# Why the free Lagrangian does not dependent on the velocity vector direction, only its speed?

For freely moving particle, It's said $$L$$ can't depend on the velocity vector, but only its magnitude.

Question: I'm looking for the contra-argument. Let's say $$L$$ depends on velocity vector. Then, how would Lagrangian be written in terms of velocity vector and why would it yield the wrong solution for freely moving particle? (no need for bringing potential in this).

Looking for the math proof. I know Euler-lagrange. If it depends on velocity vector, $$L = \frac{1}{2}m(v_i + v_j)$$ and this yields $$\frac{d}{dt}(\frac{1}{2}m) = 0$$. Would this be correct approach to prove what I'm asking? I'm trying to get the idea why Landau makes it depend on $$v^2$$. (I'm not asking why it doesn't depend $$v^4$$.) My main question is to rigorously show why it can't depend on vector, and if it did, what would it break for freely moving particle ? It mentions isotropy of space, but want to see in math proof how it's wrong.

• Hi Chemistry. I removed your 2nd question. Commented Aug 20, 2023 at 22:18
• @Qmechanic yes, saw it. You prefer me to ask as different question or ? Commented Aug 20, 2023 at 22:19
• Maybe wait until after you get an answer to the first question? Commented Aug 20, 2023 at 22:20
• @Qmechanic they were mostly connected, but ok. Commented Aug 20, 2023 at 22:21
• Lagrangian of free particle can depend on velocity vector components, see my answer here: physics.stackexchange.com/a/776969/31895 Commented Aug 20, 2023 at 23:28

The simplest thing to do is to look and see what happens if the Lagrangian depends linearly on the velocity. It turns out that this gives perfectly reasonable looking equations of motion; they just are not the equations of motion for a free particle. Suppose that we have a Lagrangian for a point particle that depends both quadratically and linearly on the velocity in three-dimensions, $$L=\frac{1}{2}m\dot{\vec{r}}\cdot\dot{\vec{r}}+\vec{A}(\vec{r},t)\cdot\dot{\vec{r}}-V(\vec{r}).$$ This is not the most general Lagrangian with at most two time derivates, but it is general enough to illustrate the dynamics of a $$\vec{v}$$-linear term. The coefficient of the $$v^{2}$$ term determines the mass, and we know how the presence of a potential would affect things, so we shall set $$V=0$$. However, we still have an arbitrary vector function $$\vec{A}$$ (which can depend on both position and time) that parameterizes the the general term that is linear in the velocity, $$L=\frac{1}{2}mv^{2}+\vec{A}\cdot\vec{v}=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2})+A_{x}\dot{x}+A_{y}\dot{y}+A_{z}\dot{z}.$$

From this we can, of course, calculate the Euler-Lagrange equations of motion. The key thing to notice is that the canonical momentum $$p_{j}=\frac{\partial L}{\partial \dot{r}_{j}}=mv_{j}+A_{j}$$ is not the same as the mechanical momentum $$\pi_{j}=mv_{j}$$. Consequently, the Euler-Lagrange equations acquire some interesting extra terms, $$0=\frac{d}{dt}\frac{\partial L}{\partial \dot{r}_{j}}-\frac{\partial L}{\partial r_{j}}=\frac{d}{dt}\left(mv_{j}+A_{j}\right)-\frac{\partial\vec{A}}{\partial r_{j}}\cdot\dot{\vec{r}}.$$ If we identify the force as the time derivative of the mechanical momentum and use the chain rule to calculate the time derivate of $$\vec{A}$$, this becomes $$F_{j}=m\dot{\pi}_{j}=\frac{\partial\vec{A}}{\partial r_{j}}\cdot\dot{\vec{r}}-\sum_{k=1}^{3}\frac{\partial A_{j}}{\partial r_{k}}\frac{dr_{k}}{dt}-\frac{\partial A_{j}}{\partial t}.$$ Writing the first two terms on the right-hand side a bit more symmetrically, this is $$m\dot{\pi}_{j}=\sum_{k=1}^{3}\left(\frac{\partial A_{k}}{\partial r_{j}}v_{k}-\frac{\partial A_{j}}{\partial r_{k}}v_{k}\right)-\frac{\partial A_{j}}{\partial t},$$ which can be recognized as the triple product expansion $$\sum_{k=1}^{3}\left(\frac{\partial A_{k}}{\partial r_{j}}v_{k}-\frac{\partial A_{j}}{\partial r_{k}}v_{k}\right)=\vec{v}\times(\vec{\nabla}\times\vec{A}).$$

