I Isotropy of the speed of light

[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.

 

[PeterDonis]

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.

I didn't say Bradley was assuming absolute motion. I said you were. Now you are not, since you correctly say "relative to the Sun". That means "in an inertial frame in which the Sun is at rest", which implies a simultaneity convention, just as everyone has been telling you.

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

No, the simultaneity of the inertial frame Bradley assumed, which, it appears from the above, you now correctly understand to be the inertial frame in which the Sun is at rest.

 

[bhobba]

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.

As Peter says, he assumed the sun was at rest in an inertial frame. It is a reasonable assumption, especially in 1728, but is now known to be false because of GR. One of the issues here is Newtonian Physics has undergone constant refinement since Newtons time. Newton used the concept of absolute space and time:
https://en.wikipedia.org/wiki/Absolute_space_and_time

It is basically mystical nonsense, but that is from our modern viewpoint. Heaven knows what future physicists will think of what we now believe.

That is why I always recommend Landau - Mechanics. This details the modern view of Newtonian Mechanics that avoids the issues with Newtons rather vague ideas. It avoids the one-way speed of light issue by defining an inertial frame as isotopic, so the problem never arises. Usually, an inertial frame is defined as a frame that obeys Newton's first law. However, that definition leaves important issues open. You need to specify a simultaneity convention such as the Einstein Simultaneity convention, which, as shown by Eddington, is equivalent to slow clock transport. Landau avoids the issue because you know in sync procedure that uses two-way light travel, it has the same speed both ways. Still, it is defining your way out of problems - the issues remain. If you search this forum, you will find a series of posts that explain Classical Mechanics fundamental basis - it is Quantum Mechanics - but I leave that for your own investigation.

Thanks
Bill

 

[meekerdb]

As Peter says, he assumed the sun was at rest in an inertial frame. It is a reasonable assumption, especially in 1728, but is now known to be false because of GR. One of the issues here is Newtonian Physics has undergone constant refinement since Newtons time. Newton used the concept of absolute space and time:
https://en.wikipedia.org/wiki/Absolute_space_and_time

It is basically mystical nonsense, but that is from our modern viewpoint. Heaven knows what future physicists will think of what we now believe.

That is why I always recommend Landau - Mechanics. This details the modern view of Newtonian Mechanics that avoids the issues with Newtons rather vague ideas. It avoids the one-way speed of light issue by defining an inertial frame as isotopic, so the problem never arises. Usually, an inertial frame is defined as a frame that obeys Newton's first law. However, that definition leaves important issues open. You need to specify a simultaneity convention such as the Einstein Simultaneity convention, which, as shown by Eddington, is equivalent to slow clock transport. Landau avoids the issue because you know in sync procedure that uses two-way light travel, it has the same speed both ways. Still, it is defining your way out of problems - the issues remain. If you search this forum, you will find a series of posts that explain Classical Mechanics fundamental basis - it is Quantum Mechanics - but I leave that for your own investigation.

Thanks
Bill

What I'm trying to understand is the operational significance of this. Suppose the speed of light is not isotropic. Could a measure of stellar aberration find the anisotropy or not? I understand that choosing the Sun as inertial reference frame implies a plane of simultaneity, but that has no operational significance since one could choose some other inertial frame. It's perfectly ok to use some special frame for a calculation so long as the answer doesn't depend on that choice.

 

[bhobba]

Suppose the speed of light is not isotropic.

Landau's answer: Then you are not dealing with an inertial frame so SR does not apply.

Ohanian's and others answer: There is no way to know that the speed of light is not isotropic unless you measure it. But measuring the one-way speed of light requires a synchronisation convention of which there is a number eg Reichenbachs synchronisation convention. Some will give, operationally, that light speed is not isotropic. However, you are making a rod to break your own back eg:
https://plato.stanford.edu/entries/spacetime-convensimul/

Landau was a smart guy. He chose his definition of an inertial frame for a very good reason. Besides, using Landau's definition of an inertial frame you can derive SR without any reference to light at all. It is simply much more elegant. There are even logical difficulties in that Einstein's second postulate says the speed of light is constant. Nothing was mentioned about direction. You would need to modify the foundational assumptions Einstein used. I think Reichenbach did it, but one must ask - why?

Thanks
Bill

 

[PeterDonis]

Suppose the speed of light is not isotropic.

As you will see if you read through this thread, there are at least two different possibilities here: anisotropy in the round-trip speed of light, and anisotropy in the one-way speed of light. As you will also see if you read through the thread discussion, the former kind of anisotropy, or its absence, is independent of any choice of frame, simultaneity convention, etc., but the latter kind is not.

Could a measure of stellar aberration find the anisotropy or not?

It obviously can't detect any round-trip anisotropy because there is no round-trip light involved.

Since one-way anisotropy depends on the choice of frame, it's not a direct observable, so one would not expect to be able to observe it.

I understand that choosing the Sun as inertial reference frame implies a plane of simultaneity, but that has no operational significance since one could choose some other inertial frame.

The choice of inertial frame used for calculations cannot change the prediction of any actual observable, no. But the usual derivation of aberration assumes that the distant star whose light is being seen is at rest relative to the Sun. In that sense the "rest frame" being used does have operational significance, since the derivation would lead us to expect observable consequences if the distant star were not at rest relative to the Sun. (However, because of how far away stars are as compared to the Earth-Sun distance, those observable consequences are much, much harder to test for than the presence of aberration itself.)

 

[Ibix]

Suppose the speed of light is not isotropic. Could a measure of stellar aberration find the anisotropy or not?

You are still missing the point.

Bradley was assuming light was a stream of Newtonian particles. His model works like being surrounded by baseball pitching machines and measuring the speed of the baseballs relative to the ground as they whizz by. There's actually no particular reason to assume that two pitching machines are even throwing at the same speed or even the same speed twice, and if a machine is mounted on a moving cart its balls' velocities relative to you won't be their muzzle velocities either. But Bradley's equipment wasn't very precise - as long as stellar drift velocities and Earth's orbital velocity and the variation in "light particle" speeds is low compared to the measured average "light particle" speed it's all expected to be lost in the noise. In principle, though, in his model he could have got a range of speeds relative to the Sun from a single star, and arbitrarily different speeds from any pair of stars. Had he been able to measure with meter-per-second precision he might well have been puzzled by the lack of variation.

But a modern analysis does not work like this. In a modern analysis the isotropy (or otherwise) of the speed of light is a direct consequence of your assumption of the isotropy of space (or the other way around - the assumptions are equivalent). If you repeat etotheipi's analysis (which can be done with simpler machinery) not defining space to be orthogonal to your worldline, then you will get an anisotropic speed of light. The isotropy or otherwise is an assumption you make.

 

[Sagittarius A-Star]

But the usual derivation of aberration assumes that the distant star whose light is being seen is at rest relative to the Sun. In that sense the "rest frame" being used does have operational significance, since the derivation would lead us to expect observable consequences if the distant star were not at rest relative to the Sun. (However, because of how far away stars are as compared to the Earth-Sun distance, those observable consequences are much, much harder to test for than the presence of aberration itself.)

I think, that is not correct. There exists no "active aberration":

It was Einstein's theory of relativity that reminded us that only relative velocities may lead to measurable effects. However, that does not imply that aberration answers the relative motion between source and observer. There is no stellar aberration due to the velocity of the source (if the emission event is given).

Source:
http://www.dierck-e-liebscher.de/publikationen/three-traps-in-stellar-aberration.html

Thus the two components of the binary star system both appear on Earth at the same angular location according to the angle transformation formula of special relativity. Of course, the Earth appears at two significantly different angles to observers on those two stars
...
It’s important to note that the direction of a plane wave arriving at the origin at a given time depends on the position of the source at the time emission, but not on the state of motion of the source, which is why the motion of the source has no bearing on the derivation of the aberration formula.

Source:
https://www.mathpages.com/home/kmath160/kmath160.htm

 

[etotheipi]

Landau's answer: Then you are not dealing with an inertial frame so SR does not apply.

Who says that SR does not apply to reference systems that are not inertial? A simple example is that of a uniformly accelerated Rindler observer, but you may also write equations for any general non-uniformly accelerated observer too

 

[Ibix]

Who says that SR does not apply to reference systems that are not inertial?

