New answers tagged

0

The general story Here's an attempt to formalize how physicists build spacetime manifolds. Let $N$ be one of $\mathbb R^n$ Some dimension product manifold of $S^1$ and $\mathbb R$ (corresponding to periodic solutions). Pick one such $N$. Now Take stress tensor $T$ defined on some open subset $U \subset N$ and some boundary conditions (e.g. falloff at ...


0

Just to offer a counterpoint: That the law of physics are the same for all observers is what is meant by "relativity". Now for any given observer, the distinction between past, present and future is not an illusion. And, there is no observer seeing spacetime from the outside, so no observer can say anything about the whole of spacetime. So the two ...


3

$I$ 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 $t_2$. ...


0

Yes, you are taking the reversal thing too seriously. If you used Kruskal-Szekeres coordinates then those coordinates don't flip from timelike to spacelike. The flip is just because you chose bad coordinates. If you had flat boring Minkowski spacetime you could pick a coordinate system where a coordinate flips from spacelike to timelike across some surface. ...


0

I dont think it's a problem, that the satellite is accelerating. Locally the time will nevertheless run slower. If the camera is directed onto a watch, you would see it ticking more slowly. If the webcam is directed on the outside, you will see the normal speed of happenings, e.g. it would show you the same number of rounds as you have seen by looking at ...


5

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 ...


2

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

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 ...


1

You have to consider the singularity and event horizon. By observing the orbits of nearby planets and such, we can calculate the approximate mass of the black hole, and thus where the singularity is. This then further gives us the Schwarzschild radius; $R_{Schwarzschild}=\frac{2GM}{c^{2}}$ This defines the event horizon, thus allowing us to then calculate ...


1

Any two points in spacetime are linked by a four-vector that physicists conventionally write as $(x^0, x^1, x^2, x^3)$, where $x^0$ is normally the timelike dimension and the other three components are spatial. If we use the usual Cartesian coordinates in flat spacetime we'd generally write the four-vector as $(t, x, y, z)$. In this case suppose the light ...


0

but as you move faster you also experience time dilation, correct? (1) Move faster relative to what? You're at rest with respect to yourself. According to you, it's the other clocks that are moving, not you. If there a million other clocks 'out there', each with a different velocity relative to you, what would it mean for you to 'move faster'? If ...


1

Short Answer, NO. Please be aware that all inertial frames are indistinguishable, so there is no "universal inertial frame". So if you are an observer ($A$) in an inertial frame ($F_A$) and see another observer ($B$) in a different inertial frame ($F_B$) which is moving to you, you beleive that you are at rest, but the observer $B$ who is at rest in his ...


1

No, and certainly not by the mechanism you describe. The "orbits" of electrons around nuclei are structures created by the electromagnetic force. This force is mediated by photons, which cannot pass out of the event horizon by definition of the event horizon. So even in the highly implausible scenario in which an atomic structure existed within a black ...


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."


-1

Which comes first for gravity: mass or space-time? Neither I'm afraid. As Einstein said, the mass of a body is a measure of its energy-content, and it's actually a concentration of energy that causes gravity. Hence a massless photon has a (very weak) gravitational field. On top of that there's a bit of an issue with spacetime as per CuriousOne's ...


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 ...


-3

No it cannot be explained why or how matter can curve space. Unfortunately like Dominect suggest many have the attitude that we should not even be concerned with the why.


1

The curvature of space-time is provided by the solutions of Einstein equation \begin{equation} R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R - \lambda g_{\mu\nu} = 8\pi G T_{\mu\nu}, \end{equation} where $R_{\mu\nu},R$ denotes respectively the Ricci tensor and Riemann scalar. It is important to note that these quantities are provided as a function of the metric ...


3

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 ...


3

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 ...


