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8

I think the problem in understanding this is the idea of "space being sucked into a black hole." The reality is matter is "sucked" into a black hole. Space is warped around the black whole, but space is not "sucked" into anything. Here's the issue. What is space? You can't touch space (or better, the space-time continuum). So, one view is that space is ...


5

If Galileo had dropped the Moon and a pebble from a very tall tower, the Moon would have fallen noticeably faster, relative to the Earth. This is true in Newtonian physics as well as GR, and it does come from the fact that the Earth falls toward the Moon too, and harder than it falls towards the pebble. The assumption that small objects do not gravitate is ...


5

When space bends, what are the lines that are being bent? The straight ones. For example: say you have two parallel laser beams traveling through space. Photons travel in a "straight line" more or less by definition, i.e. the path of a laser beam is the straightest line humans can produce, so it is a benchmark of sorts. Mathematically two parallel ...


5

I'll go into a bit more detail below but, basically, Einstein discovered that space and time can't be separated. They are part of the same thing, and gravity is actually produced be bending and warping this "spacetime," like a heavy ball sitting on a sheet of elastic. Classical (or Newtonian) Physics Time can be considered as a dimension, just like ...


4

When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe. It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are ...


4

The idea behind that quote is that you can't really separate space and time in General Relativity, which is the most complete scientific theory concerning the geometry of space and time. Instead, it works best to consider them as one integrated thing, called spacetime. To go into a little more detail, first consider galilean spacetime. Here you can think ...


4

You use the term frame of reference but we need to be careful what we mean by this. In special relativity this phrase generally means an inertial frame i.e. a frame in which Newton's first law applies. In GR we obviously can't have a global inertial frame because objects accelerate (due to gravity) whenever they are near a mass so their behaviour isn't ...


3

Proper time of an observer is time as measured by the observer's own clocks. So it's obviously frame-independent because calculating proper time of a given observer requires to use his own frame of reference.


3

The strength of gravity is given by the space-time curvature caused by all the objects in the system, in this case both the earth and the falling object. The problem is that you are ignoring the fact that the space-time curvature caused by the earth is many, many times larger than that caused by the falling object. Hence, the total curvature is to all ...


3

You are asking us for the distance of the trip in the rest frame of the photon. The problem with asking that is that there is no rest frame of a photon. A photon can never be at rest, so it has no rest frame. This is like asking what a bowl of petunias thinks about its existence as it falls to the surface. A bowl of petunias doesn't think, therefore we can't ...


3

One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if $x^0 = y^0$ the separation is spacelike. Alternatively you can work from the definition where two events are spacelike separated if (and ...


2

When we use the terms "bending" or "warping" with respect to spacetime and gravity, you have to keep in mind that these words are not being used in a literal way. Since the majority of concepts in General Relativity are far beyond what our experiences allow us to comprehend, we have come up with a few ways of picturing these concepts in our minds, none of ...


2

We need to clarify what we mean by dimensions of a mechanical system. They refer, in the standard terminology, to the number of different degrees of freedom we need to describe the kinematics of a point particle (in this case). To this extend, given any reference frame $S$, an event in the space-time is identified by its position $(x,y,z)$ and the time $t$ ...


2

Here is one solution to Einstein's field equations: $$ds^2=dt^2-dx^2-dy^2-dz^2 \text{ on } \{(x,y,z,t):x,y,z,t\in\mathbb R\}.$$ It has a global coordinate system, a global reference frame, no closed time like curves, and a famous name, Minkowski space. Our second solution is a different manifold $\mathbb R^3\times \mathbb S$, which can (but doesn't have ...


2

The field line mind picture in General Relativity is probably not useful, at least not for me, aside from in very special cases, because The dimension of the pictures you're trying to see at is too high for our everyday spatial intuition to help us much - you're trying to visualize a rank 2, $4\times 4$ tensor with 10 independent components (the metric), ...


