Re: An embarrassingly simple question about spacetime diagrams

Edward Green <spamspamspam3@xxxxxxxxxxx> wrote in message
9ca35547-1a5d-490a-b97e-aa6311122d2c@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Aug 22, 4:50 am, "Dirk Van de moortel"
<dirkvandemoor...@xxxxxxxxxxxxxxxxxx> wrote:
Edward Green <spamspamsp...@xxxxxxxxxxx> wrote in message

ad7a659b-2083-4fa3-a614-ff082cf98...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx

On Aug 21, 8:04 am, "Dirk Van de moortel"
<dirkvandemoor...@xxxxxxxxxxxxxxxxxx> wrote:
On Aug 21, 8:04 am, "Dirk Van de moortel"
<dirkvandemoor...@xxxxxxxxxxxxxxxxxx> wrote:

It's quite simple. Just think elementary analytical geometry on a
(x,y) diagram where you replace y with t.

The x'-axis satisfies
t' = 0
i.o.w.
t = v x

How does that follow? I'm afraid I don't get it.

Didn't you have analatic geometry in high school, where they
represent lines with equations?

Did I write that?
"analAtic"?
Good one :-)

Of course. What I didn't get was the logical progression from your
first equation, or line, to your second. How does t' = 0 imply
("i.o.w.") t = v x ?

Note that the t'-axis is the world line of the 'moving' observer
(the one who uses coornates (x',t') to specify events), and this
observer moves a constant velocity v in the (t,x) system.

Right. This gives you the slope of the t' axis, no? It's the slope
of the x' axis I'm having problems with.

Ok, then take my original reply, but do it in reverse order - see below (*)


You can also *directly* read it from the Lorentz transformation:
t' = g ( t - v x ) where g = 1/sqrt(1-v^2)
so, since g > 1 you have
t' = 0 <==> t - v x = 0 <==> t = v x

Nice. But from the point of view of the development I am reading,
that is cheating, since we are going to _derive_ the Lorentz
transformation from the diagram.

That would be a bit silly.
The diagram *represents* the transformation :-)


Also, I know we're presumably using dimensionless units, but shouldn't
equations still balance if we substitute conventional units?

Only in units where c = 1 (so to speak) are the axes in the (t,x)-system
orthogonal and does the light cone have a pi/4 angle with both axes.

Or if you like,
t = conventional time (T) times light speed
t = c T
v = conventional speed (V) divided by lightspeed (dimensionless)
v = V / c
so both sides have the dimension of x in
t = v x
which can also be written as
c T = V/c X
or
T = V/c^2 X
where both sides have the dimension of T.

Alright.

(*) Doing it the other way around and/but without using the
transformation for the t'-axis:

The t'-axis, being the world line of the primed observer, satisfies
x = v t
i.o.w
t = 1/v x
so, the angle between t' and x is arctan(|1/v|)
and thus the angle between t' and t is
pi/2 - arctan(|1/v|) = arctan(|v|)

For the x'-axis it's slightly different.
The x'-axis (the one you're having problems with,) satisfies
t' = 0
i.o.w, using the Lorentz transformation,
t = v x
so, the angle between x' and x is arctan(|v|)

Dirk Vdm







.

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