Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-17T17:18:12Z https://physics.stackexchange.com/feeds/question/2230 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/2230 49 Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? SMUsamaShah https://physics.stackexchange.com/users/847 2010-12-24T13:25:21Z 2018-05-27T12:28:40Z <p>I was told that the Galilean relative velocity rule does not apply to the speed of light. No matter how fast two objects are moving, the speed of light will remain same for both of them. </p> <p>How and why is this possible? </p> <p>Also, why can't anything travel faster than light?</p> https://physics.stackexchange.com/questions/2230/-/2232#2232 13 Answer by Sklivvz for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Sklivvz https://physics.stackexchange.com/users/66 2010-12-24T13:34:40Z 2014-11-03T18:06:51Z <p>In actual fact, the relative speed rule does not apply, ever.</p> <p>The <a href="http://en.wikipedia.org/wiki/Velocity-addition_formula" rel="nofollow">relativistically correct speed addition rule</a> is the following:</p> <p>$$s=\frac{v+u}{1+\frac{vu}{c^2}}$$</p> <p>When $\frac{vu}{c^2}$ is close to zero (in other words when the velocities invloved are much less than the speed of light), then the correct formula is approximated to the Galilean version $s=u+v$.</p> <p>Nothing can be faster than light, fundamentally, because as you accelerate you not only gain speed, but also mass. As you approach the speed of light, the energy given to you by the force causing the acceleration basically contributes more and more to the increase of your mass and less and less to the increase of your speed. It does this precisely so you never reach the speed of light. Instead, massless particles like photons always travel at the speed of light.</p> https://physics.stackexchange.com/questions/2230/-/2288#2288 6 Answer by sigoldberg1 for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? sigoldberg1 https://physics.stackexchange.com/users/78 2010-12-27T06:49:03Z 2010-12-27T06:49:03Z <p>As to part 2 of your question "Also why nothing can be more speedy than light?", the answer is that it's not just light. The point is that c is the <em>maximal velocity of any causal, information transmitting interaction in the universe</em>, mediated by anything travelling forwards in time (see footnote). Its just that photons, having 0 rest mass, travelling in a vacuum approach that fundamental limit, c.</p> <p>Footnote: Except maybe 'tachyons' - never seen and traveling backwards in time because they go faster than c. (Note that Norbert Wiener once pointed out that for a causal influence travelling backwards in time, we would experience it as "random", since it would apparently be an event without an antecedent cause to us).</p> https://physics.stackexchange.com/questions/2230/-/2290#2290 50 Answer by Mark Eichenlaub for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Mark Eichenlaub https://physics.stackexchange.com/users/74 2010-12-27T07:42:15Z 2010-12-27T07:42:15Z <p>The view of most physicists is that asking "How can it be that the speed of light is constant?" is similar to asking "How can it be that things don't always go in the direction of the force on them?" or "How can it be that quantum-mechanical predictions involve probability?"</p> <p>The usual answer is that these things simply <i>are</i>. There is no deeper, more fundamental explanation. There is some similarity here with the viewpoint you may have learned in studying Euclidean geometry; we need to start with some axioms that we assume to be true, and cannot justify. Philosophically, these ideas are not precisely the same (mathematical axioms are not subject to experimental test), but the constant speed of light is frequently described as a "postulate" of relativity. Once we assume it is true, we can work out <a href="http://en.wikipedia.org/wiki/Introduction_to_special_relativity">its logical consequences</a>.</p> <p>This is not to say that, in physics, postulates stay postulates. For example, many people are especially concerned about probability in quantum mechanics, and are trying to understand it based on more fundamental ideas (see <a href="http://en.wikipedia.org/wiki/Quantum_decoherence">decoherence</a> as one example). As another example, Newton's laws of motion were originally taken as unprovable postulates, but are now explained via quantum mechanics (see <a href="http://en.wikipedia.org/wiki/Ehrenfest_theorem">Ehrenfest's theorem</a>).</p> <p>At this time, the constancy of the speed of light, or more generally the principle of <a href="http://en.wikipedia.org/wiki/Lorentz_covariance">Lorentz symmetry</a>, is not justified by anything considered to be more fundamental. In fact, the assumption that it is true has been a guiding light to theoretical physicists; quantum field theory was invented by thinking about how quantum mechanics could be made to respect the ideas of relativity.