I Isotropy of the speed of light

[PeterDonis]

SR, LT and so on relies on the Einstein convention of simultaneity

The Lorentz transformation does, because the Lorentz transformation is defined as transforming between standard inertial frames. But "SR" does not, because, as I have said, you can do calculations in flat spacetime in any coordinates you like; there is no requirement that you have to do them in a standard inertial frame. Of course the transformations between coordinates that aren't standard inertial frames won't be Lorentz transformations, but so what? The physics is still the same.

 

[Sagittarius A-Star]

I think I am not the only one posting in this thread who believes that "inertial" means more than just "no pseudo-gravitational time dilation".

What else does it mean?

 

[Tendex]

Back to the isotropy of light, but now in GR's generalization, the postulate of light speed constancy is kept only locally, the metric tensor is not Minkowskian but it's still Lorentzian so it preserves locally the relation between what clocks and rulers measure in momentarily at rest local inertial frames compatible with general covariance of all general coordinate transformations as the relevant invariance.

 

[PeterDonis]

What else does it mean?

"Inertial frame" in every SR textbook I have read means a standard inertial frame, in which the one way speed of light is isotropic. Do you have a reference that uses "inertial frame" to mean an anisotropic frame?

 

[Sagittarius A-Star]

"Inertial frame" in every SR textbook I have read means a standard inertial frame, in which the one way speed of light is isotropic.

Do you have a specific example in an SR textbook?

Do you have a reference that uses "inertial frame" to mean an anisotropic frame?

Yes. Rindler differentiates between an inertial frame and a (standard) inertial coordinate system:

The basic principle of clock synchronization is to ensure that the coordinate description of physics is as symmetric as the physics itself. For example, bullets shot off by the same gun at any point and in any direction should always have the same coordinate velocity dr/dt .
...
We should, strictly speaking, differentiate between an inertial frame and an inertial coordinate system, although in sloppy practice one usually calls both IFs. An inertial frame is simply an infinite set of point particles sitting still in space relative to each other. For stability they could be connected by a lattice of rigid rods, but free-floating particles are preferable, since keeping constant distances from each other is also a criterion of the non-rotation of the frame. A standard inertial coordinate system is any set of Cartesian x,y,z axes laid over such an inertial frame, plus synchronized clocks sitting on all the particles, as described above. Standard coordinates always use identical units, say centimeters and seconds.

Source:
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

 

[vanhees71]

What you are describing is not a local inertial frame since it is not restricted to a small patch of spacetime centered on a particular event.

What you are describing is Fermi normal coordinates centered on a timelike geodesic.

Yes, and taking one at a certain point on the geodesic defines a local inertial frame.

 

[PeterDonis]

Do you have a specific example in an SR textbook?

Taylor & Wheeler, Spacetime Physics, Chapter 2 (particularly section 2.6).

Rindler differentiates between an inertial frame and a (standard) inertial coordinate system

Rindler's definition of "inertial frame" is still not your anisotropic one. His definition of "inertial frame" is just the particles, with no clocks at all.

 

[vanhees71]

But as he says, there are also (gedanken) clocks, at rest relative to each other, at each space-time point. Then you synchronize them according to Einstein's description in his paper of 1905. BTW if you have established that there are more than 3 points staying at rest relative to each other I think you already have established an inertial reference frame (at least in SR), cf. Lange's famous work within Newtonian physics, but that should work also within SR (see Laue's famous textbook on relativity).

 

[PeterDonis]

as he says, there are also (gedanken) clocks

As I'm reading what was quoted from Rindler, he treats the clocks as part of a "standard inertial coordinate system", but not as part of an "inertial frame". In other words, he is not using the same definition of the term "inertial frame" that you prefer.

 

[PeterDonis]

if you have established that there are more than 3 points staying at rest relative to each other I think you already have established an inertial reference frame (at least in SR)

I assume that by "points" you mean "freely falling objects that can be idealized as point particles". I agree that more than 3 of these remaining at rest relative to each other can establish a (local) "inertial reference frame" by Rindler's definition, since his definition just includes the particles. But your definition of "inertial reference frame" requires clocks and rulers; where are the clocks and rulers if all we have is more than 3 point particles staying at rest relative to each other? We don't even know how far apart they are and we don't have any clocks.

 

[vanhees71]

If we have more than three particles at rest to each other, and I forgot to say are not all in one plane, we can establish a spatial coordinate system and constructing three spatial tetrads forming a Cartesian basis. Then we can put clocks at a corresponding grid and synchronize them according to the standard procedure a la Einstein 1905. There you have your clocks and rulers. As you say yourself, of course you are free to use any other coordinates and use the usual rules of tensor calculus.

I still don't see, where my point of view is in any way contradicting standard definitions!

 

[Sagittarius A-Star]

Taylor & Wheeler, Spacetime Physics, Chapter 2 (particularly section 2.6).

I can find a definition of "inertial frame" in the T&W summary in section 2.10:

The free-float frame (also called the inertial frame and the Lorentz frame) provides a setting in which to carry out experiments without the presence of so-called "gravitational forces." In such a frame, a particle released from rest remains at rest and a particle in motion continues that motion without change in speed or in direction (Section 2.2), as Newton declared in his First Law of Motion.

In the above primed coordinate system, all the mentioned T&W requirements are fulfilled:

• The "without gravitational forces" I have shown in posting #147.
• The "a particle released from rest remains at rest" follows from $x'=x$, $y'=y$, $z'=z$.
• The "a particle in motion continues that motion without change in speed or in direction" follows from the following:

From $x'=x$ and x'-derivation of $t' = t + \frac{kx}{c}$ follows: $\frac{1}{U'_x} = \frac{1}{U_x} + \frac{k}{c}$

If the x-component of the velocity, $U_x$, is constant in the inertial coordinate system $(x, y, z, t)$, then $U'_x$ must be also constant in the coordinate system $(x', y', z', t')$. For the y- and z-components the same is trivial.

Rindler's definition of "inertial frame" is still not your anisotropic one. His definition of "inertial frame" is just the particles, with no clocks at all.

Yes. The primed coordinate system is laid over an inertial frame. Then the questions follows:

• Shall the primed coordinate system been called "non-standard" (and why)?
• Shall the primed coordinate system been called "non-inertial" (and why)?

 

[PeterDonis]

I can find a definition of "inertial frame" in the T&W summary in section 2.10

Which is not the complete definition. Look at section 2.6.

In the above primed coordinate system, all the mentioned T&W requirements are fulfilled

Not the requirements given in section 2.6. Those requirements explicitly include Einstein clock synchronization.

The primed coordinate system is laid over an inertial frame

The primed coordinate system is not an inertial coordinate system by Rindler's definition. His definition of an inertial coordinate system includes Einstein clock synchronization.

Then the questions follows:

• Shall the primed coordinate system been called "non-standard" (and why)?
• Shall the primed coordinate system been called "non-inertial" (and why)?

These questions are unanswerable because Rindler is no longer around to answer them, and he's the one who wrote the reference. "Unanswerable" does not mean you can just help yourself to the answer "no", which is what would be required to support your claims.

 

[Sagittarius A-Star]

His definition of an inertial coordinate system includes Einstein clock synchronization.

More precise: His definition of a standard inertial coordinate system includes Einstein clock synchronization.

Maybe, the unprimed coordinate system is the standard inertial coordinate system and the primed coordinate system is a non-standard inertial coordinate system (because it is laid over an inertial frame)?

 

[PeterDonis]

More precise: His definition of a standard inertial coordinate system includes Einstein clock synchronization.

That's the only kind of inertial coordinate system he defines.

Maybe, the unprimed coordinate system is the standard inertial coordinate system and the primed coordinate system is a non-standard inertial coordinate system (because it is laid over an inertial frame)?

