15

You can use properties of the Levi-Civita tensor namely, $\epsilon_{kij}\epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}$ so that $\vec a \times (\vec b \times \vec c) = \hat e_i \epsilon_{ijk} a_j (\epsilon_{kmn} b_m c_n )$ $ = \hat e_i ( \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm} )a_j b_m c_n $ $ = \hat e_i \delta_{im}\delta_{jn} a_j ...


12

I always have trouble with this identity, so here's a fun way to derive it in three-dimensions. It may be argued that this method is a little convoluted, but I find it much easier to remember than the Levi-Civita contraction formula, and much less tedious than working out the components! Let's call the vector $\mathbf{a \times (b \times c) = d}$, and see ...


4

A more down-to-earth approach is to prove this identity in three-dimensional space by writing it down in terms of vector components: $\vec{a} = (a_x, a_y, a_z)$, etc., and using the expression for the vector product $$ \vec{a}\times\vec{b} = \left| \begin{matrix} \hat{e}_x & \hat{e}_y & \hat{e}_z \\ a_x & a_y & a_z \\ b_x & b_y & ...


4

You should use the definition of the curvature tensor $$[\nabla_\mu, \nabla_\nu]\xi_\kappa = R_{\mu\nu\kappa}^{\ \ \ \ \ \ \lambda}\xi_\lambda.$$ Then by differentiating the Killing condition we obtain $$\nabla_\kappa\nabla_\mu\xi_\nu + \nabla_\kappa\nabla_\nu\xi_\mu = 0.$$ One can then add a term to both sides to get a commutator and insert the above ...


3

I would like to see a general quick derivation that follows on exactly in the spirit of Weinberg. Recall this involves considering a particle in a frame $\mathcal{O}$ in which the particle appears to have no velocity and also in a frame $\mathcal{O}'$ in which it appears to have velocity $$ \mathbf{v} = (\frac{dx'^i}{dt'}),$$ and then using $$dt' = \Lambda^...


3

Wikipedia gives the commutator of covariant derivatives acting on a $(2,0)$ tensor: $$\tau^{ab}{}_{;cd}-\tau^{ab}{}_{;dc}=-R^a{}_{ecd}\tau^{eb} -R^b{}_{ecd}\tau^{ae}.$$ (Carroll has the generalization to an $(n,m)$ tensor in eq. 3.68.) When one sets $cd$ to $ab$, the two terms on the right side of this cancel. No locally-flat coordinates are necessary when ...


3

Use the contraction identities of the Levi-Civita symbol. I strongly encourage you to prove these identities yourself as well, I think you will find it worth the effort in the long run.


2

Starting from your last equation, it should be clear that $$g_{\mu\nu} v^\mu f^\nu = q \delta^\alpha_\mu F_{\alpha \beta} v^\beta v^\mu = q F_{\alpha \beta} v^\beta v^\alpha,$$ since only the terms where $\alpha = \mu$ will be non-zero because of the Kronecker delta. From here it's quite trivial, since $F_{\alpha \beta}$ is an antisymmetric tensor, and $v^\...


2

In index notation, the divergence of a vector is $\partial_iA_i$ and by analogy the divergence of a tensor with two indices means either $\partial_iA_{ij}$ or $\partial_jA_{ij}$. In the case of a symmetric tensor, these are the same thing. Note that taking the divergence of a tensor with two indices produces a vector, while taking the divergence of a vector ...


2

There's no sign discrepancy at all. Start from $\Lambda^0_{\:i}=\eta^{0\mu}\Lambda_{\mu i}=\eta^{0\mu}\eta_{i\nu}\Lambda_\mu^{\:\nu}$. Since $\eta$ doesn't mix space with time in Cartesian coordinates (a fact I'll use hereafter without comment), this simplifies to $\eta^{00}\eta_{ij}\Lambda_0^{\:j}$. The next calculation shows your professor is using $-+++$, ...


2

The issue with the velocity transformation is resolved if you use the matrix inverse of the Jacobian. In your case, note that the inverse transformation you are using involves terms like $$ \frac{\partial r}{\partial x}\Bigg\rvert_\theta = \sec \theta $$ which does not make much sense since $ \theta = \theta(x,y) $ is also a function of $x$ and $y$. The ...


2

You got it a little bit wrong, but the main ideas are here. Starting from $\epsilon_{\mu \nu\rho\sigma}\partial^\rho\partial^\mu A^\nu$, you commute $\partial^\rho$ with $\partial^\mu$ without changing anything. Then, you use anti symmetry of $\epsilon$ to exchange the two indices $\mu$ and $\rho$. At this point: $$\epsilon_{\mu \nu\rho\sigma}\partial^\rho\...


2

The inner expression reads: $$ g_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} = \sum_{\text{all}(i,j)} g_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} $$ So you'd have to sum and then square root as the sum and square root operations are not interchangeable


2

It is not so readily seen, to be honest. It goes in the literature by the name "polar decomposition". The shortest argument is the one by H. Urbantke "Elementary Proof of Moretti’s Polar Decomposition Theorem for Lorentz Transformations" (here), which is a not so strightforward simplification of prof. Valter Moretti's argument here. You ...


