# Tag Info

4

By using orthogonal optical resonators, laboratory tests concerning verifying the isotropy of c have come a long way. As quoted from http://journals.aps.org/prd/abstract/10.1103/PhysRevD.80.105011 "An analysis of data recorded over the course of one year sets a limit on an anisotropy of the speed of light of $\Delta c/c \sim 10^{-17}.$ This constitutes the ...

0

If Einstein's 1905 light postulate is false (that is, if the speed of light varies with the speed of the emitter), then the question makes no sense. "Varies with the speed of the emitter" is equivalent to "varies with the gravitational potential, like the speed of ordinary falling bodies". The Pound-Rebka experiment has confirmed the latter assumption: ...

0

Further to Goodies's answer, there are two easy derivation of why. If you can see how Goodies's answer bars any massive particle from passing through the speed of light, then you need a simple derivation of why Goodies's formula is true. Method 1: The 4-position of an event in spacetime is $(c\,t,\,x,\,y,\,z)$. To get the four velocity, i.e. something that ...

3

For a massive particle to move at, or faster than, the speed of light, it would require infinite energy as shown by Einstein's relativistic equation: $$E = \gamma \cdot mc^2\quad\left(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)\\ E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\\ v\rightarrow c, E \rightarrow \infty$$ If we plug in $c$ for the velocity, ...

1

"Using the equation of the time unit I derived for time, we can say now a unit of our time for an observer will be the distance (light will cover) for a unit of time to pass for us (observer)" Is this observer meant to be the one who sees the object with the walls A and B moving at velocity v? If so, then although it's true this observer will see the length ...

20

I am wondering whether is it taken as a postulate or a proven phenomenon that c is constant irrespective of observer's speed? Either one. Both. Einstein took it as a postulate in his 1905 paper on special relativity. From it, he proved various things about space and time. The frame-independence of $c$ is also experimentally supported. This is what the ...

2

It is a well-substantiated observed phenomenon. Science deals only with provisional truths, but this hypothesis has undergone (and passed) immense amounts of scrupulous experimentation and mathematical formulation. In a Neo-Lorentzian interpretation, physics works differently in all reference frames except for one single, undetectable, privileged reference ...

1

Firstly, you need to calculate how much your hypothesised effect will change the optical delay in each of the interferometer's arms and check that you expect to see any result with your proposed experiment. Otherwise put: what are the specifications of the interferometer (arm lengths, light source requirements etc, vibrational tolerances) that will let you ...

1

I'd like to add to Chris White's excellent answer and summarise things thus: $c$ is a number that parameterises the family of all possible linear transformations that follows from Galileo's relativity with the assumption of absolute time relaxed. If you do Galileo's relativity with the assumption of an absolute time (that two relatively moving ...

6

Long ago, English speaking sailors measured horizontal distances in units of nautical miles but depths in units of fathoms. The distance in fathoms to some point $z$ fathoms deep and $x$ nautical miles along the surface is $d^2 = 1012.68591^2 x^2 + z^2$. There's nothing physical to that factor of 1012.68591. It's solely a result of using inconsistent units ...

5

The quantity $c$ is a very fundamental constant related to space and time. It is largely independent of the existence of matter or light or any other substance. It is probably more ontologically appropriate (if not pedagogically appropriate) to define it as the number that makes Lorentz transformations between reference frames work. The quantity $c$ would ...

3

This is a very complicated question and a complete answer would be very deep and very, very long. Much of it stems from the fact that the speed of light is independent of the frame of reference from which it is being observed. In this sense, it is one of few "universal constants" - that is, quantities which do not depend on the observer. Since it has units ...

0

Modified @Raskolnikov’s answer (hence community wiki because insufficient part of Incnis Mrsi to claim authorship): It is not “deduced from Maxwell's equation” alone, as a purely mathematical consequence, but also relies on surrounding physics. Maxwell's equation and wave equation for sound (even if we ignore P/S dispersion in solids for simplicity, as well ...

-1

A logical analysis of absolute motion ongoing within an absolute 4 dimensional Space-Time continuum, leads you to the awareness of the constant motion of all objects. That constant motion turned out to be the speed of light. If one analyzes the concept of constant motion of all objects located within Space-Time, and that this ongoing motion is the equivalent ...

0

Unless you have zero mass, you cannot reach the exact speed of light. But lets work with the hipothesis that you are just below light speed. Two things would happens: Doppler effect: light coming towards you would get blue shifted. That means you would be unable to see a blue object coming towards you, since you would see it as UV radiation or, depending ...

