Einstein vs Newton: The concept of inertial vs non inertial frames

  • #1

Main Question or Discussion Point

Is concept of inertial vs non inertial frame inveted in Einsten theory of relativity or Newton knows that we can see on same object from different perspective?
(newton set 3 laws for inertial frame,so did he knew for realitivtiy when view object form different perspective/frame and did he use/knew fake forces,centifugal,coriolis etc..?)
 
Last edited by a moderator:

Answers and Replies

  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
The concept of a (global) inertial frame applies to both special relativity and Newtonian physics. In fact you can mathematically prove that any theory that obeys the special principle of relativity and has an Euclidean space for any inertial observer must either be Newtonian or special-relativistic physics with the corresponding spacetime descriptions as Galilei-Newton spacetime (a fiber bundle) or an Einstein-Minkowski spacetime (a pseudo-Euclidean affine space).

Newton was well aware of the fact that acceleration of a non-inertial reference frame against the class of inertial refeference frames is observable. He was, however, pretty much worried about how to physically determine an inertial reference frame. This was a debate among physicists until the 20th century. Particularly Ernst Mach made a proposal with his famous principle how to determine Newton's "absolute space".

As it is understood today, the most comprehensive spacetime model is the pseudo-Riemannian spacetime manifold introduced by Einstein in his theory of general relativity (or maybe a Cartan spacetime to make it compatible with spin). Within GR there are no more any preferred frames of reference, but it is always possible to choose a reference frame that is locally inertial. It is constructed by the choice of a non-rotating Vierbein fixed at a free falling test mass (e.g., the ISS is a good approximation of such a frame).
 
  • Like
Likes sysprog, dextercioby, Ibix and 1 other person
  • #4
29,786
6,124
Is concept of inertial vs non inertial frame inveted in Einsten theory of relativity or Newton knows that we can see on same object from different perspective?
I don’t know the history too well, but I believe that Newton was aware of some of the issues but didn’t resolve them himself. I think the concepts were greatly clarified and formalized by other researchers between Newton and Einstein. So by the time Einstein arrived he had access to much better theoretical tools than Newton did.
 
  • Like
Likes vanhees71 and Ibix
  • #5
pervect
Staff Emeritus
Science Advisor
Insights Author
9,777
997
Is concept of inertial vs non inertial frame inveted in Einsten theory of relativity or Newton knows that we can see on same object from different perspective?
(newton set 3 laws for inertial frame,so did he knew for realitivtiy when view object form different perspective/frame and did he use/knew fake forces,centifugal,coriolis etc..?)
I'm assuming "inveted" means "invented", and not, say, "inverted". Clarify is necessary.

Newtonian mechanics has the notion of non-inertial frames, though I"m not sure historically of when this was realized or who did the work.

I can give a bit of a modern perspective, though. The modern perspective is that for the purposes of Newtonian mechanics, we simply postulate (assume) an inertial frame exists. We don't attempt to justify this in any way physically, we just assume it is true, and use logic to explore the consequences so we can compare our predictions to experiment. Starting with these assumptions, we develop the mathematical machinery, often called "general covairance", that allows us to go from the laws of physics as expressed in one frame (or more generally, set of coordinates) to arbitrary coordinates.

I'm not quite sure of the history of the development of general covariance, to be honest. I suspect Lagrangian mechanics , if you're familiar, was a step in that direction. It was certainly a strong influence on Einstein, though his approach to it differed from the modern approach. Perhaps someone else can give you more details about this - the notion of covariance seems to be at the heart of your question.

There is one important result though that you may not be aware of. In Newtonian mechanics, general covariance (and/or it's earlier predecessors) gives the result that an accelerated frame of reference appears to be basically Newtonian, but with the addition of "fictitious forces". This is no longer true in Special relativity. An accelerated frame has some noticable effect on the behavior of clocks - pseudogravitational time dilation, that are not just a matter of adding in "fictitious forces". This was influential in Einstein's work on incoporating gravity into special relativity, which eventually lead (after many many years of work on his part) to general relativity.
 
  • Like
Likes Ibix
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
There is of course no problem to introduce non-inertial reference frames in special relativity. Usually they are only local maps, not covering the entire spacetime. Examples often found in the literature are the "Rindler coordinates" referring to an observer in hyperbolic motion relative to an inertial reference frame or a rotating reference frame. Both sets of coordinates cover only a part of Minkowski spacetime of course.
 
