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I Constancy of the speed of light

  1. Sep 2, 2018 #1
    I am trying to get a better understanding about the constancy of the speed of light which is a well-established axiom of current day physics. for the start i want to understand how it is experimentally established and how these results are interpreted. My difficulty here is that this seems to be far less trivial then for other quantities given the definitions of the units/metrics the speed is measured in are themselves strongly dependent on that very quantity which makes for a weird interdependence that creates circles in my mind.

    My understanding is that one must at least hypothetically allow the possibility of ##c## not being constant in order to verify or falsify it with an experiment. But doing so implies that the current definition of the SI meter cannot be assumed to be an ‘absolute’ measure of length since it directly depends on the speed of light. Any change to it directly changes the length of the meter. Of course this definition was chosen after the constancy was well established thus in any case one needs to use a different one instead. But here comes my problem: every other definition has a strong interdependence with the speed of light albeit the type of these dependencies vary and is not clear to me. The bigger problem is that I cannot find the right wording to google an answer for myself.

    The reason why I don’t think this is a triviality is that take for example the old metric definition via the pre-SI prototype meter bar. It is a solid state and thus in rough terms a finite grid of a constant number of atoms which total size defined the meter. Properties of the grid like distances between vertices are itself mainly determined by the size of atoms it is composed of which in turn depends on the EM interaction between the electron shell and the nucleus. And since ##c## is the propagation speed of that very force it is natural to assume all atomic properties and states will be gravely affected in one way or another and with it most definitions of the two fundamental SI units (since the definitions of the second are also based on atomic states). Apart from the microscopic effects anything changing Bohr’s radius should also shrink or stretch the entire atomic grid proportionally and therefore the meter bar as a whole (or actually anything made out of atoms).

    My first naïve approach to get an understanding of the impact on atoms by an altered ##c## was that the quantum mechanical solution for Hydrogen is easily reapplied for c-modified Maxwell equations and results in the Bohr radius scaling inversely to the speed of light – i.e. a lower ##c## weakens the electric field/energy of photos and thus atom sizes grow. However this approach brings a lot of other constants into play that complicate things. For example given the relations of Planck’s quantum to the photon energy I can’t find a reason that would guarantee it remaining constant in this hypothetical circumstance. One could argue that a photons energy should reduce with ##c## under the constraint of constant frequency and if for example it scaled proportionally then atoms would shrink just in a way that both definitions of the meter (current SI and prototype bar) would remain equivalent. In a similar way definitions via wavelengths of specific atomic levels might be compromised in the same way leading to the possibility of all metric definitions (know to me) remaining equivalent in all circumstances and scale with ##c##.

    In that case no direct measurement of the speed of light would ever be able to find a different value regardless. Thus if this scenario cannot be ruled out it would require a very different approach to proof or disproof the constancy of ##c##. Or putting it the other way around: if ##c## would actually vary locally how would we experimentally detect it assuming that matter would be affected the same way as described in the scenario above?
  2. jcsd
  3. Sep 2, 2018 #2

    Mister T

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    The definition implies that meter sticks can be built in different places and at different times. They will be identical to each other, within the limits of precision of the construction techniques, only if the speed of light is the same every time and every place that anyone uses light to build one of those meter sticks.

    If one were ever to find a difference between the lengths of these meter sticks, a difference outside the bounds of the construction errors that is, it would constitute a way to falsify the principle that the speed of light is constant.
  4. Sep 2, 2018 #3


    Staff: Mentor

    The first thing you need to understand is that "constancy of the speed of light" is the wrong way to look at it. The speed of light is not a dimensionless number. The relevant dimensionless number for electromagnetism is the fine structure constant, so when you ask about the possibility of the speed of light changing, what you should really be asking about is the possibility of the fine structure constant changing.

    Since the fine structure constant affects all electromagnetic phenomena, it would affect not only the propagation of light in free space, but the interactions that determine things like the sizes of atoms. So working out the consequences of changes in the fine structure constant is not a trivial exercise. Particularly if you consider experiments such as you describe, to measure the speed of light by, for example, using a standardized meter stick (or something like that) as the measure of distance (and, for that matter, a clock whose operation probably depends on the properties of electromagnetism as the measure of time).

    Before even trying to work out for yourself how we might test such things experimentally, you should take the time to learn what has already been done, theoretically and experimentally.
  5. Sep 2, 2018 #4


    Staff: Mentor

  6. Sep 2, 2018 #5
    thanks for the link. in there i just found the part i am trying to understand but in a more general context because all metrological definitions could exhibit a similar problem given that their definition indirectly depends ##c##:


    the letter sentence is what i was trying with my original post.

    but more generally take for example the Michelson-Morley experiment for the two way isotropy of light-speed. the concept of the ether wind effectively elongating the way of light orthogonal to the wind thus creating a time difference in the arrival of orthogonal light pulses in the interferometer is a convincing method. but it is build on the premise that the interferometer itself is not affected by the ether. because for example the atoms the interferometer is composed of could take an elliptical shape in the wind elongating its one axis while squeezing the other. this change in size of the interferometer could effectively undo any time difference of the light pulses. therefore in this hypothetical scenario the experiment would always conclude a negative result regardless if an ether wind is present or not.

    basically such kind of test would require to be done with means and method that are ensured to not exhibit an identical non-isotropy.

    on the other hand this also shows a mathematical degree of freedom: using the aforementioned pre-SI meter bar as a definition of length for the MM-experiment would make the interferometer size constant by definition and therefore imply the isotropy of light mathematically - at least in a physical model based on such metrology - regardless of whether light is assumed to be isotropic or not in this though experiment. and i am not sure if such a model could even be falsified in any way as it could be mathematically equivalent and produce identical physical predictions - if its metrology satisfies the mathematical requirements of a metric i think.

    in order to measure a difference you need a metrology. and what do you do if that metrology might be such that it exactly undoes that difference (in a similar way as in the MM example above)? and what if all your alternate metrologies are equivalent to your original one?

    a metrology that is neither directly nor implicitly bound to the speed of light is hard to come by.
    Last edited: Sep 2, 2018
  7. Sep 2, 2018 #6

    Mister T

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    But you do not need a metrology to find a difference. All you have to do is compare.
  8. Sep 2, 2018 #7
    and how do you compare? lets assume you have bars that change their length depending on the part of space they are in. to check if they are of same length you need to measure them. or you could of course move them directly next to each other and compare them that way but this undoes the difference you are trying to detect. alternatively you could compare their length to something else. but that requires that something else does not change depending on its location - or at least it must not do so consistently to your bars.
  9. Sep 2, 2018 #8


    Staff: Mentor

    Yes, length contraction is indeed the explanation.
  10. Sep 2, 2018 #9


    Staff: Mentor

    It seem to me that what you are realizing is that dimensionful values are all a matter of convention. I define the length of a meter such that by definition c is constant, and by definition a meter here is the same as a meter there. I am free to do that because these are dimensionful quantities that can be arbitrarily set by a human or a committee of humans. There is, as you pointed out, no way to test it independently of our conventions. The convention defines the outcome.

    So the resolution of the problem is to look at dimensionless physical constants instead. In this case the fine structure constant. If it varies from time to time or from place to place then that could be physically detected independently of your choice of units.
  11. Sep 3, 2018 #10

    Mister T

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    That would be an interesting thing to consider. But in your original post you asked us instead to consider whether or not the speed of light is constant. If the speed of light is different in one place than in another, then they won't be the same length.
  12. Sep 5, 2018 #11
    Since the second is defined by cesium, not c or the meter, maybe OP should move the goal post there? Of course all units are arbitray, which is why the real solution is the fine structure constant.

    Also, why would length contraction “hide true results” with respect to the constancy of the speed of light? How can you have one without the other? Correct me if I’m wrong, but it appears that length contraction (and time dilation) is exactly WHY c is constant, as much as c being consant is why length contraction and time dialtion exist.

    It seems to be an if and only if relationship.

