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This is a question on Lagrangian formulation of mechanics and not Newton's formulation. So, we don't a priori take Newton's laws to be true.

This SE post has answers which brilliantly define mass explicitly using Newton's laws.

Now all the resources which I've come across, which "claim" to formulate Lagrangian mechanics (including Susskind himself) begin by postulating Lagrangian as $L(q, \dot q):={1\over 2}m\dot q^2-V(q)$, and just state $m$ to be the mass of the object.

That's either just brushing details under the rug or total ignorance! We can't just take $m$ to be granted from Newton's laws to formulate the supposedly "independent" (but equivalent to Newtonian mechanics) Lagrangian mechanics! Otherwise it'll be circular — we'd have used Newton's laws (in the form that there exists a quantity called mass for each object; see the second top-voted answer in the linked post, which takes it to be the second law) in formulating Lagrangian mechanics! We need to define mass independently here (and later need to show that this definition is equivalent to that in Newtonian formalism).

Question: So what is a mathematically precise (at least at precise as the answers in the linked post are!) definition of mass in Lagrangian mechanics?

I'll also greatly appreciate if you can show that Lagrangian and Newtonian definitions of mass are equivalent.

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  • $\begingroup$ I don’t understand why you think the Lagrangian formulation is or should be independent from Newtonian mechanics. Why would we not use Newton’s laws as at least a limiting case? $\endgroup$ – Dale Jun 25 '20 at 13:17
  • $\begingroup$ @Dale Landau formulates it independently. he even defines the mass independently from Newton's laws. But I'm unable to understand it, maybe its because of not a good translation to English? $\endgroup$ – Atom Jun 25 '20 at 13:21
  • $\begingroup$ Suggested title (v2): What is a mathematically precise definition of mass in Lagrangian mechanics? $\endgroup$ – Qmechanic Jun 25 '20 at 13:27
  • $\begingroup$ @Atom - if it is in the Landau and Lifschitz books, those are translated well. $\endgroup$ – Jon Custer Jun 25 '20 at 13:28
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    $\begingroup$ To reopen this post (v2) consider to align title and main body. They seem to ask different things. $\endgroup$ – Qmechanic Jun 25 '20 at 13:32
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Based on Landau-Lifshitz (which you should consult for more detail):

Seeking the lagrangian, you first assume that the state of the system is given completely by specifying positions and velocities of all the particles. Thus $L=L(\vec{v},\vec{r},t).$

Start by seeking free particle lagrangian and demand several symetries - homogeneity and isotropy of space and homogeneity of time. This puts lagrangian into a form $L=L(v^2)$. Then you demand yet another symmetry - symmetry under galilean transformations and you arrive at form of free particle lagrangian $L=kv^2,$ where $k$ is some constant.

Then you demand, that equations of motion of several independent free particles are supposed to be independent (because they are free). This means that lagrangian of several particle system will be just sum of one particle lagrangians, but you have freedom to pick constants $k$ for every particle independently: $$L=\sum_ik_iv_i^2$$

The last thing to do is to turn on interactions between particles. This you just assume can be done by adding some function which depends on the state of the particles: $$L=\sum_ik_iv_i^2-U(\vec{v}_i,\vec{r}_i)$$ Note this is very general demand, since you can include the kinetic part into $U$ and you have just some general function. However, the interactions are supposed to determine departure from free motion, thus keeping kinetic part out makes suggestive sense. There is, however Galilean principle of relativity. This demands, that physics should not depend on inertial frame, therefore the interactions should be infinitely propagated, as this is the only frame independent velocity in Newtonian mechanics. Therefore $U$ should not depend on the motion of particles and thus we have (after renaming of constants): $$L=L=\sum_im_i\frac{v_i^2}{2}-U(\vec{r}_i)$$

The meaning of $m_i$ is then found through the equations of motions.

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  • $\begingroup$ @ This was truly brilliant! Thanks tons! $\endgroup$ – Atom Jun 26 '20 at 19:50
  • $\begingroup$ Just one question though: Why can’t $U$ depend on relative velocities between the particles like $\vec v_i-\vec v_j$, which will be frame independent? $\endgroup$ – Atom Jun 27 '20 at 13:35
  • $\begingroup$ @Atom Interesting question. I don't understand this part very much (in L&L it was only mentioned), but note it is not velocities of particles which are supposed to be frame independent, but velocity of interaction itself - whatever that means (we are not in field theory so exact meaning eludes me). Note also, that your term would (probably?) introduce term into E-L equations which depends on acceleration of the particles, thus the force would depend on acceleration. In E-M field, it is acceleration of charge which makes interaction propagate, so perhaps this is analogical. $\endgroup$ – Umaxo Jul 13 '20 at 5:19
  • $\begingroup$ @Atom but it is really interesting topic. If I find enough time, I will look closer into this and if I'll find some answers, I'll get back to you:) $\endgroup$ – Umaxo Jul 13 '20 at 5:24
  • $\begingroup$ I really do hope that you find time! $\endgroup$ – Atom Jul 13 '20 at 6:10

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