Killtech Posted November 4 Share Posted November 4 The concept of relativity was quite the revolution for physics back in the day. But is it really something entirely new without any classical analoge? What i am struggling with is that Lorentz invariance is not particularly special to light/Maxwell, but is a rather a basic property of any linear wave equations. Let's consider a simple old classical model of an ideal uniformly distributed gas at rest in the lab frame and let's have a look at the acoustic wave equations we have for that case. Within this approximation the equation is a linear 2nd order PDE. Let's keep our minimal physical model limited to acoustics alone for now. Formally looking at the equation it becomes immediate clear that it must technically be also invariant under Lorentz group of \(c_s\) - though in this case it is the speed of sound and not of light. So if we were to use \(c_s\)-Lorentz transformed coordinates, the equation will always maintain it's form. While that may be unintuitive to mix time and space coordinates in such a scenario, we can embrace it and see where that leads us to. Note that any physical theory needs an interpretation, that translates between objects of the theory and real objects so it can be tested in experiments. Locations of events and distances in between them are a fundament of that and units are the means to express these. Since we restricted ourselves to acoustic physics only, it actually significantly limits our options to come up with a way to measure time. Remember that in order to build any kind of clock at all, we need some kind of physical oscillator as a reference available both in theory and reality. Let's look at the specification of a geodesic clock: Spoiler On 9/24/2023 at 7:21 PM, Genady said: The following theoretical construction perhaps contains the answer to your question: (let me know if you want to read the rest of this Box - it does not add anything to the construction itself) The concept is based on using bouncing light signals to construct an oscillator that serves as a clock. This concept is fully adaptable to use acoustic signals/waves. In reality it might be somewhat difficult to find a structure that reflects sound but allows the medium to pass through it unhindered, however we can practically calculate accurately what that ideal clock will measure and simulate it by means that are easier to implement. (we don't use geodesic clocks for light either but also find clocks that behave equivalently to the theoretical ideal). Let's now turn to define a measure of spatial distance. Analoge to the ideas of the SI system and considering that we have defined a unit of time, we can now use that to define the unit length as the distance an acoustic signal travels in one unit of time (times the \(c_s\) constant). Apparently those definitions now mathematically guarantee that the speed of sound must be perfectly constant, no matter what. In fact, analogue to SI system we can just define its value at will as it cannot logically even have a measurement uncertainty when expressed in those new units - as implied by construction. Yet of course the speed of sound isn't really constant, not even in an ideal gas, but be aware that this constancy is just a representation trick from a special kind of transformation and not a physical effect: the clocks and lengths we are using to express it behave very differently from their SI counterparts. As this point, we have a minimalistic \(c_s\) Lorentz invariant model of acoustic physics. It won't do any different predictions to the original classic model, if we correctly transform from the new units of time and length to their correspondence in the SI system - and note that it isn't just a simple unit trafo but a far more involved local transformation because with the exchange of the metric we have also exchanged the physical geometry from Euclidean to Minkowski which is accompanied by a reshaping of the laws of physics. In this very simple model, we can now observe that most of Einstein's gedankenexperiments conducted using light signals, work quite analoge with acoustic signals, if we chose to represent them in a structurally similar framework. In fact, if we take the perspective of a bat, a creature perceiving its environment through acoustics rather then optics, it becomes somewhat natural. Now, let's consider that the speed of sound is not actually constant because even for an ideal gas it depends on the gas pressure. This will modify the wave equations by adding the refractive index \(c_s(t,x)\) to the medium that will differ locally. For example gravity induces a gradient onto the gas pressure and consequently this will cause acoustic waves to be be bent by it. However, with the definitions of the new units we won't be able to observe the change in the speed of sound directly. Instead our definitions have made the path an acoustic signal takes a null geodesic of the new geometry. So consequently, the induced pressure gradient causes instead the new metric to deviate from the flat Minkowski space and curvature appears while the wave equation maintains its original form. The predicted effect how an wave is bent by gravity pressure will be the same in either description. Finally, one can mention that the acoustic waves discussed here are pressure waves, and because the speed of sounds dependence on the local pressure, two sound waves passing through each other actually weakly interact. Therefore a more accurate model of acoustic waves is a non-linear PDE. The point of this lengthy post is put up the question, how much of relativity is indeed special and what part of it comes from physics and which is mathematics and representation. Of course i selected only very specific aspects which show similarities and there are obviously very significant differences between acoustics and light signals in GR in general, yet we should be careful to account every difference to physics alone. Link to comment Share on other sites More sharing options...

Sensei Posted November 4 Share Posted November 4 (edited) 27 minutes ago, Killtech said: The concept of relativity was quite the revolution for physics back in the day. Newton's physics is relative.. Einstein's physics is special relative.. Edited November 4 by Sensei Link to comment Share on other sites More sharing options...

MigL Posted November 4 Share Posted November 4 Relativity has been around since Galileo ( at least ) and it doesn't need a classical analog; it IS classical. Link to comment Share on other sites More sharing options...

studiot Posted November 4 Share Posted November 4 11 hours ago, Killtech said: Since we restricted ourselves to acoustic physics only Sound waves do not obey Lorenz transformations. This is the key result implied but not explicitly stated (you have to think about it to find it) in the first postulate in Einstein's theory of relativity. 1 Link to comment Share on other sites More sharing options...

Killtech Posted November 4 Author Share Posted November 4 12 minutes ago, studiot said: Sound waves do not obey Lorenz transformations. This is the key result implied but not explicitly stated (you have to think about it to find it) in the first postulate in Einstein's theory of relativity. Oh, then i leave it open to you to prove how the linearized 2nd order PDE acoustic wave equations is mathematically not Lorentz invariant to the corresponding Lorentz group. You have to read Einsteins postulates more carefully and notice how there is no postulate about what clocks and rods have to be used. In fact that is left entirely open and instead those are constructed right from the postulates themselves. The construction i quoted does exactly that and as one can see, it works quite the same when applied to another wave signal. The reason is that Einsteins first postulate can be safely assumed for any simple physical model of waves when there is no other physics (particularly no other wave equations with a different propagation speed) that serves as a reference. That is why Einstein postulates in fact work for a much wider range of various wave equations when they are treated in isolation from other physics and if you construct the clocks and rods implied by treating the wave signals as null geodesics of some wave-specific geometry. Through the construction of proper clocks and rods, any wave equation becomes Lorentz-invariant even in general non-linear case. You can just always find a geometry which assures that. Link to comment Share on other sites More sharing options...

studiot Posted November 4 Share Posted November 4 (edited) 1 hour ago, Killtech said: Oh, then i leave it open to you to prove how the linearized 2nd order PDE acoustic wave equations is mathematically not Lorentz invariant to the corresponding Lorentz group. You have to read Einsteins postulates more carefully and notice how there is no postulate about what clocks and rods have to be used. In fact that is left entirely open and instead those are constructed right from the postulates themselves. The construction i quoted does exactly that and as one can see, it works quite the same when applied to another wave signal. The reason is that Einsteins first postulate can be safely assumed for any simple physical model of waves when there is no other physics (particularly no other wave equations with a different propagation speed) that serves as a reference. That is why Einstein postulates in fact work for a much wider range of various wave equations when they are treated in isolation from other physics and if you construct the clocks and rods implied by treating the wave signals as null geodesics of some wave-specific geometry. Through the construction of proper clocks and rods, any wave equation becomes Lorentz-invariant even in general non-linear case. You can just always find a geometry which assures that. OK consider this experiment: There is a flagpole 1000m from where I am standing with a sound pulse apparatus. My assistant is standing by the flagpole with timing apparatus. At the beginning of the experiment all is calm and still. Then storm Ciaran arrives and there is a wind directly towards the flagpole at 45 m/s. Calculate the time of transit of my sound pulse to my assistant in the cases of still air and of the storm wind and explain how this difference exhibits Lorenz invariance. Edited November 4 by studiot Link to comment Share on other sites More sharing options...

Markus Hanke Posted November 4 Share Posted November 4 13 hours ago, Killtech said: Now, let's consider that the speed of sound is not actually constant Neither is the speed of light - it explicitly depends on the permittivity and permeability of whatever medium it travels through. But that is irrelevant, since SR is only about its invariance, not any notion of constancy. Link to comment Share on other sites More sharing options...

