12

Yes, you can take a random point as your origin, but how do you know which random point you have picked? Imagine you are floating alone in empty space with nothing for a thousand km in any direction. If you pick a point somewhere around you at random to be your origin, how do you know where it is? The only way you can do it is to specify it in relation to ...


8

Let's say you've got an $\mathbb R$-vector space $V$ and some function $f:V\rightarrow \mathbb R$ which obeys $$f(a \vec v + b \vec w)=af(\vec v) + b f(\vec w)$$ for all vectors $\vec v,\vec w$ and scalars $a,b$ - that is, it's linear. Congratulations! You've constructed a tensor. The space of such object has a special name - the algebraic dual space of $...


7

Start with coordinate free: $$ t' =\gamma\Big( t-\frac{\vec v\cdot\vec r}{c^2} \Big) $$ $$ \vec r'=\vec r +(\gamma-1)(\vec r\cdot\hat v)\hat v-\gamma ct \vec v$$ and use polar coordinates to evaluate the dot products.


7

Yes, you can take a random point in space as the origin of your coordinate system and a random orientation for an associated frame of reference, and then measure distances, velocities, etc. relative to that reference frame - but the point is that this is a random choice of reference frame. Whereas we expect physically meaningful quantities, such as the ...


5

If we're talking about doing physics in $\mathbb{R}^3$, then tensors are the geometric objects that respect the symmetries of $\mathbb{R}^3$, in particular: rotations. How a tensor behaves under rotations depends on its rank, $n$, and for any rank, there are what are called the natural-form rank-$n$ tensors. A natural-form tensor is both symmetric and ...


5

I think you're being somewhat misled by your terminology and by the ambiguity of the standard, compact notation. Note that some texts use only the symbol "$\mathrm{d}/\mathrm{d}\tau$", or only "$\mathrm{D}_\tau$" (or similar symbols), for both operations you mention. Because its meaning is determined by what it is applied to. A general ...


5

Simultaneity is a human invention. No concept of simultaneity appears anywhere in the laws of physics (except simultaneity at a point, i.e. coincidence of events in spacetime). People find it useful to synchronize clocks for the same reason they find any sort of standard useful, but just like nature doesn't care whether you use liters or firkins, it doesn't ...


4

Despite the fact that spacetime cannot generally be endowed with the structure of an affine space - meaning that we cannot interpret tangent vectors as pointing from one event to another - we can preserve this idea if the events in question are infinitesimally close together. That is, we can interpret $\mathrm d \mathbf x$ as the vector pointing from one ...


4

The Lorentz transformation takes the coordinates $(x,t)$ which label an event in frame $S$ and maps them to new coordinates $(x',t')$, which label the same event in frame $S'$. The inverse transformation takes $(x',t')$ and spits out $(x,t)$. As your instructor has said, a good guess for the forward transformation for the position coordinate is to let $x'=\...


4

We can pick any point in space as an origin. However, physics doesn't care what choice we make (any choice of origin gives the same physics), since it turns out the long-range forces in physics (gravity, E&M) depend only on the relative distance between objects (in fact, since the forces depend on $r^{-2}$, not even the orientation matters). Since any ...


3

Your understanding is basically correct. To answer your question: "Why can't we just take any random point in the absolute space as origin?", we really can. The more technical issue with that is that space itself is not an immutable set of dimensions. Space itself is not static. It is expanding, and general & special relativity introduce some ...


3

inside the event horizon, the Schwarzschild coordinate 𝑟, while becoming "the time coordinate (...) also retains its geometrical significance" This is correct. But, … the r coordinate keeps measuring some sort of "distance to the origin". This is not. The geometrical significance of the Schearzschild $r$ coordinate never was as “some ...


3

$r$ is just a time coordinate inside the event horizon. $t,θ,\phi$ are the spatial coordinates. I assume what Rindler means by "retains its geometrical significance" is that the 3D surface at fixed $r$ is geometrically a cylinder of infinite length, whose cross sections are spheres of radius $r$, both inside and outside the horizon. But outside the ...


3

You can use whatever units you'd like in your coordinates. However, the infinitesimal spacetime interval $$\mathrm ds^2 = g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu$$ has dimensions of length squared. As a result, whatever choice of units or dimensions you make in your coordinates are reflected in the components of the metric. For example, if you choose ...


