How do we identify a stationary system?

[CClyde]

In his original 1905 paper “On the Electrodynamics of Moving Bodies” Einstein's concludes his thought experiment on the definition of simultaneity by stating “It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it “the time of the stationary system.” After having premised the theory on the relativity of motion, how are we to understand the meaning of this “stationary” system?

 

[PeroK]

In his original 1905 paper “On the Electrodynamics of Moving Bodies” Einstein's concludes his thought experiment on the definition of simultaneity by stating “It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it “the time of the stationary system.” After having premised the theory on the relativity of motion, how are we to understand the meaning of this “stationary” system?

There is no such thing as a "stationary" system, except as a physically meaningless label to identify one particular system that is of interest to us. Instead, we may talk about an "inertial" system. Or, in Einstein's language in 1905, "a system of coordinates in which the laws of Newtonian mechanics hold good".

The "stationary" system here is just an arbitrary inertial system. It's no more stationary physically than any other inertial system.

 

[Ibix]

Einstein arbitrarily chose one reference frames to call the stationary one. The clocks used to measure time in that system must be at rest in that system. (Edit: see the post below this one for the exact quote.)

 

[Sagittarius A-Star]

After having premised the theory on the relativity of motion, how are we to understand the meaning of this “stationary” system?

Here is the answer:

In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the “stationary system.”

 

[Nugatory]

After having premised the theory on the relativity of motion, how are we to understand the meaning of this “stationary” system?

Note that relativity of motion is not Einstein’s innovation, it’s Galileo’s contribution to our understanding of physics - Google for “Galilean relativity” for more. Einstein’s contribution was to reconcile this centuries-old completely non-controversial concept with the laws of electromagnetism discovered in the 1860s. He did so by showing that we could resolve the apparent conflict between the two by using the Lorentz transformations instead of the Galilean transformations. Both transformations are internally consistent ways of working with relativity of motion, but experiments have shown that the Lorentz transformations more accurately describe the universe we live in.

But with that said…. We get to decide what we consider to be at rest and what we consider to be moving inertially by our choice of coordinate systems. We want this to be at rest, we choose coordinates in which this is at rest, we want that to be at rest, we choose coordinates in which that is at rest.

 

[CClyde]

That’s what I hoped to hear.

So to define synchronized clocks Einstein says the two clocks synchronize if
t B − t A = t ′A − t B
Which is a valid definition, but never tested in the stationary system, just assumed true since the math works.
Then in his next thought experiment “On the Relativity of Lengths and Time”, Einstein sets a rigid rod AB in motion relative to the “stationary” system and with the use of his system of synchronized clocks, a mirror, observers and light signals, concludes the following:

“Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.”

So the motion of AB sets the clocks out of synchronization by the amounts c+v and c-v for the light paths AB and BA respectively. To be clear the length of the light paths are a simple addition and subtraction of the time it takes light to traverse a distance when the target B is moving away at v from the incident light and when the target A is moving at v toward the incident light. So my next question is:
Is the stationary system (label) in motion relative to the light path, or put another way, when AB is at rest in the stationary system, is the rigid length AB equal to the length of the light path AB?

 

[Ibix]

So my next question is:
Is the stationary system (label) in motion relative to the light path, or put another way, when AB is at rest in the stationary system, is the rigid length AB equal to the length of the light path AB?

You mean, if I have a rod that is moving in my chosen coordinate system and has length $l$ in that system, and a flash of light is emitted at one end of the rod and received at the other end, is the distance in my system between the emission and reception $l$? No it is not. It is $lc/(c\mp v)$, with the sign depending on whether the rod velocity and light travel direction are parallel or anti-parallel.

 

[CClyde]

You mean, if I have a rod that is moving on my chosen coordinate system and has length $l$ in that system, and a flash of light is emitted at one end of the rod and received at the other end, is the distance in my system between the emission and reception $l$? No it is not. It is $lc/(c\mp v)$, with the sign depending on whether the rod velocity and light travel direction are parallel or anti-parallel.

No, I’ll pose the question another way (because there re 4 different ways to ask this same question).

