High School Inertial Objects: Acceleration & Direction

[PeterDonis]

Ya, I give up.

If you want a quick summary of all the responses you've been getting, here it is:

(1) There is no such thing as "things moving inertial to each other". Things can be at rest relative to each other, but that tells you nothing about whether they are moving inertially or not.

(2) Moving inertially means an accelerometer attached to the object reads zero. This is an invariant, independent of any choice of coordinates or reference frame, and all observers will agree on the readings of a particular accelerometer.

(3) There is an absolute thing according to relativity, and it is the geometry of spacetime. The geometry of spacetime is also an invariant; it's the same for all observers and in all coordinates or reference frames.

 

[jaketodd]

That's good, thank you very much for doing that.

 

[PeterDonis]

thank you very much for doing that.

You're welcome!

 

[pervect]

Are two objects, accelerating at the same rate, and in the same direction, considered inertial to one another? If so, I will post my resulting question. If not, it's safe to disregard this thread.

Thanks,

Jake

I would say no, in both interpretations of the question that come to mind for objects acclerating "at the same rate and in the same direction" that I can think of.

"At the same rate" sounds simple, but it depends on whether you mean that their acceleration in some inertial frame of reference is the same, or whether you mean their proper acceleration is the same.

The first interpretation of "at the same rate" leads to Bell's spaceship paradox. This may be worth reading about, but seems like it's veering away from the topic the OP is interested in.

 

[PeterDonis]

I would say no

The "no" answer is obvious for the question as stated, for the reason I and others have already stated several times: there is no such thing as "inertial to each other".

I suspect that you are inadvertently reading "inertial to each other" to mean "at rest relative to each other". (I also suspect the OP of having the same confusion, which is why I've commented on this before in this thread.)

 

[PAllen]

"At the same rate" sounds simple, but it depends on whether you mean that their acceleration in some inertial frame of reference is the same, or whether you mean their proper acceleration is the same.

Aren't these the same thing? Not in magnitude, but if the acceleration of two particles is the same per some inertial coordinate system, their proper acceleration must be the same. In the Rindler case, neither proper acceleration is the same, nor coordinate acceleration in a given inertial frame.

 

[David Lewis]

… if the acceleration of two particles is the same per some inertial coordinate system, their proper acceleration must be the same.

Even if their proper accelerations are identical, the distance between the objects is frame-dependent. One observer may say the motions started at the same time, and another will say one object started moving before the other.

 

[PAllen]

Even if their proper accelerations are identical, the distance between the objects is frame-dependent. One observer may say the motions started at the same time, and another will say one object started moving before the other.

That is irrelevant to may point. My point is simply that identical acceleration in any given inertial frame implies identical proper acceleration (per simultaneity of that frame). thus the two cases are equivalent. I said nothing about distances.

If you want to allow for the case of varying proper acceleration that is associated with identical coordinate acceleration in some frame but not in another, then you cannot talk about identical proper acceleration without specifying a frame any more than you can for coordinate acceleration, because which proper accelerations you compare is frame dependent. That is, unless proper accelerations are constant, the "seemingly invariant" statement that proper accelerations are identical has no meainginIg without specifying a frame. And, if proper accelerations are identical per that inertial frame, then so are coordinate accelerations.

 

[jaketodd]

You're welcome!

Please consider the following: The two planets are moving with a non-zero impulse, and the observer starts moving, after the planets start moving, with the same, non-zero impulse. If you take the derivative of the two planets' accelerations, for a point in time, then the resulting impulse would match the impulse of the observer, and all "impulse-ometers" (if such a thing exists), would agree. However, the velocities of the planets, and the observer, would not agree! So all "impulse-ometers", for that point in time, would be reference frame invariant, even though the velocities differ. This point of view implies, that the amount of sand, flying off the planets, is reference frame variant, even though the "impulse-ometers" are reference frame invariant. The higher the velocity of the two planets, the more sand will fly off them. The only way I can see out of this, is by using the derivative of impulse (meters per second, per second, per second, per second), leading the way to "impulse-change-ometers!" Those would always agree, and be reference frame invariant. However, I think they would require spacetime bending back, and through itself.

Everyone's thoughts on this are welcome.

Thanks!

Jake

 

[jbriggs444]

Please consider the following: The two planets are moving with a non-zero impulse

What does that even mean?

 

[Dale]

@jaketodd can you clarify what distinguishes an impulseometer from an accelerometer?

