Unraveling Landau's Mechanics: Why is Space Isotropic?

[Shahi]

Hi
I am reading Landau's mechanics
So in the first chapter page 5
It reads : since space is isotropic, the lagrangian must also be independent of the direction of v , and is therefore a function only of it's magnitude
... I can't understand why

, I think Landau's book has many fans in this website , so is there anyone clarify it a bit.

 

[PeroK]

Hi
I am reading Landau's mechanics
So in the first chapter page 5
It reads : since space is isotropic, the lagrangian must also be independent of the direction of v , and is therefore a function only of it's magnitude
... I can't understand why

, I think Landau's book has many fans in this website , so is there anyone clarify it a bit.

In the context of a free particle in space, the Lagrangian involves the kinetic energy of the particle, which is a function of the magnitude of the velocity. More generally, if the Lagrangian depends on $\vec v$, then you need to take the dot product of $\vec v$ with some other quantity - such as an EM vector potential. There's can't be such a quantity in isotropic free space.

 

[Shahi]

In the context of a free particle in space, the Lagrangian involves the kinetic energy of the particle, which is a function of the magnitude of the velocity. More generally, if the Lagrangian depends on $\vec v$, then you need to take the dot product of $\vec v$ with some other quantity - such as an EM vector potential. There's can't be such a quantity in isotropic free space.

Thanks for answering
I know what the Lagrangian is but I couldn't understand the way Landau reaches this conclusion...

 

[Antarres]

Even though vector notation is used in the text, it is in general implied that Lagrangian is not defined over a vector space, but over a space of coordinates(and velocity coordinates). So, it is a functional of real valued functions, not vector functions.

However, this means that velocity of a particle enters the Lagrangian through three components: $v_x$, $v_y$ and $v_z$, assuming we're in 3D space. These components are not invariant to coordinate transformations, though(while a velocity vector is, since the change of basis compensates the changes in coordinates). So vector quantities in Lagrangian enter through components, not as full vectors(component*basis).

Now let's say we change the angle from which we look at our system(we rotate ourselves). The velocity we measure this way would be the same velocity vector, but it's components would change.
If Lagrangian would depend on this velocity direction, that means it would effectively depend on the velocity components themselves, which means that the Lagrangian would not be invariant under this change of point of view when we rotate ourselves. This would mean that our laws of motion would look different from different angles, which would apparently contradict the isotropy of the space that we assumed.
So we conclude that Lagrangian should be dependent only on quantities which are invariant with respect to these transformations, that is, it should not depend on the direction of velocity, just the magnitude.

 

[formodular]

The point is very simple: velocity is described by a vector, a vector points in a certain direction - if the Lagrangian is to be a direction-independent function of velocity the Lagrangian can only depend on the velocity vector in a way that eliminate the notion of directionality associated to that vector (so for example the dot product of velocity with some random constant vector is eliminated from consideration as the Lagrangian function will take different values when the velocity vector points in different directions). The length/magnitude of a vector is a quantity associated to vectors that does not depend on the direction the vector points in.

 

[PeroK]

The point is very simple: velocity is described by a vector, a vector points in a certain direction - if the Lagrangian is to be a direction-independent function of velocity the Lagrangian can only depend on the velocity vector in a way that eliminate the notion of directionality associated to that vector (so for example the dot product of velocity with some random constant vector is eliminated from consideration as the Lagrangian function will take different values when the velocity vector points in different directions). The length/magnitude of a vector is a quantity associated to vectors that does not depend on the direction the vector points in.

You might want to take a look at the Lagrangian for a particle in an EM field, which includes a term in $\vec v \cdot \vec A$, where $\vec A$ is the EM vector potential.

 

[vanhees71]

Take a function $$L(\vec{v})$$. Since $\vec{v}$. An infinitesimal rotation varies $\vec{v}$ by $\delta \vec{v} = \delta \vec{w} \times \vec{v}$. Now $L$ is invariant under (infinitesimal rotations) iff
$$\delta L=\delta \vec{v} \cdot \partial_{\vec{v}} L=(\delta \vec{w} \times \vec{v}) \cdot \partial_{\vec{v}} L=0$$
for all $\delta \vec{w}$. This means that
$$\vec{v} \times \partial_{\vec{v}} L=0.$$
Then you have $\vec{v} \times \vec{p}=0$ for all $\vec{v}$. This means the for all $\vec{v}$ you must have $\vec{p}=M \vec{v}$.

In polar coordinates for $\vec{v}$ this implies that
$$\vec{p}=\partial_{\vec{v}} L=\vec{e}_v \partial_v L + \frac{\vec{e}_{\vartheta}}{v} \partial_{\vartheta} L + \frac{\vec{e}_{\varphi}}{v \sin \vartheta} \partial_{\varphi} L \stackrel{!}{=}M v \vec{e}_v.$$
This implies that
$$\partial_{\vartheta} L = \partial_{\varphi} L=0.$$
This means that $L$ is a function of $v=|\vec{v}|$ only.

You can drive this further by either assuming invariance under the entire Galilei or Poincare group. Then you get uniquely the Lagrangian for a single particle being either
$$L=\frac{m}{2} \vec{v}^2$$
for the case of the Galilei group or
$$L=-m c^2 \sqrt{1-\vec{v}^2/c^2}$$
for the Poincare group, i.e., the well-known free-single-particle Lagrangians for a non-relativistic or special-relativistic free particle!

 

[Shahi]

velocity is described by a vector, a vector points in a certain direction - if the Lagrangian is to be a direction-independent function of velocity the Lagrangian can only depend on the velocity vector in a way that eliminates the notion of directionality associated to that vector (

thanks for the answering, actually there are lagrangians that depend on direction

 

[weirdoguy]

actually there are lagrangians that depend on direction

Any example?

 

[PeroK]

Any example?

A charged particle in a magnetic field.

 

[weirdoguy]

Oh in that sense. I thought Shahi was saying that in some cases Lagrangian is a vector per se. But that wouldn't mean that it depends on the direction but that it has direction. Silly me