# Tag Info

1

You seem to think of electrons and photons as point particles from relativistic mechanics that have rectilinear trajectory until they collide. This is only one of many different ways to give meaning to these two words. Within this view, if an electron collides with a photon, and if conservation of energy and momentum is assumed, the resulting state after ...

1

OK... I can't give a definitive answer to the problem. My intuition tells me that any massive particle or macroscopic mass, boosted high enough, has to look like a black hole. Why? Because it is very hard to see why/how gravity, if we believe in the equivalence principle, should be able to distinguish between kinetic energy and other forms of internal energy ...

1

A neutrino has rest-mass and travels at (near) $c$, why isn't its mass/ energy (nearly) infinite? Because it has too low rest mass or still too low speed. Neutrinos are very light particles: Their rest energy is comparable to energy of a hydrogen bond (weaker than typical chemical bound). So you can understand, that full energy must not be infinite. ...

0

Additionally to Alfred Centauri's answer I can say, that mass $m_{rel}=\gamma m$ is AUTOMATICALLY implies directional inertia, since it is not constant. Any non-constant mass causes a propulsion force. From definition of force $F=\frac{dp}{dt}=\frac{dm_{rel}v}{dt}=v\frac{dm_{rel}}{dt}+m_{rel}\frac{dv}{dt}$ we have $F_{prop}=v\frac{dm_{rel}}{dt}$ It ...

0

Whether or not it's gaining mass depends on how one defines mass. The "relativistic mass" is simply the total energy, with a factor of c squared thrown in. I recall reading that Einstein himself was against the idea of such a concept, and is quoted as saying that the only mass one should consider is the "rest mass". I would have to agree. Relativistic ...

2

No, your facts weren't gathered incorrectly, your reasoning is just incorrect. It doesn't even take knowledge of physics to answer the question, just logical reasoning. (Don't take my language as a personal insult, I'm just trying to be clear.) "It's not like time goes slow for a while, and then speeds back to "normal," so that the age of the observer once ...

4

Let's do some math, shall we? Let's call $t$ the time as measured from Earth, and let's say your engine is running with acceleration $a$ for $0 \le t \le T$. The proper time, that is, the time as measured by a clock on a ship, is given by $\tau = \int_0^T \sqrt{1-v^2/c^2}\ dt$, where $v$ is the velocity as measured from Earth. Newton's second law for ...

1

As you say, time dilation and length contraction occur when two frames of reference (observer and observed) travel at two different speeds. Both of these effects "go away" if the two frames of reference subsequently travel at the same speed; that is, time will pass at the same rate and two yardsticks will have the same length. But the EFFECTS of these ...

1

As with most apparent paradoxes in relativity, the problem is that you are neglecting to account for the relativity of simultaneity--if the rockets left at the "same time" in the frame of the station, they did not leave at the same time in the frame where rocket E was at rest during its journey across the station. Say that we place the spatial origin (x=0 ...

1

It's a trick to get the negative sign in the Pythagorean measure of distance. "Stick an $i$ here" vs "use a negative sign here" in the rules of how to figure things. It's not another different dimension nor is time a complex value. It's an x i t in the function of the interval, where $x$ is a real number. Of the several posts nominated as duplicates of ...

5

Careful with comments like "when he receives it"--simultaneity is relative, different frames will disagree about which reading on your clock happens "at the same time" that he receives the pulse. If he is 10 light years away in the frame where you were initially at rest, and you wait 10 years after sending the signal to fire your rockets, then you fire your ...

2

Time dilation does disappear as relative velocity approaches zero. The things experienced during the time experienced do not disappear; cells which have died remain dead and second hands which have ticked ahead do not reverse direction. To undo those things would require time reversal. Sine we as humans only perceive time in one direction, time reversal is ...

-2

Any change in time at all is only "permanent" because of the second law of thermodynamics and the resulting arrow of time.

10

The way I see it, time dilation is the real effect here. Length contraction (in SR) is just a consequence of the fact that the "length" of a rod is the distance between simultaneous positions of the rod's endpoints. But two observers with different velocities will have different ideas about what simultaneous is, and this means they measure different ...

8

Length contraction effects may be permanent in the same way as time dilation! You just have to choose the right example. Example: An astronaut is traveling at v=0,99 c to an exoplanet, according to Earth frame he is traveling 198 light years in 200 years. According to his frame (reciprocal gamma = 0,141) he is traveling 27,9 light years in 28,2 years. ...

4

There is an asymmetry between space and time, and that is the reason why time dilation can be permanent and length contraction can not. The reason is that you can travel back and forth in space but not in time. In turn, this is related with an asymmetry between time and space in relativity, not in the laws of motion, but in the theory itself: we cannot go ...

28

Time dilation is a comparison of rates. When an object is moving fast with respect to you, it's clock rate is slow, and when it comes to rest with respect to you its clock rate returns to normal. The time difference between the two clocks at this time is due to the accumulation due to these different time rates. That is the leftover effect of the time ...

