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171

What is a manifold? A manifold is a concept from mathematics that has nothing to do with physics a priori. The idea is the following: You have probably studied Euclidean geometry in school, so you know how to draw triangles, etc. on a flat piece of paper. In contrast to common parlance, let's take "space" to mean anything with a number of points. The ...

137

Short answer: nobody knows, but the Planck length is more numerology than physics at this point Long answer: Suppose you are a theoretical physicist. Your work doesn't involve units, just math--you never use the fact that $c = 3 \times 10^8 m/s$, but you probably have $c$ pop up in a few different places. Since you never work with actual physical ...

130

To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation. Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" ...

109

TL;DR This is a complicated question and anyone who tells you a definitive answer one way or another is either a philosopher or is trying to sell you something. I justify arguments either way below, and conclude with the AdS/CFT correspondence, in which two theories on two vastly different spacetime manifolds are in fact equivalent physically. First, let’s ...

96

Suppose we take the spacetime point of your conception as the origin, $(t=0, x=0)$, then the spacetime point for your birth would be $(t=T, x=uT)$. The time $T$ is approximately $9$ months, and we are writing the spatial position of your birth as $x=uT$ where $u$ is a velocity. The velocity $u$ can be any value from zero (i.e. born in the same spot as ...

95

I think the correct answer should be that what we call gravity is a fictional force which we experience due to living in an accelerated reference frame (as opposed to an inertial one). Unlike other forces, the force of gravity disappears by a coordinate change. If a person is in a falling elevator, they experience free fall, i.e. they feel like they are ...

89

Because what we can observe doesn't differ according to distance from us. The most important reason why we assume physical laws are not different in distant galaxies is that we can observe things in those galaxies that would not be as we observe them if the laws varied. For example, we can see light from distant parts of the known universe. The spectral ...

85

I think the question suggests you are thinking of space-time as if it were e.g. a substance, like a fluid, that we move through. That's not how we view space-time, at least in pure general relativity. But the question you ask is a deceptively simple one and it raises some complex questions. And I don't think we actually can answer them exactly because I'm ...

84

I would recommend steering clear of Schwarzschild coordinates for these kind of questions. All the classical (i.e. firewall paradox aside) infinities having to do with the event horizon are due to poor coordinate choices. You want to use a coordinate system that is regular at the horizon, like Kruskal-Szekeres. Indeed, have a look at the Kruskal-Szekeres ...

71

I'm not a quantum cosmologist, but I am an early-universe cosmologist, so I can give you my opinion after having read this paper. The article claims that Bohmian trajectories is a valid replacement for geodesics. This was claimed in the very beginning of the paper and not much is offered in the way of defense for this assumption. That's not to say that it's ...

68

This is a rewrite of Michael Brown's answer to help me get my thoughts clear, and possibly to help everyone else who's interested to get their thoughts clear too :-) Michael presents a very simple answer to my question based on the geometry of spacetime around the black hole. The key point is that the usual radius/time Schwarzschild coordinates are ...

66

Classical electromagnetic fields carry energy and momentum and therefore cause spacetime curvature. For example, the EM field around a charged black hole is taken into account when finding the Reissner-Nordstrom and Kerr-Newman metrics. The question of whether photons cause spacetime curvature is a question about quantum gravity, and we have no accepted ...

65

Let's suppress some dimensions to simplify: $$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2$$ This quantity $$\Delta s^2$$ is preserved by changes of reference frame, just as in Galilean physics the quantity $$\Delta r^2 = \Delta x^2 + \Delta y^2$$ is preserved by rotations. Notice it is also the equation of a hyperbola. Thus, the effect of a frame shift is ...

61

Although it's commonly said that fundamental particles are point particles you need to be clear what this means. To measure the size of the particle to within some experimental error $d$ requires the use of a probe with a wavelength of $\lambda=d$ or less i.e. with an energy of greater than around $hc/\lambda$. When we say particles are pointlike we mean ...

