On Mar 4, 5:40 am, "harry" wrote:
"Eric Gisse" wrote in message

...



On Mar 4, 4:47 am, "
wrote:
On Mar 4, 2:17 pm, "Artful" wrote:

wrote in message

...

I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

Seehttp://en.wikipedia.org/wiki/Minkowski_diagram
Look at the diagram in the section entitled "Minkowski diagram in
special
relativity" with the caption "In the theory of relativity both
observers
assign the event at A to different times."

Artful:

I considered it, but it seems to contradict the equations for Lorentz
transformation. I mean, when I tried to get x' and t' through the
diagram and through Lorentz transformation, I got different things. I
expressed x' and t' using trigonometry from the diagrams, and I got
some kind of ugly mess. Can you point out my mistake? Or maybe there
is an analysis of how Minkowski diagrams work somewhere on the web?

Thanks,
Ram.

Lorentz transformations have nothing to do with spacetime diagrams.
Lorentz transformations are a specific type of transformation between
inertial reference frames, and the Minkowski/space-time diagram is a
characterization of the geometry of a manifold [they generalize to
conformal diagrams] by using null paths [the paths light travel along]
which is unrelated to frame transformations.

That's wrong. Instead, I think that the intro of the above link is quite
right:

I looked at the image, saw the familiar cone, and figured it was just
another name for a space-time diagram.


"The Minkowski diagram [...] provides an illustration of the properties of
space and time in the special theory of relativity. It allows a quantitative
understanding of the corresponding phenomena like time dilation and length
contraction without mathematical equations."

Roughly, a rotation in a Minkowski diagram corresponds to a Lorentz
transformation. A different result is most likely due to either a wrong
rotation or a wrong projection. There are many detailed manuals on the web
that may be clearer and more detailed than Wikipedia, for example:http://www.physics.usyd.edu.au/super...ties/Mechanics...
(Found with Google, haven't checked it but it looks good)

Harald

On Mar 4, 5:33 am, "
wrote:

[...]

Never mind. Look at harry's response to me.

wrote in message

I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

Yes!
See http://users.telenet.be/vdmoortel/di...es/Lorentz.htm

Measuse slope of x'-axis w.r.t. x-axis. That gives you v.
Measure Ox' and Ot' (for instance in cm), multiply with
the calculated value
sqrt(1-v^2)/sqrt(1+v^2)
and you have the transformed coordinates x' and t'.
Verofy with the transformed values of the measured x and t.

See http://users.telenet.be/vdmoortel/di...es/Lorentz.png
and a somewhat interactive version
http://users.telenet.be/vdmoortel/di...es/Lorentz.htm
You can drag the point H around a bit, and also the 'top points'
of the t-axis and the t'axis.

Enjoy.
Dirk

PD wrote in message

On Mar 4, 6:08 am, "
wrote:
I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

No, you can't.

Of course you can.

The temptation is to use a Minkowski diagram like a 2D
Euclidean geometry, which would allow you to do all sorts of Euclidean
geometrical constructions (with a compass and a straight edge) and
trigonometric relations. But the relationship between x and t in a
Minkowski diagram is NOT Euclidean, and you will quickly run into
problems if you try to apply Euclidean rules to it.

It is a linear transformation, remember?
See my other reply to ram :-)

Dirk Vdm

On Mar 4, 10:52*am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
PD wrote in message

*

On Mar 4, 6:08 am, "
wrote:
I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

No, you can't.

Of course you can.

OK, you can if you are VERY careful. For example, note the effect of
the a boost, which is called a "rotation" in Minkowski space. However,
this is not your garden-variety rotation, in that both axes get
rotated, say, clockwise, to do the transformation. In fact, the x-axis
rotates one way and the t-axis rotates the other way.

All sorts of mapping issues result from this shift from Euclidean to
hyperbolic geometry. For example, trig relations get replaced by
hyperbolic trig relations (sinh, cosh, tanh, rather than sin, cos,
tan). All of these are trackable if you are careful and understand
what is fundamentally different about the Minkowski diagram from the
usual pair of axes on a flat piece of paper.