</p> <p>Although we do not have a theoretical justification for the constancy of the speed of light, we do have very accurate experimental tests of the idea. The most famous is the <a href="http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment">Michelson-Morley experiment</a>, which measured the relative speed of light in different directions to see if it was affected by the motion of the Earth. This experiment rejected the hypothesis that the motion of the Earth affects the speed of light. According to the Wikipedia article I linked, a modern version of this experiment by Hils and Hall concluded that the difference in the speed of light along directions parallel and perpendicular to Earth's motion is less than one part in $5*10^{12}$. In addition to direct tests of the speed of light, there have also been <a href="http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/experiments.html">many other experimental tests of special relativity</a>. (I haven't read this last page carefully, but, on flipping through, it looks good.) </p> <p>There are a few caveats worth mentioning. In general relativity, the speed of light is only constant locally. This means that the distance between two objects can increase faster than the speed of light, but it is still impossible for light to zip past you at a speed faster than the normal one. Also, in quantum theory, the speed of light is a statistical property. A photon may travel slightly slower or faster than light, and only travels at light speed on average. However, deviations from the speed of light would be probably be too small to observe directly.</p> https://physics.stackexchange.com/questions/2230/-/4707#4707 7 Answer by Gordon for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Gordon https://physics.stackexchange.com/users/1495 2011-02-06T19:08:42Z 2011-02-06T19:49:20Z <p>John Moffat and Moffat and Albrecht and Magueijo have variable speed of light theories where the speed varied in the early universe and is not a constant. Majueijo has a poplular book <a href="http://rads.stackoverflow.com/amzn/click/B000EXYZLI">Faster Than The Speed of Light</a> outlining his theories. IMO the book is quite outrageous and insults various people. I mention this answer for completeness only as I believe the speed of light in a vacuum is constant.</p> <p>Space can expand faster than the speed of light, but no information can be transmitted. See the <a href="http://en.wikipedia.org/wiki/Alcubierre_drive">Alcubierre warp drive</a> for some fun.</p> https://physics.stackexchange.com/questions/2230/-/4711#4711 1 Answer by Joel Rice for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Joel Rice https://physics.stackexchange.com/users/1423 2011-02-06T20:19:01Z 2011-02-06T20:19:01Z <p>That the speed of light is invariant is a property of Minkowski spacetime, and there should be plenty on that in Wiki - or search for 'geometric algebra' or Clifford Algebra.</p> https://physics.stackexchange.com/questions/2230/-/95563#95563 7 Answer by Tau23 for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Tau23 https://physics.stackexchange.com/users/38378 2014-01-27T21:51:50Z 2014-01-28T13:14:58Z <p>The speed of light it's the speed limit in the universe because in an informal sense it's infinite. If a spacecraft was built to travel at a constant 1 g acceleration it would very fast reach 99.9% the speed of light, enabling traveling through the whole observable universe in a lifetime due to the effect of time dilation. There is no rest frame for the photon in relativity but approaching it's speed makes you experience more an more a subjective close to infinite speed. So from the hypothetical 'point of view' of the photon it travels an arbitrary distance in zero time. Emission is the same point as absorbtion for a photon. Now answering why it's constant for all observers and not infinite, I have to say it comes down to the laws of causality and locality. </p> <p>Also in relativity physics, rapidity (φ) is used as an alternative to speed as a measure of motion. The equation is φ = artanh(v/c). Substituting v = c you get artanh(1) = infinity. So the rapidity of light is infinite.