57

Real particles are never completely localised in space (except possibly in the limit case of a completely undefined momentum), due to the uncertainty principle. Rather, they are necessarily in a superposition of a continuum of position and momentum eigenstates. Pauli's Exclusion Principle asserts that they cannot be in the same exact quantum state, but a ...

56

The short answer: general relativistic effects are mostly not noticeable on such a small scale (except in a few cases). For example, one common way to characterize the strength of a gravitational field is through the dimensionless number: $$A=\frac{GM}{Rc^2}$$ where $G = 6.67 \times 10^{-11} \text{ }\mathrm{m^3 kg^{-1} s^{-2}}$ is the gravitational constant ...

54

Within the context of Newtonian mechanics, there's a simple answer: velocities are not absolute, but differences in velocities are. So you can state that acceleration occurs unambiguously. In special relativity, this is a bit more complicated because of relativistic velocity addition, but all observers can unambiguously compute a "proper" acceleration for ...

51

You're using the wording "curved spacetime", but you're still only thinking "curved space" with an independent, linear time. In your curvature model, you're assuming that moving through some 3D spatial point in one spatial 3D direction will experience the same 3D path curvation independent on speed (as if you'd shoot a ball through a ...

49

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. Even when at rest in space, the apple still advances in space-time. Here is a visualization of the falling apple in distorted space-time: http://www.youtube.com/watch?v=DdC0QN6f3G4

49

Let's say space is really a lattice with spacing $\Delta x$. It turns out that this idea has more trouble with experiment than you might think, but we can plow ahead for the purposes of this question. You might propose replacing integrals in physics with discrete sums over individual lattice points, to take a concrete example let's think about the work ...

48

None of the above. Though there are many speculations about the significance of the Planck length, none is proven in any currently accepted theory. It is expected, though, that quantum gravity effects become definitely non-neglegible at the energy/distance scale set by the Planck length, so it provides a heuristic scale at which we should not expect our ...

46

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 ...

44

Charge does curve spacetime. The metric for a charged black hole is different to an uncharged black hole. Charged (non-spinning) black holes are described by the Reissner–Nordström metric. This has some fascinating features, including acting as a portal to other universes, though sadly these are unlikely to be physically relevant. There is some discussion of ...

44

Analogy: how do we know that the surface of the Earth is curved? Well, we could e.g. draw a triangle on the surface of the Earth, and check the sum of the corner angles. If the Earth was flat, you'd always find that the sum of these angles was 180°, so it would be impossible to e.g. create a triangle with two 90° corners. However, since the Earth is curved, ...

43

You say: As far as I know we have only done experiments to determine the value of h close to the Earth's surface but the value of Planck's constant determines the position of spectral lines, so we have measured it all over the universe. We do see shifts in spectral lines due to the red shift or gravitational time dilation, but these are explained by ...

42

Take the example of a hydrogen atom. The energy levels in the hydrogen atom are given by $$E_n = -2\pi^2 \frac{m_e^4}{n^2 h^2}.$$ The spacing between two energy levels determines the frequency of an emitted photon when a radiative transition is made between them: $$h \nu_{n_2\rightarrow n_1} = \frac{2\pi^2 m_e^4}{h^2}\left(\frac{1}{n_2^{2}} - \frac{1}{n_1^... 40 There are two parts to your question. First, why can we see things "46 billion light years away" if the Universe is only about 13.8 billion years old? Because the Universe is expanding. How far does a photon travel in 13.8 billion years in an expanding Universe? It depends on the rate of expansion. I'll give a simplified example to illustrate the point: ... 40 That is awesome! And it makes complete sense too! (other than a possible misusage of the word "distance"). Let's have a look at the equations of motion of you in Earth's curved spacetime, assuming that your feet are not touching the ground:$$ \frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma^{\mu}_{\nu\sigma}(x(s))\ \frac{\mathrm dx^{\nu}}{\mathrm ds}\...

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