This point of view implies, that the amount of sand, flying off the planets, is reference frame variant

Then the point of view is wrong.

 

[PeterDonis]

all "impulse-ometers" (if such a thing exists)

To know whether such a thing exists, we would have to know what you mean by "impulse".

 

[jaketodd]

To know whether such a thing exists, we would have to know what you mean by "impulse".

Impulse. I learned about it in high school honors physics class. It's the derivative of acceleration. It's how fast acceleration is changing.

Thanks, my friend. Jake

 

[jbriggs444]

Impulse. I learned about it in high school honors physics class. It's the derivative of acceleration. It's how fast acceleration is changing.

The first derivative of acceleration is more commonly known as "jerk". https://en.wikipedia.org/wiki/Jerk_(physics). It has dimensions of distance per time³
My first year physics class taught that "impulse" is a momentary transfer of momentum. It has dimensions of mass times velocity.

 

[Nugatory]

Impulse. I learned about it in high school honors physics class. It's the derivative of acceleration. It's how fast acceleration is changing.

That's not right. The derivative of acceleration is called "jerk".

Impulse is the integral of force with respect to time, so that (for example) applying 10,000 Newtons for two milliseconds is the same impulse as 20,000 Newtons for one millisecond. It's useful when analyzing collisions and sudden impacts, problems in which the exact acceleration profile is less interesting than the total momentum transfer. And because it is the integral of force over time, in principle it can be calculated from accelerometer readings.

 

[jaketodd]

The first derivative of acceleration is more commonly known as "jerk". https://en.wikipedia.org/wiki/Jerk_(physics). It has dimensions of distance per time³
My first year physics class taught that "impulse" is a momentary transfer of momentum. It has dimensions of mass times velocity.

Okay, then let's call it Jerk. So in my previous post, it would be "jerk-ometers" and "jerk-change-ometers."

Sorry about the confusion,

Jake

 

[jbriggs444]

Reading "impulse" as "time rate of change of acceleration"...

Please consider the following: The two planets are moving with a non-zero impulse, and the observer starts moving, after the planets start moving, with the same, non-zero impulse.

So we have two planets, both at rest and both with zero acceleration. In the initial rest frame, they are side by side and start accelerating simultaneously at a rate that increases uniformly from zero.

We have an observer standing to one side also at rest near the initial position of the planets. The observer begins accelerating later but also at a rate that increases uniformly from zero.

One assumes that we are dealing with proper accelerations here. Each entity experiences a uniform rate of increase in felt-acceleration over experienced-time.

Is this an accurate description of the setup so far?

If you take the derivative of the two planets' accelerations, for a point in time, then the resulting impulse would match the impulse of the observer, and all "impulse-ometers" (if such a thing exists), would agree.

As constructed, the rate of change of acceleration for all entities is constant. So yes, their "impulse-ometers" all read identically.

However, the velocities of the planets, and the observer, would not agree! So all "impulse-ometers", for that point in time, would be reference frame invariant, even though the velocities differ.

Certainly true.

This point of view implies, that the amount of sand, flying off the planets, is reference frame variant

The amount of sand that flies off is independent of velocity -- unless you imagine it blowing off in some sort of ether wind.

If you pick out a starting event and a stopping event, the amount of sand that flies off a planet between the two is a physical fact and is invariant.

The rate at which sand flies off can vary between frames because the time interval judged to have elapsed can vary between frames.

 

[jaketodd]

The amount of sand that flies off is independent of velocity -- unless you imagine it blowing off in some sort of ether wind.

It's called inertia.

 

[PeterDonis]

the derivative of acceleration

Ok, then you need to distinguish two kinds of "derivative of acceleration with respect to time":

(1) Derivative of proper acceleration with respect to the observer's proper time. We could call this "proper jerk", and it is an invariant, since proper acceleration and proper time are both invariants. Any actual observable, like how much sand is flying off of a planet, must depend only on invariants, so it would depend on this if it depended on jerk at all. (Also, a "jerk-ometer" would measure this, just like an accelerometer measures proper acceleration.)

(2) Derivative of coordinate acceleration with respect to coordinate time. We could call this "coordinate jerk", and it depends on your choice of coordinates, so it's not an invariant and no actual observable can depend on it.

Do you see the general rule here? If so, hopefully that will forestall further questions along these same lines.

