# Tag Info

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Is this the correct approach to solving the problem? The result you give for $v$ can't be correct since, within the parenthesis, you're subtracting a number from a speed squared. There's a much cleaner approach to the problem that does not require length contraction, time dilation, or the Lorentz factor. The best and most correct approach, in my ...

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Firstly, any spatial rotation isn't going to do anything interesting to the Electric Field as a vector, but its components are going to change! Restricting to Lorentz Boosts along $x$ $$\Lambda^{\mu}_{\nu}=\begin{bmatrix} \gamma &-\beta\gamma&0&0\\-\beta\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1 \end{bmatrix}$$ So ...

1

Cool! I'm working on the exact same thing. The way I proved this was since $\mathcal{L}$ and $\phi$ are both Lorentz scalars they must have the same transformation law. Therefore $$\delta \mathcal{L} = - \omega^{\mu}_{\phantom{\mu}\nu}x^{\nu}\partial_{\mu}\mathcal{L}.$$ However, note that $$\partial_{\mu} \left( -\omega^{\mu}_{\phantom{\mu}\nu}x^{\nu} ... 5 The QFT is strongly based on the group theory formalism. Often when people say about some QFT theory they primarily say about the symmetries of the theory - invariance of lagrangian of theory (or about covariance of equations of motion) under sets of transformations. The group theory formalize these statements and help to construct theories which corresponds ... 3 Landau and Lifshitz apparently like to "cancel as many ds in the numerator and the denominator as possible", perhaps for aesthetic reasons. However, to make sense of the integrals, it seems more pedagogical to express^1 everything as functions (and derivative of functions) of the parameter s, i.e., there should only appear a single ds in an integral ... 7 Yes, your confusion is wholly caused by you thinking classically ;) In a hand-wavy way, particles are certain localized excitations of the quantized fields. The QFT picture contains the particle picture in the perturbative approach known as Feynman diagrams (and, relatedly, the LSZ formalism). There, we are given the action of our theory dependent on some ... 3 The type of field that you have depends on the way that your field transforms. The fields that you encounter in quantum field theory usually are: Scalar fields, these describe spin-0 particles such as the Higgs boson. Spinor fields, these describe spin-1/2 particles, these describe for example the elementary fermions, like the leptons and quarks. Vector ... 4 Fundamental fermions like quarks and leptons are described by the spinor field, while gauge bosons like photons are described by the vector field. They together with the Higgs bosons are currently what we have in the Standard Model for elementary particles. 2 Provided that \mathcal{L} is a Lorentz scalar, the quantity \partial\mathcal{L}/\partial(\partial_{\mu}\phi) has to carry an upper index. Since \mathcal{L} is a function of \phi and \partial_{\mu}\phi, the only object that can give such an index is \partial^{\mu}\phi. Hence \frac{\partial\mathcal{L}}{\partial ... 1 Recall that when classifying representations of the Lorentz group, we consider $$\textbf{N}_{\pm} = \frac{\textbf{J} \pm i\textbf{K}}{2}, \tag{1}$$ where \textbf{J} is the angular momentum (rotation generator) and \textbf{K} is the boost generator. The generators \textbf{J} and \textbf{K} satisfy ... 1 It seems you feel that the example should be symmetrical because from each point of view (S or S'), time is running slower in the other. However, the experience is not symmetrical because as you said : S' is proper time and S is not. If you make an other "experiment" of which S is proper time, then S will measure t2 and S' will measure t2' and you will have ... -1 There's nothing physically problematic with speeds, per se, that are faster than the speed of light. For example, if you quickly rotate an Earth-bound laser such that its beam crosses over the moon, the speed of the center of the beam as it travels across the moon's surface can easily be faster than c. There's no problem with a speed greater than c in ... 2 I agree with your statements up through the claim that t'<t. That's all fine. Here I think is the issue you're running into: The quantity t' in the relationship above represents the time interval as measured in frame S'. It does not represent the number of ticks by the moving clock as measured in frame S. That's a subtle but important ... 2 So I think you should check out this problem for some useful commentary. Problem understanding sign of volume integral in Minkowski space Essentially if you look at the integral on the LHS we see it takes the form explained in the problem above$$ I^{\mu\nu}[f] = \int d^4k\, f(k^2)k^\mu k^\nu $$Where f is just 1 The integral is clearly Lorentz invariant ... 3 Your written text says "t = 1/sqrt[1-(v^2/c^2)]". If you used that equation, it's no wonder you got a nonsensical value. In the equation in the scanned image, that's a letter t in the numerator, not the digit 1. Also, although it works for this problem, the scanned image should really say something like$$\Delta t' = \frac{\Delta ...

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Here is the formal answer on your question based on particular result of Pauli theorem. Calculations are rather cumbersome, but they are general. Arbitrary fermionic field (with invariance under discrete transformations of the Lorentz group) It can be shown that each Poincare-covariant fermion field with spin $s = n + \frac{1}{2}$ and mass $m$ can be ...