I have already responded to this: Rindler doesn't say in the article you referenced, and he's not around any more to ask, so the question is unanswerable. So you can't use this reference to support your claim. That would require the reference to give an answer to the question you pose here, and it doesn't.

 

[vanhees71]

Can you again give the reference to Rindler? Just changing coordinates to non-Lorentzian ones don't change necessarily the frame of reference!

 

[PeterDonis]

Can you again give the reference to Rindler?

See post #155.

 

[Dale]

Is this way of wording what I mean by saying that the inertial frames of SR are conventional more undertandable?

Yes, that is helpful. Here is the distinction I am making. What we mean my the term “inertial frame” is a convention, and it includes the synchronization convention described by Einstein. So the definition of “inertial frame” is a convention.

The existence is not. Meaning after having defined, by convention, what we mean by the term it is still a matter of experiment to determine if such frames can accurately describe actual physical measurements of kinematics. That is the part that is not a convention.

 

[vanhees71]

As was worked out by Lange in the 1880ies for Newtonian physics, and this also applies to SR, an inertial reference frame can be operationally defined by having 4 non-coplanar free mass points staying at rest relative to each other. Then, by assumption, for any observer at rest wrt. these mass points space is Euclidean, and you can thus establish Cartesian coordinates or a Cartesian grid. Now you can use Einstein's clock-synchronization convention using a standard clock at (an arbitrarily chosen) origin to synchronize all clocks located at the grid points. In this way you can describe all physics (world lines of mass points in point-particle mechanics, classical fields in continuum mechanics and classical electrodynamics, and quantum fields in relativistic Q(F)T) in terms of the so defined Lorentzian (=pseudo-Cartesian) coordinates.

This is, of course, in a sense a convention, i.e., first assuming that for any inertial observer space is Euclidean and then synchronizing the clocks a la Einstein with light signals. Now you can of course make all kinds of measurements to test the predictions following from this convention, which mathematically of course also leads to the Poincare group as symmetry group of the space-time model, which is a Lorentzian (pseudo-Euclidean) 4D affine space and use it to construct all kinds of theories and you can test the predictions on the dynamics for these theories by measurements, and as is well known, the Minkowski space-time model passes all the tests made at an amazing precision.

It only has to be refined when the gravitational interaction is significant, and there you have to extent SR to GR, which boils down to making Poincare symmetry local (in the sense of a local "gauge symmetry").

 

[Dale]

having 4 non-coplanar free mass points staying at rest relative to each other. Then, by assumption, for any observer at rest wrt. these mass points space is Euclidean

This assumption cannot be simply made but must be experimentally tested. If you have four non-coplanar points and measure the distances and angles between the points then you can determine empirically if the geometry is in fact Euclidean or not.

 

[vanhees71]

That's what I tried to say: You assume a space-time model obeying the special principle of relativity (existence of inertial frames including the Euclidicity of space for any inertial observer). Then you can think about, how to experimentally test this assumption. First you have to establish somehow operationally an inertial frame. As shown in great detail by Lange at the end of the 19th century this can be done by having established that four non-coplanar free particles stay at rest relative to each other (under the assumption about the space-time model as mentioned above!). Then, using the (again assumed!) Euclidicity of space for all points at rest relative to the four "reference particles" you can define a "coordinate grid" (which you can choose as Cartesian for simplicity but also with any coordinates you like).

Now you also need time. For that you assume that there's some standard clock (e.g., nowadays you can simply use an atomic clock as described in the SI definition of the second) for any inertial observer at rest wrt. the reference points.

Now you need some synchronization convention, and here of course the realizability differs between the Newtonian and the SRT case. In Newtonian physics you have rigid rods and you can simply synchronize all clocks with one clock "at the origin", e.g., determined by one of the four reference points, by using a mechanical pulse along the rigid rods connecting all grid points, which instantaneously sets all clocks in motion and so synchronize it with the clock at the origin. It's clear that then it doesn't make a difference whether the clocks move wrt. each other or not, which establishes the "absolute time" of Newtonian mechanics.

