# Thoughts about Galilean transformations

• B
• ilasus
In summary, the Galilei transformations are a set of equations that describe how the coordinates of a point P change between two reference systems R and R' moving at constant speed v relative to each other. These equations include the equality t' = t, stating that time measured in R' is identical to time measured in R. However, there is a flaw in this logic, as the same observer P cannot be at rest in both reference frames simultaneously. This leads to contradictions between the equalities and inequalities in the equations.
ilasus
TL;DR Summary
Galilei transformations are a set of equations that describe how the coordinates of a point P change between two reference systems R, R' moving at constant speed v relative to each other. For example, when moving from the reference system R to R', the Galilei transformations are given by the equations:

(g) x' = x – vt, y' = y, z' = z

and when moving from the reference system R' to R, the Galilei transformations are given by the equations:

(g') x = x' + vt', y = y', z = z'

where x, y, z are the coordinates of the point P in the reference system R, and x', y', z' are the coordinates of the point P in the reference system R'. To these equations, classical physics also added the equality t' = t, which states that the time measured in the reference system R' is identical to the time measured in the reference system R. So according to the point of view of the observers in the reference system R, (g), the origin O' of the reference system R' approaches with speed v the point P, and according to the point of view of the observers in the reference system R', (g'), the origin O of the reference system R is moving away from point P with velocity -v. But if we assume that point P is actually an observer, then in the first case, the observer P is at rest at a distance x in the reference frame R, and in the second case, the observer P is at rest at distance x' in the reference frame R'. In other words, the observer P would be an "exception observer", because unlike the other observers, who are at rest in only one of the reference systems R, R', the observer P is at rest in both reference systems. Obviously, the same observer P can be at rest in two different reference frames, but not at the same time. It follows that the time t in which the origin O' approaches the point P at rest in the reference system R, is not identified with the time t' in which the origin O moves away from the point P at rest in the reference system R'. But if t ≠ t', then the positions of the point P calculated in relation to the origins of the reference systems R, R' cannot be identical either: x' ≠ x - vt and x ≠ x' + vt'. This results in a contradiction between the equalities:

(G) x' = x - vt, x = x' + vt, t' = t

and inequalities:

(G*) x' ≠ x – vt, x ≠ x' + vt', t' ≠ t

How is it explained? Or, if you think my logic is flawed, where is the flaw? Thanks.

ilasus said:
So according to the point of view of the observers in the reference system R, (g), the origin O' of the reference system R' approaches with speed v the point P, and according to the point of view of the observers in the reference system R', (g'), the origin O of the reference system R is moving away from point P with velocity -v.
You are using P to refer to two different points and/or v to refer to two different speeds but you act as if they were the same. This self-contradiction is the root of your problem.

I honestly want to understand your phrase, but I can't. How can I understand what you are saying? Can you possibly explain more simply? Or give an example?

$$\begin{eqnarray*} x'&=&x-vt\\ \frac{dx'}{dt}&=&\frac{dx}{dt}-v \end{eqnarray*}$$
Thus if some object has velocity ##\frac{dx}{dt}=\pm v## with respect to one reference frame then its velocity ##\frac{dx'}{dt}## in the other is either zero or ##-2v##. Thus your point P, which has velocity ##v## with respect to one frame cannot be the same as your point P which has velocity ##-v## with respect to the other frame. So you are using P to refer to two different points - unless you actually meant that the point P is moving with some velocity ##u=v/2##. In that case it has equal and opposite velocities in the two frames, but you are using ##v## to refer to both ##v## and ##u##.

I referred to the case "P is at rest" and not to the case "P is in motion". So we have reproduced the point of view of the observers in the reference frames R and R' for the case "P is at rest":

- Viewpoint of observers in R: P is (at rest) at distance x at time t and O' is approaching P at speed v during this time. That expresses the equation x' = x - vt.

- Viewpoint of observers in R': P is (at rest) at distance x' at time t' and O is moving away from P at speed -v during this time. That expresses the equation x = x' + vt'.

