What is the coordinate in this system then and how do they all connect to each other? I've read the mentioning of an observer and the observer's state of motion and I don't understand how that relates to a frame of reference.

Jaime Coleman
2022-05-10
Answered

What is the coordinate in this system then and how do they all connect to each other? I've read the mentioning of an observer and the observer's state of motion and I don't understand how that relates to a frame of reference.

You can still ask an expert for help

Elyse Huff

Answered 2022-05-11
Author has **15** answers

The modern concept of a frame of reference did not exist in Newtonian mechanics; the phrase reference frame was not used until the late nineteenth century. In Newton's time the fixed stars were invoked as a reference frame supposedly at rest relative to absolute space, but that is as near as it gets. Of course we now know that the "fixed" stars are not fixed. Newton resorted to the notions of absolute space and absolute time, as a starting point from which he could give meaning to his laws precisely because no better concept existed. Thus the precise answer to this question is that in the context of Newtonian mechanics you must first assume absolute space and absolute time.

Of course this was changed by Einstein. A reference frame does not refer to a coordinate system, such as absolute space (as some authors seem to think). A frame refers to physical matter. A reference frame is the matter relative to which a coordinate system is defined. We can talk of the Earth frame, the frame of the fixed stars, and when travelling by car it is natural to think of the car as the reference frame. The minimum requirement for a reference frame is that it must contain a clock, a ruler (or equivalent apparatus for measuring distance), and a physical definition for coordinate axes.

In general relativity Newtonian mechanics is re-expressed within the context of a reference frame (not a reference frame in the context of Newtonian mechanics). First we can use Newton's first law to define an inertial reference frame (replacing the need to use Newton's first law to determine "absolute space"). Then inertial frames are necessarily local and Newtonian mechanics holds in inertial reference frames.

Of course this was changed by Einstein. A reference frame does not refer to a coordinate system, such as absolute space (as some authors seem to think). A frame refers to physical matter. A reference frame is the matter relative to which a coordinate system is defined. We can talk of the Earth frame, the frame of the fixed stars, and when travelling by car it is natural to think of the car as the reference frame. The minimum requirement for a reference frame is that it must contain a clock, a ruler (or equivalent apparatus for measuring distance), and a physical definition for coordinate axes.

In general relativity Newtonian mechanics is re-expressed within the context of a reference frame (not a reference frame in the context of Newtonian mechanics). First we can use Newton's first law to define an inertial reference frame (replacing the need to use Newton's first law to determine "absolute space"). Then inertial frames are necessarily local and Newtonian mechanics holds in inertial reference frames.

Alisa Durham

Answered 2022-05-12
Author has **3** answers

A reference frame is a coordinate system that defines the position of any body and if the body is at rest or moving.

A good example is the longitude and latitude given by a GPS device. We assume that if that numbers change we are moving, the direction and speed can be related to how, and how fast they move.

From an observer out of the surface of the Earth but close to it, maybe the same frame is still useful. The ISS, at 400 km of altitude can track itself using the same system, only adding the altitude.

But from an hypothetical lunar base, it would be not pratical. They would probably use lunar coordinates to define position and movement.

A good example is the longitude and latitude given by a GPS device. We assume that if that numbers change we are moving, the direction and speed can be related to how, and how fast they move.

From an observer out of the surface of the Earth but close to it, maybe the same frame is still useful. The ISS, at 400 km of altitude can track itself using the same system, only adding the altitude.

But from an hypothetical lunar base, it would be not pratical. They would probably use lunar coordinates to define position and movement.

asked 2022-04-06

A ball is thrown at a relative velocity of $3\text{}m/s$ in $+ve$ $x-$ direction in a rocket in gravity free space. The rocket has a constant acceleration of $2m/{s}^{2}$ also in $+ve$ $x-$ direction.

Whether uniform acceleration equations of motion would apply in above case when using the rocket as a frame of reference. As an example, would the equation $s=ut+.5a{t}^{2}$ apply if rocket is taken as frame of reference?

Whether uniform acceleration equations of motion would apply in above case when using the rocket as a frame of reference. As an example, would the equation $s=ut+.5a{t}^{2}$ apply if rocket is taken as frame of reference?

asked 2022-05-13

A rotating frame of reference (since you rotate your BEC to generate these), where the energy is given by:

$\stackrel{~}{E}[\mathrm{\Psi}]=E[\mathrm{\Psi}]-L[\mathrm{\Psi}]\cdot \mathrm{\Omega},$

Where $\mathrm{\Omega}$ is the rotational velocity which you apply to the BEC.