    So why would length contraction “conspire” to give the wrong answer about c being constant? You can’t have a finite and constant universal speed limit without length and time being relative, and you can’t have length and time being relative without a finite universal speed limit (which every inertial frame agrees on). Or so it seems to me.

    Of course I could be wrong but how else can it be so in our universe where you can’t move two directions in time and where location and direction don’t change physical laws? You can work out the math for general transformations, and you’re left with the one form of transformation equations, with either an infinite universal speed limit (Galileo) or a finite one (Lorentz).

    As the universal speed limit appears to be finite, Lorentz is the correct transform, and obviously imbedded in it is length contraction and time dilation.

    Am I off base here?
  13. Sep 5, 2018 #12
    well, when looking at it alone then yes, the fine structure constant would be what i should be looking at. but that wasn't exactly my question. i am more wondering how much each of the metrological definitions already determines the result. for the current SI definition it obvious but for the older ones its not and other constants come into play if one wants to understand their interrelation with the speed of light. when looking at those it made most sense to me to consider the light-speed changing coupled with other constants strongly associated with the EM-field - everything else just seemed too odd and artificial. but looking at the fine structure written in terms of those constants the changes would then would just cancel each other out. so this, as pointed out in the discussion above, about convention and definition and to which degree the constancy of light-speed is determined on those.

    the second is lastly defined in terms of atomic states which are defined by electromagnetic interaction for which again ##c## is a fundamental quantity, along Planck's and the two electromagnetic constants. so the situation is the same as with the pre SI-meter bar definition which i mentioned in my original post.

    hmm, i think you misunderstood me here a bit. there are no "true" results to hide or conspire against. what bugs me here is that these basic metrological definitions are just set up in a way that due to their explicit or implicit dependency on ##c## they don't allow it to change one way or the another. my line of thought was that even if i consider the speed of light changing (in a specific way coupled with other constants) these definitions would just adapt accordingly and undo that change; the convention determines the result - this was however more of a technical consideration to understand how their interdependence works rather then an assumption about a physical "truth". but as metrology goes and as it was stated in this thread it is just a human chosen convention and in a sense 'arbitrary' without implications about physical "truths". they change terminology but not predictions. therefore they can only be judged in terms of practicality rather then being right or wrong.

    an interesting question would be though whether any of the statements holds if we were to use an entirely different metrology.
  14. Sep 5, 2018 #13


    Staff: Mentor

    They would all hold.
  15. Sep 5, 2018 #14

    Mister T

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    Your concern is misplaced. I already explained to you that if ##c## varied with location, for example, a meter stick manufactured over here would have a different length than one manufactured over there. Such a difference could be observed by bringing the two sticks together. Your response was to point out that it's possible that the length of the sticks might change when you move them, and do so in such a way that we are always fooled into thinking that ##c## is the same everywhere. The thing is, if such a thing were happening is there any experiment you can think of doing that might demonstrate that this is happening?

    If not, you don't have different physics. You have instead a different interpretation of the same physics.

    I kind of lost track of the statements, but changing the metrology doesn't change the physics.
  16. Sep 5, 2018 #15


    Staff: Mentor

    You're over-counting; the two electromagnetic constants and ##c## are not independent of each other, they are really only one constant; and as has already been noted, the actual electromagnetic constant is the fine structure constant, since it is dimensionless; ##c## and ##\mu_0## and ##\epsilon_0## are just reflections of different ways of choosing units.
  17. Sep 6, 2018 #16
    *portion of Killtech's quote and my quote removed because I'm only addressing this single question

    I believe they do, as no particular metrology is used to get them.

    I'm going to put in a lot of work here to show this, so I hope you actually follow (this isn't new, but I've never seen this many steps not skipped before, and believe me this was painstaking). I'm keeping this extremely general (about as general as I am capable of doing), and at no point are any specific units used. Please follow and see if you agree that at no point was any unit convention used until the very end.

