How is the Minkowski 4-space equation connected to hyperbolic functions?

[starthaus]

Wikipedia gives:

\phi1=ln(\gamma(1+\beta))
\phi2=-ln(\gamma(1-\beta))

http://en.wikipedia.org/wiki/Lorentz_transformation

There seems to a problem with way you obtain your two roots. Something is not right, but I can not put my finger on it at the moment.

You are reading the wiki page wrong, there is no \phi1 and \phi2 in the wiki solution. Wiki author gets only one of the two roots since:

-ln(\gamma(1-\beta))=ln(\gamma(1+\beta))Either way, your answer is clearly wrong. I can see that you surreptitiously changed the \beta^2 into \beta after I flagged the error. This is good but you should let people know when you correct your errors.

 

[starthaus]

Kev and stevmg

which bit of the wiki page are you looking at to get those answers for phi1 and phi2? I could see something close-ish, under the rapidity section, but I couldn't quite see that formula.

I think it's just a matter of applying the quadratic formula for roots of a quadratic equation to
<br /> e^{2\phi}-2\gamma*e^\phi+1=0<br />

i.e. letting x = e^{\phi}, substituting, and solving, to get Starhaus' two solutions for \phi.

Yes, of course, basic math. The wiki page gets only one of the two solutions.

 

[yuiop]

You are reading the wiki page wrong, there is no \phi1 and \phi2 in the wiki solution. Wiki author gets only one of the two roots since:

-ln(\gamma(1-\beta))=ln(\gamma(1+\beta))

Yep, I did misread that part.

This means your two roots are effectively:

\phi1=ln(\gamma(1+\beta)) = -ln(\gamma(1-\beta))
\phi2=ln(\gamma(1-\beta)) = -ln(\gamma(1+\beta))

so we could say:

\phi = \pm ln(\gamma(1+\beta)) \textrm{ or } \pm ln(\gamma(1-\beta))

and we can equally say:

\phi = ln(\gamma(1\pm\beta)) \textrm{ or } - ln(\gamma(1\pm\beta))

so there appears to be some ambiguity in the logarithmic definition of \phi

 

[starthaus]

Yep, I did misread that part.

This means your two roots are effectively:

\phi1=ln(\gamma(1+\beta)) = -ln(\gamma(1-\beta))
\phi2=ln(\gamma(1-\beta)) = -ln(\gamma(1+\beta))

so we could say:

\phi = \pm ln(\gamma(1+\beta)) \textrm{ or } \pm ln(\gamma(1-\beta))

and we can equally say:

\phi = ln(\gamma(1\pm\beta)) \textrm{ or } - ln(\gamma(1\pm\beta))

so there appears to be some ambiguity in the logarithmic definition of \phi

There is no ambiguity, algebraic equations degree 2 tend to have two different solutions.

 

[DrGreg]

For a given velocity (β) there is only one possible rapidity (φ), so Wikipedia is right. Of course for a given Lorentz factor (γ) there are two possible values of β, positive and negative, which is where the confusion has arisen here.

In the context quoted, γ is to be regarded as a function of β rather than the other way round.

 

[yuiop]

Getting back to steve's original question:

I was trying to find the relationship between a and \phi.

I know that sinh \phi = ct/a and cosh \phi = x/a
...
...
Now I have to let the dust settle, have you all agree what I just wrote was correct...

From the above image, you can see that:

t^2 - x^2 = constant =a^2 = (cdt^2) -dx^2

Note that the constant is not necessarily one. The hyperbolic curve is the horizontal version that represents a curve of constant proper time for various relative velocites, rather than using the vertical version you were using before that represents a curve of constant proper disnance.

Now if we use the hyperbolic identity:

\cosh^2\phi - \sinh^2\phi = 1

we can say:

a^2(\cosh^2\phi- \sinh^2\phi) = (cdt)^2 - dx^2

\cosh^2\phi - \sinh^2\phi = \frac{(cdt)^2}{a^2} - \frac{dx^2}{a^2}

From the above it natural to make the association:

\cosh^2 \phi = \frac{(cdt)^2}{a^2}

\Rightarrow \cosh^2 \phi = \frac{(cdt)^2}{(cdt)^2 - dx^2} = \frac{1}{1-\beta^2} = \gamma^2

and

\sinh^2 \phi = \frac{dx^2}{a^2}

\Rightarrow \sinh^2 \phi = \frac{dx^2}{(cdt)^2 - dx^2} = \frac{\beta^2}{1-\beta^2} = \beta^2 \gamma^2

I was trying to keep a geometric feel for all this so that when I delve into rapidity I could visualize what I was doing.