Killtech Posted November 4 Author Share Posted November 4 (edited) 57 minutes ago, studiot said: OK consider this experiment: i gave you the task to apply Lorentz trafos to the raw equations and verify that it stays invariant. Lorentz trafos are merely coordinate trafos and there is no physics involved in applying them - its just pure and simple math. 57 minutes ago, studiot said: There is a flagpole 1000m from where I am standing with a sound pulse apparatus. you are using SI lengths and therefore a different geometry. The analogy works only if you construct the units of time and length in an analogue way via acoustic signals. 57 minutes ago, studiot said: My assistant is standing by the flagpole with timing apparatus. At the beginning of the experiment all is calm and still. Then storm Ciaran arrives and there is a wind directly towards the flagpole at 45 m/s. Calculate the time of transit of my sound pulse to my assistant in the cases of still air and of the storm wind and explain how this difference exhibits Lorenz invariance. Now we have to translate everything into a different geometry first. the distance between the flag poles has to be measured first and more importantly sonic means only. So we measure how much time sound takes to go from one flag post to another and back. In calm weather that will take roughly 6 seconds. in windy conditions it's 2.61s in one direction and 3.53 in the other, so 6.04s in total - so it would seem that flags have moved away from each other. However, we still measured time with an SI clock and have to account that a geodesic sonic clock tick rate will be slightly slowed down by the windy conditions. In fact, a geodesic clocks works quite the same setup as your flags setup and thus we can calculate that it is slowed by a factor of 1.0067. Appling that to the sonic distance, we yield that in fact the flag poles did not move away from each other in soncic-meters as well. If we now also translate the SI seconds into acoustic seconds in either case, we yield that in both times it took the acoustic signal 6 sonic-seconds to go back and forth, independently of the wind. So in terms of sonic proper time, there effect of the wind is irrelevant. Tada, the magic of frame and location dependent units in relativity. we can also reverse the situation and make a similar experiment with light. instead of the calm weather take put one flag experiment into the rest frame of the barycentric coordinates, and the other in a frame moving with a given speed relative to that. Now we are stubborn and instead of providing the results of the experiments in time in proper time of the specific frame, we do it as IAU would do and give the result in TDB coordinate time. We now realize that in that time reference it took light a different amount of time to get back and forth. Edited November 4 by Killtech Link to comment Share on other sites More sharing options...

Sensei Posted November 4 Share Posted November 4 (edited) The earth is rotating. The emitter rotates with the Earth. The receiver also rotates. With a home-made apparatus, no serious experiment can be conducted. Photons emitted by the emitter, travel a curved path to reach the receiver, while the Earth rotates? The distances are too small to make measurements.. All the experiments with "speed of sound" are useless.. The field of view is only 5 km near the ground. https://www.google.com/search?q=how+many+kilometers+you+can+see+sea+horizon At the speed of light, it takes 42.5 ms for a particle to pass through the center of the Earth (12,700 km). Sound - one particle hits another particle, which hits another particle, and so on. Imagine that the Earth is rotating.. 20 minutes ago, studiot said: Calculate the time of transit of my sound pulse to my assistant in the cases of still air and of the storm wind and explain how this difference exhibits Lorenz invariance. But his name was Lorentz.. https://en.wikipedia.org/wiki/Hendrik_Lorentz Edited November 4 by Sensei 1 Link to comment Share on other sites More sharing options...

studiot Posted November 4 Share Posted November 4 3 hours ago, Sensei said: But his name was Lorentz.. What would I do without other members to correct my atrocious spellung ? Thans +1 3 hours ago, Killtech said: i gave you the task to apply Lorentz trafos to the raw equations and verify that it stays invariant. Lorentz trafos are merely coordinate trafos and there is no physics involved in applying them - its just pure and simple math. you are using SI lengths and therefore a different geometry. The analogy works only if you construct the units of time and length in an analogue way via acoustic signals. Now we have to translate everything into a different geometry first. the distance between the flag poles has to be measured first and more importantly sonic means only. So we measure how much time sound takes to go from one flag post to another and back. In calm weather that will take roughly 6 seconds. in windy conditions it's 2.61s in one direction and 3.53 in the other, so 6.04s in total - so it would seem that flags have moved away from each other. However, we still measured time with an SI clock and have to account that a geodesic sonic clock tick rate will be slightly slowed down by the windy conditions. In fact, a geodesic clocks works quite the same setup as your flags setup and thus we can calculate that it is slowed by a factor of 1.0067. Appling that to the sonic distance, we yield that in fact the flag poles did not move away from each other in soncic-meters as well. If we now also translate the SI seconds into acoustic seconds in either case, we yield that in both times it took the acoustic signal 6 sonic-seconds to go back and forth, independently of the wind. So in terms of sonic proper time, there effect of the wind is irrelevant. Tada, the magic of frame and location dependent units in relativity. we can also reverse the situation and make a similar experiment with light. instead of the calm weather take put one flag experiment into the rest frame of the barycentric coordinates, and the other in a frame moving with a given speed relative to that. Now we are stubborn and instead of providing the results of the experiments in time in proper time of the specific frame, we do it as IAU would do and give the result in TDB coordinate time. We now realize that in that time reference it took light a different amount of time to get back and forth. Wow. And here am I thinks this is easy Physics. It is easy to find the speed of sound in still air (ignoring temperature effects) as 346 m/s So in the ground frame both source and observer are stationary as is the transmission medium. So the time of flight is simply 1000/346 = 2.890 seconds. In the case including the wind, the wind carries the sound pulse forward at an additional 45 m/s so this pulse reaches the observer at 1000 / (346 + 45) = 2.558 seconds. Now a transformation is to another coordinate system and the only one that makes sense here to use is the coordinate system comoving with the wind. Here we have that the observer appears to be approaching the source at 45m/s but the sound is moving at 346m/s in the frame of the wind. So the sound will appear to travel the distance (already measured as 1000m in both systems) in 1000 / (346 +45) = 2.558 seconds. How is that a Lorentz transformation from the ground frame to the wind frame ? Link to comment Share on other sites More sharing options...