When the rod AB is at rest in the system designated “stationary”, is the length of the light path A to B, the same as the rigid measure of the rod A to B, as measured by an observer in the stationary system?

 

[Ibix]

No

Well, what do you mean by "the length of the light path" if you don't mean the distance travelled by the light?

 

[Vanadium 50]

Wasn't the question answered in #4?

If the thread is continuing under the premise "relativity is wrong because words words words" I suspect it will not end well. Just end.

 

[CClyde]

Well, what do you mean by "the length of the light path" if you don't mean the distance travelled by the light?

No. the rod is not moving.

Yes, the length of the light path is the distance travelled by the light.

 

[Dale]

Which is a valid definition, but never tested

You don’t test definitions. That wouldn’t make sense. Definitions are true by definition.

When the rod AB is at rest in the system designated “stationary”, is the length of the light path A to B, the same as the rigid measure of the rod A to B, as measured by an observer in the stationary system?

Yes.

 

[CClyde]

Wasn't the question answered in #4?

If the thread is continuing under the premise "relativity is wrong because words words words" I suspect it will not end well. Just end.

No. #4 answered the first question, as did #2, #3 and #5.

lbix was responding to my second question in #6

The thread is continuing under the premise that there are members who understand the Special Theory of Relativity and will be able to answer my questions.

 

[Dale]

The thread is continuing under the premise that there are members who understand the Special Theory of Relativity and will be able to answer my questions.

It is also continuing under the premise that you will learn from the answers from said members who understand SR.

 

[CClyde]

You don’t test definitions. That doesn’t even make sense. Definitions are true by definition.

Yes definitions are true by definition. But what they define either meets or fails the definition.

t B − t A = t ′A − t B is a valid definition, but if the times t B − t A ≠ t ′A − t B the clocks do not synchronize.

 

[CClyde]

Yes.

How do you know, everyone just told me “stationary” is a label, not a state of motion.

If the stationary system is an inertial system as everyone above and Einstein said, it is equally valid to claim it is in motion.

If it is in motion the length of the rigid measure AB will not be the same as the length of the light path.
Yes I will learn from the answers.

 

[Dale]

But what they define either meets or fails the definition.

This doesn’t make sense.

t B − t A = t ′A − t B is a valid definition, but if the times t B − t A ≠ t ′A − t B the clocks do not synchronize.

I think you are missing the point. The point is that $t_B$ is something that you can adjust, like adjusting for daylight savings time. You adjust it so that the equation is true. Then $B$ is synchronized with $A$.

 

[Nugatory]

How do you know, everyone just told me “stationary” is a label, not a state of motion.

Stationary is just a label, but so is "state of motion". Once we have chosen what we're labeling stationary (this is equivalent to choosing the origin of our coordinate system) we've also chosen what we're labeling as in motion.
We can always say that something is moving relative to something else, and we need neither coordinates nor any any notion of moving/not-moving to detect such relative motion - a laser range-finder on object A will tell us whether object B is moving or at rest relative to A.

 

[Dale]

How do you know, everyone just told me “stationary” is a label, not a state of motion.

If the stationary system is an inertial system as everyone above and Einstein said, it is equally valid to claim it is in motion.

That isn’t relevant. You specified that the rod was at rest in the same frame in which you wanted the length of the light path. That is a complete specification.

It doesn’t matter if that frame is the stationary frame or any other frame. Nor does it matter if that frame is moving with respect to some other frame of interest.

If it is in motion the length of the rigid measure AB will not be the same as the length of the light path.

In motion in which frame? The length of the light path in which frame? You need to make a complete specification.

In your previous question you stated “When the rod AB is at rest in the system designated ‘stationary’, is the length of the light path A to B, the same as the rigid measure of the rod A to B, as measured by an observer in the stationary system?” So it was answerable.

Here you specified neither the frame in which the rod’s motion was described nor the frame in which the length of the light path was desired. So it is not answerable.

 

[FactChecker]

How do you know, everyone just told me “stationary” is a label, not a state of motion.

If the stationary system is an inertial system as everyone above and Einstein said, it is equally valid to claim it is in motion.

You should try to be careful with your terms. You mean the unaccelerated motion of an inertial reference frame. Among inertial reference frames, any one of them can be considered "stationary".

 

[Ibix]

No. the rod is not moving.

Yes, the length of the light path is the distance travelled by the light.

So the rod is not moving in my frame of reference. Then the distance travelled by the light is, in my frame of reference, the same as the length of the rod.

 

[malawi_glenn]

The thread is continuing under the premise that there are members who understand the Special Theory of Relativity and will be able to answer my questions.

Have you tested that premise? Or are you taking it as a definition? ;)

 

[Ibix]

How do you know, everyone just told me “stationary” is a label, not a state of motion.

Calling a frame "stationary" is just a label. You could apply it to any other frame and nothing would change except nomenclature. However, two objects (e.g. a rod and me) can be stationary with respect to each other.

Say there's a car coming towards you. It is perfectly ok to label yourself as stationary and the car as moving, or the car as stationary and you as moving towards the car. Both views are perfectly legitimate. But once you've decided which thing you're calling "at rest" you no longer have freedom to decide that the other thing is also at rest - otherwise we could avoid car accidents simply by asserting that we aren't moving and neither is the other vehicle, which would be silly.

So Einstein is free to call one of his frames stationary if he likes. Once he's picked a frame, though, he's decided what he means by stationary and is no longer free to declare things in other states of motion to be stationary too.

Einstein's paper is not really the best source from which to learn relativity, by the way. It isn't aimed at beginner physicists and his exposition has a lot of rough edges that have been knocked off in more modern treatments.

 

[PeroK]

How do you know, everyone just told me “stationary” is a label, not a state of motion.

If the stationary system is an inertial system as everyone above and Einstein said, it is equally valid to claim it is in motion.

If it is in motion the length of the rigid measure AB will not be the same as the length of the light path.

Yes I will learn from the answers.

Your problem is acually not with SR but with the concept of using two different inertial reference frames to analyse the same set of events. All of your questions and doubts would apply equally in classical, Newtonian physics. The only factor that is different in SR from Newtonian physics is that the speed of light is assumed to be $c$ in both frames. All the rest (defining a "stationary" system; setting a uniform rod in motion; considering synchronised clocks attached to that rod) is elementary physics.

Many students over the years who have come to PF have had the same problem. It's nothing specific to SR that is the problem. It's the whole concept of having two reference frames in relative motion.

 

[Ibix]

It's the whole concept of having two reference frames in relative motion.

I second this. Many people's intuition is incredibly bad at separating measures made with one frame from those made with another and taking care when switching frames for analysis. In everyday life you can get away with it, but it's fatal when attempting to learn relativity.

 

[A.T.]

The only factor that is different in SR from Newtonian physics is that the speed of light is assumed to be $c$ in both frames.

@CClyde This difference and the core of SR are nicely visualized here:

 

[vanhees71]

Note that relativity of motion is not Einstein’s innovation, it’s Galileo’s contribution to our understanding of physics - Google for “Galilean relativity” for more. Einstein’s contribution was to reconcile this centuries-old completely non-controversial concept with the laws of electromagnetism discovered in the 1860s.

"Non-controversial" is a bit a joke. It's one of the many battle places between Newton and Leibniz, i.e., Newton as a strong proponent of absolute space, while Leibniz emphasized the "relativity" of all motion. Of course, once more Leibniz was right ;-)).

 

[CClyde]

I think you are missing the point. The point is that tB is something that you can adjust, like adjusting for daylight savings time. You adjust it so that the equation is true. Then B is synchronized with A.

A is a light source B is a mirror, so t A, t B, and t ′A are the times of emission, reflection and return respectively.
You can adjust the distance between and the times of A and B, but that is irrelevant to the purpose of the definition, and testing it, which it to find out if the time for light to travel from A to B is equal to the time for it to travel from B to A.

This is necessary to communication accurate local times between A and B, in order to synchronizing their clocks.