</p> https://physics.stackexchange.com/questions/2230/-/142330#142330 5 Answer by Hypnosifl for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Hypnosifl https://physics.stackexchange.com/users/59406 2014-10-20T22:18:29Z 2014-10-23T13:23:00Z <p>The fact that different observers in relative motion can measure the same light ray to move at a speed of c has to do with the fact that each observer defines the "speed" in terms of distance/time on rulers and clocks at rest <em>relative to themselves</em>. It's crucial to understand that different observers use different rulers and clocks to measure speed, because in relativity each inertial observer will see the rulers of other inertial observers to be shrunk (<a href="http://en.wikipedia.org/wiki/Length_contraction" rel="nofollow">length contraction</a>), and the clocks of other inertial observers to be running slow (<a href="http://en.wikipedia.org/wiki/Time_dilation" rel="nofollow">time dilation</a>) and to be out-of-sync with one another (<a href="http://en.wikipedia.org/wiki/Relativity_of_simultaneity" rel="nofollow">relativity of simultaneity</a>). Each observer can be imagined to measure the speed using a pair of clocks at different positions along a ruler (the clocks synchronized in their own frame using the <a href="http://en.wikipedia.org/wiki/Einstein_synchronisation" rel="nofollow">Einstein synchronization convention</a>), measuring the time T1 on the first clock as the light wave passes it, and the time T2 on the second clock as the light passes it, and then if their ruler shows the clock to be a distance D apart, this observer concludes the speed of the light ray was D/(T2 - T1).</p> <p>But now consider how the rulers and clocks of this observer will look in my frame, if I see the observer to be moving at some velocity v along my x-axis (with the ruler parallel to the x-axis). From my perspective, the ruler which the moving observer used to measure the distance is shrunk by a factor of $\sqrt{1 - v^2/c^2}$ due to length contraction, the time between ticks on the clocks of the moving observer expands by $1 / \sqrt{1 - v^2/c^2}$ due to time dilation (or equivalently, in $T$ seconds of time in my frame I only see the moving observer's clock tick forward by $T \sqrt{1 - v^2/c^2}$), and the rear clock's time-reading is ahead of the front clock's reading by $vL/c^2$ due to the relativity of simultaneity, where $L$ is the distance between the clocks in the observer's own frame, as measured by their own ruler.</p> <p>Let's look at a numerical example. Say that the ruler is 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor $1 / \sqrt{1 - v^2/c^2}$ (which determines the amount of length contraction and time dilation) is 1.25, so in my frame the ruler's length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by $vL/c^2$ = (0.6c)(50 light-seconds)/$c^2$ = 30 seconds.</p> <p>Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.</p> <p>Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.</p> <p>If you want to also consider what happens if, after reaching the front end of the moving ruler at 100 seconds in my frame, the light then bounces back towards the back in the opposite direction towards the back end, then at 125 seconds in my frame the light will be at a position of 75 light-seconds on my ruler, and the back end of the moving ruler will be at that position as well. Since 125 seconds have passed in my frame, 125/1.25 = 100 seconds will have passed on the clock at the back of the moving ruler. Now remember that on the clock at the front read 50 seconds when the light reached it, and the ruler is 50 light-seconds long in its own rest frame, so an observer on the moving ruler will have measured the light to take an additional 50 seconds to travel the 50 light-seconds from front end to back end.</p> https://physics.stackexchange.com/questions/2230/-/144940#144940 0 Answer by Sean for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Sean https://physics.stackexchange.com/users/37452 2014-11-05T11:08:03Z 2014-11-05T11:08:03Z <p>People may say things such as "... we do not have a theoretical justification for the constancy of the speed of light... ", or say " These things simply are, therefore there is no deeper, more fundamental, explanation. ". Thus such people in general accept the effect, yet they have no desire to find the cause.</p> <p>However, it is to be noted that if you analyze the simple idea of absolute motion taking place within an absolute 4 dimensional structure known as Space-Time, you soon end up independently deriving all of the Special Relativity equations, and of course you simultaneously acquire a full understanding of Special Relativity. If interested, see proof of this simple analysis of motion at <strong><a href="http://goo.gl/fz4R0I" rel="nofollow">http://goo.gl/fz4R0I</a></strong></p> <p>By starting at the absolute foundation, this in turn reveals the absolute cause. </p> <p>Thus by having seen the absolute cause, no matter how fast two observers are moving across space via two different velocities, it becomes crystal clear as to why the speed of light will remain the same for both of these observers.