B Trying to understand Lorentz Transformations

[Ibix]

People may have worked on this for a long time. You should not assume that it was intuitively obvious to the casual observer.

Indeed. The mathematical tools to derive the Lorentz transforms from symmetry arguments existed at least as far back as Newton. Nobody did it (edit: or as PeroK points out below, nobody published it) for around 250 years.

 

[PeroK]

Hmmm. Are you saying that as long as the equation is giving the right results, I should just learn it and use it and not get too bogged down in how it originated?

In many ways, SR can be the opposite of that. Everything ought to make sense and, in the end, you wonder how no one thought of it before Einstein.

The problem here is perhaps one of mathematical experience and technique. For example, you see posts on here where some typical mathematical operation (squaring both sides of an equation, say) is seen by the inexperienced as a great leap of imagination.

At the next level, you are viewing a set of linear equations as something almost mystical that a mathematician dreamed up out of nowhere. But, once you have a bit of mathematical experience, it's the most obvious thing to do.

At the most basic level you are asking: why on Earth would anyone ever come up with the idea of analysing a possible linear relationship between time and space coordinates?

And our answer is, well no one did for several hundred years, but why not try? Maybe someone in 1850 did try but was too appalled by the consequences and didn't even publish!?

There is an element of hindsight, of course, that the great symmetries of time, space, energy and momentum etc. have been studied in depth and we know that SR could have been built on these.

I know some texts on SR simply produce the Lorentz Transformation as a postulate. That would be too much like pulling a rabbit out of a hat for me. But, developing it from symmetry and homogeneity principles seems very natural to me, even if it relies on a fairly modern view of physics.

It's good that you want to understand why this approach is valid. But, you must also try to develop the scientific capabilities of your brain. And sometimes that means going with the flow and seeing where it leads. It feels to me sometimes that there's a part of your brain that wants to reject this stuff and, obviously, you do want to learn it. So, perhaps you yourself have to start beating up on the reactionary part of your brain - instead of asking us to do it

 

[stevendaryl]

In the equation y = mx + b, the first time I read it and the description for it it made sense to me right away. It makes sense because I visualize that what it's describing is a straight light which isn't curved, and I understand why we use each of the terms. The Lorentz equation for time, though, doesn't jump out at me in the same way as being intuitive.

I would say that if you have two different inertial coordinate systems systems in one spatial and one time dimension, $(x,t)$, and $(x',t')$ we can describe their "disagreement" using four factors:

1. They may disagree about what objects are at rest.
2. They may disagree about the length of objects.
3. They may disagree about the time between two events.
4. They may disagree about which events are simultaneous (take place at the same time).

(There are actually a couple of other sources of disagreement that I'm going to ignore: 5. They can disagree about the direction in which $x$ increases. 6. They can disagree about the origin of the $(x,t)$ coordinate system.)

Effect 1 leads to the transformation for $x'$ involving both $x$ and $t$.
Effect 4 leads to the transformation for $t'$ involving both $x$ and $t$.
Effects 2 and 3 are matters of scaling, in a sense.

Galilean transformations only involve effect 1.

Why would different inertial coordinate systems disagree about which events are simultaneous? Well, you have to go through, as Einstein did, the thought process of asking how do you establish that distant clocks are synchronized. One possibility is that you bring two clocks together, synchronize them, and then take them to different locations. But how do you know that moving clocks around doesn't affect them? Another possibility is that you use some signal with a known speed (such as light) to synchronize distant clocks, and take into account the transit time. But if two different coordinate systems disagree about whether objects are at rest, then they will disagree about transit time, as well (it takes longer to catch up to a moving object, if it's moving away, and it takes less time to catch up to a moving object, if it's moving toward you).

So if you're willing to buy that clock synchronization might be relative to an inertial coordinate system, then effects 2 and 3 follow. The way that you would measure the length of a moving object is by noting that one end is at location $x$ at the same time that the other end is at location $x+L$. You subtract those two coordinates to get a length $L$. But if different coordinate systems disagree about whether clocks are synchronized, then they will disagree about whether the two ends were measured at the same time. So they will disagree about the length of the moving object. There is a similar reason that different inertial coordinate systems may disagree about the time between events.

 

[robphy]

(Hoping not to divert the thread... but replying to a side comment by PeroK...)

At the most basic level you are asking: why on Earth would anyone ever come up with the idea of analysing a possible linear relationship between time and space coordinates?

And our answer is, well no one did for several hundred years, but why not try? Maybe someone in 1850 did try but was too appalled by the consequences and didn't even publish!?

If anyone was able to do that, it would have been Felix Klein.
https://de.wikisource.org/wiki/Felix_Klein

In 1872, he developed his Erlangen Program
https://en.wikipedia.org/wiki/Erlangen_program
which essentially laid the foundations for Minkowski spacetime.
(Klein essentially unified elliptic, hyperbolic, and Euclidean geometry via projective geometry.
The scheme was extended to the "Cayley-Klein Geometries" (9 in 2-dimensions),
of which Minkowski spacetime, Galilean spacetime, and the DeSitter spacetimes are included (but apparently not interpreted physically).

Here's a curious comment from Toretti's Philosophy of Geometry from Riemann to Poincaré, p.129
https://books.google.com/books?id=EcLrCAAAQBAJ&printsec=frontcover&dq=torretti&hl=en&sa=X&ved=0ahUKEwio6bP235bhAhUkUt8KHZydCAIQ6AEIKDAA#v=onepage&q=finite sequence of rotations&f=false

The passage in Klein that Toretti refers to is
on p.189 of Klein's Vorlesungen uber Nicht-Euklidische Geometrie (1926)
(which [if i decoded the preface correctly] is partly based on lectures from 1892 and 1893
[possibly these handwritten lectures https://archive.org/details/nichteuklidische01klei ],
plus additional material the editors put together). In that passage,

(update: here are the figures Klein refers to... looks like worldlines in Minkowski and Galilean spacetimes. Note how the lines bunch up near the [light-cone] edge in 114.)

I transcribed the section

Die Erfahrung zeigt uns, daß wir stets durch eine endliche Anzahl derartiger Drehungen in die Nähe der Ausgangslage zurückkommen, ja sogar über sie hinausgelangen können. Diese Eigenschaft ist aber bei der hyperbolischen und parabolischen Messung von Winkeln nicht vorhanden, da wir dort durch die entsprechende Abtragung untereinander kongruen ter Winkel nie über bestimmte Grenzlagen herauskommen können (vgl. die Abb. 114 und 120). In einer Maßbestimmung, die in der Außenwelt anwendbar ist, muß also sowohl die räumliche, wie auch die ebene Winkelmessung elliptisch sein.

and fed it into Google translate (bolding mine)

Experience shows us that we can always come back by a finite number of such rotations in the vicinity of the starting position, and even get beyond them. However, this property is not present in the hyperbolic and parabolic measurement of angles, since we can never get beyond certain boundary layers there by the corresponding removal of congruent angles (see Figures 114 and 120). In a measure that is applicable in the outside world, so both the spatial, as well as the level angle measurement must be elliptical.

In hindsight, the hyperbolic and parabolic measurement of angles [between radial lines... thought of as inertial worldlines] refer to rapidity in Minkowski spacetime [so velocity v=c\tanh\theta ] and in Galilean spacetime [here, where, it turns out, Galilean-rapidity coincides with velocity [up to a constant carrying units] ].
So, the point is:
the foundations were there.. but maybe not yet realized in the physical world.

(As I said, these lectures were published posthumously in 1926 and drawn on material from 1892 and 1893 onward...
but it's not clear how far onward. Was this passage before 1905 (when Einstein published) or 1908 (when Minkowski formulated "spacetime geometry")?)From the Wikisource link, it might be interesting to read the transcription of
https://de.wikisource.org/wiki/Über_die_geometrischen_Grundlagen_der_Lorentzgruppe (1910)
(which Google Translate translates to "About the geometric foundations of the Lorentz group")
which can be read via Google Translate [in the Chrome browser].

My $0.02.
(Sorry for intruding.)