So we have the final force law that is derived from the Lagrangian $$L$$ with an arbitrary linear dependence on the velocity vector, $$\vec{F}=\vec{v}\times(\vec{\nabla}\times\vec{A})-\frac{\partial A_{j}}{\partial t}-\vec{\nabla}V,$$ where I have reinserted the force due to a potential term $$V$$. This expression for the force is actually nothing more than the Lorentz force law, written in terms of the scalar and vector potentials, $$\vec{F}=q\left[\vec{E}+\left(\vec{v}\times\vec{B}\right)\right],$$ where the electric and magnetic fields are are $$\vec{E}=\frac{1}{q}\vec{\nabla}V-\frac{1}{q}\frac{\partial A_{j}}{\partial t}, \quad \vec{B}=\frac{1}{q}\left(\vec{\nabla}\times\vec{A}\right).$$ In other words, apart from an overall factor of the charge, $$V$$ and $$\vec{A}$$ are the standard scalar and vector potentials of electrodyamics.

The upshot of this analysis is that a term in the Lagrangian that depends linearly on $$\vec{v}$$ is not generally consistent with the dynamics of the particle being free. A constant $$\vec{A}$$ has no effect on the physical force law, but having an $$\vec{A}$$ term that depends on the position of the particle means that the particle will be subjected to magnetic forces. Similarly, if $$\vec{A}$$ depends explicitly on time, there will be electric forces.

• Thanks for the great analysis. Imagine now that in the thought process, Landau might have come up with linear dependence on velocity, but we don't have $A$ at all. If so, $L$ would be a vector. For example: $L= \frac{1}{2}m(\vec v) = \frac{1}{2}m(v_{x\vec i} + v_{y\vec j} + v_{z\vec z})$. If so, $L$ ends up being a vector. can't euler lagrange be used here ? I know if L is a vector, action would be a vector, but by varional calculus, wouldn't we still arrive at euler lagrange ? if we could, $\frac{\partial L}{ \partial v_x} = \frac{1}{2}m$ and $\frac{d}{dt}(\frac{1}{2}m = 0$ => 0=0. Incorrect? Commented Aug 21, 2023 at 7:13
• @Chemistry The Lagrangian is simply not a vector, any more than your weight is a vector; that's just not how it is defined in the formalism. If the Lagrangian had multiple components (like a vector), each component would produce a separate Euler-Lagrange equation, and they would not generally be consistent. The only want to form a quantity that is linear in $\vec{v}$ and a scalar is to take the dot product of $\vec{v}$ with another vector.
– Buzz
Commented Aug 21, 2023 at 19:57

For simplicity, consider time-independent Lagrangians. A (first-order) Lagrangian for a single body with position $$𝐫$$, and velocity $$𝐯$$, both as functions of time with $$𝐯 = d𝐫/dt$$, will be expressible as a function $$L(𝐫,𝐯)$$ of the position and velocity.