</p> <p>Unfortunately, even though one can fully encompass Special Relativity via ones independent investigation of the absolute foundation, as a characteristic of Special Relativity this absolute foundation can not be detected. Thus such exploration is confined to being limited to a one way trip. Since these absolutes are undetectable, they were considered to be of no importance, thus they were quickly thrown out the window. Relativity was then accepted as being the all important rather than the absolute being considered to be the all important.</p> <p>The lessor was considered to have clearly outsized the greater !</p> <p>Thus when absolute explanations are being asked, absolute answers in general are not being given, since absolutes are consider to be of no importance.</p> https://physics.stackexchange.com/questions/2230/-/147188#147188 3 Answer by Quantumplate for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Quantumplate https://physics.stackexchange.com/users/42704 2014-11-17T09:12:25Z 2014-11-17T09:12:25Z <p>Consider sound waves. If you have a car with a siren does the car's speed affect the speed of sound? No, instead sound always travels at the same speed (let's ignore for now the fact that the speed of sound can change, e.g. by the density of the medium it is travelling through).</p> <p>Varying the car's speed will cause a <a href="http://en.wikipedia.org/wiki/Doppler_effect" rel="nofollow">Doppler Effect</a> where the sound waves compress (or expand depending on direction) and the frequency of the wave is effected (e.g. the pitch is higher as the source of the sound approaches you or lower as it travels away from you).</p> <p>Essentially you are creating the sound wave as you move through the medium that it is travelling in. This means that as you create it the waves are bunching up in front and expanding out behind. It's even possible to travel faster than the speed of sound where you are out racing the wave just created (and it's all behind you, nothing in front).</p> <p>Light behaves in a similar fashion (well kind of). We know from experiment that it is always measured with the same speed (in a vacuum) regardless of the speed at which it was emitted (technically the speed of the material that it was emitted from). This shouldn't really be that surprising as this is how sound waves behave too.</p> <p>The speed of the material emitting the light will effect the frequency of the light (e.g. if it's moving away from you the waves will spread out and it will be red shifted, or if it's moving towards you the waves will bunch up and be blue shifted).</p> <p>In summary:</p> <ul> <li><p>Light only has one speed (the speed we measure it at or C). So far nice and simple. This figure is not really special or surprising (it has to be something).</p></li> <li><p>It's speed is independent of the material or frame of reference it is emitted from (similar to how the speed of sound is independent of the speed of the siren/car).</p></li> <li><p>More on the speed of sound. It is independent of the speed of it's source because there is a decoupling from the source of the wave (compression/expansion of air molecules) and its propagation through the medium (air molecules bouncing against each other).</p></li> <li><p>Objects (e.g. planes) can travel faster than the speed of sound. This is not the case with light. Nothing can travel faster than the speed of light.</p></li> </ul> <p>Why is matter restricted to slower than light speed travel? Simply put it requires more and more energy to accelerate mass the closer you get to the speed of light. To get to the speed of light requires infinite energy and you can't go past infinite (to go faster).</p> <p>So one of the reasons Relativity is so confusing is that the speed of light is constant regardless of your frame of reference. Imagine lots of flying ambulances and you could only measure another ambulance by its sound. An ambulance flying towards you might have a higher frequency sound wave but you couldn't tell if that was due to your motion towards the sound wave or the motion of the ambulance emitting it (no absolute frame of reference to help you out). </p> <p>I'm not sure how much the analogy of sound waves holds up to light waves but hopefully it gives you a more intuitive idea how this can occur.