# I Applying Conservation Law To Euler-Lagrange Equation

#### space-time

In my most recent thread, I discussed the conservation law involving the 4-velocity vector:

gab(dxa/dτ)(dxb/dτ) = -c2

Now, I've read that you can apply this law to the Euler-Lagrange equation in order to get some equations that are apparently equivalent to the geodesic equations.

Now here is the Euler-Lagrange equation:

(∂L/∂xa) - (d/dτ)(∂L/∂(∂xa/∂τ)) = 0

Now, I have read that to find the Lagrangian L, you do this:

L = gab(dxa/dτ)(dxb/dτ)

This however, is the exact same conservation law as mentioned in the beginning.

This would mean that L = -c2 , which is just a constant.

If you plug L = -c2 into the Euler-Lagrange equation, then you literally just get 0 = 0 (which is not useful at all).

How then, is the conservation law mentioned in the beginning supposed to be used with the Euler-Lagrange equation in order to solve the geodesic equations?

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#### Pencilvester

In my most recent thread, I discussed the conservation law involving the 4-velocity vector:

gab(dxa/dτ)(dxb/dτ) = -c2

Now, I've read that you can apply this law to the Euler-Lagrange equation in order to get some equations that are apparently equivalent to the geodesic equations.

Now here is the Euler-Lagrange equation:

(∂L/∂xa) - (d/dτ)(∂L/∂(∂xa/∂τ)) = 0

Now, I have read that to find the Lagrangian L, you do this:

L = gab(dxa/dτ)(dxb/dτ)

This however, is the exact same conservation law as mentioned in the beginning.

This would mean that L = -c2 , which is just a constant.

If you plug L = -c2 into the Euler-Lagrange equation, then you literally just get 0 = 0 (which is not useful at all).

How then, is the conservation law mentioned in the beginning supposed to be used with the Euler-Lagrange equation in order to solve the geodesic equations?
Saying $0=0$ in that specific way is indeed not useful, but that doesn’t mean that anything that we know to equal 0 is not useful. For example, you mentioned two expressions L could be equal to, and only plugged the trivial one in to the Euler-Lagrange equation. Did you try plugging the other expression in?

#### Orodruin

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You cannot just plug that into the EL equations in that fashion. If you do you are substituting something that holds for an on-shell solution into the action and then varying it (which implies also considering off-shell solutions).

You must first write down the on-shell conditions, ie the EL equations. Once you have those, you can start substituting conservation laws (which typically are first integrals of the EL equations) into the EL equations themselves.

In your particular case, the first integral is the Bianch identity. When solving the set of differential equations that you obtain from the EL equations, the Bianchi identity follows. You can then replace one of the differential equations by the Bianchi identity. You cannot insert the Bianchi identity before you have derived the set of differential equations because without them you have no way of showing that it holds (on-shell).

#### haushofer

I believe d'Inverno discusses this particular issue in chapter 6 or so.

#### vanhees71

Gold Member
In my most recent thread, I discussed the conservation law involving the 4-velocity vector:

gab(dxa/dτ)(dxb/dτ) = -c2

Now, I've read that you can apply this law to the Euler-Lagrange equation in order to get some equations that are apparently equivalent to the geodesic equations.

Now here is the Euler-Lagrange equation:

(∂L/∂xa) - (d/dτ)(∂L/∂(∂xa/∂τ)) = 0

Now, I have read that to find the Lagrangian L, you do this:

L = gab(dxa/dτ)(dxb/dτ)

This however, is the exact same conservation law as mentioned in the beginning.

This would mean that L = -c2 , which is just a constant.

If you plug L = -c2 into the Euler-Lagrange equation, then you literally just get 0 = 0 (which is not useful at all).

How then, is the conservation law mentioned in the beginning supposed to be used with the Euler-Lagrange equation in order to solve the geodesic equations?
This is a bit a tricky business. You can use different Lagrangians for the free particle. In common use are the following three:

(a) (1+3) formalism (not manifestly covariant): You just use the coordinate time of a coordinate system, and the free Lagrangian reads
$$L=-m \left (\frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} t} \right)^{1/2} = -m \sqrt{c^2-(\mathrm{d}_t \vec{x})^2}.$$

Now you realize that this is a homogeneous function of degree 1 in the three-velocities $\mathrm{d}_t \vec{x}$. Thus the action is parametrization invariant (and by the way in contradistinction to the Lagrangian also Lorentz invariant!). So you can use

(b) manifestly covariant parametrization-invariant formalism

You introduce an arbitrary parameter $\lambda$, which you can consider as a Lorentz scalar. Then the Lagrangian is
$$L=-m \left (\frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} t} \right)^{1/2}.$$
This leads to somewhat inconvenient equations, but it's clear that this inconvenience gets less severe, when for the equations of motion you choose an affine parameter. For massive particles you can choose the particle's proper time, which is just proportional to the Minkowkian arclength of the worldline.

Of course you cannot use the proper time as parameter to begin with. This would give you a constant free-particle Lagrangian.

(c) You simply use the most convenient Lagrangian ever. For an arbitrary parameter $\lambda$ you simply assume
$$L=-\frac{m}{2} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \lambda}.$$
Then you realize that the "Hamiltonian" (it's an extended version of the Hamiltonian, not necessarily the usual one used, e.g., for canonical quantization, which uses the (1+3)-form (a) and is not manifestly covariant)
$$H=p_{\mu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda}-L=L,$$
where
$$p_{\mu}=\frac{\partial L}{\partial (\mathrm{d}_{\lambda} x^{\mu})},$$
is conserved, because $L$ doesn't depend explicitly on $\lambda$!

This means in this Lagrangian $\lambda$, though treated as a free parameter, is automatically an affine parameter, and for massive particles you can choose the value $L=-m c^2$, i.e., choosing $\lambda$ as the proper time $\tau$ when solving the equations of motion. Another advantage of this form of the free Lagrangian is that it can also be used to describe the case of massless particles (which is sometimes a nice heuristics to derive ray-optical approximations by assuming a naive photon picture, though it's conceptually utterly flawed).

Interactions with an external field can be easily included by writing down additional interaction parts for the Lagrangian, which are again homogeneous functions of rank 1 in $\mathrm{d}x^{\mu}/\mathrm{d} \lambda$. Since these do not contribute to the Hamiltonian, still $L=\text{const}$ along the solution of the Euler-Lagrange equations and thus $\lambda$ is still automatically an affine parameter. The most simple example is the interaction Lagrangian for a charged particle in an electromagnetic field, described by its four-potential $A^{\mu}$,
$$L_{\text{int}}=-\frac{q}{c} \frac{\mathrm{d}x^{\mu}}{\mathrm{d} \lambda} A_{\mu}(x),$$
for which the variation of the action is also gauge invariant, as it must be.

Note that (b) and (c) can be used without problems in GR, again with (c) the most convenient choice to derive free-particle motion (i.e., the motion if only gravitation is present) as well as the generalization of motion in other external fields (like the example with electromagnetic fields).

For more details, also for an alternative argument for (c), see

"Applying Conservation Law To Euler-Lagrange Equation"

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