    First, assume that Newton's law of inertial holds good when an object is instantaneously at rest with respect to a coordinate system. Or rather, such a system exists for any object. Not an unreasonable assumption (basically we're assuming the existence of inertial rest frames. We okay so far?). So let the inertial frame S exist, and let x, t be the coordinates of one object, and let S' be the inertial frame of another object moving with speed v in the positive x direction relative to S, and let x',t' be its coordinates. Still no choices of units or anything else.

    Next arrange for the two systems to coincide at the origin (as this is just choosing reference frames, it is generally valid).

    These two fames must be related linearly in general by the form dx' = Adx + Bdt and dt' = Cdx and Ddt, with A,B,C and D as constants for a given value of v. All that's assumed here is isotropy and the law of inertia holding. Still no metrology.

    For simplicity, we're assuming that the second object is at rest in S' at the origin of the x',t' coordinates, so x' = 0. Therefore 0 = Adx + Bdt, and hence dx/dt = -B/A = v, which means B = -Av.

    Next do some algebraic manipulation:

    Ax = x' - Bt
    Ax = x' - B(t' - Cx)/D
    Ax = x' - Bt'/D + BCx/D
    Ax - BCx/D = x' - Bt'/D
    ADx - BCx = Dx' - Bt'
    (AD - BC)x = Dx' - Bt'

    and doing a similar thing with t:

    Dt = t' - Cx
    Dt = t' - C(x' - Bt)/A
    Dt = t' - Cx'/A + BCt/A
    Dt - BCt/A = t' - Cx'/A
    ADt - BCt = At' - Cx'
    (AD - BC)t = At' - Cx'

    Now consider the first object moving from S'. It's moving in the other direction, so
    x/t = [(Dx' - Bt')/(AD-BC)]/[(At' - Cx')/(AD - BC)]
    x/t = (Dx' - Bt')/(At' - Cx')
    x/t = (Dv' - B')/(A' - Cv')

    x/t = (Dv' - B')/(A' - Cv')

    but since the first object is at rest in S, we can let x/t = 0, giving

    0 = (Dv' - B')/(A' - Cv')
    Dv' + B = 0
    v' = dx'/dt' = B/D

    Hence, B = -Dv, and then drawing from v = -B/A, it must be that,

    B = -D(-B/A)
    B = D(B/A)
    1 = D/A
    A = D

    So go back and plug B = -Av and A = D back in:

    (AD - BC)x = Dx' - Bt'
    (A2 + vAC)x = Ax' + Avt'
    (A2 + vAC)x = A(x' + vt')

    Now the only way for this to work both ways in a physically meaningful way (where the transformation looks the same except swapping the sign on the v) is for (AD - BC) = (A2 + vAC) to equal 1.

    So, solve for C:
    A2 + vAC = 1
    vAC = 1 - A2
    C = (1 - A2)/(vA)

    Then, a very general linear transformation between inertial coordinates that has the reciprocal nature we would expect from a real transformation can be found:

    (go back to the very first thing, but I won't write it in differential form):

    x' = Ax + Bt
    x' = Ax - Avt
    x' = A(x - vt)


    t' = Cx + At
    t' = At + Cx (allowed because of the commutative property... and done because this is usually how it is written)
    t' = At + x(1 - A2)/(vA)
    t' = At - x(A2-1)/(vA) (factored out a -1 from the parentheses)
    t' = At -vx(A2-1)/(v2A) (factored out a v from the parentheses, same reason as above)
    t' = A(t - vx(A2-1)/(v2A2) (factored out an A, because again this will be how it is usually written)
    t' = A(t - vx(A2-1)/(v2A2)

    So, the two basic, very general transformation equations are:

    x' = A(x - vt)

    t' = A(t - vx(A2-1)/(v2A2))

    Now to make things easier to read, we can let (A2 - 1)/(v2A2) = k.
    For the x coordinate, solve for A, because it's A that is the coefficient multiplying what's in the parentheses, and we want this to look neater:

    (A2 - 1)/(v2A2) = k
    A2kv2 = A2 - 1
    A2 - A2kv2 = 1
    A2 (1 - kv2) = 1
    A2 = 1/(1 - kv2)
    A = 1/√(1 - kv2)

    Then quickly looking to simplify t' using k and the solving of A:

    t' = A(t - vx(A2-1)/(v2A2)
    t' = A(t -vxk)
    t' = A(t - kvx) (again used commutative property because that is how this will end up looking as it is usually written)
    t' = (t - kvx)√(1 - kv2)

    So, in easier to see form, the two transformation equations are:

    x' = (x - vt)/√(1 - kv2)


    t' = (t - kvx)√(1 - kv2)

    So, this is an extremely GENERAL set of transformation equations. It doesn't tell a whole lot about the universe, because k can be anything.