Now the argument that the term $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ came from?

I know that in a rotating frame of reference you have that $\overrightarrow{v}={\overrightarrow{v}}_{r}+\overrightarrow{\mathrm{\Omega}}\times \overrightarrow{r}$ . If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:

$\frac{1}{2}I{\mathrm{\Omega}}^{2}=\frac{1}{2}J\mathrm{\Omega}.$

This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?

$\stackrel{~}{E}[\mathrm{\Psi}]=E[\mathrm{\Psi}]-L[\mathrm{\Psi}]\cdot \mathrm{\Omega},$

Where $\mathrm{\Omega}$ is the rotational velocity which you apply to the BEC.

Now the argument that the term $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ came from?

I know that in a rotating frame of reference you have that $\overrightarrow{v}={\overrightarrow{v}}_{r}+\overrightarrow{\mathrm{\Omega}}\times \overrightarrow{r}$ . If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:

$\frac{1}{2}I{\mathrm{\Omega}}^{2}=\frac{1}{2}J\mathrm{\Omega}.$

This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?

asked 2022-05-09

Imaging that the speed of that spacecraft is almost the speed of light "${c}^{-1}\text{}\text{m/s}$". But no acceleration. Meaning it's an inertial frame of reference right? Will you be able to throw a ball in the direction of movement? What about the direction inverse to the movement? Does that means that there is a direction of movement in an inertial frame of reference?

asked 2022-05-08

The reason the galaxy does not spin around me (relative to me) at a trillion times the speed of light when I do a pirouette is the inertial frame of reference to which I am confined along with the rest of humanity.

Where does this frame of reference end?

To the best of my limited understanding, gravity doesn't just stop at some point. Anything that has mass has gravity; gravitational waves (or whatever) extend from the mass spherically. They weaken as they get further away from the source, they get pushed around by other, more powerful, gravity waves, they become progressively weaker as the distance increases, but do they ever vanish completely, i.e. get reduced to absolute zero? Or do they become weaker and weaker forever, without zeroing out? And if so, where does a frame of reference end?

Where does this frame of reference end?

To the best of my limited understanding, gravity doesn't just stop at some point. Anything that has mass has gravity; gravitational waves (or whatever) extend from the mass spherically. They weaken as they get further away from the source, they get pushed around by other, more powerful, gravity waves, they become progressively weaker as the distance increases, but do they ever vanish completely, i.e. get reduced to absolute zero? Or do they become weaker and weaker forever, without zeroing out? And if so, where does a frame of reference end?

asked 2022-04-26

Can you identify the direction of your movement? Let's say ship moving nose forward, or tail forward? Is there a sense of "direction of movement" in such an inertial frame of reference?

asked 2022-05-15

We have two particles $A$ and $B$ flying with different speed. Now the rest frame for $A$ and $B$ is not the same. But is the inertial frame the same? Or would be the inertial frame the same if we have a third fix point $C$ and we would observe them both from $C$?

For me it seems logical to have $3$ frame of references. From the point of view of $A$, from the point of view of $B$ and from an independent point of view $C$. Is this correct?

For me it seems logical to have $3$ frame of references. From the point of view of $A$, from the point of view of $B$ and from an independent point of view $C$. Is this correct?

asked 2022-04-12

I have one vector and its components given a particular frame of reference (${A}_{x},\phantom{\rule{thinmathspace}{0ex}}{A}_{y},\phantom{\rule{thinmathspace}{0ex}}{A}_{z}$). I have a second vector's components that is computed in a different frame of reference (${B}_{x},\phantom{\rule{thinmathspace}{0ex}}{B}_{y},\phantom{\rule{thinmathspace}{0ex}}{B}_{z}$) and I need the dot product between these two vectors. I can't simply take the dot product (i.e.,$\mathbf{A}\cdot \mathbf{B}={A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}$) because the two vector's components are computed with different orientations in mind.

Given that there is no way to orient one vector to match the frame of reference of the other vector, is there a way I can still get a dot product?

Given that there is no way to orient one vector to match the frame of reference of the other vector, is there a way I can still get a dot product?