# Tag Info

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"Since the ship observer sees the photon travel the same distance and amount of time regardless of direction, the second observer must also see it travel the same time in both directions". This is a wrong assumption. It's not time alone but the spacetime interval which is the same for both. Try to compute it in all those experiments (between the events 'the ...

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There is no frame of reference that has speed c relative to any other frame of reference. This is well known in Special Relativity. Thus, if it were the case that either spaceship were at speed c in some frame of reference, there is no "synchronizing clocks", there is no "spaceship A sees..." or "spaceship B sees..." because there is no frame of reference ...

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Let us start from classical physics. The fundamental invariance postulate in classical physics is that all physical laws describing an isolated physical system in an inertial reference frame are invariant in form under the action of Galileian (Lie) group. That group is made of $4$ (Lie) subgroups. (1) spatial translations, invariance under space ...

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There is. Have a look at http://en.wikipedia.org/wiki/Noether%27s_theorem. Isotropie is "coupled" with conservation of angular momentum, homogenity of space to conservation of momentum and homogenity of time to conservation of energy.

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According to the following paper and commentaries, general relativity can be derived from a standard model matter field equation combined with some other consistency criteria. If new matter such as dark matter is found then the given procedure could give a new theory of gravity or it might just lead back to general relativity. How quantizable matter ...

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From what I've been told by my mentors there were a few things that happened. Poincare was extremely generous person, he routinely attributed his contributions to others left and right. For instance, he contributed the transformations to Lorentz, while Lorentz himself admitted he didn't do them in the form in which Poincare presented them, and didn't see ...

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Special relativity is used in the SM formulation. It is kinematics, so somehow more basic than interactions between bodies. A QFT derivation of General Relativity has been the Holy Grail of the field for many years. In the early times, Feynman, Dirac, and the others tackled this problem, but after decades of failures it was more or less considered ...

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There is not a universal rest frame. There is, however, a galactic rest frame. Because you can look up at the stars, falsely assume that they do not change, and count your rotations that way. However, that method is only as reliable as the premise that the stars don't move, which they do slightly. The Hafele-Keating experiment used a variant of this, ...

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Landau's hint goes as follows: Four-velocity is $$u^{i} = \frac{dx^{i}}{ds} = \gamma_{v}(1,\frac{1}{c}\mathbf{v}),$$ where $$\gamma_{v} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$ We have used $$ds = \frac{cdt}{\gamma_{v}}.$$ To proceed, we note that $$\dot{\gamma}_{v} = \frac{\gamma^{3}_{v}}{c^2}\mathbf{v}\cdot\dot{\mathbf{v}},$$ therefore $$w^{i} = ... 1 Please leave both barn doors open and take a picture when the runner and the pole are completely inside. The fortitude of the pole or the doors are nonsensical to this paradox. What is important is that the concept of simultaneity in different inertial reference frames are different. What the observer at rest relative to the barn sees is, the runner and ... 0 Assuming the electrons are moving at the same velocity the beam would look the same because the relative velocity is zero between the electrons. However, according to the scientist(lab frame), the beam gets shrunk (and according to the electrons the scientist gets shrunk). 0 Newton's first law (really Galileo's law of inertia): This works as well in Special Relativity as it does for Newton. If an object has a constant relative speed near the speed of light and no force acts on it, it keeps moving at a constant relative speed. Check. Can't go faster than light, though. Newton's second law: F=ma Acceleration depends on ... 0 Remember that you must always specify the inertial frame of the observer. Other than that, your question makes perfect sense. The "closing" velocity of the two photons approaching each other will be 2c only in an inertial (stationary) observer at rest relative to the center point. To an observer "riding" with one of the photons, (either one), the closing ... 4 From Wikipedia: The number 5 is a relic of old notation in which \gamma^0 was called "\gamma^4". 2 I don't know is there is a historical reason I just assumed that was just no to mistake it with \gamma^3 which would be called \gamma^4 if your Lorentz indices run from \mu=1,2,3,4 instead of the usual \mu=0,1,2,3. 0 The world lines of both clocks pass through two particular events ('points' in spacetime), the event of your leaving the home and the event of your returning. The worldline of the clock at home is straight while the worldline of the clock in your pocket must be curved due to the acceleration you undergo during your near light speed trip out and back. A ... 0 Let's say we have two clocks. Let's call the (coordinate) times of these two clocks t_A, and t_B, respectively. I leave one at home and keep one in my pocket. Then, I started running [...] then come back to my house. If I compare those two clocks how would they differ in time? If it is also given (corresponding to the comment by the OP ... 0 The angle changes because the rod contracts only in the direction of motion. This means if we broke the rod into two components (like a vector)$$\vec{L}=L_{x}\hat{x}+L_{y}\hat{y}$$only the L_{x} would "feel the length contraction" since its the only part in the direction of motion. So the new length vector would be ... 2 Because the angle depends on the length in the x direction and the length in the y direction, specifically the ratio between them. Therefore, if the x direction is contracted while the y direction remains unaffected, changing the ratio between the two, the angle will change too. 0 Technically they wouldn't differ in time because to come back to your house you would have to decelerate and accelerate, which is under the realms of general relativity or as seen in comments below (thanks to @dgh) we can use comoving frames. You could use two frames one moving towards school and one away from school and Lorentz transform between the two, or ... 0 [Note: Since the statement of the question, including the recent "Edit" after comments, already generally describes how to obtain the solution, my following answer is largely limited to pointing out some remaining mistakes; at least unless requested otherwise.] The observed wavelength for the photon coming towards the observer ... more specificly: ... 0 The question of what is the velocity of a photon relative to another photon does not make sense. Neither it does asking what is the velocity of anything relative to a photon. This is because in special relativity we only have the concept of a velocity defined for a massive observer, which is defined from the four-velocity$$ u^\mu = \frac{d x^\mu}{d\tau} $$... 6 The mass m in the formula is NOT the rest mass m_0 and therefore dependent on the velocity:$$ m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma m_0 $$This means you cannot simply take the normal mass of a nitrogen atom and put it in there, if you assume a speed v \neq 0. The c in the formula doesn't mean that the particle travels at light ... 6 Your calculation is wrong because E=mc^2 doesn't mean that the object has velocity c. To make my answer useful i will give a very brief overview of Dynamics at higher velocities which is a consequence of Special theory of relativity. At higher speeds(of order of c) Newtonian mechanics is not valid. The linear momentum is defined as:$$\vec p=m_0\gamma ...

