# Tag Info

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If the Lagrangian is non-singular so that the Legendre transformation to pass from Lagrangian to Hamiltonian formalism is well defined, the answer is Yes. If a quantity is conserved in view of Noether's theorem in Lagrangian formulation, passing to the Hamiltonian formulation it turns out to be the generator of a canonical transformation that preserves in ...

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As Kyle says, $\nu$ is just a (free) index. You can use any letter. More precisely, $x^\mu$ is the $\mu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. And $x^\nu$ is the $\nu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. So you can see that the vector is the same, $\mathbf{x}$, you just name the components with a different ...

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Let me present a slightly different perspective to Alfred's answer, although I'm basically saying the same thing. I suspect you've got hung up on the idea that velocity causes the relativistic effects like time dilation, but the underlying cause is something different. All the weird effects in SR are caused by a fundamental symmetry of spacetime, which is ...

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We need to untangle this a bit but first: the cause of time dilation is the geometry of spacetime which is such that there is an invariant speed c. Now, remember that velocity or speed is not a property of an object; there is no absolute rest. Further, consider the case of three objects in uniform relative motion with respect to each other. If I choose ...

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The problem is that when you write $r$ and $t$ in the equation for the time dilation you are using the Schwarzschild radial and time coordinates, which are part of a system of coordinates that works well far away from the black hole but fails at and within the event horizon. To make things a bit clearer we'll rewrite your equation as: $$t_0 = ... 5 The value r = \frac{2GM}{c^{2}} defines a special surface in the Schwarzschild spacetime called the event horizon. Observers inside this radius cannot be stationary with respect to points very distant from the black hole, and they cannot communicate with any observer outside this radius, so the notion of time dilation doesn't make sense for them. 7 If you define "now" to be all those points in space and time that have hypothetical, pre-synchronized, stationary clocks that read the same time as your clock, then there "currently" exists a hypothetical observer somewhere, who is moving relative to us, for whom "now" includes Earth, circa 1900. But these notions of "now" are different for the two ... 1 The first statement is very much true. Light moves a finite, if very fast, speed. Even ignoring any movement/ relativistic effects this simply means that observers closer to earth will see it in it's most recent state. It may sound strange for light, but we see exactly the same phenomenon in sound, an observer noticeably closer to the source of a sound will ... 0 No. Not that we know of unless we change the current laws of physics to allow faster than light travel or imaginary mass 2 Is there any significance in saying an observer as an imaginary entity? Yes. From Wikipedia: Physicists use the term "observer" as shorthand for a specific reference frame from which a set of objects or events is being measured. Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is ... 3 Your teacher is correct that the mass of an object if it is moving with very high energies appears to increase according to the formula , it is called the "relativistic mass" . Where E is the energy of the particle and c the velocity of light. But each elementary particle ( these are concepts that apply to elementary particles to start with) is ... 0 Short answer: If you're talking about the spatial components of force, then yes, through a Lorentz transformation on the force four-vector. By "real force" I assume you mean a non-inertial force, so that you are computing force from an inertial frame, or, more generally, in a freefall frame - i.e. a locally flat Minkowskian tangent space to the spacetime ... 0 Well, take the electromagnetic force...it has been shown that the induced magnetic field around moving charges is a relativistic reference frame effect. See this post How Special Relativity causes magnetism Therefore a static electric force in one frame becomes a magnetic force in another. However, from a Newtownian view, we either allow pseudoforces and ... 3 As suggested in the comments, "lowering an index" is just coordinate notation for the isomorphism \flat:TM\to T^*M between defined by$$X^\flat (Y) = \langle X,Y\rangle\text{,} where $X$ and $Y$ are arbitrary vectors. I've tried using the definition of the metric ${g_{\alpha\beta}=\hat{\mathbf{e}}_\alpha\cdot\hat{\mathbf{e}}_\beta}$ where ...

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Also I've searched for it in books like Carroll's or Lawden's, but it's given pretty much as if it would be a definition. Because it is. No need for differential geometry, linear algebra is sufficient here: At a given point of space-time, the tangent space is just a vector space, the cotangent space its dual (ie the space of real-valued linear ...

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Not to mention that the force diminishes with distance, making small changes more useless the further you seek to communicate

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At the moment, it's impossible to use gravitational waves for communication purposes. If no gravitational waves have been observed, how do yuo expect to use them?

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