# What does it mean if a transformation is 'linear'?

1. Jul 12, 2009

### Nanyang

Many authors seem to start deriving the Lorentz transformations (for a motions only in one direction) by first stating that the transformation equations have to be linear, and I am always lost at this part. What do they mean by that? How does "a uniform rectilinear motion in K must also be uniform and rectilinear in K' " explain that?

Many thanks in advance for the help and apologies if this has been raised many times.

2. Jul 12, 2009

### facenian

well, I'm no expert on this but I think I know what it means to be linear, let $x_i (i=0,2,3,4)$ be the space time coordantes, then if the new coordenates are $x^{'}_i$ we have
$$x^{'}_i=\sum_k a_{ik}x_k$$
where a_ik do not depend on the space time coordenates.
However I do have a question on the linearity issue and I think I will initiate a thread on that.

3. Jul 12, 2009

### malawi_glenn

facenian is correct, but one can also have a shift in constant, e.g.

$$x\acute{}_i=\sum_k a_{ik}\,x_k \,+ \, b_i$$

4. Jul 14, 2009

### Tac-Tics

The term linear comes from linear algebra.

A transformation T is called linear if

(1) aT(u) = T(au)

(2) T(u+v) = T(u) + T(v).

Property (1) means that if you double the distance in one frame, then the distance doubles in the other.

Suppose the direction "up" transforms to "north" and "left" translates to "west". Property (2) then says walking up and to the left in one frame is the same as walking north and to the west in the other.

Linearity is very, very useful. It has a powerful synergy with the notion of a vector basis. Linear functions are convenient for calculations, too, because every linear transformation can be thought of as an NxM matrix. To apply a transformation, you simply do matrix multiplication between this matrix and the vector you want.

5. Jul 19, 2009

### Nanyang

Thanks for the replies, I think I know what being linear means now.

However, I still do not understand how does "A uniform and rectilinear motion in K must also be uniform and rectilinear in K' " imply that the transformations are linear. I have read the other thread started by facenian on this issue but is scared away by all the math that I have never seen before.

So is there a much less mathematical way of putting it, for a 'beginner', without loss of any physics? Or is it like what is claimed in facenian's thread that the statement "A uniform and..." cannot be accounted for this linearity issue and that it was a guess that turned out to be correct?

Last edited: Jul 19, 2009
6. Jul 20, 2009

### facenian

According to discussions on thread "Question on linearity....." I learned that it is true that there exist transformations that are not linear and preserve uniform an rectlinear motions but it seems that for reasons of smoothness transformations that are no linear mus be descarded. I'm afraid that to understand the details of all this a little math is needed.

7. Jul 20, 2009

### Fredrik

Staff Emeritus
If the goal is to derive linearity from some well-defined axioms that are motivated by our intuition about physics, I don't think it gets any easier than my post #56 in the other thread. It is however possible to use a different set of axioms. There are other axioms that are just as "intuitive" and in a sense weaker than mine. See e.g. the article that atyy linked to on page 1. (It's a very good article, but it requires a lot of "mathematical maturity", meaning that even though all the information you need is in there, it's still difficult to understand it if you're not used to this way of thinking).