Principle of relativity for proper accelerating frame of reference

In summary: No, what I'm saying is that, under the same circumstances, the laws of physics in the two frames will be the same.No, what I'm saying is that, under the same circumstances, the laws of physics in the two frames will be the same.
  • #71
Here is the German

1628104650920.png


In The Collected Papers of Albert Einstein, Vol. 2, this is translated as

1628104773639.png


In particular, in the original German we have "relativ zueinander" which translates to "relative to each other".

The English translation found in the book The Principle of Relativity does not contain the phrase "relative to each other".

Another English translation is found in Arthur I. Miller's book Albert Einstein's Special Theory of Relativity:

"The laws by which the states of physical systems undergo changes are independent of whether these changes of state are referred to one or the other of two coordinate systems moving relative to each other in uniform translational motion."
 
Last edited:
  • Like
Likes dextercioby and cianfa72
Physics news on Phys.org
  • #72
cianfa72 said:
constant speed w.r.t. the Earth, it should be.
Yes.

cianfa72 said:
So in Newtonian mechanics there is actually no way for an exact (logically unexceptionable) operational definition of 'uniform translatory motion' (i.e. inertial motion), I believe.
I would agree, yes. In my view, as I've said, this was one of the issues with Newtonian mechanics that was fixed in General Relativity.
 
  • Like
Likes vanhees71
  • #73
PeterDonis said:
Unfortunately I don't know Italian. But that does bring up a relevant point: Einstein's original 1905 papers were in German; what @Dale linked to earlier in the thread was a translation into English. I don't have a reference handy to the original papers in German, but it would be interesting to see how the principle of relativity postulate is stated in the original German version.
This paper is open access:

https://onlinelibrary.wiley.com/doi/10.1002/andp.19053221004

The passage in question seems to be the first paragraph in Sect. 1 (after the introduction):

Es liege ein Koordinatensystem vor, in welchem die Newtonschen mechanischen Gleichungen gelten. Wir nennen dies Koordinatensystem zur sprachlichen Unterscheidung von später einzuführenden und der Präzisierung der Vorstellung das "ruhende System".

Ruht ein materieller Punkt relativ zu diesem Koordinatensystem, so kann seine Lage relativ zu letzterem durch starre Maßstäbe unter Benutzung der Methoden der euklidischen Geometrie bestimmt und in kartesischen Koordinaten dargestellt werden.
The English translation (from wikisource) reads
Let us have a co-ordinate system, in which the Newtonian equations hold. For verbally distinguishing this system from another which will be introduced hereafter, and for clarification of the idea, we shall call it the "stationary system."

If a material point be at rest in this coordinate-system, then its position in this system can be found out by a measuring rod by using the methods of Euclidean Geometry, and can be expressed in Cartesian co-ordinates.
The translation is not too literaly but I think faithful contentwise.

Then in paragraph 2 he states the famous "postulates"

Die folgenden Überlegungen stützen sich auf das Relativitäts-
prinzip und auf das Prinzip der Konstanz der Lichtgeschwindig-
keit, welche beiden Prinzipien wir folgendermaßen definieren.

1. Die Gesetze, nach denen sich die Zustände der physi-
kalischen Systeme ändern, sind unabhängig davon, auf welches
von zwei relativ zueinander in gleichförmiger Translations-
bewegung befindlichen Koordinatensystemen diese Zustands-
änderungen bezogen werden.

2. Jeder Lichtstrahl bewegt sich I am ,,ruhenden“ Koordi-
natensystem mit der bestimmten Geschwindigkeit V , unabhängig
davon, ob dieser Lichtstrahl von einem ruhenden oder be-
wegten Körper emittiert ist.
The tranlation in the said Wiki:
The following considerations are based on the Principle of Relativity and on the Principle of Constancy of the velocity of light, both of which we define in the following way.

1. The laws according to which the states of physical systems alter are independent of the choice, to which of two co-ordinate systems (having a uniform translatory motion relative to each other) these state changes are related.

2. Every ray of light moves in the "stationary" co-ordinate system with the definite velocity V, the velocity being independent of the condition, whether this ray of light is emitted by a body at rest or in motion.
I think, when read in context Einstein is very clear that he assumes what we call "inertial frames of reference" for his (special) principle of relativity.
 
Last edited:
  • Like
Likes dextercioby
  • #74
cianfa72 said:
ok, let's go back to the analysis in SR of the two accelerated spaceships with the same proper acceleration (i.e. accelerometers at rest in each of them measure the same constant proper acceleration). Just to take it simple we neglect their lenghts.

From the point of view of an SR inertial frame they undergo a motion with the same coordinate acceleration. Since the transformations from the inertial frame to end up in each of the two accelerated frames (coordinate systems) - in which each spaceship is at rest respectively - are the same then using the same argument as in post #51 we can conclude that physics laws have to be the same in both spaceships (i.e. same set of equations).
What about this claim, does it sound right ?
Thanks.
 
  • #75
I don't think that this is correct generally. The special principle of relativity tells us that (a) there is a distinguished class of reference frames, called "inertial frames", where Newton's Lex Prima holds. Newton's Lex Prima operationally (!) defines what "inertial frame" means, i.e., together with the assumption that in an inertial frame of reference space is Euclidean, you can empirically check whether a given frame of reference is an inertial frame by using force-free bodies and checking that they always move with constant velocity wrt. the reference frame under investigation.
 
  • #76
vanhees71 said:
I don't think that this is correct generally. The special principle of relativity tells us that (a) there is a distinguished class of reference frames, called "inertial frames", where Newton's Lex Prima holds. Newton's Lex Prima operationally (!) defines what "inertial frame" means, i.e., together with the assumption that in an inertial frame of reference space is Euclidean, you can empirically check whether a given frame of reference is an inertial frame by using force-free bodies and checking that they always move with constant velocity wrt. the reference frame under investigation.
but the point is: how can we say that no force is acting on a body (i.e. which is - if any - the operational definition/rule to use to establish a body is actually a force-free body ?)
 
  • #77
Well, that's the subtle point. We can only try our best to avoid all interactions with other bodies/fields. The problem for sure are gravitational interactions, which cannot be shielded, but this problem is solved by general relativity with the qualification that there are no global but only local inertial reference frames.

According to our current understanding of cosmology, I'd say the best locally inertial reference frame is the rest frame of the cosmic microwave background radiation.
 
  • #78
cianfa72 said:
What about this claim, does it sound right ?
To be strictly correct, you should use the term "coordinate chart" instead of "frame".
 
  • #79
cianfa72 said:
how can we say that no force is acting on a body
Special relativity uses the definition you gave earlier: no force is acting on the body if an accelerometer attached to the body reads zero.

The part that requires general relativity is how the presence of gravitating masses affects things. Special relativity assumes that you can construct global inertial frames using bodies on which no forces are acting. But that turns out not to be possible in the presence of gravitating masses. General relativity gives you a way to handle that case.
 
  • #80
PeterDonis said:
To be strictly correct, you should use the term "coordinate chart" instead of "frame".
So far we had a thread about the topic coordinate chart vs frame of reference.
In this case which is the reason to use strictly the term coordinate chart ?

Thank you.
 
Last edited:
  • #81
cianfa72 said:
In this case which is the reason to use strictly the term coordinate chart ?
Because the mathematical argument you made uses the concept of mathematical transformations, and that concept applies to coordinate charts: you're transforming from one system of coordinates (an inertial one) to another (a non-inertial one), and the fact that it's the same mathematical transformation in both cases is what grounds the argument.
 
  • Like
Likes cianfa72 and vanhees71
  • #82
PeterDonis said:
Because the mathematical argument you made uses the concept of mathematical transformations, and that concept applies to coordinate charts: you're transforming from one system of coordinates (an inertial one) to another (a non-inertial one), and the fact that it's the same mathematical transformation in both cases is what grounds the argument.
I see..as discussed so far Reference frame vs coordinate chart, you are really making clear that what is actually changing is just the 'map' (coordinate system/chart) and not the 'territory' (Minkowski spacetime in this specific case).

So basically you rule out other possible meanings for the term 'reference frame' such as frame field or 'set of physical rulers, clocks, apparatus etc. used to performs measurements in a physical lab'.
 
Last edited:
  • #83
cianfa72 said:
So basically you rule out other possible meanings for the term 'reference frame' such as frame field or 'set of physical rulers, clocks, apparatus etc. used to performs measurements in a physical lab'.
Only for the particular argument you made earlier in the thread, since that argument involved mathematical transformations, and for those you need coordinate charts. You can't even formulate the argument in terms of frame fields or measurement apparatus in a lab.
 
  • Like
Likes vanhees71
  • #84
An accelerating frame is not in rest with reference even to itself. So how can we compare forces or relativistic effects between two co-moving but accelerating frames.
 
  • #85
There's a problem with things like "same acceleration profile" and "same relative velocity" and "rigid acceleration". They are not mutually consistent. The front an back of a ship that accelerates, change relative velocity because time dilation is different at the front and back. "Rigid acceleration" is ill defined...nothing is rigid in relativity.

To say physics is the same in all inertial reference frames is right. To say it's the same in all reference frames is true too, except the reference frame has to be small enough that there is no significant difference in acceleration within the reference frame...it can't be extended to a global reference frame.
 
  • #86
guptasuneet said:
An accelerating frame is not in rest with reference even to itself.
I don't know where you are getting this from. It is perfectly possible to have an accelerating frame in which all objects with constant spatial coordinates are at rest relative to each other. The canonical example is Rindler coordinates in Minkowski spacetime.
 
  • Like
Likes cianfa72 and vanhees71
  • #87
meekerdb said:
To say physics is the same in all inertial reference frames is right. To say it's the same in all reference frames is true too
Not in special relativity. In SR the laws of physics in non-inertial frames do not look the same as they do in inertial frames.

You have to go to General Relativity to obtain a framework in which the laws of physics look the same in all frames.

meekerdb said:
nothing is rigid in relativity.
This is not true either. It is perfectly possible to have a body in rigid motion in relativity. It is just that the requirements for that are not quite what one's Newtonian intuition would think.
 
  • Like
Likes valenumr and vanhees71
  • #88
PeterDonis said:
It is perfectly possible to have an accelerating frame in which all objects with constant spatial coordinates are at rest relative to each other. The canonical example is Rindler coordinates in Minkowski spacetime.
At rest relative each other: is actually involved a physical procedure that each of them (i.e. let me say each of those objects) has to use to check it is actually at rest w.r.t. each of the others ? Thanks.
 
Last edited:
  • #89
cianfa72 said:
At rest relative each other: is actually involved a physical procedure that each of them (i.e. let me say each of those objects) has to use to check it is actually at rest w.r.t. each of the others ? Thanks.
yes, constant round trip signal times.
 
  • Like
Likes cianfa72, DrGreg, PeterDonis and 1 other person
  • #90
cianfa72 said:
At rest relative each other: is actually involved a physical procedure that each of them (i.e. let me say each of those objects) has to use to check it is actually at rest w.r.t. each of the others ? Thanks.
Any feedback about this? Thank you.
 
Last edited:
  • #91
cianfa72 said:
Any feedback about this? Thank you.
Yes, post #89. (Which is a way to determine you are using "stationary" coordinates in a stationary spacetime.)
 
  • Like
Likes cianfa72
  • #92
cianfa72 said:
Sorry, I do not see post #89. I don't know why it goes directly from #88 to #90 skipping #89.
Odd. Do you need to refresh the page?

This is post #89 :
PAllen said:
cianfa72 said:
At rest relative each other: is actually involved a physical procedure that each of them (i.e. let me say each of those objects) has to use to check it is actually at rest w.r.t. each of the others ? Thanks.
yes, constant round trip signal times.
 
  • Like
Likes cianfa72
  • #93
DrGreg said:
This is post #89 :
okay now I see it, thank you.
That is the unique test (let me say 'The test') to check other bodies are at rest w.r.t. you ? In other words: is 'body at rest relative w.r.t. you' defined that way from an operational point of view?
 
Last edited:
  • #94
cianfa72 said:
okay now I see it, thank you.
That is the unique test (let me say 'The test') to check other bodies are at rest w.r.t. you ? In other words: is 'body at rest relative w.r.t. you' defined that way from an operational point of view?
SR case:
There are several operational definitions of at mutual rest. They all agree in inertial frames. They do not agree in noninertial frames. Consider:

A. No mutual Doppler shift.
B. Mutual round trip signal times constant over time.
C. An object considered rigid in the every day sense continues to span two bodies over time. While in SR, there are fundamental limitations on rigidity, a case where it is plausible to talk about a stable mutual rest is one case where it still makes sense.

So, in inertial frame, all of these agree. However, in a Rindler frame, e.g. an accelerating rocket, B and C agree, while A does not. Further, A does not make physical sense in this case. Consider a given uniformly accelerating Rindler observer. Then another world line at Doppler rest from it, extended to the past and future, will necessarily touch the reference observer. So two different bodies at alleged mutual rest come to touch each other.

GR case

In general, all of these definitions disagree. However, in a stationary spacetime, B and C will agree. This includes the region around an isolated planet or star, whether rotating or not.

Also, in an FLRW cosmology, all three agree (C must be replaced by a mathematical analog, since real light year long massless rods are too hard to come by) to very high precision out to substantial distances. Further, they don‘t agree at all with mutual rest defined as those observers seeing isotropy. Instead, such comoving observers are all found to be moving apart per these notions of mutual rest.

[edit: mixed up letters fixed, pointed out by @PeterDonis ]
 
Last edited:
  • #95
PAllen said:
SR case:
There are several operational definitions of at mutual rest. They all agree in inertial frames. They do not agree in noninertial frames.
Sorry, why frames are actually involved ? The definition of relative rest between bodies should involve only them, I believe.

Are you thinking of those bodies at rest w.r.t. a given reference frame (inertial vs non-inertial) ?
 
Last edited:
  • #96
cianfa72 said:
Sorry, why frames are actually involved ? The definition of relative rest between bodies should involve only them, I believe.