I gather that whether flat spacetime in non-inertial coordinate systems was part of SR or GR was not always agreed the way it is today. Einstein initially developed SR in inertial frames, and the study of non-inertial frames lead into the development of GR. Some old sources therefore regard Rindler coordinates to be part of GR - or at least not SR. The modern view is that flat spacetime means SR and curved spacetime means GR, and coordinates are up to you. But there was some disagreement along the way (terminology only, effectively).

I don't have Landau so I can't comment on that in particular. It's not a young text, though, so it's more than possible it uses older conventions.

 

[vanhees71]

Of course SR (as well as Newtonian mechanics) is valid also when described in non-inertial reference frames. It obviously works well for the standard mechanics 1 examples, i.e., the Foucault pendulum on the rotating Earth and the Euler equations of the spinning top (Newtonian physics). For SR the standard example coming to my mind is the Sagnac effect. SR is generally covariant as is any description of a differentiable manifold (here a fixed pseudo-Euclidean affine space, Minkowski space).

GR is the relativistic theory of the gravitational interaction, and only when you take into account gravity you have a curved spacetime (in the standard formulation a torsion-free pseudo-Riemannian/Lorentzian manifold).

I highly recommend to read the introductory paragraph of vol. 2 of Landau and Lifhitz, which from the very beginning puts the equivalence principle in the right context in stressing that true gravitational fields can never be eliminated by any choice of coordinates, because the non-vanishing curvature tensor is non-vanishing independent of the choice of coordinates. In freely falling local frames you always have tidal forces which you cannot "transform away". That's why nowadays one speaks about "microgravity" rather than "weightlessness" when it comes to describing experiments, e.g., on the ISS.

 

[bhobba]

I don't have Landau so I can't comment on that in particular. It's not a young text, though, so it's more than possible it uses older conventions.

It's got to do with the definition of an inertial frame. Landau defines it as a frame where all points, instants in time and directions are equivalent. It's a Classical Mechanics text and is fundamental to Classical Mechanics and Relativity. Most texts define it simply as a frame Newtons First Law holds in - but careful analysis shows that is not really precise enough without going into the details - another thread discussed it:
https://www.physicsforums.com/threads/what-do-Newtons-laws-say-when-carefully-analysed.979739/

I expressed it badly, saying SR does not apply. You can use generalised coordinates and SR to analyse it. It is how the twin paradox is resolved in SR:
https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

But notice how it is done - by breaking the frame into a large number of inertial frames of 'infinitesimal' size - i.e. a local inertial frames. Actually, you do the same in GR, where one can always find a coordinate system where locally (i.e. in an infinitesimal region) it is inertial. Then we have Lovelocks Theorem to derive GR - but there is the added issue of space-time being curved, so you need such a powerful theorem.

Thanks
Bill

 

[etotheipi]

@bhobba you don't need to splice together lots of instantaneous inertial frames, you can introduce non-inertial frames completely in their own right.To any observer in general motion with non-vanishing arbitrary four-acceleration $a^a$ you may define a vector field $(e_{\mu})^a$ of basis vectors along the the worldline which define coordinates with respect to this local frame. If these are such that $d(e_{\mu})^a/dt = 0$ along the worldline then the observer is inertial, and his/her four-acceleration and four-rotation both vanish

Again a simple example of an accelerated observer is a Rindler observer $\mathscr{R}$, one for whom $a:=\sqrt{a_{c} a^{c}}$ is a constant and $\omega^a = 0$. His/her worldline is restricted to a plane $\Pi \subseteq \mathbf{R}^4$; given a second inertial observer $\mathscr{O}$ of frame $(\tilde{e}_{\mu})^a$ this can wlog be taken to be $\Pi = \mathrm{span}((\tilde{e}_{0})^a, (\tilde{e}_{1})^a)$, i.e. uniform acceleration along the $x$-direction.

The four-velocity of $\mathscr{R}$ satisfies $u^a = \cosh{(at)} (\tilde{e}_{0})^a + \sinh{(at)} (\tilde{e}_{1})^a$ and differentiating we find that $a^a = a [\sinh{(at)} (\tilde{e}_{0})^a + \cosh{(at)} (\tilde{e}_{1})^a]$. The accelerated frame basis $({e}_{\mu})^a$ and inertial basis $(\tilde{e}_{\mu})^a$ are related by a time-dependent Lorentz transformation $\Lambda(t)$ i.e. \begin{align*}({e}_{0})^a(t) &= \cosh{(at)}(\tilde{e}_{0})^a + \sinh{(at)} (\tilde{e}_{1})^a \\({e}_{1})^a(t) &= \sinh{(at)}(\tilde{e}_{0})^a + \cosh{(at)} (\tilde{e}_{1})^a\end{align*}and thus the coordinates associated with the Rindler observer and the inertial observer satisfy\begin{align*}\tilde{t} &= (x + a^{-1}) \sinh{(at)} \\ \tilde{x} &= (x + a^{-1}) \mathrm{cosh}(at) - a^{-1}\end{align*}

 

[Dale]

Suppose the speed of light is not isotropic. Could a measure of stellar aberration find the anisotropy or not?

No, it could not. Reichenbach proved that his simultaneity convention reproduces all experimental results of relativity.

In fact, from a more modern perspective Reichenbach’s result is not strong enough. All measurements are scalars, so they are agreed on by all coordinates. They are thus independent of all synchronization conventions, not just Reichenbach’s.

 

[Dale]

It's got to do with the definition of an inertial frame. Landau defines it as a frame where all points, instants in time and directions are equivalent.

That is, IMO, a more elegant and better definition. It also highlights why inertial frames are useful and why sometimes it is useful to not use them.

The whole point of picking a coordinate system is to simplify the math. Often a problem has no analytical solution in one coordinate system, but choosing coordinates that respect the same symmetries as the system can simplify the problem to obtain an analytical solution. So the easiest coordinates to use are ones that have the same symmetries as your object.

The laws of physics are homogenous and isotropic (locally), so it makes sense that they will have a particularly nice form in coordinates that share those properties. The utility of the inertial frame is that it has the same symmetries as the laws of physics.

Different frames can be used, and they will get equivalent answers. But using poorly adapted coordinates complicates the math substantially. The isotropy of the one way speed of light is thus just a convention, but it is a useful convention precisely because the laws that govern EM are isotropic.

 

[vanhees71]

No, it could not. Reichenbach proved that his simultaneity convention reproduces all experimental results of relativity.

In fact, from a more modern perspective Reichenbach’s result is not strong enough. All measurements are scalars, so they are agreed on by all coordinates. They are thus independent of all synchronization conventions, not just Reichenbach’s.

I'm not so sure. Usually when you derive the symmetry group of spacetimes for which by assumption the special principle of relativity holds you have to assume the Euclidicity of space for inertial observers to end up with the Galileo and the Poincare groups as the only possibilities. Then experiment has to decide which one is the better, and there the Poincare group and thus SRT "wins".

I'm pretty sure one can find also work about the question, whether there are space-time models with the special principle of relativity valid but weakening the assumption about the symmetries of space for any inertial observer and how they look in detail, but I'm not aware of papers about this.

Then of course we have GR, where the special principle of relativity, holds only locally, i.e., you gauge the Poincare symmetry of special relativity (leading to a slightly more general spacetime with curvature an torsion, i.e., a Einstein-Cartan manifold though).

 

[PeterDonis]

There exists no "active aberration":

Not the way the author is defining "active aberration", no: he is simply making the obvious point that once a particular light ray is emitted from the source, nothing the source does after that emission can affect how that particular light ray is observed at the receiver.

However, we are not talking about just one light ray. We are talking about a source that is emitting a continuous stream of light rays, and a receiver that is receiving that continuous stream of light rays. That continuous stream of light rays, as observed by the receiver, will be different for different states of motion of the source relative to the receiver. Aberration is one aspect of the difference; Doppler shift is another.