0

Depends on your view on "spacetime". In the classical view on space and time, time is completely independent from space, expressed e.g. in the Galilean transformations which transform space and time independently. https://en.wikipedia.org/wiki/Galilean_transformation Thus you can just visualize it as $\mathbb{R}^3$, which is not hard to sketch. In the ...


2

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 ...


1

Yes, but by definition. Not by any meaningful physics. Imagine a path through 3-space. You can define the path by a function of time that returns a position. ${\bf f}(t)=(x(t),y(t),z(t))$. Then the velocity as a function of time is ${\bf v}(t)={\bf f}'(t)$. Easy. You could do the same thing through 4-space, by describing a path parameterized by some other ...


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 ...


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

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: ...


0

If we were discussing electromagnetic forces, you might be asking whether electromagnetism causes the forces or vice versa. To make this less of a chicken-egg question, I'd assume "electromagnetism" here means the existence of nonzero values for electric charges and the electric and magnetic fields. The electromagnetic force on a particle of charge $q$ of ...


1

As general relativity says, gravity is nothing but a geometry of space time. The bending of spacetime is caused by mass and energy distributions, which we see as gravity. Both accounts to the same thing. The spacetime is affected by mass as well as condensed energy. It could successfully explain why there is gravity and almost everything related to that. But ...


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

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 ...


-1

In reference to the given participants $A$ and $B$ there can be another participant $M$ (uniquely) identified as the "middle between" $A$ and $B$ by the conditions that $M$ and $A$ had been and remained at rest to each other (which can be expressed in terms of spacetime interval values, following the prescription linked in the OP), likewise $M$ and $B$ had ...


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.


-1

Conditions for participants $A$ and $B$ having been and remained "at rest to each other" are (1): The events in which $A$ took part are straight to each other; and likewise, separately: the events in which $B$ took part are straight to each other; i.e. explicitly $$\forall~\varepsilon_{AF}, \varepsilon_{AJ}, \varepsilon_{AP} \in \mathcal E_A : $$ ...


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

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 ...


1

Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect, Physicists are hopeful that gravity will be quantized similar to the other three forces, and that gravitational waves are a confluence of gravitons. There exists the classical general relativity where the ...


3

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 ...


0

Well, by a process of elimination, if you have your time orientation (a vector field $\tau^\mu$), and you have two equivalence classes, for $g(X,\tau) > 0$ and $g(Y, \tau) <0$, the only remaining possibility for a third class is that $g(Z,\tau) = 0$. But any vector tangent to a timelike vector will be spacelike. It can be proven thusly : if you have ...


0

Well, in your very ideal system and from the classical point of view, yes the dices (considering they as an electromagnetic neutral system) will be in touch at some time independent of where are they at the beginning.


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 ...


0

Let's start focusing on Bosonic String Theory. There are two kind of strings: open and closed. The closed strings can freely propagate in the 26 dimensional space of the theory. The 26 arises from a precise calculation, in order for the theory to be consistent, Lorentz invariant and other stuff. In the spectrum of the closed string there is always a ...


0

The general relativity model of the universe is a mathematical model, and in this all of time and space are defined in frameworks. Going from one framework to another requires going through GR transformations. The variable of time within the framework of a planet close to a large sun, for example, is defined and stable within it. There is no slowing or ...


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 ...


-6

"Does a black hole really slow down time?" No. Gravitational time dilation is an absurd concept. General relativity predicts that gravitational time dilation occurs even in a HOMOGENEOUS gravitational field ("the homogeneous gravitational field is the gravitational field which, in every point, has the same gradient of the potential. Such a field is produced ...


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.


3

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 ...


0

The premise "vacuum is not empty" could explain the Hubble redshift, in a STATIC universe: http://www.eleceng.adelaide.edu.au/thz/documents/davies_2001_cha.pdf "As pointed out by DeWitt, the quantum vacuum is in some respects reminiscent of the aether, and in what follows it may be helpful to think of space-time as filled with a type of invisible fluid ...


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 ...



Top 50 recent answers are included