    To gain some insight into what k means, we'll have to use a third coordinate system; the first two have speed v between them, so let the third have speed u between the second set. This will make the transformation from the first to the third a combination of the other two, this time with two k's, kv and ku. The new coordinate system is S'', and it's coordinates are given by x'' and t''. So, using substitution (like a composite function) - note this is so messy I'm going to use LaTex:

    ##\frac{x''}{t''} = \frac{x - \frac{u + v}{1 + k_v uv}t}{\frac{1+k_u uv}{1+k_vuv}t - \frac{k_uu + k_vv}{1 + k_vuv}x} ##

    Now, because this has to follow the same form as the prior transformation, which had t' = (t - kvx)/√(1 - kv2), the coefficient on t in the denominator has to be 1. Otherwise the transformation would not hold generally for all inertial systems, which would make it inconsistent with this entire derivation.


    ##\frac{1+k_u uv}{1+k_vuv} = 1##

    which MUST mean that ku = kv = k.

    For similar reasons, the coefficient on the t in the numerator must be the speed between first and third coordinate systems. Let this speed be denoted w. Thus,

    ##w=\frac{u + v}{1 + kuv}##

    Therefore the transformation between the first system and the third is:

    ##\frac{x''}{t''} = \frac{x -wt}{t - \frac{ku + kv}{1 + kuv}x} ##

    ##\frac{x''}{t''} = \frac{x -wt}{t - k\frac{u + v}{1 + kuv}x} ##

    ##\frac{x''}{t''} = \frac{x -wt}{t - kwx} ##

    and so you can see this holds generally, which means there is only one k to worry about.

    So, whatever k is, ultimately you can make it anything you want with a suitable choice of coordinates/units. I'm not making any particular choice other than looking at the three important values everyone considers for every math problem like this: -1, 0 and 1. Anything else is really irrelevant as it would just be a translation of some form.

    So if you let k = 0,

    x' = (x - vt)√(1 - kv2)
    x' = (x - vt)√(1 - 0⋅v2)
    x' = (x - vt)

    t' = (t - kvx)√(1 - kv2)
    t' = (t - 0⋅vx)√(1 - 0⋅v2)

    out pops the Galilean transformation.

    If you let k = -1 you get this:

    x' = (x - vt)√(1 - kv2)
    x' = (x - vt)√(1 - (-1)v2)
    x' = (x - vt)√(1 + v2)

    t' = (t - kvx)√(1 - kv2)
    t' = (t - (-1)vx)√(1 - (-1)v2)
    t' = (t + vx)√(1 + v2)

    Interesting, but I'll leave this alone for now.

    If you let k = 1, you get this:

    x' = (x - vt)√(1 - kv2)
    x' = (x - vt)√(1 - v2)

    t' = (t - kvx)√(1 - kv2)
    t' = (t - vx)√(1 - v2)

    No assumptions about specific measuring devices were made here, and whatever the case, obviously v cannot be g̶r̶e̶a̶t̶e̶r̶ ̶t̶h̶a̶n̶ greater than or equal to 1. In terms of mathematical structure, this is, of course, the Lorentz transformation.

    As far as I can tell, this is pretty close to as general as you can get, and it leaves us with three options: the Galileo transformation, some weird thing that I've never seen before that likely has no correlation with reality (for k = -1), and the Lorentz transformation. For k = -1, again, I've never personally seen anything at all that follows that form, and I'm assuming no physical measurement has ever given us anything like that, so I am discounting it. If you have a good, physical reason for keeping it, please share.