Sensei Posted November 4 Share Posted November 4 Just now, studiot said: What would I do without other members to correct my atrocious spellung ? Thans +1 ..let me guess - you would turn on the language checker in your web browser...? Just now, studiot said: And here am I thinks this is easy Physics. ..I think it is different here.... It is easy for a layman.... and difficult for a non-layman.... ps. All this stuff, theoretical experiments, at the speed of sound, are not enough accurate.. Like I said, the Earth, rotates.. Link to comment Share on other sites More sharing options...

Boltzmannbrain Posted November 4 Share Posted November 4 18 hours ago, Killtech said: Note that any physical theory needs an interpretation, that translates between objects of the theory and real objects so it can be tested in experiments. I have good news and bad news. The bad news first: you do not understand Einstein's theory of general relativity. The good news is that if you read and understand what the posters are saying, you will see how different the theory is from classical mechanics. Link to comment Share on other sites More sharing options...

Killtech Posted November 4 Author Share Posted November 4 48 minutes ago, studiot said: Wow. And here am I thinks this is easy Physics. It is easy to find the speed of sound in still air (ignoring temperature effects) as 346 m/s So in the ground frame both source and observer are stationary as is the transmission medium. So the time of flight is simply 1000/346 = 2.890 seconds. In the case including the wind, the wind carries the sound pulse forward at an additional 45 m/s so this pulse reaches the observer at 1000 / (346 + 45) = 2.558 seconds. Now a transformation is to another coordinate system and the only one that makes sense here to use is the coordinate system comoving with the wind. Here we have that the observer appears to be approaching the source at 45m/s but the sound is moving at 346m/s in the frame of the wind. So the sound will appear to travel the distance (already measured as 1000m in both systems) in 1000 / (346 +45) = 2.558 seconds. How is that a Lorentz transformation from the ground frame to the wind frame ? For me it would make more sense to stick to the frame of the flags in either case, but we can also switch to the wind frame. Now let's make things a bit simpler and choose a medium such that the speed of sound is \(c_{s}=\frac{1\text{000}}{3}\frac{m}{s} \) in SI. Let's assume the wind speed in the scenario 2 is at \( v=\frac{150}{3}\frac{m}{s} \). Let's remember the linear acoustic wave equations is \( \triangle\boldsymbol{p}-c_{s}^{-2}\frac{\partial\boldsymbol{p}}{\partial t^{2}}=0 \) in the rest frame of the medium. Let us now do an acoustic Lorentz transformation of that equation into a frame that moves with \(v\) relative to the medium (i.e. where we would feel the wind). so therefore we have \( \beta=\frac{v}{c_{s}}=\frac{150}{1000}=0.6 \) and \(\gamma=1.25 \). It may be intuitive to use Galilean transformed coordinates for such low speeds, but let's do choose the crazy acoustic Lorentz coordinates instead with \(x'=\gamma(x-vt)\) and \(t'=\gamma(t-\frac{vx}{c_{s}^{2}}) \). Now rewriting the acoustic wave equation into the new exotic acoustic coordinates yields \( \triangle'\boldsymbol{p}(\boldsymbol{x}',t')-c_{s}^{-2}\frac{\partial\boldsymbol{p}(\boldsymbol{x}',t')}{\partial t^{'2}}=0 \). So in fact we find that if we use these type of coordinates, it becomes entirely irrelevant if there is any wind present or not. That's the whole point of Lorentz trafos as back in the day Poincaré and others studying Lorentz aether theory observed. However, we changed to a frame where now the flags are moving relative to our frame, therefore we have to account for that and within the new coordinates the (coordinate) distance between them contracts by the factor \( \gamma^{-1} = 0.8\). In the 2nd scenario we can assume that we were actually using exotic acoustic Lorentz coordinates for the flag frame, since those assure us that we can use the wave equations just so as if there was no wind present. These are very different from what we would classically use to discuss the problem, specifically mixing time and space at very low velocities already. Transforming to the wind frame by an acoustic Lorentz trafo will just reverse the trafo we had to use to get into the flags rest frame, so it brings us back to the normal classical coordinates of the problem. When there is no wind and we are in the rest frame of the medium the coordinates agree with our usual choice and the waves take \(\Delta t=\frac{2\cdot 1000}{c_{s}}=6\) where \(\Delta t\) is in coordinate time \(t\). If there is wind we can do the same and yield the same result, however \(\Delta t'\) is now provided in the quite different coordinate time \(t'\) compared to the calm weather case. Keep in mind that even if we are in the same frame, we use different coordinates to remove the wind from the equation. Now let's consider transforming to the frame of the wind. While the distance between the flags are contracted, the flags are still moving so the signal has to cover a distance of \(L\gamma^{-1}+v\Delta t_{1}\) in one direction and \(L\gamma^{-1}-v\Delta t_{2}\) in the other. So these new coordinates make it quite complicated to do the calculation in the rest frame of the medium. Link to comment Share on other sites More sharing options...