Are you thinking of those bodies at rest w.r.t. a given reference frame (inertial vs non-inertial) ?
You can replace frames with statements about the bodies, or just about one of the bodies considered as a 'reference'. Then, instead of inertial frames, substitute inertial motion (zero proper acceleration) for the reference body. Instead of Rindler frame, substitute uniform acceleration read by an accelerometer for the reference body. Similar statements can be made for the GR cases.
 
  • Like
Likes vanhees71
  • #97
PAllen said:
You can replace frames with statements about the bodies, or just about one of the bodies considered as a 'reference'. Then, instead of inertial frames, substitute inertial motion (zero proper acceleration) for the reference body. Instead of Rindler frame, substitute uniform acceleration read by an accelerometer for the reference body. Similar statements can be made for the GR cases.
Sorry, not sure to grasp what you mean with 'reference body'. Do you refer to the body w.r.t. define if every other body is at rest ?
 
  • #98
PAllen said:
SR case:
There are several operational definitions of at mutual rest. They all agree in inertial frames. They do not agree in noninertial frames. Consider:

A. No mutual Doppler shift.
B. Mutual round trip signal times constant over time.
C. An object considered rigid in the every day sense continues to span two bodies over time. While in SR, there are fundamental limitations on rigidity, a case where it is plausible to talk about a stable mutual rest is one case where it still makes sense.

So, in inertial frame, all of these agree. However, in a Rindler frame, e.g. an accelerating rocket, A and C agree, while B does not. Further, B does not make physical sense in this case.
I think you mean A does not agree and does not make physical sense, correct? A is Doppler rest.
 
  • #99
PeterDonis said:
I think you mean A does not agree and does not make physical sense, correct? A is Doppler rest.
Yes, mixed up my letters. I’ll fix it.
 
  • #100
cianfa72 said:
Sorry, not sure to grasp what you mean with 'reference body'. Do you refer to the body w.r.t. define if every other body is at rest ?
Yes.
 
  • #101
cianfa72 said:
My point is the following: Galileo principle of relativity applies not only to inertial frames but even to not-inertial constant (proper) accelerated frames having constant relative velocity (it definitely makes sense in the context of Newtonian mechanics).

Then what about in the context of SR ? I was trying to single out two (proper) accelerated frames (spaceships) having constant relative velocity to ask if we can continue to apply the principle of relativity even to them.
Even in classical physics you can go through some really simple 2d math. Define some inertial observers (i.e constant velocity), and then define a subject that undergoes acceleration for a period of time. All energy and momentum of every observer will be conserved for inertial observers for the starting and ending time of the accelerated subject, but the same will not hold true when using calculating total energy and momentum of observers from the accelerated subjects reference frame before and and after the acceleration.

sorry if that's wordy. I can formalize it with math if you want, and its just basic algebra.
 
  • #102
cianfa72 said:
This formulation is different from the principle of relativity as stated by Galileo. AFAIK the Galileo formulation makes no assumption about the state of motion of the frames involved (proper accelerated or not).

In other words Galileo principle of relativity applies as well even if the first frame is proper accelerated (i.e. bodies at rest in it have got the same proper acceleration as measured by accelerometers attached to them) and the second frame is moving with constant relative velocity w.r.t the first frame.

The laws of physics in both frames will be the same even if, of course, they are not in the 'simplest' form as in any inertial frame.

Maybe this difference with SR is 'summarized' by the keyword 'special', I guess... :oops:
I don't think that's correct if I understand what you are claiming. Even in classical physics, you can't make energy and momentum conserved from an accelerating frame.
 
  • #103
PeterDonis said:
Special relativity uses the definition you gave earlier: no force is acting on the body if an accelerometer attached to the body reads zero.

The part that requires general relativity is how the presence of gravitating masses affects things. Special relativity assumes that you can construct global inertial frames using bodies on which no forces are acting. But that turns out not to be possible in the presence of gravitating masses. General relativity gives you a way to handle that case.

My humble observation: if an observer can compute energy and momentum and see that those values are conserved, then they are inertial. If these values are not conserved, the subject is non-inertial.
 
  • #104
cianfa72 said:
okay now I see it, thank you.
That is the unique test (let me say 'The test') to check other bodies are at rest w.r.t. you ? In other words: is 'body at rest relative w.r.t. you' defined that way from an operational point of view?
I think you mean inertial, not accelerating. Or am I missing a subtle point?
 
  • #105
valenumr said:
if an observer can compute energy and momentum and see that those values are conserved, then they are inertial.
This is obviously false for curved spacetimes; in curved spacetimes, either there is no valid global energy and momentum at all (in non-stationary spacetimes), or, in stationary spacetimes, observers who have constant energy and momentum, viewed globally, are not inertial (for example, "hovering" observers in Schwarzschild spacetime, with constant kinetic and potential energy, have nonzero proper acceleration).
 
  • Like
Likes vanhees71

Similar threads

Back
Top