 

[Histspec]

With those assertions, the most general transformation we can possibly write is$$\begin{eqnarray*}t'&=&At+Bx+Cy+Dz\\
x'&=&Et+Fx+Gy+Hz\\
y'&=&It+Jx+Ky+Lz\\
z'&=&Mt+Nx+Oy+Pz\end{eqnarray*}$$where the capital letters are constant. We are not making any assumptions about these constants. Many of them may be zero - but we'll prove that, we won't assume it.

All the off-diagonal elements including $y$ or $z$ must be zero from symmetry.

The above transformation is already a complete Lorentz transformation (in matrix form $\mathbf{x=g}\mathbf{x}'$ or $\mathbf{x'}=\mathbf{g}^{-1}\mathbf{x}$) leaving invariant $-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ by using the conditions

$$\begin{matrix}\mathbf{g}^{{\rm T}}\mathbf{A}\mathbf{g}=\mathbf{A}\\
\mathbf{A}=\mathrm{diag}(-1,1,1,1)
\end{matrix}$$

One can see, that the Lorentz transformation and its group can be understood as a branch of linear algebra.

 

[Histspec]

If anyone was able to do that, it would have been Felix Klein.
https://de.wikisource.org/wiki/Felix_Klein

In 1872, he developed his Erlangen Program
https://en.wikipedia.org/wiki/Erlangen_program
which essentially laid the foundations for Minkowski spacetime.

Indeed. Like many other mathematicians in the 19th century, Klein analyzed the transformations leaving invariant the interval $-x_{0}^{2}+x_{1}+x_{2}$ or $-x_{0}^{2}+x_{1}+x_{2}+x_{3}$ which he related to hyperbolic geometry (or planes/spaces of constant negative curvature).
https://en.wikipedia.org/wiki/History_of_Lorentz_transformations#Klein_(1871–1897)

For instance, in his 1896-lecture (pp. 13–14) on the theory of the top, he specifically identified one of these coordinates with time (even though later in the book, he assured the readers that he didn't imply any "metaphysical" interpretation of his formulas on non-Euclidean geometry):

$$\begin{matrix}x^{2}+y^{2}+z^{2}-t^{2}=0\\
=(x+iy)(x-iy)+(z+t)(z-t)=0\\
x+iy:x-iy:z+t:t-z=\zeta_{1}\zeta_{2}^{\prime}:\zeta_{2}\zeta_{1}^{\prime}:\zeta_{1}\zeta_{1}^{\prime}:\zeta_{2}\zeta_{2}^{\prime}\\
\frac{\zeta_{1}}{\zeta_{2}}=\zeta\rightarrow\zeta=\frac{x+iy}{t-z}=\frac{t+z}{x-iy}\\
X^{2}+Y^{2}+Z^{2}-T^{2}=0=\ \text{etc.}\\
\zeta=\frac{\alpha Z+\beta}{\gamma Z+\delta}\rightarrow\begin{matrix}\zeta_{1}=\alpha Z_{1}+\beta Z_{2}, & \zeta_{1}^{\prime}=\bar{\alpha}Z_{1}^{\prime}+\bar{\beta}Z_{2}^{\prime}\\
\zeta_{2}=\gamma Z_{1}+\delta Z_{2}, & \zeta_{2}^{\prime}=\bar{\gamma}Z_{1}^{\prime}+\bar{\delta}Z_{2}^{\prime}
\end{matrix}\\
(\alpha\delta-\beta\gamma=1)
\end{matrix}\quad\begin{array}{c|c|c|c|c}
& X+iY & X-iY & T+Z & T-Z\\
\hline x+iy & \alpha\bar{\delta} & \beta\bar{\gamma} & \alpha\bar{\gamma} & \beta\bar{\delta}\\
\hline x-iy & \gamma\bar{\beta} & \delta\bar{\alpha} & \gamma\bar{\alpha} & \delta\bar{\beta}\\
\hline t+z & \alpha\bar{\beta} & \beta\bar{\alpha} & \alpha\bar{\alpha} & \beta\bar{\beta}\\
\hline t-z & \gamma\bar{\delta} & \delta\bar{\gamma} & \gamma\bar{\gamma} & \delta\bar{\delta}
\end{array}$$

This type of Lorentz transformation in terms of linear fractional (Möbius) transformations and spin transformations became important in relativistic quantum theory beginning in the 1920ies.