studiot Posted November 4 Share Posted November 4 You seem to have forgotten my comment from Einstein. 8 hours ago, studiot said: This is the key result implied but not explicitly stated (you have to think about it to find it) in the first postulate in Einstein's theory of relativity. I see I owe you an apology here: it is actually the second postulate. Quote Einstein : On the Electrodynamics of Moving Bodies ..And also introduces another postulate, which is only apparantly irreconcilable with the former, namely that light is always propagated in empty space with a definite velocity c, which is independent of the state of motion of the emitting body. What he does not describe is the effect of motion of any transmission medium, which is different from other waves where this motion has to be taken into account. This is the key difference with light (in empty space). It has no transmission medium. So c will be invariant in all frames. there is nothing to add to it. An acoustic wave between A and B however, depends also on the motion of the transmission medium. Link to comment Share on other sites More sharing options...

Killtech Posted November 4 Author Share Posted November 4 12 minutes ago, studiot said: What he does not describe is the effect of motion of any transmission medium, which is different from other waves where this motion has to be taken into account. This is the key difference with light (in empty space). It has no transmission medium. So c will be invariant in all frames. there is nothing to add to it. An acoustic wave between A and B however, depends also on the motion of the transmission medium. Then let's account the version of Lorentz Aether Theory as corrected by Poincaré and which in that final form is equivalent to SR - which was developed from it. In that scenario, light in vacuum is assumed to move through a medium, the aether. Therefore it is almost the same as for acoustic waves. In that theory, there can be an aether wind in vacuum just as in our acoustic case. It still turns our that it does not matter because using a special type of coordinates, the wind can be transformed away, just as it can be done for acoustics. However, that invariance to the wind holds only in specially selected coordinates of time and space. Lorentz and Poincaré have shown that even if light was a wave propagating through a luminuferious aether, it would still have the exact same physical behavior that it has in SR. The Michelson Morley experiment does yield a null result in LAT just the same. The big jump from LAT to Einsteins SR is the declaration that these special coordinate times and spatial distances are not just an obscure mathematical transformation but the actual time observed by some clocks and actual distances observed by some rods. Einstein postulates of relativity translate into just that. Maybe Poincarés thoughts on the subject can help you understand what Einstein postulates do: https://en.wikipedia.org/wiki/Lorentz_ether_theory#Principles_and_conventions if you do want to argue that the wind cannot be transformed away in acoustics by proper coordinates, i would encourage you to try to show that for the transformation i used in my post before. Link to comment Share on other sites More sharing options...

studiot Posted November 4 Share Posted November 4 8 hours ago, Killtech said: Now we have to translate everything into a different geometry first. the distance between the flag poles has to be measured first and more importantly sonic means only. So we measure how much time sound takes to go from one flag post to another and back. In calm weather that will take roughly 6 seconds. in windy conditions it's 2.61s in one direction and 3.53 in the other, so 6.04s in total - so it would seem that flags have moved away from each other. This sound authoritative because it contains number. But when I analyse those numbers I find disturbing discrepancies. Firstly in calm weather the outward and return journey should be identical so exacly half of 'roughly 6'. In fact the double journey takes 2000 / 346 or 5.780 seconds. Now this double journey should cancel out any effect of medium motion so if I add up your to outward and return times I make 2.61 + 3.53 = 6.14, not 6.04 as you have. Either way this is quite different from my value. Furthermore adding my outward and return times 1000 / (346 + 45) + 1000 / (346 - 45) = 2.55754 + 3.32226 = 5.88 seconds. Comfortingly close to my undisturbed double journey value. Link to comment Share on other sites More sharing options...

