# Lorentz transformation

1. Jul 29, 2007

### bernhard.rothenstein

Consider, please, that you have derived the Lorentz transformation for the space coordinates
x=g(V)[x'+Vt'] (1)
Taking into account that as a result of the Einstein clock synchroniation procvedure x/c=t, x'/c=t' we obtain dividing both sides of (1) by c
t=g(V)[t'+Vx'/c^2].
If that is time saving it is physically correct as well?

2. Jul 29, 2007

### Meir Achuz

How can you derive (1) without also deriving the t equation at the same time?

3. Jul 29, 2007

### Fredrik

Staff Emeritus
You got the correct result, but you need to strengthen your argument somehow if you want it to be a derivation of the 1+1-dimensional Lorentz transformation. You need to explain why the result holds for arbitrary (t',x') when you're using a relationship (x=ct) that only holds for events on the light cone.

4. Jul 29, 2007

### bernhard.rothenstein

Thanks. If you accept results derived without using the Lorentz transformations from thought experiments (say length contraction) then consider the relative position of I and I' as detected from I when the synchyronized clocks of that frame read t. Let E(x,0,t) amd E'(x',0,t') two events taking place on the overlapped axes as detected from I and I' respecively. Taking place at the same point in space when tghe clocks of the two frames read t and t' respectively they represent the same event.
It is obvious that
x=Vt+x'g(V)
and we have it. Google and arXiv offers many such derivations.
If you dislike thought experiments, I respect your point of view!

5. Jul 29, 2007

### bernhard.rothenstein

lorentz transformations

Thanks. Please be more explicit concerning your last sentence. I consider that E(x,0,t) and E'(x',0,t') represent the space time coordinates of the same event that takes place at the same point in space when the synhronized clocks of the two frames read t and t' respectively. Your point of view is important for me.

6. Jul 30, 2007

### Meir Achuz

I have advised you not to use "obvious" in a scientific discussion.
I guess you don't know all the "obvious" jokes.
What is "obvious" about putting gamma in with no explanation?

7. Jul 30, 2007

### bernhard.rothenstein

Sorry it should be
x=Vt+x'/g(V) (1)
where x'/g(V) accounts for length contraction derived from a thought experiment, Obvious in my first language is "evident". So I consider that it is "obvious" that the length of a rod can be expressed as a sum of two of its components, all measured by observers of the same inertial reference frame in that case I. I have tried to answer your question concerning the fact that it is possible to derive the Lorentz transformation for the space coordinates without deriving that for time coordinates. Please excuse me if I was not able to do that.

8. Jul 30, 2007

### Meir Achuz

Even written correctly, I recommend that, as you point out, "evident" is too much like "obvious". In any event, I hope we can agree that the rod does not actually change its length. Then Lorentz would be right and Einstein (and I) wrong. A transformation of coordinates is different than a change of physical length.

9. Jul 30, 2007

### Fredrik

Staff Emeritus
Yes, I understood that (t,x) and (t',x') are the coordinates of the same event in two frames. But that event is supposed to be arbitrary, right? And the coordinates of most events don't satisfy x=ct. So if you use that x=ct, then how do you know that the end result holds for arbitrary (t,x)?

10. Jul 30, 2007

### bernhard.rothenstein

Thanks. (x-0) and (x'-0) is the transformation of a length and not of a coordinate.

11. Jul 30, 2007

### bernhard.rothenstein

Thanks. The two equations
x-0=V(t-0)+(x'-0)/g(V) and
x'-0=(x-0)/g(V)-V(t'-)
impose the condition that the two events take place at the same point in space being expressed as a function of distance and time intervals.

12. Jul 31, 2007

### Fredrik

Staff Emeritus
Perhaps I'm not making myself clear. The 1+1-dimensional Lorentz transformation is a function that maps R2 onto itself. It's a function that given the coordinates of an arbitrary event, tells you the coordinates of that same event in another frame. An event with coordinates (t,ct) is not arbitrary. In fact, the set of all such events is a line through the origin.

You started with an equation (1) that holds for all (t,x). Then you used x=ct to derive equation (2). So how can you be sure that equation (2) holds for all (t,x)?

13. Aug 2, 2007

### bernhard.rothenstein

x=g(V)(x'+Vt') (1)
x'=g(V)(x-Vt) (2)
I think that among other facts (x,ct) and (x',ct') define the locations and the readings of two clocks of two reference frames instantly located at the same point in space. Being synchronized a la Einstein we have x=ct and x'=ct'. Dividing both sides of (1) and (2) we obtain directly the LT for the time coordinates.
So the same (x,ct) and (x',ct') are related by
x=kx'
t=kt'
where k is the Bondi factor. Am I raight at that point?
Meir asked me on this thread if we can derive only the LT for the space coordinates. My answer was yes.
With simple algebra, combining (1) and (2) we obtain the LT for the time coordinates.
Regards

14. Aug 2, 2007

### Fredrik

Staff Emeritus
I'm sorry, but your comments are just making me very confused. That stuff about clock synchronization makes no sense to me.

I understand that you have somehow obtained "half" the Lorentz transformation (for the spatial coordinate only) and would like to use it to derive the other half (for the time coordinate).

But as I've said, the Lorentz transformation doesn't just hold when x=ct. It holds for all (t,x).

Do you, or do you not, agree with the following two statements?

