Spatial homogeneity condition for a free particle Lagrangian

[cianfa72]

TL;DR Summary

How the spatial homogeneity condition enters in the definition of Lagrangian for a free particle

Hi, reading "Mechanics" book by Landau-Lifshitz, they derive from spatial homogeneity that the Lagrangian $L$ of a free particle cannot explicitly depend on spatial coordinates $q$ in an inertial frame.

However my point is as follows: suppose to consider the Lagrangian $L= \frac 1 2 m{\dot x}^2 + x$ where $x$ is the one dimensional spatial coordinate in a given inertial frame.

Fixed $x(t_0)$ and $x(t_1)$, one can evaluate the action $S$ for different (timelike) paths starting and ending at those fixed events. Suppose that one gets a stationary action for a given path $\alpha(t)$.

Now if one translates the initial and final spatial coordinates by the same constant $c$, then $\alpha(t) + c$ gives a stationary action also for this new scenario (even though the stationary value of the action will be different).

Even though the above Lagrangian doesn't give the right solution/path for a free particle, however, does it imply that the spatial homogeneity condition is still fulfilled for it ?

 

[PeterDonis]

suppose to consider the Lagrangian $L= \frac 1 2 m{\dot x}^2 + x$ where $x$ is the one dimensional spatial coordinate in a given inertial frame.

This supposition is self-contradictory. An inertial frame is a frame in which a free particle has constant velocity; but it is easily shown from the Euler-Lagrange equation for your $L$ that in your frame, a free particle has a constant acceleration in the negative $x$ direction. So your frame is not an inertial frame.

 

[renormalize]

TL;DR Summary: How the spatial homogeneity condition enters in the definition of Lagrangian for a free particle

Hi, reading "Mechanics" book by Landau-Lifshitz, they derive from spatial homogeneity that the Lagrangian $L$ of a free particle cannot explicitly depend on spatial coordinates $q$ in an inertial frame.

However my point is as follows: suppose to consider the Lagrangian $L= \frac 1 2 m{\dot x}^2 + x$ where $x$ is the one dimensional spatial coordinate in a given inertial frame.

Can you clarify your question? Adding the potential $x$ to the free-particle Lagrangian makes it manifestly not spatially-homogeneous. Are you asking something about how this affects the value of the stationary action?

 

[cianfa72]

Can you clarify your question? Adding the potential $x$ to the free-particle Lagrangian makes it manifestly not spatially-homogeneous. Are you asking something about how this affects the value of the stationary action?

Perhaps I wasn't clear.

The Lagrangian is a function that describes a system in a given inertial frame (in this case a free particle). Given a Lagrangian, to me spatial homogeneity means/implies that if one translates the spatial coordinates of the start and end events by the same values (say $x_0,y_0,z_0$ in 3D) one should get the same solution path as the previous scenario (since such a path gives in both cases a stationary action).

My proposed Lagrangian (although it isn't the right Lagrangian for a free particle in an inertial frame since it doesn't get the right answer) nevertheless seems to me to fulfill such a condition.

 

[Orodruin]

in this case a free particle

No. Your particle is not a free particle, it is moving in a non-constant potential $-x$.

 

[JimWhoKnew]

My proposed Lagrangian (although it isn't the right Lagrangian for a free particle in an inertial frame since it doesn't get the right answer) nevertheless seems to me to fulfill such a condition.

Your Lagrangian describes vertical motion in a Newtonian uniform gravitational field. Experiments like throwing a ball up and re-catching it, behave similarly whether conducted at sea level or on top of a hill. In that sense, it is homogeneous.
In this view, however, the frame is not inertial. You can discuss instead the uniform electric field inside an infinite capacitor (ignoring acceleration radiation), and get the same sense of homogeneity.

 

[cianfa72]

Your Lagrangian describes vertical motion in a Newtonian uniform gravitational field. Experiments like throwing a ball up and re-catching it, behave similarly whether conducted at sea level or on top of a hill. In that sense, it is homogeneous.
In this view, however, the frame is not inertial.