</p> https://physics.stackexchange.com/questions/2230/-/147204#147204 2 Answer by Žarko Tomičić for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? Žarko Tomičić https://physics.stackexchange.com/users/64108 2014-11-17T10:26:21Z 2014-11-17T10:26:21Z <p>There is some confusion about two points...first, independence of the speed of light on the speed of its source is nothing surprising. It is the same as independence of the speed of water waves on the speed of a boat. Nothing funny there. BUT, independence of the speed of light on the reference frame of two observers in relative motion to the same source, that is the funny stuff. I guess that you dont care about answers like because of Lorentz transformations or similar stuff. Maybe you are wondering why would anyone (Einstein for example) come to this idea. Thing is, its because of Michaelson-Morley experiment and because of Maxwell equations. MM experiment gave us proof that this is true, and Maxwell equations gave us motivation. In these equations speed of light appears as a speed of electromagnetic waves. So, it comes to be something of a natural constant. It plays a special role as you can see, same as Planck constant or gravitational constant. Natural laws must be the same in all reference frames so the speed of light has to be the same to. There is also the fact that if you could travel at the speed of light you would observe a stationary light wave, and that is just NOT possible. There is no actual proof for all of this, just strong gut feeling. </p> https://physics.stackexchange.com/questions/2230/-/297787#297787 0 Answer by JakobH for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? JakobH https://physics.stackexchange.com/users/37286 2016-12-10T08:26:35Z 2016-12-10T16:47:03Z <p>Unfortunately, "<em>Why</em>" questions almost always get here at StackExchange the answer "<em>Because that's how nature is!</em>". Unsurprisingly the top answer tells you exactly that.</p> <p>There are many physicists here at the site who believe that it's not the goal of physics to answer "Why" questions, but rather to fit models to observations. /rant</p> <p>However, there <strong><em>are</em></strong> ideas for answers to several of the most important "Why" questions. These ideas are currently not experimentally verified and therefore aren't canonical answers. As long as there is no experimental proof, these ideas are just that: ideas. Nevertheless, I always enjoy reading about such ideas and maybe you do, too. </p> <p><strong>So here's an idea that explains why nothing can travel faster than with speed of light:</strong> </p> <p>Spacetime is not continuous, but discrete and there is a minimal length $l_m$ and a minimal time interval $t_m$.</p> <p>If this is the case, nothing can move with a speed larger than $\frac{l_m}{t_m}$. In order to move faster than $\frac{l_m}{t_m}$ and object would need to travel the minimal distance $l_m$ in a time interval shorter than <strong>minimal</strong> time interval $t_m$. </p> <p>In addition, this maxium speed has the same value in all frames of reference, because the minimal lenght and minimal time interval are the same in all frames of reference. If this wouldn't be the case there would be a preferred frame of reference which is in disagreement with the <a href="https://en.wikipedia.org/wiki/Principle_of_relativity" rel="nofollow noreferrer">principle of relativity</a>.</p> <p>So, do we actually explain anything with this idea? Now, of course, we have to answer the question: Why should spacetime be discrete and why should there be a minimal length and time interval?</p> <p>First of all, this idea is not as strange as it may sound if you hear it the first time. A discrete spectrum with some minimal amount is exactly what we are used to work with in quantum theory. Hence, we can formulate the idea differently and say: spacetime is quantized. Natural candidates for the minimal length and minimal time interval are the <a href="https://en.wikipedia.org/wiki/Planck_length" rel="nofollow noreferrer">Planck length</a> and <a href="https://en.wikipedia.org/wiki/Planck_time" rel="nofollow noreferrer">Planck time</a>.</p> <p>It is an attractive idea on its own to think that spacetime consists of "spacetime-atoms", i.e. minimal building blocks. <a href="https://en.wikipedia.org/wiki/Loop_quantum_gravity" rel="nofollow noreferrer">For example</a>, "the main output of [Loop Quantum Gravity] is a physical picture of space where space is granular. The granularity is a direct consequence of the quantization. It has the same nature as the granularity of the photons in the quantum theory of electromagnetism and the discrete energy levels of atoms. Here, it is space itself that is discrete. In other words, there is a minimum distance possible to travel through it."</p> <p>I'm not an expert on any of this, but I think this is very neat way to explain why there is a maximum speed. Maybe we can even turn things around and say that the observation that the speed of light is the maximum speed is a strong hint towards the idea that spacetime is quantized. </p> <p>This line of thought is similar to the conclusion that the observed quantization of electric charge is a strong hint towards the idea of a <a href="https://en.wikipedia.org/wiki/Grand_Unified_Theory" rel="nofollow noreferrer">Grand Unified Theory</a>. </p> <p>The main caveat to keep in mind is that here is currently no experimental proof that spacetime is quantized. (And, of course, no experimental proof of a Grand Unified Theory.)