# Tag Info

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Particles with zero rest-mass can only move at the speed of light, while massive particles can never reach it. This is a fact from special relativity and is independent of the source of said mass. The fact that the Higgs mechanism makes certain particles massive is not intrinsically related to this.

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According to Wald's GR [...] e.g. in a case where there are several massive bodies in relative motion--there exists no natural set of curves whose comparison with geodesics could be used to define gravitational force. Why not? Why not, indeed. What could be more "natural" than to make the required comparison ("with geodesics") for each participant ...

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First, note that there is no unified theory of QFT and gravity, so talking about geodesics and about the Higgs is really not possible within the framework of our current theories. Nevertheless, the confusion here seems to stem somehow from the idea that all particles are "initally" massless, and "then" the Higgs comes along and gives them mass. This idea of ...

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This is discussed in section 4.3 in my 1984 edition. The quote supplied can't really be understood in isolation - you need to consider the whole section. Wald's point is that in general relativity there are no inertial observers because in general spacetime is nowhere flat. In Newtonian physics or special relativity acceleration can be measured relative to ...

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Ok I solved case 2. Let $t'=\tan^{-1}(\omega t)$ and $x'=x\sqrt{1+t'^2}$. Also observe that $x=A\cos \omega t + B\sin \omega t$ to represent simple harmonic motion. We get $x'=A+Bt'$. So only case 3 remains open. Also as u can see, I just did guesswork, without proper techniques for how to solve this in general.

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If you are allowing non-linear transformations of space-time (as you have done in your Case I), then I don't see why you can't write any motion of an object as x' = 0. i.e. Suppose the motion of the object in the (x,t) coordinates can be written as f(x,t) = 0 for some non-linear function f. Then, by setting x' = f(x,t), t' = t, the motion in (x',t') can be ...

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The Lorentz transformation simplifies to the classical transformation for all your cases since you took $c$ as infinity. There is no time dilation. The transform is just an identity multiplication for all 3 cases with: $$x' = x - vt$$ and $$t' = t$$ Now that you are in the classical world in one dimension, your first case is merely a tilted line with ...

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The singularity comes from the scale factor $a(t)$: $$ds^2 = -dt^2 + [a(t)]^2 ( dr^2 + r^2 d \Omega^2)$$ By solving the Friedmann equations for the scale factor we know that: $$a(t) = a_0 t^{\lambda}$$ where $\lambda$ is some positive number that depends on the matter-radiation ratio of the universe. At $t=0$ the scale factor becomes $a(0)=0$. So at ...

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This is an extended comment on Valter's answer, so please upvote his answer not this one. In Relativity (General and Special) there is no unique way to divide spacetime into space and time. Different observers, using different coordinate systems, will disagree about whether a four vector is just a displacement in time or just a displacement in space. So to ...

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In a certain sense (regime) acceleration is caused by the curvature of time more than the curvature of space. Actually, the curvature is of the spacetime so that, making rigid distinctions has no much sense. However, if you consider the motion of a particle free falling in a region of spacetime, the equation of its story is the geodesical one: ...

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Wald is a first rate relativist, and as such he is phrasing the concept of general covariance in terms of purely geometrical quantities, rather than resorting to the somewhat imprecise notion of coordinate transformations. In the discussion on pg. 57, he goes on to give an example of what it means to violate the principle of general covariance. In his ...

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I'm hardly a GR expert, so if you want a more technical analysis I'm sure others will be able to give you one. However, the answer to your apparent questions is fairly straight forward. It is not the curvature of space or the curvature of time that causes accelerations, it is the curvature of space-time. We live in a four dimensional universe (ignoring ...

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General covariance basically means you can change your coordinate system arbitrarily and express the laws of physics in the new coordinates. Because of this freedom, the relationship between coordinate distances, angles, etc. and physical distances, angles, etc. is variable and is expressed by the metric. So the quoted statement is basically saying that ...

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I'm not a GR expert (and there are some on this site), but I'll try to answer this. How does an object on the surface of earth stay fixed? Since there is no concept of "attraction" and only space time why cannot an object keep moving around on the surface of earth. The answer here is that there generally aren't any forces moving the object to the side. ...

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you should look for the theory of "Covariant Quantum Mechanics" originally introduced by M. Modugno and J. Jadcyk, in particular the "special algebra of quantizable functions". There is not so much literature since the theory is mathematically quite hard to enter, but afterwards it is worth. I have worked in that field for several years. The basic idea is to ...

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According to quantum mechanics the time evolution of the universe is described by a path integral that will sum over all histories. If we consider a robot whose processor runs at a clock cycle of $\tau$ to simplify things, then all the possible time evolutions during that period of $\tau$ will contribute to explain the robot's observations, including the ...

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Fortunately for experiments in physics we have better proxies than the accuracies of our five senses. We have detectors and computers and .... With these tools a theory of how the universe is made has been developed, from elementary particles with the theory of quantum mechanics building up the observables around us, to the astrophysical models that fit ...

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Quite a philosophical approach. There is still the reliance on our four other senses in order to make sense of our physical world, however the same approach can be imply to those senses also with the delay in neurological impulses. One must also take into account, as you would call it, the in between frames of other people's perceptions, as well as those ...

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Lets make it 10 minutes of travel, because I'm lazy. At 100x light speed you travel 1000 light minutes from Earth, so looking at earth you see images from 10:10 minus 1000 minutes which is 10:10 minus 16.6 hours. Which I am also too lazy to work out, but you get the idea?

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1.When they travel to the watery planet, they say that 1 hour on this planet is 7 yrs om earth. How is this possible? Is the planet moving at a speed close to c? Or does strong gravitational field influence time? Sure. This is gravitational time dilation. It's due to the gravitational field of the black hole. You can calculate it using $$\frac{d ... 0 Despite being a physicist, not only do I not "run away screaming", but I rather think that if there is any certainty in my life, it is my being conscious - the rest are just temporary beliefs. I have written an essay titled "Who am I in the block-universe". https://drive.google.com/file/d/0B94tMBt9zxIHOXlVSmx3MWhtdlE/view?usp=sharing It DOES NOT solve the ... 0 It is Minkowski spacetime. Both \bar{p}=\left(\begin{array}{c}E\\{\vec p}\end{array}\right) and \bar{x}=\left(\begin{array}{c}t\\ \vec x\end{array}\right) are invariant 4-vectors in spacetime. A 4-vector is composed of a time component and a spacial component (with 3 sub-components related to ordinary 3-space). By invariant we mean that their magnitudes ... 0 There are already a few answers that explain the mathematics behind it, but since you've said that your "math knowledge is quite limited" I'll try and break it down into simpler terms. You're already familiar with a 3D vector dot product, and it seems your confusion arises from dot products of a four-vector. Now what they didn't teach you when they taught ... 0 In Minkowski spacetime you assign four coordinates to your events: x=(x^0,x^1,x^2,x^3) - in this notation x^0 =ct and the other three coordinates are the spatial ones. Suppose these are the coordinates of a point in spacetime reached by a light ray which started at the origin, then you have that: (x^0)^2=c^2 t^2 = \sum_{i=1}^3 (x^i)^2 \Rightarrow s^2 = ... 1 That's the definition of the dot product in Minkowski space-time. To be clear, any space-time is endowed with a metric. Standard {\mathbb R}^3 that you may be familiar with has a metric \delta_{ij} = \text{diag}(1,1,1), i=1,2,3. Given two vector \vec{v} = (v^1,v^2,v^3) or v^i for short and similarly w^i, the dot product is defined as$$ \vec{v} ...

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Is space-time warped around the spaceship? No, not measurably. The time dilation effect does not arise from any curvature of spacetime. Space-time is warped by gravity (the stronger the gravitational field the slower time goes). The parenthetical statement is not really a correct description of what spacetime curvature is. Spacetime curvature is ...

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I tink your question is mostly answered by the answers to the question Universe being flat and why we can't see or access the space "behind" our universe plane?. The rubber sheet analogy gives the impression that there is an extra dimension in which spacetime is curved. Curvature in an extra dimension is called extrinsic curvature. However in ...

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Eddington–Finkelstein coordinates use the same position coordinates as Schwarzschild coordinates, only the time coordinate is transformed, so first consider how to define Schwarzschild coordinates in a physical way. This pdf explains a way of defining the position coordinates in section 9.1.1: • We may assign a practical definition to the radial ...

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Coordinates are not physical. They are entirely up to you. In the Eddington Finkelstein coordinates, the lines of constant u theta and phi are null radial lines going out to infinity while r is the coordinate such that the surfaces swept out by the spherical symmetry have an area of 4 pi r^2. (Ie, take a point of the spacetime. Apply a spherical symmetery ...

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From the photon's perspective, there is no distance between the quasar and your eye. There is thus no path for it to have followed: for it, the curvature of space-time is not even a straight line; it is a single point. There is no photon's perspective. See Does a photon in vacuum have a rest frame? . Any particle which has mass can follow the same ...

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You don't need a massless photon to (at least asymptotically, in a limiting sense) make up zero spacetime distances. The faster a massive particle moves, the less spacetime it travels (not space alone, not time alone, but spacetime). Put it mathematically, $$E^2=p^2c^2+m^2c^4$$ and you can see that the value of m gets overshadowd by a large momentum, when ...

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You're considering this in the noninertial frame of the earth, which makes it more confusing. In GR, we consider free-falling frames of reference to be the inertial frames. In such a frame, the ball's center of mass follows an inertial path, which looks like a straight line in spacetime. Meanwhile, the spot painted on the ball follows a parallel path ...

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Which rotational state? When the ball goes up, then falls, it obey the energy conservation. It goes up to a point where all its energy becomes potential, then returns and its potential energy transforms into kinetic. From the beginning the ball has no rotation movement, its angular velocity is zero, and so the energy of rotation. There is complete ...

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Have a look at my answer to How to explain centripetal force in terms or relativity because much of the discussion there is relevant. Consider what we mean by a tidal force. Suppose you're floating around in space and you arrange a number of marbles around you so they lie on the surface of a perfect sphere. Now monitor the shape of the surface marked out by ...

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E=m*c**2 is not the defining equation of relativity. The theory is called special relativity and the equation is a derived part of the results of the theory. It is the result of Lorenz transformations on moving systems, which do take care of space and time in addition to energy.

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In special relativity, mass / energy has no influence on spacetime. However, in general relativity, the curvature of spacetime is directly related to energy, or equivalently, mass. The Einstein field equation $$R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu}$$ includes the stress-energy tensor, which ...

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