1. Unless you prove that the equations you obtain hold for all (t,x) in R2, you can't claim to have performed a derivation of the Lorentz transformation.

2. The set of points (t,x) that satisfy x=ct is just a one-dimensional subspace of R2.

15. Aug 2, 2007

### country boy

Bernhard, you can see what Fredrik is talking about by putting t'=x'/c into (1). You get only a transformation between x and x'. That's not the general Lorentz transformation.

Another way of looking at it is that by imposing speed-of-light relations between space and time, you have made it impossible to deal with simultaneous events located at different places.

16. Aug 2, 2007

### bernhard.rothenstein

Thanks. Please tell me how do you convince someone about the way in which the LT reflect Einstein's clock synchronization procedure.

17. Aug 2, 2007

### bernhard.rothenstein

Lt

Thanks for the patience with which you answer my questions. I learn a lot of them.
I propose the following basis for a discussion to which I invite all the participants on the forum.
Consider that the addition law of parallel speeds can be derived without using the LT. Am.J.Phys. presents papers which prove that fact. Consider a light clock at rest in the I(0) inertial reference frame. Let C(0) be the clock located at its origin O(0) and on the mirror where from the initial light signal starts. When the light signal returns there C(0) reads t(0)=2d/c C(0) moves with u relative to I and with u' relative to I'. Reading t(0) C(0) is located in front of a clock C of I which reads t and in front of a clock C' of I' that reads t'.
Time dilation makes that
t=t(0)/sqrt[1-(u/c)^2] (1)
t'=t(0)/sqrt[1-(u'/c)^2] (2)
Expressing the right side of (1) as a function of u' via the addition law of parallel speeds and taking into account (2) we obtain
t=t'g(V)(1+Vu'/cc)=g(V)[t'+Vx'/cc) (3)
taking into account that when detected from I clock C(0) has advanced with x=ut whereas when detected from I' it has advanced with x'=u't'.
t and t' represent the readings of the clocks C and C' when they are located at the same point where clock C(0) is located.
Multiplying both sides of (3) with u we obtain
x=ut=g(V)t'(u'+V)=g(V)(x'+Vt') (4)
where V represent the relative speed of I' and I g standing for gamma.
In order to go farther with our discussion (if you are ready to do it) please comment the derivation presented above.
Regards

18. Aug 3, 2007

### Fredrik

Staff Emeritus
Bernard, you need to put a lot more effort into your explanations. I just don't understand what you're saying half the time. And when I do understand what you're saying, I often don't understand why you're saying it.

I think I read that paper a long time ago. It proved that the velocity addition law must take the form (u+v)/(1+Kuv), where K is a constant that can't be derived from symmetry considerations alone. It must be determined experimentally. If this K is non-zero, then 1/sqrt(K) is a quantity with dimensions of velocity that's the same in all frames.

Why is this relevant? The velocity addition law can also be derived from the fact that the speed of light is the same in all frames. You are using that the speed of light is the same in all frames below, so why is it important that you can derive velocity addition (with an unspecified K) without it?

This makes me think of a space-time diagram, with the world-line of this clock coinciding with the time axis. I don't understand the notation "I(0)".

You are considering a light clock, so I assume that you have made the identification c=1/sqrt(K) at this point.

I don't understand what this means. The clock is at the origin and on a mirror? Is the mirror at the origin too? Are you talking about one of the mirrors inside the clock? Then why just mention one of them? I don't understand the "(0)" notation. And how is the light clock oriented? Is it perpendicular or parallell to the motion of the second observer that you introduce later?

Now I picture a world line that goes from (0,0) to (1,d) to (2,0). (I'm choosing c=1). I suppose you must have meant that one of the mirrors of the light clock is at x=0 and the other at x=d. Why didn't you just say so?

So let's continue describing things from the clock's point of view. The world line of I is a line through the origin with slope -1/u, and the world line of I' is a line through the origin with slope -1/u'. But if I try to draw this diagram, I won't know which line represents I and which line represents I'.

This is very very confusing. You seem to be saying that the world line of a second clock C goes through the point (t(0),0)) (in the first clock's frame) and has a slope -1/u. (And that the same thing goes for a third clock C', but with u' instead of u). But that would be completely pointless, and it would contradict how you use t and t' below.

You're bringing time dilation into this without proof, but I guess that's OK, since it can be derived from the existence of a universal speed, and that seems to be one of your starting assumptions.

What you're doing here is calculating the time coordinates of the event (t(0),0) in frames I and I' respectively. But why am I telling you that? You should be explaining these things yourself.

The first part of what you say doesn't make sense. u is not a function of u', so gamma(u) can't be either.

I'm going to stop here. There are just too many things in your derivation that I can't make sense of.

19. Aug 3, 2007

### robphy

In my opinion, every "derivation" of the Lorentz Transformation should be capable of being formulated in terms of geometric relations on a Minkowski spacetime diagram [which, by the way, is about to turn 100 years old].

I'm not saying that the motivation or presentation of the derivation for the intended audience must use the spacetime diagram... but, it seems to me, should be expressible unambiguously in terms of a spacetime diagram. When formulated with a spacetime diagram, the physical and mathematical interpretations of the assumptions and results are probably easier to see and understand.

Last edited: Aug 4, 2007
20. Aug 3, 2007

### bernhard.rothenstein

lorentz transformations

thanks Frederik
as I see our problem starts and ends with your point of view concerning the addition law of parallel speeds. Where did you publish it?
Regards