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If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed. Similarly, to A it APPEARS that B's clock is moving slower, but B sees his own clock moving at normal speed. [...] Is this true or false Foremost: it is improper. (In a ...

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Is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed. Similarly, to A it APPEARS that B's clock is moving slower, but B sees his own clock moving at normal speed. This is true. If the above is true, then ...

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Time dilation actually occurs with objects that are moving at some portion the speed of light relative to any observer "that is to say" relatively stationary to that object compared with light speed; However to the "OBJECT" in relative motion, it would of course see this effect on the stationary observer as the one's clock moving slow. but that is only ...

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[...] as defined by Einstein in his book "Relativity the special and general theory") it [simultaneity] is defined as a matter of observation. Einstein's literal prescription can be read for instance here: http://en.wikisource.org/wiki/Relativity:_The_Special_and_General_Theory/Part_I#Section_8_-_On_the_Idea_of_Time_in_Physics To summarize in my own ...

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Alfred Centauri. In science the only theories that count are those that can be potentially confirmed by experimental measurement. Everything else is philosophy/religion/fiction. In this experimental setup we have only two readings and each of them show that the clocks are synchronized at both places. This indicates that no time dilation has occurred ...

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An approach alternative to that discussed by David Bar Moshe is to start from a different coordinate system in the Rindler wedge $W_R$: $$ds^2 = e^{2y}(−g^2dt^2+dy^2)$$ here $t, y \in \mathbb R$. The relation with the standard spatial coordinate in $W_R$ is $x=e^y$, where $x>0$ is related with the alternate form of the (same) metric: $$ds^2 = -g^2 x^2 ... 1 The actual question in this question, is a good physics question. Freely interpreted, it basically asks if SR effects, in particular time-ordering of spacelike separated events, make it difficult or impossible to simulate physics. The answer to that is no. An "external" Simulator (be it a particle physicist or the hypothetical people simulating our ... 3 Essentially by definition (due to Wigner), one-particle Hilbert spaces of elementary particles support unitary strongly continuous irreducible representations of Poincaré group. Conversely, any multi-particle Hilbert space, with either fixed or undefined number of particles either identical or distinguishable, cannot be irreducible under the action of the ... 4 No, you do not want representations of the diffeomorphism group for the same reason that you do not want representations of the gauged Lie group in Yang-Mills. The diffeomorphisms are a gauge symmetry, not a real symmetry of the theory. Gauge transformations act trivially on physical states, they map one redundant description of a state onto another. They ... 1 The difference is that Poincare invariance is a global symmetry, so it acts nontrivially on the physical states. This has real physical consequences; for instance, if you act with a translation operator on the state of a particle localized at the origin, you get the state of a particle localized at some position other than the origin. Poincare invariance ... 8 Newton's third law is really a special case of the conservation of momentum. Suppose you have two rigid bodies with momenta \mathbf{p}_1 and \mathbf{p}_2. If they only interact with each other, then \mathbf{p}_1 + \mathbf{p}_2 is constant, since total momentum is conserved. Differentiating this gives \frac{d\mathbf{p}_1}{dt} + \frac{d\mathbf{p}_2}{dt} ... 0 Phase space is Lorentz invariant. You can prove this by writing an integral over d^4 x d^4 p and doing a Lorentz transformation, but there's a nice short proof in Padmanabhan's book "Gravitation", p.26 : For an observer moving with four-velocity ui, the proper three-volume element is given by d3V = u0d3x which is a scalar invariant. To prove this, note that ... 4 The problem with this sort of scheme is that Alice has no control over the results of her measurements, since those are random. This means that she can control which basis Bob's spin is projected on, but she cannot control which of the basis states gets chosen. Bob will then see a random mix of results which turns out to contain no trace of what Alice was ... 0 This is a twin paradox in disguise. Suppose you rearrange your experiment slightly to have the two observers start at the same place but travelling in opposite directions, then they slow to a halt and accelerate back towards each other and pass each other at the starting point. The problem is that if they synchronise their clocks at the start how can they ... 