• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,734 comments
1,470 users with positive rep
818 active unimported users
More ...

  If a person and a weighing scale are being pulled in opposite direction by the same force, is their rest frame inertial?

+ 0 like - 1 dislike

Imagine a person and a weighing scale accelerating through empty space towards each other, with the person being pulled by a downward force F (being exerted by some distant, unknown source $S$) and the scale being pulled upward by a force of the same magnitude F (being exerted by a different distant and unknown source $S'$).

Later they collide and the person ends up standing on the scale, but they stop moving since both are being pulled against each other with the same force F.

In this case, do we say that the person's rest frame (and by extension the scale's rest frame since it's also at rest w.r.t. the person) is inertial or non-inertial?

[My thoughts so far: before hitting the scale, if the person used an accelerometer, he/she would detect an acceleration and conclude his/her rest frame was non-inertial. But after hitting the scale, he/she would no longer be accelerating due to the distant source $S$, since the force is balanced by the scale pushing against the person. Hence no reading on accelerometer and the person would conclude that his/her rest frame is inertial. Am I correct in saying this?]

asked Apr 25, 2020 in General Physics by sphyrch (0 points) [ revision history ]

This is not graduate-level, voting to close. 500+ rep users can upvote the closevote here.

1 Answer

+ 0 like - 0 dislike

From the point of view of general relativity, inertial frames are those local reference frames which are in free-fall and non-rotating. "Local" reference frames because the free-fall condition may only be realised locally.

You pose the question referring to unknown sources \(S\) and \(S'\)of the forces acting on person and scale, respectively. If the force on the person is gravitation, e.g. if \(S\) is a distant star or a planet, then, prior to hitting the scale, the person is in free-fall (the person considered small enough compared with the scale of variation of the star's gravitational field for differences of the gravitational field across the person's extension, i.e. tidal forces, to be negligible). The person in free-fall, if non-rotating, is in an inertial system.

The scale cannot be shut off from gravity, but let's assume \(S'\) to be such that it overcompensates the gravitation on the scale. For example, a giant spring mounted on \(S\) could accelerate the scale away from the star/planet towards the person. Before making contact with the person, the scale is not in free-fall, and therefore not in an inertial system (it is accelerated relatively to a free-fall system, thus relatively to an inertial system).

After person and scale hit (and stick together instead of rebouncing) the person is at rest relative to the scale. The movement relative to the sources is a different question. As the scale still is pushed by the spring, the person now, via the scale, is also pushed by the spring. The person no longer is in free-fall, but accelerated relatively to a free-fall system; therefore, the person on the scale is not in an inertial system.

(By the way, something similar happens when you stand on the ground or actually step on a scale.)

Consider a different setup:

Assume you do your experiment in a laboratory which can be assumed in free-fall and non-rotating in the above sense. The laboratory frame is an inertial frame. Assume that person and scale are accelerated towards each other by springs, such that after making contact the person-scale system is at rest in the laboratory. Then, while person and scale are accelerated towards each other they are accelerated relatively to an inertial system and neither person nor scale are in an inertial system. Once person and scale have hit and are at rest in the laboratory frame they are in an inertial system.

answered Apr 26, 2020 by Flamma (90 points) [ no revision ]

Thanks for the very clear answer! So is this how the working principle behind an accelerometer specified? Inertia is the property of matter to resist motion, and any change in its state of motion is caused by a physical force that imparts a proper acceleration to it. So if anyone with an accelerometer undergoes proper acceleration, then so will the accelerometer which is in its simplest form a block attached to a spring. Block undergoes acceleration -> spring is compressed/stretched. Would I be correct in saying that or is there a caveat?

@sphyrch: Inertia certainly is not a property of matter to resist motion. It might perhaps be said it is a property of matter to resist a change of the state of motion. Such a statement is somewhat imprecise, however.

Within Newtonian mechanics \(\vec F=m\vec a\) makes this more precise. And if you relativistically consider a particle with charge \(q\) in an electromagnetic field \(F^{\mu\nu}\), then \(m{{\rm D}U^\mu\over {\rm D}\tau}=qF^{\mu\nu}U_\nu\), with \(m\) the particle's mass, \(\tau\) its proper time and \(U_\nu\) its 4-velocity. So all other parameters being the same, the mass affects how much acceleration you get from a given force.

As for your spring-block accelerometer:
I think you state it the wrong way round. Assume a person carrying the block on a spring; assume that person is accelerated (but assume no external force on the block). Then at first the person will be moving relative to the block; this will compress/stretch the spring; the compressed/stretched spring exerts a force on the block; the block accelerates.

For analogous reasons you get pressed into the seat of a car if the car accelerates (you need to accelerate along with the car, therefore the seat has to push you), space travellers get pressed into their seats at the launch of a spacecraft, and you, standing on the ground, experience weight, because you get accelerated relative to a free-fall system defined by the local gravity of Earth.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights