# What kind of 'rotation' is the Gaililean transformation?

1. May 4, 2010

On Wikipedia

http://en.wikipedia.org/wiki/Lorentz_transformation

Thus, the Lorentz transformation can be seen as a hyperbolic rotation of coordinates in Minkowski space, where the rapidity φ represents the hyperbolic angle of rotation.

My question is:

How can the Gaililean transformation been seen, that is, as what kind of 'rotation'?

2. May 4, 2010

### Mentz114

If you mean $x'=x+vt$, that is not a rotation, but a translation that depends on velocity.

$$\left[ \begin{array}{cccc} 1 & 0 & 0 & v_x \\\ 0 & 1 & 0 & v_y \\\ 0 & 0 & 1 & v_z \\\ 0 & 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} x \\\ y \\\ z \\\ t \end{array} \right]$$

3. May 4, 2010

### George Jones

Staff Emeritus
Galilean transformations are analogous to Poincare transformations. Galilean transformations are generated by spatial rotations and spatial translations. Poincare transformations are generated by Lorentz rotations and spacetime translations.

4. May 4, 2010

I am sorry. I know perfectly well what a Galilean transformation is. I also know exactly well how the transformed reference frame moves with respect to the original reference frame in variation with v. I just want to know the name for the Galilean transformation as a certain type of 'rotation' similar to the 'hyperbolic rotation' as a name for the Lorentz transformation. I put rotation between quotes, because I am well aware it is not a rotation as, for example, a circular rotation is.

5. May 4, 2010

### Phrak

Hu? How do you figure?

6. May 4, 2010

I do not have the slightest idea what you want to say with your quotation:

"A man ought to read just as inclination leads him; for what he reads as a task will do him little good." Samuel Johnson

Reading is always: trying to understand what has been said (written) and that could be a very difficult task whether you like it or not.

7. May 4, 2010

You still do not get it. I simply want as an answer something like: ... - rotation and one word for the dots.

8. May 4, 2010

There is not any objection to descripe the Gaililean transformation as a transformation in spacetime.

9. May 4, 2010

### Ich

parabolic.

10. May 4, 2010

### Phrak

This sort of stuff seems to be coming up lately.

I don't think anyone knows what you want, exactly. If you want to 'hyperbolically' transform Euclidean space, use shears and rescaling matrices. No rotations and no translations. With a proper combination of shears and rescaling the end of a vector will follow a hyperbola such as x2 - y2 = c.

11. May 4, 2010

### tiny-tim

The Lorentz transformation is called a 'hyperbolic rotation', as distinct from an ordinary circular rotation, because it doesn't repeat … a 'hyperbolic rotation' can get bigger and bigger indefinitely, while a circular rotation keeps returning to the identity (zero rotation).

This is like a hyperbolic space of constant curvature (the constant is negative), which goes on for ever, as opposed to a spherical space of constant curvature (the constant is positive), which doesn't.

In between those two spaces, we have flat spaces of constant curvature (the constant is zero).

So I suppose, by analogy, you could call the Galilean translation a flat rotation, or a rotation centred at infinity.

(but nobody does)

12. May 4, 2010

### George Jones

Staff Emeritus
Then I have no idea what you want.
When I wrote my post, I was rushing to catch my bus to work, and talking to my daughter. I should have written "Galilean transformations are generated by spatial rotations, spacetime translations, and Galilean boosts. Poincare transformations are generated by Lorentz transformations and spacetime translations."

13. May 4, 2010

### Phrak

Nah. One couldn't possibly be distracted by a little girl demanding daddy's attention. :)

Thanks for the clarifications.

14. May 5, 2010

You wrote: "... you could call the Galilean translation a flat rotation ...". I think this is a wonderful answer, very close to the answer I was looking for. I am writing an introductory text on SR for high school students and I will use this expression to describe the Galilean translation mathematically. I will mention that this expression is coined by you. For that, however, I need your real name.

15. May 5, 2010

### tiny-tim

tiny-tim is my real name​

if you want any biographical details, you can mention that i obtained a degree in fishics at the university of the bowliverse, and since then have done research into fish-decimal calculus under sir isaac the newt

16. May 5, 2010

Thanks a lot for your information. I will use your name and affiliation. At the moment the text is still in Dutch, however, I am about to write the text in English. If you already want to look to the text in Dutch you can go to:

I still have to implement the name coined by you. I have to think how to translate 'flat' in Dutch. By the way, you can very well see how the framework (x',t') moves with respect to (x,t) if you vary v on:

http://webphysics.davidson.edu/applets/Minkowski/Minkowski_FEL.html

If you create a green point, then you can see how the green point moves parallel to the x-axis. That's why I like the name flat rotation.

17. May 5, 2010

Hi Tim,

I was looking for the website of the university of the bowliverse (or is it the University of Bowliverse), but could not find it.
I would be most pleased, if you could give me the address and the address of the department you are working.

18. May 6, 2010

### Cyosis

Hint: Goldfish, bowl, universe.

19. May 7, 2010

I do not have the slightest idea what you mean unless you are joking. That's oké with me.

20. May 7, 2010

### Ben Niehoff

As someone earlier mentioned, it would be called a "parabolic rotation". A parabola being the special case on the border between hyperbolic and circular (elliptical).

Constantly rotating points trace out circles. Constantly accelerating points (i.e., with constantly increasing rapidity) in Minkowski spacetime trace out hyperbolas. Constantly accelerating points in Galilean spacetime trace out parabolas.

21. May 8, 2010

### tiny-tim

"parabolic"

Hi Ben!

I don't like "parabolic" in this context.

Yes, if you vary the eccentricity of a conic (eg by tilting the plane that cuts a cone), you do get a circle which turns into a really elongated ellipse which turns into a parabola which turns into a really elongated hyperbola which turns into a "square" hyperbola.

But if instead you vary the curvature (or the radius), you get an increasingly larger circle (no ellipses), which turns into a plane, which turns into a "square" hyperbola with increasingly smaller major axis, which turns into a letter "X".

The latter procedure, with its constant curvature, much better represents the difference between spherical Galilean and Minkowski space.
I don't see how acceleration comes into it.

The original question arose out of the Lorentz transformation being described as a hyperbolic rotation (of coordinates in Minkowski space) …
no acceleration involved! … only a symmetric rotation of the whole space!