# Tag Info

## Hot answers tagged spacetime

41

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: $$\frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma^{\mu}_{\nu\sigma}(x(s))\ \frac{\mathrm dx^{\nu}}{\mathrm ... 11 What GR says is correct: the straight line between, say, London today and London tomorrow is not the curve that spends all the time between in London: whether it actually passes through the centre of the Earth I'm not sure, and it depends on how fast you are moving as well as where you are. The caveat is that the straight line (geodesic) not the shortest ... 7 It makes sense as a "visual" description. In GR, free particles with mass move on time-like geodesics. A common description of geodesics are such curves that locally minimalize path length, but this desciption comes from Riemannian geometry, not Lorentzian geometry, which GR is. In Lorentzian geometry, timelike geodesics are those that locally maximalize ... 6 Your intuition that the Einstein equations are equations for the metric tensor, not for the manifold is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your ... 6 That quote requires some modification for it to make sense: "General Relativity basically says that the reason I am sticking to the floor is that the path of maximal aging between 'here now' and 'here tomorrow' is through Earth's center." 5 Yes, this statement is true, in the sense that the four-velocity u^\mu = (\gamma, \gamma \vec{v}) always satisfies$$u^\mu u_\mu = 1$$as you can check using the definition of \gamma. (I'm setting c=1.) Therefore the magnitude of the four-velocity is always equal to the speed of light. However, this statement can be really misleading. It's true that ... 4 The proper time of a time-like curve is its length. 3 If you remove a closed subset from a valid spacetime, then the result is a valid spacetime. You can put the Lorentzian metric on the original spacetime, then simply restrict it to the part that you keep. So it's still a 4d manifold without boundary. It's still Lorentzian and so on. You could even imagine the original manifold as some charts and transition ... 3 I read a few lines about general relativity and [... an equation for] the eigentime of a time-like curve. I suppose that this is referring to an equation similar to$$\tau A_J^Q := \int_0^1~dt~\sqrt{g[~\dot\gamma, \dot\gamma~]},$$where A denotes a particular participant ("material point", "principal identifiable individual"), the quantity being ... 3 Let's analyse the evolution of the curvature in the \Lambda\text{CDM} model. If \rho_R, \rho_M, and \rho_\Lambda are the densities of radiation, matter and dark energy, and$$ \rho_c = \frac{3H^2}{8\pi G} $$is the critical density, then we can define$$ \Omega_{R} = \frac{\rho_{R}}{\rho_{c}},\quad \Omega_{M} = \frac{\rho_{M}}{\rho_{c}},\quad ...

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Even if Alice and Bob are both first to measure their spin (according to their respective reference frames), two spins entangled into the singlet state will still give opposing results. That's what quantum mechanics predicts. Finding out that the entangled spins gave agreeing results would falsify a prediction of quantum mechanics. People would be very ...

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In an expanding universe, the farther away an object is, the faster is it's recessional velocity relative to an other object at distance $r$: $$v_{rec}=H(t) \cdot r$$ with $H(t)$ as the current Hubble parameter. On the other hand, for every given distance there is also an escape velocity $$v_{esc}=\sqrt{\frac{2\cdot G\cdot M}{r}}$$ If you want your 2 ...

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Spacetime isn't an object, like some sort of elastic jelly, that galaxies move through churning it up as they go. Spacetime is a mathematical object that we use for calculating observables in relativity. So there isn't any sense in which a singularity twists up spacetime. The treatment of singularities is quite subtle in relativity. We describe spacetime as ...

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Should physics make an ultimate answer to the why questions? In my opinion, it is not the physicist's aim, and moreover it is beyond the scope of physics. Physics mostly builds theories as our tools to understand and predict some aspect of the surrounding infinitely complicated world. Sometimes we come to a theory that is of such a generic applicability ...

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$I$ has a clear physical meaning if $I\lt 0$ – which is a significant percentage of the spacetime, so to say: $$I = -c^2\Delta t_{\rm proper}^2$$ where $\Delta t_{\rm proper}$ is the time measured by clock that moves by a constant velocity (without acceleration); and that visits the point $(x_1,y_1,z_1)$ at time $t_1$ and $(x_2,y_2,z_2)$ at time $t_2$. ...