Sorry, the frame is inertial. However the particle isn't free (it is subject to the uniform vertical gravitational field).

You can discuss instead the uniform electric field inside an infinite capacitor (ignoring acceleration radiation), and get the same sense of homogeneity.

Ok, so is that the right sense of spatial homogeneity or are there others ?

 

[JimWhoKnew]

Sorry, the frame is inertial. However the particle isn't free (it is subject to the uniform vertical gravitational field).

In the "gravitational view" the particle is falling freely and the frame is not inertial^*
(either accelerated or in a uniform gravitational field). In the "capacitor view", frame is inertial and particle is not free.

^*What are your requirements for a frame to be inertial?

Ok, so that is the right sense of spatial homogeneity ?

I don't know what's "right" and "wrong" in this context. I can say that the "physics" in both views posses the kind of translational invariance that you've argued in OP.

 

[cianfa72]

In the "gravitational view" the particle is falling freely and the frame is not inertial^*
(either accelerated or in a uniform gravitational field). In the "capacitor view", frame is inertial and particle is not free.

^*What are your requirements for a frame to be inertial?

Yes, I wasn't discussing the general relativity point of view. In this context uniform gravitational field and uniform electric field inside a ideal capacitor are actually the same.

I can say that the "physics" in both views posses the kind of translational invariance that you've argued in OP.

Exactly. If translation invariance is the same as spatial homogeneity, then I don't grasp the requirement from Landau-Lifshitz to have a Lagrangian independent from spatial coordinates.

 

[PeterDonis]

The Lagrangian is a function that describes a system in a given inertial frame

But, as I've already pointed out, the frame you used is not an inertial frame.

the frame is inertial.

No, it isn't. See my post #2.

 

[JimWhoKnew]

Exactly. If translation invariance is the same as spatial homogeneity, then I don't grasp the requirement from Landau-Lifshitz to have a Lagrangian independent from spatial coordinates.

What section in L&L ?

An irresponsible guess: the translational invariance applies to motion and forces, but if you consider potentials, they change from point to point (the force at $x+C$ is the same as at $x$, the potential isn't).

 

[cianfa72]

But, as I've already pointed out, the frame you used is not an inertial frame.

Sorry, why not ? As @Orodruin pointed out the Lagrangian $L= \frac 1 2 m{\dot x}^2 + x$ describes a particle in an inertial frame in which there is a non-constant potential $-x$ (i.e. in an uniform field).

 

[PeterDonis]

Sorry, why not ? As @Orodruin pointed out the Lagrangian $L= \frac 1 2 m{\dot x}^2 + x$ describes a particle in an inertial frame in which there is a non-constant potential $-x$.

By that criterion any frame is an inertial frame, which makes the term vacuous.

The more pertinent question would be what L&L defined "inertial frame" to mean.

 

[PeterDonis]

The more pertinent question would be what L&L defined "inertial frame" to mean.

Looking at p. 5, it looks like L&L define an "inertial frame" to be a frame "in which space is homogeneous and isotropic and time is homogeneous". By this definition, it seems to me that the frame in which the OP Lagrangian is written is not an inertial frame, because, while it is spatially homogeneous (the acceleration due to the non-constant potential is the same everywhere), it is not spatially isotropic: the acceleration has a definite direction.

In the next paragraph, L&L infer restrictions on the form of the Lagrangian of a particle that is "moving freely"; but they do not define what "moving freely" means independently of those restrictions. @Orodruin is saying that a particle subject to the non-constant potential in the OP Lagrangian is not "moving freely", which by implication would mean that the frame that Lagrangian is written in is an inertial frame--but, as above, I don't see how that is consistent with L&L's definition of "inertial frame". That definition does not say space is isotropic in an inertial frame except for the effects of a non-constant potential in the Lagrangian.