In SRT there is instead the postulate of the "constancy of the speed of light" (if you follow Einstein's way) or the existence of an "invariant limiting speed" (if you follow the more modern group-theoretical approach that the special principle of relativity is also consistent with Lorentz rather than Galileo transformations). Then you use light signals (or any signal which propagates at the invariant limiting speed) to synchronize all clocks at rest relative to the clock at the origin as described by Einstein.

Now you have established an inertial frame, a coordinate chart (however you like to call it), and you can build models (like Maxwell's equations for electromagnetism, point-particle or continuum mechanics, etc. etc.) to make predictions for observable facts given the above physically realized inertial frame and check whether they are consistent with the real-world observations. In this sense the above "kinematics" are indeed testable by experiment and of course must be tested to establish that they are (at least in some realm of validity) a space-time model in accordance with Nature.

 

[Dale]

Then you can think about, how to experimentally test this assumption.

Ah, I see. For the purposes of this thread I was reserving the word "assumption" for something that is a convention which can simply be assumed and there is no possible experimental test. You are using it as a synonym for "hypothesis", which is indeed an assumption. Particularly for this thread a hypothesis is a wholly different sort of thing than a convention, even though both can reasonably be called assumptions.

 

[vanhees71]

Well, you can call it a hypothesis too. It's just another word. The difficulty of this age-old problem about the "kinematics" indeed is that on the one hand you need some mathematical model to establish a quantitative description of space (locations of bodies) and time intervals, which you indeed must "assume" or "hypothesize" (however you want to call it) but on the other hand of course to make it a valid physical model/theory you have to find a way to realize the mathematical model with real-world measurement devices. At the end you consider the space-time model valid (maybe within some limited realm of validity as the Newtonian space-time model or SRT in view of the most comprehensive today known space-time model of GR) if it leads to consistent descriptions of the phenomena.

The problem is that you cannot make "assumptions" (in your meaning of the word) without already having a "hypothesis" of a space-time model but on the other hand have to test this very hypothesis doing experiments.

 

[Dale]

This is not nearly as difficult an issue as you are making it out to be. For the purposes of this thread there are two scientifically different types of assumptions that you can make. One is an assumption which has no experimental consequences, and the other is an assumption which has experimental consequences.

Since the former have no experimental consequence you are free to choose any value constrained only by mathematical consistency. No experiment can ever possibly contradict such assumptions. I will call these assumptions “conventions” for clarity.

The second type has experimental consequences, so changing them changes your experimental predictions. Thus experiments can contradict these assumptions. I will call these assumptions “hypotheses” for clarity since making and testing them are central to the scientific method.

To determine if an assumption is a convention or a hypothesis all you need to do is to make different assumptions and see if your experimental measurements would differ.

For example, we can assume that the charge on an electron is negative or positive and we can assume that the charge on a proton is negative or positive. There are four possible permutations of those assumption. Examining the predicted experimental outcomes we find that there is no difference between -e with +p and +e with -p. We also find that there is a predicted difference between -e with +p and +e with +p. So the fact that they have a different sign is a testable hypothesis while the sign itself is an untestable convention.

Similarly with the geometry. That the one way speed of light is c is a convention, it can be changed without changing the predicted experimental results. That the spatial geometry is Euclidean is a hypothesis, changing it changes the experimental results.

It is well understood that many such assumptions (both conventions and hypotheses) will be made in the course of any scientific experiment. Nevertheless, the distinction between conventions and hypotheses is clear and rather straightforward. That you must make assumptions in no way prevents you from distinguishing which assumptions are testable and which are not.

 

[Killtech]

To determine if an assumption is a convention or a hypothesis all you need to do is to make different assumptions and see if your experimental measurements would differ.

That said, isn't the isotropy of light actually a convention and not a hypothesis? I asking this because i find that trying to make a self-consistent hypothetical model which assumes otherwise always ends up in a logical contradiction when attempting to model an experiment measuring the properties of light: it will still show the isotropy contrary to the assumption. The Michelson-Morley experiment null result explanation via aether theory is an example.