The temptation is to use a Minkowski diagram like a 2D
Euclidean geometry, which would allow you to do all sorts of Euclidean
geometrical constructions (with a compass and a straight edge) and
trigonometric relations. But the relationship between x and t in a
Minkowski diagram is NOT Euclidean, and you will quickly run into
problems if you try to apply Euclidean rules to it.

It is a linear transformation, remember?
See my other reply to ram :-)

Dirk Vdm

PD wrote in message

On Mar 4, 10:52 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
PD wrote in message



On Mar 4, 6:08 am, "
wrote:
I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

No, you can't.

Of course you can.

OK, you can if you are VERY careful. For example, note the effect of
the a boost, which is called a "rotation" in Minkowski space. However,
this is not your garden-variety rotation, in that both axes get
rotated, say, clockwise, to do the transformation. In fact, the x-axis
rotates one way and the t-axis rotates the other way.

All sorts of mapping issues result from this shift from Euclidean to
hyperbolic geometry. For example, trig relations get replaced by
hyperbolic trig relations (sinh, cosh, tanh, rather than sin, cos,
tan). All of these are trackable if you are careful and understand
what is fundamentally different about the Minkowski diagram from the
usual pair of axes on a flat piece of paper.

Since in Euclidean terms we don't allow imaginary angles, we
shouldn't really call this a "real" rotation ("real" like in The Real
Numbers).
I made this thing some time ago. It just took some straightforward
standard high school level analytic geometry with lines, slopes,
and intersections, and as you can see, even if you're not careful,
once you have that scale factor, one can easilily "read" the
transformed coordinates from the Minkovski diagram :-)

Dirk Vdm

Dirk Van de moortel wrote in message

wrote in message

I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

Yes!
See http://users.telenet.be/vdmoortel/di...es/Lorentz.htm

Measuse slope of x'-axis w.r.t. x-axis. That gives you v.
Measure Ox' and Ot' (for instance in cm), multiply with
the calculated value
sqrt(1-v^2)/sqrt(1+v^2)
and you have the transformed coordinates x' and t'.
Verofy with the transformed values of the measured x and t.

See http://users.telenet.be/vdmoortel/di...es/Lorentz.png
and a somewhat interactive version
http://users.telenet.be/vdmoortel/di...es/Lorentz.htm
You can drag the point H around a bit, and also the 'top points'
of the t-axis and the t'axis.

Enjoy.
Dirk

Found my notes.

Take the standard drawing of the Minkovski diagram with perp.
x-axis and t-axis, and do some kid's analytic geometry....

The t-axis is represented by t = 1/v x.
The x-axis is represented by t = v x.

Take a point H with coordinates (X,T).
Take line through H, parallel with t'-axis:
t - T = 1/v ( x - X )

Intersection of this line with the x'-axis gives:
x = (T-X/v) / (v-1/v)
t = T + 1/v ( (T-X/v)/(v-1/v) - X )
The distance between (0,0) and this intersection is given by
D = sqrt( x^2 + t^2 ) .
Simplify this to
D = sqrt( (1+v^2)/(1-v^2) ) 1/sqrt(1-v^2) ( X - v T )
= sqrt(1+v^2)/sqrt(1-v^2) X'
where
X' = 1/sqrt(1-v^2) ( X - v T ) Lorentz!

So, to find X', just multiply this distance D with the scale factor
sqrt(1-v^2)/sqrt(1+v^2)

Likewise for the T'.

Dirk Vdm

On Mar 4, 7:21*pm, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
PD wrote in message

*



On Mar 4, 10:52 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
PD wrote in message



On Mar 4, 6:08 am, "
wrote:
I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

No, you can't.

Of course you can.

OK, you can if you are VERY careful. For example, note the effect of
the a boost, which is called a "rotation" in Minkowski space. However,
this is not your garden-variety rotation, in that both axes get
rotated, say, clockwise, to do the transformation. In fact, the x-axis
rotates one way and the t-axis rotates the other way.

All sorts of mapping issues result from this shift from Euclidean to
hyperbolic geometry. For example, trig relations get replaced by
hyperbolic trig relations (sinh, cosh, tanh, rather than sin, cos,
tan). All of these are trackable if you are careful and understand
what is fundamentally different about the Minkowski diagram from the
usual pair of axes on a flat piece of paper.