To put it another way: the usual derivation of aberration assumes that the incoming light rays from the source are all parallel, all coming from exactly the same direction, relative to the Sun. This is equivalent to assuming that the source is at rest relative to the Sun. (Strictly speaking, as far as aberration is concerned, it could be moving either directly towards or directly away from the Sun, since that would only show up in Doppler shift, not aberration.) If the source were moving (strictly speaking, moving transversely) relative to the Sun, the direction from which the light rays were coming, relative to the Sun, would not be constant; it would change as the source moved. (The change would have light travel time delay built in, so the direction observed "now" at the Sun would be the direction the source was one light-travel time ago; that addresses the issue the author you referenced was concerned about.) This would change what was observed.

 

[Sagittarius A-Star]

Not the way the author is defining "active aberration", no: he is simply making the obvious point that once a particular light ray is emitted from the source, nothing the source does after that emission can affect how that particular light ray is observed at the receiver.

In the second link I posted, a calulation for a binary star system is done. It shows, that also the transversal velocity of a star (in the frame of the receiver) while that emissions can't effect, how that particular light ray is observed at the receiver.

The different angles in the two sender frames translate to the same angle in the receiver frame.

However, we are not talking about just one light ray. We are talking about a source that is emitting a continuous stream of light rays, and a receiver that is receiving that continuous stream of light rays.
...
If the source were moving (strictly speaking, moving transversely) relative to the Sun, the direction from which the light rays were coming, relative to the Sun, would not be constant; it would change as the source moved.

Yes, that would have been a problem for Bradley. Reason: He made a transformation between two observer frames, but with a time offset of several months (The Earth can't be at rest in both relevant frames at the same time.).

I think the constraint is, that the star is far enough away, but not, that it is at rest relative to the sun.

 

[PeterDonis]

In the second link I posted,

...it says:

"the “line of sight” to the Earth is undergoing aberration as viewed from the revolving star, just as the “line of sight” to the star is undergoing aberration as viewed from the revolving Earth"

That is what I am saying.

 

[PeterDonis]

I think the constraint is, that the star is far enough away, but not, that it is at rest relative to the sun.

The former constraint is what makes violations of the latter constraint have effects that are too small to be observed.

 

[Dale]

Usually when you derive the symmetry group of spacetimes for which by assumption the special principle of relativity holds you have to assume the Euclidicity of space for inertial observers to end up with the Galileo and the Poincare groups as the only possibilities.

Sure, but changing your synchronization convention does not alter any of the properties of spacetime. The spacetime still has the same symmetries, like Killing vectors. The coordinates just do not have the same symmetries as the spacetime.

It is like analyzing a sphere using Cartesian coordinates. Nothing you do with the coordinates changes any of the symmetries of the sphere, but the coordinates simply don’t reflect them. Similarly here, spacetime remains isotropic despite the anisotropy of our coordinates.

 

[vanhees71]

Sure, there's nothing preventing us from using any spacetime-coordinates we like in SR. In general it leads to the description of SR in wrt. non-inertial reference frames. The spacetime geometry (pseudo-Euclidean affine space) is independent of any choice of coordinates, which is what makes it a geometry.

 

[Sagittarius A-Star]

Sure, there's nothing preventing us from using any spacetime-coordinates we like in SR. In general it leads to the description of SR in wrt. non-inertial reference frames. The spacetime geometry (pseudo-Euclidean affine space) is independent of any choice of coordinates, which is what makes it a geometry.

https://www.physicsforums.com/threads/isotropy-of-the-speed-of-light.1000066/post-6469408

 

[vanhees71]

What has this to do with my statement above? This has been discussed at length. It's a simple example for defining a non-inertial reference frame.

 

[Sagittarius A-Star]

What has this to do with my statement above? This has been discussed at length. It's a simple example for defining a non-inertial reference frame.

Sorry, I saved it already unintentionally while editing.

I think, calling it a "non-inertial reference frame" could be misleading, because that could imply the existence of fictuous forces. It is not a standard inertial coordinate system.

