What is the Mathematical Definition of an Inertial Frame?

[Trying2Learn]

This question concerns inertial frames.

I am aware that an inertial frame is one that is not accelerating.
I am aware of an alternative definition: it is one on which no forces are applied.
(Yes, they are the same thing.)
I am also aware of the d'Alembert "forces" that appear when a frame is not inertial.
And I have been satisfied with this for some time.

However, now I am hoping for a more mathematical definition of an inertial frame (but I admit I may be on the verge of confusing myself with an irrelevant issue).

First, go here to the Introduction of Frankel's text on the Geometry of Physics
http://assets.cambridge.org/97811076/02601/frontmatter/9781107602601_frontmatter.pdf

And go to the very last page.

There... right there he asserts that a basis can be obtained by unitized derivatives of the coordinate functions.
(I wish I could post the very next page, but you can get the idea).

I like that...

Now I am wondering if I can extend this to a definition of an inertial frame when the frame bases are obtained by partial derivatives of the coordinate functions that are parameterized by time.

I can sort of reason it for the case of rotations.

But could someone explain how to assert that a frame that is "accelerating in a straight line is not inertial," by using the notion of a frame derived from coordinate functions.

 

[Orodruin]

I am aware of an alternative definition: it is one on which no forces are applied.

You cannot apply a force on a coordinate system. You apply forces on objects.

But could someone explain how to assert that a frame that is "accelerating in a straight line is not inertial," by using the notion of a frame derived from coordinate functions.

Just look at the derivatives of the coordinate frames.

 

[Trying2Learn]

Just look at the derivatives of the coordinate frames.

Yes, but that is where I am having this block. I cannot figure out what to show. And the words are very important.
You said "look at the derivatives of the coordinate FRAMES."
Do you mean to say "look at the derivatives of the coordinate FUNCTIONS?"

If so, could you show me? I just have this block and cannot make the leap to see it.

 

[Orodruin]

Do you mean to say "look at the derivatives of the coordinate FUNCTIONS?"

No. The coordinate frame is the set of coordinate basis vectors. Your coordinates naturally define a set of basis vectors. If those basis vectors change from event to event then your frame is non-inertial.

 

[Trying2Learn]

No. The coordinate frame is the set of coordinate basis vectors. Your coordinates naturally define a set of basis vectors. If those basis vectors change from event to event then your frame is non-inertial.

OK then. I can see how if the frame rotates that the basis vector change.

But if the frame accelerates linearly in ONE DIRECTION, how does that change the base vectors?

 

[Trying2Learn]

OK then. I can see how if the frame rotates that the basis vector change.

But if the frame accelerates linearly in ONE DIRECTION, how does that change the base vectors?

THAT IS IT! that is what I am trying to ask. Now I have zeroed in on my question:

If a frame accelerates linearly in ONE DIRECTION, how does that change the base vectors (and show me, mathematically, not by physical appreciation)?

 

[Orodruin]

OK then. I can see how if the frame rotates that the basis vector change.

But if the frame accelerates linearly in ONE DIRECTION, how does that change the base vectors?

The basis vectors are spacetime 4-vectors, not spatial vectors in three dimensions.

 

[Trying2Learn]

The basis vectors are spacetime 4-vectors, not spatial vectors in three dimensions.

Yes, but is this not, essentially, killing a fly with a sledgehammer?

I understand this path, but that is not my issue.

You are in a car. The car accelerates. You lean back -- fictitious forces, etc. I can see that.
So, CONTINUING that same logic, but in a CLASSICAL sense, how does the linearly accelerating frame change the base vectors (without recourse to 4D space-time)?

 

[Orodruin]

without recourse to 4D spact-time

Spacetime is 4-dimensional also in classical physics. The spatial basis does not change, but the time direction does.

 

[Trying2Learn]

Spacetime is 4-dimensional also in classical physics. The spatial basis does not change, but the time direction does.

Yes, I get that. But I really feel there should be a simpler explanation.

If there exists an angular velocity, the directions of the base vectors change (this is simple and I can see it)
If there exists a linear acceleration... then...?? (How do the base vectors change?)

 

[Orodruin]

If there exists a linear acceleration... then...??

Then you simply cannot deduce this by looking only at the spatial vector basis as it appears only in the time direction.

 

[Trying2Learn]

Then you simply cannot deduce this by looking only at the spatial vector basis as it appears only in the time direction.

