# Tag Info

66

The speed $c$ that is constant is so when measured locally relative to a freefalling frame (i.e. one for which all points follow spacefime geodesics wrt to the metric $g$). Local means that the frame's extent must be "small" enough that it can be thought of as flat: think of this as zooming in on the spacetime manifold, which is a smooth object, with enough ...

61

The time delay depends on the direction the wave is travelling. If it is travelling along the line connecting Livingston and Hanford then the delay time would indeed be the Livingston-Hanford distance divided by $c$: However suppose the wave was travelling normal to the line connecting the two detectors. In that case the wave would arrive at both of them ...

33

Suppose you are floating in a river, and you have with you a model boat, called the SS Lightray, that can do 3 m/sec through the water. When you set the boat travelling upstream as far as you're concerned it is doing 3 m/sec. But I'm standing on the bank watching the river flowing at 1 m/sec, so when I look at your boat I see it travelling at a net speed of ...

16

It didn't. The key is that the wave is not point-like, it has an extended "wavefront". The wavefront/event just arrived at the two locations with that delay, which gives some information about its direction. You might also be interested in: Is it really possible to break the speed of light by flicking your wrist with a laser pointer?, which discusses ...

11

Imaginary things can "travel" faster than light A shadow or a light spot can seem to travel faster than light, because it's not a particular physical thing, but a series of separate things, separate physical particles emitted at different time and at different locations. Imagine that you have launched a lot of tiny bots into space with a very accurate ...

7

I think you've said a lot of things that are correct, you've just come to the wrong conclusion. You're right that that it will take us three seconds to SEE the shadow moving across the moon, because that's how long it takes the light to get there and back. But what we'll see is not the shadow slowly moving across the moon over a period of three seconds. ...

6

To explain it in layman's terms, without using advanced concepts: Space is warped in the "inside" of the black hole (that is, under the event horizon) so much, that it behaves completely different than what we perceive hear on Earth. The "outward direction" simply does not exist. For example, here on Earth, we can go in the three spacial dimensions in both ...

5

The force between the charges goes to zero. To see this, work in the frame of one of the charges. From its perspective, the other point charge is moving rapidly away, and the field of a moving charge is weaker along the direction of motion, as shown below. One cheap way of seeing this is to pretend the field lines have been "length contracted". For ...

5

This is an exact value. Meter is defined by the speed of light. Meter is a distance that the light travels during $\frac{1}{299792458}$ of a second.

4

To understand this you need to understand what we mean by speed. If I want to measure positions and times I need to set up a coordinate system. For example I can take my stopwatch and my metre rule and construct some Cartesian axes $t$, $x$, $y$ and $z$, then I can describe every point in spacetime by its position in my coordinates $(t,x,y,z)$. Once I have ...

4

Fizeau in his paper page 92 wrote the following (rough translation) with regard to his experimental set up: The first telescope was placed in the belvedere of a house [an architectural structure sited to take advantage of a scenic view] in Suresnes , the second on the hill of Montmartre, a distance of approximately 8633 meters The disc with seven hundred ...

3

What kind of information are we talking about in superluminal information transport? It might be more intuitive to call it superluminal communication. For the prohibited kind of superluminal information transport, you need to be able to achieve the following. At the start of the protocol, A holds a classical bit which either has value 0 or 1, and which is ...

3

To a very rough approximation we can say that frequencies of speech are selected by standing waves in the speakers mouth, larynx etc. If they breath helium the speed of these standing waves increases but their wave length, being constrained by dimensions of their body, remains the same. This results in higher frequency sounds produced. (think $f=v/\lambda$...

3

Not through our normal sight. First of all, doppler shift will totally change the color of everything, objects ahead of us shifting towards blue/ultraviolet, and these left behind becoming more red/infrared. Then, Lorenz Contraction combined with changed time and distance it takes light reflected off objects (and separately, emitted by objects) to reach our ...

3

In a vacuum all frequencies and amplitudes of light travel at the same speed of c = 299 792 458 m/s. Frequency is equivalent to colour. Amplitude relates to intensity. When light travels in material mediums (air, water, glass, etc) it travels at a slower speed v < c which depends on frequency. The ratio of c/v is what we measure as the refractive ...

3

It is known that moving a lantern will result in a spot or projected light that travels faster than light is the screen is far enough. However, this effect and similar ones do not refer to the motion of actual objects at faster than light speeds, nor this process allows transfer of information faster than light.