 

[etotheipi]

In general you do have inertial force terms for arbitrarily accelerating observers, in analogy to classical mechanics! I do not have time to derive it now so here is just the result. Definitions: an observer of four-velocity $(u_O)^a$ and four-acceleration $(a_O)^a$ assigns a spatial momentum $P^a = p^a - E(u_O)^a$ to a particle $A$ whose four-momentum is $p^a = mu^a$ and whose energy as measured by $O$ is $E = - p_a (u_O)^a$. Let the four-vector from $O(t)$ to $A(t)$ be $X^a$, then write\begin{align*}

v^a &= D^{\mathrm{FW}} X^a - (\omega \times X)^a \\

&= D^{\mathrm{FW}} X^a - g^{ae} \epsilon_{bcde} (u_O)^b \omega^c X^d \\ \\

&= \frac{\mathrm{d} X^a}{\mathrm{d}t} - (a_O)_b X^b (u_O)^a - g^{ae} \epsilon_{bcde} (u_O)^b \omega^c X^d

\end{align*}so that $v^a$ is the measured spatial velocity of $A$ by $O$, and all the components are with respect to $O$'s local basis. Now write $q^a = \gamma mv^a$. The general formula is:\begin{align*}

\frac{\mathrm{d}q^a}{\mathrm{d}t} = D^{\mathrm{FW}} P^a - m\dot{\gamma} (\omega \times X)^a - \gamma m\left\{ 2 (\omega \times v)^a + (\omega \times (\omega \times X))^a + (\dot{\omega} \times X)^a \right\}

\end{align*}Writing $F^a = \frac{\mathrm{d} q^a}{\mathrm{d}t}$ will show how the inertial force terms arise. In all these formulae the notation $(x \times y)^a = g^{ae} \epsilon_{bcde} (u_O)^b x^c y^d$.

 

[Sagittarius A-Star]

In general you do have inertial force terms for arbitrarily accelerating observers, in analogy to classical mechanics!

Yes, but that discussed frame (with un-isotropic one-way light speed) is not the rest frame of an accelerated observer.

And thanks for providing this good example for my claim in above posting #234.

 

[etotheipi]

But there are no dynamical equations in that post, so why do you mention inertial forces?

 

[Sagittarius A-Star]

But there are no dynamical equations in that post, so why do you mention inertial forces?

In there is an equivalent equation, showing, that there is no pseudo-gravitational time-dilation. This implies that there are no inertial forces.

 

[etotheipi]

I still don't see. Writing the metric in non-inertial coordinates in general results in some quadratic form $\mathrm{d}s^2 = g_{ij} \mathrm{d} \tilde{x}^i \mathrm{d} \tilde{x}^j$, yes. What is it to do with inertial forces? These arise when you analyse the dynamical equations of some particle.

 

[Sagittarius A-Star]

What is it to do with inertial forces? These arise when you analyse the dynamical equations of some particle.

I showed in there, that there can be no inertial forces. For that proof, I don't need to distiguish, if there is a test particle or not.

 

[etotheipi]

Okay, well I don't understand. To me it just looks like you wrote the time transformation. But it's quite possible I don't know enough.

 

[Sagittarius A-Star]

Okay, well I don't understand. To me it just looks like you wrote the time transformation. But it's quite possible I don't know enough.

I wrote the time transformation, yes, but I also concluded from the independence of the x'-coordinate: "There is no pseudo-gravitational time-dilation". And without a gradiend of a pseudo-gravitational potential, there can't be an inertial force.

@vanhees71
I think, your formulation in #233 "non-inertial reference frame" could be interpreted as:
A1) non-accelerated frame(with non-Einstein-synchronization) without test particle -> no inertial force
A2) non-accelerated frame (with non-Einstein-synchronization) with test particle -> no inertial force
B1) accelerated frame without test particle -> no inertial force
B2) accelerated frame with test particle -> inertial force.

So, as I said, it could imply the existence of fictuous forces (case B2), but which cannot appear in the discussed frame (cases A1, A2).

 

[PeterDonis]

I think, your formulation in #233 "non-inertial reference frame" could be interpreted as:
A1) non-accelerated frame(with non-Einstein-synchronization) without test particle -> no inertial force
A2) non-accelerated frame (with non-Einstein-synchronization) with test particle -> no inertial force
B1) accelerated frame without test particle -> no inertial force
B2) accelerated frame with test particle -> inertial force.

What do you mean by "without test particle" vs. "with test particle"? The definition of a frame does not have anything to do with test particles.

 

[vanhees71]

The equation of motion for a free particle in flat Minkowski space time of course always leads to a straight world line, i.e., uniform motion as it must be. That's of coarse also true in Newtonian mechanics. It's nothing else than the 1st Newtonian postulate.

