Determination of Lorentz transform from euclid geometry

[atto]

A first stage of the determination.

We have a body of length L = AB, which moves along the x-axis with velocity v,
say that coming to us.

A -------- B v <---
|
| h - vertical distance
| ./ - a light converges to us with the speed eq. c
| /
O ---------> x

At some time t = 0, we see the point A at the x = 0;
which means, of course, that the point A is actually in a different place,
because the time of flight of light is: dt = h/c, that is, the body moved to the x = -vdt - it's obvious?

But we are interested how we see at time t = 0 the whole body,
which is where we see the other end - the point B?
x_B (t = 0) = ?

 

[Doc Al]

Realize that the Lorentz transformations relate calculated positions and times after the travel time of light is accounted for, not raw observations of what some particular observer literally "sees". (Usually one assumes observers everywhere in a frame, so that light travel time can be neglected.)

 

[atto]

Realize that the Lorentz transformations relate calculated positions and times after the travel time of light is accounted for, not raw observations of what some particular observer literally "sees". (Usually one assumes observers everywhere in a frame, so that light travel time can be neglected.)

You can assume such method of time sync., ie.:
t = 0 is a time of the emission moment of a light (at a point),
instead of the reception moment by the observer.

I intend only to recognize the true nature and validity of the Lorentz transformation, using only elementary geometry; and this should work good... after all we can see the moving bodies from a distance,
under some conditions, which are not so difficult to recognise.

 

[Doc Al]

I intend only to recognize the true nature and validity of the Lorentz transformation, using only elementary geometry;

You'll need more than just geometry--you'll need some physics.

 

[Barry911]

Lorentz transformations are rotations in hyperbolic 4-S of Minkowski.

 

[atto]

You'll need more than just geometry--you'll need some physics.

What we are missing... what we still need to determine the observed position of the point B, for the given conditions?

 

[atto]

Lorentz transformations are rotations in hyperbolic 4-S of Minkowski.

And what of that?
You might as well say that the complex numbers lie in the complex plane... the real numbers on a line, vectors in the vectors space, ect.

 

[Dale]

I don't see how, even in principle, you can hope to derive $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$ from $ds^2=dx^2+dy^2+dz^2$

 

[Barry911]

What of that?
Minkowski space is more than a mathematical object cf.- "real number", "vector". Minkowski space has the
physicality of S.R. Note the confusion of "Dale Spam" above. In principle you cannot derive the interval equation from an elliptical space its source is hyperbolic thus the minus sign.

With all due respect,

Barry911

 

[ghwellsjr]

A first stage of the determination.

We have a body of length L = AB, which moves along the x-axis with velocity v,
say that coming to us.

A -------- B v <---
|
| h - vertical distance
| ./ - a light converges to us with the speed eq. c
| /
O ---------> x

At some time t = 0, we see the point A at the x = 0;
which means, of course, that the point A is actually in a different place,
because the time of flight of light is: dt = h/c, that is, the body moved to the x = -vdt - it's obvious?

But we are interested how we see at time t = 0 the whole body,
which is where we see the other end - the point B?
x_B (t = 0) = ?

I agree with your determination of where point A is when it is viewed at t=0 by O but unless you know that the length of the body is contracted and by how much, how can you determine where point B is at t=0?

I believe this is the physics that Doc Al said you would need in post #4.

 

[pervect]

And what of that?
You might as well say that the complex numbers lie in the complex plane... the real numbers on a line, vectors in the vectors space, ect.

I'm not sure what you're trying to do, the remark about hypervolic rotations was terse, but got to the essence of the issue - at least one staement about what you were trying to do.

To go into a bit more detail on this approach:

Consider rotations in a 2d Euclidean space, with cartesian coordinates (x,y). You can write:

x' = cos(theta) x - sin(theta) y
y' = -sin(theta) x + cos(theta) y

Rotations preserve distances i.e x^2 + y^2 = x'2 + y'2

Now consider a lorentz boost in a 2d Minkowski space, with coordinates (t,x) You can write

x' = cosh(rapidity) x - sinh(rapidity) c t
c t' = -sinh(rapidity) x + cosh(rapidity) c t

See the wiki article on rapidity for more info on rapidity, http://en.wikipedia.org/wiki/Rapidity

Note that tanh(rapidity) = v/c

The lorentz boost transformation above preserves the Lorentz interval, i.e x^2 - (ct)^2 = x'^2 - (c' t)^2, just as the rotation preserved distance. It's easy to verify both statements with a bit of algebra.

Notice that for a light beam either x = ct or x = -ct. The lorentz interval is just (x-ct)(x+ct), so a zero Lorentz interval between two events is a mathematical characteristic of them being connected by a light beam.

WHen you preserve the Lorentz interval, you also preserve the nature of lightlike separation, with light moving at a constant velocity c. This happens because the Lorentz interval between two events is zero if they are connected by a lightbeam, and if two events are connected by a lightbea, the Lorentz interval is zero, both conditions are equivalent, as either one implies the other.

Because the Lorentz transform preserves the Lorentz interval, it insures that any two events that are connected by a lightbeam moving at a velocity c in (t,x) coordinates are connected by a lightbeam moving at velocity c in (t', x') coordinates.

Thus the self consistency of Lorentz transformations should be obvious at this point. As others have remarked, starting with this mathematical insight, you also might want to do some physics, The mathematical existence and motivation of the transform is hopefully is clearer now (?), but the work to apply them to physics still needs to be done.

 

[atto]

I agree with your determination of where point A is when it is viewed at t=0 by O but unless you know that the length of the body is contracted and by how much, how can you determine where point B is at t=0?

I believe this is the physics that Doc Al said you would need in post #4.

Distance from A to B is equal L: |AB| = L;
It can be rather easily measured, thus we have no problem with this.

In this stage we don't know yet whether the length L is a contracted version of some: L' = k*L, where k is unknown; just the L is given.

This is a further stage... now we want just to predict the observed position of the point B.

At this stage, we are interested in: how will look the moving object? And that's all;
ie. we have to solve a problem of visualization the moving object.

 

[atto]

OK. I solved something already and discovered an equation - the condition of the visibility a moving point x at the same time of reception:

$h^2 + (x - vt)^2 = (h - ct)^2$

for x = 0 we get: t = 0, this is the point A at x = 0, thus we observe the body at a time moment:
$t_o = h/c = const$

The t parameter is an emisson time moment at a point x of the body: x = [0, L].

for any other x <> 0 we get a negative t, this just means we see the image from the past - the more distance to the point of emission, the more time needed for a light to reach us at the same time moment of observation: t_o.

The equation is a hyperbola wrt the parameters (x,t), thus we probably have some good symptom of the Lorentz transform.

 

[Doc Al]

I believe this is the physics that Doc Al said you would need in post #4.

Exactly. (Or the equivalent.)

 

[ghwellsjr]

I think you are trying to analyze what is called the Terrill effect which indicates that an object like a rod will appear elongated as it approaches and contracted as it recedes. But I don't think your equations or your idea has anything to do with the Lorentz Transform or with the actual visualizations. It's a very complex subject and threw a lot of people off for a long time.

 

[Barry911]

Euclidean geometry and Lorentz transformations

Hello Atto:

The point of my post on Lorentz transformations was that in the end one
needs the curved space time of Lobochevsky to give the transformations physical
substance. Perhaps the cleverest derivation of the transformations is that of
Sir Arthur Stanley Eddington in his "Space, Time and Gravitation" BTW still must
reading for relativity fans.
Eddington used as a model for derivation nothing more than swimmers traveling up and
down a stream and across a stream. Initially the stream has velocity 0 relative to the
shore then he considers the swimmers (longitudinal and transverse making complete
circuits). The Lorentz transformations fall out when you calculate the RATIO of the
end net velocities! It's a fun thing just remember that the observer role is the stream not
either of the swimmers.

 

[atto]

But I don't think your equations or your idea has anything to do with the Lorentz Transform or with the actual visualizations.

For what reason do you think so?

The condition of the 'visibility' of a body is only one:
the light from the body parts must reach us at the same moment of time.

And this is just the equation of the hyperbola i wrote.

 

[atto]

By the way: someone probably is looking for the power of the devil in this transformation.
Sorry, but there is nothing in it beyond a simple geometry.

 

[ghwellsjr]

For what reason do you think so?

The condition of the 'visibility' of a body is only one:
the light from the body parts must reach us at the same moment of time.

And this is just the equation of the hyperbola i wrote.

Doesn't the hyperbola indicate that the perceived length of the object is symmetrical as it approaches and as it recedes?

 

[atto]

Doesn't the hyperbola indicate that the perceived length of the object is symmetrical as it approaches and as it recedes?

No. The observed image of a moving object isn't symmetric.