2

If you plot grid-lines on a graph what are the grid-lines? Well, on a typical 2D plot with x increasing to the right and y increasing upward, the horizontal grid-lines are the loci of constant y and the vertical grid-lines are the loci of constant x. In 3D the grid lines are the loci of constant x and y for the ones that run parallel to the z-axis and of ...


2

I think you are mixing up two different concepts, which is muddying the waters. Firstly, relativity (both special and general) is a geometrical theory and the proper time for an observer has a precise definition as the length of a world line along which the observer travels (give or take a factor of $c$). This length is calculated using the metric. As ...


1

Your first line is incorrect, gravity is real, just jump up and you will confirm that. Frames of reference are used to find how gravity affects our measurements of spacetime. Your second line is correct. Objects, from photons to planets, moving in general relativity, obey the geodesic equation, so a curved line can be a "straight line", that is, if you ...


1

In general relativity there might not be a general frame of reference that will look the way an inertial frame of reference looks in special relativity. And the fundamental deep down reason is that we didn't assume there had to be, thus it didn't have to happen. Whether a particular solution to Einstein's equation has one or not is up to experiment to ...


1

According to what I have read, we have measured the universe to be flat More or less. I'm happy enough with the WMAP results that indicate that the universe is flat. To be blunt I never thought it could be anything other than flat. the shape of the universe is directly related to the mass-energy density. That's what they say. But IMHO two out of three ...


1

Light will not travel on a geodesic in the background spacetime, as light is an electromagnetic field and has stress-energy and so itself warps the spacetime around it. However if the light induced curvature is small compared to the curvature already there, then it won't change things much, so will go on an approximately geodesic route. Same for you ...


1

Just some minor remarks: (i) An expanding universe filled with matter and a zero spatial curvature are not in contradiction. Please read up about the Friedmann equations and the Robertson-Walker metric. (Sorry, but I think that's the way to understand cosmology properly.) (ii) Flat geometries are not necessarily infinite (non-compact). Take for example a ...


1

This is really one google search away, see e.g. page 26 (marked 64) here. As already noted by John Rennie, Penrose diagrams are not suited for the analysis of Kerr CTCs because they show a $\phi = const., \theta=\pi/2$ slice of the global structure. The $r<0$ region is however accessible only through $\theta \neq \pi/2$. The Boyer-Lindquist coordinates ...


1

I"m pretty sure that this discussion does appear in Hawking and Ellis, though I admit that it's been a while since I looked. It's not done through a Penrose diagram, though. The argument really comes down to the fact that for sufficiently small $r$, $d\phi$ is timelike. But, by construction, the orbits of $\phi$ are closed curves. When $d\phi$ is ...


1

Is gravitational radiation a form of mass/energy that can form momentum? We think so. We haven't detected any gravitational waves as yet, but there's a high degree of confidence that gravitational waves are real. If it causes ripples in space-time, would space-time be susceptible to forming vortices if massive bodies passed each other at close distances ...


1

I think Gennaro covered, this, I'll give a layman's explanation. In spacetime there are four general dimensions, three of space and one of time. Why is it that other dimensioned qualities seem to be rarely considered as part of spacetime? For example, why isn't speed part of spacetime, forming a five-dimension spacetime-speed manifold? Objects in ...


1

There is a tiny bit more going on than the otherwise excellent answer by zeldrege suggests. Imagine that you wish to probe an unspecified object to examine its structure. If we use light to look at the structure of an object, we need to have its wavelength smaller than the size of the details we wish to look at. Probing an object that has a (linear) size ...


1

John, have a look at the simple inference of time dilation due to relative velocity. If you and I are identical twins, and you take a fast out and back trip, when you come back we agree that you've experienced less time than me. As you pointed out, we can relate this to the Lorentz factor and write: $$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ ...


1

The reason why the commutation relations between a field and its conjugate at equal times are of the form $$ \left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y}) $$ is only to mirror and copy the canonical hamiltonian commutation relations $[q_i,p_j]=i\hbar\,\delta_{ij}$. No causality is involved, rather it is somehow ...



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