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For simplicity, let's ignore the expansion of space in general relativity and just consider how things work for two galaxies moving apart in the flat spacetime of special relativity. Suppose a galaxy is moving away from the Earth at 0.6c, and it's emitting a radio wave that in its own rest frame has a wavelength of 4 light-nanoseconds (a light-nanosecond the ...


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You can draw this yourself. Make the vertical axis time and the horizontal axis space. The lines of simultaneity for various observers are all the lines with slopes of less than 45 degrees (in absolute value). So pick any two points that are connected by a line of slope greater than 45 degrees (or equivalently, draw any line with slope greater than 45 ...


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So in the galaxy's reference frame it is stationary, so from its point of view it should emit un-shifted light with energy $E$ and speed $c$. This is correct. The part you have confused is: Regardless of how fast we are going with respect to the originating galaxy, the photon moves at speed $c$ and has energy $E$ when we observe it. The observed ...


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"Regardless of how fast we are going with respect to the originating galaxy, the photon moves at speed c and has energy E when we observe it." Not true. The photon moves at c, but it will have a shifted energy E' < E (if the relative velocity is moving away from us). The emitter doesn't "know" that it's moving, and it doesn't need to. The point is that ...


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It is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems. Correct, they are equivalent (assume no GR, and no QM) in the sense that if the integral versions hold for any surface/loop then the differential versions hold for any point, and if the differential versions hold for every ...


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I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the $p_3$ direction so that $p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$ $$ \delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3}$$ $$ = ...


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Any Lorentz transformation will leave $p^2$ unchanged, hence the mass of the Dirac $\delta$ will remain at $m^2$. Observe that, for physical reasons, one usually only considers the component of the (full) Lorentz group that is connected to the identity. Any transformation in this proper subgroup leaves the sign of $p_0$ (which by the spectral condition is ...


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Did physicists immediately realize Newtonian mechanics was incorrect after special relativity was published? Of course not. Physicists did not immediately realize that Einstein's description of electromagnetism (the second part of his 1905 paper on special relativity) was correct after special relativity was published. I can't think of a single ...


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does the ship appear to move slower as it travels a greater percent speed of light? When the ship moves faster, it appears to move faster. When the ship travels faster and appears to travel faster, also the twin inside the ship travels faster and appears to travel faster. If we carefully adjust the speed of the ship to be 4 km/s faster, that will ...


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There is research on going on this The accelerators at the Helmholtz-Zentrum Dresden-Rossendorf (HZDR) provide ions – that are electrically charged particles – of most of the chemical elements available for experiments in materials research over a wide energy range. What is quite unique in the world is the fact that even very heavy metallic ions – beside ...


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Several misconceptions in your question. Let's focus on one statement: "For the twins paradox to be plausible, one of the twins must reach "99.995% the speed of light" (Lorentz factor of γ = 100 ) without being atomized." No. That's not true, for several reasons. Firstly, it is a RELATIVE speed, and relative to the frame of reference of a cosmic ray ...


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The Planck length is a quantum effect (among others). SO the answer is: yes, speeds in the universe are limited by c, and also by the Planck length, indirectly. However, this latter limit has never been measured nor achieved, so that in practice, the limit by c is sufficient. All Planck values are limits. There is no way to get a measurement result lower ...


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Why is the scalar product of four-velocity with itself -1 The scalar product is invariant In the coordinate system in which the object is (momentarily) at rest, the only non-zero component is the temporal component. See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$. Then, (setting $c = 1$) we have $$\frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0 ...


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The derivatives with respect to $\tau$ very much are numbers, but they are not all $1$. Consider your worldline as a curve $\gamma$ parameterized by $\lambda$. We have \begin{align} \gamma : \mathbb{R} & \to \mathbb{R}^4 \\ \lambda & \mapsto (x^0, x^1, x^2, x^3). \end{align} At any point in your worldline you have a position $(x^0, x^1, x^2, x^3)$, ...


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First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices. In this parameterization, the four velocity takes the form \begin{equation} ...


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While we are getting closer to speed of light our length in the direction of the movement is according to Lorentz transformation getting shorter. This are two misconceptions here. One is that the way this is written implies that velocity is absolute. This is not the case. The "relativity" in relativity theory means exactly the opposite. Velocity is ...


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But we can not (even theoretically) consider length shorter than Planck length. This is a popular misconception. Treating Planck units as special is really more numerology than anything else. For example, the Planck mass is about the mass of a single biological cell. Does that mean physics doesn't apply to anything smaller (or is it larger?) than a ...


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Suppose you are observer $O$, and I am observer $O'$. Suppose you, I, and the photon all cross paths at a time we can both agree to call $0$ and a place we can both agree to call home. Eventually, the photon passes Joe's diner, which I say is $L$ miles from home, and --- if we choose units so $c=1$ --- it does so at time $t=L$. According to you, Joe's ...


