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1

Just to complement John Rennie's answer, one can always perform a Lorentz transformation to a coordinate system such as the particle is at rest for a given time. It's called instantaneous rest frame (IRF). This frame changes point to point, unless the particle's velocity is constant. In such a frame, we have $ds^2 = -c^2d\tau^2,$ where $\tau$ is the ...

1

There is, in a sense, a way to 'guide' oneself to the equations of motion based on the symmetries. The form of mechanics most suitable for this purpose is Hamilton's principle - the system takes a path for which the action has a stationary value for variations with fixed endpoints: $$\delta S=0$$ $S$ is generally expressed as (under some parametrization of ...

0

It depends on how you define mass. I like to think of it as mass is just rest mass. I mean the mass you weight on a scale when nothing is moving. On different media light moves slower not because it gains mass but because of its interaction with the atoms in the media. The photons get absorbed and reemited in such a manner that when you sum the waves for ...

1

First, we will look at the energy of a free relativistic particle of (rest) mass $m$ moving with velocity $v$: $$E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $E=mc^2$ when $v=0$. We now consider a few cases: $m\ne0$: In this case, $E\rightarrow\infty$ as $v\rightarrow c$. Therefore, a massive particle that at any point of time is moving at less than ...

0

The mistake is that in the equation under "One checks that", the term $-c^2$ shouldn't be there. Instead, one should insert minus the squared value of the $t$-derivative of $c$, and the derivative of a constant is zero. This is assuming that $t$ is really a different name for $s$, an arbitrary parameter along the curve, and the letter $c$ means that the ...

1

Every particle needs to have energy to be a particle (if it had none it wouldn't even exist). Since energy is equivalent to mass and therefore gravitates I would say YES, all particles that have a speed less than the speed of light must also have mass. Because the speed of the particle is less than the speed of light an observer could travel with the same ...

2

I recently read a paper on the possible kinematics: http://scitation.aip.org/content/aip/journal/jmp/9/10/10.1063/1.1664490 It states that under the 3 assumptions they made, there were more then 10 possible Lie-Algebras (while they discarded one by heuristic arguments)

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The electric and magnetic fields transform like a second rank tensor not a four vector. I would suggest you look in Jackson classical Electrodynamics for the transformations. One observer will see a moving magnetic field which will have an electric field component. At low velocity this will just be interpreted as magnetic induction.

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The equation by David Z can only be applied for particles having mass not for photons or other massless particles. The general equation is $$E^2-p^2=m^2,$$ where $E$, $p$ and $m$ are energy, momentum and mass of the particle resp. And this mass is rest mass, there is nothing like relativistic mass in modern books.

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Relativistic momentum is given by calculating Newtonian momentum, but instead of using mass, we use relativistic mass. Therefore, starting with $p = m_0v$ To become correct, we use relativistic mass $p = \dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}v$ Or, in a more attractive form $p = \dfrac{m_0v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

9

The modern answer I know a body's momentum is equal to the product of its velocity and mass. No, it's not. At least, not in general. $p=mv$ is an approximate formula that works well for massive particles or particle-like objects traveling at slow speeds. But for fast-moving objects, or massless objects, or things that don't even count as objects (I'm ...

0

I don't know a reference but it isn't too hard to show that proper acceleration is indeed $d \phi / d\tau$. First start from the more common definition of proper acceleration which is $du/dt$. Where $u$ is the proper velocity, $t$ is the coordinate time. Applying the chain rule several times we get: $$\frac{du}{dt} = \frac{du}{d\phi} \frac{d\phi}{d \tau} ... 0 Perhaps a better way is to redefine your energy equation to:$$ m \ c^{2} \left( \gamma_{rms} - 1 \right) = \frac{3}{2} k_{B} \ T $$where \gamma_{rms} is the Lorentz factor of the rms speed of the relativistic gas. This can be shown to be:$$ \left( V_{rms} \right)^{2} = c^2 \left[ 1 - \left( \frac{3 \ k_{B} \ T}{2 \ m \ c^{2}} + 1 \right)^{-2} \right] ...

0

There are two separate problems here that I can help you with - terminology, and math. First problem: you're not using the term "proper time"...erm...properly. The proper time of an object is the time that a clock attached to that object would read - nothing more, nothing less. So if you're wearing a watch right now, it will tell you your proper time ...

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It looks like the answer is negative: ...

2

In special relativity particles that move faster than the speed of light are commonly called Tachyons and there is some research about the properties of such particles. Nevertheless, most physicists regards them as unphysical, because their proper time is imaginary.

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No there is no time dilation. Both arms are part of your frame of reference.