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Spacetimes with two spatial and one time dimension are known as 2+1D spacetimes, and there are lots of questions on this site about them. They are popular in quantum gravity studies because quantum gravity is a lot simpler in 2+1D. However the physics of 2+1D spacetimes is very different to our 3+1D spacetime. For example stable planetary orbits are only ...

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Since the traveller is circling Earth, he/she is actually accelerated. So his/her frame of reference is not inertial (if we take that of the beholder to be inertial). The two points of views are then not equivalent, unlike those in the twins paradox. If I'm not mistaken, one may show that the beholder will "see" the traveller be slower, and the traveller ...

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If cosmological expansion applies on the scale of the earth moon system, then in some short period of time $\delta t$ the distance between the earth and moon increases from $r$ to $r+\delta r$. So the force of gravity between the bodies changes to: $F+\delta F=\frac{GMm}{(r+\delta r)^2}\approx\frac{GMm}{r^2}\left(1+\frac{\delta ... 2 I beleive you have a huge confusion. The Minkwoski metric only gives geometrical properties of the spacetime, and it basically states that the spacetime is not euclidean. What you are asking is similar to asking that Newton laws and classical mechanics comes from the fact that we live in an euclidean world (without time component). The first postulate: ... 2 From the Minkowski metric plus the first assumption of Special Relativity, the rest may be derived. That is, if it is assumed that physical laws are the same in all inertial reference frames, which was first proposed by Galileo, and taken up by Newton; plus the Minkowski metric for spacetime, we automatically obtain the invariant spacetime interval, ... 2 The truth of the matter is that we don't know. Gravitons are a theory slowly becoming accepted but there is simply little evidence to support them as a viable theory on how gravity works. Gravity is still just the magical notion that was devised centuries ago. We only know that gravity is not constant and it is the only force not conforming to the theory ... 2 Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect I conceive the question as one referring to General Relativity (GR) which is an entirely classical theory, and not to a (thus far missing) theory of Quantum Gravity. The point of view of General Relativity is ... 2 Yes, it can. Curvature (whatever measure for it you use, Riemann tensor, Ricci tensor, Ricci scalar, you name it) is a function of spacetime, and hence of time. 2 An example of the sort of system you describe would be the Earth and the Moon, with the Earth playing the part of your mass$m_1$and the Moon$m_2$. Neither of these are point masses, but courtesy of Gauss' law we know that the gravitational field of a sphere is the same as the gravitational field of a point mass provided you are farther away than the ... 2 Per Einstein's theory, they both are one and the same thing. There is some, yet not known property of space, and mass, that causes this phenomenon. Even if the property becomes known today, the question will be why that property exists? 2 This is one of the most misunderstood things about entanglement, which is that it doesn't matter who goes first. Neither measurement actually affects the other one, contrary to the intuitive implications of "wave function collapse". Entanglement is correlation, not causation. 2 Your question has nothing to do with entanglement. You might as well ask this instead: Physics predicts that two positive charges will repel each other. Suppose I bring two positive charges into close proximity and find that they attract each other instead. How can this contradiction be resolved? Or you could posit any other experimental result that ... 1 I can't comment on string theory, but in quantum field theory the U(1), SU(2), etc symmetry groups are local gauge symmetries. They are not a symmetry of the spacetime in which the symmetry is formulated. So whether the spacetime is discrete or not makes no difference to the local gauge symmetry. As far as I know the physical significance of the local gauge ... 1 Suppose I'm orbiting the Earth. The spacetime curvature is controlling my motion i.e. I move in a circle centred on the Earth rather than a straight line because the spacetime in my vicinity is curved. This is an example of Wheeler's statement - the mass of the Earth curves spacetime and the curvature tells me how to move. Now suppose I throw a ball I'm ... 1 Exactly as an ideal clock at rest with the observer (here pictured as a timelike curve) measures the proper time of the observer, ideal rulers at rest with the observer measure the distances in the rest space of the observer. Mathematically these rulers are pictured as an orthonormal basis made of$3\$ vectors normal to the unit tangent vector to the ...

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