Also, why is this matrix referred to as a rotation of axes? What is rotated?

If you look at the diagram below:

You can see that the x' axis has been rotated anticlockwise while the t' axis has been rotated clockwise relative to the x and t axes respectively. The amount of rotation of the axes depends on the relative velocity between the F and F' frames. The relative velocity in the above diagram is 0.6c so the proper time t' =1 translates to t=0.8 and the proper distance x' =1 translates to x=0.8. Note that point P = (x',t') = (0,1) remains on the horizontal hyperbola in the left diagram, for all relative velocities and the point P = (x',t') = 1,0) remains on the vertical hyperbola in the right diagram, for all relative velocities.

Diagrams from http://hubpages.com/hub/Minkowski-Diagram

 

[starthaus]

For a given velocity (β) there is only one possible rapidity (φ), so Wikipedia is right.

That would be \phi=arctanh(\beta) since tanh(\phi)=\beta

Of course for a given Lorentz factor (γ) there are two possible values of β, positive and negative, which is where the confusion has arisen here.

There is no confusion and there is no connection to the positive and negative \beta. The two solutions arise from the other definition of \phi, i.e.

\phi=arccsh(\gamma)

In the context quoted, γ is to be regarded as a function of β rather than the other way round.

True but it has nothing to do with the issue being discussed.

 

[Dale]

How do I relate the \gamma from the Lorentz transforms [1/\sqrt[]{(1 - v^2)/c^2}] back to the a²? There is a relation which would change the a² for different v's (or different \gamma 's) but is "lost in translation." How do I get it back?

Is this what you are asking:

dt^2 - dx^2 = -da^2 = d\tau^2

\frac{d\tau^2}{dt^2} = 1 - \frac{dx^2}{dt^2}

\frac{d\tau}{dt} = \sqrt{1 - v^2}

\gamma = \frac{dt}{d\tau} = \frac{1}{\sqrt{1 - v^2}}

 

[yuiop]

Wikipedia gives \varphi = artanh(\beta) = \ln\left(\frac{\sqrt{1-\beta^2}}{1- \beta} \right)= \ln\left(\frac{1}{\gamma(1-\beta)}\right)

so it would seem that the rapidity \varphi = arctanh(\beta)= 1/\phi. Curiously, Wikipedia defines both \phi and \varphi as the rapidity on different pages.

http://en.wikipedia.org/wiki/Rapidity

http://en.wikipedia.org/wiki/Lorentz_transformation

[EDIT]\varphi = arctanh(\beta) \ne 1/\phi. Slipped up again. See post 43.

 

[starthaus]

Wikipedia gives \varphi = artanh(\beta) = \ln\left(\frac{\sqrt{1-\beta^2}}{1- \beta} \right)= \ln\left(\frac{1}{\gamma(1-\beta)}\right)

so it would seem that the rapidity \varphi = arctanh(\beta)= 1/\phi. Curiously, Wikipedia defines both \phi and \varphi as the rapidity on different pages.

http://en.wikipedia.org/wiki/Rapidity

http://en.wikipedia.org/wiki/Lorentz_transformation

No, if you do the math carefully:

\varphi = \phi

 

[yuiop]

No, if you do the math carefully:

\varphi = \phi

Please demonstrate.

 

[starthaus]

Please demonstrate.

arctanh(\beta)=arccsh(\gamma)

Both wiki pages you cited say the same exact thing, you have just misread what they say.

 

[yuiop]

arctanh(\beta)=arccsh(\gamma)

Both wiki pages you cited say the same exact thing, you have just misread what they say.

Using the inverse hyperbolic function (http://en.wikipedia.org/wiki/Inverse_hyperbolic_function)

\varphi = \ln\left(\frac{1}{\gamma(1-\beta)}\right) = -\ln(\gamma(1-\beta)) = \ln(\gamma(1+\beta))

So DrGreg and Wikipedia are right when they say there is one unique solution to the rapidity when it is defined by \varphi, which is more tightly defined than \phi which has two solutions (as you have shown). There is a subtle difference.

