# Assuming Lorentz transform is affine

1. Dec 12, 2015

### Erland

Alonso has in fact raised a relevant issue. Given any clock synchronization in the embankment frame, there is a unique pair of points at the embankment with coordinates x=a and x=b, and a unique time t=T, such that at t=T in the embankment frame, these points are next to the back end and the front end of the train, respectively, and x=0 in the embankment frame is next to the midpoint of the train.
In this case, it is assumed that a=-b, that is, that the midpoint between the two points at t is next to the midpoint of the train. But in fact, this cannot be proved from the two postulates alone. We need also use that the Lorentz transformation is linear to prove this (more precisely, affine, if the origins (0,0) are not mapped to each other), and this linearity cannot be proved from the postulates alone. I think that sweet springs with "homogeneity" and FactChecker with "uniformity" mean the same thing.
To derive the linearity from more fundamental or intuitively obvious principles is a very tricky issue. See https://www.physicsforums.com/threa...formations-are-the-only-ones-possible.651640/

2. Dec 12, 2015

### FactChecker

In the larger context of curved space-time, I don't know if it is even correct to talk about linearity. I would call the linearity a simplifying assumption, rather than something to be proven from basic principles. But that does not hurt the logic of relative simultaneity. The distances can be made arbitrarily small so that any non-linearity can be ignored.

3. Dec 12, 2015

### Erland

You are undoubtedly correct, FactChecker, but here we are dealing with Special Relativty, in which there is no curvature. But even in this case, the linearity cannot be proved from the two postulates alone.

4. Dec 12, 2015

### Staff: Mentor

Yes, it can be proved from the postulates. Specifically, the postulates describe transformations between inertial frames. Inertial frames map straight lines to straight lines. Transformations which map straight lines to straight lines are affine.

The property you say is not proven is, in fact, contained in the word "inertial" in the postulates.

5. Dec 12, 2015

### FactChecker

The logic of relative simultaneity holds in any case. The Lorentz contraction is first order and any non-linearity, if it exists, is second order. The effect of any non-linearity can be made arbitrarily small compared to the Lorentz contraction by considering small distances. At high speeds, the Lorentz contraction can be made large and the local effect of non-linearity can be made small.

6. Dec 12, 2015

### Erland

https://www.physicsforums.com/threa...formations-are-the-only-ones-possible.651640/
whether transformations which map straight lines to straight lines must be affine. I accept that this is true, although it seems exceptionally hard to prove. Someone dug up a proof which was completely unintelligible. (Btw, it is false in one dimension.)

2. Inertial systems are usually characteritzed by the requirement that the laws of physics take the same form in all of them. How does this imply that transformations between inertial systems must map straight lines to straight lines?

7. Dec 12, 2015

### Staff: Mentor

The principle of inertia is that free objects travel in a straight line at constant speed. This is a straight line in spacetime.

Inertial transforms preserve which objects are inertial, and therefore map straight lines through spacetime in one frame to straight lines through spacetime in another frame.

8. Dec 12, 2015

### Ibix

An unaccelerated object moves in a straight line. If a straight line maps to a curve then an unaccelerated object is accelerating in the other frame.

9. Dec 12, 2015

### Erland

DaleSpam and Ibix, your argument seems very convincing, but there is a possible loophole. We are talking about transformations which map all straight lines to straight lines (in 4d-spacetime). Therefore, your argument is only valid if all straight 4d-lines represent possible paths of a physical objects. If this is true, then physical objects can move with any speeds, including speeds greater than c (light speed), even infinite speed (a line in spacetime with constant time coordinate). But we cannot assume this, and it also turns out to be wrong.
So, your argument does only prove that lines representing possible motions of physical objects are mapped to other lines, and this might not (and does not) include all lines.

10. Dec 12, 2015

### Staff: Mentor

I think this is a non issue. We haven't discovered tachyons (and many believe that they do not exist) but there is no reason that you could not have inertial tachyons and no reason that an inertial transform should be restricted from mapping a tachyonic straight line to a straight line in another frame.

11. Dec 12, 2015

### Ibix

Is there anything wrong with this argument: a rod that is straight in some inertial frame lies along a straight spacelike line. If a straight spacelike line does not map to a straight line then the rod is curved in some other frame, and we need an explanation for that.

12. Dec 12, 2015

### Staff: Mentor

You would have to make it an inertial rod, because otherwise it can indeed be curved in other frames.

