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It's not quite correct to take $\Lambda \rightarrow \infty$, even at the end of the calculation. That comes from ancient mistaken notions that the field theory under consideration needs to describe physics upto arbitrarily small distances. The modern way to think about this is that you're making your theory agnostic of the value of $\Lambda$. A theory is ...


1

One reason for the box is the Fourier expansion of field in stable macroscopic condition (thermal radiation, cavity oscillations) works well only for finite volume. For infinite volume, the Fourier integral of such stationary field is problematic, because the field function is not L2 integrable.


1

Quantizing in a finite volume is not specific to the electromagnetic field, and it is not a necessity, neither for the electromagnetic field nor for any other. It is generally more well-behaved to quantize in a finite volume because no infrared-like divergences appear from allowing arbitrarily low momenta (since no arbitrarily long wavelengths fit into the ...


1

The answer: $$2\partial^\sigma A^\rho.$$ It's not the chain rule but the product rule you use. It is as if the tensor $$\partial_\mu A_\nu$$ were the variable you are differentiating with respect to. To understand the indices, consider a simpler example: $$\frac{\partial}{\partial x^i}(x^j x_j) = \frac{\partial}{\partial x^i}(x^j x^k g_{jk})= \delta^j_i x_j ...


0

The $p$ in the Fourier transform of the (free real scalar) field $$ \phi(x) = \int \left(a(\vec p)\exp(-\mathrm{i}px) + a^\dagger(\vec p)\exp(\mathrm{i}px)\right)\frac{\mathrm{d}^3p}{2p^0} $$ is a number, it is essentially a change in coordinates on $\mathbb{R}^n$ like every Fourier transform. The canonically conjugate momentum $\pi(x) := \frac{\partial ...


1

Static magnetic fields by themselves have no momentum, you need an electric field and a magnetic field to have momentum. Also, the momentum comes from the total fields. So even if you think of two magnetic fields $\vec B_1$ and $\vec B_2$ and two electric fields $\vec E_1$ and $\vec E_2$ the momentum density is $\epsilon_0\left(\vec E_1+\vec ...


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I'm not quite familiar with this subject. Nevertheless, I suggest you to read E. M. Lifthitz, and L. P. Pitaevskii, Statistical Physics, part 2, translated by J. B. Sykes, and M. J. Kearsley (Pergamon Press Ltd., Oxford, 1980) section 24.


3

Of course, conservation laws for particle number, momentum, and energy follow directly from the classical equations of motion. However, by fluid dynamics we mean more than that. We mean that the conservation laws can be expressed in terms of coarse grained, hydrodynamic variables, such as the density, the energy density, and the fluid velocity. For example, ...



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