# Intertial frames

1. Sep 25, 2015

### Krishankant Ray

According to the postulates of Einstein theory, laws of physics are same in all inertial frame. What about non- inertial frames? Why they can't be same in non-inertial frame?

2. Sep 25, 2015

### Orodruin

Staff Emeritus
You are referring to the special principle of relativity which is at the foundation of SR. The general principle of relativity extends this to arbitrary frames.

3. Sep 25, 2015

### bcrowell

Staff Emeritus
This is more like Einstein's hazy original interpretation of GR, which was wrong. https://www.physicsforums.com/threads/distinction-between-special-and-general-relativity.827721/

Neither SR nor GR needs to be formulated in terms of frames of reference. Frames of reference are an optional thing that don't even need to be discussed if you don't want to.

They can, and it doesn't require GR. For example, electromagnetism can be formulated in the language of tensors, and then the laws have the same form in any coordinate system, regardless of whether the coordinate system even has an interpretation in terms of a frame of reference.

4. Sep 25, 2015

### Orodruin

Staff Emeritus
5. Sep 26, 2015

### loislane

Not really. All tensors are not created equal, and their invariance properties depend on the way they are defined and their domain of application as proper tensors, thus you may have cartesian tensors, pseudo-euclidean tensors, when they exceed their domain of application they become pseudotensors, meaning that they no longer follow tensor transformation laws strictly. The tensorial formulation of EM is in terms of differential forms. This type of tensors only require a smooth manifold and an affine connection, when the connection is flat as is the case in the relativistic formulation of EM in flat spacetime, such manifolds are equivalent to affine spaces with flat geometry, of wich Minkowski space is an example.

This kind of tensors, a generalization of Euclidean or affine tensors that substitutes the invariance under coordinate transformations from orthogonal transformations to the also linear Lorentz transformations are not tensors anymore when going out of the domain where they are defined in, be it smooth manifolds with affine connection or the added condition of a flat connection as in electromagnetism in flat spacetime. They are restricted in their tensor condition to certain coordinate systems, like cartesian or skewed coordinates, not arbitrary ones.
This doesn't mean curvilinear coordinates cannot be used in an affine space like Minkowski, just that in the presence of curvilinear coordinates nonlinear terms will appear in the form of christoffel components not transforming as a tensor even if the geometry is flat.
Conversely there is no direct consequence from the restriction of the tensorial form to smooth manifolds with connection when going to EM in curved spacetime since the general coordinate transformations in curved spacetime refer to those up to diffeomorphisms of smootn manifolds.

Noninertial frames like for instance rotating frames involve the kind of nonlinear coordinate transformations that are not included in that set of coordinate transformations, much to the dismay of Einstein himself who thought for many years that he had managed to include absolute acceleration in his general theory of relativity

6. Sep 26, 2015

### martinbn

@loislane, that doesn't make much sense.

7. Sep 26, 2015

### Orodruin

Staff Emeritus
It does make sense, but perhaps not at the level the OP is looking for ...

8. Sep 26, 2015

### loislane

As written your argument isn't very helpful(as in specifying what parts make some sense and wich not much sense to you), is it?
What specifically did you not understand?

9. Sep 26, 2015

### vanhees71

I also don't understand why this shouldn't make sense. Perhaps it's a discrepancy in language?

10. Sep 26, 2015

### bcrowell

Staff Emeritus
Loislane's #5 reads to me as a farrago of correct statements, incorrect statements, irrelevancies, historically based misconceptions, and imprecise use of language. If we want to go into this, a thread at the "I" level is probably not the place for it.

11. Sep 26, 2015

### vanhees71

These levels get on my nerves ;-). Why is this not appropriate for an "I" level thread? It's a quite basic question, how you understand "tensors". For physicists a tensor has components which transform under a certain class of transformations from one reference frame to another. There are thus tensors with respect to rotations, other tensors with respect to Lorentz transformations and so on.

Also for physicists reference frames are very important. There may be advantages for some calculations to use the mathematicians' notion of a frame-independnet formulation of the theory, but when it comes to observations in the real world you always introduce a reference frame since you use real-world measurement apparati, which define a frame.

