# Tag Info

1

For (1), the event $a'$ is co-located in time, i.e., simultaneous with event $a$ and thus the associated interval is space-like (the distance through space is greater than the distance through time in any frame). There is no world line (time-like curve) that includes the events $a$ and $a'$ so, the event $a'$ cannot be the event "receive signal initiated at ...

2

Calculate the spacetime interval $$\Delta s^2 = -c^2\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2\text{.}$$ In this sign convention, $$\begin{cases} \Delta s^2 < 0\text{,}& \mbox{timelike separation}\\ \Delta s^2 = 0\text{,}& \mbox{lightlike separation}\\ \Delta s^2 > 0\text{,}& \mbox{spacelike separation}\\ \end{cases}$$ Someone ...

0

No, there is no violation. Both of the observers are in inertial reference frames, and Newton's Laws of Motion hold in inertial reference frames. The only problem would be with *non-inertial * reference frames, but there is no problem now. The difference in time would be possible, especially if the observer and object are travelling in opposite directions.

0

Actually, third law won't apply solely to particles in Special Relativity because you need to include the field into conservation of total momentum.

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So If I am stationary, the moving observer may say that my measurement is wrong that means Newtons third law is violated? Is it right? No that isn't right. Each observer is in an inertial reference frame and each observer will independently see that Newton's third law applies.

3

but they assume a "pull down" force themselves. The images of flat sheets "pulled down" where the planets are do not reflect the fact that the curvature of spacetime is an intrinsic curvature that is measured by geodesic deviation. What has been done, in order to help visualize the spatial curvature, is to take a two dimensional spatial slice and ...

2

Massive objects distort spacetime, as described by the Einstein Field Equations. In turn, this causes particles to accelerate: the GR equivalent of $\mathbf{F}=m\mathbf{a}$ are the geodesic equations: $$\frac{\text{d}^2x^\alpha}{\text{d}\lambda^2} + \Gamma^{\alpha}_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\lambda}\frac{\text{d}x^\nu}{\text{d}\lambda} = 0,\qquad ... 3 Why don't you want to assume gravity? Gravity it is an experimental fact, a starting point for doing physics. General Relativy is a geometrical theory of gravity, built on the basis of Special Relativity and always having in mind that it should recover the non-relativistic Newtonian theory of the gravitational field. The "pull down" is a deviation of the ... 0 You don't have to look at other frames of reference to find out if yours is an inertial frame. With respect to your frame of reference, that is using your cartesian coordinate axes and your clock, if you determine that Newton's first law is valid then yours is an inertial frame of reference. Let's consider two examples. (1) You are sitting inside a ... 0 I'm not asking for a definition of a tangent vector. I'm asking what criterion you can use to decide whether a certain object can be described as a tangent vector. For example, how do we know in this coordinate-free context that the four-momentum can be described as a vector, but the magnetic field can't? If I understand your clarification correctly, ... 0 Mathematicians have their axioms to define what a vector is, physicists start with a vector as a physical quantity that has a magnitude and a direction. Or at least, this is how Feynman defines it in volume 1, 11-4 of his lectures on physics. These two properties belong to the object and can't possibly depend upon the coordinates used to label them. Edit: ... 3 There are 4 common definitions of tangent vectors, some of which make use of coordinates only casually or even not at all. Definition via transformation laws There's a somewhat technical one preferred by some physicists (those who value calculation rules over geometric insight - shut up and calculate, you probably know the type): A vector is just an ... 1 From MTW's "Gravitation" (via Google Books): Updated answer to edited question: For example, how do we know in this coordinate-free context that the four-momentum can be described as a vector, but the magnetic field can't? I'm reminded of a relevant section from "A First Course in General Relativity" by Schutz. In section 4.4 on the ... 3 Honestly, this coordinate-free GR stuff (Winitzki's pdf in particular) looks like GR as would be taught by a mathematician--very similar to do Carmo's text on Riemannian geometry. In classic (pseudo-)Riemannian geometry, vectors are defined as derivatives of affine parameterized curves, covectors as either maps on vectors to scalars or as gradients of scalar ... 0 About faster than light... I know (in fact I am currently yet studying) different extensions of relativity. Some options naturally arise: 1) Yes, Ben... Sudarshan's (and Recami's) Meta-relativity is one "option", somewhat oldfashioned. Problems: tachyons have not been observed in Nature yet. metarelativity paper metarelativity paper 2012 2) Carlos ... 1 Nice question. I don't understand the Lorentz-violating possibilities very well, so I'll only try to comment on Lorentz-invariant theories. The classic papers are Tolman 1917, Bilaniuk 1962, and Bilaniuk 1969. Bilaniuk 1969 can easily be found online by googling, and gives a good overview. Tolman proposed a causality paradox involving tachyons, known as ... 0 Suppose two bicycle chains on sprockets and a frame in a J-shape were pulled inwards in a mirror image of each other? They would cancel out both linear and rotational forces and mass would be transfered linearly "round the U bend" without a linear opposite reaction(?) The structure could then be rotated to a new position, and the process repeated. 1 So at atomic level for same charges that repel eachother(electrons etc.), if thought of as a slingshot effect, elastic repulsive collision and buoyancy fields etc. makes repulsion a pseudo or dummy. as for like gravity is doing it all, then it will be more insightful or not? Not. To describe all the available data the theory has to be much more ... 2 The slingshot effect is in fact only a transfer of angular momentum from one body to another. So for example if you're using Jupiter for a swing by manoeuvre with your satellite, you do nothing more than slow down Jupiter (in the Sun-Jupiter-(satellite) system) and speed up your satellite. Of course because of the huge difference in mass, you only see a ... 2 Let's recap: upper indices are vectors (x^\mu), the inner product on Minkowski space is given by g_{\mu \nu} so "dual vectors" have lower indices x_\nu = g_{\nu \rho} x^\rho. Then you see that a matrix (in the sense of linear map between vectors) has one upper and one lower index, because it maps a vector to another vector:$$x^\mu \mapsto ...