</p> <p>Nevertheless, the idea that spacetime is quantized offers such a beautiful explanation for something that otherwise is simply an experimental fact that it seems worthwile to push forward in this direction. (Similar, the way a Grand Unified Theory explains why electric charge is quantized is so beautiful that many physicists believe that the basic idea of a Grand Unified Theory is correct, although there is currently no experimental proof.)</p> https://physics.stackexchange.com/questions/2230/-/408541#408541 0 Answer by kakaz for Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? kakaz https://physics.stackexchange.com/users/1012 2018-05-27T12:28:40Z 2018-05-27T12:28:40Z <p>It is relatively easy to get constant speed of light as constant of nature from Galilean equivalence of inertial observers ( GP) principle. </p> <p>GP says that any two observers, moving wit (any) constant speed relatively to each other, are equivalent, that is, they both describes physical reality in the same way.</p> <p>At particular it means they share the same kinematics principles, and they are using isomorphic coordinate frames. That's is if observer O has this coordinate system $(t,x)$ and observer O' has $(t',x')$, they are related. If we consider the simplest relationship, linear dependence, it is described by the following equations valid for observer O' :</p> <p>$$x' = at + bx$$ $$t' = ct + dx$$</p> <p>As observer O is equivalent to observer O' and physical reality observed from coordinate frame O is exactly the same, <em>the same relations must be valid for observer O</em> :</p> <p>$$x = at' + bx'$$ $$t = ct' + dx'$$</p> <p>Please, notice, the same coefficients a,b,c,d are used! </p> <p>If O and O' has synchronized clocks, we may combine both descriptions, trying to obtain further requirements for coefficients a,b,c,d, <a href="http://www.jakubw.pl/faq/fizyka/node31.html" rel="nofollow noreferrer">as described ( in Polish), here.</a> Assume that O and O' moves with relative speed V.</p> <p>We have to define ( unknown) function d( V) </p> <p>$$x = d( x' + Vt' )$$ $$t = d( t' + x' V \frac{1-d^{-2}}{V^2})$$</p> <p>Now, if we assume there is third observer O'' which moves with speed of U relatively to observer O' we may ask, what is her speed for observer O? Using above formulas we obtain the following result: $$O("U+V") = \frac{U+V}{1+UV\frac{1+d^{-2}(V)}{V^2}}$$</p> <p>If we set that $d(V) =1$, we will have Galilean physics restored. But there is no argument to do this!</p> <p>We where using observer O as a basis, and observer O' moving with speed V relatively to it. Another observer O'' was moving with the speed of U relatively to O', and we obtain related speed of O'' valid for observer O. Speed U was speed of O'' in frame O' which was "dragged" by frame O' wirth "draging speed" of V. </p> <p>What if we would start with observer O'' and calculate the same? As O and O' are equivalent, the only difference is, that this time, "dragging speed" will be U whilst another one, V would be relatively to O'. In other words, U and V exchange its role. It means, that relative speed $O("U+V")$ and $O''("V+U")$ Has to be the same! Observers are equivalent, remember!</p> <p>So we may write:</p> <p>$$\frac{U+V}{1+UV \frac{1+d^{-2}(V)}{V^2}} = \frac{V+U}{1+VU\frac{1+d^{-2}(U)}{U^2}}$$</p> <p>After rearranging terms we obtain the following formula: $$\frac{1+d^{-2}(V)}{V^2} = \frac{1+d^{-2}(U)}{U^2}$$</p> <p>There is only one way to fulfill the following equation: $f(x)=f(y)$. $f(x)$ has to be constant! So the whole fraction.</p> <p>We obtain the final result: $$\frac{1+d^{-2}(V)}{V^2} = C$$ where from dimensional analysis cames that constant C has dimension of 1/ velocity^2. It has to be universal constant assuming validity of GP, and strict Galilean physics is restored when C is equal to 0.</p> <p>Function $d(V)$ is as follows: $$d(V)=\frac{1}{\sqrt{(1-CV^2)}}$$</p> <p>In this way we obtain Lorentz transformations assuming GP and following general rules with linear relationship between three inertial observers. </p> <p>If we want to know value of ( unknown yet) constant C we should perform various experiments. But we may notice, that Lorenz transformations leaves Maxwell equations unchanged, and it gives us relation between our constant C and speed of light in vacuum: $$C=\frac{1}{c^2}$$</p> <p>Reasoning above was performed ( and published in 60'ties) by polish physicist Andrzej Szymacha.</p>