# Tag Info

36

You can always embed a (spacetime) manifold in a sufficiently high-dimensional space (if you have a $d$ dimensional manifold it can be embedded in a space of $2d$ dimensions). But that doesn't specify which space it is - it could be any sufficiently high dimensional space. So assuming it is embedded doesn't tell you anything at all. Hence it is simpler to ...

18

I would say the answer is just the scientific principle of parsimony: if an empirically inconsequential entity can be dropped from a theory, it is preferable to drop it. As you have pointed out, the embedding space is not needed mathematically, and thus it also does not impact the empirical claims of the theory. Hence, parsimony tells us to drop it. If you ...

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

Even if spacetime is embedded in something bigger, we don't have access to it or any way of making observations of it. This tells you that our theories should only be formulated using quantities that can be computed "intrinsically", without reference to any embedding. So Riemann curvature is in, but mean curvature or second fundamental form are out....

8

I have a lot of sympathy with this question, because I think it is true that physicists have been a little too strong on the idea that embedding is "wrong". Embedding is a well-defined mathematical idea that gives the same predictions for observable phenomena as does a treatment of the same manifold without embedding. So it is not ruled out by ...

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 ...

6

The net gravitational force on any mass inside a spherical shell is zero. This is a consequence of Newtons' shell theorem. As per this link it states "In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to ...

6

The event horizon' radius of black holes is an infinite-redshift surface (a one-way surface where particles can never escape to infinity). It can be computed analytically (or at least numerically) by finding the largest (real) positive root of the inverse of the component $g_{rr}$ of the metric tensor, i.e., by solving $$\frac{1}{g_{rr}}=0.$$ It is common to ...

5

There is no absolute time. Time always passes at the normal rate for you, your clock, and everything in your frame of reference no matter what your conditions. A distant observer considered to be in a different frame of motion, acceleration, and / or gravitational conditions will also perceive their own time perception and clock as running normally. Yet each ...

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 ...

4

While any $n$-dimensional manifold can be embedded in $R^m$ for some $m>n$ that doesn't mean the metric in the embedding space corresponds to the metric on the manifold. Of course, usually we do embeddings so the induced metric corresponds to the intended metric for the manifold but this is not always feasible. One wormhole manifold that is allowed by GR (...

4

First, imagine a stopwatch at rest that measures a time interval $\Delta t$. This concretely means there are two events: event 1 is that we start the stop watch at time $t_1$ and position $x_1$, and event 2 is that we stop the watch at time $t_2=t_1+\Delta t$ while the watch is still at position $x_2$. The spacetime interval in the stopwatch frame is then \...

4

Your teleportation device (which is an extreme form of superluminal travel) would indeed allow you to travel backward in time, assuming (a) special relativity is true and (b) you were able to use this device in any reference frame to teleport between any two points at the same time (with "at the same time" defined by the reference frame you are in)....

4

General relativity is not the only possible theory of gravity, it is just the simplest such theory - or at least the simplest such theory that works. For many decades people have suggested modifications to GR by tweaking it or adding extra features and there is a long list of such theories in the Wikipedia article Alternatives to general relativity. The ...

3

Q: Can the astronaut tell at what speed is he moving? and how do you define speed in that case when you have no outside references. Can you refer to the speed relative to the empty space? No. This is already the case in Newtonian mechanics. Speed is always relative to some reference frame. Q: Can you identify the direction of your movement? Let's say ship ...

3

There is only one correct resolution to the twin "paradox" regardless of your setup. There is no need for hand-waving or use a special set of coordinates (such as the Rindler coordinates OP mentions). No. The only real definition for the time elapsed on someone's person clock is the proper time of their world-line. I would go as far as to argue ...

3

Actually if you look through Tong's notes, the derivation he is presenting does not consider the metric to be the fundamental variable (it sort of doesn't make sense to talk about isometries of a dynamical metric), but rather the worldline of a particle moving on a fixed background metric. In particular, after Eq 4.32, Tong writes down the action \begin{...

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

This may not have the detail you require, but in Sean Carroll's book From Eternity to Here, he uses the following analogy: "In fact, there is a much more intuitive way of representing a wormhole. Just imagine ordinary three-dimensional space, and "cut out" two spherical regions of equal size. Then identify the surface of one sphere with the ...

3

That Alien has moved in on our street - or someone very similar! Bet the Alien has better manners though. Anyway, about the video: The physics facts seems fine and yes it would be a good way to talk about heights. For presentation it would be best to try and improve the sound quality, speak more slowly and clearly. Perhaps its the different accents, but for ...

3

To understand this you must first have a clear picture on the difference between intrinsic curvature and extrinsic curvature. Imagine an ant that lives on a sphere. To the ant the sphere looks flat because he is so tiny. But if he does certain experiments he might deduce that the sphere is actually curved. For example if he walks around a large enough ...

2

Are gravitational waves produced only when a mass accelerates? Acceleration of a mass is a necessary but not sufficient condition for the energy to transform to gravitational waves. A symmetric accelerating mass does not radiate, If an object changes shape asymmetrically, the spacetime ‘dents’ travel outwards like ripples in spacetime called ‘...

2

On a position vs time graph (a Spacetime diagram), the point $(x', ct')=(0,1)$ lies on a hyperbola centered at the origin and physically represents "one tick" of an inertial astronaut's watch, where the astronaut travels with velocity $v=\beta c$. That point (event P) on the hyperbola can be written as (\gamma v/c, \gamma) =\left(\frac{\beta}{\... 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 Note that the Lorentz transformation must be applied to both x and t. So you'll end up with a new equation x' = ct', and c = \frac{x}{t} = \frac{x'}{t'}. Both observers agree that the light travels at speed c, but in general they disagree on the distance it travels and the time it takes. 2 Assuming you are talking about radial motion the limit is when the proper acceleration for a stationary observer is equal to a_0. This is calculated in twistor59's answer to What is the weight equation through general relativity? The proper acceleration at a distance r is given by: a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}  Note that ...

2

Welcome to Stackexchange. You will find many questions here regarding rotation curves of galaxies. No, warped spacetime isn't the right tree to bark up because the galaxy is non-relativistic, so Newtonian gravity is perfectly adequate. Yes, alternative theories of gravity (such as MOND) can explain this one evidence for dark matter as well but fail given the ...

2

I recommend you to edit your video and avoid making bold and wrong claims like "nothing in quantum physics is experimentally proven". I will leave you some links to learn what planck length is really about, they should be easy enough to be understood by 10th grader Wikipedia Forbes South Wales Fermilab

2

The Euclidean analogue of your question is "What are some uses of the length (or square-length) of a line segment on the plane?" In special relativity, one uses a spacetime diagram, with the underlying Minkowski metric. The time dilation problem is essentially relating "the spacetime interval (the elapsed proper time) along a straight ...

1

By "distance" I presume you mean spatial distance only. This is a frame dependent concept, so clearly there is no unique answer to your question. However, if by "distance" you mean distance in spacetime according to the Minkowski metric then that's another story. The worldline of A is described by a line $A=A(\tau)$ in 4-space, and that ...

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