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Your statement: "Since the speed of light is constant, the light coming from the clock must travel a longer distance to reach my eye as it moves away. This would make time appear to slow down? If, on the other hand, the clock is moving towards me, the distance the light must travel to reach my eye becomes shorter and shorter, thus time would appear to speed ...

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A photon is fired across the width of a space ship a distance of one unit “ud”. The space ships velocity v relative to a fixed observer is 0.8*c The observer sees the photon travel a distance of cT and knowing the width to be one unit of distance calculates that 1ud = T*(c^2-v^2)^0.5 1ud = Tc*(1-(v/c)^2)^0.5 1ud = 0.6 Tc A photon is now fired forward in the ...

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In special relativity there is a distinction between 'experiencing events' and the concept of an observer: Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from In ...

1

No matter in which direction the clock moves (away or towards you) , the time will slow down in the clock. Time will always dilate as the clock moves faster and faster, but will be apparent to a human eye only once it reaches speeds close to the light speed. This is not taking into effect the doppler shift which is merely the increase/decrease in the ...

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Relativity is not needed: If you replace the clock with a strong laser which fires a ray of light every second with an atomic clock, you know that the ticks will not slow down because every tick will be followed by another after a second from the laser's perspecive . But how can you arrange that with the fact that the light move further and further away ? ...

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Analyzing one moving clock from the perspective of one stationary person will be inadequate to derive special relativity from. With just that set-up, you aren't actually using the key fact that the speed of light is the same for all observers – all you're actually using is just the fact that the speed of light is finite. With just taking into account that ...

1

It isn't possible to comment on your viewpoint that relativistic mass is "an effect of the potential difference between reference frames", because that statement is too vague to be either true or false. In particular, I can't even tell whether you think that the relativistic equations involving mass are correct but can be expressed in an illuminating way ...

3

It is, in principle, measurable by a balance scale. In practice, this is difficult to do, because how do you user your scale when the thing your trying to weigh with it is flying away at close to lightspeed? If, however, you confine the object to a circular track so that you can keep it positioned over the scale for an arbitrary length of time to make the ...

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[...] measurements of the speed of light, especially in the context of the invariance of $c$. The proposition of wanting to measure the speed of light (in the sense of "signal front speed", a.k.a. "speed of light in vacuum"), comparing "speed of light values" between different trials, is arguably absurd. Because: quantitative geometric (or kinematic) ...

3

Your experiment is a bit more complicated than you might think because there are two effects at play. I would guess you're really interested in the effect of time dilation, but there is also a slowing because the distance between you and the moving person is increasing with time. Suppose I'm counting at one number per second, and I count "one" as I pass you ...

1

In relativity the notion of simultaneity is relative to the obeserver. While one observer (the one "standing") see all the rays of light arrive at the same time, another observer (the one going near the speed of light) will see one ray arriving before another. The paradox here is you think simultaneity can be defined in an absolute way independent of the ...

4

Here's a qualitative argument for why special relativity is a special case of general relativity. When you first learn special relativity it tends to be introduced using the two postulates that Einstein started with. This is a perfectly good basis for SR, but it causes no end of intuitive problems for students. Just search this site for examples. It also ...

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Yes, special relativity is a special case of general relativity. General relativity reduces to special relativity, in the special case of a flat spacetime. I.e., general relativity reduces to special relativity, in the special case of gravity being negligible, for example in space far from any objects, or when considering a small enough piece of space in ...

3

Actually given that the first postulate says that all physical laws are the same in all inertial frames, you could replace the second postulate by the postulate: "Maxwell's equations are the physical laws for electromagnetism". From Maxwell's laws you can derive that the speed of light in vacuum has a specific, constant value, in SI units ...

2

As the highest rated non-selected answer to the question "Has anyone ever measured the one way speed of light perpendicular to the Earth at the Earth's surface?" correctly states, the one way speed of light is an unmeasurable quantity. The only measurement that can be performed is to measure the round-trip speed of light. In other words, you need ...

0

Since 1983, the meter has been defined to be the length of the path traveled by light in a vacuum in 1/ 299,792,458 of a second. So light in vacuum travels at precisely 299,792,458 m/s by definition. Measurements involving the speed of light are at this point refining precisely how long a meter is, not refining what the speed of light is. That being said, ...

2

Maxwell's theory had predicted that the speed of light varies with the speed of the observer. Initially (prior to Fitzgerald and Lorentz advancing the ad hoc length contraction hypothesis) the Michelson-Morley experiment was compatible with the assumption that the speed of light varies with the speed of the light source (as predicted by Newton's emission ...

1

Einstein did not prove this postulate ; he simply asked "what if it is true?". He had very good reasons for asking that question. His efforts to answer the question challenged a whole raft of "beliefs" about time and space, none of which were based on proof either ; they were (up until then) assumed true by so-called "common sense" alone. He made ...

1

The way I think of it, as a non physicist who quite likely has a few things wrong, is as follows: time actually passes slower for Alfred, and thus there is no difference in the speed of light. If you say, "Why does time pass slower for Alfred and not Bernard, after all, motion is relative, right!?" Well, this stumped me for a long time as a non-physicist, ...

