26

Indeed, any vector can be resolved in terms of two components (in $n$-dimensional space in terms of $n$ components). For this being possible the components should be linearly independent, i.e. in your case they should not be parallel. The advantage of using two orthogonal/perpendicular components is that their scalar product is zero, which simplifies the ...


7

Vadim's answer resolves the Q very nicely. I would just like to add that given a set of n vectors in a vector space you can always construct a set of n orthonormal vectors which can then be used as a basis. This is well known as the Gram–Schmidt orthogonalization process.


6

Suppose you are attempting to find a point on a map given a starting point. Which do you prefer: 4 km north and 3 km east of your current location 7 km north and 4.2 km southeast of your current location We can use whatever non-parallel vectors we want to describe an offset in two dimensions. Is it now clear why north and east are preferable to north and ...


6

OP's example is clever. However in physics, the constraint function $f$ is often implicitly assumed to obey various regularity conditions, e.g. differentiability, which OP's example fails. For details, see e.g. this related Phys.SE post.


5

No. A system with $n$ degrees of freedom is said to be under a holonomic constraint if imposition of the constraint reduces the number of degrees of freedom by $1$ , to $n-1$. Every holonomic constraint can be captured by some relation of the form $f(q_1...q_n,t)=0$, but not every constraint that can be captured in this form is necessarily holonomic. You ...


5

Gauge freedom is a freedom to redefine the fields of your theory in a way that doesn't change the physics. More specifically in a way that does not change the Lagrangian from which your fields were derived. It is not just any freedom that you have to choose something. Typically the gauge fields end up appearing as Lagrange multipliers in the Lagrangian ...


3

I don't know if it is the best way to look for such coordinate (probably not), but for general metric $$ds^2=g_{tt}dt^2+2g_{tr}dtdr+g_{rr}dr^2$$ on given timelike subspace you can look for coordinate transformations $t\rightarrow t'(t,r), r\rightarrow r'(t,r)$ that takes metric into the form $$ds^2=2g_{t'r'}dt'dr'+g_{r'r'}dr'^2.$$ The coordinate $t'$ will ...


2

Hawking and Ellis appear to confuse a map with a manifold. The spacetime manifold consists of coordinate systems which can be determined by measurement. It exists only in so far as it can be physically measured. It does not extend to singularities, and does not even extend beyond an event horizon. Given a boundary on a manifold (e.g. an event horizon) there ...


2

The clocks are synchronised in the S' frame but not the S frame. This is not a contradiction. You have to deconstruct 'synchronisation'. If two clocks, one near an observer and one far away, are synchronised this does not mean that the observer sees them indicating the same time, because of the time the image of the clock takes to reach them. If the far ...


2

Since there is already another answer, this is just meant to provide a different perspective on the matter. For clarity, it might help to think about Kaluza-Klein theory, which brings the two kinds of gauge symmetry into the same picture, so that we can appreciate the difference. We start with an arbitrary five-dimensional metric $G_{MN}$, then postulate ...


2

There are two main cases of separating into components: standard coordinates system, and separating into normal and parallel. In a standard coordinate system, there is a set of coordinates, and points are given in terms of those coordinates. Not all coordinate systems are vector spaces; for instance, the longitude latitude system isn't, since the Earths ...


2

You are doing it correctly. In reference to your comment, to find the minkowski metric in cylindrical coordinates, it is easiest to transform the line element so: $$ ds^2=g_{\mu\nu}dx^\alpha dx^\beta \tag{1}, $$ Where usually the minkowski metric is denoted by $\eta_{\mu\nu}$ Which is $$ ds^2 = -c^2dt^2+dx^2+dy^2+dz^2 \tag{2} $$ In Cartesian. If you ...


1

A more geometric approach is to consider the $2n+1$-dimensional contact manifold ${\cal M}$ with coordinates $(q^i,p_j,t)$. The Hamiltonian action functional is $$S_H[\gamma]~=~\int_I \gamma^{\ast} \Theta, \qquad \Theta~=~p_j \mathrm{d}q^j -H \mathrm{d}t, \tag{1}$$ where $\gamma:I\to {\cal M}$ is a curve. This action formulation (1) is world-line (WL) ...


1

In a canonical transformation, the new hamiltonian could have nothing to do with the initial hamiltonian, it just have to preserve Hamilton's equations. So in the new variables $(Q,P,t)$ you have to have that $$\dot{Q} = \frac{\partial K}{\partial P} \qquad \dot{P} = -\frac{\partial K}{\partial Q}$$ where $K$ is the new Hamiltonian. Whenever this happens ...


1

This is mainly because the X and Y axes are also perpendicular to each other and therefore to measure quantities along these axes the vectors are resolved into X and Y components.


1

As $p$ and $q$ do not depend functionally on one another $$ \frac{\partial q_i}{\partial p_j} = 0$$ and also $$ \frac{\partial p_i}{\partial q_j} = 0$$ for all $i,j$


1

I think there is an misunderstanding on your side. $$\frac{\partial q_j}{\partial q_i} = \delta_{ij},$$ where $\delta_{ij}$ is the so-called Kronecker delta. I hope this helps so far.


1

In general given a metric $g_{\mu\nu}$, an orthonormal basis $e^a$ is such that, $$\mathrm ds^2 = g_{\mu\nu}\mathrm dx^\mu \mathrm dx^\nu = e^a e^b \eta_{ab}$$ where $\eta = \mathrm{diag}(1,-1,-1,\dots,-1)$. So, if you write out the left-hand side, and the right-hand side, you get an equation you have to solve. More explicitly, $$g_{\mu\nu}\mathrm{dx}^\mu \...


1

First of all, the equation you write for $\Phi'(x')$ is not correct. The reason is that in principle you are allowed to use the same generator multiple times in the product since the order matters (incidentally, since $\Phi$ is a scalar primary, the action of $D$ and $K$ would vanish). An analogy is the Euler angles where you can parametrize a rotation as a ...


1

Yes, indeed you can use your knowledge of the scalar Helmholtz equation. The difficulty with the vectorial Helmholtz equation is that the basis vectors $\mathbf{e}_i$ also vary from point to point in any other coordinate system other than the cartesian one, so when you act $\nabla^2$ on $\mathbf{u}$ the basis vectors also get differentiated. This forces you ...


1

Conceptually it is rather straightforward, however, implies some computations. The circumference $C$ is: $$C=\oint ds = \oint a \sqrt{\sinh^2 u + \sin^2 v}\sqrt{1+\left(\frac{du}{dv}\right)^2} dv $$ Because of $\sinh^2 u + \cos^2 v =1$ via $\sinh^2 u = \sin^2 v$ we can already simplify: $$C =a \oint \sqrt{2\sin^2 v}\sqrt{1+\left(\frac{du}{dv}\right)^2} ...


1

I cannot comment because of my low reputation hence I am writing this as an answer I must say that in the latter transformation $$\Delta t'=\gamma \Delta t -\frac{V}{c^2}\gamma \Delta x$$ where $\Delta x=V \Delta t$ (The event describing the presence of observer in motion at time = $\Delta t$) this implies $$\Delta t'=\gamma \Delta t -\frac{V^2}{c^2}\...


1

There are various ways to see this: Consider the Hamiltonian formulation. Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Hamiltonian $H(q(t_0),p(t_0),t_0)$ is a function of both the instantaneous position $q^i(t_0)$ and the instantaneous momentum $p_k(t_0)$ at the instant $t_0$. Here $q^i(t_0)$ and $p_k(...


1

Consider the definition of length. We may define it as the difference of the spatial coordinates between two points with respect to a reference frame, when it's temporal coordinates are the same. Basically length is the distance between two points when they are at the same time coordinate. Thus the relationship between length with respect to two different ...


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