The issue seems to be build into a fundamental convention of measurement: in order to be able to measure anything at all one needs to define measures and these ultimately must be defined in terms of real physical entities. Now as it happens both the measure of length and that of time is very directly defined through purely electromagnetic interactions - i.e. light. but if you measure something in terms of multitudes of itself don't expect it to be able to vary.

The SI definitions are conventions which don't just define some units but more importantly the basic physical objects of reference in multitudes of which everything is measured. But you could just as much use a different convention i.e. acoustic lengths (within the domain where it's well defined) that instead uses the speed of sound as a physical reference. Of course both measurement and the theory need to be conducted using the same convention. So if you measure the speed of sound using acoustic lengths it becomes constant while therefore the (acoustic) speed of light won't be anymore - therefore a theory using that convention has different equations of motions while maintaining consistency with measurement i.e. both combined ultimately yield just an equivalent description.

Riemann actually noticed and discussed that issue in his habilitation work. He then abstracted the conventions of measurement into the word of a metric (i.e. the metric tensor; and it that context it is useful to look up the Greek origin of the word). he uses the word magnitude as the physical reference that can be carried forward to compare other magnitudes with.

 

[meekerdb]

That the speed of light in vacuo is 299,792,458m/s and isotropic is a definition now. But it's based on lots of empirical evidence. Although Michel-Morely, Foucault, and Fizeau measured round-trip light speed, Roemer and Bradley didn't.

 

[hutchphd]

Roemer and Bradley didn't.

But of course there is no particular weight to this data because the tacit assumption of simultaneity must be made to draw conclusions about isotropy.

 

[Dale]

That said, isn't the isotropy of light actually a convention and not a hypothesis?

Yes, it is a convention.

i find that trying to make a self-consistent hypothetical model which assumes otherwise always ends up in a logical contradiction

Reichenbach has already investigated this topic well. His synchronization convention is self consistent and produces an anisotropic one way speed of light.

 

[meekerdb]

But of course there is no particular weight to this data because the tacit assumption of simultaneity must be made to draw conclusions about isotropy.

What simultaneity is assumed in Bradley's measurement?

 

[Ibix]

What simultaneity is assumed in Bradley's measurement?

Bradley considers stellar aberration due to the orbital speed of the Earth. In modern terms he is therefore adopting the simultaneity convention of the Sun-centered inertial frame in which the Earth's velocity is defined.

 

[meekerdb]

His result is frame independent because he compares abberration six month apart. It doesn't matter what the Sun's inertial frame is.

 

[Ibix]

His result is frame independent because he compares abberration six month apart. It doesn't matter what the Sun's inertial frame is.

He clearly assumes that the Earth has equal and opposite velocities at those six month intervals. That's the Sun centered frame.

This is, in fact, something @PeterDonis pointed out in #32, replying to an earlier post of yours in this thread.

 

[bhobba]

That said, isn't the isotropy of light actually a convention and not a hypothesis?

No - it follows directly from the properties of an inertial frame. Classical mechanics, EM etc., are all assumed to use an inertial frame. Of course, it is just a conceptualisation, but a beneficial one. It is an assumed abstraction. It is generally assumed here on the earth, for many (probably even most) practical purposes, it can be considered an inertial frame. If you want to call it a convention and not an abstraction, go ahead - it is just semantics and a very unproductive thing to argue about. But understand what is going on, it is a 'convention' widely used in many areas of science. In fact, the POR is really Newton's first law in disguise. To remind you what the POR says - the laws of physics are the same in an inertial frame or a frame moving at constant velocity relative to an inertial frame. SR assumes the POR - hence it fundamentally assumes you are dealing with inertial frames, so the speed of light must be isotropic. Now, if you want to base SR on different postulates, go ahead, but the standard treatment using the POR is just so elegant you would need a good reason to do so. BTW you can do it, e.g. as someone mentioned, you can look at SR as a limiting GR case. But then you face the difficulty of justifying GR without assuming SR first. I do not even know if anyone has done that.