 

[Nugatory]

Hmmm. Are you saying that as long as the equation is giving the right results, I should just learn it and use it and not get too bogged down in how it originated?

It depends on what you're trying to accomplish. "How it originated" is a question for the history of science, and that history is full of wrong turns and unnecessary detours.

Thus, if your goal is to understand the physics you'll want to learn from a good modern treatment that starts with the equations that best capture what we've learned with the benefit of decades or centuries of hindsight.

If your goal is to understand how humanity built this awe-inspiring intellectual structure from the raw materials of math and experiment (and for the non-specialist this question may be more important) then you'll want to study how it originated. But even then you're best off starting with a solid grasp of the modern treatment. It's a lot easier to follow the writings of Marco Polo or Lewis and Clark if you have access to a modern map; and if you don't understand the geometric formulation of special relativity you will be unable to see that it is hidden and struggling to emerge from the earliest papers.

 

[NoahsArk]

Thank you for the responses.

Based on the feedback seems like I need to work on solving problems involving the transformations in order to try and build my intuition more.

 

[Herman Trivilino]

Making up your own problems to solve can be even more helpful.

 

[sweet springs]

The SR relation, square of invariant distance,

c2t2−x2=c2(t1)2−(x1)2​

I would say this relation assures that Lorentz transformation is linear for x,y,z and t. For example if the transformation includes $x^2$, term $x^4$ should appear in the above formula. That's a mess.

 

[arydberg]

You seem to have difficulty grasping the fact that SR is totally outside your conception of reality. You are trying to make sense out of it. You cannot. x and t have values in a lab frame. When viewed from a moving frame we get x' and t'. They do not agree with x and t

 

[NoahsArk]

I'm now stuck on how to derive the LTs. This is the part I get stuck on which is shown in this video at 11min 15 sec:


When the Taylor and Wheeler book gets to that part of the derivation I also get stuck.

Basically, what he is saying is that if you have an equation of this form: aT² + bX² = cT² - dX², then you have to conclude that the coeficients "a" and "c" must be equal, because you have a T² on both sides of the equation, and you also have to conclude, since there are X²s on both sides, that "b" and "d" must also be equal. This is not true- you can have many values of "a" which are different from "c", and the equation will still be equal so long as "b" and "d" are also different by the right amounts.

Please let me know what I am missing.

 

[FactChecker]

If the coefficients, a, b, c, and d, are constants rather than functions of T and/or X, then you must have a=c and b=d. I think the initial assumption was that they are constants.

 

[sweet springs]

This is not true- you can have many values of "a" which are different from "c", and the equation will still be equal so long as "b" and "d" are also different by the right amounts.

Yes, if you carefully choose special values of X and T, i,e. tag of special event happening in special place and special time for given a, b, c and d.
But the teacher is so ambitious that tag of any place any time events, e.g. my writing this article here now, your reading this article there later, my death in some place and in sometime in near future, your birth, etc., follow the same formula with same a,b,c and d, to have correspond tag of different number in another IFR. It means that for any values of X and T the equation must hold.

The formula in the lecture something like
FT^2+GXT+HX^2=c^2T^2-X^2 so
(F-c^2)T^2+GXT+(H+1)X^2=0
Here all the coefficients of X^2, XT,T^2 should vanish so that this equation holds even if X and T may take any value tag for any events.

 

[Ibix]

Basically, what he is saying is that if you have an equation of this form: aT2 + bX2 = cT2 - dX2, then you have to conclude that the coeficients "a" and "c" must be equal, because you have a T2 on both sides of the equation, and you also have to conclude, since there are X2s on both sides, that "b" and "d" must also be equal. This is not true- you can have many values of "a" which are different from "c", and the equation will still be equal so long as "b" and "d" are also different by the right amounts.

$$\begin{eqnarray*}aT^2+bX^2&=&cT^2-dX^2\\
(b+d)X^2&=&(c-a)T^2\end{eqnarray*}$$
You are correct that, for any chosen $X$ and $T$ this can be made true by picking carefully chosen values of $(b+d)$ and $(c-a)$. However it will not be true for (almost) all other values of $X$ and $T$. So such transforms would not be globally applicable - and the whole point is to find a relationship that works for any $X$ and $T$.