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Given that you imposed $t= \gamma t'$, the obtained relation between $L$ and $L'$ is $$ \tag 1 L = \gamma L'.$$ Your reasoning is now, if I understand, that (1) is inconsistent with the phenomenon of length contraction because $L$ is the distance measured in the reference frame $\mathcal O$ and $L'$ is the one measured in the reference frame $\mathcal O'$ ...


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In theory, a massive particle can be accelerated asymptotically close to the speed of light. But one can never actually reach the speed of light because the kinetic energy of such a (massive) particle would be infinite. This is because the mass of the particle itself become heavier at higher speeds due to relativistic effects. Specifically: $$\text{Kinetic ...


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Well, by massive, I assume you mean objects that have non-zero rest mass. In that case, it would take infinite energy for that object to reach the speed of light. However, their speed would get closer and closer to the speed of light as more energy is put in, until their speed was practically (but not exactly) the speed of light. Additionally, the smaller ...


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In the $O$ frame in time $t$,the light will travel a distance $ct$. For an interval $t$ in the $0$ frame,time interval in $0'$ will be $t'=t/γ$. In this time interval light will move a distance of $ct'$ in the $0'$ frame. Now as we can see $ct$ is greater than $ct'$, means distance travelled by light in the $0$ frame is more(as $t$ is greater than $t'$?) ...


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All clocks obey the time dilation formula, therefore all durations measured by a clock obey the time dilation formula. All time differences definitely do not obey the time dilation formula. To figure out the time that it takes for something to travel some distance you might use two clocks. In that case you are not measuring the time with a clock, you are ...


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No contradiction, because time dilation and length contraction, even if dilating and contracting, are working in the same sense: For this purpose I recommend to use an example which is simpler than yours (I think it is not useful to consider the light ray beside O and O'. If you are asking a question try first to reduce it as much as possible). Example: A ...


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Be careful. L and L' are the lenghts of the paths followed by the photon. The contraction of lenghts due to relativistic effects is an effect that you experience when you can measure simoultaneously the lenght of something. Take into account the following situation: ...


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In this answer we'll basically repeat Leandro M.'s good answer using formulas. Let us for simplicity use units where $c=1=\hbar$. Consider a spinless relativistic complex scalar field $$\tag{1} \left(\partial_t^2-\partial_x^2+m_0^2\right)\phi(x,t)~=~0 $$ in 1+1 spacetime dimensions. The Lagrangian density for a spinless relativistic complex scalar field ...


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So, I can sum it as: In relation to what do we say that something is faster than some other thing? We just arbitrarily decide that something is motionless. Very often we pretend that earth is the motionless thing, although earth revolves around the sun. We know for sure that earth is not motionless, so we can conclude that earth for sure experiences ...


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Special Relativity only applies to non-accelerating inertial reference frames, thus special relativity cannot be used to model your example. But why do we say that the man is moving faster than earth? why is it not that earth is moving faster than the man? That's a subtle misunderstanding of relativity - the critical factor is the frame of reference of ...


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Has this kind of phenomena ever been used to measure SR properties ? Yes. A similar apparatus, called a Fizeau apparatus, has been around for a long time... It uses a rotating cog rather than a falling box (for obvious reasons). http://en.wikipedia.org/wiki/Fizeau%E2%80%93Foucault_apparatus EDIT (re comment): Yes, lasers can also be used to measure ...


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Imagine for each observer there is a grid in space that maps how far light travels in an amount of time. You can measure an object at rest in this grid by how long it takes light to travel its length and back again (this is important to measure it both ways). Since light speed is a constant, the object needs to change shape to keep its 'light ...


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As far as I know no-one has ever directly measured the length of a relativistic object. However indirect measurements of Lorentz contraction have been made at the RHIC. This collides nuclei together at relativistic speeds and analyses the resulting shrapnel. The nuclei are far too small to see, but from the results of the collisions we can infer that the ...


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Remember that any inertial frame is equally valid. You've been looking at the muons and mountain experiment from the POV of a person standing on the ground (or) on the mountain. In that frame the muon cover a distance equal to the mountain's height and take several times the muons (uncontracted) lifetime to go that distance, but they make it because the ...


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When you have a transformation from one frame to another, this tells you how measurements with meter sticks and stopwatches in one frame relate to measurements with meter sticks and stopwatches in another frame. If you had one transformation based on lightspeed, you'd get a relationship between measurements with meter sticks and stopwatches in one frame ...


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My gut feeling tells me things should have energy because of their charge, like they have energy because of their mass. You are not the first one who had this idea. There was indeed a concept calles "electromagnetic mass", where the electrostatic energy $E_{em}$ of a charged particle at rest would be $$E_{em}=\frac{e^{2}}{2\cdot r}$$ (where $e$ is the ...


