# Tag Info

6

This was a reference to the apparent measurement that neutrinos travel faster than light. FTL travel can be used to travel back in time (though the procedure for doing so is somewhat involved).

0

Please see the Usenet Physics FAQ which gives a lot of background about this and similar questions. In particular, let me quote part of the question "Is Faster-Than-Light Travel or Communication Possible?" which addresses your question. 4. Rigid Bodies If you have a long rigid stick and you hit one end, wouldn't the other end have to move ...

0

No pole is entirely incompressible. You push some particles at one end of the rod, they move, then push the next particles along, all at the speed of light or less, so causality is never violated.

0

By definition, these equations are true in any frame. Linear momentum is $p=\gamma mv$, energy is $E=\gamma mc^2$, rest energy is $E_{0}=mc^2$ and kinetic energy is $T=(\gamma-1)mc^2$. We've been given that in this frame $p=mc$. From this, we must conclude that $p=\gamma mv = mc$. Correct? $p=\gamma m v$ true in any frame. This means $\gamma v = c$, ...

1

$$\newcommand{\k}[1]{\left( #1 \right)}$$ \begin{align} \nabla\cdot{\vec{J}\over r}={1\over r}\k{\nabla\cdot\vec{J}}+\vec{J}\cdot\k{\nabla\k{{1\over r}}}\\ \nabla'\cdot{\vec{J}\over r}={1\over r}\k{\nabla'\cdot\vec{J}}+\vec{J}\cdot\k{\nabla'\k{{1\over r}}}\\ \nabla'\k{{1\over r}}=-\nabla\k{{1\over r}} \end{align}

1

All inertial observers (including A) will see light move at exactly $c$. So both A and B see the light move at $c$. But from A's frame, it appears that the light moves with a speed relative to B of $(c - 0.8c) = 0.2c$. Likewise the light in the other direction would be observed by A to be moving relative to B at $(c + 0.8c) = 1.8c$ As long as you are ...

1

You write: "From the Spaceship’s point of view, the first signal was sent from Earth when it was 4.8 Spaceship light years away. Thus it would take 4.8 Spaceship years to reach the Spaceship." Be careful. I, on earth, send out a light signal at the same time (according to me) that the space ship takes off. According to you (on the now-moving spaceship), ...

4

Review of the idea we're talking about So let's review the mechanism by which this idea works. (In my view it makes little sense to critique the idea if we don't have a detailed understanding of it.) The idea is that, you've got currents on wires, and those wires (I'll call them wire-1 and wire-2) consist of nucleons and electrons. All of the nuclei are at ...

-1

The speed of light can be derived from simple classical mechanics using the equations of Maxwell. This page offers such a derivation, utilising simple vector calculus. The speed of light in vacuum is seen to be $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ Where $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of vacuum. The answer to the second ...

0

Rapidity has a number of related advantages, as you and the accepted answer have pointed out. It also has a geometric interpretation. See this SE answer (but note that that answer answers a somewhat different question).

0

Rapidity is especially meaningful in the context of Galileo's relativity principle. That is, if you ride an inertial frame, and use the same physics to impart the same impulse to yourself to accelerate to a new, colinear moving frame, then this "standard physical manoeuvre" will always lead to the same rapidity change relative to your initial frame. As a ...

3

If the velocity of the aether wind is a sizeable fraction of c, the apparent velocity of c will depend strongly and obviously on the direction in which the measurement is taken. Since this is not true, the aether wind velocity must be quite small, which requires a sensitive instrument to detect the effects. It was exactly this range of possible wind speeds ...

0

The speed of the earth in its orbit about the sun is about 30 km/s. Michelson assumed that the speed of the earth through the rest frame of the ether was of this order.

4

So, special relativity says that every frame is as good as any other frame, and there is no absolute frame of reference. All good. And special relativity is experimentally falsified and general relativity uses the word general to cover the case when inertial frames are only local and not global, such as in the universe we live in. Suppose there is ...

3

If you ask around about magnetic fields, you will read seemingly-authoritative articles which say magnetism is a consequence of length contraction. They should say that magnetic fields and magnetic forces are a unique and required element to add to electric fields and forces in order to make it relativistically invariant. Which is a slightly different ...

0

We don't know why; it's a physical fact that's been established by experiment. One could suppose after all, any finite speed was possible; and this is what Newtonian Mechanics allows.

-4

To make a long story short, when we write $c = 299792458$ meter/second, then the number 299792458, while in principle just an arbitrary number that defines the meter relative to the second, can be interpreted as a physical measure of the human body just like the number 25 appears in the expression 25 kg/meter^2 for the threshold BMI that separates overweight ...