Under these conditions, I reasoned that if P is in fact an observer, then observer P is at rest both from the point of view of the observers in R and from the point of view of the observers in R'. This means that P is an "exception observer", since it is the only observer considered at rest by both the observers in R and those in R'. Do you agree that there is such an "exception observer"?

ilasus said:
I referred to the case "P is at rest" and not to the case "P is in motion"
Ok. In my last post set ##dx/dt=0## and see what ##dx'/dt## is. Hint: it is not also zero. Thus you are referring to two different points and calling them both P.
ilasus said:
Do you agree that there is such an "exception observer"?
Of course not. Define frame R to be the rest frame of a set of traffic lights and P to be a car (or the location of a car) at rest in this frame. Define R' to be the rest frame of a truck driving through the lights at 30mph and P to be a car being towed by the truck, so at rest in R'. Can those two cars be the same car, both stopped at the lights and being towed at 30mph?

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You seem to be think that there is an absolute reference frame where things not moving are at rest in all other reference frames. This is not the case. An object can only be at rest within its own rest frame.

By the velocity v of frame R' with respect to R, we mean $$v=\left(\frac{\partial x}{\partial t}\right)_{x'}$$By the velocity -v of frame R with respect to R', we mean $$-v=\left(\frac{\partial x'}{\partial t'}\right)_{x}$$

Vanadium 50, malawi_glenn, Ibix and 1 other person
1. Please allow me to use those notions that Galileo knew, so to refer to the definition of speed that Galileo knew and not to the definition of speed that you refer to.

2. We are talking about logic. So please don't say what you "seem" to believe about what I believe, but analyze my statements accurately and with scientific rigor. Do you think the sentences I wrote are logical? Did I correctly express the point of view of the observers in R? Did I correctly express the point of view of the observers in R'? Do you think I am wrong somewhere? Thanks for understanding.

phinds and weirdoguy
Thread locked temporarily for Moderation...

ilasus said:
We are talking about logic
No, we are talking about physics. Physics uses logic as a tool, but physics is not the same as logic. You can make statements which are logically impeccable but still get the physics wrong.

That said, it seems to me that you have made errors in both respects. See below.

ilasus said:
Did I correctly express the point of view of the observers in R? Did I correctly express the point of view of the observers in R'? Do you think I am wrong somewhere?
No, no, and yes.

ilasus said:
x, y, z are the coordinates of the point P in the reference system R, and x', y', z' are the coordinates of the point P in the reference system R'.
So far, so good.

ilasus said:
To these equations, classical physics also added the equality t' = t, which states that the time measured in the reference system R' is identical to the time measured in the reference system R
Ok here.

ilasus said:
So according to the point of view of the observers in the reference system R, (g), the origin O' of the reference system R' approaches with speed v the point P, and according to the point of view of the observers in the reference system R', (g'), the origin O of the reference system R is moving away from point P with velocity -v.
Wrong in two ways. First, you have not given enough information to say how the point P is moving in either R or R'. For that, you would actually have to give specific functions x(t), y(t), z(t) describing how point P is moving in R, or specific functions x'(t'), y'(t'), z'(t') describing how point P is moving in R'. Given a set of functions in one frame, you can then use the Galilean transformations to obtain the corresponding set of functions in the other frame. But you can't deduce the functions in either frame from the Galilean transformations alone.

Second, your claim about the relationship between the motion of P in R and the motion of P in R' is wrong. For a simple counterexample, consider the case where point P is at rest at the spatial origin of R, i.e., x(t) = y(t) = z(t) = 0 for all t. Then the Galilean transformation gives x'(t') = - v t', y'(t') = z'(t') = 0. So here, the origin of R is point P, i.e., it "approaches" P with speed 0, and the origin of R' is moving away from point P at speed v (because point P is moving away from the origin at speed -v).

ilasus said:
if we assume that point P is actually an observer
Then you have changed nothing about the physics. An "observer" is just something that is located at some point. Saying P is an observer doesn't magically change how P moves in any frame.

ilasus said:
the observer P is at rest in both reference systems.
This is impossible, as has already been pointed out.

ilasus said:
Obviously, the same observer P can be at rest in two different reference frames, but not at the same time.
Wrong. A point P is at rest in one frame only. Its state of motion can't change with time, so the frame it is at rest in at any given instant is the frame it is always at rest in.