7

There is actually an equivalent to "total elapsed proper time" along time-like curves in spacetime (which can represent the worldlines of particles moving slower than light), and that is the "proper distance" along a space-like curve (which cannot be any real particle's worldline). See the spacetime wikipedia article for more on time-like vs. space-like, ...

12

It's my understanding that when something is going near the speed of light in reference to an observer, time dilation occurs and time goes slower for that fast-moving object. According to the 'something', it is the observer's clock that runs slower and it is the observer's rulers that are contracted. That is to say, the time dilation and length ...

2

Putting CuriousOne's comment into an answer, In the theory of relativity, time dilation is an actual difference of elapsed time between two events as measured by observers either moving relative to each other or differently situated from gravitational masses. Wikipedia I see that such a definition might be misleading as it talks about time ...

0

Time dilation is just that: time dilation and not simply clock time dilation (although the former implies the latter). Any process or chain of events will take longer when observed from a relatively moving, inertial frame. We can, for example, put unstable particles into high speed rings and their lifetime will increase by the factor ...

1

The explanations involving clocks ... Textbook examples or explanations of relativity that involve clocks are often about time-dilation. ... Is valid only if the clocks is ticking ... These examples/explanations generally assume that any clock they mention is a working clock. It doesn't have to be a clock that ticks. It could be any type of clock ...

0

If I understand correctly, your question is why would time dilation for any other type of clock. Light is an electromagnetic wave, all electromagnetic waves travel at the speed of light. If, say, you have a mechanical watch it has a spring, atoms in which exchange electromagnetic signals, which travel at the speed of light. Therefore a mechanical watch can ...

0

we don't have time contraction, it's the length contraction. and we have just one of them in any reference frame but not the 2 of them together.

2

The only naturally occurring symmetry breaking radiation of this kind is the CMB. Unless you are talking about charged particles of more than approx. 1e19eV energy (in the CMB rest system), the effects are negligible, as far as I know. For those ultrahigh energy particles, however, this so called Greisen–Zatsepin–Kuzmin limit (GZK limit) forms a cosmic fog ...

