# Tag Info

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I would like to post this like a comment to the answer above, but I am not permitted to do so. At the above answer the word "axiomatic" is used at much ambiguous ways I could say very naive at best. A real axiomatic theory must define, or assume as primitive notions, ALL the terms it uses and ALL the claims it does and that approach do not do that. If you ...

28

I assume you used the formulae $f_o = fs\sqrt{\frac{1+v/c}{1-v/c}}$ for the clocks ahead of you and $f_o = fs\sqrt{\frac{1-v/c}{1+v/c}}$ for the clocks behind you. Those formulae do imply a singularity for the clock that is closest to you. Which equation to use? The answer is neither. Those expressions assume the travel is along the line of sight to the ...

3

Replacing $GL(4, \mathbb R)$ with $M(4, \mathbb R)$ we have $${\cal V}:= \{M \in M(4, \mathbb R)\:|\: \mbox{if x\in {\cal T}_+, then Mx \in {\cal T}_+}\}\tag{1}\:,$$ where $${\cal T}_+ := \{x \in \mathbb R^4 \:|\: x^T\eta x \geq 0 \}$$ so that we can equivalently restate the given definition as {\cal V}:= \{M \in M(4, \mathbb R)\:|\: \mbox{if x\in ... 2 If you consider a straight-on trajectory, there really will be a discontinuity. That is the same as with the audible Doppler effect. There is a smooth drop in frequency of a fire truck's siren passing you on the street. The reason is, that there is a certain distance between you and the truck at the closest point. If that were not the case, i.e. the siren ... 3 The signal from the clock moving towards you is the Doppler shifted version of the value you "know" it to be - that is, first slow it down by gamma (clock moving relative to your frame of reference), then speed it up by Doppler shift. Ditto, with sign reversed, for clocks behind you. Now the clock moving at right angles shows what you expect and there is no ... 0 "Length contraction", is the measure of a umproper length of an object, in a inertial frame O' which has a relative velocity to a inertial frame O, where the coordinates of the object are fixed in this frame O More precisely, one measures a proper time \Delta \tau' in O' (so at x'=0), which corresponds to the synchronisation with the beginning ... 0 To make the example (and the question) clear, the particles should be on different frames, where when the 2 frames are at rest and coincide, the particles have the given separation. Then when one frame moves relatively (or accelerates) to the other with a given velocity (what velocity exactly?) the Lorentz-contraction appears but only with respect to the ... 0 There's the article from Ohio State University http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html which explains quite well why the clocks on a GPS satellite are faster by about 38 microseconds every day. The article then claims that not compensation for these 38 microseconds per day would cause a GPS to be off by about 11 km per day, plainly ... 4 Rob's explanation of how we know is bang on, but I wanted to address a part of your question that might point to a basic misunderstanding. What is special relativity inside the nucleus? Everything is always relativity. Everything. Always. All those Newtonian equations like T = \frac12 m v^2 for the kinetic energy can be properly understood as ... 5 The relationship between nuclear masses and mass differences and binding energies has been confirmed by many decades of careful nuclear spectroscopy. It's possible to measure an atom's mass by purely mechanical means: you ionize the atoms, accelerate them to a known energy, and use a magnetic field to measure their momentum. This lets you come up with an ... 0 In general it depends on the history of the acceleration. For the case of constant acceleration your question can be answered quite simply, and I've described the calculation in the Q/A How long would it take me to travel to a distant star? or see John Baez's article on the relativistic rocket for a list of useful related equations. The derivation I used is ... 0 No. According to special relativity's postulates, the speed of light in a vacuum is the same for all observers, regardless of the motion of the light source. You may want to read the relativistic Doppler effect. It refers to the change in frequency of light due to motion of observer and/or source. 1 An analogy (possibly from Griffiths' textbook): imagine you're out in desert. Many miles away you see a truck moving on the highway. In which case is it easier to judge its speed: if it's moving radially towards or away from you, so that you have only its change in apparent size to go on, or if it's moving laterally across your line of vision? 4 The key sentence is "from the frame of reference of the accelerator, the circumference must stay the same." So from the accelerator frame, no length is changing, just accelerator's time passes by more quickly by a factor of gamma than in the particle's frame. This is what allows the particles to "travel extra distance" than would normally be possible in ... 0 Can you clarify the set-up of the lamp system? You say that the lamp turns on if the two touchers are touched simultaneously. But that implies that in frame 1, the two space-time events (event L1L2, pair of touchers on the left touching, and event R1R2, pair of touchers on the right touching) L1L2 and R1R2 are simultaneous. Which means that they are ... 1 If all the lengths in this diagram are as measured by the observer on set 1 then only lamp 1 will light up, but not lamp 2. When the left "touchers" on each set align in frame 1 then the right touchers will too, because they are both 10 meters separated from the left ones. However, the touchers on set 2 are further than 10 meters apart according to observer ... 0 The key point in writing an action for spinors is the existence of a Clifford algebra (expanded by the gamma matrices)\{\gamma^a,\gamma^b\} = 2\eta^{ab}\mathbf{1},$$where the index a runs from 0 to 3 (or 0 to D-1, where D is the spacetime dimension). The whole basis for the algebra is given by the \gamma's and all possible products... due ... 0 What I find interesting is that one event can be seen as two events by another. Can it not ? If one bolt of lightening struck the ground near the little red man, and the light expanded outward until eventually striking mirrors that are located at the ends of a train passing by, the light will eventually return to the little red man via these reflections. ... 1 "The speed of light is constant, no matter how an observer moves relative to the light." This is true. The light is moving independently, thus an observers actions of movement, make no changes to it. However, the interesting part is that the speed of light is always "Measured" as the speed of light no matter how an observer is moving relative to that ... 0 Now since the coil moves through this region of space, it should therefore possesses an induced electric field as well. I don't follow your reasoning here. The electric and magnetic fields are reference frame dependent; the fields 'mix' in a certain way via the Lorentz transformation. In the reference frame in which the magnet is at rest, the ... 0 In Faraday's experiment, the relative velocity between the coil and the magnet define whether a change of induced magnetic field flux occurs around the coil or not. So if you consider them at rest, and when you consider them moving at the same velocity relative to one another, then there's no change of magnetic field flux felt by the coil. Be careful there ... 5 Your teacher is correct. The link you give, gives the correct definitions of the various "mass" concepts in special relativity. m_0 here characterises an entity that is being described by a four momentum vector in the special relativity framework. In vector spaces, the vector has an invariant length, (otherwise mathematically it would not be a vector ... 5 It's a matter of terminology. You can define an object's mass to be proportional to its energy in the reference frame you observe it from, which includes kinetic energy in the mass, or you can define it to be proportional to the object's energy in its own rest frame, which excludes kinetic energy. Or you can do both, which is what physicists originally did ... 3 A personal point of view is that you may consider that Lorentz transformations apply primarily on momenta, and not primarily on (infinitesimal or not) space-time coordinates. This is, of course, a "strong" postulate. If you assume (some additional postulates are needed there) that transformations are linear, and that there is a rotation invariance, you ... 5 The answer is - velocities just do not add like that when we are getting close or even equal to the speed of light c. The whole body of Special theory of relativity concerns itself with what are the consequences of postulating the speed of light to be constant in all frames of reference. That is, we find transformations between frames of reference by ... 0 I'm missing something in this argument: what basic inconsistency will arise if the events were simultaneous in both the frames? The one-way speed of light will not be c in all inertial frames of reference. Essentially, the clocks at rest in each frame are synchronized according to the Einstein synchronization convention. Indeed, the Lorentz ... 1 The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues. If an object is moving with speed u, and you then apply a boost v, the object's new speed is not u+v but rather (u+v)/(1+uv/c^2). This is always less than c. Therefore it's not possible to ... 1 First, from the point of view of O. The lightning strikes at points A and B happen simultaneously. Light propagates away from those points, and since O is halfway between A and B, the light fronts reach him at the same moment (equal distances and equal velocities gives equal times). Now, for things as O' sees them. The important thing to ... 0 In special relativity, you are better off thinking of things as four-vectors, rather than three-vectors. In that case, you generalize momentum to 4-momentum {}^{1}(and you take time derivatives with respect to the clock of the spaceship). Then, you have \bf F = \frac{d{\bf p}}{dt} Since momentum is given by p = \frac{mv}{\sqrt{1-v^{2}/c^{2}}}, and ... 2 Consider$$M_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 ...

