B Trying to understand Lorentz Transformations

[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.