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The assumption of linearity is based on the observation that measurement results between reference frames depend only on their relative motion, not their absolute position (which doesn't exist anyway). The derivation is a bit involved, so I can understand why it was glossed over. Good on you for noticing the gloss. In general, we start with $$x' = F(x,t; ... 0 They have to be linear transformations, otherwise the origin of the coordinate system would not be arbitrary as required by the homogeneity of space. Another way to see this is to consider a particle moving along a straight line in one frame of reference. Under a non-linear transformation, this straight line would become curved, meaning that the particle is ... 0 OK, I get your answer, but as this is a homework question and you have really not shown any attempt to get the solution I cannot show you it. The doppler equation you have is fine: I would write it as \omega = \omega_{\rm star} \gamma (1-v) where we are using units with c=1. You then need to use a standard relationship between acceleration in the two ... 0 If we suppose that \Psi is confined we must have \Psi=0 in the external region, so in the border of internal region we must have \Psi=0 and \frac{d\Psi}{dx}=0. But with our flat U, stationary K-G equation say that \Psi is sinusoidal or exponential, so this requisites cannot be satisfied sumultaneously. Conclusion: the confinement of \Psi is ... 0 Since the spaceship is moving and according to special relativity light doesn't feel any sort of "kick" on account of the motion of the spaceship, The absence of a "kick" you're referring to is related to the speed of light, not its direction. If you run very quickly and point the flashlight in front of you, the light's speed, as measured by any ... 1 If the spaceship is moving uniformly with respect to someplace else and not accelerating, then you are in an inertial frame. In such a frame there is no measurement you can do to tell that you are the one moving with respect to that other place. Similarly, if you are in a train with no windows that is moving at constant speed and you throw a ball towards the ... 1 I sure hope my assumption is correct that you're just needing help understanding where an equation that you encountered in your book came from, rather than this having been a homework problem that you were supposed to do. I fully support the policy here of not doing people's homework problems for them. From the starting point of ... 3 Perhaps a slightly different way of thinking about this would be in terms of the Poynting vector. The force exerted is given by$$F = \frac{1}{c} \int {\bf S} \cdot d{\bf A}, $$where {\bf S} is the Poynting vector and in your terms S = c E. In this case I assume everything is at normal angles, so no need to worry about that. The Poynting vector is ... 0 These are two separate questions. 1) Does light move at c in non-inertial frames? First to get a bit pedantic, I assume you mean the coordinate velocity. As light does not have a well defined proper velocity, this seems to be the most reasonable way to interpret your question here. While rewriting Maxwell's equations in a non-inertial frame, and then ... 2 1) Why don't we consider finite dimensional representations of this group? As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation. 2) Why associate the Lorentz group to fields? The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already ... 0 As an ex-theoretical chemist, not claiming any expertise in this field: Even though mass-energy of near light-speed neutrinos, individually, might still be very-very small, there are likely to be many-many more of them (~10^9?) than electrons (used in another answer, above, to make a comparison). Can we therefore make a back-of-the-envelope estimate of the ... 0 You are probably thinking about the moments of a velocity distribution function. Suppose you had some probability distribution that was a function of space, time, and velocity or f = f\left( \mathbf{x}, \mathbf{v}, t \right), where f has the units of # per unit volume per cubic velocity. From this we can define things like the number density:$$ ...