1

Consider the following results: From the definition of scalar product of four vectors, $$\tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$ The usual dispersion relations: $$\tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$ The velocity $\textbf{v}_i$ in terms of momentum and energy: $$\tag{3} ... 0 You can't speak about "time contraction" if events are not occurring at the same space position. What you noticed is the relativity of simultaneity, which basically, indeed, depends both on space and time. You can even show that two events that happens at the same time at two different places in an inertial frame are such that they will happen one before the ... -2 The Schrodinger equation is non relativistic and yes, it will give different solutions in different frameworks. As the wavefunctions will be different, their square, which will give the probabilities of finding the state at a given (x,y,z) in time t, will be different. The relevant equations for relativistic situations are the Klein Gordon for bosons and ... 1 Assuming a direct hit - so traveling through about 50 km of atmosphere - at 0.1 c that would take about 2 ms if it didn't get slowed down too much by the atmosphere. What about drag force? Let's assume a radius r, density \rho, mass m = \frac43 \pi r^3 \rho. If it is a sphere, it experiences a drag force F=\frac12 \rho_a v^2 C_d A. Putting \rho_a=1 ... -1 The reason that the Bucket knows that it is spinning is that the Universe has a horizon in much the same way that the Earth has a horizon. The Earth's Horizon is curved line that is there not only because of Perspective Geometry, but also because the Earth is curved. Because the Earth has a two dimensional surface the height tilts back away from the ... 3 Your starting point is incorrect. You say: The point is, Rindler's observer shows us that the "action" of an accelerated observer on space-time is non trivial (there exists a black hole behind a uniformly accelerated observer). You're correct that there is a singularity, but it is only a coordinate singularity. The Riemann tensor is everywhere zero in ... 0 For Wheeler's delayed choice experiment, the 'choice' of a photon is said to be delayed from the reference frame of the experimenter. There are a couple of things to point out here. First the delayed 'choice' is not an actual thing (e.g. the photon doesn't go backwards in time through the slit and make a different choice), it arises from a common ... -2 This is not true. I will prove that we are traveling at the speed of light. Step 1:$$E=Fd=mad=mv/t\cdot d=mv\cdot v=mv^2$$but$$E=mc^2$$so$$c^2=v^2$$but if object is at rest then C cannot be zero(it is a constant) and that simply implies that object at rest are traveling at the speed of light! 2 In relativity there's no objective frame-independent way to compare the rate two clocks at different locations are ticking--different coordinate systems can give different answers (ultimately this is due to the relativity of simultaneity). There is also no frame-independent notion of speed, so you can't say in any objective sense that clocks moving at high ... 2 This is the answer for the "continuum" bit of your question or what happens if Bob leaves for Vogon on a sublight spaceship. I've assumed that Alice and Charlie are one light year away from each in the rest frame of either and that they are at rest with respect to each other (ie the separation will remain one light year for the entire experiment). In ... 1 Alice: 2 years Bob: No time has passed Charlie: 2 years There are a couple of things to point out: First two planets aren't likely to be stationary with respect to each other, they will be rotating at different rates around different suns that are themselves in a solar system with it's own movement. But for a realistic scenario (e.g. another planet in ... 0 There is no length contraction in circular movements! Length contraction is only in the direction of movement. A cyclic universe would be e.g. a 3-dimensional universe curved in a fourth dimension, or if a cyclic dimension is curved in a second dimension Thus length contraction could happen only locally, where the curving is not perceivable. 7 Why do you assume the length needed to "span the cyclic space" would be the same in different frames? With cyclic universes it helps to think of them as equivalent to an infinite universe where matter just repeats cyclically, see the "tiling diagram" with the bee and the spider on this page. From this perspective, it's clear the width of a given tile shrinks ... 1 All you need to do is conserve energy and momentum in the lab frame. Firstly you conserve energy in lab frame: $$E_{\gamma 1} + E_{\gamma2} = E_{\pi} = 1.3GeV$$ Then you work out what the pion's momentum was (still in the lab frame) using the mass-energy-momentum relation where the E_\pi is the total kinetic and mass energy: ... 0 Would receptor A be hit by the light after 1 second? Yes because they are separated by (1 light second) in the same frame of reference. If so, would that prevent receptor B from being hit by the light? Not necessarily, it depends on the size (thickness) of the ray. If receptor A displaces receptor B below the line of contour of the ray, then ... 1 I believe that one could rephrase the question as "if the limit of the speed of sound in a medium must be the speed of light in vacuum, what does that mean for the limit on rigidity of an object?" Speed of sound is given by$$c=\sqrt{\frac{E}{\rho}}$$- it depends on both density and Young's modulus. I would consider "rigidity" to be just the modulus, and ... 0 The simple answer is to point out that the Riemann curvature tensor and its various contractions are invariants i.e. they do not depend on the coordinates that you are using. For a non-rotating observer in flat spacetime the Riemann tensor vanishes everywhere, and therefore for the rotating observer the Riemann tensor must also vanish everywhere. Spacetime ... 2 "First, is my problem formulation correct with respect to special relativity, and second am I correct in how I'm solving the problems?" Your equations are perfectly ok. However, your scenario 1 seems suspect. Why would the sum of the rest masses be conserved in a relativistic inelastic collision? Scenario 2 seems fine to me. I like your quest for ... 0 Yes, when we heat an object its mass increases. The complete equation is$$E^2=(pc)^2 + (mc^2)^2$$And from this equation if a system has zero momentum (p=0), then it has energy E=mc^2 When you heat an object the molecules or atoms begin to vibrate, rotate with more kinetic energy. But this doesnt increase the momentum of the system (the object is a ... 2 All internal energy such as thermal, rotational, and internal potential energy contributes to the rest mass of an object. In fact the vast majority of the mass of an atom is due the internal energy between quarks that make up the nucleus rather than the rest mass of the quarks themselves. So yes, a hot objected has greater rest mass and would weigh more ... 2 Problem 1: Oftentimes, for collision problems with relativity, it helps to look a the problem in a frame where the total momentum is zero. For this problem, we create a new observer traveling at such a speed that Mass 1 and Mass 2 are heading toward each other at the same speed (since they have the same mass). We'll call this speed v_{cm}. The velocities ... 1 If you think of a right triangle, the base of the triangle is the mass, the height of the triangle is the momentum, and the energy or relativistic mass is the hypotenuse. A felt acceleration in the lab corresponds to a change in velocity (\frac{momentum}{mass}). This is the tangent of the base angle of the triangle. It appears to the lab occupants like ... 1 As mentioned by Qmechanic, the answer to your questions is no. However, assuming space-time is oriented, we have the following: For any pseudo-Riemannian metric g, there exists a normalized time-like one-form h^0 and a Riemannian metric g^R so that$$ g = 2h^0\otimes h^0 - g^R  This yields a locally Euclidean topology compatible with the manifold ...

4

The acceleration you'd measure in a lab with an accelerometer would be the proper acceleration, which in relativity (unlike Newtonian physics) is distinct from the coordinate acceleration in some inertial frame (though at any given moment, the proper acceleration is equal to the coordinate acceleration in the inertial frame in which the object has an ...

1

Consider the case that $p$ is space-like. Then, there is a frame in which $p = (0,\sqrt{p^2},0,0)$. Now, define $p_+ = (\sqrt{p^2},\sqrt{p^2/2},0,\sqrt{p^2/2})$ and $p_- = (-\sqrt{p^2},\sqrt{p^2/2},0,-\sqrt{p^2/2})$. Obviously, $p_\pm$ are light-like and $p = \frac{1}{\sqrt{2}}(p_+ + p_-)$. Proceed analogously for the time-like case.

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