Anyway, Steve. Are we any closer to answering your original question?

 

[starthaus]

Wikipedia gives \varphi = artanh(\beta) = \ln\left(\frac{\sqrt{1-\beta^2}}{1- \beta} \right)= \ln\left(\frac{1}{\gamma(1-\beta)}\right)

so it would seem that the rapidity \varphi = arctanh(\beta)= 1/\phi. Curiously, Wikipedia defines both \phi and \varphi as the rapidity on different pages.

http://en.wikipedia.org/wiki/Rapidity

http://en.wikipedia.org/wiki/Lorentz_transformation

[EDIT]\varphi = arctanh(\beta) \ne 1/\phi. Slipped up again. See post 43.

Yes, you have a hard time getting basic math right.

 

[stevmg]

To starthaus, kev, yossell, DrGreg, DaleSpam et al - You have all done yeoman's work in explaining this to me. I will have to now download all these posts so that I can read them on paper (easier to do than computer scree and I can jump back and forth between posts quicker.)

We need an area where we can put drafts of posts so that we can work on them (LaTeX is tedious) and not be forced to shove them onto the PF.

starthaus - It is not the Wikipage that is wrong, it is my brain that is wrong. Takes time to get my brain right - which it eventually does. I will have to download all these replies onto paper so I can peruse them and it will be easier to go back and forth between posts than it is on computer. I need to understand the hyperbolic relationships for me to understand your acceleration paper which uses this. Otherwise, it is just "monkey see, monkey do."

kev - how did you upload those graphics:

https://www.physicsforums.com/showpost.php?p=2812721&postcount=36:

without them appearing as thumbnails, as they did in my post:

https://www.physicsforums.com/showpost.php?p=2811664&postcount=17

I see you dowwnloaded them from a website, which I do not have, but I do have images. Are there any web-accessible programs or whatever to generate these kinds of graphs or did you get them from other web postings?

The vertical hyperbolas you posted - are they the usual hyperbolas associated with these kinds of problems. Surely, the asymptotic light cone is. Which way do we go? Up and down, or sideways? Don't make fun of me, now, and you all know what I mean.

I WILL eventually get this on a "cognitive" (internalized) level.

 

[yuiop]

kev - how did you upload those graphics:

https://www.physicsforums.com/showpost.php?p=2812721&postcount=36:

without them appearing as thumbnails, as they did in my post:

If you can find a suitable image on the web then right click on the image and select "properties" and copy the url address from there. Wrap the link address in [ img ] [ /img ] tags (without the spaces) or use the insert image feature in the PF post editing tools. If the images are made by you, then you need to post them on your own website or on a website like photobucket that let's you store images on the web. You can also upload images to your PF blog, open up the thumbnail to view it and grab the link address from the properties. (not sure if PF would encourage you to do this, as you would be using up their storage space, but it is technically possible )

 

[stevmg]

Here's one for the books, I hope, even if it's Mad Comic Books.

starthaus has stated and in books like Taylor/Wheeler 2nd ed Spacetime Physics it is stated that:

\exists a \phi such that cosh (\phi) = \gamma

Wherever I have read, that has sort of been a jump off assumption proven backwards by then showing consequences of that assumption to be true. To wit, v\gamma = sinh (\phi).

Now, what bugs the crap out of me is that I have never seen (not that I had such a great library at my disposal) an intuitive yet accurate demonstration that, in fact, the above assertion is directly true.

Without going to the e's \pm/2 etc. which are the standard textbook definitions of sinh and cosh it would be neat if we could have a geometric visualization of what a hyperbolic function is. I have attached a .pdf file scanned from a calculus review book, very thin, from 1959 which was first published in 1957 just such a demonstration in a file Hyperbolic functions.pdf. It is The Calculus by C.O. Oakley from the Barnes and Noble College Outline Series, pages 198-200. No plagiarism here as I am giving him (it is a he) full credit.