-1

First, we do not know why the speed of light is what it is. Of course its numerical value depends on the particular unit system, but the basic fact remains that light has some particular speed, and we cannot explain that any further. This is similar to other physical constants such as $\hbar$ and $G$. Dimensionless constants such as $\alpha$, the fine ...

0

Your confusion arises from an incorrect reading of event coordinates. It is a frequent beginner's mistake, so it's worth it to trace its roots step by step. Here's how: 1) Representing events: Every point in the Minkovski diagram represents an event. The same point represents the event both in the unprimed and in the primed frames. That is, we do not ...

0

Here's the right way to say this (and to highlight the symmetry between the outbound and inbound journeys). Label the origin $O$. Label the point where the voyager returns to earth $P$. At time $C$, the earthbound observer says "$C$ minutes have passed since my friend departed at time $O$. But right now, his clock shows time $A$. Therefore his clock is ...

8

Special relativity is indeed the origin of magnetism.* As you correctly note, the actual speeds involved in conduction currents in real-life conductors is at crawling pace and it is much smaller than the speed of light. However, most objects tend to be neutral and therefore exert no electrostatic forces on each other, in which case the magnetic contribution, ...

-6

I had read here that magnetism arises from a current because of the special relativistic effect associated with the speed of the moving charges in that current. Yes, you can read about that in apparently authoritative sources, but it just isn't true. It doesn't make sense either, because charged particles move rotationally in a magnetic field, and the ...

1

It's because of Lorentz contraction. From the reference frame of the wire, the charge appears to be moving. From the reference frame of the charge, however, it's the wire that's moving. If the wire is moving at all, then Lorentz contraction will shrink the wire in that reference frame a minuscule amount. This results in an unbalanced number of charges in ...

0

The Minkowski diagram captures time dilation even without invoking the Minkowski distance, provided we account for the different units/scales along the unprimed and the primed axes, see for example "Minkowski diagram in special relativity" (Wikipedia link). The calculation is exactly equivalent to that using the Minkowski distance as explained by WillO, but ...

3

This is not possible. It violates the law that there is no preferred frame. If you were in the moving train, you'd see only the relative velocity between you and other objects. Your rulers, and clocks would all be completely normal to you, so there is no way to solve for your absolute speed. You could find other peoples relative speed by comparing ...

1

No, there is nothing that is "keeping light from going faster". The local velocity of light in vacuum can not be different than the standard $c=3 \times 10^8{ m\over{sec}}$. There are two parts to my answer. 1) When light passes near you in vacuum, you will always measure the standard $c=3 \times 10^8{ m\over{sec}}$ using your local meter stick and ...

3

I wonder why light does not have infinite speed. It does not have infinite speed because the experiment of Michelson and Morley has proven that it has the same constant speed in any reference frame. Moreover, any experiment on electromagnetic waves shows the presence of retarded potentials, that is, events need a certain time to propagate in space once ...

0

If the speed of light were infinite then the laws of physics would be non-local. The way things work in our universe is that the state of a system in the future only depends on the present state of itself an its local neighborhood. E.g., the future state of the Earth one year ahead depends only on the present state inside a bubble of one light year ...

2

The speed of light is determined (in terms of other fundamental constants) by Maxwell's equations. In particular, the speed of light $c$ must satisfy $c^2=1/(\mu_0\epsilon_0)$, where $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of the vacuum. Because neither $\mu_0$ nor $\epsilon_0$ is equal to zero, the speed of light cannot be infinite, ...

1

A person traveling at 0.86 $c$ would not contract in his own frame. He would not contract at all. His body would be the same as if he were traveling at 0.0 m/s. He would not be crushed. Distant galaxies are receding from us at relativistic speeds. They look just fine. Their internal physics is the same as for our own galaxy. Throw that book in the ...