0

Imagine an observer watching a moving rocket carrying on it both, a light source and a clock. If on the rocket the clock is synchronized with the frequency of the light being emitted So we suppose not just any clock being carried on the rocket, but a good clock which is characterized (properly) by some particular constant frequency $f_r$; as is the ...

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The answer is simple: Maxwell's equations. Maxwell published his electromagnetic theory in the 1860s. This generated a huge schism in physics. Maxwell's electromagnetism was in direct conflict with Newtonian mechanics. There is no allowance in Maxwell's electrodynamics for the speed of the emitter or the speed of the receiver. The speed of light is constant ...

1

There are many possible ways of stating the equivalence principle, which are all logically related but not exactly equivalent.[Sotiriou 2007] One way of stating the e.p. is that GR becomes SR on small scales. From this statement of the e.p., I think it should be pretty clear that SR can't be used to prove the e.p. If principle A says that theory B is a good ...

2

Again, imagine those on the rocket are measuring the distance of one wavelength comparing it with some high precision ruler, then will not the observer watch the ruler contract and then necessarily the wavelength of the light along with it, a contradiction to the original conclusion? No. This is an example of the kind of confusion that can result if one ...

0

For an observer with 4-velocity $u^\mu$, the density of 4-momentum is $u_\nu T^{\mu\nu}$. If $u^\nu = (1,0,0,0)$, then indeed the 4-momentum is $p^\mu = T^{\mu 0}$ and this is a 4-vector. Now let primed indices correspond to components in some other frame. Then $$p^{0'} = L^{0'}_\mu p^\mu = L^{0'}_{\mu} T^{\mu 0}.$$ But it is not the case that $$p^{0'} ... 1 One really doesn't talk about components of the Stress-Energy tensor being momentum densities. Instead, you think about projections on vectors. So, the energy density observed by a timelike observer with 4-velocity u^{a} is given by$$j^{a}=T^{a}{}_{b}u^{b}$$Now, when you boost, you have two choices. You can either boost to a new reference frame, and ... 0 According to the Einstein's Special Theory of Relativity, The mass of a moving object is observed greater by a stationary observer depending on the relative speed of the object with respect to the observer.$$m=\left(\frac {m_0}{\sqrt{1 - v^2/c^2}}\right)$$So yes, the Kinetic Energy is stored as relativistic mass(not rest mass) 0 Your solution to distant instant communication is indeed appealling, but it has its shortcomings. The pole vibrates and thus produces sound waves within the pole;sound ofcourse traverses at a much lesser velocity compared to light,thus what ever force is exerted by person A,it will produce sound waves which will reach B in ages(considering the ... -1 Even if the answers from CuriousOne, Terry Bollinger, Mr.WorshipMe are correct, the historical answer is not yet given. For instance, the invariance of the speed of light was not a problem, since this concept was not known before Einstein... who introduced it to define simultaneity ! As referred to the original Einstein's paper, the motivation for the ... 1 The "slowing" of light in a medium can be entirely explained using a classical wave-based approach. An incoming EM wave wiggles the electron clouds around the atoms in the material. These electrons clouds re-emit a much weaker EM wave having a very small amplitude. This re-emitted wave is 90-degree phase shifted from the original wave but superposes with ... 0 Let's allow your rope to have fantastical properties but not violate conservation of energy or causality. Then if the ends of the rope are within the event horizon, your rope could conceivably not stretch. But if the ends are outside the event horizon: The rope must expand. There is no such thing as a rigid object in special relativity, because the concept ... 5 There is a definition that \left( \frac{m}{2}, \frac{n}{2}\right) representation is equal to spinor tensor$$ \psi_{a_{1}...a_{m}\dot{b}_{1}...\dot{b}_{n}}, $$where \psi_{\dot{b}} transforms as complex conjugation of \psi_{b}. Why do we assume that \left( \frac{1}{2}, 0\right) and \left( 0, \frac{1}{2}\right) represent spinors? You can think about ... 1 Emmy Noether discovered a fundamental connection between symmetries and conservation laws, embodied in her famous theorem. In simple terms, Noether's theorem is that For every symmetry in a physical system, there must be a conserved quantity. The proof requires neither Lorentz invariance nor causality. By applying Noether's theorem, we find that ... 3 First consider the spatial component of the equation above, you get$$ \frac{d}{d\tau} \left( \gamma(v)\overrightarrow{v} \right) = 0 $$But \gamma(v) = \frac{dt}{d\tau} so you have$$ \frac{d^2 t}{d\tau^2} v + \gamma(v) \frac{d\, \overrightarrow{v}}{d\tau} = 0 $$Now since$$ \frac{d \, u}{d\tau} = 0 \implies \frac{du^0}{d\tau} = \frac{d^2 t}{d\tau^2} = 0 ...

2

A negative binding energy would make the vaccum unstable. For example suppose a virtual electron and positron pop out of the vacuum. This costs energy to create the particles, but if their binding energy could be greater in magnitude than their rest masses then they could bind to form an energy state lower than the vacuum from which they were created. The ...

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