1

If you define "length", without exception, as a constant multiple of the time between two events, it will behave no different than the time, really. Instead, try this for a definition: the time taken for light to go from $M$ to $A$ (say), with a correction for the motion of $A$ during this time, so that we get the length at the "original" time, when light ...

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This definition works but only in the frame of the object to be measured. More generally, if you add time as a 4th dimension then a distance is defined between two events and not two points in space. That's probably why you can't generalize your definition to a moving object.

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4-momentum transfers accordingly to Lorentz transformation, and for a photon travelling along Y axis ($\nu$ is frequency) it has such form $$p^\mu=(h\nu, 0,h\nu,0)$$ Lorentz transformation for observer moving along $x$ axis is: $$p^0\rightarrow \frac{p^0-\beta p^1}{\sqrt{1-\beta^2}},p^1\rightarrow \frac{p^1-\beta p^0}{\sqrt{1-\beta^2}}, p^2\rightarrow p^2, ... 5 That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass m (\geq 0) can have an arbitrary momentum p (in some direction, with magnitude \geq 0). The energy of such a particle is$$ E = \sqrt{m^2c^4 + p^2c^2}$$The velocity of a particle is equal to$$ v = \frac{pc^2}{E} $$When m = 0, E ... 0 It would be really difficult to get experimental evidence of length contraction of an extended object. A measuring device attached to the object would contract along with it, and even if measurements were recorded and then read after the object stopped (relative to an observer's inertial frame), they would show no contraction. Thus, length contraction can ... 0 In special relativity, one cannot speak of 'stationary' and 'moving' frames. You can only talk about motion as relative to another frame. If two frames are moving with a reasonably high relative velocity, observers in either frame will measure clocks in the other frame to run slower. As for the relation:$$t_0 = \frac{t}{(1-\frac{v^2}{c^2})^\frac12}$$... 0 I take two mirrors parallel and facing each other. For simplicity, I'll first discuss the case that these two mirrors were and remained at rest to each other. A light pulse is being infinitely reflected between the two mirrors M_1, M_2. E_1, E_2 are the events when the pulse is reflected by M_1, M_2 [...] Accordingly (and as far as I ... 2 It is worthwhile to note that \Delta t' = \gamma \Delta t only for events with the same x coordinate. The more general relation (i.e. Lorentz transformation) is:$$\Delta t' = \gamma \left(\Delta t - \frac{v}{c^2}\Delta x \right)$$where we assume that v is the x-component of the velocity of the observer moving with respect to the mirrors, so that ... -5 Your question is about what will be the speed of elektron? Is this sum of speeds of electron and the ship, or not? If it is a sum of these speeds, speed of elektron will be more than speed of light ? If not, then there will be a time which is slowing down, so,"can the electron reach to command in circuits? Is not it? Firstly the speed of electron won't be ... 2 No, because of the large scale. Doing things like this only seems instantaneous. The speed of a push on this object is actually the speed of sound in the object. 22 Answering this requires a bit of a preamble, so bear with me ... Any observer can construct a coordinate system to locate points in spacetime. By construct a coordinate system I mean they choose a ruler for measuring distance and a clock for measuring time, then they choose the directions of their three spatial dimensions. Even before Einstein, coordinate ... 4 Einstein says this, not because your watch is some ancient artifact with power over time, but because in the theory of relativity, time is relative. We can not longer say that the time is blah blah o clock everywhere. The time is different at different points. Therefore, your time is the time on your clock, and this is the "correct" time for your reference ... 1 It will maintain its speed. It's the same as every other object. Also, making it reach the speed of light will take infinite energy. If you actually calculate out the amount of energy to go slower, you get a complex number. 0 You asked: "Does one imply another?" No. Neither implies the other. However, I think there are benefits to first being clear what the ideas are, particularly since I think each idea actually already assumes an arrow of time. In the first case, you start with an arrow of time that only earlier times affect later times, and then end up strengthening that to ... 0 Because equations 1 and 2 are for light signals only, and light signals satisfy those very unique equations because the speed of light is constant. 0 To have particles that move faster than light requires violations of Lorentz invariance. So, to investigate the effects of this, one first needs to build a model in which Lorentz invariance is not an exact symmetry that is consistent with the known physics. This has been done here. Vacuum Cherenkov radiation is then indeed a predicted effect if charged ... 0 Wavelength, speed and frequency of light: V=c/lambda, v being the frequency. Light emitted or radiated from a MONOCHROMATIC source has a range of wavelengths and velocity, the frequency is invariant, The speed of light actually is slowed down. That's why we define refractive index as c/v. Example: Light will travel in air having a refractive index 1.00029 ... 1 Is it possible to travel ONLY through Time and not Space? Well, in a sense, we can't help but travel through time, but in the way that I think you mean, one way to do it is to hover just above the event horizon of a black hole - preferably a super-massive one so the tidal effects wouldn't be a problem. Park a spaceship at 1/10th of 1% greater ... 3 The problem with your question is that velocity is relative so there is no absolute way to say whether something is travelling through space or not. Any observer can set up some time and space coordinate system (t, x, y, z) to measure positions of spacetime points. The observer can then use these coordinates to measure changes in position with time i.e. ... 1 Well, a rocket traveling at close to the speed of light would be very hard to see at all cause it would go from a moon's distance in one direction to a moon's distance in the other direction in a little over 1 second, and seeing a rocket as far away as the moon would be difficult - but I'm thinking that's not what your asking, so lets pretend that we have a ... 2 Mass-energy equivalence (Rest energy = rest mass * c^2) can be applied to a moving body: Total energy = relativistic mass * c^2 From the perspective of a stationary observer, the total energy in a moving body is greater than the rest energy in a stationary body. As velocity approaches c, it is much greater. The ensuing explosion would be much bigger than ... 0 Ah, yes, I think I got an answer. With moving particle energy also moves in the same direction. It was strange for me that we have r-component of \vec{S} but it seems to be normal. 1 It depends on the direction of movement of your rod. If it travels perpendicular (its main axis is perpendicular) to its velocity, there is no length contraction. On the contrary, if it travels parallel to its velocity, you can apply the standard length-contraction formulae of relativity (cv http://en.wikipedia.org/wiki/Length_contraction) 2 As I recall, covariant refers to how an object transforms when you boost to another inertial frame. An example would be the relativistic 4-momentum P^{\mu}. Invariant refers to quantities which are unchanged under boosts to different frames. For example the product P^{\mu}P_{\mu}=m has the same numerical value in any frame. Sometimes a relativistic ... 0 You should understand how potentials work and how they relate to forces. It is not as simple as saying that information travels at the speed of light. As I will show below, potential can change instantaneously everywhere (but that doesn't mean that information is traveling faster than light. Lets begin with the Maxwell's equations:$$ \nabla \cdot ...

