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6

A metric structure $g$ and a symplectic structure $\omega$ are two very different structures, although sometimes they can co-exist in a compatible way. Unlike a symplectic structure, there are no Jacobi-like identity and no Darboux-like theorem for a metric structure. There exists a unique torsionfree metric connection $\nabla$ on a pseudo-Riemannian ...


1

So, in classical mechanics, we know nothing of this strange "spacetime" and its metric. We know only that systems are described by $n$ continuous generalized coordinates $q^i$ with a certain range, and we take the manifold $\mathcal{M}$ consisting of all possible different $\vec q$ as our starting point. Note that there is no metric, no form, nothing on ...


3

I'm pretty sure I know the answer to this question, even though this question provides very little context for what the various tensors are, it gives a couple expressions without including the important surrounding equation for context, and it includes an equation with a typo. First of all, as it stands, the first equation has unbalanced indices. I assume ...


1

To calculate explicitly the curvature and geodesic equations for the conical spacetime you need an explicit metric. The metric $ds^{2}=dr^{2}+r^{2}d\phi^{2}$ describe a conical spacetime in the range of definition of the coordinates $(r,\phi)\in (0,\infty)\times [0,2\pi-\alpha)$. You can notice that this metric describe a flat spacetime in the domain of ...


6

I think that it is only necessary to use the cyclic identity. Contracting both sides with the Levi-Civita, we should have $$0 = (R_{abcd} + R_{adbc} + R_{acdb}) \varepsilon^{abcd} \tag{1}.$$ Let $S = R_{abcd}\varepsilon^{abcd}$. Then $R_{adbc}\varepsilon^{abcd} = -R_{adbc}\varepsilon^{acbd} = R_{adbc}\varepsilon^{adbc} = S$ where the last step is renaming ...


9

I was always told that to find whether or not a field is conservative, see if the curl is zero. This is almost always true, but not always true. I have now been told that just because the curl is zero does not necessarily mean it is conservative. Correct! To illustrate what's going on, let's do an example. Conside the following vector field: ...


2

I) The (Lagrangian) canonical conjugate momentum $$\tag{1} p_i ~:=~\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}^i} $$ transforms as a one-form/co-vector $$\tag{2} p_i~=~ p^{\prime}_j \frac{\partial f^j(q,t)}{\partial q^i} $$ under (possibly time-dependent) position coordinate transformations $$\tag{3} q^i~\longrightarrow~ q^{\prime j}~=~f^j(q,t)$$ ...


2

The objects that we call vectors in Euclidean space are, in more general situations, actually two possible different types of objects. If you change coordinates ($q_i\to q_i'$), the components change with different rules, according to how the Jacobian $\Lambda_{ij} = \frac{\partial q'_i}{\partial q_j}$ (using your notation) appears. 1) Vectors (or tangent ...


1

It may be a mistake in Eq.4. Your function $T(r,z)$ does not depend on $\xi$ and has nothing in principle to do with triangles. Since you are only interested in values of $T$ for $(r,z)$ pairs that satisfy Eq.1, you could write $T$ as a function of $r$ and $\xi$, OR as function of $z$ and $\xi$. However, until you do, the partial derivative of $T$ with ...


1

The stationary action principle and the Euler-Lagrange (EL) equations are very broad and general constructions. The field variables in the variational principle could in principle map into some generic manifold $M$. On the other hand, Euler-Poincare (EP) equations appear in the special situation where the manifold is a Lie group $M=G$, and the action is ...


3

The main difference between Hamilton and Lagrange/Newton mechanics is, that it happens directly on the phase space, i.e. any point on your manifold already fully determines the state of your system. Intuitively, you realize this by specifying position and momentum coordinates. On a mathematical level, the world we see is some smooth manifold (a priori not ...


4

When we say that they are unit vectors, we mean that the proper length is equal to one. The proper lengths of the two vectors are $$\gamma_{ab} t^a t^b=1,\quad \gamma_{ab}n^a n^b=1$$ and should be equal to one, i.e. $1\to 1$, at all times. (In the Minkowski signature, one of these squared lengths is minus one, but that won't change anything about the text ...


6

Despite my comment, on second look your second equation, attributed to Lambourne, is always identically zero. This is because you multiply the symmetric tensor $$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}$$ against $R^{\mu}{}_{\nu\beta\gamma}$, and the riemann tensor is antisymmetric on those last two indices, and tracing a symmetric tensor ...


3

EDIT2: I've just seen Jerry Schirmer's answer, which confirms to me that Lambourne's equation is incorrect. I would mark his as the accepted answer. I'll leave the below for reference, though it is entirely false! Indeed, as pointed out by Peter in the comments, Lambourne uses the same convention as MTW for the Riemann tensor (see equation 3.35). Funnily, ...



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