# Tag Info

4

Yes. I think Randall Munroe put it perfectly in this comic: The rubber sheet analogy does not tell you much about actual gravity.

3

Well, I suppose we could add time as a 4th. However, apart from that there is no experimental evidence whatsoever for more than 3 spacial dimensions. String Theory and its friends remain theories with negligible experimental evidence in their favor.

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The black axes give the frame of the meter stick. The black vertical axis is the worldline of the left end of the stick and the parallel black line is the worldline of the right end of the stick. Your condition 1) says that the spatial distance from A to B, measured in the black frame, is shorter than the spatial distance from A to B, measured in the ...

3

Comments to the question (v1): In Newtonian mechanics with Newtonian gravity, a body can have orbital angular momentum wrt. a reference frame. A non-point-mass can also have spin angular momentum. Bodies can exchange angular momentum via tidal forces. In GR, it possible to assign angular momentum to certain space-time regions (such as e.g. the Kerr ...

2

Yes, the statement is correct. One proof I am familiar with was given by V.Fock (of Fock states, etc) in his book "Theory of Space, Time, and Gravitation". See Chap.1, Sec.8 therein (pg.20, bottom). In modern notation the idea is as follows: In inertial frame $S$ parametrize straight lines as ($x_0 = ct$) $$x_i = \xi_i + \beta_i s, \;\;\; i=0,1,2,3$$ ...

2

I found the Weinberg passage, but to quote it I need to do it in an answer (too long). So here it goes. We have seen in this chapter that the nonvanishing of the tensor $R_{\lambda \mu \nu \kappa}$ is the true expression of the presence of a gravitational field. We also saw in Chapter 1 that Gauss was led to introduce the Gaussian curvature $K = -R/2$ as ...

2

What you have done here is a Galilean transform, that is a non-relativistic transformation. Take your final result (which is quite correct): $$t' = \frac{\sqrt{\beta^2 + \alpha^2}}{\sqrt{\eta^2 + \mu^2}} \tag{1}$$ We know that the vertical velocity is $\eta$, so the vertical distance moved in our time $t$ is given by: $$\beta = \eta t$$ We also know ...

2

All of your questions have no good answer at this point in time. All of them are being researched by physicists, cosmologists, and theorists. We don't even know whether our universe is the only universe or whether what lies beyond the visible universe is just more space like the kind we can see. Our best theories of space and time (Einstein's General ...

2

Questions like this are complicated because you have to be clear what you mean by time. The simplest definition is that time is what is shown on a clock, so if I was holding some hypothetical clock that had been reset to zero at the Big Bang my clock would currently be showing 13.799 billion years i.e. the age of the universe. The question then becomes what ...

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You need to accelerate and slow down the flywheel. In order for the system to be self contained, the energy has to come from within your device, The total energy of the device does not change with the accelerating flywheel since the energy has to come from something else in the device.

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Alice makes the observation of her choice. Bob makes the observation of his choice. The pair of observations has an outcome with probability distribution determined by the initial joint state of the two particles. The spacetime locations of the two observations, and the states of motion of the observers (relative to each other or anything else) have ...

2

I can't say that I have ever seen any attempts at simulating non-causal spacetimes (the closest I've seen is the simulation of fields upon such spacetimes). A few non-causal spacetimes do admit a time slicing, by the way, although by definition not all of these slices are achronal. Just solving it like any other PDE might be an avenue worth exploring, but ...

2

When you talk about a point in space, you're talking about a specific set of $(x, y, z)$ coordinates. Of course there's no use to talking about a point in space unless something is happening there, e.g. $(0, 6, 0)$ is the cannonball's starting location". An event is the same idea in $3+1D$ spacetime- it's a specific set of $(t, x, y, z)$ coordinates. ...

1

The problem is that no "intuitive" explanation can capture what gravity is actually about, because if it could, then general relativity should itself be intuitive. The rubber-sheet analogy is in my view not a totally misleading analogy for what it wants to show (namely that masses curve spacetime), but it tackles the wrong problem - the main problem being ...

1

Relativistic mass will not increase the gravitational pull, the gravitational force depends on the rest mass of an object.

1

As ACuriousMind stated in his comment, light also follows geodesics. For a Lorentz-metric, such as the metric of spacetime in GR, $g_{ab}$, a tangent vector $X^a$ is spacelike if $g_{ab}X^aX^b>0$, timelike, if $g_{ab}X^aX^b<0$ and lightlike or null, if $g_{ab}X^aX^b=0$ (assuming $(-+++)$ signature, if opposite signature is used, then these signs are ...

1

Let's suppress some dimensions to simplify: $$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2$$ This quantity $$\Delta s^2$$ is preserved by changes of reference frame, just as in Galilean physics the quantity $$\Delta r^2 = \Delta x^2 + \Delta y^2$$ is preserved by rotations. Notice it is also the equation of a hyperbola. Thus, the effect of a frame shift is ...

1

A higher speed does not equal a higher mass. This is something commonly taught to beginning students of relativity as an explanation of relativistic momentum because it is easy to understand (but does lead to misconceptions). Also, momentum must be transferred to slow down the flywheel.

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Time dilation: linear or exponential or other? Other $$\Delta t' = \gamma\Delta t = \frac{\Delta t}{\sqrt{1 - \frac {v^2}{c^2}}}$$ Lorentz factor $\gamma$ as a function of speed (in natural units where $c=1$) - Image by Zayani CC BY-SA 3.0

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The world lines exists independent of the frame you choose. That is, Minkowski space-time is an affine space (like the euclidean space $\mathbb E^n$, not to be confused with $\mathbb R^n$) where there are no frames. Here you can "draw" world lines, and doesn't matter that there is none inertial frames yet. Then, when you select the frame you are actually ...

1

One thing you have to note is that speed is relative, Clock A would see clock B moving from A's point of reference, and B would see A moving in B's reference, so you shouldn't be using the word "stationary" in this context. Both the clocks would see the other clock tick slower, B would see A's future only if it returns back to A, this makes it obvious to A ...

1

No. Photons do not experience the passage of time, as they are traveling at the speed of light. Remember when they discovered that neutrinos must have mass? Originally it was thought that neutrinos traveled at the speed of light, but then it was discovered that neutrinos change their flavors over time, which means that time must pass for neutrinos, which ...

1

I don't know the "formal" proof, but here is my proof: Time dilation and length contractions are given to us by the Lorentz transformations by: t’ = t/(1-v2/C2)1/2  and d’ = d/(1-v2/C2)1/2 (in other words “same” or proportional to each other) where: t = distance/length traveled through the T dimension in observers own frame of ...

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There is no %100 proof in science; at least not for good science. It's always a question of being the most accurate / descriptive / useful theory. For example, Newtonian gravity is 'true' to the extent that it is very effective in a huge diversity of situations. General Relativity (GR) includes all of the accuracies of Newtonian Gravity, and then also ...

1

There's a very real phenomenon called 'Gravitational Lensing', in which light is bent from its original trajectory by a massive enough cluster of matter (which curves the space-time around it). Moreover, it's bent by a different amount than predicted by a simply application of Newtonian ideas, as kindly pointed out by Rob Jeffries. Is this evidence enough? ...

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No, the straight beam will not magically turn into a curved beam. I suspect you have a slightly confused idea of what the curvature of spacetime means physically. Basically it means that a freely moving body will appear to accelerate relative to some distant observer. Conversely if we want stop the body from accelerating then we have to apply a force (i.e. ...

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