0

The gamma matrices do not change if one does not apply a change of representation (e.g., chiral -> standard) along with the Lorenz transformation. Recall that you can write Dirac's equation in any frame with gamma-matrices in the same (e.g., chiral) representation. If you change the representation of them by using an invertible matrix $\gamma^\mu \to ... 1 I'll revise my answer, since my last answer was too specific and was deleted You can start with an equation relating relativistic kinetic energy to velocity:$E_k = m_0c^2\left[ \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right]$For these types of problems where you can't use a calculator, you need to be able to estimate the order of magnitude of the ... 0 Yes, As the following article explains jupiter sized objects at speed close to light 2 Another set of experiments which support$E=mc^2$are Compton scattering experiments. The mass-energy of the electron is an important quantity in analyzing these events, and the results are consistent across a wide range of energies for the primary photon and scattering angles. The energy of the secondary photon is given by$$E_{\gamma '}= ... -2 It is not carried away at all. It is now stored in the nuclear bond. 3 Is it carried away as momentum imparted on the [product] atom? Is it carried away in neutrinos? Is it carried away as gamma rays? All of these can happen, and in general nuclear reactions will output their energy via a combination of these. The specific combination, of course, depends on the specific reaction. Also, if neutrinos are massless, can ... 6 Emilio's answer was also the first that came to my mind, but I was not quick enough to post. However, even more precise experiments come from particle accelerators, and similar devices. https://en.wikipedia.org/wiki/Tests_of_relativistic_energy_and_momentum The power of the magnets in the LHC is determined by the relativistic mass of the particles going ... 5 There is a huge mass of experimental evidence that confirms the mass-energy equivalence. The clearest example that it happens is, of course, nuclear power, both in its explosive and civilian forms. If you want a detailed breakdown of all the experiments that corroborate it, I would recommend the entire archive of Physical Review C or a similar ... 2 All of special relativity is captured by spacetime diagrams like the one you've drawn. The lorentzian (or, if you prefer, minkoskian) distance from the blue point to$P$is 1.25, meaning that a clock traveling along the green worldline will record 1.25 ticks between those points. The lorentzian distance from the blue point to the pink point is 1, meaning ... 1 You are using the non-relativistic expression for energy to ask a question where relativity is required. That's creating your apparent contradiction. See, for example, https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation. Second, your question doesn't seem to have anything to do with gravity specifically. There is a gravitational force between ... 0 If you're lucky then$\tau$is the proper time already, and the answer is$\tau_1 - \tau_0.$If you're not lucky then it is best to change symbols so that we don't get confused, you have a general parametric path$\vec r(p)$related to a time coordinate$t(p);$for a small change$\delta p$the ship travels a distance$|\vec r'(p)| \delta p$in a time ... 0 I think we can divide most potential solutions into 3 broad categories depending on what sort of reference is used: Some property inherent to your body or brain: As mentioned in rob's answer, the most obvious is probably to use one of several second counting methods, or a song, drum beat or similar that you have experience performing at a fixed pace. Some ... 0 Because in Bob's reference frame, the light clock has moved sideways a certain distance during the time of flight of the photon. So to hit the "top" mirror, the photon must have travelled some horizontal distance (depending on Alice's speed) as well as the vertical distance; the total displacement is a diagonal. BTW, the diagrams on the page you linked to ... 1 If your hypothetical stranded astronaut is able to use her own head-to-sole height as a length reference, I would expect her to count seconds by muttering "mississippi one, mississippi two, mississippi three" the way she has been doing since playground days. If your astronaut is a musician she might recall a piece of music for which she knows the ... 1 You've stated that you'd recreate an SI length unit$\text{m}$(meter) from knowledge of your own height. So you've got a reasonably accurate ruler. Create a small angle pendulum with length$L$. Use this clock to measure the speed of light (in vacuum). Call this$c_p$(measured with the planet's pendulum period). The ratio of$c$(measured in SI units ... 1 In order to understand this, you need to understand to concept of shear stress. Consider the diagram below: In this diagram a bloc of viscous fluid is held between two parallel plates of surface area$A$and with distance between the plates of$h$. We now cause the upper plate to move at a constant speed$v$, while keeping the bottom plate perfectly ... 1 I don't know if I have to use (Lorentz) metric anywhere here. No you don't. This question is wholly about linearity and the definitions of tensors. Choose a basis$\{\hat{e}_j\}_{j=0}^N$where my index runs from nought to$N$to be in keeping with standard notation in relativity, but that is the only link: this question is general. Our tensor ... 0 The basic answer is yes, you are understanding things right. The "low-speed effect" is that if a ruler of size$2L$with a set of clocks is moving past you at low speed$v$, and the clock at the center shows time$t$, and those clocks are in sync in the moving frame, then they appear out-of-sync in your frame: if the moving ruler measures$x$relative to ... 1 What the commenters are referring to is that the Lorentz factor$\gamma$really only deviates significantly from$1$at speeds that are fairly close to$c$, the speed of light.$\large{\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}$. With$v\$ the relative speed between observers. The Lorentz factor pops up in relativistic calculations all the time and is a ...

1

I would not ordinarily answer a question like this, because these questions usually come from people who have made no effort. It's clear from your comment on m4r35n357's answer that you are an exception to that rule, so I'm happy to provide the spacetime diagram. You will find that it pays to get good at drawing these; as m4r35n357 says, they are always ...

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