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Objects have a property called "electric charge". This electric charge decides how strong a force they feel when close to other electrically charged objects. The electric charge of an object is more or less independent of inertial mass. So given a large, fixed, electrically charged object, you can make a small electrically charged test object feel ...


10

Gravitons are theorised by looking at the linearisation of a perturbation of curved spacetime and it turns out that it is a massless spin-2 particle. Hence as one commenter has pointed out, it presupposes curved spacetime. However, having derived the graviton in curved space, we can consider it in flat space. According to Feynmans Lectures on Gravitation, if ...


7

gravity isn't actually a force ... I'm curious definitionally. Ok, so “isn’t actually a force” is more of a pop-sci summary than something definitionally correct in General Relativity. In Newtonian mechanics the gravitational force is mediated by the gravitational field which gives the gravitational acceleration at each event. In general relativity the ...


7

These are all fine questions, but it is important to get first a proper intuition. Gravitons are to gravity what photons are to electromagnetic field, or what phonons (quasi-particles of sound) are to deformation. They are all excitations of some field (quantum field, to be precise). "Excitation" implies there is some default state of the field, ...


6

This is an interesting question that could really be formulated as follows: When collapsing an object that has the ratio $a^* = cJ/(GM^2)$ larger than one, can an over-extremal black hole form? There is no rigorous proof, but the answer seems to be no, at least in reasonable physical situations. The simple reason for that is that when you have an object ...


5

We don't have a quantum theory of gravity yet, so the honest answer to your questions is "nobody knows". But I think most people would guess that the quantum theory of gravity would bear a similar relationship to curved spacetime as the quantum theory of electrodynamics (QED) bears to Maxwell's picture of electric and magnetic fields. That is, ...


5

There is a simple geometric interpretation of the quantity that is called the age of the universe. Spacetime has a large-scale shape. It looks something like this: Later times are at the top, earlier times at the bottom. You shouldn't take this too literally because it isn't a proper embedding (I flipped the sign of the metric to make it Euclidean, and ...


5

No. The age of the universe does not depend on any referential system. In order to measure time, you need some physical quantity that's changing to measure time against. In the case of cosmology, it's the time perceived by a typical observer based on the expansion parameter $a$ --see below. In a manner of speaking, it's the time that follows the galaxies in ...


4

In addition to other good answers, the equivalence of inertial and gravitational masses is equivalent to the experimental fact that all masses fall at the same speed. Science knows that fact to be true since Galileo, and it seems obvious to us because we learned elementary physics long ago, but without doing the experiment it's actually far from obvious that ...


4

"Do we have any evidence that flow of time is not relative in quantum scale?" No, we do not. In fact, quite the reverse; Wikipedia is not always reliable! Firstly, the basic quantum equations simply assume that time is absolute (Newtonian) and not relative, they make no claim that this accurately represents reality. This is what the Wikipedia ...


3

It's imporant o keep track of what is a vector, and and what are just numbers. The components of vectors, tensors etc are numbers, and the covariant derivative of a number-valued function is just the ordinary derivative. In particular the array of numbers ${\omega^a}_{b\mu}(x)$ are just number-valued functions, so $$ \nabla_\nu {\omega^a}_{b\mu} =\partial_\...


3

This answer will confirm that QFT does propose that quantum fields can be in configurations that break the energy bounds believed to hold for ordinary matter, such as the weak energy condition and dominant energy condition. I think it is misleading, however, to use the terminology "negative mass" for this situation, at least in many examples, as I ...


3

I believe it is just a matter of notation inconsistency, common in physics. Manifolds are by definition locally Euclidean (i.e. locally look like $\mathbb{R}^n$, the $n$ dimensional Euclidean space). There is no metric given a priori. In Special Relativity, we also define a metric of Lorentzian signature, and call this structure Minkowski spacetime. This ...


3

William Clifford, a British physicist and mathematician, inspired by Riemann's notion of a manifold, a generalisation of non-Euclidean geometry, published in 1876 a small note in the Proceedings of the Cambridge Philosophical Society titled The Spacetime Theory of Matter. Here he wrote a full fifty years before Einstein formulated General Relativity, the ...


3

It is likely that the talk was about multiboundary wormhole/black hole geometries in (2+1)–dimensional gravity. Remember that BTZ black hole could be seen as a factor space of $\text{AdS}_3$ spacetime by a discrete group of isometries generated by a single element. Multi-black hole geometries are factors $\text{AdS}_3/Γ$ by more complex discrete subgroups $...


2

The velocities in cosmology that can exceed $c$ don't have much in common with the velocities in special relativity that can't. Even in special relativity, if you define a "recessional speed" similar to the way it's defined in cosmology, it can exceed $c$ even though the usual special-relativistic relative speed of the same objects is less than $c$....