1

IMHO it's important to look hard at the ontology of what's actually there and take care to distinguish between reality and abstraction. For example: I was reading about the light cone in relativity... Relativity is just about the best-tested theory we've got. I "root for relativity". But I will say this: a light cone is an abstract thing. You cannot point ...

0

IMHO, of course Time has singular direction for all real life events. E.g. living bodies, human bodies, etc. age and cannot be still or start moving the opposite direction. (The Curious Case of Benjamin Button is a fiction. Lets not mix facts with fiction here.) Causality principle is true and so is the second law of thermodynamics. But the former is more ...

1

The Pauli matrices anticommute, so the product of two of them has to be antisymmetric in its indices. The only antisymmetric 2-tensor in two dimensions is the Levi-Civita symbol $\epsilon$. Hence, you can "guess" the structure of the product in question up to a constant without calculating anything explicitly.

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Regarding your assertions: Events $\varepsilon_{AJ}$ and $\varepsilon_{BK}$ were simultaneous in the inertial frame of participants $A$, $B$, $M$. This is a perfectly reasonable statement and it is the sort of language used in everyday physics. Participant $M$ was the middle between $J$ and $K$, in the inertial frame of participants $A$, $B$, $M$. ...

1

I would like to bring the ladder paradox here to explain simultaneity of events.A ladder (an inertial frame) is moving horizontally with a relatively high constant speed with respect to a garage (another inertial frame). The garage has an open door where the ladder can not actually enter if the ladder was at rest in the garage's frame but that is not ...

1

You might be thinking of the Unruh Effect. It says that observers undergoing high acceleration see an increased vacuum temperature. Mandatory Wikipedia link: http://en.wikipedia.org/wiki/Unruh_effect Remember that according to special and general relativity, observers traveling at a constant velocity (speed and direction) all have to observe the same ...

1

You start with $$E' = \frac{E-up}{\sqrt{1-u^2/c^2}} = \\ \frac{E-up}{\sqrt{(1-u/c)(1+u/c)}}.$$ You know that $p = \frac{E}{c}$, so $$\frac{E-uE/c}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E(1-u/c)}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E\sqrt{1-u/c}}{\sqrt{1+u/c}} = \frac{E\sqrt{c-u}}{\sqrt{c+u}},$$ now using $E = hf$, $$f' = f\sqrt{\frac{c-u}{c+u}}. ... 3 Let us first replace the Minkowski metric tensor$$\eta~=~\eta_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$$with a more general constant metric tensor$$g~=~g_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}.$$Note that the raised EM tensor$$F^{\mu\nu}~:=~g^{\mu\lambda} F_{\lambda\kappa}g^{\kappa\nu} depends on the (inverse) metric. The Maxwell ...

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