The hyperbolas in the graphs are of the form x² - y² = a². The foci of these hyperbolas are along the abscissa and these hyperbolas are symmetric. The positive x-intercept is at (a, 0). The coordinates of the foci are irrelevant for this discussion. By the Einstein-Minkowski 4-space equations:

x² - c²t² = x'² - c²t'² or x² - c2^t2 = a constant, say a²

This matches the equation in this textbook literally as is (if you use x for and ct for y).

From here on let us use c = 1 and v = \beta = "the old v"/c

Now, from that Oakley .pdf file \exists a \phi such that cosh (\phi) = x/a

If we look at \gamma we get \gamma = 1/(1 - v²)^(1/2)
Remember, I am using v instead of \beta because it is easier to type and in some textbooks v is used as a fraction of c.

This is "backwards" as this ratio has the constant (1) on the top and the variable [(1 - v²)^(1/2)] on the bottom.

That's easy to fix:

[1/[(1 - v²)^(1/2)]/[1/1] is still the same RATIO. Shortening it, we get
[1/[(1 - v²)^(1/2)]

substituting that for x
we get x² = [1/[(1 - v²)]. This is the same as \gamma² or "gamma squared"
and, in this case, a = 1 or a² = 1

Now that hyperbola is x² - y² = 1² = 1

y² = x² - 1

Moving on along

y² = [1/[(1 - v²)] - 1 = v²/[(1 - v²)]

or y = v/[(1 - v^2(1/2) = v\gamma or "v times gamma"

y also = sinh (\phi) from Oakley. So, now we have a geometric "proof" that \gamma = cosh (\phi) and
v\gamma or "v times gamma" = sinh (\phi).

From above, x = \gamma = [1/[(1 - v²)^(1/2)] = cosh (\phi)
Both of these are useful in the matrix multiplication representation of the Lorentz transformations.

From the algebra above we have shown that x²/1² - y²/1² = 1² or cosh² (\phi) - sinh² (\phi) = 1. This is a well known identity from hyperbolic trigonometry.

This reasoning above is not a tautology ("circular logic") because that same identity in hyperbolic trigonometry is proven in other ways (see Oakley, p. 200 in the attached .pdf)

 

[starthaus]

Here's one for the books, I hope, even if it's Mad Comic Books.

starthaus has stated and in books like Taylor/Wheeler 2nd ed Spacetime Physics it is stated that:

\exists a \phi such that cosh (\phi) = \gamma

Wherever I have read, that has sort of been a jump off assumption proven backwards by then showing consequences of that assumption to be true. To wit, v\gamma = sinh (\phi).

Now, what bugs the crap out of me is that I have never seen (not that I had such a great library at my disposal) an intuitive yet accurate demonstration that, in fact, the above assertion is directly true.

Yes, you have seen https://www.physicsforums.com/blog.php?b=1911 but you forgot. I showed you how tanh(\phi)=\beta is derived. This is equivalent with cosh(\phi)=\frac{1}{\sqrt{1-tanh^2(\phi)}}=\gamma.

 

[Dickfore]

If you differentiate x^{2} - c^{2} \, t^{2} = a^{2} with respect to t, considering x a function of t, then you ought to obtain:

<br /> 2 x \, \dot{x} - 2 \, c^{2} \, t = 0<br />

or

<br /> \beta \equiv \frac{\dot{x}}{c} = \frac{c \, t}{x}<br />

The definition of \gamma is:

<br /> \gamma = (1 - \beta^{2})^{-\frac{1}{2}} = \left[ 1 - \left(\frac{c \, t}{x}\right)^{2} \right]^{-\frac{1}{2}} = \left(\frac{x^{2} - c^{2} \, t^{2}}{x^{2}}\right)^{-\frac{1}{2}} = \frac{x}{a}<br />

 

[stevmg]

Getting back to steve's original question:
If you look at the diagram below:

You can see that the x' axis has been rotated anticlockwise while the t' axis has been rotated clockwise relative to the x and t axes respectively. The amount of rotation of the axes depends on the relative velocity between the F and F' frames. The relative velocity in the above diagram is 0.6c so the proper time t' =1 translates to t=0.8 and the proper distance x' =1 translates to x=0.8. Note that point P = (x',t') = (0,1) remains on the horizontal hyperbola in the left diagram, for all relative velocities and the point P = (x',t') = 1,0) remains on the vertical hyperbola in the right diagram, for all relative velocities.