We assume it's differentiable (so we can carry out the usual calculations for the least action principle) with differential coefficients: $$𝐟 = \frac{∂L}{∂𝐫}, \hspace 1em 𝐩 = \frac{∂L}{∂𝐯}.$$ Then, when subject to a variation, we get the following expression $$ΔL = 𝐩·Δ𝐯 + 𝐟·Δ𝐫$$ with the variation in $$𝐯$$ subject to the condition: $$Δ𝐯 = Δ\left(\frac{d𝐫}{dt}\right) = \frac{d(Δ𝐫)}{dt}.$$ As usual, we integrate this by parts, applying this condition, to obtain: $$𝐩·Δ𝐯 = 𝐩·\frac{d(Δ𝐫)}{dt} = \frac{d(𝐩·Δ𝐫)}{dt} - \frac{d𝐩}{dt}·Δ𝐫,$$ resulting in: $$ΔL = \frac{d(𝐩·Δ𝐫)}{dt} + \left(𝐟 - \frac{d𝐩}{dt}\right)·Δ𝐫.$$

When applying this to the action integral $$S = \int L dt$$, the total differential of $$𝐩·Δ𝐫$$ integrates out to a boundary term. For the least action principle, the boundary values $$Δ𝐫$$ are zero, so one is left with just $$ΔS = \int \left(𝐟 - \frac{d𝐩}{dt}\right)·Δ𝐫 dt$$ and from this, you get the equation of motion $$𝐟 = \frac{d𝐩}{dt}.$$ by requiring that $$ΔS = 0$$ for otherwise-arbitrary $$Δ𝐫$$.

You'll recognize this as the force law, with $$𝐟$$ representing the force and $$𝐩$$ the momentum; and this is a generic framework for deriving that law. The role that the Lagrangian plays is to produce a set of relations for the dynamic quantities $$𝐟$$ and $$𝐩$$ in terms of the kinematic quantities $$𝐫$$ and $$𝐯$$. In general, both the force and momentum will be functions of position and velocity, but the force law will have the same form - as expressed above.

You may think of the relation established by the Lagrangian, between the dynamic quantities and the kinematic quantities as being the "constitutive law" for the dynamic quantities. So, different Lagrangians give you different constitutive laws, but the same equation for the dynamics - the force law.

If you require the Lagrangian to be independent of position, then it reduces to a function of the form $$L(𝐯)$$. Then you have $$𝐟 = 𝟎$$, and you obtain - as a result - the law of inertia: that the momentum be constant over time. The momentum, in this case, will be a function of the velocity. So there, the constitutive law has the form: $$𝐟 = 𝟎, \hspace 1em 𝐩 = \frac{∂L}{∂𝐯}(𝐯).$$ The force is zero and the momentum is some function of velocity.

If you require the Lagrangian to also be independent of direction, then its dependence on $$𝐯$$ must reduce to a dependence on the scalars that can be derived from $$𝐯$$. The most general scalar is a function of $$v^2$$; your $$v^4$$ is included in that as a case-in-point, expressible as $$\left(v^2\right)^2$$. More conveniently, write it as a function of $$I = ½ v^2$$. In that case, Lagrangian reduces to a function of the form $$L(I)$$, its differential coefficient may be defined as $$m = \frac{∂L}{∂I},$$ and from this, we find $$ΔL = m ΔI = m 𝐯·Δ𝐯.$$ This leads to the constitutive law: $$𝐟 = 𝟎, \hspace 1em 𝐩 = m𝐯,$$ and you may recognize the coefficient $$m$$ as the mass. For this Lagrangian, the mass is a function of $$I$$, i.e. of $$v^2$$ - it's speed-dependent.

For the non-relativistic physics of Newton, it is constant, so that $$L(I)$$ is a first-order polynomial in $$I$$: $$L(I) = L(0) + m I$$. The constant $$L(0)$$ has no significance in the action integral, but you can think of it as the Lagrangian for all the other components of the system that the body in question is a part of.

In relativity, it is speed-dependent and has the form $$m(I) = \frac{m_0}{\sqrt{1 - 2I/c^2}} = \frac{m_0}{\sqrt{1 - (v/c)^2}},$$ where $$c$$ is the vacuum speed of light, the body is required to have a speed $$v < c$$, and $$m_0 = m(0)$$ is its mass at zero speed: or its rest mass. One Lagrangian which generates this relation is: $$L = \frac{2m_0I}{1 + \sqrt{1 - 2I/c^2}} = \frac{m_0v^2}{1 + \sqrt{1 - (v/c)^2}}.$$ In the limit $$c → ∞$$ that reduces to the form $$L = mv^2/2$$ seen in Newtonian physics, with a constant mass $$m = m_0$$.

