# [Solutions]Ch 1 Special Relativity and Flat Spacetime

1-6

https://petraaxolotl.wordpress.com/chapter-1-special-relativity-and-flat-spacetime/

9. For a system of discrete point particles the energy-momentum tensor take the form

$T^{\mu \nu}=\Sigma_a \frac{p_{\mu}^{(a)} p_{\nu}^{(a)} }{p^{(a)}}\delta^{(3)}(x-x^{(a)})$,

where the index a labels the different particles. Show that, for a dense collection of particles with isotopically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor

$T^{\mu\nu}=(\rho +p)U^{\mu}U^{\nu} + p\eta^{\mu\nu}$.

Doing average over 4-volume,

${\bf T}^{\mu\nu}=\dfrac{1}{\Delta V_4}\int_{\Delta V_4}T^{\mu\nu}d V=\dfrac{1}{\sqrt{-g} d^3x^i dx^0}\int_{\Delta V_4}T^{\mu\nu} \sqrt{-g} d^3x^i dx^0$.

Then a) in ${\bf T}^{\mu\nu}$ only delta-functions depend on x, b) metric determinant g is a macroscopic quantity, is constant over selected volume and can also be taken away from the integral.

${\bf T}^{\mu\nu}=\dfrac{1}{d^3x^i dx^0}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}\int_{\Delta V_4} \delta^{(3)}({\bf x}-{\bf x}^{(a)}) d^3 x^i dx^0 = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}$

In the last expression the sum is taken over the particles which have worldlines passing through $\Delta V_4$.

Consider symmetry,

${\bf T}^{0 0} = \dfrac{1}{d^3x^i}\sum_a p_a^0 \equiv \rho$

${\bf T}^{i 0} = \dfrac{1}{d^3x^i}\sum_a p_a^i$. Because of isotropy, ${\bf T}^{i 0}= 0$.

${\bf T}^{i j} = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^i p_a^j}{p_a^0}$. The sum should produce a symmetric macroscopic 3-tensor of second order. But all 3-tensors are defined by 3 eigenvectors. Because no preferred direction exits and no preferred directions correspond to the case when the matrix has all eigenvalues equal, that is when the matrix is proportional to kronecker delta. The coefficient of proportionality is the pressure: ${\bf T}^{i j} \equiv P \delta^{ij}$.

Replace ：

$\delta ^{ij}\to \eta^{ij}+U^i U^j$

$T^{00}\to \rho U^0U^0$,

then we get the result.

10. Using the tensor transformation law applied to $F_{\mu\nu}$, show how the electric and magnetic field 3-vectors E and B transform under

(a) a rotation about the y-axis;

(b) a boost along the z-axis.

(a) $\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} 1&0&0&0 \\ 0&cos\theta&0&sin\theta\\ 0&0&1&0\\ 0&-sin\theta&0&cos\theta \end{array}\right)$

(b)$\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} cosh\theta&0&0&-sinh\theta \\ 0&1&0&0\\ 0&0&1&0\\ -sinh\theta&0&0&cosh\theta \end{array}\right)$

$\overline{F}^{\mu\nu}=\Lambda F \Lambda^T=\left(\begin{array}{cccc} 0& - E1*cosh(x) - B2*sinh(x)& B1*sinh(x) - E2*cosh(x)& E3*sinh(x)^2 - E3*cosh(x)^2\\ E1*cosh(x) + B2*sinh(x)& 0& B3&- B2*cosh(x) - E1*sinh(x)\\ E2*cosh(x) - B1*sinh(x)& -B3& 0& B1*cosh(x) - E2*sinh(x)\\ E3*cosh(x)^2 - E3*sinh(x)^2& B2*cosh(x) + E1*sinh(x)&E2*sinh(x) - B1*cosh(x)& 0 \end{array}\right)$.

13. Consider adding to the Lagrangian for electromagnetism an additional term of the form $\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}$

(a) Express $\mathcal{L}'$ in terms of E and B.

(b) Show that including $\mathcal{L}'$  does not affect Maxwell’s equations. Can you think of a deep reason for this?

(a) $\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=\tilde{\epsilon}_{ijk}(F^i F^{jk}-F^{i0}F^{jk}+F^{ij}F^{0k}-F^{ij}F^{k0})=4F^{0i}F^{jk}\tilde{\epsilon}^{ijk}=-8\vec{E}\vec{B}$

(b) The expression in (a) can be expressed by 4-dimension divergence:

$\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=4\partial_i[\tilde{\epsilon}^{ijlm}A_k (\partial_l A_m)]$

Integrate $\mathcal{L}'$ over 4-dimension space, according to Stokes theorem, we can get the additional term in Lagrangian $L$. This term vanishes when varying action $S$