worked for the extremely low speed of the water, at most 7m/s. This was the first indication that something peculiar was going on that compound speeds are not adding according to the only method then known, that of Newton. Shift of Platforms ---------------- In the above examples we worked with 'piggybacked' platforms. That is, an airplane was the m1 frame, a rocket fired from that plane was the m2 frame. The speed of the rocket was added (by either Newton or Einstein equations) to that of the plane. If we have the two platforms m1 and m2 both referenced to the stational platform we have a rotation of the basic equation. In Newton, the two speeds would be subtracted to get the speed of m2 relative to m1. Leaving out the algebra to do this rotation, we end up with V[m2/m1] = (V[m2/s] - V[m1/s]) / (1 - (V[m2/s] * V[m1/s] / c^2)) This can be terribly confusing for the novice at relativity. Suppose we have two spaceships running parallel with speeds of 0.9*c and -0.7*c. (They are offset slightly so there is no actual headon collision.) Newton gets the closing speed as (0.9*c)-(-0.7*c) = (1.6*c). By the Einstein formula, we have V[m2/m1] = (V[m2/s] - V[m1/s]) / (1 - (V[m2/s] * V[m1/s] / c^2)) = ((0.9) - (-0.7)) / (1 - (0.9) * (-0.7) / (1^2))) = (1.6) / (1-(-0.63) / (1)) = (1.6) / (1.63) = (0.9816) The weird thing is that from the stational frame we really do clock the closing speed to be 1.6*c! It is from the one frame experiencing the other that the speed is collapsed to lass than c. All of the foregoing is so contrary to common sense that many home astronomers valiantly try to dismiss it as some voodoo maths or spurious illusion. Yet it is a real phaenomenon and is a direct result of the distorted spacetime perceived by observers in relative motion.