# Deriving Maxwell's equations from the Lagrangian

• offscene
In summary, the conversation discusses how to take the partial derivative of the Lagrangian with respect to ##\partial(\partial_\mu \mathcal{A}_\nu)## using the relations ##\frac{\partial}{\partial(\partial_\mu \mathcal{A}_\nu)}(\partial_i \mathcal{A}_j) = \delta^i_\mu \delta^j_\nu## and the chain rule. After expanding and using the chain rule, the partial derivative of the first term in the Lagrangian is found to be ##-\partial^\mu \mathcal{A}^\nu## and the partial derivative of the second term is ##\partial_\rho \mathcal{A}
offscene
Homework Statement
Given ##\mathcal{L} = -\frac{1}{2}(\partial_\mu \mathcal{A}_\nu)(\partial^\mu \mathcal{A}^\nu)+\frac{1}{2}(\partial_\mu \mathcal{A}^\mu)^2##, compute ##\frac{\partial{\mathcal{L}}}{\partial(\partial_\mu \mathcal{A}_\nu)}##.
Relevant Equations
Euler-Lagrange equations of motion.
This isn't a homework problem (it's an example from David Tong's QFT notes where I didn't understand the steps he took), but I am confused as to how exactly to take the partial derivative of the Lagrangian with respect to ##\partial(\partial_\mu \mathcal{A}_\nu)##. (Note the answer is: ##-\partial^\mu \mathcal{A}^\nu+(\partial_\rho \mathcal{A}^\rho)\eta^{\mu \nu}##)

Last edited:
Write ##(\partial_\mu \mathcal{A}_\nu)(\partial^\mu \mathcal{A}^\nu) = (\partial_\lambda
\mathcal{A}_\rho)(\partial_\sigma \mathcal{A}_\tau) \eta^{\sigma \lambda} \eta^{\tau \rho}## and use the relations ##\frac{\partial}{\partial(\partial_\mu \mathcal{A}_\nu)}(\partial_i \mathcal{A}_j) = \delta^i_\mu \delta^j_\nu## to show that the partial of the first term of ##\mathcal{L}## with respect to ##\partial_\mu \mathcal{A}_\nu## is ##-\partial^\mu \mathcal{A}^\nu##.

Similarly write ##(\partial_\mu \mathcal{A}^\mu)^2 = (\partial_\rho A_\sigma \cdot \eta^{\rho \sigma})^2## and use the chain rule to get that the partial of the second term of ##\mathcal{L}## with respect to ##\partial_\mu \mathcal{A}_\nu## equals ##\partial_\rho \mathcal{A}^\rho \cdot \eta^{\mu\nu}##.

Greg Bernhardt, offscene and topsquark
Euge said:
Write ##(\partial_\mu \mathcal{A}_\nu)(\partial^\mu \mathcal{A}^\nu) = (\partial_\lambda
\mathcal{A}_\rho)(\partial_\sigma \mathcal{A}_\tau) \eta^{\sigma \lambda} \eta^{\tau \rho}## and use the relations ##\frac{\partial}{\partial(\partial_\mu \mathcal{A}_\nu)}(\partial_i \mathcal{A}_j) = \delta^i_\mu \delta^j_\nu## to show that the partial of the first term of ##\mathcal{L}## with respect to ##\partial_\mu \mathcal{A}_\nu## is ##-\partial^\mu \mathcal{A}^\nu##.

Similarly write ##(\partial_\mu \mathcal{A}^\mu)^2 = (\partial_\rho A_\sigma \cdot \eta^{\rho \sigma})^2## and use the chain rule to get that the partial of the second term of ##\mathcal{L}## with respect to ##\partial_\mu \mathcal{A}_\nu## equals ##\partial_\rho \mathcal{A}^\rho \cdot \eta^{\mu\nu}##.
Thank you so much for your help, I have a question after using the chain rule on the second term. After expanding as you suggested and using the chain rule, I get: =##(\partial_\rho \mathcal{A}_\sigma \eta^{\rho \sigma}) \cdot \eta^{\rho \sigma} \delta_\mu^\rho \delta_\nu^\sigma## but this means we must substitute ##\rho=\mu## and ##\sigma=\nu## everywhere to satisfy the delta, however this gives ##(\partial_\mu \mathcal{A}^\mu) \cdot \eta^{\mu \nu}## which isn't the same as the answer right? I'm struggling to find where I'm going wrong here.

Note that ##\eta^{\mu\nu}## are constants, so they have zero derivatives. The partial derivative of ##(\partial_\rho \mathcal{A}_\sigma \cdot \eta^{\rho\sigma})^2## with respect to ##\partial_\mu\mathcal{A}_\nu## is $$2(\partial_\rho \mathcal{A}_\sigma \cdot \eta^{\rho \sigma})\cdot \partial_\mu\mathcal{A}_\nu(\partial_i \mathcal{A}_j \cdot \eta^{i j}) = 2(\partial_\rho \mathcal{A}_\sigma \cdot \eta^{\rho \sigma})\delta_{\mu i} \delta_{\nu j} \eta^{ij}$$ Can you reduce the latter expression further?

offscene

## What is the Lagrangian for the electromagnetic field?

The Lagrangian for the electromagnetic field is given by $$\mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu}$$, where $$F_{\mu\nu}$$ is the electromagnetic field tensor. This tensor is defined as $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$, with $$A_\mu$$ being the four-potential of the electromagnetic field.

## How do you derive the Euler-Lagrange equations for the electromagnetic field?

To derive the Euler-Lagrange equations for the electromagnetic field, we start with the Lagrangian $$\mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu}$$. The Euler-Lagrange equation for the field $$A_\mu$$ is given by $$\partial_\nu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)} \right) - \frac{\partial \mathcal{L}}{\partial A_\mu} = 0$$. Since $$\mathcal{L}$$ does not explicitly depend on $$A_\mu$$, the second term drops out, leading to $$\partial_\nu F^{\nu\mu} = 0$$, which is one of Maxwell's equations in the absence of sources.

## What role does the field tensor $$F_{\mu\nu}$$ play in Maxwell's equations?

The field tensor $$F_{\mu\nu}$$ encapsulates the electric and magnetic fields in a covariant form. Its components are related to the electric field $$\mathbf{E}$$ and the magnetic field $$\mathbf{B}$$ by $$F_{0i} = E_i$$ and $$F_{ij} = -\epsilon_{ijk} B_k$$. Maxwell's equations can be derived from the properties of $$F_{\mu\nu}$$ and its dual tensor $$\tilde{F}^{\mu\nu}$$, leading to both the homogeneous and inhomogeneous Maxwell equations.

## How does the principle of least action apply to deriving Maxwell's equations?

The principle of least action states that the equations of motion for a system are obtained by finding the stationary points of the action $$S = \int \mathcal{L} \, d^4x$$. For the electromagnetic field, the action is $$S = \int -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} \, d^4x$$. By applying

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