From the standpoint of General Relativity, of course, the concept of "inertial frame" in Newtonian mechanics is fundamentally inconsistent anyway, because gravity is treated as a force but it cannot be felt. GR uses a different concept of inertial frame which has a direct physical meaning: bodies with zero proper acceleration (which is a direct observable) have worldlines which are straight lines. The OP has said they don't want to discuss the GR viewpoint here, but without it I'm not sure it's possible to get a satisfactory answer to the underlying issue the OP's question raises.

 

[cianfa72]

By that criterion any frame is an inertial frame, which makes the term vacuous.

The point is not about the criterion for the definition of inertial frame.

The more pertinent question would be what L&L defined "inertial frame" to mean.

L&L define an inertial frame a frame of reference in which the space is homogeneous and isotropic and time is homogeneous.

My question is: is really required that the spatial homogeneity imply a Lagrangian for a free particle independent from spatial coordinates in an inertial frame ?

 

[PeterDonis]

The point is not about the criterion for the definition of inertial frame.

Yes, it is. Your OP question, which you repeat again in this post of yours, explicitly asks about an inertial frame.

L&L define an inertial frame a frame of reference in which the space is homogeneous and isotropic and time is homogeneous.

Yes, as I've already said. There is no need to repeat it.

My question is: is really required that the spatial homogeneity imply a Lagrangian for a free particle independent from spatial coordinates in an inertial frame ?

Which requires us to know what the L&L definition of an inertial frame is.

I don't think you are thinking clearly about the question you are asking.

 

[PeterDonis]

My question is: is really required that the spatial homogeneity imply a Lagrangian for a free particle independent from spatial coordinates in an inertial frame ?

Based on L&L's presentation, I would say that L&L define "free particle" to mean that the Lagrangian only depends on the square of the velocity in an inertial frame. They call this an "inference", but, as I noted before, they do not define what "free particle" means independently of the condition I just gave. So the only way L&L give us to know whether we have a "free particle" is to look at the Lagrangian in an inertial frame and see what it depends on. (In the next section they use Galilean invariance to show that in fact the free particle Lagrangian in an inertial frame must be $\frac{1}{2} m v^2$. Which of course means that any other Lagrangian, including the one in your OP, cannot possibly be the Lagrangian of a free particle in an inertial frame.)

 

[cianfa72]

Based on L&L's presentation, I would say that L&L define "free particle" to mean that the Lagrangian only depends on the square of the velocity in an inertial frame. They call this an "inference", but, as I noted before, they do not define what "free particle" means independently of the condition I just gave. So the only way L&L give us to know whether we have a "free particle" is to look at the Lagrangian in an inertial frame and see what it depends on.

Indeed, they claim at p. 5 that a free particle is one subject to no external action. However this is a sort of "circular thinking" since it lacks of a definition of "no external action".

Coming back to the OP, I don't see the necessity for the Lagrangian of a free particle to be independent from the spatial coordinates in an inertial frame.

 

[PeterDonis]

I don't see the necessity for the Lagrangian of a free particle to be independent from the spatial coordinates in an inertial frame.

You're missing the point: they are basically defining a "free particle" to be one whose Lagrangian depends only on $v^2$ in an inertial frame. And they are basically defining "inertial frame" to be one in which the Lagrangian of a free particle depends only on $v^2$. In other words, they are giving you an interlocked set of definitions that can only be taken together as a whole; it is not really possible to split them apart and ask why just one of them is the way it is.

 

[PeterDonis]

Coming back to the OP

Your argument in the OP is not valid, as has already been stated.

 

[cianfa72]

Maybe I'm wrong, however at p. 5 they seem to derive the form of the Lagrangian of a free particle in an inertial frame from space homogeneity and isotropy and time homogeneity in a such frame.

 

[PeterDonis]

at p. 5 they seem to derive the form of the Lagrangian of a free particle in an inertial frame from space homogeneity and isotropy and time homogeneity in a such frame.

Go back and read my previous posts in the thread again. I have already addressed this.

 

[PeterDonis]

space homogeneity and isotropy

These terms are also not defined except in terms of the properties they imply for an inertial frame and the Lagrangian of a free particle. So they are also part of the interlocking set of definitions I referred to.

