# Tag Info

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The full form of Einstein's Mass-Energy equivalence relation looks like this: $$E^{2} = m^{2}c^{4} + m^{2}p^{2}$$ where $m$ is the object's rest mass (the mass of the object when it is at rest relative to the frame of reference). But you can get this from the normal $$E = mc^{2}$$ if you realize that mass is not constant - i.e. the equation should really ...

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Time is that which is measured by clocks. How clocks behave when they are being moved relative to each other is simply a collection of experimental facts. At no time do we need any light to do those experiments, and it totally doesn't matter to moved clocks if we are moving them during the day or the night. So what, exactly is your question? Is it why ...

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If by "slowing down" you mean the time it takes for the sound to travel the headphone-ear distance, then I guess not. Meaning the time between production-detection $\Delta t$ of the sound wave is the same in all inertial reference frames, because in a frame relative to the air, sound travels faster than in a frame stationary with respect to air. But in the ...

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I am not sure, but I think that it does not slow down because your headphones travel with you, at the same speed.

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As long as your music player is traveling with you, the music would slow down (as seen by a stationary observer). But you would not notice anything different. That's the beauty of relativity. Your "internal" clock would slow down as much as the music player's clock, and they remain relatively in synch.

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There is geometrical significance. You are sooo close. You are in Euclidean space, but you should be in hyperbolic space. As @fqq points out, you have stumbled upon rapidity, a parameter in hyperbolic geometry that is the analog of angle in Euclidean geometry. In Euclidean geometry an angle (in radians) is a parameter that measures the Euclidean length ...

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Deriving the Lorentz Transformation equations clearly can be done via the math to math approach, but understanding can thus still be at a reach from you since math is an external tool, external from the mind. A logical analysis of motion soon produces a geometric representation of motion. The geometric representation is then quickly used to produce the full ...

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"Most probably in reality there are some extremely complex laws and equations which makes this question more complicated." Not really. The equations are rather straightforward. Let's measure velocities in units of the speed of light and let's denote the velocity of $B$ as observed by $A$ as $v_{BA}$, the velocity of $A$ as observed by $C$ (the ...

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Consider two observers $O$ and $O'$ moving with velocity $v$ with respect to each other. Both of them will use a photon bouncing off a mirror to define any kind of duration, so it is easy to show that their duration will satisfy the relation $$\Delta t' = \gamma \Delta t \,,$$ and a similar argument with measuring any relative distance $\Delta l$ between ...

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Special relativity is often introduced to students using light clocks because this is a reasonably accessible way to understand that phenomena like time dilation and length contraction must occur. However you should not be mislead into thinking that we use light clocks to define special relativity. The fundamental principle of special relativity (and in fact ...

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That light has a fixed velocity in vacuum comes from observations . In order to fit the data Lorenz transformation were imposed on the rigorous mathematical model for electromagnetism, Maxwell's equations. It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to ...

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In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no". There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the ...