Thanks
Bill

 

[Dale]

But understand what is going on, it is a 'convention' widely used in many areas of science. In fact, the POR is really Newton's first law in disguise.

The assumption/convention/hypothesis terminology was introduced by me for the purpose of clarity in this thread. He is using the terminology correctly in that context. It is a convention, regardless of how widely used.

Note that the isotropy of the one way speed of light is a convention related to the second postulate, not the first.

 

[Buzz Bloom]

In fact, the POR is really Newton's first law in disguise. To remind you what the POR says - the laws of physics are the same in an inertial frame or a frame moving at constant velocity relative to an inertial frame. SR assumes the POR - hence it fundamentally assumes you are dealing with inertial frames, so the speed of light must be isotropic. Now, if you want to base SR on different postulates, go ahead, but the standard treatment using the POR is just so elegant you would need a good reason to do so.

Hi @bhobba:

I tried to find the meaning of "POR" in this thread and also in the Internet, but I failed. Please tell me what "POR" means.

Regards,
Buzz

 

[Nugatory]

I tried to find the meaning of "POR" in this thread and also in the Internet, but I failed. Please tell me what

Principle of Relativity.

 

[bhobba]

Note that the isotropy of the one way speed of light is a convention related to the second postulate, not the first.

Then how can the frame be isotropic if the one-way speed of light is not the same? I think there is an issue here about definitional terminology. My definition is as found in Landau - Mechanics. The other more common definition is it is a frame where Newtons First Law holds. I however find that definition imprecise. It may be that the extra precision of Landau is 'equivalent' to the convention of the one-way speed of light being the same?

Thanks
Bill

 

[PeterDonis]

His result is frame independent

The measurements of aberration are frame independent, yes.

But the calculation of the Earth's velocity from the measurements of aberration is not frame independent. The calculation requires adopting a frame, and its result gives Earth's velocity in that frame. As has already been pointed out, the frame Bradley adopted (though AFAIK he wasn't explicit about this) is an inertial frame in which the Sun is at rest.

 

[Dale]

I think there is an issue here about definitional terminology. My definition is as found in Landau - Mechanics. ... It may be that the extra precision of Landau is 'equivalent' to the convention of the one-way speed of light being the same?

I suspect that is so. The Reichenbach simultaneity convention is reasonably well-known, but I have never seen it in any textbook. And I don’t blame them, I wouldn’t put it in a textbook either. It would take a lot of effort to explain it well, and the benefit to the reader is as close to nothing as I can imagine. It is a tough concept with no utility. So although it is correct it simply isn’t in any textbook I am aware of.

Then how can the frame be isotropic if the one-way speed of light is not the same? I think there is an issue here about definitional terminology. My definition is as found in Landau - Mechanics. The other more common definition is it is a frame where Newtons First Law holds.

So Newton’s first law is compatible with anisotropy. If something is moving in a particular direction at a particular speed then Newton’s first law says it will continue in that same direction at that same speed. It says nothing about the comparison of two different objects going in different directions.

In geometrical terms, Newton’s first law says that force-free objects have straight line worldlines. There is no requirement that there be any particular relationship between those lines in different directions, merely that they be straight. In particular, different directions can be “scaled” differently and still be straight.

 

[meekerdb]

He clearly assumes that the Earth has equal and opposite velocities at those six month intervals. That's the Sun centered frame.

This is, in fact, something @PeterDonis pointed out in #32, replying to an earlier post of yours in this thread.

He may have assumed that, but it's not necessary. So long as you assume the speed of the Sun is small compared to the speed of light, its speed cancels out. And in the actual case the assumption is confirmed by the result.

 

[PeterDonis]

So long as you assume the speed of the Sun is small compared to the speed of light, its speed cancels out.

You do not appear to grasp the fact that there is no such thing as absolute speed. There is no such thing as "the speed of the Sun is small compared to the speed of light" in any absolute sense. That can only be true relative to a particular choice of frame. (Relative to cosmic rays coming into the solar system, the speed of the Sun is not small compared to the speed of light, for example; neither is the speed of the Earth, for that matter.)