3

The BIPM (Bureau International de Poids et Mesures) defines the meter as the distance traveled by light in $\frac{1}{299792458}$ seconds. The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second. The speed of light as $299792458$ m/s is therefore exact and not a measured value. Similarily, ...

3

To see what's going on, it's enough to do this in two dimensions, with the Lorentz form $\pmatrix{-1&0\cr 0&1\cr}$. (I've set $c=1$.) The Lorentz group is the group that preserves this form. A typical element is $$\pmatrix{\pm\sec\theta&\tan\theta\cr \tan\theta&\pm\sec\theta\cr}$$ where $\theta$ runs through the open interval from $-\pi/... 3 That light moves with a fixed speed in vacuum, in all reference systems is an experimental fact. Maxwell's equations fit so well all macroscopic electromagnetic data that the speed of light is fixed is not under question. It is inherent in the construction of the classical theory. Light is made up by a zillion of photons. Photons are elementary particles in ... 2 That is the case for central gravity fields. You can derive Newtonian gravitation from a spacetime metric $$ds^2 = c^2d\tau^2 = c^2\left(1 - \frac{2GM}{rc^2}\right)dt^{2} - dr^2 - r^2d\Omega^2$$ where the only metric element different than unity is$g_{tt} = c^2 - GM/r$. The relevant Christoffel symbol is$$\Gamma^r_{tt} = \frac{1}{2}g^{rr}\frac{\partial ... 2 One may calculate the time that light (electromagnetic waves) need to go through the arms of the interferometer. If the speed of light relatively to the aether is$c$, then$c$is also the approximate speed through the arm that is approximately orthogonal to the velocity$\vec v$of the Earth (or interferometer) relatively to the aether. Well, the speed is a ... 2 To answer this, You can't add velocities as in Newtonian mechanics. In relativity, If frame B is moving at velocity u and C is moving at velocity v both with respect to an inertial frame A, then relative velocity between B and C is$\frac{u+v}{1+ uv/c^2}$. So substituting your arguments here would now leave us with relative speed less than c. Just because ... 2 The observed wavelength does change, and this is called the doppler effect. But the speed does not change. The statement "...the wavefront of a short pulse emitted like this will reach a given distance along y before it will along x" does not follow from any logical reasoning. Following the same logic you will conclude that when the source is next to the ... 2 According to all observational evidence (including the original Michelson-Morley experiment) speed of light is constant in all directions. The confusion comes from misinterpretation of the picture you attach. I propose to understand it as follows. Say, you have a lightsource that emits pulses at a given frequency. Each pulse propagates at a constant speed ... 2 No. It means that the energy content of a particle at rest whose mass is$m$is equal to$mc^2$. In general,$E=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}$represents the energy content of a particle of mass$m$moving with a speed$v$. If it is at rest, i.e.$v=0$, then$E=mc^2$. 2 No. It means that when one is accounting for the total energy of the system one must account for the energy contribution of the mass. It also means that the mass isn't necessarily conserved in a system, while energy and momentum will be. In special cases, energy and/or momentum may even be constant in an defined subsystem. The quantity$E^2- p^2c^2$of a ... 2 No, it doesn't. It just means that particle of mass$m$can be converted to$mc^2$of energy. You can find a good popular article on that topic here: https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence 2 Keep in mind that frequency is both the wave property that is preserved in a change of medium (both wavelength and velocity change) and the physical property of sound waves that we experience as pitch. So the frequencies you hear in both cases are ones produced by the vocal cords. Nor do we expect the gas environment of the vocal cords to have a large ... 2 Let's be a little more clear about what we mean here. We can go down to a radioactive beam facility (these exist), or to a muon storage ring like Muon g-2 (now at Fermilab), or even climb a steep mountain and use cosmic muons (Halfdome is a scenic location, let's imagine doing it there).1 In any case, call the lifetime of the beam particles$\tau$. (For ... 2 The answer is that it has nothing to do with light, c, black holes, event horizons or relativity. It is simply escape velocity. We know that two bodies attract each other with a force$f = G\frac{M m}{d^2}$and we know that something is in a stable orbit when its potential energy$PE = -GMm/d$matches its kinetic energy$KE = m{v^2}/2 \$. Solve these two for ...

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