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The physical reason IS the constancy of the velocity of light... since I'm writing in a tablet the answer won't be complete, but expect to get you to the mathematical cross-road. Constancy of velocity of light implies that $$\frac{d|\vec{x}|}{dt} = c, \quad\Rightarrow\quad d|\vec{x}| = c\,dt.$$ Since $d|\vec{x}| = \sqrt{dx^2 + ... 4 In relativistic heavy-ion collisions, you're usually interested in reactions with much higher energy than the kilo-eV scale of a few dozen bound electrons. A typical RHI collision will have something like 10,000 particles in the final state, so a few dozen bound electrons in the initial state won't make much difference either. What does make a difference is ... 2 Not only can it be done, it is being done at several facilities. See e.g. http://alicematters.web.cern.ch/?q=lhc-heavy-ion-program-begins. LHC trumps RHIC's energy by orders of magnitude. It can accelerate heavy ions to 2.76 TeV per nucleon pair, compared with RHIC’s 200 GeV. Since a nucleon has a rest mass of approx. 1GeV, these nuclei are highly ... 1 The (average) mass of the Calcium ion$\text{Ca}^+$is around$m_{ca} = 40 amu = 40 \times 1.661 \times 10^{−24} g = 66.44 \times 10^{−24} g$. That means that to accelerate (from rest) an ion of Calcium to a speed of$\frac{3}{4}c$the energy needed would be: $$\Delta E_{ca} = (\frac{1}{\sqrt{1-(\frac{3}{4})^2}}-1)m_{ca}c^2 = 0.512m_{ca}c^2$$ which is ... 1 "How do we know that clocks slow down relative to each other?" Experimentally. This has been observed many times in the lab. The same answer is true for ANYTHING in physics and science in general. We only know that it is true, because we have experimental evidence for it. 1 One aspect that comes to my mind is the concept of causality. Superluminal propagation would allow for a violation of this principle, creating various paradoxical phenomena. 0 No, since by the principle of relativity: A body in constant velocity motion cannot determine whether it is in motion in a certain direction or whether everything else is in motion in the other direction. No physical experiment can determine this hence for all purposes a body in motion will simply claim that it is at rest while the other body is in motion. ... 1 I'll make a few general comments first and then try to answer "What I am looking for is a clear and simple explanation on the exact purpose 'c square' serves in this equation." First off, while$E=mc^2$is still widely used in the media, it is not widely used amongst physicists. That equation uses the concept of relativistic mass, which is more than a bit ... 0 But magnetic field at that point can have only on "CONSTANT VALUE". Not true. If a particle is acted on by some combination of electrical and magnetic forces in one frame of reference, then in another frame of reference, it will be a different combination of electrical and magnetic forces. It's possible to have a force that's purely electrical in one ... 1 Rather than rehash the equations which have been referenced already, I wanted to first cover the Michelson-Morley experiment. Michelson, Morley and even Lorentz were actually able to do a considerable amount of work on the prediction of the expected existence of the aether wind. The foundations of the underlying equations where strong by this point. The ... 0 i would say the intuition is the simple observation (by Einstein, Lorentz, Poincare and others) of these 2 things: Velocity of light ($c$) is$c$-onstant accross interial frames (extrapolated result from Maxwell-Lorentz equations) The velocity of light ($c$) is an upper limit on every other velocity a material body or signal can achieve (in effect$c$... 0 This is how you derive the equation for time dilation. The metric used in special relativity is the Minkowski metric: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$ and the basic principle of special relativity is the the line element$ds$is an invariant, that is all observers in all inertial frames will measure it to have the same value. Suppose we are ... 6 To me personally, the most intuitive way of understanding SRT is to always keep in mind that, in SRT, the interval $$ds^2=-c^2dt^2+dx^2=-c^2d\tau^2$$ has to be invariant. From this simple formula, everything seems to flow naturally. In particular, it is easy to see the form of the Lorentz factor arise from here, by using$\frac{dx}{dt}=v\to dx=vdt$. Using ... -1 There is a philosophical question of what happens to your shadow, after it would travel faster than the speed of light. This is easily realized with the thought experiment of an object is moving around a point source of light. Given that the radial velocity of a rotating body is$\omega = vr$, then the shadow projected at a distance$r$can then be ... 2 If we assume (1) the 'stick' (rod) does not contract according to an observer fixed at any point on the rod and (2) any point on the rod has constant proper acceleration and (3) the rod is momentarily at rest at$t=0$in some inertial frame of reference then every point on the rod is a Rindler observer which means that every point on the rod has a ... 3 You're correct that different parts of the object have to accelerate differently. The details are pretty complicated—when you push one end of an object, transient pressure waves bounce around inside it, and it eventually settles down into a new equilibrium state, which happens to be moving and (in the original reference frame) shorter than before. A simpler ... 11 Your shadow, surely, is not just the apparent darkening of a 2D region of a diffuse surface you happen to be standing in front of; it is the entire volume of space that your presence is preventing light from reaching, the extrusion of your silhouette from the light source out to infinity (for a point light source, or possibly to a point a finite distance ... 17 Note: This is basically item 3 in jkel's answer. If you move at an appreciable fraction of the speed of light, then your shadow can appear to be "trailing" you, although it will always be "attached" to your feet if you're on flat ground. Suppose a person is moving in the direction shown below and that there are plane waves coming in from an angle. The ... 21 Here is how you can "run faster than" (or at least, get away from) your shadow: you jump at sunset (I just realized 15 minutes after posting that this is the point that @jkej's answer made as possibility #2) Your shadow will detach from your feet, and it will "run away" from you. In the frame of reference of the shadow, you are running away from it. ... 