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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

With 4-position $$ x^\mu = \left(ct,{\bf r}\right) $$ the 4-velocity is defined as ($\tau$ is proper time) $$ v^\mu = \frac{d x^\mu}{d \tau} = \gamma \frac{d x^\mu}{dt} = \gamma \left(c,{\bf \dot{r}}\right) $$ Taking the dot product of $v$ with itself gives: $$ v^\mu v_\mu = \gamma^2 \left(c^2 - {\bf \dot{r}}^2\right) = c^2 = \left(c ...


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 ...


0

Actually it was that maxwellian electromagnetism had no problems, in contrast to the newtonian classical mechanics framework. Theory of Relativity alters the Newtonian framework not the Maxwellian framework. i would say that even if Einstein hadn't invented SR, someone else would (as indeed many others notably Poincare, Lorentz et al) were alredy on the ...


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 ...


4

I think I have understood the problem. Since I want to compute the length in the moving reference frame $R'$, I must assume that $t_2' = t_1'$ but not $t_1=t_2$, because I want to measure the position of the two ends of the bar at the same time in the moving frame. The calculation is as follows: $$ x_2-x_1 = \gamma(x_2'+ut_2')-\gamma(x_1+ut_1) = ...


1

No. My answer is negative, even if I confirm the statements of other answers: "The first thing is almost completely arbitrary, especially in full general relativity. The second thing is an unambiguous result of an experiment."(Jerry Schirmer) "In Einsteinian relativity all observers can still agree on a number of facts, they are just ...


1

There are apparently some extensions of the Standard Model that allow for Lorentz symmetry to be violated, although from what I understand the symmetry is broken by spontaneous symmetry breaking which means the theory would have been symmetric in the era when the forces were unified, and the symmetry was broken by a random decay to different vacuum state. So ...


0

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 ...


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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 ...


1

Since this is also true for sound propagating in air No, actually, it isn't. There's something crucial that you've left out. Look more closely at the 2nd postulate: light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. Now, this holds for the emitter of the light ...


2

If I understand correctly you are asking how observer dependent is electromagnetic radiation. The first thing is that non uniformly accelerated charges are described in a inertial frame by Larmor's formula and Abrahm-Lorentz force which take into account the radiated field and the recoil on the particle. Now in Newtonian mechanics and special relativity ...


1

If one thinks in terms of spacetime and world lines, the answer is plain to see. While each particle has a world line which must be within the light cone of any event along the world line the distance between the two particles is not a physical entity and does not have a world line. Thus, it is not directly constrained by the 2nd postulate. Put another ...


0

Here's another prospective. The temperature o an object (particle) is a function of its energy. Theoretically there is no limit to the energy we can keep adding into a system. However objects emit radiation that is dependent on their temperature. Object with higher temperature emit radiation with shorter wavelength. According to quantum mechanics the ...


1

(Edit) Notice one thing, Geremia. When two particles finally meet, or when a material body (spaceship in my second example below) and a photon meet (or even when two material bodies meet), none of them separately have exceeded the speed of light. Each of them travel at either c or $v<c$, and they simply meet on their way toward each other. So the closing ...


0

Both equations (for the instantaneous field of a charge moving with constant velocity $v$) are correct. (Well, maybe the primes should be swapped in the second equation, so that the unprimed frame is that in which the charge is moving.) The first figure is not an accurate representation of the first equation: as Jan Lalinsky stated, the field lines should ...


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.


1

A particle moving at the speed of of light does not experience time, as it has no rest frame. Furthermore, a particle cannot continuously accelerate and eventually reach the speed of light, since massless particles can only move as fast as light. They either move at the speed of light or do not exist at all.


2

Or I have to say object is really measured to shrink? That's the right answer. Imagine a thought experiment. Let's say we are shooting another sequel of the Speed movie, 5 (I lost track :) ). So there is a bomb on the board of a spaceship, and a terrorist says you cannot slow down below the speed of 0.5c, because if a mechanism that involves a ...


0

Einstein spells out in On the Electrodynamics of Moving Bodies how "time of occurrence" is defined, in terms of local readings on a network of clocks at different positions in space, which have previously been synchronized using the Einstein synchronization convention. If you are an inertial observer, imagine you have a 3D grid of rulers at rest relative to ...


1

The point at rest in $k$ moves as $(x(t),y(t),z(t)) = (x_0+vt,y,z)$ in the frame in which $k$ itself moves with $v$ in positive $x$-direction. Thus, we have to set $x'(t)=x(t)-vt = x_0 +vt-vt$ to obtain a stationary $x'(t) = x_0$ coordinate for a tupel $(x',y,z)$.


0

$$ dx = cos (\Omega t) dx' -x' \Omega sin (\Omega t) dt - sin (\Omega t) dy' -y' \Omega cos (\Omega t) dt$$ It's basically just the product rule and the chain rule.


7

Relativistic mass is obsolete. See Why is there a controversy on whether mass increases with speed? . Therefore this is not a question that modern physicists would consider of interest. Furthermore, the usual motivation given for using the relativistic mass convention is that it lets you use Newton's second law without modification, but Newton's second law ...


2

Your last expression (4) is equal to (2), you just have to realize what does it say. $\lambda$ isn't $\tau$ and $$u^c = \frac{d\xi^c}{d \tau} = \frac{d\xi^c}{d \lambda} \frac{d\lambda}{d \tau}$$ If you look back to your Lagrangian and how it was derived, you should be able to say what is $d\lambda/d\tau$. To be very explicit, the action of a free ...


0

The universe doesn't have to "know" that any more than it has to "know" that a tilted ladder should "become" shorter. The key effect is the relativity or simultaneity. The length is defined as the distance of the two end points at the same time. Due to relativity of simultaneity, the same time in the observer's frame corresponds to two different times in ...


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: $$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$ ...


0

The root cause is the fact that the speed of light is the same value for all observers, whether they are moving towards the source or away from it. The consequence is that people travelling at different velocities relative to each other measure things differently. As to why c = const, nobody knows.


0

Because space is not Euclidean. It's that simple. IMO This also precludes the mediation of gravity by particles.


2

Here is/was my attempt at a solution to this problem - thanks to the contributions of others here, I am now reasonably sure this is correct. The setup is two plane, polarised waves travelling in opposite directions (let's say $x$). e.g. $${\bf E} = E_0 \sin (kx + \omega t)\hat{\bf j} + E_0 \sin(kx - \omega t) \hat{\bf j}$$ The associated magnetic field ...


0

There is something you should be careful of regarding Liouville's theorem. If there are momentum-dependent forces, then Liouville's theorem changes because phase density is no longer incompressible. Suppose we define $f_{s}$ = $f_{s}(\mathbf{x},\mathbf{p},t)$ $\equiv$ the particle distribution function of species $s$, which is non-negative, contains a ...


3

This isn't a direct answer, but adds to Ben Crowell's answer and Curious Kev's answer and is simply something to help you have confidence in the correctness of your physical reasoning, which is perfectly sound. Think of your standing waves in a box with perfect mirrors at either end, initially at rest relative to an observer $A$, and then you give the box ...


2

It's just a drawing convention. Rather than "vertical", time is orthogonal to the x-axis. The reason it is not shown vertical is because the paper surface is 2D, and the author uses the vertical axis for drawing the altitude with respect to the ground. Just recall the way you draw the 3D axis. Here, the author uses X (horizontal),Y (vertical) and time ...


1

As long as you don't forget that Andromeda galaxy is cca. 2.5million light years away, then there should be no paradox at all. The observers are only seeing different slices of history that took place 2.5million years ago (plus/minus 1day), the decisions has already been made long time ago. It's like when you take a newspaper from a pile, more recent papers ...