Diagrams from http://hubpages.com/hub/Minkowski-Diagram

kev -

I cannot make heads, tails, x's, y's, t's or anything out of those diagrams.

I know Galilean transformations are "shear." That is - like a box squished to the left or right with non-perpendicular axes. Is that a form of rotation in itself? Or does the abscissa and ordinate axes remain perpendicular under Galilean transformations?

Is it the Lorentzian transformations that twist the ordinate axis off perpendicular to the x-axis?

This thumbnail (I still haven't figured out how to get a full sized picture uploaded) is what I am talking about. In this case, the y-axis is twisted or rotated off perpendicular.

If it were just a Galillean transformation, the y-axis would still be perpendicular but the "world lines" would be off perpendicular...

Am I right?

Remember, step-by-step I learn...

 

[stevmg]

Yes, you have seen https://www.physicsforums.com/blog.php?b=1911 but you forgot. I showed you how tanh(\phi)=\beta is derived. This is equivalent with cosh(\phi)=\frac{1}{\sqrt{1-tanh^2(\phi)}}=\gamma.

You are correct. I didn't use that derivation as I was trying to show what someone starting from total scratch would come up with - even if they did not know of that well known relationship you stated (and derived.) However, your derivation is logical and actually could be placed in "the chain of thinking" from "my beginning" to the final conclusion.

Steve

 

[stevmg]

If you differentiate x^{2} - c^{2} \, t^{2} = a^{2} with respect to t, considering x a function of t, then you ought to obtain:

<br /> 2 x \, \dot{x} - 2 \, c^{2} \, t = 0<br />

or

<br /> \beta \equiv \frac{\dot{x}}{c} = \frac{c \, t}{x}<br />

The definition of \gamma is:

<br /> \gamma = (1 - \beta^{2})^{-\frac{1}{2}} = \left[ 1 - \left(\frac{c \, t}{x}\right)^{2} \right]^{-\frac{1}{2}} = \left(\frac{x^{2} - c^{2} \, t^{2}}{x^{2}}\right)^{-\frac{1}{2}} = \frac{x}{a}<br />

I love that! That's freakin' neat! Good on you!

Of course it took me 25 years to figure out that \dot{x} = dx/dt but I'm a little slow.

Steve

 

[yossell]

.
I know Galilean transformations are "shear." That is - like a box squished to the left or right with non-perpendicular axes. Is that a form of rotation in itself? Or does the abscissa and ordinate axes remain perpendicular under Galilean transformations?

Is it the Lorentzian transformations that twist the ordinate axis off perpendicular to the x-axis?

I think I wouldn't worry about the fact that the Lorentz transformations are described as a rotation. I think that's only based on a formal (i.e. mathematical) similarity between matrices for Lorentz transformations and the matrices for rotations in standard Euclidean space, where the metric is positive definite (i.e. no negative contribution from any of the axes). If you're thinking in images, it's better to imagine something like a shear, as in Kev's diagrams - but with some funny stretching.

 

[Mentz114]

I think I wouldn't worry about the fact that the Lorentz transformations are described as a rotation. I think that's only based on a formal (i.e. mathematical) similarity between matrices for Lorentz transformations and the matrices for rotations in standard Euclidean space, where the metric is positive definite (i.e. no negative contribution from any of the axes).

True. Rotations mix spatial dimensions so x'=f(x,y) , y'=h(x,y) under a rotation. For boosts, space and time are mixed, t'=L(t,x), x'=L(t,x), hence apparent time-dilation and spatial contraction.

 

[stevmg]

I think I wouldn't worry about the fact that the Lorentz transformations are described as a rotation. I think that's only based on a formal (i.e. mathematical) similarity between matrices for Lorentz transformations and the matrices for rotations in standard Euclidean space, where the metric is positive definite (i.e. no negative contribution from any of the axes). If you're thinking in images, it's better to imagine something like a shear, as in Kev's diagrams - but with some funny stretching.

Linear algebra is a long way away from me...

Is it possible to give me a "quick and dirty" description of the kinds of transformations.

I know what "translation" is - that's moving the x or y origin to a new place but with the ordinate and abcissas still at right angles and no change in direction of either major axis.

Rotation was what we used to do in Analytic Geometry to the equations of conics to get rid of the xy term in the general conic equation Ax² + Bxy + Cy² + Dx + Ey + F = 0. In this case, we would rotate both axes and the new equation would have no xy term in it. We would then translate both x and y - wise to get a standard conic equation. The new x-axis and y-axis wre still perpendicular but angled off from the original x-y axes.

Now I have seen diagrams where the axes are no longer perpendicular (i.e., "squished") - is that "shear?" Is that what happens in Lorentz transforms? In Galilean transforms, don't the axes stay the same but the world lines (the t-vectors) even if straight are not at \pi/2 to the x or distance axes provided there is a non trivial v. I would presume that Galilean transformations would not alter the x, y, z versus t perpendicularity as these dimensions are independent of each other and no change in x, y or z does anything to t while in relativity, changes in x, y or z (or any other linear dimension if they exist) DOES alter the t component in itself.

So, we have translation - I think I covered that, rotation in the Analytic Geometry sense, but unexplained are rotation in the "squishing" sense or reflection, whatever that is.

Oh, btw there are alterations of axes where one changes the bases of each unit so that one can stretch or compress a dimension. Log paper or exponential paer are examples of that, I guess.

Can you give me a really q&d go-over of that? Am I on the right sheet of music?

 

[stevmg]

True. Rotations mix spatial dimensions so x'=f(x,y) , y'=h(x,y) under a rotation. For boosts, space and time are mixed, t'=L(t,x), x'=L(t,x), hence apparent time-dilation and spatial contraction.

Thanks for the reply. Please look at post #55 and you will see that I am more basic than that and I need to know the basics before we use terms like "boost" &tc. I really need the basics here but it will come back as I did take linear algebra probably before you were born.

Let's start with one thing... what is a "shear" when it comes to changing the axes? When is it used?

stevmg

 

[stevmg]

Getting back to steve's original question:

If you look at the diagram below:

You can see that the x' axis has been rotated anticlockwise while the t' axis has been rotated clockwise relative to the x and t axes respectively. The amount of rotation of the axes depends on the relative velocity between the F and F' frames. The relative velocity in the above diagram is 0.6c so the proper time t' =1 translates to t=0.8 and the proper distance x' =1 translates to x=0.8. Note that point P = (x',t') = (0,1) remains on the horizontal hyperbola in the left diagram, for all relative velocities and the point P = (x',t') = 1,0) remains on the vertical hyperbola in the right diagram, for all relative velocities.

Diagrams from http://hubpages.com/hub/Minkowski-Diagram

kev - I got to the website and saw all those diagrams that the article had. Now, how did you put that into your note? There are numerous diagrams on that website and you only used a couple.

stevmg

 

[stevmg]

Yes, you have seen https://www.physicsforums.com/blog.php?b=1911 but you forgot. I showed you how tanh(\phi)=\beta is derived. This is equivalent with cosh(\phi)=\frac{1}{\sqrt{1-tanh^2(\phi)}}=\gamma.

starthaus -

I didn't realize you were referring me straight to that article you wanted me to see.

I haven't yet gotten into your article yet...

That's why I was looking for the visual or geometric approach so I could better internalize the concepts. I am satisfied now with what I have done for my benefit unless someone can point out an error in the math.

stevmg

 

[yossell]

That's why I was looking for the visual or geometric approach so I could better internalize the concepts.

Kev's diagrams pretty much say it all. Perhaps they say too much. Let's just look at the first diagram. You may know a lot of this already, but maybe you'll find it helpful.

The first diagram plots life (space-time life) for inertial (unaccelerated) observer O who starts at the origin and (in his own frame). The t-axis is the blue line going up plots O's life as time ticks by - since there's no change in the x coordinate along this line (and we're suppressing y and z directions), this line shows O not going anywhere, just letting the time tick by. The blue notches along this diagram represent ticks of the clock, each notch a unit of time.

The x-axis is, in this diagram, the events that are simultaneous with t = 0, for Observer O. It's an unusual way of thinking about the x axis, but it's very helpful in space time. Each notch in this axis represents a unit distance. The horizontal blue line drawn at t = 1 represents all the events that are simultaneous with whatever happens at the point (1 0), according to O.