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You are asking about time dilation. Using $$\Delta t' = \gamma \Delta t$$ where $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ We can rearrange for $v$ to get: $$v = \sqrt{c^2 - \frac{\Delta t^2}{\Delta t'^2} c^2 }$$ or maybe better $$\frac{v}{c} = \sqrt{1-\frac{\Delta t^2}{\Delta t'^2}}$$ to get the speed as a fraction of the speed of light. Plugging in ...


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I can think of a light beam as a pulsating stream of water from a hose traveling at the speed of light. If there's a hole in the side of my space ship and the hose of streaming water is pointed directly, perpendicular to my ship's direction of travel, at the hole, then only a portion of the pulsated water will enter the hole. Now concerning the portion of ...


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This is an effect known as time dilation. In this post, I will be taking material from the excellent book, Einstein Gravity in a Nutshell, by A. Zee. Figure 1 will be the basis for the argument. We bounce a photon around to create a clock. It is postulated that the speed of light is the same in all frames. In the rest frame, the time it takes for light ...


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You do not have to shoot the photon at an angle in your example, this is a result of the principle of relativity. The photon appears to travel diagonally to an outside observer (an observer at rest relative to the moving mirrors). If you move along with the mirrors the photon will still appear to move perpendicular to the mirrors. You can also imagine ...


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There are two frames which I will call O and O'. Suppose I'm someone in O. By that, I suppose, you mean I'm someone at rest in $O$. Are the lorentz transformations a means by which to calculate coordinates on O' in a way that only frames like me would calculate? The Lorentz transformations transform the coordinates (of events) in $O$ to the ...


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It is worth reminding yourself what "a frame of reference" means: an agreement on how to measure when and where some event happened. For the purposes of special relativity we generally only talk about inertial frames which have some constant velocity (less that $c$) with respect to one another. If you have the coordinates $(x,y,z,t)$ you know how far over, ...


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As stated at http://en.wikipedia.org/wiki/Minkowski_space "In 1905, with the publication in 1906, it was noted by Henri Poincaré that, by taking time to be the imaginary part of the fourth spacetime coordinate √−1 ct, a Lorentz transformation can be regarded as a rotation of coordinates in a four-dimensional Euclidean space with three real coordinates ...


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What is Minkowski spacetime? Being some particular sort of "spacetime", it is foremost some set of "events" (i.e. certain identifiable participants actually having met each other, or at least the thought-experimental idea of such a meeting of certain identifiable participants). A set of events is a "Minkowski spacetime" due to satisfying further ...


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I am writing this answer to add some extra things to sahin's post, which he didn't expand upon. By now you know that Minkowski space(time) is basically a four-dimensional vector space (space that is "flat", though obviously in lieu of metric structures, flatness is hard to define, though Minkowski space obviously does have a metric structure, the term ...


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I believe you know about the three dimensional Euclidean space which in general represented by $x, y$ and $z$ coordinates. Euclidean space has the important property of being flat. Now, you can think of the Minkowski space as a four dimensional space which is flat! The fourth coordinate is generally chosen to be time $t$, so people call it not "space" but ...


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His clothes, the music player and so on might of course move with him (supposing they would be able to withstand the rapid acceleration and decelleration involved, which I doubt), but the music player would have to play the music much faster all of a sudden for it to appear the same. It's entirely possible of course for Quicksilver to have a specially ...


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A tachyon is a particle with an imaginary rest mass. This however does not mean it "travels" faster than light, nor that there's any conflict between their existence and the special theory of relativity. The main idea here is that the typical intuition we have about particles -- them being billiard ball-like objects -- utterly fails in the quantum world. It ...


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Comments to the question (v1): We will not discuss tachyonic states here, because they are pathological and signal an instability of the theory. Then $$\tag{1} p^{\pm}~\equiv~\frac{p^0 \pm p^1}{\sqrt{2}}~\geq~0 $$ is manifestly non-negative, since the energy $p^0\geq |p^1|$. In the light-cone formalism $p^{+}>0$ is strictly positive, since the special ...


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Time will flow the same way as in the video for the last spaceship whether or not the other two spaceships are there. The purpose of having three spaceships in the video is to make it easier to reason about what's happening to each observer's experience of space and time. If you watch the part at 8:22, you'll see how you can treat all spaceships as if they ...


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The definition of the Lorentz force is fundamental already, it can't really be derived. However, a useful treatment/thought experiment is to consider the electric/magnetic fields due to a current carrying wire in the stationary frame and a frame moving parallel to the wire. In the stationary frame there is only a circulating magnetic field. In the moving ...


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derive the equation … in a more fundamental way? There is no more fundamental way. That equation defines the magnetic field: it is the field that produces forces in moving charges such that the force is proportional to the charge and to velocity. Also, the force must be perpendicular to a given direction which can vary from point to point. More ...



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