# B Trying to understand Lorentz Transformations

#### 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

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

#### bhobba

Mentor
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

Last edited:

#### Dale

Mentor
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

Mentor
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 gotta love Landau

Thanks
Bill

Last edited:

#### Dale

Mentor
Landau is one of the few authors to define inertial frames using symmetry
That is interesting. What is his symmetry-based definition?

#### bhobba

Mentor
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

Last edited:

#### vanhees71

Gold Member
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

Staff Emeritus
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

Mentor
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

Mentor
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

Staff Emeritus
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

Gold Member
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

strangerep said:
... 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

Gold Member
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.

"Trying to understand Lorentz Transformations"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving