Isotropy of the speed of light

[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*}

 

[Ibix]

So if Bradley chose a star at azimuth, so \begin{align*}

\cos{\varphi}=0

\cos{\bar{\varphi}}=\sin({\bar{\varphi} - \varphi})=w.

\end{align*}

...because @etotheipi has assumed isotropy in his definition of $w$.

 

[meekerdb]

...because @etotheipi has assumed isotropy in his definition of $w$.

Where did he assume isotropy? Stellar aberration could be used to directly test isotropy. Suppose Bradley had pointed two telescopes anti-parallel (neglecting the obstruction of the Earth) and measured the speed of light from stars opposite on the celestial sphere. He could have gotten two different measures of the speed of light.

 

[Ibix]

Where did he assume isotropy?

His $u^a$ and $w^a$ are orthogonal. That would not be the general case if he were adopting a non-isotropic simultaneity convention.

 

[Ibix]

He could have gotten two different measures of the speed of light.

And, depending on the simultaneity criterion he used in his analysis (as @Nugatory notes he would assume isotropy because why wouldn't he?) he could get different values.

 

[etotheipi]

yes for any $p$ on the worldline of say $O$ I took $w^a$ to lie within the vector subspace of $\mathbf{R}^4$ consisting of all vectors metric-orthogonal to $u^a(p)$

 

[meekerdb]

yes for any $p$ on the worldline of say $O$ I took $w^a$ to lie within the vector subspace of $\mathbf{R}^4$ consisting of all vectors metric-orthogonal to $u^a(p)$

But does that change the answer?

 

[etotheipi]

But does that change the answer?

to what question?

 

[Nugatory]

Yes, Bradley's formula is wrong.

In the sense that any purely classical computation is wrong, yes. However, he still came quite respectably close to one-half the two-way speed of light because he was working with conditions in which the error was small.

But does that change the answer?

Yes, but not by much - surely less than the margin of error in his measurements.

 

[meekerdb]

In the sense that any purely classical computation is wrong, yes. However, he still came quite respectably close to one-half the two-way speed of light because he was working with conditions in which the error was small.
Yes, but not by much - surely less than the margin of error in his measurements.

So you're saying he measured the one-way speed of light, but he relied on some assumptions that, though false, were close enough that he got a good answer. What assumptions were those? Weren't they just that the Sun's motion was inertial and the Earth's speed relative to it was small compared to c? Bradley deliberately chose stars near the zenith to observe, so that was not an assumption.

 

[Nugatory]

What assumptions were those?

Galilean relativity - this was the 18th century and he was doing classical Newtonian physics. Thus his calculation failed to include relativistic effects that are included in @etotheipi’s four-vector calculation. Of course every Newtonian calculation will have a small error of this type, and that doesn’t stop us from using Newtonian physics when relativistic effects are negligible. That’s how Bradley could use the “wrong” formula and still come up with the right answer to the mints of experimental accuracy.

However, this assumption and error is largely irrelevant to this thread. Of much greater importance is that he assumed isotropy and as @Ibix says in #294 a simultaneity convention as well.