Thread unlocked to give the OP one opportunity to respond. @ilasus, please think carefully before responding. Simply re-asserting anything you have already asserted, or making general claims like "I'm using logic", will add nothing to the thread and will simply result in the thread being closed permanently. You either need to understand and accept the errors that have been pointed out, and retract your claim that the Galilean transformation is somehow wrong, or you need to actually engage with the criticisms that have been made and show why they are not valid. (And any attempt at the latter will face a very, very difficult burden of proof.)

I did not claim that the Galilean transformations are wrong. I stated that my interpretations of the equations x' = x - vt and x = x' + vt', presented in post #5, lead to a contradiction. And I wanted to know if my logic is wrong and why. Thank you for the answers.

ilasus said:
I stated that my interpretations of the equations x' = x - vt and x = x' + vt', presented in post #5, lead to a contradiction. And I wanted to know if my logic is wrong and why.
Your problem (as in your last three threads (1, 2, 3), now I've reviewed them) is that your setup is contradictory. One object cannot be at rest in two frames, but you keep setting up a problem where you define your object P as at rest in two frames.

The contradiction is right there in your scenario specification - it's nothing to do with the coordinate transforms, Galilean or otherwise.

DrClaude and Vanadium 50
ilasus said:
I did not claim that the Galilean transformations are wrong.
Not in so many words, but that is what this claim...

ilasus said:
I stated that my interpretations of the equations x' = x - vt and x = x' + vt', presented in post #5, lead to a contradiction.
...implies.

ilasus said:
I wanted to know if my logic is wrong and why.
In other words, you have no substantive response to the explanations you have already received in this thread. That being the case, this thread is now closed. Thanks to all who participated.

SammyS

## What are Galilean transformations?

Galilean transformations are a set of equations in classical mechanics that relate the coordinates and time of events as observed in two different inertial frames of reference moving at a constant velocity relative to each other. They are named after the Italian scientist Galileo Galilei and are used to transform the position and time coordinates from one frame to another.

## How do Galilean transformations differ from Lorentz transformations?

Galilean transformations assume that time is absolute and the same for all observers, which works well at low velocities compared to the speed of light. Lorentz transformations, on the other hand, take into account the constancy of the speed of light and the relativity of simultaneity, making them applicable in special relativity. Lorentz transformations replace Galilean transformations when dealing with high velocities close to the speed of light.

## What are the basic equations of Galilean transformations?

The basic equations of Galilean transformations for a frame moving with a constant velocity $$v$$ along the x-axis relative to another frame are:$$x' = x - vt$$$$y' = y$$$$z' = z$$$$t' = t$$These equations show how the coordinates (x, y, z) and time (t) in one frame relate to the coordinates (x', y', z') and time (t') in another frame.

## Why are Galilean transformations important in classical mechanics?

Galilean transformations are important in classical mechanics because they provide a simple and intuitive way to relate the observations made in different inertial frames. They are foundational in understanding how velocities and positions transform under relative motion, which is crucial for solving many problems in Newtonian mechanics. They also illustrate the principle of relativity, which states that the laws of physics are the same in all inertial frames.

## What are the limitations of Galilean transformations?

Galilean transformations have limitations when dealing with high velocities close to the speed of light. They do not account for the effects of time dilation and length contraction predicted by special relativity. As a result, they fail to accurately describe the behavior of objects moving at relativistic speeds. In such cases, Lorentz transformations must be used to provide an accurate description of the physical phenomena.

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