 

[malawi_glenn]

Why do you still think that a definition needs to be tested?

 

[Ibix]

You can adjust the distance between and the times of A and B, but that is irrelevant to the purpose of the definition, and testing it, which it to find out if the time for light to travel from A to B is equal to the time for it to travel from B to A.

You can't test this. It's a definition of what "synchronised" means if the speed of light is isotropic, that clock A sees clock B lag exactly as much as B sees A lag. You simply adjust one of the clocks (i.e., adjust $t_B$ as Dale says) until this is the case.

What you can do is to work through the implications of the claim that such synchronization is possible simultaneously in multiple frames (that is, derive the Lorentz transforms) and see if it leads to self-contradiction (it doesn't) or to predictions that are falsified by experiment (ditto).

 

[PeterDonis]

This is necessary to communication accurate local times between A and B, in order to synchronizing their clocks.

No, you have it backwards. You have to synchronize the clocks at A and B first in order to communicate accurate local times between A and B according to a particular inertial frame. How you synchronize the clocks at A and B determines which inertial frame the times on the clocks correspond to.

 

[Sagittarius A-Star]

A is a light source B is a mirror, so t A, t B, and t ′A are the times of emission, reflection and return respectively.
You can adjust the distance between and the times of A and B, but that is irrelevant to the purpose of the definition, and testing it, which it to find out if the time for light to travel from A to B is equal to the time for it to travel from B to A.

This is necessary to communication accurate local times between A and B, in order to synchronizing their clocks.

The isotropy of the one-way-speed is only a convention. In his popular book from 1916, Einstein wrote in part 1, section 8:

That light requires the same time to traverse the path $A \rightarrow M$ as for the path $B \rightarrow M$ is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity."

Source:
https://en.wikisource.org/wiki/Rela..._I#Section_8_-_On_the_Idea_of_Time_in_Physics

 

[Dale]

Yes I will learn from the answers.

If you wish to learn from the answers, as you stated that you did, then please pay attention. When I say “you are missing the point” I am not inviting you to argue. I am telling you that the past 118 years of scientists have learned something about this and you have missed the lesson that we have learned. Whole books and many follow-up papers have delved into this topic in exceptional depth, and you did not understand the concept correctly. To learn you will need to abandon your erroneous concept and learn the correct one.

You can adjust the distance between and the times of A and B, but that is irrelevant to the purpose of the definition,

No. As I stated, the purpose is to establish how two distant clocks can be synchronized. Yes, you can also adjust the distance, but doing so does not change the synchronization status. It is only adjusting $t_B$ that will change the synchronization.

and testing it, which it to find out if the time for light to travel from A to B is equal to the time for it to travel from B to A.

We are not testing the invariance of $c$, that is already postulated. We are exploring the practical consequences and utility of the invariance of $c$. The useful consequence of the invariance of $c$ is an operational method for synchronizing distant clocks.

 

[Ibix]

How did Einstein synchronize his clocks in the stationary system unless he “assumes” a privileged, absolute state of rest?

He synchronised his clocks by adjusting one of them until they both agree on the lag of the other. Any pair of clocks that are mutually at rest can do this. Two pairs of clocks synchronised this way that are not at rest with respect to the other pair (i.e. clocks A and B move with speed v and are synchronised by their own measure, clocks C and D move with speed u and are synchronised by their own measure) will not agree on the same synchronisation.

Some of you are trying to teach Einstein, not me.

No - you have not understood what Einstein is saying, so you see conflict between what we say and your misunderstanding of what he said.

I think we have already agreed this equality is not found in their measures. And no matter how much we adjust the times on either clock, the time of a light path from A to B will never be the same as the time of a light path B to A when AB is a rigid rod in motion.

Nobody agrees with you on this, not here and not Einstein.

 

[Sagittarius A-Star]

So my question #2 is:
How did Einstein synchronize his clocks in the stationary system unless he “assumes” a privileged, absolute state of rest?

Only clocks at rest in the "stationary system" can be synchronized with reference to the "stationary system".