    If not, that means there are ONLY two possibilities: either the Galileo transformation or the Lorentz transformation. Which depends upon your choice for k. We've yet to speak of any specific method of measurement.

    Now, on to that part. In both the Galileo and Lorentz case there is an absolute speed limit. It's easy to see in the Lorentz case: the absolute speed limit is v = 1 (note, again, no specific units decided). In the Galileo case, there appears to be no limit imposed by the equation itself, which means the limit is infinity.

    So again, we have two choices: a FINITE universal speed limit or an INFINITE universal speed limit. If the speed limit is finite, the Lorentz transformation is the right one, and it is irrelevant which choice of units you use, as again, this entire derivation was done without making any unit assumptions or measuring conventions. Only assumptions about isotropy and Newton's laws holding instantaneously in a reference frame were made.

    So... experimentally speaking, is there a finite fastest speed? If the answer is yes, then the Lorentz transformation is the correct one, irrespective of whatever units or measuring convention you use.

    *edited when I realized the obvious that v cannot EQUAL 1, let alone be greater than it, for the case where k = 1. No division by zero and no imaginary numbers allowed here.
    Last edited: Sep 6, 2018
  18. Sep 6, 2018 #17
    Quick note about my last post and the "discounted" case of k = -1. There is no real value that would cause you to divide by zero or get an imaginary number for v, but if the universal speed limit is infinite then you end up with indeterminate forms (∞/∞), so I'll take that as another reason to discount it, besides the fact that no measurement has ever actually shown that transformation to be correct (as far as I am aware), while countless have confirmed Galileo (in the low speed limit) and Lorentz (in the high speed limit).
  19. Sep 6, 2018 #18
    Nice! I did a similar thing for my own purposes based on Reflections on Relativity. One thing that I added that was not explicitly mentioned in that article is a simple demonstration of the invariance of the spacetime interval, like:

    t'^2 - x'^2 = \frac{(t^2 - x^2) - v^2(t^2 - x^2)}{1 - v^2} = t^2 - x^2

    Not as many steps as you might have put, but it might be worth considering adding something along these lines to your text for regular re-use ;)
  20. Sep 6, 2018 #19
    I’m pretty sure I remember someone posting that last week, and it was actually the germ of my last post. Looking at it now, they did skip a LOT of steps, and it looks like there is a tiny typo* (but then I might be the one who is wrong), but that link is part of why I considered doing all this mind numbing work.

    But the point I’m hoping is made (and one of our guys with a grad degree can say for sure) is that it doesn’t matter which way you go about doing it: you’re still left with only two transformation equation options (really one, with one particular choice about the universal speed limit distinguishing them), and this occurs completely irrespective of unit or measurement conventions, as far as I can tell.

    *It looks like they left off a t’ in this:

    (AD - BC)t = -Cx + At’

    having it as (AD - BC)t = -Cx + A

    But I might have followed incorrectly. If they didn’t make a typo, then I messed something up in my post, because I had (AB - BC)t = At’ - Cx. But if so I would have to have made a “double mess up” to correct for it, because I ended with the Lorentz transformation.

    EDIT- actually I think the typo is they meant for the x’ to be v’. They (presumably) were at this step:

    x/t = (Dx' - Bt')/(At' - Cx')

    and divided by t’ in the numerator and denominator, but forgot to change the x’ to v’. That’s what I did anyway, without skipping steps, so I’m assuming that’s right. Doing that and setting x/t = 0 gives v’ = B/D. Of course it doesn’t matter because with x/t = zero when you multiply by the denominator it goes away anyway.

    Anyway that is a great source all around.
    Last edited: Sep 6, 2018
  21. Sep 8, 2018 #20
    Hey, I know this should have been obvious, but I just realized that k = 1/c2. I don’t know why that surprised me but it’s pretty cool. If c → ∞ then k = 0 and we have the Galileo transform, and the square keeps the units right. More insight every day. Love it. At some point, however, I’m going to have to stop this stuff and actually learn to use these transformation laws to solve physics problems though, lol.
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