63

That's because in thermodynamics we sometimes use the same letter to represent different functions. For example, one can write the volume of a system as $V=f_1(P,T)$ (a function of the pressure and the temperature) or as $V=f_2(P,S)$ (a function of the pressure and the entropy). The functions $f_1$ and $f_2$ are distinct in the mathematical sense, since they ...


42

Infinitesimal volume elements do not have to be cubes. Some familiar examples come from typical solids of revolution problems from calculus 1/2. Typically one discusses using either the "disk/washer" or "cylindrical shells" methods for finding the volume of the solid. As you can guess, the former method uses infinitesimally thin disks/...


28

It's simply flat space in Boyer-Lindquist coordinates. By writing $\begin{cases} x=\sqrt{r^2+a^2}\sin\theta\cos\phi\\ y=\sqrt{r^2+a^2}\sin\theta\sin\phi\\ z=r\cos\theta \end{cases}$ you'll get good ol' $\mathbb{M}^4$.


23

This can only be answered by pointing out the difference between special and general relativity. There is the historically motivated definition, which is still in widespread use in popular and semi-popular accounts, according to which special relativity only deals with inertial frames and coordinates (1a), while general relativity deals with accelerated ...


21

Your comments (and to a lesser extent, your Question) indicate a severe confusion about ever having an infinitesimal volume. You never construct an infinitesimal volume. Infinitesimal volumes appear at the end of a limiting process. Where do the infinitesmial rectangular parallelepipeds you are discussing appear? They appear in the limit of an iterated ...


18

You should really think about the variables we use as being like coordinates on some manifold, the configuration space (roughly the same as the phase space, I won't be careful about the distinction). In this language, changing variables is equivalent to changing coordinates on this manifold. The action is some scalar function on this space, and we can take ...


17

For a given spacetime the metric tensor may be written in both (globally) diagonal and non-diagonal forms depending on what coordinates we choose. For example for flat spacetime one diagonal form (not the only diagonal form) is the Minkowski metric: $$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$ But we could choose to write the metric using rotating cylindrical ...


15

This is presumably a flat spacetime described in funny coordinates. You can check this by calculating the Riemann tensor to see if it's zero. If I was going to do this, I would code it in the open-source computer algebra system Maxima, using the ctensor package.


14

Why do we have to revise the definitions of momentum and force in special relativity? I am going to downplay the theoretical reasons, because they aren’t what’s important. What’s important is that the Newtonian definition of momentum is useless because experimentally in relativistic particle collisions it isn’t conserved. When you change the definition to $\...


12

A more geometric approach to the Hamiltonian formulation 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. From this it ...


11

The key point here is of course that this is done locally about a point defined to be the origin of the new coordinate system: as the book you've linked says, your equation for $\text{d}z$ keeps us on the surface of the sphere if we are displaced by small amounts $\text{d}x$ and $\text{d}y$ from an arbitrary point on the sphere. The book then goes on to ...


10

Your presumption that partial derivatives can be uniquely specified by a single coordinate is false. Consider the function $f : \mathbb{R}^2 \to \mathbb{R}$, $$ f : (x, y) \mapsto x + y \qquad \left(\frac{\partial f}{\partial x}\right)_y = 1 \, . $$ But I can replace the second coordinate by $z = x+y$. Now we have $$ f : (x, z) \mapsto z \qquad \left(\frac{\...


9

I hope you are familiar with Cartesian Coordinate System According to this system, we take any vector directed downward and leftward to be negative, and any vector directed upward or rightward to be positive. The signs represent the direction of the vector(*). Since gravity always acts downward, towards the center of the Earth, we take it to be negative. ...


9

In relativity we deal with spacetime which is a 4-dimensional structure called a (pseudo-Riemannian) manifold. It includes space and time together. Meaning that time is just another direction in spacetime, perpendicular to the three dimensions of space. In the manifold we have a metric which describes all of the geometric properties of spacetime. It ...


9

To answer my own question, $dx^a$ is indeed the dual basis vectors for one-forms, i.e. $dx^a \equiv \textbf{e}^a$. The wedge product for one-forms is defined as $$\textbf{e}^a\wedge \textbf{e}^b = \textbf{e}^a\otimes\textbf{e}^b-\textbf{e}^b\otimes\textbf{e}^a.$$ Using this on Zee's definition, we get $$ \begin{align} \frac{1}{2!}t_{ab}dx^adx^b\equiv\frac{1}{...


8

Different coordinate systems have different kinds of volume elements; The volume elements are a consequence of how the grid lines of the coordinate system are set. Volume element can be generated by nudging the parameters which describe points in the space by infinitesimal amounts and figuring out the volume of the region generated as a consequence. This is ...


8

The concept of particles also makes sense for non-inertial observers and for observers in curved spacetime, as long as we remember that real observers are local and that the particle concept is only approximate. (In this answer, "local/localized" doesn't mean localized at a point. It only means localized in some small neighborhood.) The ...


8

Two facts are sufficient to make Minkowski tensor globally diagonal: Symmetry of the metric tensor $g_{ij} = g_{ji}$, the simple reason for this can be found in the answer to this question Why is the metric tensor symmetric?. For every symmetric matrix one can switch to the basis of eigenvectors, where it is diagonal. Therefore, it can be diagonalized at ...


8

Flat spacetime has, as you say, a zero Riemann tensor and the Riemann tensor will be zero in all coordinate systems. However the metric will look different in different coordinate systems. For example the Minkowski metric is: $$ ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2 $$ while the Rindler metric is: $$ ds^2 = -\left(1 + \frac{a}{c^2}x \right)^2 c^2 dt^2 + dx^2 ...


7

You need more information. Knowing how the components of the metric tensor transform for a specific chart transition does not uniquely specify how the coordinates themselves transform. Specific to your example, which axis $(x,y,$ or $z)$ did you choose for your polar axis? Where did you choose $\phi=0$? Did you choose $\phi$ to run conventionally (in the ...


7

If you assume flat spacetime and cartesian coordinates, then there's no problem with defining an affine structure on spacetime, and indeed this is often done by various authors who call $x^\mu$ the coordinates of a spacetime displacement vector. The situation is a bit more complicated in curvilinear coordinates, because the same vector will have different ...


7

The physics does not depend on the choice of coordinates: a straight line is a straight line. Of course the actual description of a straight line (i.e. the actual equation to be satisfied by the points on a straight line) is not the same in different coordinate system (see for instance this webpage). The form of the equation of motion will also change, but ...


7

ANSWER - Parts I & II $\texttt{C O N T E N T S}$ $\texttt{Abstract}$ $\boldsymbol\S\texttt{ A. Proper homogeneous Lorentz transformations : }\Lambda$ $\boldsymbol\S\texttt{ B. Lorentz boosts : }\mathrm L\left(\mathbf{v}\right)$ $\boldsymbol\S\texttt{ C. The decomposition : }\Lambda = \rm L\left(\mathbf{v}\right)\mathcal R $ $\boldsymbol\S\texttt{ D. The ...


6

If you're dealing with a classical Lagrangian as you have mentioned, then the numerical value is definitely not conserved! For example, let's work with just the kinetic term. If a particle of mass $m$ is moving relative to me at some velocity $\dot{x}$, then it has Lagrangian $$ L_\text{me}= \frac{1}{2} m\dot{x}^2. $$ Now, if I move into the frame of the ...


6

In general relativity, the matter source is the stress-energy tensor $T_{ab}$, a quantity that is lorentz-covariant in special relativity, so under a change of reference frame, $T$ changes. It is set explicitly equal to another covariant term made up of spacial curvature, $R_{ab} - \frac{1}{2}R g_{ab}$, so any shifts between energy and momentum will be ...


6

In the provided link, the author on page 36 states that the metric is locally Euclidean around the chosen point $A$ (the north pole). Following the author's orientations on figure 2.2, the coordinates of the north pole are $$A = (0,0,a)$$ Locally around this point then means that the $x$ and $y$ coordinates do not go far away from $A$, that is, far away from ...


6

Firstly, I'd like to mention that for the equation $$t'=\alpha t-\beta x$$ to be dimensionally consistent $\alpha$ needs to be a dimensionless constant, but $\beta$ needs dimensions of inverse velocity (time / length). The transformation ought to depend on $v$, but we can't simply put $v$ in the denominator because that's undefined when $v=0$. One possible ...


6

The "principle of general covariance" as it is usually stated is a pretty vacuous statement. Basically any theory can be made generally covariant. The biggest hurdle is that special relativity uses an affine space (Minkowski spacetime is basically a special pseudo-Euclidean affine space) to model spacetime, and affine spaces admit position vectors ...


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