# Tag Info

2

Yes you would need a new representation. The reason is the Lorentz group is connected with the group of translations and as you point out the time-translation is different. If $K^i$ is the generator of boosts in the $i$ direction and $P^j$ the generator of space translations (i.e. the linear momentum) $$[K^i,P^j]=iH\delta_{ij}$$ So if the interacting ...

0

You are right, in the frame where $O$ is at rest, the time $P1$ and $P2$ take to travel across $O$ will be the same, call these times $t_1=t_2$ But it is also true that in $S$, $P1$ will take an longer time to cross $O$ than $P2$. How is this possible? There are various ways to begin understanding how. Here is one. Go back to the rest frame of $O$, and ...

1

Another important example: the Pythagorean Sum defined by $f(x) = \sqrt{x}$. As Doetoe says, you're simply transporting the binary operation + to the domain of $f$ if $f$ is e.g. bijective. If your domain is a real interval, your question actually describes the following: Which operation is calculated by the addition of affine co-ordinates implemented by ...

1

Yash, Imagine S is the Sun sending two photons, P1 and P2, and the object O is represented by two asteroids, O1 and O2, equidistant from each other all the time and from S at $t_0$ - they are moving in the same direction at the same velocity (c/2). So for some time one will be moving towards the Sun and one away from it. So you are right that the speed of ...

1

Let $X$ be a set with a binary operation $\star:X\times X\to X$, which could be a group structure or whatever you want. Let $f:X\to Y$ be a mapping to some other set and let $F(x,y) := f(x\star y)$ (your middle expression is this one with the operation +). If $f$ is sufficiently nice, e.g. a bijection, then your construction transports the binary operation ...

1

The Bohr model is not an accurate model, electrons do not move in circles with a statistical distribution of circles. Even if they did, just because two charges have the same velocity does not mean the other one sees the other as static. The force one charge $q_1$ feels do to another charge $q_2$ depends on the position and velocity right now of the charge ...

0

Simpler setup. Have two people with a meter stick between them, all mutually at rest. The are in the same frame so they synchronize their watches. Then at the same watch time they send a signal to each other and hold a mirror and bounce the other one's signal back and then when their own light comes back they look at their watch. If their watch says it ...

0

Another take on Omar Nagib's Answer is that one justifies the use of velocity $-v$ for A relative to B whenever we use $v$ for B relative to A by what is sometimes (rather vaguely and unhelpfully) the reciprocity relationship. As Omar notes this relationship is not particular to special relativity but also applies to the Galilean velocity addition (infinite ...

1

The answer to 1) is that when you say there are two observers in a relative motion $v$, Then by definition it follows that each observer claims he's at rest and it's the other observer who's moving with speed $|v|$. Or to put it another way, consider the relativistic velocity combination formula: $w=\dfrac{u-v}{1-\dfrac{uv}{c^2}}$ $u$ is the velocity of ...

0

Additional possibility: when light travels through a medium such as water, it slows down substantially. If light leaves water and enters air, it does in fact speed up when it enters the air. I have no idea how to calculate the time required for this "speed up" to happen, but if the time interval can be calculated or measured, you can calculate an ...

0

Imagine sitting on an airplane and playing with a laser pointer. As you fire the laser beam upward perpendicular to the motion of the plane you see it go straight up to the ceiling. The same would be true on a spaceship from the point of view of its passengers. As for the people watching this happen on Earth, they will see you fire the photon at an angle ...

1

The speed of light is constant in a vacuum. However, it can change direction in the presence of gravity so in a sense it does accelerate.

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So a train passenger has synchronized two clocks at both ends of the train. Then you, a person standing on a platform, synchronize two other clocks at both ends of the train. Let's say the two clocks at the front of the train read the same time. Now the difference of the clocks at the rear of the train is the synchronization "error". It's your ...

2

The whole problem is really about knowing what the words mean. An event is a time and a place together as single object. For instance the event where a light sends its first, last, or only pulse. Or the event where a beam or particle touches something and bounces. Anything you can describe with a time and place together is an event. And different people ...

1

As you know, dealing with "simultaneous" events in relativity is tricky. If I think two things happen at the same time, you may not. However, it is guaranteed that if if I think two things happen at the same time and the same place, you will agree. That's because these two things are just the same spacetime point. In your frame, we have the simultaneous (in ...

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$$E=\sqrt{(m_0c^2)^2+(p c)^2}$$ And time dilation is given by $$T={{T_0} \over {\sqrt{1-v^2/c^2}}}$$ Keep in mind that the photon is the force carrier for the electromagnetic force. Now, what happens if the speed of light reduces to just $1.1v_s$ where $v_s$ is the speed of sound? Nothing at all! Sound waves at the atomic level are caused by the electro ...