12. Sep 26, 2015

### bcrowell

Staff Emeritus
"I" is defined as undergrad level. Tensors, diffeomorphisms, etc., are not typically discussed in undergraduate courses. The OP has already received correct and incorrect answers at the level s/he indicated s/he would be able to understand. I'll start a separate thread.

13. Sep 26, 2015

### bcrowell

Staff Emeritus
14. Sep 26, 2015

### stevendaryl

Staff Emeritus
In the modern way of thinking about it, the "laws of physics" can be expressed in a way that doesn't depend on any particular rest frame or coordinate system. However, I think that what Einstein meant was this: (Considering only a single spatial dimension, for simplicity) If you graph the position $x$ of a particle as a function of time $t$, then in an inertial frame, the graph of a "free" particle makes a straight line. The graph of a light signal also makes a straight line, and furthermore, the speed (the slope of the line) has the value $c=300,000,000\ m/s$ (or whatever the precise number is). In a noninertial coordinate system, this will no longer be true; the "speed" of a particle need not be constant (if by "speed" you just mean the slope of the graph of $x$ versus $t$). So the behavior looks different in a noninertial frame.

The modern way of thinking about these things distinguishes between physical acceleration (also called "proper acceleration") and coordinate acceleration.

15. Sep 26, 2015

### bcrowell

Staff Emeritus
Stevendaryl's #14 is nice -- and three cheers for an answer that's at the level requested by the OP.

However, I would say this in a different way. There is a distinction between inertial motion and noninertial motion, and that distinction is a clear one in SR and GR. However: (1) the relativistic definition of inertial motion is free-falling motion, which is different from the Newtonian definition; and (2) Einstein was simply wrong when he tried to describe this by assigning some foundational role to frames and the form of the laws of physics in various frames.

This may also be of interest: https://www.physicsforums.com/threads/distinction-between-special-and-general-relativity.827721/

16. Sep 27, 2015

### vanhees71

Here I disagree. Physics only works in frames. You have to define a frame, when measuring distances and time intervals. This is done by using "clocks an rods" in a very general sense, i.e., any devices which measure space and time. Physics is about observations and measurements, and as soon as you take that into account you define a frame. To measure something you need to compare the quantity you want to measure with some normal, defining your units! If this is not I level, then you can't do physics on I level!

17. Sep 27, 2015

### robphy

I'm not sure what "only works" means.

In my opinion, there is a distinction between the "laws" of physics and "measurements by observers" that one might do to express and interpret the laws.
By analogy, a law might be the statement $\vec C=\vec A+\vec B$. Introducing a set of axes, I could express this in various coordinate systems... resulting in expressions that describe the "law" as well as "specifics about the observer doing the measurement". At this level, one might be interested in finding ways to compare specific-observer measurements (i.e., find transformations that map one set onto another)... but, while important in doing measurements in the lab, it clutters the essence of the law itself.

To me, it seems more like:
from the myriad of measurements that observers make, we abstract ("tease out") a common feature that expresses the "physical law" (without regard to observers). Hopefully, this can be done cleanly so that, if we wish, we could re-insert an observer in order to provide a way to predict what that observer would measure.

Last edited: Sep 27, 2015
18. Sep 27, 2015

### bcrowell

Staff Emeritus
This is not really true. In principle, all measurements in relativity can be reduced to what a geometer would call incidence relations, meaning basically intersections. For example, if two particles collide, their world-lines intersect, and this is an observable fact. You can have, e.g., clocks without having a frame of reference. A frame of reference is a very high-tech, elaborate thing that is the result of a program of surveying by techniques such as Einstein synchronization. It's an optional convenience.

19. Sep 27, 2015

### vanhees71

A clock defines a reference frame, namely its restframe.

20. Sep 27, 2015

### Staff: Mentor

This is not correct. We have already discussed this previously. The GPS system implements a frame called the earth centered inertial frame. In this frame none of the components of the GPS system are at rest.