-3

When scientists say the escape speed of a black hole is greater than the speed of light, they mean that if you leave the surface of black hole at a speed which is near the speed of light, and during the trip you do not get extra boost of speed, then eventually you will fall back. That is absolutely true. But if you get a speed boost during the trip, you can ...

1

Relativity consists of special relativity (SR) and its generalization general relativity (GR), which includes gravity. In SR, the history of an object is described by its world-line through spacetime. Every event has future and past light cones, and a material object such as a person's body is limited to "traveling" from the past light cone to the future ...

0

Remember that the situation you describe is exactly the same as you remaining stationary while the mirror moves past you at relativistic speeds. The reflection from a moving mirror is analysed in this article. If the mirror is parallel to it's direction of motion ($\phi$ = 0) the normal rules apply i.e. the angle of incidence is equal to the angle of ...

0

Plot the imaginary copy of yourself on the other side of the mirror, by the laws of geometrical optics, for every moment of time. *) Then consider that copy a real body you observe, and apply all known relativistic effects to its apparent image, taking into account both your and its motion. In the case of a parralel flat mirror and uniform motion, all those ...

2

Yes, the answer is actually very simple: While you increase the speed, the required amount of energy increases - because with the speed, the objects mass increases. And, to get to the light speed, you'd need infinite amount of energy, and the object itself would have an infinite mass. You may know that photons, which do move with lightspeed, have zero ...

7

One way to think of a "moving shadow" is by following the last photon that was allowed through. In that case, the speed of a shadow is exactly the speed of light. On the other hand, you could also define the speed of a shadow as the speed of the boundary between dark and light. In that case there is no thing that's actually moving, so there's no bound on ...

2

A scalar is invariant under rotation A pseudo-scalar is also invariant under a proper rotation but changes sign with parity. A vector is not invariant under a general rotation (only invariant under rotation around a rotational axis parallel to the vector), but rather transforms according to multiplication with a rotation matrix.

0

Something moving such that it "measures the proper time" is simply something moving from point 1 at time 1, to point 2 at time 2, as seen in S (you have the data). Recall the basic definition of velocity.

1

I don't know a definitive answer to your (really good) question, but here is a quote from an old textbook I have by Christian Moller ("The Theory of Relativity"): Shortly after the advent of the relativity theory, Planck, Hassenoerl, Einstein and others advanced separately a formulation of the thermodynamical laws in accordance with the special ...

0

My wild guess (no offense, though): (0) it's not meant entirely as a trick question, but (1) there's a small but nevertheless important mistake in how you understood and presented the statement of the problem; because it seems a lot more sensible if the specification "(b)" is instead: (b) At the moment the green (left) car passes Marvin, the ...

0

Although I'll end up with the same answer as Stan I think it's nice to see how to do this using the Lorentz transformations. We'll take the observer to be moving left to right at velocity $v$, and the front of the train is at $x$ = 0 and the end at $x$ = 500m. I'll call the length of the train $L_0$ for consistency with Stan's answer. The events marking when ...

0

You should know that Lorentz-FitzGerald contraction is (in units of $c = 1$): $$L = L_0\sqrt{1-v^2}\text{,}$$ where you are given that $L_0 = 500\,\text{m} = 1.6678\,\text{$\mu$s}$ and that in the inertial frame the contracted length $L$ passes by the observer in $\Delta t = 780\,\text{ns}$. Therefore $v = L/\Delta t$ and: $$L^2 = ... 0 This is a trick question. All the measurements you are doing are in Marvin's frame, so no transformations to any other frames are required. The time required is just the distance in Marvin's frame divided by the relative speed in Marvin's frame. 1 I actually attended a lecture recently on a problem similar to this although the stress energy tensor we were handed was slightly different (possibly due to generalization to curved spacetime with metric  g_{\mu \nu} ). We had that:$$T_{\mu \nu}=\partial_{\mu} \phi \partial_{\nu} \phi - \mathcal{L} g_{\mu \nu}  The Lagrangian used is that of GR (In ...

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