Killtech Posted November 4 Author Share Posted November 4 11 minutes ago, studiot said: Furthermore adding my outward and return times 1000 / (346 + 45) + 1000 / (346 - 45) = 2.55754 + 3.32226 = 5.88 seconds. Comfortingly close to my undisturbed double journey value. 1000 / (333+50) = 2.611 1000 / (333-50) = 3.533 i don't see the value in using odd numbers for this particular example, so i rounded them for convenience. i did this also for the speed of sound by assuming a medium that has exactly that speed so the back and forth time is exactly 6. and yes, i slipped up at one digit when adding up the numbers, sorry. Link to comment Share on other sites More sharing options...

studiot Posted November 4 Share Posted November 4 (edited) 49 minutes ago, Killtech said: i don't see the value in using odd numbers for this particular example, so i rounded them for convenience. 2.611, 3.533 and 1.0067 are no odd numbers ? Pity if instead you didn't post the speed of sound you employed and the fact that you changed my figure of 45 m/s to 50 m/s. That would have saved me looking up the speed of sound on Google and accepting the first reasonable academic answer listed. Of course resulting in pointless exchanges of posts. You still haven't shown how your claim that this method of analysis (Which is classical ) supports a Lorentz transformation into any inertial frame I choose. I choose to work in the ground frame (as you did) and tranform this into the wind frame (as Arago et al did) and I arrive at the fact that the time of transit is dependent on the wind speed, as otheres did. If you think there is a Lorenz tranformation then it should work and give these figures for the transformation to the wind frame. So what is x' = ?(x, t) t' = ?(x, t) in your system ? For interest here is Professor Beiser's alternative and simpler explanation as to why other waves are not a good analog (which was my point). The behaviour of light waves is unique. Edited November 4 by studiot 1 Link to comment Share on other sites More sharing options...

Killtech Posted November 4 Author Share Posted November 4 (edited) 28 minutes ago, studiot said: Pity if instead you didn't post the speed of sound you employed and the fact that you changed my figure of 45 m/s to 50 m/s. That would have saved me looking up the speed of sound on Google and accepting the first reasonable academic answer listed. Of course resulting in pointless exchanges of posts. Sorry. Bad habit of a mathematician not to care about actual numbers, but rather how we can get to them. 28 minutes ago, studiot said: You still haven't shown how your claim that this method of analysis (Which is classical ) supports a Lorentz transformation into any inertial frame I choose. I choose to work in the ground frame (as you did) and tranform this into the wind frame (as Arago et al did) and I arrive at the fact that the time of transit is dependent on the wind speed, as otheres did. Let's first acknowledge that of course that treatment you quoted is entirely correct and so is the result. However, if we are willing to use a different concept of space time, there is an alternative approach to the same problem. 28 minutes ago, studiot said: If you think there is a Lorenz tranformation then it should work and give these figures for the transformation to the wind frame. So what is x' = ?(x, t) t' = ?(x, t) in your system ? i wrote it down explicitly in this post (i hope the latex expressions do survive the quoting. if not, please go to my quoted post [Edit: the quoting broke the "\frac" latex command]): 3 hours ago, Killtech said: Now let's make things a bit simpler and choose a medium such that the speed of sound is cs=10003ms in SI. Let's assume the wind speed in the scenario 2 is at v=1503ms . Let's remember the linear acoustic wave equations is △p−c−2s∂p∂t2=0 in the rest frame of the medium. Let us now do an acoustic Lorentz transformation of that equation into a frame that moves with v relative to the medium (i.e. where we would feel the wind). so therefore we have β=vcs=1501000=0.6 and γ=1.25 . It may be intuitive to use Galilean transformed coordinates for such low speeds, but let's do choose the crazy acoustic Lorentz coordinates instead with x′=γ(x−vt) and t′=γ(t−vxc2s) . Now rewriting the acoustic wave equation into the new exotic acoustic coordinates yields △′p(x′,t′)−c−2s∂p(x′,t′)∂t′2=0 Edited November 4 by Killtech Link to comment Share on other sites More sharing options...