18 Depending on exactly what you mean by away from your shadow, I can think of a number of methods: Position yourself in the shadow of some object larger than you. This would result in your shadow disappearing altogether; the ideal solution in my opinion, but perhaps considered cheating by the likes of Mr Elephant. Position yourself in such a way that your ... 4 This is borderline philosophical stuff, but... If you want to put some distance between you and your shadow, you have to fly. Actually, you have to put some distance between and the opaque surface right below you, so swimming in a tank would do the trick as well. This will disconnect you from your shadow. If you have a jetpack, or if you are in an ... 1 The problem is best analyzed in a reference frame in which both sets are moving, toward each other, at the same speed: this is a reference frame (with reference to Fig. 1) moving leftward at speed$v = 50 km/s$. In this frame relativistic contractions are identical, thus the circuit will close. Also, it will be closed for a time long with respect to the ... 1 Preliminary remarks: if a book say that invariance of$c$is a direct consequence of M-M experiment, stop reading it. Answer to your question: as to kinematics the answer to your question is yes (but I don't know how to carry on with dynamics): we can derive relativistic kinematics from different postulates than the one of invariance of$c$. Consider these ... 0 Your first idea would work if it was possible to accelerate a rocket to the speed of light rather than only being able to approach the speed of light. But a person on a rocket moving at$.99c$relative to Earth will still look out his window and see photons zipping past at the speed of light, not at$.01c$. Since it would require infinite energy supplied to ... -1 1st Law of Motion Every object in a state of consistent motion tends to remain in that state of motion unless an external force applied to it. 2nd Law of Motion It is pertaining to the relationship between an object’s mass, its acceleration, and the applied force. In this law, the direction of the force vector is the same as the direction of the ... 1 The crucial point here is, that a reference frame where photons are at rest simply cannot be defined in special relativity: There is no Lorentz transformation which transforms you from a given inertial system into a reference frame or a space-time where some photon is at rest. This, however, means that concepts such as travelled distance and proper time do ... 2 As my answer to How do photons know they can or can't excite electrons? explains, when a photon interacts with matter it is no longer just a photon. The photon/matter system has to be described by a wavefunction that describes both and that isn't separable into a photon bit and a matter bit. When the interaction is strong, e.g. in Bose-Einstein ... 0 Photons are not considered as observers which would be able to observe their proper time. But if you would do so, their hypothetical proper time would be zero. You obtain this result by multiplying the time observed by any real observer with reciprocal gamma (which is 0 for v=c), see http://en.wikipedia.org/wiki/Time_dilation. However, you may not forget ... -1 Why do i have the sense that this can have a meaningful physical interpretation (even if velocities are both$c$)? First lets assume that both velocities are$c$or$1$in the question's notation. Then the relativistic velocity addition formula simply gives:$1_A \oplus 1_B = 1_B \oplus 1_A = 1_{tot}$, correct? Yes, does the fact that this is mathematicaly ... 0 Einstein (i think) used a simple example. Travelling upon a photon and looking himself in a mirror. Would he be able to look at his mirror-image or not? Einstein (and Lorentz et al) said yes, as such light (should) travels with same constant speed$c$in all (inertial) frames of reference. Then read Einstein's original 1905 paper (On the electrodynamics of ... -1 To be honest, we were somewhat surprised when reading your posts. You talk of "gravitomagnetism". This is the theory of Oliver Heaviside which he published in 1893 and it is the Maxwell-analogy for gravity. However, it is known that the general relativity theory is based upon a totally different set of premises than gravitomagnetism. Also, it is known that ... 0 Yes. See Principles of Electrodynamics by Melvin Schwartz. He derives all electrodynamics including Maxwell's equations from Coulomb's Law and Special Relativity. 0 The extra mass-energy that goes into the spinning ball has to come from somewhere. If it comes from observer B, then his mass decreases by the same amount that the mass of the ball+string increased. Incidentally, observer A seems irrelevant here. Lorentz invariance means that you can pick a single reference frame and use that. You don't need more than one ... 0 Sorry to disappoint you, but the second "article" you are citing is the usual pseudo-science nonsense published by the crank crowd. I would suggest you pick up half a dozen good textbooks on relativity and study the introductory chapters until you understand the experimental difference between relativity and ether models. It's the only way to get past the ... 1 To see that this is the simplest possible non-relativistic quantum field theory for fermions, it's useful to derive the dynamics. The canonical momentum for$\psi(x,y,z)$is the Lagrangian's derivative with respect to$\partial_\tau \psi(x,y,z)$– and it is$\psi(x,y,z)^\dagger$(up to signs and$\pm i$which depend on conventions). At any rate, the ... 1 A photon has no mass. Ever. It has momentum - and there is a relationship between its energy and momentum, and because we know its speed, we can pretend it has mass$m = \frac{p}{c}$. But mass of something traveling at the speed of light is not something we can relate to things not traveling at the speed of light. When light travels in a medium with a ... 1$E$=$mc^2$A much better expression is$E^2 = (mc^2)^2 + (pc)^2$, where$m$is the "mass" (also known as "intrinsic mass", also known "rest mass", but most physicists nowadays just use "mass") of the particle and$p$is the particle's momentum. This reduces to$E=mc^2$in the special case of a particle with zero momentum, but it also reduces to$E=pc\$ ...

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And so photons have mass No - photons don't have mass - they have momentum. And energy. But just because energy is equivalent to mass, doesn't mean they have mass. And they can only travel at the speed of light. A photon cannot travel at any other speed - so you cannot apply the Lorentz transformation to it. The Lorentz transformation applies to "rest ...

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