You must solve the equation, ie. write it in an explicit form: t = t(x),
and then you just compute the times for two symmetric points, ie: x+ = a, x- = -a.

And the observed position of a moving point x, is just a position of the point at a time moment of emission:

x_o = x - vt; where t is the computed time of emission for a point x.

 

[ghwellsjr]

No. The observed image of a moving object isn't symmetric.

You must solve the equation, ie. write it in an explicit form: t = t(x),
and then you just compute the times for two symmetric points, ie: x+ = a, x- = -a.

And the observed position of a moving point x, is just a position of the point at a time moment of emission:

x_o = x - vt; where t is the computed time of emission for a point x.

Then why don't you just do it and show us how it looks as the object approaches, passes and recedes?

 

[atto]

OK. Here you are computed the two points:

For simplicity, let: c = 1; h = 1;

and: v = 0.8, gamma = 1.66666666...

The solutions:

approaching: x = 1: t = -1.201
receding : x = -1: t = -0.270

Thus the observed positions are just that:
+1' = 1 + 0.8*1.2 = 1.96
-1' = -1 + 0.8*0.27 = -0.784

 

[ghwellsjr]

OK. Here you are computed the two points:

For simplicity, let: c = 1; h = 1;

and: v = 0.8, gamma = 1.66666666...

The solutions:

approaching: x = 1: t = -1.201
receding : x = -1: t = -0.270

Thus the observed positions are just that:
+1' = 1 + 0.8*1.2 = 1.96
-1' = -1 + 0.8*0.27 = -0.784

Can you please show all the details of the derivations? Pretend like I'm really ignorant.

 

[atto]

I forgot a version compatible with a 'contracted' body.

When the body is physically shortened then the point x is actually in the position: x/gamma = x*0.6, for v = 0.8;

therefore we must compute for x = +/-0.6, instead of +/-1:
x = 0.6; t = -0.312
x =-0.6; t = -0.12

and now we observe:
+1' = 0.6 + 0.8*0.312 = 0.85
-1' = -0.6 + 0.8*0.12 = -0.5

-------
The equation:
$$y^2 + (x-vt)^2 = (y-ct)^2$$
and we simply solve for the variable t:

$y^2 - 2yct + (ct)^2 - y^2 - x^2 +2xvt - (vt)^2 = 0$
$t^2(c^2-v^2) - 2t(yc - xv) - x^2 = 0$$t = \frac{(yc-xv) \pm \sqrt{x^2(c^2-v^2) + (yc-xv)^2}}{c^2-v^2}$

for y = 1, c = 1 this is:
$t = \frac{(1-xv) - \sqrt{x^2(1-v^2) + (1-xv)^2}}{1-v^2}=\frac{(1-xv) - \sqrt{x^2-2xv+1}}{1-v^2}$

 

[ghwellsjr]

I forgot a version compatible with a 'contracted' body.

When the body is physically shortened then the point x is actually in the position: x/gamma = x*0.6, for v = 0.8;

You've lost me. I thought you were trying to determine the x-location of point B when the image of point B and the image of point A both arrived simultaneously at point O at time t=0. We already agreed that point A is at some positive x-value under this circumstance. It would help if you would describe your variable in those terms. I can't even tell if this is supposed to be for point B or point A.

therefore we must compute for x = +/-0.6, instead of +/-1:
x = 0.6; t = -0.312
x =-0.6; t = -0.12

Did you use your equation below to determine the values of t? Why are there two of them? Does one of them apply to point B and the other to point A?

and now we observe:
+1' = 0.6 + 0.8*0.312 = 0.85
-1' = -0.6 + 0.8*0.12 = -0.5

What are the primes for? And what do these apply to?

-------
The equation:
$$y^2 + (x-vt)^2 = (y-ct)^2$$
and we simply solve for the variable t:

$y^2 - 2yct + (ct)^2 - y^2 - x^2 +2xvt - (vt)^2 = 0$
$t^2(c^2-v^2) - 2t(yc - xv) - x^2 = 0$


$t = \frac{(yc-xv) \pm \sqrt{x^2(c^2-v^2) + (yc-xv)^2}}{c^2-v^2}$

for y = 1, c = 1 this is:
$t = \frac{(1-xv) - \sqrt{x^2(1-v^2) + (1-xv)^2}}{1-v^2}=\frac{(1-xv) - \sqrt{x^2-2xv+1}}{1-v^2}$

Please provide a detailed explanation of each step along the way and also an overview of what you are doing.