Now look at the steeper red line. This is the space-time path of some OTHER inertial observer, o', travels. O' is inertial too - he travels at a constant velocity hence he cuts out a straight line in space-time. The faster he travels, the more angled his world-line. But the speed of light is a limiting factor, so the angle that possible observers can travel is bounded: namely by lines that represent the speed of light. Often, units are chosen so that light can be represented by lines that lie at 45 degrees to the axis.

Ok - so that steeper red line represents the path of O' - it's HIS time axis, commonly written as his t' axis. But what about HIS x - axis? Well, his x-axis are those events that HE regards as simultaneous. These events appear on the same map - both O and O' are privy to the same events - they just disagree about time and simultaneity. Well, when you follow the Lorentz transformations, it turns out that the x-axis of O' is tilted up - so that it becomes the less tilted of the two red lines. So that red line marked x' axis is the events that observer O thinks happen at the same time as events at t' = 0. You can see that, apart from the origin, O and O' disagree about which events are simultaneous with which. Again, lines drawn parallel to this red line represent lines of simultaneity in the frame of O'.

If we're using units where light travels paths at 45 degrees, then for any frame F, the t axis and the x-axis make the same angle with the 45 degree line - fold the paper along a 45 degree line, and the t-axis and x-axis of a frame get mapped onto each other. Now, the faster something travels, the closer it is to a 45 degree angle - the closer O judges it as getting to the speed of light - and the more its x-axis is tilted over. If you were to imagine continuing the process, the x and t axes collapse onto each other at the speed of light.

This tells us the angles - but we would also like to know how the clocks and lengths of O' work - we would like to draw HIS notches - that is, we want to calibrate his clocks and lengths. What does HE regard as a tick? What does he regard as a unit distance? Well, because of time dilation, and lorentz contraction, the notches that he draws are different from ours. I think kev's second diagram represents the notches as seen by the second observer.

All that drives this is the Lorentz transformations. They're just linear transformations - it's just a matter of matrix multiplication, so, though they're abstract, you're probably more familiar with them than you realize.

As usual, this took longer to say - hope there aren't too many typos...

 

[starthaus]

Linear algebra is a long way away from me...

Is it possible to give me a "quick and dirty" description of the kinds of transformations.

I know what "translation" is - that's moving the x or y origin to a new place but with the ordinate and abcissas still at right angles and no change in direction of either major axis.

Rotation was what we used to do in Analytic Geometry to the equations of conics to get rid of the xy term in the general conic equation Ax² + Bxy + Cy² + Dx + Ey + F = 0. In this case, we would rotate both axes and the new equation would have no xy term in it. We would then translate both x and y - wise to get a standard conic equation. The new x-axis and y-axis wre still perpendicular but angled off from the original x-y axes.

Now I have seen diagrams where the axes are no longer perpendicular (i.e., "squished") - is that "shear?" Is that what happens in Lorentz transforms? In Galilean transforms, don't the axes stay the same but the world lines (the t-vectors) even if straight are not at \pi/2 to the x or distance axes provided there is a non trivial v. I would presume that Galilean transformations would not alter the x, y, z versus t perpendicularity as these dimensions are independent of each other and no change in x, y or z does anything to t while in relativity, changes in x, y or z (or any other linear dimension if they exist) DOES alter the t component in itself.

So, we have translation - I think I covered that, rotation in the Anaslytic Geometry sense, but unexplained are rotation in the "squishing" sense or reflection, whatever that is.

Oh, btw there are alterations of axes where one changes the bases of each unit so that one can stretch or compress a dimension. Log paper or exponential paer are examples of that, I guess.

Can you give me a really q&d go-over of that? Am I on the right sheet of music?

Here is a crash course.

1. Rotation in 2D plane is:

x&#039;=x*cos(\phi)+y*sin(\phi)
y&#039;=-x*sin(\phi)+y*cos(\phi)

You can convince yourself that this is correct by trying to plot a few images.

2. Lorentz transforms :

x&#039;=x*cosh(\phi)-(ct)*sinh(\phi)
(ct)&#039;=-x*sinh(\phi)+(ct)*cosh(\phi)


Because the Lorentz transforms look like the 2D transforms for rotation they are called, by abuse of language, "rotations in the hyperbolic plane". This is where the "rotation" comes from.