The only solution that applies for any $X$ and $T$ is where $(b+d)=0$ and $(c-a)=0$.

 

[NoahsArk]

Thank you for the responses. It's starting to make more sense but I'm still a bit stuck:

If the coefficients, a, b, c, and d, are constants rather than functions of T and/or X, then you must have a=c and b=d.

Even if they are constants, why can't they have different values?

Here all the coefficients of X^2, XT,T^2 should vanish so that this equation holds even if X and T may take any value tag for any events.

The only solution that applies for any X and T is where (b+d)=0 and (c−a)=0.

Say b = 2 and d = 4 and c = 9 and a = 3. Now the equation holds true for all T and X values, even though b + d and c - a do not = 0.

 

[Ibix]

Even if they are constants, why can't they have different values?

Because then they are not constants. They change with position and/or time.

Say b = 2 and d = 4 and c = 9 and a = 3. Now the equation holds true for all T and X values, even though b + d and c - a do not = 0.

No it doesn't. That gives you $6X^2=6T^2$ which only holds true for $X=\pm T$. If you put in $X=6$ and $T=22$ (for example) your values do not produce an equality.

 

[stevendaryl]

Say b = 2 and d = 4 and c = 9 and a = 3. Now the equation holds true for all T and X values, even though b + d and c - a do not = 0.

The original equation was:

$aT^2 + b X^2 = c T^2 - dX^2$

This has to hold for all $X$ and $T$. So it has to hold in the case $T=0$ and $X=1$. In that case, we have:

$a \cdot 0 + b \cdot 1 = c \cdot 0 - d \cdot 1$

So we conclude: $b=-d$

It also has to hold when $X=0$ and $T=1$. So:

$a \cdot 1+ b \cdot 0 = c \cdot 1 - d \cdot 0$

So we conclude: $a = c$.

 

[Dale]

Please let me know what I am missing.

The equation needs to hold for all values of X and all values of T.

 

[FactChecker]

Suppose $a,b,c,d$ are constants and $(b+d)X^2=(c-a)T^2$ for all values of $X , T$. Pick any set of values, $X_1$ and $T_1$. We have $(b+d)X_1^2=(c-a)T_1^2$.

Now leave $X$ equal to $X_1$ and let $T_2 = T_1+1$.

Then $(c-a)T_1^2=(b+d)X_1^2=(c-a)T_2^2=(c-a)(T_1+1)^2=(c-a)(T_1^2 + 2T_1+1)$.

This gives $0 = 2(c-a)(T_1+1)$. So either $c=a$ or $T_1=-1$.

 

[Ibix]

it has to hold in the case $T=0$ and $X=1$.
<snip>
It also has to hold when $X=0$ and $T=1$.

@NoahsArk - just to stress that @stevendaryl could have picked any numbers here. There's nothing special about the zero and the one - they're just easy to work with.

 

[NoahsArk]

It makes sense now thank you! The major flaw in my example was that I was assuming X and T are equal!

 

[bhobba]

The simplest such transformation is a linear transformation, which is the form chosen in the derivation you cited.

Since inertial frames are homogeneous in space and time it must be linear. If the origins coincide at t=0 then if we consider small values of Δx and Δt and using matrix notation f(x+Δx, t+Δt) = A f(x, t) where A is a matrix.

But because of homogeneity regardless of x and t A is the same By breaking the x and t into the sum of a large number of small Δ's you get f(x,t) = A (x,t).

I post it a lot but IMHO the following is the most illuminating derivation of the Lorentz Transformations:
http://physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Thanks
Bill

 

[Dale]

Since inertial frames are homogeneous in space and time it must be linear.

Are you sure about that? I thought that being affine was sufficient, but I admit that I do not have a reference for that.

 

[bhobba]

Are you sure about that? I thought that being affine was sufficient, but I admit that I do not have a reference for that.