1

The components $\Lambda^i_{\,\,\,j}$ cannot be uniquely determined. The best way you can motivate the form of these components are given in bolbteppa's answer. Perhaps this is the best you can do. However, this can still feel like cheating, especially when you generalize the results from $(v,0,0)$ to the case of $\mathbf{v}$. So, this answer will complement ...


1

As Weinberg says there in that section (page 29), only $\Lambda^0_{\ 0} = 1$ and $\Lambda^{i}_{\ 0} = \gamma v_i$ are uniquely determined - the other $\Lambda^{\alpha}_{\ \beta}$ are not uniquely determined (the reason for this being that if $\Lambda^{\alpha}_{\ \beta}$ carries a particle from rest to velocity $\mathbf{v}$, then so does $\Lambda^{\alpha}_{\ \...


1

Yeah that’s all that there is to it essentially. $\delta$ is a kind of matrix — it is a tensor with two indices; so a square matrix. The whole point of this step is to put an expression with contracted indices into a matrix form, where matrices are multiplied. That’s why the reordering happens — the author wants to have the last index of the previous symbol ...


1

I don't assume this will be a complete answer, just several suggestions that I hope could be helpful. So you are using the $(-,+,+,+)$ metric, I'm using the opposite $(+,-,-,-)$, sorry. Consider that a generic Lorentz transformation is subjected to the following \begin{gather*} \Lambda^T \mathbb{G} \Lambda = \mathbb{G} \end{gather*} where $\Lambda=({\Lambda^\...


1

Note that $\partial_\mu\partial_\nu A^{\mu\nu}=\eta_{\alpha\mu}\eta_{\beta\nu}\partial^\alpha\partial^\beta A^{\mu\nu}=\partial^\mu\partial^\nu A_{\mu\nu}$, upon renaming indexes on the last step. By similar reasoning, $A^\mu_{\;\,\mu}=A_{\mu}^{\;\,\mu}$ Also, note that $\eta^{\mu\nu}\eta_{\mu\nu}=\delta^\mu_{\mu}=4$, so the last term should be $2\partial^\...


1

Note that we expect there to be $n^4$ components to start out with for an arbitrary $(4,0)$ tensor $T^{abcd}$ in $n$ dimensions. (and in general a generic $(m,0)$ tensor in $n$ dimensions should have $n^{m}$ components) (a) Start with the antisymmetric case where $A^{abcd} = - A^{bacd}$. Notice that for any $a=b$ we end up having $A^{aacd} =0$, which is sort ...


1

In Minkowski space $\epsilon^{0123}=-\epsilon_{0123}$ but people differ in which of these two expressions they take to be $+1$, so it's a good idea to always state your conventions. Similarly with $p_\mu$: It is always true that $p^\mu= (E, {\bf p})=m V^\mu =mdx^\mu/d\tau$, but $p_\mu=(-E,{\bf p}) \to -i\hbar \partial_\mu$ only in the $(-,+,+,+)$ ...


1

The $\sigma^i\sigma^j\partial_i\partial_j$ will lead to $$(\partial_x\sigma^x+\partial_y\sigma^y+\partial_z\sigma^z)(\partial_x\sigma^x+\partial_y\sigma^y+\partial_z\sigma^z)=\partial_x^2 1+\partial_y^21+\partial_z^21$$ The $\sigma^i\sigma^j$ terms with $i\neq j$ vanish as Pauli matrices anticommute and therefore these terms should cancel out. This result is ...


1

Here is another proof. The covariant derivatives satisfy the Jacobi identity $$ [\nabla_\mu,[\nabla_\nu,\nabla_\kappa]]+ [\nabla_\nu,[\nabla_\kappa,\nabla_\mu]]+[\nabla_\kappa,[\nabla_\mu,\nabla_\nu]]=0. $$ This can be verified directly, but it is also known that pretty much any associative algebra will satisfy the Jacobi identity, and the elements $\nabla_1,...


1

You can show that in two steps. Show that if a Lorentz tensor vanishes in one Lorentz frame, it vanishes in all Lorentz frames. This is quite simple so I won't do it. This means that you switch to any frame and calculate this identity since what you want to get is zero, which, if true, is valid in all reference frames. Thus, switch to the normal ...


1

This article describes string theory. In physics, string theory is a theoretical framework in which the point-like particles of particle physics are replaced by one-dimensional objects called strings. String theory describes how these strings propagate through space and interact with each other. On distance scales larger than the string scale, a string ...


1

This is just notation. We write $\vert 1\rangle\langle 0\vert \otimes \vert 1\rangle\langle 0\vert$ as $\vert 11\rangle\langle 00\vert$. It generalizes exactly as you would expect for multiple qubits e.g. $\vert 000\rangle = \vert 0\rangle\otimes\vert 0\rangle\otimes\vert 0\rangle$


Only top voted, non community-wiki answers of a minimum length are eligible