# Lorentz Transformation without length?

1. Nov 7, 2008

### Saw

The Lorentz Transformation

tA = [tB+vxB/c^2]/sqrt(1-v^2/c^2)

can be simplified to

tA = tB (1-v)/sqrt(1-v^2),

1) Refer to v as a fraction of c. Thus every time we write v/c in the original formula, we write v, in the understanding that these v units are units of c.

2) Refer to length in light-seconds = x/c. Thus every time that the original formula writes x/c, we write x, in the understanding that these x units mean x light-seconds.

This way, in the numerator, the sync factor vx/c^2 can be broken down into (x/c)(v/c). Expressed with this convention, x/c becomes x light-seconds and v/c becomes v = the corresponding fraction of c.

In the denominator, the expression sqrt(1-v^2/c^2) becomes sqrt (1-v^2).

Thus the LT looks as follows:

tA = (tB + xB*v )/ sqrt(1-v^2)

3) But it seems as if xB could also be replaced by tB. If a point is deemed to be xB light-seconds away from the origin of the coordinate system B, it is because it is assumed that light takes tB seconds to reach that place as measured in B frame. In fact, if xB in our notation is (x) km / (c) km/s and c is the unity, the expression is equivalent to tB.

Thus the LT adopts this look:

tA = (tB + tB*v )/ sqrt(1-v^2) = tB (1+v)/ sqrt(1-v^2)

Although I cannot do the intermediary algebra, it appears that these other expressions give the same result:

tA = tB * sqrt [(1+v)/(1-v)]

tA = tB * sqrt(1-v^2) /(1-v)

I cannot think of any practical situation where x in km is needed. Is this right or have I missed anything?

2. Nov 7, 2008

### Fredrik

Staff Emeritus
If you e.g. want to use seconds as a unit of length, I don't have a problem with that, but you can't just say that x=t. Those variables are the coordinates of an event. An event is a point in a 2-dimensional manifold (4-dimensional if we include y and z), and x=t defines a line in that 2-dimensional manifold. Most points are not on that line.

3. Nov 7, 2008

### ZikZak

You've missed the fact that x and t represent displacements along two different axes of the coordinate system. They differ in both direction and magnitude.

4. Nov 7, 2008

### Saw

Yes. Thanks to both. I had in the back of my mind an example where two observers A and B synchronise their clocks when they meet and symultaneously send light signals to the front of B car, where light activates a detector that triggers an event. Relative v is 0.5 c. The light reflects back; when A and B receive it, they read their clocks (3.46s and 2s, respectively) and conclude that the event took place at tA=1.73 s and xA=1.73 light-seconds and tB=1 s and xB = 1 light-second. I suppose that this coincidence time = length, in this particular example, is due to the fact that the event takes place where light was. If the event had taken place somewhere else, not in the vicinity of the light signal (for example, the bomb explodes in B hands when her clock reads 1 s), then tB is not enough information to derive tA: we should also take into account that xB = 0 and would get 1.15 s for tA, instead of 1.73 s. So length matters, unless you know that length equals time because the event has taken place by a light signal sent by the obsever.

This puts me a little closer to understanding spacetime diagrams and the idea that light's trajectory is painted with a line that has a 45º angle. This is a "privilege" of light, isn´t it? Any inertial frame (i.e., moving with constant velocity) will travel equal paths in equal times and that will deserve that its trajectory is painted as a straight line. But only light manages to get units of the same size in both axes. This reflects the fact that, in practice, both the space and the time coordinates of an event are measured with the same instrument (light)and the same method (the Einstein convention). Could we say then at least that, in SR, time and length are measured with the same instrument and the same unit? Of course, as you have made me see, an event may be closer in time, but farther in length or viceversa. But can we say that the "line" (x=t) against with which you measure its position in both axes is light? I know this is not very precise, but how does it sound to you?

5. Nov 7, 2008

### ZikZak

Yep, pretty much! Light rays and only light rays travel at 45 degree angles in spacetime diagrams.

Feel free to measure time and length in the same units. Usually, we measure them both in meters, but you can use seconds if you like.

6. Nov 7, 2008

### bernhard.rothenstein

Events are generated by propagating light signals and by moving tardyons. Consider that in the I' (your B) reference frame a light signal starts to propagate at a time t from its origin O.
After a given time of propagation it generates the event E(x=ct;t=x/c). Performing the Lorentz transformation of its time coordinate to the I frame (your A) the results is
t=g(t'+Vx'/cc)=g(1+V/c)t'=gx'(1+V/c)x'/c. It is a good exercice to derive the results using a relativistic space time diagram where the world line of the light signal propagates under an angle of 45 degress with the axes of the diagram.