0

I think that if light had 110% the $v$ of sound the, that would be the max velocity in the universe. So, if we say that everything in this world moves at the same speeds as in ours and the only change is that of $c$, we would have relativistic speeds in this world and as a consequence relativistic phenomena. Well, this world can not be the same as our, ...

0

The other answers are correct, but I think there is one notion that hasn't yet been explored. The rotation group is compact. The Lorentz group is noncompact. Take any element $X\in\mathfrak{so}(N)$ from the Lie algebra $\mathfrak{so}(N)$ of the former and $e^{\tau\,X}$ is a periodic function. Not so for the Lorentz group: no finite sequence of finite ...

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It's important to state exactly what one means by "aether" when saying that aether theories are discredited. Specifically, the notion of medium that one can in principle detect one's motion relative to is what has been ruled out by experiment. Mediums such as water for acoustic waves fall into this category: the acoustic wave equation changes its form when ...

1

Sorry, but the Michelson-Morely did not prove that there is no ether wind. It proved that, if there is an ether wind, it is at rest with respect to the surface of the earth, regardless of the motion of the earth with respect to the rest of the universe. While it is extremely difficult to imagine how this could possibly be, since it requires that the ether ...

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From Wikipedia: James Clerk Maxwell began working on Faraday's lines of force. In his 1861 paper On Physical Lines of Force he modelled these magnetic lines of force using a sea of molecular vortices that he considered to be partly made of aether and partly made of ordinary matter - light and monopoles. He derived expressions for the dielectric constant and ...

1

Einstein didn't actually get rid of the aether. He said the luminiferous aether was redundant when he was doing special relativity in 1905. But later when he was doing general relativity, he described space as an aether. See his 1920 Leyden Address. He said this: "Recapitulating, we may say that according to the general theory of relativity space is endowed ...

0

You can easily understand this if you read about relativity of simultaneous events. Events which occur to be simultaneous in one frame of reference are not simultaneous in any other frame of reference. You can not calculate the time left for you to react based on your intuitions. You should carry out the calculations strictly using the relativistic ...

0

Let's say that the Earth is stationary, the two rockets have the same speed v and the refrence planes are: K for Eath, K' for rocket1 and K'' for rocket2. We can separate the problem in to two parts. 1st part: The rocket goes from earth to a distance $L$(in K). Because the time $t0$ is the same, at distance $L$ the time $t1'$ is equal to $t1'=γ(t1 - ... 3 From a perspective on the spaceship, the radar beam travels away from me at the speed of light, so I might imagine that I would have ample warning of an object positioned at rest one light-hour away. One light-hour away in which frame? If that one light-hour away is from the perspective of the rest frame of that object, you don't have much time at all. ... 1 If the pulse of light is going directly down, it will miss the base of the mast. There ought to be nothing confusing at all about the speed of light being independent of the speed of the source. Light has an E=hf wave nature. Sound has a wave nature too, as do ocean waves and seismic waves. The speed of the waves depend on the properties of the medium. In ... 0 What you have implicitly done is to create a single preferred frame of reference, the one in which the sailboat's velocity and the velocity of the beam of light are referenced. That is very contrary to relativity theory. The ball is just a distraction here, as is gravitation. Instead of that sailboat with an impossibly tall mast, imagine a spaceship with ... 3 Per Bort's comment, this is easier to think about from the spaceship's frame rather than the debris's frame. A rock is traveling toward your spaceship at .99 times the speed of light. You send out a pulse of light that intercepts the rock when it's one light-minute away from the ship. The pulse bounces back and arrives at the ship 1 minute later. The ... 0 As stated in the comments: Even though the radar signal may come back at the speed of light at any frame, it would require an enormous distance for you to react to any resting object. If your computer could re-calculate the routes in$\frac{1}{c} \textrm{ s} $(s.i system) it would take you at least$1\textrm{ m}$to have enough space to manouver, since the ... 1 In the MM experiment, it does not actually matter whether the light beams are exactly perpendicular: what matters is that the light traveling along one "leg" would experience the ether drag head-on, while the other would experience it from the side. As the table rotates, the effect of the ether (if it exists) would shift from one arm to the other, and would ... 4 The short answer is yes. An object moving at non-relativistic velocity$\vec{v}$in a weak gravitational field will have a proper time$\Delta \tau$elapse that is related to the time$\Delta t$on distant clocks (far from the Earth) by the equation $$\frac{\Delta \tau}{\Delta t} \approx 1 - \left( \frac{1}{2} \frac{v^2}{c^2} + \frac{G M_E}{r c^2} \right) ... 0 Michelson and Morley compare the speed detected when moving at two different speeds (going with the rotating earth and perpendicular to the direction of the rotating earth). It observed no difference. If an astronaut moved in two different ways and compared the speed detected when moving at those two different speeds they should also get no difference. ... 1 Not quite. Galilean relativity, which has absolute time, is definitely causal; you can't have future events influence past ones because all observers agree on all times. However, it is interesting how much relativity you can get with incomplete axioms. There's a derivation on page 38 here that shows you can get the Lorentz transformations, but with a ... 0 Based on this post: Is causality a formalised concept in physics and the content within, the issue of causality seems to me to be a philosophical as much as a physical question. So I would guess there is no clearcut answer to your question. Have a read, see what you think, and see will you end up as confused as everybody else seems to be regarding the ... 1 Both give the correct time. The frequency of a pendulum is:$$ \nu = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$$Whatever your height above the Earth, if you measure$g$and measure$l$then use the calculated frequency as a measure of time, the pendulum will correctly measure time and its measurements will agree with the atomic clock. If you fail to take account ... -3 Twin paradox modified for a periodic universe I do wish people would keep speculations out of physics lessons. The Planck mission found no evidence of any kind of toroidal topology. There's no evidence at all that we live in "spatially periodic" universe. Besides, the old game was Asteroids, not Pac Man. Can part (b) be solved only using special ... 0 Does a ticking watch have more mass? Yep. As does a wound-up watch, or a hot watch, or a watch that's higher up in a gravitational field. It claims that a ticking watch has more mass then a non ticking watch due to the intrinsic KE, PE and thermal energy of the watch's internal movements that manifests itself as part of the watch's mass. Is this ... 0 What's wrong with it is that they didn't define the two watches very clearly. It's obvious to assume they meant identical watches (my watch is more massive than my wife's, ticking or not) but they don't specify the conditions of the watches. Really though, we only need one watch and an arbitrarily accurate scale to show what's going on. Let's say our watch ... 0 It is correct to say that we measure the mass of moving objects to be greater than an object at rest relative to our reference frame. Another way of saying moving object is- an object with kinetic energy. Think about it this way, if moving objects didn't appear to gain mass then as you give them more kinetic energy they could accelerate to the speed of ... 1 It's not so much that it is ticking that is crucial here but the fact that the watch is in a higher energy state that in its, say unwound state. This fact increases the rest mass of the watch by an amount$\Delta E/c^2$, where$\Delta E$is the potential energy input to elastically stress the spring and thus wind the watch. A wound, broken, unticking watch ... 0 It would certainly seem that Alice is younger, assuming that Bob is the one in the inertial reference frame described at the beginning of the problem. The logic would bethat as measured in Bob's frame,$\Delta x_B = 0$,$\Delta x_A = R$($R$here being the "radius" of the Universe; in other words, Alice has "come back" to Bob's location), and$\Delta t_A = ...