Sorry to push like this, but I do not understand your response. It is almost a negative response. It should not be this complicated. There must be a simple statement.

 

[Orodruin]

You are insisting that you should be able to tell that a frame is accelerating by just looking at the spatial basis vectors. This is wrong. If you want to deduce this from looking at the basis vectors of your coordinate frame, you must look at the time direction.

 

[Trying2Learn]

You are insisting that you should be able to tell that a frame is accelerating by just looking at the spatial basis vectors. This is wrong. If you want to deduce this from looking at the basis vectors of your coordinate frame, you must look at the time direction.

OK, now a light has finally been turned on, in the darkness.

Could you elaborate in a few more sentences? NOW I think YOU am closing in on what I need.

 

[Dale]

Yes, but is this not, essentially, killing a fly with a sledgehammer?
...
in a CLASSICAL sense, how does the linearly accelerating frame change the base vectors (without recourse to 4D space-time)?

He answered your question. Why should you object if the answer feels like a sledgehammer or uses spacetime? Frankly, this is an absolutely unacceptable post. You have no basis on which to dismiss his perfectly valid answer like that. If you know of a simpler answer (I don’t) then why are you asking the question? If you don’t know of a simpler answer then on what basis are you characterizing it this way?

The proper response would have been a hearty “thank you” instead of a rude dismissal of a good answer (the only answer as far as I know)

 

[Trying2Learn]

He answered your question. Why should you object if the answer feels like a sledgehammer or uses spacetime? Frankly, this is an absolutely unacceptable post. You have no basis on which to dismiss his perfectly valid answer like that. If you know of a simpler answer (I don’t) then why are you asking the question? If you don’t know of a simpler answer then on what basis are you characterizing it this way?

The proper response would have been a hearty “thank you” instead of a rude dismissal of a good answer (the only answer as far as I know)

He is closing in on my answer. Yes, I am, pushing. Why do you see disrespect when none is intended. Please let him continue.

 

[Orodruin]

I will do this in one spatial dimension since the generalisation is obvious. Consider an inertial frame with coordinates $t$ and $x$. Now introduce a moving origin described by $x_0(t)$ in this frame and introduce the accelerated coordinate system $x' = x - x_0(t)$, $t' = t$. The event vector expressed in the original inertial frame is $\vec X = t \vec e_0 + x \vec e_1$. The coordinate frame in this new system is given by (note that $x = x'+x_0(t')$)
$$
\vec e_0' = \frac{\partial\vec X}{\partial t'} = \vec e_0 \frac{\partial t}{\partial t'} + \vec e_1 \frac{\partial x}{\partial t'} = \vec e_0 + v_0(t') \vec e_1, \qquad
\vec e_1' = \frac{\partial\vec X}{\partial x'} = \vec e_1.
$$
The spatial basis vector $\vec e_1'$ is constant, but $\vec e_0'$ depends on $t'$.

 

[Trying2Learn]

I will do this in one spatial dimension since the generalisation is obvious. Consider an inertial frame with coordinates $t$ and $x$. Now introduce a moving origin described by $x_0(t)$ in this frame and introduce the accelerated coordinate system $x' = x - x_0(t)$, $t' = t$. The event vector expressed in the original inertial frame is $\vec X = t \vec e_0 + x \vec e_1$. The coordinate frame in this new system is given by (note that $x = x'+x_0(t')$)
$$
\vec e_0' = \frac{\partial\vec X}{\partial t'} = \vec e_0 \frac{\partial t}{\partial t'} + \vec e_1 \frac{\partial x}{\partial t'} = \vec e_0 + v_0(t') \vec e_1, \qquad
\vec e_1' = \frac{\partial\vec X}{\partial x'} = \vec e_1.
$$
The spatial basis vector $\vec e_1'$ is constant, but $\vec e_0'$ depends on $t'$.

Ah HA! THAT IS IT! That was the step I was too dense to see. I see it now.

(If I choose to thank you, I will do so privately because I am stubborn and Dale is waiting for me to genuflect which I will not do. Give me a moment.)

THat was what I was looking for! Than... ;-)

 

[Dale]

Why do you see disrespect when none is intended

”killing a fly with a sledgehammer” is disrespectful. Particularly since you labeled the question as an “A” question and thus specifically requested sledgehammers! If you didn’t intend disrespect then reconsider your wording and carefully consider the level of response you request.