As in Newtonian mechanics, it's however common slang for centuries that there are "inertial forces in a non-inertial reference frame". That's of course a bit misleading in the sense that there are not really "forces" but just "covariant derivatives of vector components" with respect to (in SR proper in Newtonian mechanics absolute) time.

It's also easy to derive from the action principle (here in the most simple "square form"). To that end one uses
$$L=-\frac{1}{2} \dot{x}^{\mu} \dot{x}_{\mu}=-\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the $x^{\mu}$ are arbitrary space-time coordinates and $g_{\mu \nu}$ the Minkowski-metric components wrt. the corresponding holonomic basis. The dot means derivative with respect to an arbitrary world-line parameter, $\lambda$, which is automatically an affine parameter, because $H=L=\text{const}$, because $L$ doesn't depend on $\lambda$ explicitly.

The equations of motion (leading to nothing else than the said straight lines in Minkowksi space) are
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}}-\frac{\partial L}{\partial x^{\mu}}=0.$$
This gives
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} (g_{\mu \nu} \dot{x}^{\nu})-\frac{1}{2} \frac{\partial g_{\alpha \beta}}{\partial x^{\mu}} \dot{u}^{\alpha} \dot{u}^{\beta}=0.$$
This is nothing else than
$$\ddot{x}^{\mu}=-{\Gamma^{\mu}}_{\alpha beta} \dot{x}^{\alpha} \dot{x}^{\beta} \qquad (*)$$
with the standard Christoffel symbols
$${\Gamma^{\mu}}_{\alpha \beta}=\frac{1}{2} g^{\mu \gamma} (\partial_{\alpha} g_{\gamma \beta} + \partial_{\beta} g_{\alpha \gamma}-\partial_{\gamma} g_{\alpha \beta}).$$
Then you can say the right-hand side of (*) are the "fictitious/inertial forces" as in the analogous case in Newtonian mechanics.

That's of coarse just semantics and doesn't contain too much physics ;-).

 

[Sagittarius A-Star]

What do you mean by "without test particle" vs. "with test particle"? The definition of a frame does not have anything to do with test particles.

I added this distinction only because of etotheipi's concern in posting #239.

 

[PeterDonis]

I added this distinction only because of etotheipi's concern in posting #139.

Ah, I see. That concern was about the distinction between the metric and the geodesic equation; the latter is where "inertial forces" (which is really a misnomer) appear. But neither of those things are a frame, and defining a frame automatically gets you both an expression for the metric and an expression for the geodesic equation. There is no such thing as a "frame without test particle" where you only have the metric and no geodesic equation.

 

[Sagittarius A-Star]

Ah, I see. That concern was about the distinction between the metric and the geodesic equation; the latter is where "inertial forces" (which is really a misnomer) appear.

To my understanding, the geodesic equation follows automatically from the metric, because you can derive the geodesic equation by applying the principle of maximum proper time (which is compatible to the action principle) to a free test particle.

 

[PeterDonis]

To my understanding, the geodesic equation follows automatically from the metric

Yes, it does. But the two are still conceptually distinct: the metric describes the geometry of spacetime, the geodesic equation describes the curves that freely moving objects (objects that feel no force) follow.

 

[Dale]

The dot means derivative with respect to an arbitrary world-line parameter, λ, which is automatically an affine parameter, because H=L=const, because L doesn't depend on λ explicitly.

By the way, as an aside. Do you know of a particularly easy or convenient way to make sure that $\lambda$ is not just any affine parameter but is specifically the proper time (specifically for a timelike worldline). I know that I can take $\lambda$ and go back to the metric to convert it to proper time, but can I do anything to the Lagrangian or the Euler equations to make it happen automatically?

 

[PeterDonis]

Do you know of a particularly easy or convenient way to make sure that $\lambda$ is not just any affine parameter but is specifically the proper time (specifically for a timelike worldline)

Any affine parameter along a timelike worldline is proper time. Converting from one affine parameter to another is just rescaling the unit of time and/or picking a new "zero point". There is no one particular one that is "the" proper time; they all are.