If you put the body in a system, where it has interaction with other parts of the system, then the above arguments on symmetry with respect to location and direction don't apply anymore, though they may apply to the system as a whole. An example of a Lagrangian that includes such interactions and is a second order polynomial in $$𝐯$$, but generic as a function of position $$𝐫$$ is: $$L(𝐫,𝐯) = L_0(𝐫) + 𝐋_1(𝐫)·𝐯 + L_2(v^2/2),$$ which combines a Lagrangian $$L_2(I)$$ for a free inertially-moving body with a velocity-dependent Lagrangian $$L_0(𝐫) + 𝐋_1(𝐫)·𝐯$$ for the interaction. Here, the corresponding momentum - deemed the "canonical momentum" $$𝐩 = \frac{∂L}{∂𝐯} = 𝐋_1 + m𝐯,$$ differs from the "kinetic momentum" $$m𝐯$$, with the inclusion of a positionally-dependent term that might be considered as a kind of "potential momentum": $$𝐋_1(𝐫)$$. Cases in point include the interaction with the electromagnetic field $$L_0(𝐫) = -e φ(𝐫), \hspace 1em 𝐋_1(𝐫) = e 𝐀(𝐫),$$ where $$φ$$ is the electric potential and $$𝐀$$ and $$e$$ the electric charge of the body; or more generally with its generalization: a gauge field $$L_0(𝐫) = -\sum_a{e_a φ^a(𝐫)}, \hspace 1em 𝐋_1(𝐫) = \sum_a{e_a 𝐀^a(𝐫)},$$ that may have two or more sets of components indexed by $$a$$; e.g. the weak force has 3, the electro-weak force, which electromagnetism has been subsumed within, has 4.

As a footnote: in his treatments of electromagnetic theory in the 1860's and 1870's, Maxwell called $$𝐀$$ the "electromagnetic momentum", so he did actually think of it as something connected to momentum; here: a potential momentum per unit charge.

This answer continues the line of arguments of my Phys.SE answer here, where it was argued that the Lagrangian $$L(\vec{v})$$ is a function of velocity $$\vec{v}$$ only.

To implement isotropy in space as a quasi-symmetry of the Lagrangian consider an infinitesimal rotation $$v^i\quad \longrightarrow\quad v^{\prime i}~=~v^i+ \epsilon^{ij}v^j, \tag{1}$$ where $$\epsilon^{ij}=-\epsilon^{ji}$$ is an infinitesimal antisymmetric matrix. (Technically it belongs to the Lie algebra $$so(3)$$ of the 3D rotation group $$SO(3)$$.)

So the infinitesimal change of the Lagrangian $$\Delta L~:=~L^\prime-L ~=~ \frac{\partial L}{\partial v^i}\epsilon^{ij}v^j\tag{2}$$ should be a total time derivative $$\frac{\mathrm{d}F}{\mathrm{d}t} \tag{3}$$ even off-shell. Since eq. (2) only depends on the velocity $$\vec{v}$$, it follows that $$F~=~\vec{a}\cdot\vec{q}\tag{4}$$ is a linear function of position $$\vec{q}$$. We now decompose the Lagrangian $$L(\vec{v})~=~L_1(\vec{v})+L_{\neq 1}(\vec{v}) \tag{5}$$ into a linear and a non-linear part. OP's example belongs to the linear part $$L_1(\vec{v})$$. The linear part $$L_1(\vec{v})$$ is a total time derivative, so we can w.l.o.g. assume that $$L_1(\vec{v})=0$$ is zero. Then the function $$F=0$$ also becomes zero.

So we can w.l.o.g. assume that rotations are implemented as strict symmetries (rather than just quasi-symmetries) of the Lagrangian, i.e. the Lagrangian $$L=L(v^2)$$ depends only on the speed.