Killtech Posted November 6 Author Share Posted November 6 (edited) from these answers i got here, i think i have a better idea how to rephrase my initial post to make the concept much clearer. It will require some knowledge of the historic development of SR and its predecessor LAT. On 11/4/2023 at 2:25 PM, Markus Hanke said: Neither is the speed of light - it explicitly depends on the permittivity and permeability of whatever medium it travels through. But that is irrelevant, since SR is only about its invariance, not any notion of constancy. of course i meant the vacuum case, where light does not have a medium. However, i was applying the analogy of relativity to sound in the same way as the historic predecessor to SR did, the Lorentz Aether theory, when light was still described as having a medium in vacuum, the luminoferious aether. Below is the discussion what acoustics are invariant to. On 11/5/2023 at 12:25 AM, studiot said: If you think there is a Lorenz tranformation then it should work and give these figures for the transformation to the wind frame. So let's rephrase the whole concept: Let's start again and proceed by baby steps at the risk of stating the obvious, but whatever. The linear approximated acoustic wave equation is \(\partial_{x}^{2}p-c^{-2}\partial_{t}^{2}p=0\) where \(c_{s}=\frac{1}{3}\frac{km}{s}\) is the speed of sound in the medium we use (close enough to air). This is of course the version with only one spatial dimension, but for simplicity it is enough for now. The equation should hold in the rest frame of the sonic medium. Let's compare this model with the historic start of SR, when light in vacuum was still modelled as traveling through a luminoferious aether medium. Apart from having an equation for a transversal wave instead of a longitudinal one, that model is quite similar to the acoustic case. And indeed a wave in a medium was what inspired Lorentz approach. However, the peeps of old found out quickly, that these type of equations are invariant under a special type of coordinate transformations, the Lorentz trafos. Coordinates are just a math tool and have not much to do with physics per se, so let's ask if we can play an analogue trick for acoustics. in fact a trafo like \(t'=\gamma_{s}(t-vc_{s}^{-2}x)\) and \(x'=\gamma(x-vt)\) with \(\gamma_{s}^{-2}=(1-\beta_{s}^{2})\) and \(\beta_{s}=vc_{s}^{-1}\) (note that here cs is the speed of sound!!) transforms the wave equation into \(\partial_{x'}^{2}p(x',t')-c^{-2}\partial_{t'}^{2}p(x',t')=0\) , (i don't think i need show this explicitly, since the change in variables works exactly the same as in relativity theory) i.e. sound remains invariant under these kind of coordinate changes. The \(c_s\) in that equation however isn't the real speed of sound anymore but an artificial coordinate speed of sound. More on that later. Let's call these acoustic Lorentz transformations, short A'rentz trafos. In order to give a little more life to this math curiosity, let's consider @studiot simple experiment where we have one observer and a wall on the ground 1km away from each other. The observer emits a sound wave, it is reflected by the wall and an echo returns to the observer. The time is measured how long it takes to go there and back again... not the hobbit, just his sound. This setup will be discussed under two situations, in calm weather with no wind speed and in windy conditions with a speed of \(v=\frac{1}{5}\frac{km}{s}\) For reference, let's do it classically first, that is in calm weather it takes the echo \(\Delta t=c_{s}^{-1}\cdot(1km+1km)=6s\) to get back. When there is wind it's \(\Delta t=(c_{s}+v)^{-1}1km+(c_{s}-v)^{-1}1km)=(\frac{15}{8}+\frac{15}{2})s=\frac{15+60}{8}=9.375s\) Now let's do the wind case differently. Exploiting the A'rentz invariance of sound, we can start in the wind frame (rest frame of the medium) where we have the wave equation for and just transform to the ground/observer frame, right? For that we get \(\beta_{s}=vc_{s}^{-1}=\frac{3}{5}\) and \(\gamma^{-2}=(1-\frac{9}{25})=\frac{16}{\text{25}}\), thus \(\gamma_{s}=\frac{5}{3}\). Appling that ensures the equation keeps its form same as in the calm weather case. Unfortunately, there is a bit more to do, because if the distance between observer and wall in the wind frame was \(\Delta x=1km\) and moving with \(v\), then the new coordinates, which are its rest frame, will therefore undo an A'rentz length contraction, so \(\Delta x'=1\gamma_{s}=\frac{5}{3}\) and hence the signal will need a travel time of \(\Delta t'=2\Delta x'c_{s}^{-1}\). However, notice that the time is given in the coordinate time \(t'\) and not \(t\) and therefore we have to transform it back, thus \(\Delta t=\gamma_{s}\Delta t'=2\gamma_{s}^{2}c_{s}^{-1}=\frac{2\cdot 25\cdot 3}{16}=\frac{75}{8}=9.375s\). I think I best to stop at this simple point for now, so we can check that we all agree that these calculations are correct, and importantly that both the classic calculation and the acoustic analog of the relativistic formalism do yield the same result. Furthermore let's observe that the Lorentz invariance allows to remove the rest frame of the medium from the calculation by a proper choice of coordinates for any simple wave equations, not just for light and its aether. More on that later. ps if i gonna do the latex here, i need a working preview of the formulas i write. HOW DO I GET THAT?!? Edited November 6 by Killtech Link to comment Share on other sites More sharing options...

Killtech Posted November 13 Author Share Posted November 13 (edited) Help me out a bit. The post above demonstrates how one can remove the wind from the equation using the Lorentz formalism. Going further to the case of a medium with a refractive index, that is if we have a sound equation like \(\partial_{x}^{2}p-c(x)^{-2}\partial_{t}^{2}p=0\) with non-constant speed of sound \(c(x)\). I want to find coordinates such that it reverts back to the previous case where c was constant. Looking at the known solutions for that case i figured a coordinate trafo like \(t'\rightarrow t+T(x)\) will do the trick where the additional term fulfils \((\nabla T)^{2}=n(x)^{-2}\) the eikonal equation with \(n(x)=\frac{c(x)}{c_{0}}\) the refractive index. And it almost works but doing the change of variables i get an additional term \(\partial_{x}^{2}T\partial_{x}p\) when doing the second derivative of \(\frac{\partial p}{\partial x}=\frac{\partial t'}{\partial x}\frac{\partial p}{\partial t'}\) due to the product rule. So instead i was thinking to define the spatial coordinate \(x'\) in term of the rays \(x'(s)\) implied the the eikonal equation, that is \(\frac{d}{ds}n\frac{dx'}{ds}=\nabla n\). Such coordinates will only work locally, since due to the possibility of lensing effects initially parallel rays may intersect. The idea is that \(x'\) resembles the shape of null-geodesic in GR, i.e. it is intended as an analog to geodesic coordinate. Anyhow, written in terms of coordinates that follow the rays of the wavefronts, i don't see how the wave equation itself could look any different then in the trivial case. Edited November 13 by Killtech Link to comment Share on other sites More sharing options...

KJW Posted November 14 Share Posted November 14 On 11/4/2023 at 10:03 AM, Killtech said: What i am struggling with is that Lorentz invariance is not particularly special to light/Maxwell, but is a rather a basic property of any linear wave equations. Let's consider a simple old classical model of an ideal uniformly distributed gas at rest in the lab frame and let's have a look at the acoustic wave equations we have for that case. Within this approximation the equation is a linear 2nd order PDE. Let's keep our minimal physical model limited to acoustics alone for now. Formally looking at the equation it becomes immediate clear that it must technically be also invariant under Lorentz group of cs - though in this case it is the speed of sound and not of light. So if we were to use cs -Lorentz transformed coordinates, the equation will always maintain it's form. While that may be unintuitive to mix time and space coordinates in such a scenario, we can embrace it and see where that leads us to. I see two problems with this: 1: The wave equation for electromagnetism but with the speed of sound replacing the speed of light may not be the correct wave equation for describing the propagation of sound. In this wave equation there is no dependency on the velocity of the medium, whereas sound waves have a fixed velocity relative to the medium, which may differ from the velocity relative to the observer if the medium is in motion relative to the observer. 2: If the Lorentz transformation but with the speed of sound replacing the speed of light is applied to space and time coordinates, the resulting new coordinates are no longer space and time coordinates. Indeed, it is not obvious what these new coordinates represent. Note that it is Minkowskian spacetime that is invariant to Lorentz transformations, not just the electromagnetic wave equation, so that applying a Lorentz transformation to space and time coordinates results in new coordinates that are also space and time coordinates. Link to comment Share on other sites More sharing options...

studiot Posted November 14 Share Posted November 14 1 hour ago, KJW said: If the Lorentz transformation but with the speed of sound replacing the speed of light is applied to space and time coordinates, the resulting new coordinates are no longer space and time coordinates. ... so that applying a Lorentz transformation to space and time coordinates results in new coordinates that are also space and time coordinates. Perhaps you would like to rephrase this to resolve the apparent contradiction ? Link to comment Share on other sites More sharing options...

joigus Posted November 14 Share Posted November 14 So is it clear now that the sound equation is not Lorentz invariant, as v_{s} is not a universal constant, nor is it a Lorentz scalar, or are we still discussing that? Link to comment Share on other sites More sharing options...

KJW Posted November 14 Share Posted November 14 3 hours ago, KJW said: Note that it is Minkowskian spacetime that is invariant to Lorentz transformations, not just the electromagnetic wave equation, so that applying a Lorentz transformation to space and time coordinates results in new coordinates that are also space and time coordinates. In this case, actual Lorentz transformations (with the speed of light in a vacuum). Link to comment Share on other sites More sharing options...

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