2

How is relativity related to anti-particles? As far as I know relativity doesn't say anything about antiparticles. But particle physics does. Have a look at the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". An electron ...

1

If I measure the speed of a particle in the lab and then write down in my notebook the value I measured, the number an observer in a different inertial frame reads from my notebook will be the same (although the numerals may be Doppler shifted, length contracted, etc.). In the same way, the name of my cat is "Mittenz" independent of any choice of ...

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Yes, any (measured) momentum value (of someone or something specific, with respect to a specific system of suitable participants) is invariant, i.e. unchanged by any and all coordinate transformations, such as Lorentz transformations; and so is any (measured) value of velocity, or of speed (of someone or something specific, with respect to a specific system ...

3

I would like to hear a deeper explanation of what we believe anti-matter to be, why it annihilates with matter and how this relates to relativity. This is the table of elementary particles deduced from innumerable measurements: Each particle has a characteristic mass and several characteristic quantum numbers. To each particle there corresponds an ...

1

The statement that a positron is like an electron moving backwards in time is in itself perfectly explainable with classical physics. As the charge of the particles is opposite, the force caused by the electrical and the magnetic field i.e. q(E + v x B) will be opposite. So, fields accelerating electrons, will decelerate positrons at the same rate and vice ...

1

If you with your rods and clocks are in free fall (ie: your metric is the Minkowski diag(-1,1,1,1) ) in a vacuum and the light ray passes near you, you will always measure the standard speed c= 2.99792458 E+8 m/sec. However, the speed of light is observed to be different if the observer and his rods and clocks are in a different gravitational environment ...

1

The answer to your question depends on fine definitions. Locally the speed of light is always the same; more precisely, the universal, Lorentz invariant speed $c$ (which is also the maximum speed of a cause-effect relationship and experimentally observed to be the same as the speed of light) is constant. This means that any measurement of light speed in any ...

2

We can write total energy $E$ two ways: $$E^2=p^2c^2+m^2c^4 \\ E=T+mc^2,$$ where $T$ is kinetic energy. Eliminating $E$ from those two equations will give you the desired result.

2

Hint: $T = E - E_0 = m\gamma c^2 - mc^2 = mc^2(\gamma -1)$ and $p = |\vec p| = m\gamma |\vec v| = m\gamma v$

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