2

You interpretations of conditions 1,3 and 4 are spot on. However, in concrete terms condition 2 means that the length contraction formula for a ruler in S' being viewed from S only holds true if the ruler is stationary in S' (ie the positions of both its ends are pined down at the same time t').


2

$\pm C >0 $ iff the signature is $(\pm,\mp,\mp,\mp)$, respectively. $\pm C = (\frac{d\tau}{d\lambda})^2$, where $\tau$ is proper time. Apart from the signature, $C$ is arbitrary. It is not necessarily constant, and not necessary the same for different timelike curves, even if it is a geodesic.


2

Additional comment regarding the use of the Lorentz transformation for this problem. SEE BELOW. UPDATE: (A note on notation, you use capital letters to refer to points in "space", which trace out worldlines. In my original answer, I referred to "P" as an event (akin to a point in the diagram). In your notation, I should have referred to ...


2

Just a few comments to add to the discussion. (Based on my answers in Meaning of simultaneity in special relativity and Space time diagrams : Length contraction ) By "SR", I mean the standard textbook formulation of Special Relativity with a Minkowski metric on $R^4$. The metric tensor provides a definition of orthogonality, which can be defined ...


2

Based upon what OP has said in the comments, let me say that the argument they give appears to be correct as far as I can see. I was initially worried that they ignored the functional dependence of the particle position in the transformation, but this seems not to be the case. Indeed, it appears that OP's concern is sourced from the observation that the ...


2

If the two observers are moving relative to each other, then in general events that seem simultaneous to one will not seem simultaneous to the other. That is because their respective time axes will be tilted relative to each other, which means that surfaces of constant time in one frame will be out of alignment with surfaces of constant time in the other. It ...


2

Two defns which turn out to be equivalent: A tensor of rank $p$ is that which acts on $p$ vectors to produce a scalar invariant. A tensor is that which behaves mathematically in the same way as an outer product of vectors or a sum of outer products of vectors. As you see, no need to mention contravariant, covariant, and no need to mention components or ...


2

Note the words "conformal transformation" can mean slightly different things in different places. But conformal transformations, in the sense that it's used in conformal field theories, are not just coordinate transformations. Instead they are a simultaneous coordinate and field transformation such that the metric is left invariant. This often ...


2

It is best to never let anyone get away with saying that all physical theories are generally covariant. Gravity is, but that's about it. If you want to consider passive co-ordinate transformations, you are just performing a substitution in the integral and \begin{align} S = \int d^dx \mathcal{L}[\phi(x)] = \int d^dx^\prime \left | \frac{dx^\prime}{dx} \right ...


2

Vector spaces don't have an origin, but they have a null vector, $\vec 0$, satisfying: $$ \vec a + \vec 0 = \vec a $$ for all $\vec a$ in the vector space. But $\vec 0$ is not a place. Affine spaces have an origin ${\bf O}$ so that an origin plus a vector defines any point: $${\bf P} = {\bf O} + \vec p $$ and difference between two points is a vector: $${\bf ...


1

It seems that be that you are automatically using a fixed set of coordinates, which is not how you should be thinking about the metric. Mathematically by construction, the metric, $g$, is a geometrical object, specifically a 2 rank tensor. As such we are free to choose which local coordinates we want to use to represent it. In the context of general ...


1

Assume we are working in special relativity with $D+1$ dimensional spacetime. Suppose we have an inertial coordinate system $(x^{\mu})$, so objects that don't interact with anything move in uniform motion according to these coordinates. If we can determine another coordinate system where the metric $g$ is known, and if we know how to transform between the ...


1

I think your questions aren't well-posed since you aren't giving definitions of your coordinates (or are making implicit assumptions). I will show how radar experiments (in the spirit of the Bondi k-calculus) will define a metric $ds^2=dt^2-dx^2$. (This is based on my answer https://physics.stackexchange.com/a/508251/148184 to Minkowski Metric Signature .) (...


1

The two positions of the origin of S that you have determined, namely -3.75ly and -2,4ly, differ because they relate to the position of the origin of S at two different times. The first figure, namely -3.75ly, is the position of the origin of S at time t=5yr. The second figure, namely -2.4ly, is the position it would be at t'=4yr. The inconsistency arises ...


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