 

[Nugatory]

So long as you assume the speed of the Sun...

Any time we are speaking of the speed of an isolated object, in this case the sun, we are assuming a particular reference frame and its implied simultaneity convention.

 

[meekerdb]

What simultaneity is relevant to Bradley's meausurement? The top and bottom of his telescope?

 

[meekerdb]

You do not appear to grasp the fact that there is no such thing as absolute speed. There is no such thing as "the speed of the Sun is small compared to the speed of light" in any absolute sense. That can only be true relative to a particular choice of frame. (Relative to cosmic rays coming into the solar system, the speed of the Sun is not small compared to the speed of light, for example; neither is the speed of the Earth, for that matter.)

Bradley was measuring the speed of light compared to the orbital speed of the Earth relative to the Sun. There's no assumption of absolute motion there. Are you denying that he measured the speed of light?

 

[etotheipi]

sorry to interject, who is Bradley?

 

[Ibix]

sorry to interject, who is Bradley?

An astronomer who made an early measure of the speed of light by considering stellar aberration - Wiki has some discussion https://en.wikipedia.org/wiki/Aberration_(astronomy).

 

[etotheipi]

Ah okay, thanks. From the looks of it I'm fairly certain Bradley's formula is wrong but I'll check it l8r.

 

[etotheipi]

Yes, Bradley's formula is wrong. Let $O$ and $\bar{O}$ be two observers with four-velocities $u^a$ and $\bar{u}^a$. Let $e^a = w^a / \sqrt{w_b w^b}$ be a unit vector parallel to the velocity of $w^a$ of $\bar{O}$ with respect to $O$ (and vice versa for $\bar{e}^a$). Let a photon have velocity $n^a$ w.r.t. $O$ and $\bar{n}^a$ w.r.t. $\bar{O}$. Let $k^a = u^a + n^a$ and $\bar{k}^a = \bar{u}^a + \bar{n}^a$ be tangents to the photon worldline and thus equal up to some constant of proportionality i.e. $k^a = \zeta \bar{k}^a$. Let $\varphi$ is the angle between the photon velocity and $\bar{O}$'s velocity both as measured by $O$ (and vice versa for $\bar{\varphi}$). Write $e_a n^a = - \cos{\varphi}$ and $\bar{e}_a \bar{n}^a = - \cos{\bar{\varphi}}$.

Then orthogonally decompose $u^a = \gamma(\bar{u}^a - w \bar{e}^a)$. We also may orthogonally decompose $e^a = \gamma(\bar{e}^a - w \bar{u}^a)$. Finally define a vector $\eta^a = n^a + \cos{\varphi} e^a$ such that\begin{align*}

k^a = u^a + n^a &= \gamma(\bar{u}^a - w \bar{e}^a) + \eta^a - \cos{\varphi} e^a \\

&=\gamma(\bar{u}^a - w \bar{e}^a) + \eta^a - \cos{\varphi} \gamma(\bar{e}^a - w \bar{u}^a) \\

&= \gamma(1+ w \cos{\varphi}) \bar{u}^a - \gamma (w + \cos{\varphi}) \bar{e}^a + \eta^a = \zeta \bar{k}^a = \zeta (\bar{u}^a + \bar{n}^a)

\end{align*}so then $\zeta = \gamma(1 + w \cos{\varphi})$ by comparing coefficients of $\bar{u}^a$. Equating the other orthogonal piece of both sides gives\begin{align*}
\zeta \bar{n}^a &= \eta^a - \gamma(w + \cos{\varphi}) \bar{e}^a \\

\bar{n}^a &= \frac{1}{\zeta} \eta^a - \frac{\gamma}{\zeta} (w + \cos{\varphi}) \bar{e}^a

\end{align*}and contracting with $-\bar{e}_a$ will give $\cos{\bar{\varphi}}$, i.e.\begin{align*}

\cos{\bar{\varphi}} = -\bar{e}_a \bar{n}^a &= -\frac{1}{\zeta} \bar{e}_a \eta^a + \frac{\gamma}{\zeta} (w + \cos{\varphi}) \bar{e}_a \bar{e}^a \\

&= 0 + \frac{w + \cos{\varphi}}{1 + w \cos{\varphi}} \\

&= \frac{w + \cos{\varphi}}{1 + w \cos{\varphi}}

\end{align*}

 

[Nugatory]

Are you denying that he measured the speed of light?