 

[atto]

You've lost me. I thought you were trying to determine the x-location of point B when the image of point B and the image of point A both arrived simultaneously at point O at time t=0. We already agreed that point A is at some positive x-value under this circumstance. It would help if you would describe your variable in those terms. I can't even tell if this is supposed to be for point B or point A.

Yes, the equation describes the simultaneous observation for any point at a line y = h, thus especially for the points A(0,h) and B(L,h).

But the observed image is not fixed in t = 0, but it's: t_o = h/c.
The parameter t is here the moment of emission at a point (x,y) of the moving line - a body.

Please provide a detailed explanation of each step along the way and also an overview of what you are doing.

The equation is the condition of visibilty:
$d(x,t)/c = dt = t_o - t$

and in this case: $d(x,t) = |r(x,t)| = \sqrt{h^2 + (x-vt)^2}$

and we assume the moment of observation: t_o = h/c;
ie. the point of a moving body x = 0 is in the nearest distance at a time t = 0;
this is just a boundary condition of the equation, or the definition of a time coordinate.

 

[TrickyDicky]

A first stage of the determination...

Brilliant, you should publish this in Nature, but..er...you do realize that you are using something more than Euclidean geometry to obtain lorentz transformations,i.e. finite lightspeed c, right? And the difference between using time as a parameter vs a dimension?, in the first case you use the Euclidean scenario like it was done by Einstein and everyone before 1908, when Minkowski(pseudoeuclidean) spacetime(the second case with t as a dimension) was first presented.

 

[atto]

Brilliant, you should publish this in Nature, but..er...you do realize that you are using something more than Euclidean geometry to obtain lorentz transformations,i.e. finite lightspeed c, right? And the difference between using time as a parameter vs a dimension?, in the first case you use the Euclidean scenario like it was done by Einstein and everyone before 1908, when Minkowski(pseudoeuclidean) spacetime(the second case with t as a dimension) was first presented.

I think in the Minkowski spacetime the product c x t is transformed, not t - time alone, thus this is completely equivalent to the euclidean version.

Namely: the time part of Lorentz transform:
$(ct)' = \gamma (ct - xv/c)$
and in the euclidean space there is exactly the same relation.

 

[Barry911]

In relativistic optics an observer does not see any distortion of an object in an inertial frame rather the frame
passing the observer with a relative velocity appx. c, sees a rotation of the frame toward the observer
when approaching and away from the observer on moving away. Its also argued that all clocks in the
frame appearing elliptical but interpreted by the observer as circular cancels the observation of time dilatation.
Cosmic censorship with a vengeance!
For a long time I failed to realize that a frame with velocity V approaching and passing an observer has
a positive angular momentum and must be treated by GR.

Thanks

Barry

 

[atto]

In relativistic optics an observer does not see any distortion of an object in an inertial frame rather the frame passing the observer with a relative velocity appx. c, sees a rotation of the frame toward the observer when approaching and away from the observer on moving away. Its also argued that all clocks in the frame appearing elliptical but interpreted by the observer as circular cancels the observation of time dilatation.

I have seen such rotated images, but I don't know how these was calculated, because it's rather impossible.

For example, when we observe a cube at angle of 90 degrees:

|C| - a moving cube
|
| any light rays falls almost perpendiculary
|
|
O

Thus it's evident that the cube could not be rotated in any way.
We just can see the one frontal face, and nothing more, because the face covers other parts of the cube.

And the geometry of observed images does not depend on time, thus the time dilation is completely irrelevant here.

 

[Dale]

I have seen such rotated images, but I don't know how these was calculated, because it's rather impossible.

If you don't know then isn't it a little premature to claim the impossibility? After all, it isn't a violation of the laws of physics, just a strange optical phenomenon in the midst of many strange phenomena.

 

[atto]

If you don't know then isn't it a little premature to claim the impossibility? After all, it isn't a violation of the laws of physics, just a strange optical phenomenon in the midst of many strange phenomena.

I can quickly check it out.
Sufficient to calculate the eight moving points - two squares one after another.

The equation is the same, but with two distances: one h, and a second h+a.

http://www.math.ubc.ca/~cass/courses/m309-01a/cook/terrell1.html

It looks incorrectly - rather impossible.

There unnecessarily complicate the matter by analyzing the moving observer.
It is much easier to calculate for the stationary one.

Then would be no problem with an aberration of light,
which there is probably not considered at all, and perhaps for this reason such unrealistic results.