 

[jaketodd]

Ok, then you need to distinguish two kinds of "derivative of acceleration with respect to time":

(1) Derivative of proper acceleration with respect to the observer's proper time. We could call this "proper jerk", and it is an invariant, since proper acceleration and proper time are both invariants. Any actual observable, like how much sand is flying off of a planet, must depend only on invariants, so it would depend on this if it depended on jerk at all. (Also, a "jerk-ometer" would measure this, just like an accelerometer measures proper acceleration.)

(2) Derivative of coordinate acceleration with respect to coordinate time. We could call this "coordinate jerk", and it depends on your choice of coordinates, so it's not an invariant and no actual observable can depend on it.

Do you see the general rule here? If so, hopefully that will forestall further questions along these same lines.

No, I don't understand what you mean. The "jerk-ometers" would all agree, yet the velocities would differ, creating two different realities, dependent on reference frame. The "jerk-ometers" are invariant because they rely on the geometry of spacetime, which is universal, as you have said.

 

[jbriggs444]

No, I don't understand what you mean. The "jerk-ometers" would all agree, yet the velocities would differ, creating two different realities, dependent on reference frame. The "jerk-ometers" are invariant because they rely on the geometry of spacetime, which is universal, as you have said.

The velocity is a coordinate dependent quantity. Accordingly, as @PeterDonis points out, no observable quantity depends on it.

 

[jaketodd]

The velocity is a coordinate dependent quantity. Accordingly, as @PeterDonis points out, no observable quantity depends on it.

If jerk and acceleration can be invariant, in the context of spacetime geometry, then so can velocity, because the derivative of velocity, is acceleration, and the derivative of acceleration is jerk. They are all self-contained within spacetime geometry. Perhaps "proper velocity?"

 

[A.T.]

If jerk and acceleration can be invariant, in the context of spacetime geometry, then so can velocity, because the derivative of velocity, is acceleration

The derivative of velocity, is coordinate acceleration, which is frame dependent. What is frame invariant is proper acceleration.

 

[jbriggs444]

If jerk and acceleration can be invariant, in the context of spacetime geometry, then so can velocity, because the derivative of velocity, is acceleration, and the derivative of acceleration is jerk. They are all self-contained within spacetime geometry. Perhaps "proper velocity?"

First you would have to define "proper velocity". Doing it as the integral of proper acceleration over proper time runs into problems -- you would be taking an infinite sum of infinitesimal vectors drawn from different vector spaces.

 

[Ibix]

If jerk and acceleration can be invariant, in the context of spacetime geometry, then so can velocity,

To add to A.T.'s comment, velocity turns out to be related to the angle between your worldline (your path through spacetime) and the worldline of whatever you chose as stationary. But proper acceleration (not coordinate acceleration) is a measure of the curvature of your worldline - it has no dependence on any other worldline, so is grame independent.

 

[jaketodd]

The derivative of velocity, is coordinate acceleration, which is frame dependent. What is frame invariant is proper acceleration.

First you would have to define "proper velocity". Doing it as the integral of proper acceleration over proper time runs into problems -- you would be taking an infinite sum of infinitesimal vectors drawn from different vector spaces.

Once again, proper velocity might do the trick. "Proper velocity equals velocity at low speeds." That's according to: https://en.wikipedia.org/wiki/Proper_velocity

 

[PeterDonis]

If jerk and acceleration can be invariant, in the context of spacetime geometry, then so can velocity, because the derivative of velocity, is acceleration, and the derivative of acceleration is jerk.

Wrong. Again you are not paying careful attention to the distinction I made in my previous post. See below.

Perhaps "proper velocity?"

Here's the problem: there is no such thing as proper velocity. That is, there is no invariant quantity that is a velocity, corresponding to proper acceleration or its derivative with respect to proper time.

(There are some technicalities involved here, but they are beyond the scope of a "B" level thread.)

 

[jbriggs444]

Once again, proper velocity might do the trick. "Proper velocity equals velocity at low speeds." That's according to: https://en.wikipedia.org/wiki/Proper_velocity

So, what does this have to do with the rate at which sand drops off of accelerating planets?

 

[PeterDonis]

proper velocity might do the trick

Nope. The name "proper velocity" for this quantity is misleading; it is not an invariant the way proper acceleration or proper jerk are invariants. Don't rely on Wikipedia as a source for learning physics.

 

[Dale]

It's called inertia.

Yes, it is called inertia, and it behaves the way @jbriggs444 described, not the way you described.