 

[cianfa72]

By that criterion any frame is an inertial frame, which makes the term vacuous.

Sorry, reading again it, I believe I missed your point: which is the criterion you were talking about ?

 

[PeterDonis]

which is the criterion you were talking about ?

@Orodruin's post implied that we can call any frame an inertial frame, by simply saying that if the Lagrangian for a particle in that frame is anything other than $\frac{1}{2} m v^2$, the particle is not a free particle. That makes the term "inertial frame" vacuous.

 

[cianfa72]

by simply saying that if the Lagrangian for a particle in that frame is anything other than $\frac{1}{2} m v^2$, the particle is not a free particle. That makes the term "inertial frame" vacuous.

By the above is anything other than $\frac{1}{2} m v^2$, you actually mean if the particle is described by a Lagrangian other than $\frac{1}{2} m v^2$ in that frame, the particle is not a free particle.

 

[PeterDonis]

By the above is anything other than $\frac{1}{2} m v^2$, you actually mean if the particle is described by a Lagrangian other than $\frac{1}{2} m v^2$ in that frame, the particle is not a free particle.

I said "if the Lagrangian for a particle in that frame is anything..." What is unclear about that?

 

[cianfa72]

Ok, got it (sometimes I'm in trouble with english). Thanks

 

[Orodruin]

I mean, you cannot tell from just the form of a Lagrangian if the coordinate corresonds to the coordinates of an inertial frame (although it may be possible to rule it out depending on the kinetic term). Like, the Lagrangian discussed here can be either that of a free particle in a non-inertial frame or that of a particle moving under an external force in an inertial frame. The Lagrangians for the two are equivalent.

 

[PeterDonis]

the Lagrangian discussed here can be either that of a free particle in a non-inertial frame or that of a particle moving under an external force in an inertial frame. The Lagrangians for the two are equivalent.

In neither of these cases, however, would it be valid for addressing the question the OP is asking about L&L's claim, since that claim is specifically about the Lagrangian of a free particle in an inertial frame.

 

[cianfa72]

See also this thread from PSE. The first answer, basically addresses my question.

 

[PeterDonis]

See also this thread from PSE. The first answer, basically addresses my question.

I'm not sure it does, since that answer implies that the proposed "free particle Lagrangian" in the original question is in fact a "free particle Lagrangian". But it isn't, for the same reasons that the one in your OP isn't--at least, not if we assume it is written in an inertial frame, which is a key qualifier in L&L's presentation that is not mentioned at all in the PSE thread.

 

[cianfa72]

I'm not sure it does, since that answer implies that the proposed "free particle Lagrangian" in the original question is in fact a "free particle Lagrangian". But it isn't, for the same reasons that the one in your OP isn't--at least, not if we assume it is written in an inertial frame, which is a key qualifier in L&L's presentation that is not mentioned at all in the PSE thread.

Sorry to insist on this point: the proposed Lagrangian there, namely $$L = \frac{1}{2}mv^2 + 4x^3v_x$$ gives rise to the same stationary action solution as $L = \frac{1}{2}mv^2$ in the same frame (i.e. given two fixed start and end events the path between them with stationary action is the same). However the former depends on spatial coordinate $x$.

From L&L p. 5, a free particle Lagrangian written in an inertial frame should not depend on $x$ coordinate based on spatial homogeneity requirement that applies in an inertial frame.

The above Lagrangian fulfills the "translation invariant" condition, however according to L&L it is not a "valid" free particle Lagrangian written in an inertial frame since it does depend on spatial coordinate $x$.

Sorry to repeate this, but this is the point unclear to me.

 

[PeterDonis]

the proposed Lagrangian there, namely $$L = \frac{1}{2}mv^2 + 4x^3v_x$$ gives rise to the same stationary action solution as $L = \frac{1}{2}mv^2$ in the same frame

What are you basing this on? The Euler-Lagrange equations are not the same.

 

[PeterDonis]

The above Lagrangian fulfills the "translation invariant" condition

How? $x^3$ is not translation invariant.