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We need the complex conjugated variable $\bar{u}$ because the Hamiltonian $H(u,\bar{u})$ is typically not holomorphic in $u$. The normalization of the CCR$^1$ $$\{u,\bar{u}\} ~=~-i \tag{1}$$ explains the $\sqrt{2}$-normalization in $$ u=\frac{q+ip}{\sqrt{2}}.\tag{2}$$ Recall that Hamilton's equations can be written in terms of the Poisson bracket $$\dot{u}...


5

If the metric has a zero determinant at some point or surface, then the metric is changing signature at that point, e.g., from $+---$ to $0---$. It's conceivable that spacetime would actually do this. However, it would violate the equivalence principle, and the standard mathematical machinery of GR breaks down in this situation, because everything is ...


3

Shouldnt they be equal, given that spacetime interval is invariant? I must have made an error in my reasoning but i dont see where, can someone help me? Yes, they should be equal. The reason that they are not equal is because you incorrectly performed the Lorentz transform. I will show you where you made the mistake, but I will also show how I go about such ...


3

The top diagram shows the velocities in the lab frame. Particle $A$ is moving east at speed $x$ and particle $B$ is moving west at speed $y$. I'm taking the east direction to be positive, so the velocity of $A$ is positive and the velocity of $B$ is negative. To find the velocity of $B$ relative to $A$ we have to transform to the frame where $A$ is ...


2

... in a euclidean space, the tangent space is the manifold. I think this is where your confusion lies. A flat Euclidean manifold $M$ still has a distinct tangent space $T_P$ at each point $P$, and a vector at $P$ lives in the tangent space $T_P$. Each tangent space happens to resemble $M$ iself but this is just coincidence. The vectors in a vector space on ...


2

A tensor doesn't have a magnitude, and it doesn't have a component along a particular direction. Those are properties of vectors. A tensor in three dimensions has 9 components, each of which corresponds to two directions. For example, in spherical coordinates, a tensor $T$ could have a component $T_{r\theta}$.


2

In the usual coordinate system where $p_t^2-(p_x^2+p_y^2+p_z^2)$ is Lorentz invariant, a boost in the $t$-$z$ plane is \begin{gather} p_t\to p_t\cosh\theta+p_z\sinh\theta \\ p_z\to p_z\cosh\theta+p_t\sinh\theta \end{gather} with $$ \cosh^2\theta-\sinh^2\theta = 1. $$ In light-cone coordinates with $p_\pm = p_t\pm p_z$, this implies \begin{align} p_+&\...


2

Given a manifold $M$ and a coordinate system $x^\mu$, the metric tensor $g_{\mu \nu}$ is given as the scalar product of the coordinate basis vectors $e_\mu = \partial_\mu$, that is $g_{\mu \nu} = e_\mu \cdot e_\nu$. Of course you can define any basis vectors or any basis one-forms. Usually that defines a tetrad, i.e. a local set of basis vectors or basis ...


1

Your confusion arises from mixing two notions of a derivative. In particular, a partial derivative and a total derivative along a curve. As you note, as such, there is obviously no codependence between two different coordinates. This means that if you move purely in the direction of one of the coordinates on a manifold, all the other coordinates remain ...


1

In this context, a tensor is an abstract operator on Euclidean space $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3$. In other words, a tensor takes in a vector $\mathbf{v} \in \mathbb{R}^3$ and maps it to another vector $\mathbf{v}' = T(\mathbf{v}) \in \mathbb{R}^3$. The components of a vector depend on which basis we are using, and the same applies to a tensor....


1

Suppose body $A$ is going to the right with a speed $x$ and another body $B$ is going to the left at a speed $y$. The motion of the bodies can be represented as vector diagram 1. To both motions add a velocity to the left of equal magnitude to that of the velocity of body $A$, ie stopping body $A$, as shown in vector diagram 2. On adding the two ...


1

With any extended object such as a physical pendulum, you are considering not the force on a single particle but on all of the particle (atoms, molecules, whatever) that make up the object: right off the bat, you're considering the positions of $10^\text{something}$ particles. But that's not the whole story, because these particles exert forces on each ...


1

Typically, when picking a tetrad basis, you can't choose a tetrad that is ALSO a coordinate basis. In the case of a non-diagonal metric like the Kerr solution in Boyer-Linquist coordinates, you'll end up with tetrad vectors that have more than one component. Obviously, there are many choices you can make, but ordinary linear algebra tells you how you can ...


1

The tetrad is not a coordinate system. It is a set of four (orthonormal) vectors defined at each point in spacetime. Since at each tangent space it works as a basis, you can define components of vectors $V^{(\mu)}$ with respect to it$^1$, but it makes no sense to talk of coordinates $x^{(\mu)}$. Whenever you need to convert a vector between "ordinary" ...


1

Mentions of cosmology in references do not at all limit the generality of the definitions and instead provide a reason to look for distance measures between astronomical sources suitable for “general spacetimes”: when the distances are becoming cosmological there are fewer assumptions that could be made on the structure of spacetime. Here is a source that ...


1

OP's first formula applies to the Lagrangian formalism in configuration space. In the context of canonical transformations (CT), it must be replaced with the corresponding formula in Hamiltonian phase space, i.e. the generating function $F(q,p,t)$ is also allowed to depend on momenta $p$. Since the new phase space variables $(Q,P)$ are supposed to depend on ...


1

What seems to be the common consensus in physics is that a gravitational field does not have a stress energy tensor due to the equivalence principle, but rather a psuedotensor. A more modern (at least since 1982 work of Penrose) perspective is that gravitational energy–momentum is a quasi-local quantity rather than simply local, i.e. associated to a closed ...


1

Is this psuedotensor frame dependent? Can I choose a frame in a gravitational field that doesn't experience gravitational stress energy? Yes and yes. In any local free-falling frame, the gravitatioal field is zero, so any gravitational stress-energy would have to vanish.


1

Ben Crowell's answer correctly covers the case in which by "space" you mean the full-dimensional spacetime. However, the induced metric on a submanifold of a perfectly well-behaved spacetime can be singular. For example, as discussed in this answer, the induced metric on a null hypersurface is always singular.


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