Does the Galilean transform rely on 2 events?

[jimmy4554564]

Homework Statement

$∆x = ∆x′ + v ∆t$

Relevant Equations

$∆x = ∆x′ + v ∆t$

From my limited understanding the Galilean transform has 2 frames but 4 four perspectives. For example x is the stationary frame when using
$∆x = ∆x′ + v ∆t$ and x' is moving. When using $∆x' = ∆x - v ∆t$ and x' is stationary and x is moving.

Now lets use the example of $∆x = ∆x′ + v ∆t$ . Delta x' represents the event in (x')'s frame and if the object is moving it will have a time and velocity.
By 2 events I mean $x'$ position and $v ∆t$ represents the second event the first object moving.

Is this explanation correct?

 

[Orodruin]

No. Or at least it is impossible to tell based on your description.

Two events and their respective coordinates in two different frames are involved. I do not understand what you think would be ”4 perspectives”

 

[jimmy4554564]

Take the picture below. $∆x = ∆x′ + v ∆t$ . x sees x' moving . I have 2 perspectives here. I kind of just invented the term perceptive. While if I switch to $∆x' = ∆x + v ∆t$ x' is stationary and x is moving.
I would also need 2 graphs where velocity is negative and is traveling in the opposite direction and another graph where x was stationary.

So to keep it short perceptive = the number of graphs.

I realize there are 3 graphs but I think 2 of them could be combined in the picture below.
Should I site where I got the picture from?

 

[PeroK]

Take the picture below. $∆x = ∆x′ + v ∆t$ . x sees x' moving . I have 2 perspectives here. I kind of just invented the term perceptive. While if I switch to $∆x' = ∆x + v ∆t$ x' is stationary and x is moving.
I would also need 2 graphs where velocity is negative and is traveling in the opposite direction and another graph where x was stationary.

So to keep it short perceptive = the number of graphs.

I don't really understand what you are asking. In this case, we have something like:

Train frame: a passenger is walking down the train - a dispacement of $\Delta x'$ (Perhaps she walked from one seat to another seat on the train. The seats are $\Delta x'$ apart.)

Ground frame: the passenger is moving with the train and walking down the train. In this frame her displacement is $\Delta x$, which is a combination of the movement of the train and the movement of the passenger relative to the train.

The Galilean transformation allows you to express $\Delta x$ in terms of $\Delta x'$ and the movement of the train, which is $v \Delta t'$. Your diagram illustrates this:

View attachment 319699

In short:$$\Delta x = v\Delta t' + \Delta x'$$Is that clear?

 

[Orodruin]

I have 2 perspectives here. I kind of just invented the term perceptive.

That is generally going to be a bad idea as nobody else will understand what you are talking about if you do not use accepted terminology.

The images you are showing are illustrating why the transformation takes the form it does. The opposite illustration will just be a redrawing of the same thing.

 

[jimmy4554564]

Please correct me if I am wrong but each graph represents a stationary frame when using the graph.

My mistake was assuming one graph is moving and the other is stationary. Instead each graph represents stationary from there perceptive and the other graph is moving.

So basically $x = x' + vt$ the LHS != RHS. But $stationary = moving$ and moving becomes stationary when I move $vt$ and vice versa.

Now this confuses me I always thought the left and right side have to be equal
For example $-1 = -1$ but if I am not mistaken in the Galilean transform $stationary= moving$, then moving = stationary. How is this possible?I want to repeat I understand that moving and stationary have different values I always thought that an equal sign has to have the same value.

If I made mistakes correct me.

Thank you everyone.

 

[PeroK]

Please correct me if I am wrong but each graph represents a stationary frame when using the graph.

Each frame is inertial. Stationary and moving are relative. There is no such thing as an absolutely stationary frame. The surface of the Earth is not absolutely stationary.

My mistake was assuming one graph is moving and the other is stationary. Instead each graph represents stationary from there perceptive and the other graph is moving.

Stationary and moving are relative. Think of left and right.

So basically $x = x' + vt$ the LHS != RHS. But $stationary = moving$ and moving becomes stationary when I move $vt$ and vice versa.

Now this confuses me I always thought the left and right side have to be equal
For example $-1 = -1$ but if I am not mistaken in the Galilean transform $stationary= moving$, then moving = stationary. How is this possible?

I don't understand this.

I want to repeat I understand that moving and stationary have different values

moving and stationary are relative terns. They are not numbers.

 

[jimmy4554564]

Each frame is inertial. Stationary and moving are relative. There is no such thing as an absolutely stationary frame. The surface of the Earth is not absolutely stationary.

I just meant that in the S frame the y axis and x = 0 represent a stationary frame and the other points are a different frame. This is the same for the S' frame. The S' frame the y' axis and x' = 0 represent a stationary frame and the other points are a different.

I don't understand this.

I just meant that $x = x' +vt$ . Lets say the values are $x' = 5m + (10m/s) (10 sec)$ $x' = 105m$.
While $x' = x - vt$ , $x = 105m - (10m/s) (10 sec)$ , $x = 5 m$ . I guess my confusion was thinking x is the exact same as x' .

Did I make any mistakes ?

 

[PeroK]

I just meant that in the S frame the y axis and x = 0 represent a stationary frame and the other points are a different frame.

Each point is in all frames of reference. The x and y axes are the coordinate axes.

This is the same for the S' frame. The S' frame the y' axis and x' = 0 represent a stationary frame and the other points are a different.

The S' frame is moving relative to the S frame.

I just meant that $x = x' +vt$ . Lets say the values are $x' = 5m + (10m/s) (10 sec)$ $x' = 105m$.
While $x' = x - vt$ , $x = 105m - (10m/s) (10 sec)$ , $x = 5 m$ . I guess my confusion was thinking x is the exact same as x' .

Did I make any mistakes ?

If the frames are moving relative to each other along the x axis then $x \ne x'$.

 

[jimmy4554564]

Thanks for the help we are in agreement i just worded it a little wonky.