Trying to understand Lorentz Transformations

[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.

 

[Mister T]

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 $v$.

 

[bhobba]

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).

It is a fudge. A better justification is the most reasonable relativistic Lagrangian is C*dτ where τ is the invariant proper time and C a constant. C is defined as -m (units the speed of light 1) so you get -m*dτ = -m*√(1-v^2) dt. Expanding in a power series and keeping the first two terms (ie powers in v above 2 are negligible) you get -m + 1/2*mv^2 - since a constant changes nothing in a Lagrangian you get L= 1/2*mv^2 so you see where the 1/2 comes from. But there is still a fudge - why define mass as -C. No easy answer here I think.

Thanks
Bill

 

[bhobba]

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

its because speeds are much less than c as explained in my post where its derived from the relativistic Lagrangian.

Thanks
Bill

 

[stevendaryl]

its because speeds are much less than c as explained in my post where its derived from the relativistic Lagrangian.

Thanks
Bill

Oh, so it's just in the nonrelativistic limit, all higher powers drop out.

 

[SiennaTheGr8]

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).

It is a fudge. A better justification is the most reasonable relativistic Lagrangian is C*dτ where τ is the invariant proper time and C a constant. C is defined as -m (units the speed of light 1) so you get -m*dτ = -m*√(1-v^2) dt. Expanding in a power series and keeping the first two terms (ie powers in v above 2 are negligible) you get -m + 1/2*mv^2 - since a constant changes nothing in a Lagrangian you get L= 1/2*mv^2 so you see where the 1/2 comes from. But there is still a fudge - why define mass as -C. No easy answer here I think.

Thanks
Bill

I'm not an expert in analytic mechanics, but I thought these kinds of "fudges" (multiplying a Lagrangian by a constant, be it $1/2$ or $-1$) don't really need justification per se. Isn't the $-1$ just conventional so that the "stationary" action is a "least" action? And isn't the $1/2$ just to make the analytic approach line up with the traditional definition of kinetic energy? (Couldn't the traditional definition of low-limit kinetic energy itself be tweaked to $mv^2$ if some other quantities were likewise redefined?)

In other words, my impression was that this sort of thing is aesthetic in a sense, and that a suitable Lagrangian "works" regardless of how you scale it (i.e, the important thing is that it's proportional to proper time). Am I way off here?

 

[vanhees71]

Well, concerning the factor 1/2 in $L_0=m \dot{\vec{x}}^2/2$ for the non-relativistic and the expression $L_0=-m c^2 \sqrt{1-\dot{\vec{x}}^2/c^2}$ in the relativistic case, it can of course be derived via Noether's theorem from taking into account the full 10-dim. Galileo or Poincare group, respectively. Of course, in both cases $m$ is the invariant mass, which is the same quantity in both Newtonian and relativistic physics. It enters by the definition of the corresponding generator for boost transformations.

 

[strangerep]

... 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 v2

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

It's because Landau [p7] is working with the non-relativistic case only (and hence we're perhaps drifting off-topic in this thread which is about Lorentz transformations). But, briefly, since $L$ can only depend on $v^2$, an infinitesimal Galilean boost gives an infinitesimal change in $L$ which turns out to be a total derivative, hence doesn't affect the EoM.

 

[strangerep]

[...] I thought these kinds of "fudges" (multiplying a Lagrangian by a constant, be it $1/2$ or $-1$) don't really need justification per se. Isn't the $-1$ just conventional so that the "stationary" action is a "least" action?

Imho, it's more fruitful to understand that "extremal" of the action is more fundamental for physics. If an infinitesimal change to the Lagrangian made the new EoM totally different (non-isomorphic) from the original EoM, that's a recognizably different physics, with different experimental outcomes.

There is a related "robustness principle" (sometimes called "Lie-stability" in a group context) which notes that the theoretical constants we measure in experiments only have a finite experimental accuracy. A theory which predicts big changes in experimental behaviour due to a very small change in the Lagrangian is unlikely to be a good theory of physics.

And isn't the $1/2$ just to make the analytic approach line up with the traditional definition of kinetic energy? (Couldn't the traditional definition of low-limit kinetic energy itself be tweaked to $mv^2$ if some other quantities were likewise redefined?)

In this case, the $1/2$ is more important, and changing it would up-end large amounts of classical mechanics theory. E.g., from basic definitions of position, velocity and acceleration, one can derive easily that, for a constant acceleration $a$ applied to a body which has initial velocity $v_0$ until the body has covered a distance $x$, the following relationship applies: $$2ax ~=~ v_f^2 - v_0^2 ~,$$where $v_f$ is its velocity when it has covered the distance $x$. The factor of 2 on the LHS is the origin of the factor of $1/2$ in the definition of kinetic energy (after both sides have been multiplied by $m$ and the notion of "$\Delta$energy(work)= Force $\times$ distance'' is introduced in order to talk about conserved quantities.

IoW, sometimes a factor's origin is intrinsic to the math, whereas in other cases you can indeed insert a fudge factor arbitrarily for later convenience.

 

[SiennaTheGr8]

But isn't the $1/2$ nevertheless "optional" (I hesitate to say "arbitrary") in a sense? I understand that it comes from the work–energy principle, but you could in principle still tweak definitions of other quantities to make kinetic energy $mv^2$, right? Not that there would be any good reason to do so, of course! My point question, rather, concerns whether one really needs an (analytic) justification for slapping a fudge factor onto a Lagrangian. That is, is it not justification enough here that scaling by $1/2$ turns the Euler–Lagrange equation into the EoM with exactly the form we expect from the classical/Newtonian approach?

 

[strangerep]

[...] is it not justification enough here that scaling by $1/2$ turns the Euler–Lagrange equation into the EoM with exactly the form we expect from the classical/Newtonian approach?

If by "here", you mean "in Landau's treatment", I still raise the objection that he does not introduce it via the general method for choosing a Lagrangian, i.e., $L = T - V$ (kinetic energy - potential energy).

I daresay you could insert compensating numerical factors all over the place, including the potential energies, i.e., re-scale all energies, since multiplying a Lagrangian by a constant doesn't change the EoM. And if such rescaling doesn't affect any physically measurable results then, sure, choose a scale which is most convenient.

 

[vanhees71]

Of course, you can multiply the Lagrangian by a (non-zero) constant an still getting the same equations of motion, but what should this be good for?

The standard definitions of the physical quantities we are used to is indeed usually the most convenient choice. The one exception is the choice of SI units in electromagnetism which is quite aphysical, but again, it's also convenience for practitioners to deal with "nice numbers" (not too big and not too small) when dealing with voltages and currents in everyday electrics and electronics.