5

Square the proper time $d \tau = dt / \gamma$, multiply by $c^2$, rearrange, and take the square root: $$\left(\frac{d\tau}{dt}\right)^2 = \gamma^{-2} = 1 - \left(\frac{\bf{v}}{c}\right)^2 \Rightarrow \sqrt{\left(c \frac{d\tau}{dt}\right)^2 + {\bf{v}}^2} = c$$ The proper time $d\tau$ is the time elapsed in the frame moving with respect to the lab frame, in ...

3

I realize that in essence there is no object which can be considered as "not moving in space". No object at all is moving in space if you are taking the point of view of its reference frame! The law of conservation of energy is requiring that the energy of its mass (e = mc2) is "transported through time", or in other words, that time is passing for ...

4

Yes it is - well, sort of. The coordinate invariant form of velocity is the four velocity, and the magnitude of any four velocity is always $c$ (or $1$). So even if you are stationary in space in your chosen coordinate system the magnitude of your four velocity is still $c$. Whether moving at the speed of light on the time axis is a good way to state this ...

1

The fact that different observers in relative motion can measure the same light ray to move at a speed of c has to do with the fact that each observer defines the "speed" in terms of distance/time on rulers and clocks at rest relative to themselves. It's crucial to understand that different observers use different rulers and clocks to measure speed, because ...

0

The problem is that your analysis is all done from the perspective of the frame that measures the box to be moving at 0.1c -- in this frame, it's true that the time for the light to get from the source to the wall is different from the time for the light to get from the wall back to the source. But if these same events are measured by someone inside the box ...

0

It appears from your edit that you added the velocity of the ship and the speed of light classically. However, the velocity of the ship and the speed of light need to be added relativistically, using a Lorentz Transformation. Here's a quick way to know something is amiss with your moving ship: You state that the stern observer sees a wave propagation speed ...

0

The speed of a photon, and all massless particles, is essentially always $c = 299792458\space m/s$. The Doppler effect is entirely contingent on the fact that the velocity is the same throughout. We have a source, $S$, emitting light at a certain speed: ...

1

As stated in the comments: The Doppler shift is a change in observed frequency due to relative speed difference. However, the speed with which the signal propagates is the speed of light.

2

At what speed does our universe expand? This question doesn't make sense in the form in which it was posed. To see why, let's start by thinking about how we know the universe is expanding. The expansion of the universe was originally discovered by Lemaître and Hubble, who found that the redshifts of galaxies were proportional to their distances from ...

-1

I haven't done this in a long time, but my understanding from Feynman's QED is that the speed of a photon is unknown (Heisenberg) - photons traveling in a vacuum are around the speed of light +-, but at any instant the speed differs due to uncertainty. The photons going faster and slower than light speed cancel in the same manner that photons bouncing off a ...

4

Comment to the question (v4): It seems relevant to mention that there in principle could be a difference between the universal speed limit constant $c$ (which is usually casually referred to as the speed of light in vacuum), and the actual speed of light in vacuum if the photon has a rest mass, at least from an experimental point of view. Of course, no ...

11

In quantum mechanics a particle can be treated as a wave and a wave can be treated as a particle. This is the notorious wave particle duality. I won't go into this any further here because it's been discussed to death in lots of previous questions. Search this site for wave particle duality if you're interested in finding out more. Anyhow, assuming I ...

3

If rest mass does not change with v then why is infinite energy required to accelerate an object to the speed of light? The momentum of a material particle, a conserved quantity, is theoretically and experimentally a non-linear function of velocity given by $$\vec p = m \frac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ which goes to infinity as $v ... 1 In relativity the rest mass is the mass of an object measured from a reference frame in which it is at rest. But this is not the mass involved in acceleration or inertial mass. Inertial mass, or the opposition of the body to the change of movement (directional or in magnitude), will grow with the speed of the body: $$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$ ... 3 The beginning of the chapter, the author does use$c\$ for the Lorentz transformations (cf., Equation (13.1)). \begin{align} A_0&=\gamma(A_0'+(v/c)A_1') \\ A_1&=\gamma(A_1'+(v/c)A_0') \\ A_2&=A_2'\\ A_3&=A_3' \end{align} Shortly after Equation (13.1), the author lists several enumerated remarks. In particular is #4: Lest we get tired ...

1

Here is a slightly different take on this using the boundary conditions for electromagnetic fields at an interface. A key boundary condition, that is derived from Faraday's law, is that the component of the E-field tangential to the boundary must be continuous. So take an EM wave travelling at normal incidence with the electric field solely in a direction ...

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