Since in Euclidean terms we don't allow imaginary angles, we
shouldn't really call this a "real" rotation ("real" like in The Real
Numbers).
I made this thing some time ago. It just took some straightforward
standard high school level analytic geometry with lines, slopes,
and intersections, and as you can see, even if you're not careful,
once you have that scale factor, one can easilily "read" the
transformed coordinates from the Minkovski diagram :-)

Dirk Vdm

Dirk -- You are a godsend. That Java thing rocks. I tried some values,
and I checked it and it gives the same result as the Lorentz
Transformation. However, I tried to get x' and t' analytically from
the graph, but it didn't come out like the x' and t' that the app
said. Do I have a computation mistake, or did I not understand how x'
and t' are retirieved? I'll tell you how I retrieved them: For x', for
example I took the point x' on the graph and calculated its distance
from O. Is that what I'm supposed to do?

Thanks,
Ram.

wrote in message

On Mar 4, 7:21 pm, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
PD wrote in message





On Mar 4, 10:52 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
PD wrote in message



On Mar 4, 6:08 am, "
wrote:
I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

No, you can't.

Of course you can.

OK, you can if you are VERY careful. For example, note the effect of
the a boost, which is called a "rotation" in Minkowski space. However,
this is not your garden-variety rotation, in that both axes get
rotated, say, clockwise, to do the transformation. In fact, the x-axis
rotates one way and the t-axis rotates the other way.

All sorts of mapping issues result from this shift from Euclidean to
hyperbolic geometry. For example, trig relations get replaced by
hyperbolic trig relations (sinh, cosh, tanh, rather than sin, cos,
tan). All of these are trackable if you are careful and understand
what is fundamentally different about the Minkowski diagram from the
usual pair of axes on a flat piece of paper.

Since in Euclidean terms we don't allow imaginary angles, we
shouldn't really call this a "real" rotation ("real" like in The Real
Numbers).
I made this thing some time ago. It just took some straightforward
standard high school level analytic geometry with lines, slopes,
and intersections, and as you can see, even if you're not careful,
once you have that scale factor, one can easilily "read" the
transformed coordinates from the Minkovski diagram :-)

Dirk Vdm

Dirk -- You are a godsend. That Java thing rocks. I tried some values,
and I checked it and it gives the same result as the Lorentz
Transformation. However, I tried to get x' and t' analytically from
the graph, but it didn't come out like the x' and t' that the app
said. Do I have a computation mistake, or did I not understand how x'
and t' are retirieved? I'll tell you how I retrieved them: For x', for
example I took the point x' on the graph and calculated its distance
from O. Is that what I'm supposed to do?

See my last message where I calculated the scale factor
for X' when X and T are known.
I'll leave the calculation of T' as an exercise. Let me know
if you get stuck :-)

Dirk Vdm

On Tue, 4 Mar 2008 04:08:59 -0800 (PST),
" wrote:

I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

[Hammond]
Only an amateur would try to use a Minkowski diagram
(rotation diagram) to study a Lorentz transformation.
The EXPERTS use OBLIQUE COORDINATE DIAGRAMS.... you've
probaly seen these in professional publications.
The Minkowski diagram does not show "true lengths" and as
someone pointed out you CANNOT use Euclidean Geometry in the
diagram.... and it does NOT give you ture picture of the
transformation.
The EXPERTS use the LOEDEL (oblique) diagram.
Understanding this diagram is equivalent to MASTERING SR,
and it can be done in a few minutes. It turns out that the
Lorentz Transformation is IDENTICAL to the coordinate
transformation between two OBLIQUE coordinate systems... one
obtuse and the other acute. Euclidean geometry applies, the
scales are true length,... the whole thing is a miracle!
The CLASSIC text on this is SHADOWITZ'S book, which only
costs $6.95 in the ppbk Dover edition....one of the all time
best book bargains ever! DON'T LEAVE HOME WITHOUT IT!!!


SPECIAL RELATIVITY, Albert Shadowitz, 1968, Dover ppbk,
ISBN 0-486-65743-4 only $6.95 a few years ago.

By the way, you can Google the Loedel diagram but all you
get is amateur crap explanations..... for chrissakes spend
the $6.95 and buy Shadowitz'a classic book and wrap up your
studies of SR in a couple of hours!

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