4 First and foremost, draw the spacetime diagram. In the reference frame of the flash bulb, the other two clocks are always synchronized since those clocks have, at all times, the same speed as each other. If you trace the wordline of the two accelerated clocks (in the frame of the flash-bulb), you'll find they are congruent and so, the proper time along the ... 1 The statement is a consequence rather than an assumption. It is limiting the domain of event pairs to those which can be considered causally connected. In a sense it is requiring that there be the possibility that the event A be detectable in the future of event B if it is to be considered causal. The light-cone from event A defines the boundary of that ... 2 The dynamics of a classical point particle moving in the background of any curved space-time is always Hamiltonian (with respect to the canonical symplectic form), thus automatically satisfying the Liouville’s theorem. This is because the action functional is given by the integral of the line element:$$ I = -m \int ds = -m \int \frac{ds}{dt}(q, v) dt = ...

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Now for an easy answer: Physics moving (at a constant speed) is physics standing still. From one person looking at another, they will generally disagree about how fast the other is moving and what time it is, but the underlying formulas of physics are unchanged. The space and time part is a really strange idea, but just imagine what the hell this place ...

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I've heard that a spacecraft could never exceed the speed of light because it's (relativistic) mass quickly approaches infinity and therefore there could never create a big enough rocket to propel it faster and faster. In fact, the spacecraft could never even reach, much less exceed the speed of light. I think that you'll agree that the ...

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This is more an extended comment than an answer. Given the principle of relativity (along with a handful of almost axiomatic assumptions), it is indeed possible to derive a general coordinate transformation that involves an invariant speed. Which leaves the conjecture that the invariant speed is the measured speed of light, c, a matter of empirical ...

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The problem is- The rocket is not 'fighting' with any force-field, it is 'fighting' with the very nature of space-time. So unless we have something of zero rest mass 'things' will tend to infinity. And yes the thrust will increase but space-time will distort( following Lorentz transformation, no GR effect here) in such a way that reaching 'c' 'tests' our ...

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Both Newton and Einstein say that all the laws of physics remain same in a frame at rest with respect to the observer and also in a frame moving with constant velocity. Now Maxwell showed that speed of EM wave in vacuum equals 'c' is a law of physics. But this law was inconsistent with the Galilean transformation, hence Lorentz transformation were ...

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This question is really about history and what was known to the protagonists in your tale and at what time. As a principle, relativity was embraced every bit as fully by Newton and Galileao as it was by Einstein - it's just that Einstein had a few more experimental results he had to gather into relativistic thinking. As in dgh's answer the whole point of ...

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Physics occurring in one spacecraft traveling fast (at a uniform speed) is the same as physics in another spacecraft "standing still". In fact, the whole point is that these words - traveling fast and standing still - are relative. All the things you are describing like relativistic mass are only apparent to observers in other reference frames. The person in ...

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First, to be clear, events are points in spacetime that exist independent of any coordinate system as is the interval associated with two events. So, we can say, without introducing a coordinate system, that the interval associated with two events is timelike, lightlike, or spacelike In the timelike case, we can say that one event is later than the other, ...

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The two postulates of STR doesn't say that any signal cannot move faster than light. This superficial appearance is deceiving. After all, the postulates involve notions such as inertial frame and speed; therefore they require and presume definitions of how to measure whether or not two participants (such as a "sender $A$" and a "receiver $B$") ...

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