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So let's say these electrons go really really fast like 0.999999997 times the speed of light. We know the upper bound on the neutrino mass is less than 1 eV (the kinetic energy of taking one electron through a one volt potential difference) from experiments. So if we plug in to find the total energy of the neutrino we find. $$E_{relativistic} = \gamma ... 1 The OPERA experiment reported: the speed of neutrinos is consistent with the speed of light within the margin of error The margin of error was around 3 \times 10^{-6}c, so their result restricted the speed of the neutrinos to be at least 0.999997c and less than 1.000003c. For comparison, the speed of the protons in the LHC is 0.999999991 c, so ... 0 They do not actually travel at the speed of light. Instead, they travel at speeds extremely close to the speed of light. 0 According to my understanding your reasoning is correct. Concerning your subtle points: First, when you say that the distance between points x and y is equal to d meters, you (if otherwise not specified) mean that it is the distance in your reference frame, measured by a ruler at rest. And when you say that the distance is t light seconds, this ... 0 One day visiting a science demo years ago, there was a display of the effects of a standard drinking straw put through a block of wood simulating a tree trunk. The display was in regards to the velocity of thrown objects from a tornado. A tornado can increase the velocity of otherwise harmless object to travel through just about anything. It can ... 2 All matter is made up from atoms. While it's very hard to accelerate a spaceship to nearly the speed of light we can collide two atoms (well, two nuclei) at speeds approaching the speed of light. This experiment is done at the RHIC and also at the LHC in its heavy ion mode. So what happens when we collide nuclei at nearly the speed of light? Well, the ... 0 I believe anything with relativistic velocities and much mass (let alone a spaceship) would quickly be destroyed by "junk" in space (when I say "junk", I include stray hydrogen atoms.) To get an idea of the energy released when something like that hits something (like a planet), in Randall Monroe's latest book What if, he states that a baseball with a ... 1 When a straw or a hose punches through a tree, that isn't due to the object having a far greater mass due to the tornado, because it doesn't. At the 318 MPH speed of an F-5 tornado, the relativistic mass of an object is only about 1.00000146 times as big as its rest mass, a negligible difference. A more useful explanation is that the ability to punch ... 1 The traveling astronaut is younger. The situation is not reversible between both astronauts because the traveling astronaut is submitted to an effect which is similar to acceleration because he is following the curvature of space in the fourth dimension. The solution must consider the geometric/ topological constellation. And topologically, the traveling ... 2 In our frame, the light beam approaches the rear end of the train at speed c+v, because both are moving and in opposite directions. Note that this does not mean anything is traveling faster than light. Similarly, the light beam approaches the front of the train at speed c-v. We want to choose the position of emission so that they arrive at the same ... 0 I didn't see anyone mention the practical reason to use an approximation for energy. It is that in most problems you will be computing differences in energy. In that case, for small velocities, you can not only go the approximation {m v^2\over 2}+m c^2, but if you are also not converting mass to energy or vice-versa, you can drop the m c^2 as well ... 6 I am hoping someone could explain, rather in-rigorously, the use of the word inconsistent. Essentially, in this context, inconsistent means the two theories give different, incompatible answers to the same question. In this specific case, the question is: If light (an electromagnetic wave) is measured to propagate at speed c in an inertial frame ... 2 It is usual to make a distinction between proper rotation as opposed to all rotations (which include those that change handedness). The matrices of proper rotations have determinate +1 while those of improper rotations have determinate -1. In terms of groups the proper rotations are \mathrm{SO}(3) while all rotations taken together are the orthagonal ... 2 [...] my question concerns two relatively moving observers. Can one of them distinguish that the other has adopted the oppositely handed coordinate frame? What information does observer A have about observer B? If all A knows about B is B's state of motion, and if he assumes that B is going to choose coordinates in which he is at rest at the origin, ... 12 Inconsistency between two theories just means that there are statements that one theory says are true, and the other says are false. An easier example than the one you're asking about is the inconsistency between Newtonian mechanics and special relativity. Newtonian mechanics says that if you keep applying a force to a material object, it will eventually go ... 2 A direct answer: Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. The equation E^2 = (mc^2)^2 + (pc)^2 does work for all values. As other answers have explained, at low speeds (or equivalently, low momenta), E = \frac{1}{2}mv^2 + mc^2 gives approximately the same ... 4 If the body's speed v is much less than c, then the equation reduces to That is an imprecise wording, and the cause of your confusion, I believe. The correct description is that the second equation is an approximation of the first, and it will be closer to accurate the lower v is. The approximation is important for understand the connection ... 7 Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. Yes, it would be nice if we could find an equation that was "universal". In the case of mass m, energy E, and momentum p, there is such a universal equation:$$E^2 = (mc^2)^2 + (pc)^2$$Einstein, always wanted to ... 0 Frames per second and meters per second are fundamentally different units, and comparing their numbers is a textbook example of comparing apples to oranges. What if I measure light in miles per second: 186{,}000\ \mathrm{miles/sec}? Or centimeters per second: 3\times10^{10}\ \mathrm{cm/s}? Which framerate do you take correspond to the "speed of light": ... 0 Is it possible for a high-speed camera to shoot video at 299,792,458 FPS? A rate of 299,792,458 frames per second would create a movie in which, between subsequent frames, light illuminating the scene happens to propagate one meter. So nothing really special occurs at this particular frame rate, and no laws of physics prohibit a camera shooting at this ... 0 Simply, no. We can not get film to travel near the speed of light. That said, you could record a packet of light, thus "record" light, using another technique. If I understand it right, they re-expose the "film" over and over again. They do this some thousand times and each exposure adds more brightness until the images becomes visable. I do believe the ... 1 The above answers are all good, but I want to add something else. You can derive the energy-momentum relation from at least principle, the action is A=-m\int \sqrt{1-v^2}{\rm d}t . Lagrangian is \cal L$$=-m\sqrt{1-v^2}$, then you can get momentum$p$from derivative with respect to$v$(this is the real momentum, not$mv$). Energy$H=E=pv-\cal L$, just ... 5 If you put$p = \gamma mv $where$\gamma = \frac{1}{\sqrt{1 - \beta^2}}$and$\beta = \frac{v}{c}$in Einstein's equation, you get$E^2 = (pc)^2 + (mc^2)^2 = (\gamma mvc)^2+ (mc^2)^2 = (\frac{c^2}{c^2 - v^2})v^2(mc)^2 + (mc^2)^2= (\frac{v^2}{c^2 - v^2})(mc^2)^2 + (mc^2)^2 = (\frac{c^2}{c^2 - v^2})(mc^2)^2 = (\gamma mc^2)^2 \therefore E = ...

