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The necessity of anti-particles was first noticed when trying to construct quantum mechanical descriptions of particles that obey the relativistic energy-momemntum-mass $m^2c^4 = E^2 - (\mathbf{p}c)^2$ relationship. The Schrödinger equation is intuited from a combination of de Broglie's rules $E = hf$ and $p = h/\lambda$ and the classical Hamiltonian $E = ... 0 Calculating this is based on the concept of retarded time and retarded variables. The retarded time is:$ t_r = t - \frac{|\vec{S} - \vec{s}|}{c} $The retarded field would then be:$ \vec{f} \left( t \right) = a \int _{\vec{s} \in C} \frac{x\left(\vec{s}, t_r \right)}{{|\vec{S} - \vec{s}|}^2} \hat{\left( \vec{S} - \vec{s} \right)} \, |dC| $Or ... 1 Relativity just requires "constant speed of light in vacuum". It makes no claims about the speed of light in a medium. When you are moving relative to water, you will observe a different speed of light depending on your relative velocity. But you will still have all the other effects of relativity at work - such as time dilation. 1 Let us say you have two frames of reference; frame$F$and frame$F'$such that$F'$is moving at velocity$v$in the positive$x$direction of$F$. Given a space time event that occurs at$(ct,x,y,z)$in frame$F$the Lorentz transform helps us to find the space-time coordinates$(ct',x',y',z')$of that event in frame$F'$. If, however, you know the event ... 2 You can determine the charge of an electron from a static measurement in one frame. Another frame could determine the charge of an electron from a static measurement in their frame. And they might agree or disagree. We postulate they agree, but we had three options: We could postulate that whether or not something is an electron depends on your frame ... 0 I am assuming you have 2 particles facing each other, and that they are approaching each other ? First, as mentioned elsewhere on this page, "..a particle moving at the speed of light does not experience time, and thus is unable to make any measurements." Instead, let's change the particle #1 that you are sitting on to having a specific velocity that is ... 1 I don't know the "formal" proof, but here is my proof: Time dilation and length contractions are given to us by the Lorentz transformations by: t’ = t/(1-v2/C2)1/2 and d’ = d/(1-v2/C2)1/2 (in other words “same” or proportional to each other) where: t = distance/length traveled through the T dimension in observers own frame of ... 2 What you have done here is a Galilean transform, that is a non-relativistic transformation. Take your final result (which is quite correct): $$t' = \frac{\sqrt{\beta^2 + \alpha^2}}{\sqrt{\eta^2 + \mu^2}} \tag{1}$$ We know that the vertical velocity is$\eta$, so the vertical distance moved in our time$t$is given by: $$\beta = \eta t$$ We also know ... 2 You can think of this question as someone a distance$2d$ahead of you releasing a pulse of light (the man in the mirror). In that way, the problem simplifies to "how far can light travel in$0.80 \mu s$?" Solving this will give you the value of$2d$($2d = ct$). So divide by$2$to get$d$, and the calculation is simply$d = ct/2 = 120m$. EDIT: I should ... 0 So in a four-dimensional orientable space we have a$[0\;4]$orientation tensor $$\epsilon_{\alpha\beta\gamma\delta} = - \epsilon_{\beta\alpha\gamma\delta} = - \epsilon_{\gamma\beta\alpha\delta} = - \epsilon_{\delta\beta\gamma\alpha}$$Usually with respect to some basis we choose$\epsilon_{0123} = 1$or so to finish off the specification of the whole ... -1 Don't forget that angular momentum is a psuedo vector. Recall what that means about properties under reflection, for example. 2 No your argument is not correct. Firstly, velocities do not add linearly like 3-vectors in Euclidean space: the relativistic sum of two velocities always has a speed of less than$c$if both the velocities' magnitudes are less than$c$(no matter what their direction). Secondly, photons have no rest frame: that's a basic property of things that have zero ... 1 The equation you quote: $$t' = t\sqrt{1-\frac{3GM}{rc^2}} \tag{1}$$ gives the time relative to an observer at infinity. You want the time relative to an observer on the Earth's surface. You need to calculate: $$t_\text{satellite} = t\sqrt{1-\frac{3GM}{r_\text{satellite}c^2}}$$ and: $$t_\text{Earth} = t\sqrt{1-\frac{2GM}{r_\text{Earth}c^2}}$$ ... 0 You start your question by saying: From my understanding, the above statement can be deduced by imaging a light-clock in a moving train. but this is not really true. When we're teaching relativity to students we have a tendancy to use diagrams of light clocks because it's something physical that students can have a feel for. Unfortunately it gives the ... 2 As you correctly note, you need to prove$t_B > t_A$is preserved by Lorentz transformations. It's not clear to me whether your final answer demonstrates this (since the sign of$x_b - x_a$isn't totally obvious), but I think you're on the right track. I might have responded with the following argument. The invariant interval $$ds^2 = -dt^2 + dx^2 ... 0 If you are using scalar speeds, the issue becomes clear: V and V' always equal each other because they both represent the rate of change of separation between two objects, which is a property of the two-object system, not of either object individually. In fact, every observer watching S and S' will calculate V and V' to be equal, regardless of ... 