7

The MacLaurin series expansion of the Lorentz factor is $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\left(\frac{v^2}{c^2}\right)^2 +\; ... $$ So, for $v\ll c$, the first two terms give a good approximation to the actual value. It turns out that the MacLaurin expansion of another function has the same first two ...


6

$1/\sqrt{1-v^2/c^2}$ is only approximately equal to $\sqrt{1+v^2/c^2}$ for small $v$. As $v \to \pm c$, the error grows arbitrarily large. The former is always right, the latter is not. This is just the same as how $1/(1-x) \approx 1 + x$, where the approximation is only good for small values of $x$. If the cost of a product goes down by $5\%$ ($x = 0.05$), ...


0

So as Team Barn says that the pole is grabbed simultaneously, there is no stretching or compression from their perspective-only Lorentz contraction. Actually, this can't be the case since instantaneous contraction would contradict special relativity. In the barn inertial frame of reference, the pole is initially at rest and has a length of $L_0$. ...


2

I think you are considering two different situations: 1) Team Pole passes through the barn at constant velocity, simultaneously (in Barn's frame) grab the pole, and continue on with the same constant velocity. In this case, your second calculation is correct. The pole's length does not change from Team Barn's perspective. The pole remains length $L_0$ ...


0

If you were to try to grab a real pole at both ends at a significant fraction of the speed of light, the result would be similar to a nuclear explosion. The final state would be a bunch of ionized gas expanding a tens of thousands of km/s from the center of the system. As Jerry Schirmer said, you can't simply neglect the dynamics of the impact in this case. ...


0

There's going to be a distortion of the pole during the acceleration period that goes away. The pole bearers should observe that they are holding a pole with length $L_{0}$, assuming that it's rigid. Note that it will start out length contracted from their potential, though, so during your $\Delta t$, they will stretch the pole out to its rest length.


1

One expects the energy stored in the capacitor to transform like the zeroth component of the four-vector $(U,\vec p)$. In its rest frame the field configuration around the capacitor has $$(U,\vec p)_\text{rest}=(U_0,\vec 0),$$ and by the Lorentz transformation the moving observer will see $$(U,\vec p)_\text{moving}=(\gamma U_0, \gamma\vec\beta U_0),$$ where ...


5

Yes, the concept of a standing wave is frame dependent, and a standing wave selects a specific frame just like the concept of a rest frame. My initial answer was similar to BenCrowell's, and I made the same oversimplification mistake that his initial answer does. The are no spatial coordinates for the standing wave (even in its rest frame) in which the ...


1

The interpretation is that two events being simultaneous as measured in frame $S$ doesn't imply that the events are simultaneous in frame $S'$. Which events count as being "simultaneous" depends on the frame of reference. This is known as the relativity of simultaneity. Added clarification due to comment: The coinciding of the origins is an event, call ...


1

You have to be careful about the difference between speed and velocity. Saying that two clocks are moving at the same speed is different from saying that the relative speed between the two clocks is zero. For example, as measured in some inertial frame of reference, two clocks can be moving at the same speed but in opposite directions, in which case their ...


1

Yes, charge doesn't change in a Lorentz transformation. That's precisely why charge density must change in a Lorentz transformation. If in the lab frame, a length $L$ of a wire has a (stationary) charge density of $\lambda$ on it, the total charge on the wire is $Q=\lambda L$. In a frame of reference in which due to length contraction the wire is ...


0

First off Gamma is $\frac{1}{\sqrt{1-v^2/c^2}}$ but that's beside the point. While by itself it looks okay, if you plug i into a Lorentz transformation everywhere there's a velocity you will also need to either have your time component complex or one of your spatial components as a complex number. This results in that component reversing sign meaning that ...


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 ...


5

The statement from the Wikipedia articles is, as written, wrong. The EM field tensor - as a tensor - does change under change of reference frames. It is covariant, but not invariant under the Lorentz group, while the electric and magnetic field are neither, but they are covariant under the rotation group. The electric and magnetic fields are ordinary, ...



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