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Special equations of the kind you mention are useful as they elucidate limiting behaviors. Your example is no different from a statement like: the hypotenuse $c$ of a right triangle with legs $a$ and $b$ with $b \ll a$ is given by $c = a + \frac12 b^2/a$. The Pythagorean relation $c^2 = a^2 + b^2$ is valid for any right triangle, but the special case $b ... 3 First, your findings are correct and are found by Einstein. However the fact that a new term appears is only showing that our previous knowledge was incomplete. The new term expresses the Energy related to the rest mass of the body which was not considered before. And it makes sense because before, in equating the Energy of a body only were included those ... 11 First, the non-relativistic equation $$E= mc^2 + \frac{mv^2}{2}$$ is equivalent to its second power, $$E^2 = (mc^2)^2 + m^2 c^2 v^2+ \frac{m^2v^4}{4}$$ If$v/c\ll 1$, then the last term is much smaller than the previous two, and the first two terms on the right hand side are equivalent to the correct relativistic $$E^2 = (mc^2)^2+ (pc)^2$$ which ... 0 time delation is very simple phenomenon.Time delation occurs because the speed of light is same for all observer in same media.so the rate of time experienced by the observer changes with respect to object moving near to speed of light because two events in space time having different time origin with respect to each other never coincide with each other. 2 If the curve is a geodesic then in the coordinate system of an observer moving along the geodesic coordinate time and proper time are the same. That's because in the freely falling observer's coordinates$dx = dy = dz = 0$and therefore$ds^2 = -c^2dt^2 = -c^2d\tau^2$. This makes proper time a natural way of parameterising the curve because it's just the ... 2 The ground state of Dirac sea is full of particles with negative energy, according to Dirac equation. When a particle/electron absorbs energy and excites to a positive energy state with energy level$E$, it will leave a hole with energy$E-E\prime$(if the total energy absorbed is$E$) , which is positive according to energy conservation. So the hole seems ... 3 I) OP wrote (v2): I know that$\mathcal{L}$is a functional not a function. Actually for local theories the Lagrangian density $$\mathcal{L}(\phi(x), \partial\phi(x), \partial^2\phi(x), \ldots, ;x)$$ is a function of$\phi(x)$,$\partial\phi(x)$,$\partial^2\phi(x)$,$\ldots$, and$x$. In contrast, the action$S[\phi]=\int \!d^dx~\mathcal{L}$is a ... 1 You could say that$\partial_{\mu}\mathcal{L}=0$only if the fields it applies to are treated as independent variables and not as functions of$x^\mu$. I.e. the Lagrangian does not explicitly refer to coordinates, but if you were to consider the fields as$\phi = \phi(x^\mu)$, then obviously$\partial_{\mu}\mathcal{L}\neq 0$. The more rigorous way to say ... 1 If I remember well, this section of Peskin & Schroeder uses the Wick rotation to solve integrals of type $$\int \frac{1}{k^n + ...} \cdot ...$$ Where$k^n = k \cdot k \cdot k ... (\rm n \, times)$. By performing the Wick rotation we suddenly get a spherically symmetric problem which enables us to use the well known tricks for such a case. But notice ... 1 Poincare invariance simply means that the Lagrangian does not change under a Poincare transformation: $$\mathcal{L}(x)\rightarrow\mathcal{L'}(x)=\mathcal{L}(\Lambda^{-1}x).$$ Note that I have only written down Lorentz transformations, for simplicity. We now say that$\mathcal{L}$is invariant if $$\mathcal{L}(x)=\mathcal{L}'(x).$$ This does not mean ... 0 Poincare invariant lagrangian should be invariant under Poincare transformation i.e it should consist of scalars under P.T.Example QED lagrangian. 0 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 ... 0 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} ...

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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 ...

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