2

"A body's mass increases when it acquires velocity" is false. Velocity is relative, so any apparent increase in mass is entirely in the eyes of the observer moving relative to the body, i.e. it's an artifact of the coordinate system. That's why modern relativity textbooks generally avoid "relativistic mass" and only talk about "...


2

Gravitation is not a force. If you stand on the Earth then you are accelerated upward by the electromagnetic force. There is no force pulling you towards the Earth. This upward acceleration is what makes you see your weight when standing on a scale. On a heavier planet your weight increases. But by scaling the scale you always get the same value for your ...


2

I think John Rennie's answer is needlessly confusing. Gravitational time dilation is not due to differences of gravitational potential energy. It isn't an "effect" at all. It's just a geometric property of worldlines in certain manifolds. If two ships separated by 2 km on the equator both sail north to the 30th parallel, they'll be about 1 km apart ...


2

The value of $\kappa$ different from the value of $1$ (which is what we have in the usual form of Schwarzschild solution) corresponds to the deficit (for $\kappa<1$) or excess of solid angle. If we consider a sphere of constant radius $r=R$ when $R\gg M$, the distance from the central object (black hole) is approximately $\kappa^{-\frac12} R$, while the ...


1

Note that from the relation $e^\lambda{}_{a;\nu} = \omega_a{}^b{}_\nu e^\lambda{}_b$ you give you can deduce by contracting with $e_{\lambda c}$ $$e^\lambda{}_{a;\nu}e_{\lambda c} = \omega_{ac\nu}$$ Note, however, that I am using the definition of the covariant derivative that takes tetrad indices $a,b,c$ as mere labels and thus the covariant derivative of ...


1

The insightful answer by A.V.S. gives most of the information here. I wish merely to add that a major shaping factor for a theory of gravity was not just that it be covariant and reduce to Newtonian the right limit, but that it would have the equivalence principle built in. I think if you insist on the latter then you will be led to Einstein's G.R. (or else ...


1

No, there is not such general rule that you are looking for. But you can count the maximum number of independent Christoffel symbols in a general space. In fact, for each independent component of the metric tensor, there are, at most, $N$ distinct Christoffel symbols. Let me first start with an example. If you consider a two-dimensional Cartesian coordinate ...


1

As a comment by @knzhou suggests, $P^μ$ cannot be (proportional) to 4-momentum. But it could be some sort of “matter density” current. What really would not be consistent is the “Lorentz force law” that is put here by hand only. Force law is not an independent equation (even though Einstein originally thought so) but follows under limiting procedure of self ...


1

The derivative of a curve on a manifold is a vector field along the curve. It's not possible to differentiate a vector field along a curve without using a connection. This is usually left implicit in the covariant derivative. Hence the first derivative of a curve is the usual one but to take the second derivative we need to use the covariant derivative.


1

Just, to make it more clear (as mentioned in the comment), first prove this $${P_{\alpha \mu }}{u^\mu } = ({u_\alpha }{u_\mu } + {g_{\alpha \mu }}){u^\mu } = {u_\alpha }\underbrace {{u_\mu }{u^\mu }}_{ = - 1} + \underbrace {{g_{\alpha \mu }}{u^\mu }}_{ = {u_\alpha }} = 0,$$ and hence, you get this $${P_{\alpha \mu}}(\rho + P){ _{;\nu }}{u^\mu }{u^\nu } =\...


1

Classical mechanics has a more or less axiomatic framework. Here, gravitational and inertial mass is represented by the same concept and the same symbol, $m$. If we decide to disambiguate the concepts by representing them by different symbols, say $m$ for inertial mass and $M$ for gravitational mass and also by different concepts of mass, the resulting ...


1

Truly excellent question. All "why would we expect" questions are inherently subjective, and there is no single correct answer to this question. But I'm going to give a very heterodox (and I'm sure unpopular) answer: I would argue that you are correct, and in fact there never was any reason to expect that every object would have one inertial mass ...


1

From what I've understood, you seem to be making a mistake with the $h_{ab} \Delta K$ term, with $K$ defined as $K= h^{ab} K_{ab}$. The indices in $K$ are summed, not the same as free indices of $S_{ab}$, so you need to make them different. For example, witing the Israel Junction Condition in full and using $\Delta h_{ab} = 0$ gives $$ S_{ab} = \Delta K_{ab}...


1

I never really fully understood what motivated general relativity or why the Newtonian concept of gravity was considered problematic. The main problem, and certainly the issue which most motivated Einstein, is that Newtonian gravity is not consistent with relativity. Relativity places an upper limit on all causal influences, but Newtonian gravity has no ...


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