0 In tensor notation, \Lambda^\mu{}_\nu is the matrix that performs the Lorentz transformations. Now, since p^\mu is a tensor, under an arbitrary Lorentz transformation, it transforms as$$ p^\mu \to p'^\mu = \Lambda^\mu{}_\nu p^\nu $$Consider now a particle moving entirely in the x direction, with velocity v_1. Its 4-momentum is$$ p^\mu = \gamma_1( ... 0 This is so because the distance between those two frame of references measured from both perspectives should be the same at any given point of time. Speed is just the 'change of distance' measured b/w those two frames per unit time, which should obviously also be the same from both . -1 Since Alvin is in a spaceship he will of course be moving faster then Jack standing on Earth. Time does affect things with gravity and speed so Alvin watch would be going slower because the faster you go the slower time will pass for you. Explanation -1 According to the special relativity, there is a formula describing the relation of particle mass with its speed: $$m = {m_0 \over \sqrt {1- ({v \over {c}})^2}}$$ where$m_0$is the mass of the particle when it is still----rest mass. So as long as a particle moves, no matter how large or small the velocity is, the mass of the particle$m$will vary and be ... 0 You have to bear in mind that you have 2 different reference frame. Let's call$S$to Jack's frame (on Earth) ans$S'$to the Alvin's frame (on the ship). From$S$you see that 5 seconds on Earth passes when 4 seconds passes for the ship (and viceversa in$S$). But, from$S'$, you see that 4 seconds for the ship means 3.2 seconds on Earth. The key point ... 6 Ignore the diagram. It's complex because it includes all the various bits of electronics required for the experiment and that confuses the issue. The experiment is just a variant of Mössbauer spectroscopy. The experiment uses a$^{57}$Fe source that emits a gamma ray with an energy of 14.4 keV. Since this energy matches the spacing in the energy levels of ... 2 Yes, the statement is correct. One proof I am familiar with was given by V.Fock (of Fock states, etc) in his book "Theory of Space, Time, and Gravitation". See Chap.1, Sec.8 therein (pg.20, bottom). In modern notation the idea is as follows: In inertial frame$S$parametrize straight lines as ($x_0 = ct$) $$x_i = \xi_i + \beta_i s, \;\;\; i=0,1,2,3$$ ... 2 Why$m^2$in front of$\phi^2$and why is$m$the mass? Fist of all, from dimensional analysis the prefactor to the$\phi^2$term in the Lagrangian must have mass-dimension$^12$in$3+1$dimensions since the Lagrangian has mass-dimension$4$and$\phi$has mass-dimension$1$. This just tells us that we can write the term as$m^2\phi^2$where$m$is ... -2 Wow 'll try and sort this out. Light travels at a velocity that is dictated by a relationship of properties of 'freespace' (empty) called permittivity and permeability. it is the ability to reach and pull for lack of a simplier explaination ... Lasers 'cut' because they heat up and either melt or disintegrate complex molecules. Once melted the excess can be ... 1 Your confusion might originate in what it means a simultaneous measurement. The fact that "Light from farther end leaves before it does from the closer end" is irrelevant. That would be true even if the object were at rest. One way to make a simultaneous measurement to determine the length is to have clocks along the path of the rod, at rest relative to ... -1 The length of the rod will always appear smaller in your frame than in the proper frame, the mathematical relation between both lengths is given by: $$L=\gamma L_{o} \,\, ,$$ Where $$\gamma=\sqrt{1-\Big(\frac{v}{c}\Big)^{2}} \,\, .$$ I'm not sure if I've understood very well your question, but I think it can answer your question. 0 ...what will happen if it attains the speed of light? That's not something you have to worry about. Saying that it takes an infinite amount of energy is the same thing as saying, "No matter how much you increase the kinetic energy of a particle, you will never observe it to move as fast as light." Nothing in the universe has an "infinite" amount of ... 1 How can we say that difference in energy between 99.99% and 100% of speed of light is infinite? The relationship between energy and velocity of an object is not linear but instead follows this equation$E^2 = m_0 ^2c^4 + p^2c^2$(1) where$p = \gamma m_0 v$and where$\gamma\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} $and as you see, as$v$... 1 Since I can't comment, I can't ask for specification. Most probably problem is following. In the lab system$\Lambda$with total energy$E= 10~GeV$flies along x axis and after some time decays into proton and pion. In the$\Lambda$rest frame, which is oriented the same way (x, y, z coordinates), those daughter particles must fly back to back isotropically, ... 3 The black axes give the frame of the meter stick. The black vertical axis is the worldline of the left end of the stick and the parallel black line is the worldline of the right end of the stick. Your condition 1) says that the spatial distance from A to B, measured in the black frame, is shorter than the spatial distance from A to B, measured in the ... 0 The geometry for a spinning ball changes. If you consider the circumference as small line segments, the fact that they are spinning means that the line segments are moving in the direction of motion and exhibit Lorentz contraction while the radii are not foreshortened since they are moving perpendicular to the spin. You are now dealing with non Euclidean ... 