13. Dec 12, 2015

### Ibix

So: a rod that is straight and moving inertially in an inertial frame lies along a straight spacelike line. If a straight spacelike line does not map to a straight spacelike line the rod becomes curved in another frame and we would need forces to appear to explain this.

14. Dec 12, 2015

### Erland

Well, we could possibly say that since tacyons are supposed to not interact with other objects/fields, if the law of inertia applies to them, they must move with constant speeds in straight lines, if they exist at all, that is...
But I don't think that this interpretation is the only logically possible one.
The problem is that there is no really stringent definition of inertial system (at least noone I know of).

Perhaps one way would be to define the class of Inertial systems as a class of reference frames which is closed under the family of transformations which map lines to lines, and such that the laws of physics holds in one of them (recall that also Einstein in his 1905 article singled out one of them and called it "stationary").
But this is not fully stringent either, since it invokes "the laws of physics" and that is quite vague...

15. Dec 12, 2015

### Staff: Mentor

This is true, but not really a problem IMO. The two postulates are fairly "casual" statements, intended to convey the physical intuition behind the theory.

To do proofs you establish the rigorous mathematical structures that are most convenient and then informally identify them with things in the postulates. We have considerable freedom there, so we use it to make things easier later.

It doesn't have to be the only logical interpretation. We can pick for convenience whenever there is more than one logical possibility.

Last edited: Dec 12, 2015
16. Dec 12, 2015

### sweet springs

Preservation of the spacetime interval between any two events, infinitesimally $$dS^2=dS'^2$$, and uniformity of spacetime which means no special point cooridinate and no special direction appears results Lorentz transformation whose only parameter is magnitude of velocity divided by light speed. This is my interpretation from Landau textbook.

17. Dec 12, 2015

### strangerep

Heh, I arrive late to this conversation, but... what's wrong with simply specifying it via non-acceleration? Then the transformations we're interested in are those which map solutions of $$\frac{d^2x^i}{dt^2} = 0$$ among themselves. That leads one to fractional-linear transformations, and (many pages of) further analysis yields both a universal speed constant, and a universal length constant (related to the cosmo constant). But we've had this conversation before in that older thread you linked earlier.

18. Dec 13, 2015

### bcrowell

Staff Emeritus
We normally assume in SR that spacetime has the structure of a vector space. We can construct a basis for this space that consists only of future-timelike vectors. Therefore preservation of the straight-line property for timelike lines implies the same thing for lightlike and spacelike lines.

The broader issue in this discussion is that issues like these can never be resolved using an axiomatic system that is formulated as informally as Einstein's 1905 axiomatization of SR. A nice treatment that takes some care with these fundamentals is Bertel Laurent, Introduction to spacetime: a first course on relativity.

Another thing to realize is that a transformation can have the property of preserving geodesics while not having the property of being a linear operator on the coordinates, or vice versa. For example, we can apply a Lorentz transformation in Rindler coordinates.

19. Dec 13, 2015

### Erland

How does this follow?

20. Dec 13, 2015

### bcrowell

Staff Emeritus
Take a two-dimensional space for simplicity. When we say that an operator $\Lambda$ is linear when it acts on vectors $e_1$ and $e_2$, we mean that for any real numbers $u$ and $v$, $\Lambda(ue_1+ve_2)=u\Lambda(e_1)+v\Lambda(e_2)$. If this holds for vectors $e_1$ and $e_2$ that are linearly independent, then you can easily check that it holds for the whole space.

21. Dec 13, 2015

### Erland

Yeah, but the problem was to prove that the Lorentz transformation is linear.

22. Dec 13, 2015

### bcrowell

Staff Emeritus
No, I wasn't addressing the general problem. I was addressing a subproblem, which was the issue you raised in your #9. That's why I quoted your #9 in my #18. You objected that linearity on timelike lines didn't imply linearity on all lines. Linearity on timelike lines does imply linearity on all lines.

23. Dec 13, 2015

### Staff: Mentor

Excellent point! So simple and clear in retrospect that I cannot believe I didn't see that earlier.

24. Dec 13, 2015

### Erland

You mean something like this: Given two linearly independent vectors: If all lines which are parallell to any of those are mapped to lines, then so does all lines parallell to any linear combination of these vectors? Ok, I think that is correct...

25. Dec 13, 2015

### sweet springs

If a rod contracted before, a rod contracts now and will contract in future.
If a rod here contracts, a similar (be careful to the direction) rod there also contracts.
If 1 m rod contracts to 0.8 m rod, 10 m rod contracts to 8 m.
Such anytime, anywhere and same proportion conditions seem reasonable.