  • #7
PeterDonis
Mentor
Insights Author
2019 Award
30,195
9,354
Both sets of coordinates cover only a part of Minkowski spacetime of course.
This is true of Rindler coordinates, but is it true of rotating coordinates such as Langevin coordinates? AFAIK those cover all of Minkowski spacetime; they just don't have a timelike "time" coordinate in all of Minkowski spacetime.
 
  • Like
Likes vanhees71
  • #8
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Interesting. Do you have a definition of these coordinates? I Thought langevin observers move on helical curves, and the map covers only the part of space time with ##\omega R<c##.
 
  • #9
PeterDonis
Mentor
Insights Author
2019 Award
30,195
9,354
Interesting. Do you have a definition of these coordinates? I Thought langevin observers move on helical curves, and the map covers only the part of space time with ##\omega R<c##.
Do you mean the Born chart, as shown in this Wikipedia article?

https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

If so, while the article does give limits of ##0 < r < 1 / \omega## for the radial coordinate (it's using units in which ##c = 1##), I'm not sure that's actually required for it to be a valid chart. As far as I can tell, the chart is still one-to-one for ##r \ge 1 / \omega##; the ##t## coordinate is simply no longer timelike there, so the frame field that is given for the Langevin observers is not a valid frame field there (since it no longer has a timelike vector).
 
  • Like
Likes vanhees71
  • #10
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Yes, that was what I had in mind.
 
  • #11
29,786
6,124
This is true of Rindler coordinates, but is it true of rotating coordinates such as Langevin coordinates? AFAIK those cover all of Minkowski spacetime; they just don't have a timelike "time" coordinate in all of Minkowski spacetime.
I think that depends on whether by “reference frame” you mean “coordinate chart” or if you mean “tetrad”/“frame field”. The coordinate chart covers all of Minkowski spacetime, but the tetrad cannot
 
  • #12
PeterDonis
Mentor
Insights Author
2019 Award
30,195
9,354
I think that depends on whether by “reference frame” you mean “coordinate chart” or if you mean “tetrad”/“frame field”.
Yes, this is a fair point, but note that the specific quote from @vanhees71 that I originally responded to said "coordinates". :wink:
 
  • Like
Likes Dale
  • #13
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Now I'm confused. I hope we'll not have again a debate about reference frames, but mathematically the Born coordinates provide a chart which covers only a part of Minkowski space. In the notation of Wikipedia the pseudometric reads
$$\mathrm{d}s^2 = -(1-\omega^2 r^2) \mathrm{d} t^2 + 2 \omega r^2 \mathrm{d} t \mathrm{d} \varphi + \mathrm{d} z^2 + r^2 \mathrm{d} \varphi^2.$$
The ranges of the coordinates are ##t,z \in \mathbb{r}##, ##0<r<1/\omega##, ##\phi \in [-\pi,\pi]##.

I use the usual physicists' sloppy slang and call these coordinates.
 
  • #14
29,786
6,124
Now I'm confused. I hope we'll not have again a debate about reference frames, but mathematically the Born coordinates provide a chart which covers only a part of Minkowski space. In the notation of Wikipedia the pseudometric reads
$$\mathrm{d}s^2 = -(1-\omega^2 r^2) \mathrm{d} t^2 + 2 \omega r^2 \mathrm{d} t \mathrm{d} \varphi + \mathrm{d} z^2 + r^2 \mathrm{d} \varphi^2.$$
The ranges of the coordinates are ##t,z \in \mathbb{r}##, ##0<r<1/\omega##, ##\phi \in [-\pi,\pi]##.

I use the usual physicists' sloppy slang and call these coordinates.
It is a perfectly valid coordinate chart for ##0 < r < \infty##
 
  • #15
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Well, this is not the standard point of view, because obviously ##\mathrm{det} g=0## for ##\omega r=1##. I think this is again a matter of semantics. I just refer to a chart in Minkowski space, which must be compatible with the entire structure of the manifold, including the fundamental form. In regular points of the map its components must be a matrix with signature ##(3,1)## which is not the case for ##\omega r>1##.

[EDIT: Wikipedia is of the same opinion, i.e., it's a chart covering not the entire Minkowski space]

https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart
 
Last edited:
  • #16
29,786
6,124
That isn’t a requirement for a coordinate chart. All that is required for a coordinate chart is that the chart is a smooth and invertible mapping between an open set of events in the manifold and an open set of points in R4. The metric is not relevant. See p 37 at: https://arxiv.org/abs/gr-qc/9712019

There are many valid coordinate charts which do not have 3 spacelike and 1 timelike coordinate basis vectors.
 
  • #17
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
What you discribe is a coordinate chart for a differentiable manifold without additional structure. Minkowski space is a pseudo-Euclidean affine manifold and thus you want coordinate charts which are compatible with this structure, but I guess that's again a fruitless discussion about pure semantics rather than any physics or math content. So let's stop here...
 
  • #18
29,786
6,124
What you discribe is a coordinate chart for a differentiable manifold without additional structure.
Yes, the additional structure is not necessary for defining coordinate charts.

Btw, I take Carroll over wikipedia. And regarding Born, if he proposed the same restriction as Wikipedia, then it is important to recognize that it is fine to define a chart as being any open subset of another chart. So the restriction on the range of ##r## is valid, but not necessary.
 
  • #19
DrGreg
Science Advisor
Gold Member
2,290
895
... obviously ##\mathrm{det} g=0## for ##\omega r=1##.
I think you have forgotten the off-diagonal term. By my reckoning$$
\mathrm{det} \, g = \begin{vmatrix}
-(1 - \omega^2 r^2) & 0 & \omega r^2 & 0 \\
0 & 1 & 0 & 0 \\
\omega r^2 & 0 & r^2 & 0 \\
0 & 0 & 0 & 1
\end{vmatrix} = \begin{vmatrix}
-(1 - \omega^2 r^2) & \omega r^2 \\
\omega r^2 & r^2
\end{vmatrix} = -r^2 \, ,
$$ strictly negative except for a coordinate singularity at ##r = 0##.
 
  • Like
Likes PeterDonis
  • #20
29,786
6,124
except for a coordinate singularity at r=0.
Good point. I had forgotten that. So indeed it doesn’t cover the whole manifold. I believe it cannot cover not just the point at ##r=0## but also ##\phi = \pi##. But large ##r## is fine
 
  • #21
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Argh. Yes you are right. Calculating the eigenvalues with Mathematica also shows that ##g## has the right signature everywhere, except at ##r=0## (which is just the usual coordinate singularity of spatial cylinder coordinates). So indeed one has a chart covering almost all of Minkowski space.

Now I'm puzzled. How is then the standard result that the non-inertial reference frame of an observer rotating with constant velocity around an axis with distance ##r## to be explained? The Wikipedia article seems correct to me

https://en.wikipedia.org/wiki/Born_coordinates
 
  • #22
29,786
6,124
How is then the standard result that the non-inertial reference frame of an observer rotating with constant velocity around an axis with distance r to be explained?
What specifically about it are you wanting to explain?
 
  • #23
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
The fact that it is usually said that the Born coordinates only cover the part of spacetime given by ##\omega r<1##.
 
  • #24
DrGreg
Science Advisor
Gold Member
2,290
895
How is then the standard result that the non-inertial reference frame of an observer rotating with constant velocity around an axis with distance ##r## to be explained?
Even though it makes sense as a mathematical coordinate system for all ##r > 0##, it has the physical meaning of representing observers on a rotating disk only for ##r < 1/\omega##. Beyond that radius the ##t## coordinate is either null or spacelike.
 
  • #25
29,786
6,124
The fact that it is usually said that the Born coordinates only cover the part of spacetime given by ##\omega r<1##.
Probably it is because people usually confuse tetrads/frame-fields and coordinate charts. The corresponding tetrad covers only ##0 \le \omega r < c##, but the coordinate chart is valid ##0<\omega r < \infty##.

This is one situation where the distinction between a tetrad and a coordinate chart is important. As a result, the typical ambiguous usage of "reference frame" is problematic. Hence my comment in post #11.
 
Last edited:
  • Like
Likes PeterDonis

Related Threads on Einstein vs Newton: The concept of inertial vs non inertial frames

Replies
14
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
10
Replies
227
Views
20K
  • Last Post
Replies
11
Views
2K
Replies
3
Views
897
Replies
3
Views
873
  • Last Post
Replies
16
Views
2K
Replies
14
Views
5K
Replies
7
Views
2K
Replies
3
Views
22K
Top