1 The transformation under time reversal of the forms electrondynamics is subtle because the gauge field 1-form$A = A_\mu \mathrm{d}x^\mu$and the field strength$F = F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu$are not the correct physical objects to transform. This may be seen by observing that the Maxwell equations are$\mathrm{d}F = 0$and ... 1 Cubero et al. 2007: Thermal equilibrium and statistical thermometers in special relativity (http://arxiv.org/abs/0705.3328) came to the conclusion that 'temperature' can be statistically defined and measured in an observer frame independent way. With fully relativistic 1D molecular dynamics simulations they verified that the temperature definition ... 1 I think most of you are confusing what contracts because of motion. Not the distance between the muon and the Earth, nor the distance between the point in which the muon began its motion and its destination. Length contraction deals with the contraction "of the moving object" (that is the muon's length, if we could talk of it). If you could see an airplane ... 1 A reference frame is equivalent to a choice of coordinates. So, choosing an accelerated frame in Minkowski space is equivalent to choosing a specific coordinate system on Minkowski space. Most importantly, this means that there is not genuine curvature in an accelerated frame, i.e. it is fundamentally different than gravity. The equivalence principle ... 0 If Instead of that positively charged cat we use an iron nail (which is also moving in sync with the current like that cat which was moving previously), You are asking what if you had a stationary neutral wire with a current and you moved a piece of iron so that it moved at the rate the mobile charges moved why would it react? If you go back to the ... 0 Here's one way to think of it. The (Earth) timing of the laser celebration (10 Earth years) was decided BEFORE the rocket left Earth, and the rocket crew knows that the Earth clock is running slower, so their own clock reading must be GREATER than Earth clock reading (> 10 years). This logic may not alleviate the paradox of mutual time dilation, but it ... 1 In that case, time dilation still occur, of course. In order to show this using t=d/v, you'd have to take into account the space contraction in the direction of motion. Mathematically, if d is the height of the clock, then the time taken from a photon at the bottom to reach the top of the clock isn't$\frac{d+vt}{c}$but$ \frac{d/\gamma+vt}{ c}$. When you ... 0 The two parallel mirrors, A and B, are traveling together such that the axis of reflection is parallel to the direction of travel, and in such a direction that A trails behind B as both travel at the same velocity near the speed of light relative to our observatory. A laser attached to A fires a pulse at B which returns at some time$x/c$later. Because B is ... 1 yes, time dilation still occurs. The reason is that the mirrors are there to provide an intuitive view of how/why time dilation occurs, not to to create it. This time dilations still occurs regardless of the existence of the mirrors or the direction of motion. But the drawing will no longer have explanatory power. -1 In the first case light travels in a larger distance than the one of the mirrors due to the horizontal transportation.Given that the speed of light is constant,time dilation occurs.In the second case the distance that light travels is the same (the transportation of the system and the velocity of light are parallel) ,the speed of light off course is constant ... 0 No. The Uncertainty Principle has to do with the act of measuring. Basically, you cannot simultaneously measure both position and momentum to an arbitrary degree of accuracy. The more accurately you meausre one, the less accurate your measurement of the other becomes. The uncertainty in momentum , as far as I know, won't result from your not knowing when ... 1 Let's suppress some dimensions to simplify: $$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2$$ This quantity $$\Delta s^2$$ is preserved by changes of reference frame, just as in Galilean physics the quantity $$\Delta r^2 = \Delta x^2 + \Delta y^2$$ is preserved by rotations. Notice it is also the equation of a hyperbola. Thus, the effect of a frame shift is ... 3 Let's say we're sending a scouting mission to an Earth-like planet 100ly away to see if it's suitable for colonization. We could send the scouts out at near light speed, and due to time dilation they could easily survive the trip without dying of old age. If we send them out at .9c then the entire round trip will only take 20 years in their frame of ... 0 In the frame of the now stationary electron you could put a cylindrical shell of conducting material around the wire. Since the wire has a net charge and has an electric field pointing from it, the conductor, if isolated will redistribute its charge to have an equal and opposite charge on the inside surface of the shell and an equal charge on the outside ... -4 I was reading about Supplee's paradox, which is about whether a relativistic projectile, subject to uniform gravitational acceleration, would float or sink underwater. However the solution of the paradox seems a little unclear to me, as given in the link. It isn't very clear, is it? Given certain assumptions about how to treat the gravitational ... 1 First of all, the speed of the object (relative to you, or relative to anything else) does not appear in the equation$E=mc^2\$, so the equation cannot possibly say anything about what happens as the speed of the object increases. Second, the equation in fact applies only to an object at rest. The correct equation for a moving object is ...