He measured the aberration and then calculated the one-way speed of light (more pedantically, the ratio of that one-way speed to the speed of the Earth relative to the sun) from these measurement results using assumptions that are equivalent to isotropy.

These assumptions are so plausible that it be would perverse not to make them (and that wouldn’t occur to any 18th-century physicist, so Bradley neither noticed nor stated them) but they are still assumptions.

 

[meekerdb]

Yes, Bradley's formula is wrong. Let $O$ and $\bar{O}$ be two observers with four-velocities $u^a$ and $\bar{u}^a$. Let $e^a = w^a / \sqrt{w_b w^b}$ be a unit vector parallel to the velocity of $w^a$ of $\bar{O}$ with respect to $O$ (and vice versa for $\bar{e}^a$). Let a photon have velocity $n^a$ w.r.t. $O$ and $\bar{n}^a$ w.r.t. $\bar{O}$. Let $k^a = u^a + n^a$ and $\bar{k}^a = \bar{u}^a + \bar{n}^a$ be tangents to the photon worldline and thus equal up to some constant of proportionality i.e. $k^a = \zeta \bar{k}^a$. Let $\varphi$ is the angle between the photon velocity and $\bar{O}$'s velocity both as measured by $O$ (and vice versa for $\bar{\varphi}$). Write $e_a n^a = - \cos{\varphi}$ and $\bar{e}_a \bar{n}^a = - \cos{\bar{\varphi}}$.

Then orthogonally decompose $u^a = \gamma(\bar{u}^a - w \bar{e}^a)$. We also may orthogonally decompose $e^a = \gamma(\bar{e}^a - w \bar{u}^a)$. Finally define a vector $\eta^a = n^a + \cos{\varphi} e^a$ such that\begin{align*}

k^a = u^a + n^a &= \gamma(\bar{u}^a - w \bar{e}^a) + \eta^a - \cos{\varphi} e^a \\

&=\gamma(\bar{u}^a - w \bar{e}^a) + \eta^a - \cos{\varphi} \gamma(\bar{e}^a - w \bar{u}^a) \\

&= \gamma(1+ w \cos{\varphi}) \bar{u}^a - \gamma (w + \cos{\varphi}) \bar{e}^a + \eta^a = \zeta \bar{k}^a = \zeta (\bar{u}^a + \bar{n}^a)

\end{align*}so then $\zeta = \gamma(1 + w \cos{\varphi})$ by comparing coefficients of $\bar{u}^a$. Equating the other orthogonal piece of both sides gives\begin{align*}
\zeta \bar{n}^a &= \eta^a - \gamma(w + \cos{\varphi}) \bar{e}^a \\

\bar{n}^a &= \frac{1}{\zeta} \eta^a - \frac{\gamma}{\zeta} (w + \cos{\varphi}) \bar{e}^a

\end{align*}and contracting with $-\bar{e}_a$ will give $\cos{\bar{\varphi}}$, i.e.\begin{align*}

\cos{\bar{\varphi}} = -\bar{e}_a \bar{n}^a &= -\frac{1}{\zeta} \bar{e}_a \eta^a + \frac{\gamma}{\zeta} (w + \cos{\varphi}) \bar{e}_a \bar{e}^a \\

&= 0 + \frac{w + \cos{\varphi}}{1 + w \cos{\varphi}} \\

&= \frac{w + \cos{\varphi}}{1 + w \cos{\varphi}}

\end{align*}