2

In what sense does the answer make sense? Does it describe some physical phenomenon? You've done the algebra right, and gotten a result, but I don't think your inputs (imaginary velocity) mean anything physically. So the answer can't be expected to mean anything physically.

1

Actually, "unitary representation" is meant with respect to the spinors, which do not form a finite-dimensional space and therefore allow a unitary representation of the proper Lorentz group. The action is defined by $D(\Lambda)\psi(x)=U(\Lambda)\psi(\Lambda^{-1}x)$, and you can simply calculate that this is unitary on your spinor space. However, this does ...

4

The thrower's height doesn't change i.e. it is the same in both the reference frames of the thrower and the bug. That's because distances normal to the direction of motion are not changed by Lorentz transformations. In the bug's frame the thickness of the thrower decreases, so the thrower is flattened in the direction of motion, but the height is unchanged. ...

2

This is taking too far the concept of relativistic mass. Einstein himself was not fond of this concept, according to his quote on this wikipedia article. If I understand your question correctly, you find an apparent paradox pursuing the consequences of plugging the relativistic correction factors ($\gamma$) in both sides of Kepler's third law. On one side, ...

2

Well, the definition of the causality relations are on Wikipedia. But they do not do what you think it does: Events that causally precede another event are merely the events that could have influenced the event they precede (assuming that no causal influence is possible that is faster than light) - the causal relation does not make any statement about "how ...

1

how is the detailed connection between the statement that the length measurement has to be simulanous and the quoted derivation? Sally didn't measure a length, she measured a time and, from that, calculated a length. Both Sally and Sam agree that their relative speed is $v$ so Sally can calculate the distance between the ends of the platform by ...

0

I think in the derivation above length was not measured by measuring both ends of the moving rod (platform) simultaneously. Einstein gives an example of a simultaneous measurement in his popular book explaining special relativity:

2

If two events both have a spacetime interval of zero, can they both be said to be happening “now”? There is an interval associated with any two events but there is not an interval associated with an event. From the Wikipedia article "Spacetime": In spacetime, the separation between two events is measured by the invariant interval between the ...

1

I'm not sure this is a complete answer to your question, but thinking about special relativity that way will get you into trouble. Essentially, that way of interpreting special relativity attributes all of its weirdness to signal delay. Here's how I think you're interpreting the barn door experiment: The ladder is put stationary in the barn and is found to ...

0

OK, just one more try to end this stupid question. There IS a way to formulate physics using light rays as your basis: a double null coordinate system${}^{1}$. If you have a ray moving in the $+x$ direction, define the two coordinates $$2\xi = t + x\;\;\quad\quad\quad2\eta = t - x$$ Then, the metric becomes $$ds^{2} = -4d\xi \,d\eta + dy^{2} + dz^{2}$$ ...

2

According to Einstein's theory of mass-energy equivalence, if the photon is a particle of pure energy, and if $E=mc^2$, then the photon is theoretically traveling at $c^2$; not $c$ You have neglected dimensional analysis: E $\to$ Joule = $\frac{\rm kg\,m^2}{\rm s^2}$ m $\to$ kilogram = $\rm kg$ which means that $c^2$ has units of m$^2$/s$^2$, not ...

9

Actually you're quite correct, though possibly not in the way you expected. Ordinary velocity isn't an invariant because obviously different observers moving at different speeds will measure different velocities. However there is an invariant form of velocity called the four velocity that is an invariant under special relativistic (i.e. Lorentz) ...

1

Instead of using existing spacecraft, let's use a photon rocket powered by the gamma rays emitted by matter anti-matter annihilations in its reacor. Where does the anti-mater come from? We will produce it using solar energy. We'll use giant solar panels that generate a huge voltage in vacuum which leads to Swinger pair production. Moving away from the Earth ...

1

From ScienceMuseum: Apollo 10 holds the record as the fastest manned vehicle, reaching speeds of almost 40,000 km per hour (11.08 km/s or 24,791 mph to be exact) during its return to Earth on 26th May 1969. Using the formula (as above). After traveling for 40 years, you would be a little over 0.86 seconds younger. Added: I did some calculating and ...

2

Almost none. Let's be much more generous than your idea of human-carrying craft. Let's just use the fastest probe. The Helios II craft, after nearing the sun, reached a heliocentric speed somewhere near 70 km/s. Obviously, its speed was more due to the gravitational influence of the sun than its engines. $$t = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}}$$ ...

2

So let's just say that the spacecraft can accelerate until it's moving away from the Earth at the speed of the fastest currently-existing spacecraft First, note that the fastest speed, relative to Earth, that a spacecraft has obtained is an exceedingly small fraction of the $c$ and, thus, one should not expect significant time dilation. For ...

0

Well, obviously it would be the speed of light times two. However, this can be misleading. Equations are all fine and dandy, but if you do not understand them, they are not of complete help. Imagine that you have the following... 1) A 300,000 km long spaceship which is at rest in space. 2) A clock is located at each opposite end of the spaceship, and these ...

0

Your notion seems to be based on the thinking that light is a bunch of photons, and a photon is some kind of weird particle that travels at the speed of light, like some tiny spaceship. Then you ask, how can this tiny spaceship violate physical laws? What makes it so special? But a photon isn't a particle in any classical sense. It's not like a tiny ...

0

Special relativity states: ... I'll select and discuss the given statements in some particular order (which may be called "in order of simplicity of discussion") ... [...] The observer is (anything [...]) Right. Synonymous to "observer" or "anything", in the context of the theory of relativity, there are also the descriptions "material point" or ...

3

Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...

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