Yes affine is sufficient - ie mapping straight lines to straight lines. I am just influenced by Landau who uses symmetry in his definition of inertial frames and my demonstration follows directly from that. I have't read Landau's Classical Theory of Fields for a while so am not sure how he does it.

Added later - just to elaborate Landau is one of the few authors to define inertial frames using symmetry. Most simply say its a frame where Newtons first law holds - this naturally leads to using affiine transformations. Its just a different, but equivalent way of doing it. I personally am not a fan of that definition of an inertial frame - but that is another story that requires it's own thread. Besides - you got to love Landau

Thanks
Bill

 

[Dale]

Landau is one of the few authors to define inertial frames using symmetry

That is interesting. What is his symmetry-based definition?

 

[bhobba]

That is interesting. What is his symmetry-based definition?

An inertial frame is one that is homogeneous in space and time and isotropic in space. It's in the first couple of pages of Landau Mechanics. Its very useful because it imposes restrictions on Lagrangian's which via Noether implies conservation laws.

It resolves an issue with Newton's First Law - namely a particle continues to move at constant velocity in a straight line unless acted on by a force. That's usually the definition of an inertial frame ie one that obeys Newtons First Law. So far so good - but then you have Newtons Second Law - F=ma from which the first law follows. Together they are on the surface vacuous (not really - but that is another story - I had a long forum discussion with John Baez about it - before I thought it vacuous) - nonetheless when you think about it its not straightforward.

You can make it a lot clearer by means of Landau's definition. Its easy to show any two inertial frames must move at constant velocity relative to each other - but what of the converse - is a frame moving at constant velocity to an inertial frame also inertial. The real content of the first law is the answer is yes. To return to the motion of particles, formally you also need the Principle Of least Action and its easy (well to someone like Landau it's easy - to a mere mortal like me you just sit there and wonder how he figured it out) to show in an inertial frame where the Lagrangian has the same symmetry properties the frame its Lagrangian is L= (mv^2)/ 2. That's how you define a free particle and of course it moves at constant velocity in a straight line. If it doesn't then it's not free and we say its acted on by a force defined as the generalized force derived from the Lagrangian.

This is all done in Landau - Mechanics and one reason why I am so enamored by it - for me it leaves Goldstein far behind. Not all agree though:
https://physics.stackexchange.com/questions/23098/deriving-the-lagrangian-for-a-free-particle
'I think the best advice one can give you is don't read Landau & Lifshitz. They are great books for reference, but practically impossible to learn from. If you're into analytical mechanics then Goldstein is a good place to start, or Arnold, if you're more interested in the mathematical aspects.'

For me Landau is beauty incarnate - it converted me from math to physics and I think every physicist, like the Feynman Lectures, should get a copy early on and read it regularly - it grows on you more and more. What is it MIT says (about Feynman)- devour it.

Interestingly once you start to understand Landau and the path integral approach to QM you realize, strangely, the real basis of classical mechanics is QM.

Thanks
Bill

 

[vanhees71]

Well, for me both Landau and Goldstein are even topped by Sommerfeld concerning the didactics. I've still not found a better textbook (series) on classical physics at all. The only severe shortcomings are that it (a) misses to mention Noether's theorems and (b) uses the $\mathrm{i} c t$ convention of the Minkowski pseudometric :-(.

 

[strangerep]

An inertial frame is one that is homogeneous in space and time and isotropic in space.
[...]
To return to the motion of particles, formally you also need the Principle Of least Action and its easy [...] to show in an inertial frame where the Lagrangian has the same symmetry properties the frame its Lagrangian is L= (mv^2)/ 2. [...]

Well,... leading up to his eq(4.1), Landau really only shows that a Lagrangian which is independent of spatiotemporal position, and independent of direction, must be at most linear in $v^2$, and not involve $x$ or $t$ explicitly. Inserting the factor of $m$ is fine -- to give it dimensions of energy. But his factor of $1/2$ is a fudge (imho).

 

[stevendaryl]

Well,... leading up to his eq(4.1), Landau really only shows that a Lagrangian which is independent of spatiotemporal position, and independent of direction, must be at most linear in $v^2$

Is there an intuitive summary of this conclusion? It seems strange to me that you couldn't have higher powers of $v$.