# Tag Info

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No. The charge enclosed by a surface is really just the number of protons inside minus the number of electrons inside all multiplied by the charge of the proton (which is a constant). If the electric field has a large density in one region then it has less in another so that total flux over a closed surface is still the net enclosed by the whole surface is ...

1

Entanglement experiment performed in one frame of reference guarantees that the two measured results are synchronized in this frame of reference. If we try to perform the same entangled experiment while the two ends of the same length fiber optic lines are attached to two frames of references moving with a constant relative velocity, the two measured results ...

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Imagine a typical path. It starts at event $A$=$(ct_A,x_A,y_A)$ and ends at event $B$=$(ct_B,x_B,y_B).$ (Note I'm using $w=ct$ as the fourth dimension so we can measure everything in meters.) In the frame of the fluid this typical path has length $l_0.$ In both frames the separation is null. We can actually choose the origin of the fluid frame and your ...

1

Perhaps it's more illuminating to look at the whole thing in a spacetime diagram. we have the earth frame with coordinates $(t,x)$, and its trajectory through spacetime is the blue line. The trajectory of the spaceship is the red one. Straight worldlines are inertial frames of reference, curved or non-straight worldlines are non-inertial frames of ...

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The axis of simultaneity, or in other words, the set of events which are simultaneous as measured in the rest frame of the ship, does indeed change suddenly when we turn back. This is because it depends on your reference frame. There isn't a single inertial frame that stays with the ship for the whole journey; you can either accept that the frame is ...

1

First let us cut away some of the meat and the bones of your question. Let us forget about thermodynamics for a minute and classical physics; however we will need GRAVITY (not Newton’s version). Instead of calling it Gravity let us call it General Relativity or (GR). We can still use the term Gravity, but when discussing Quantum Mechanics it’s better to ...

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If you are abit more careful about making the statements, then "both perspectives" are actually correct. Let me be more concrete and explain. Let $E$ be the one who stays on Earth and $S$ be the one who is on the spaceship. The first issue you have is you said "let's say seven years passes on Earth" - this is an ambiguous statement: from whose perspective? ...

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No. A heavier object will gain more kinetic energy when falling than a lighter object but it has a higher potential energy to start with. It is actually not kinetic energy that is needed to fall but potential energy that is converted in kinetic energy while falling. Generally, while the gravitational force on an object depends on it's gravitational mass, so ...

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First of all I would like to thank you for letting me participate of this discussion. I've been trying to get an answer for this question myself for at least six years now. Ever since I've realized that the experiment hasn't been done. Then one day in the same year (2009) I found a video on YouTube made by a German Blacksmith named Martin Grusenick. We've ...

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I'm struggling to identify whether a scalar is a Lorentz-scalar. E.g: ∂iAii∈1,2,3. How do I determine if this is a Lorentz-scalar or not? If got the same problem with tensors. How do I differentiate between a tensor and a Lorentz-tensor? By "tensor" do you mean a tensor in three space dimensions? If so, then a tensor is never a Lorentz tensor. A Lorentz ...

3

Absolutely not. Both you and the people on Earth would see each other moving $7$ times slower. To repeat myself: you would see them going $7$ times slower and they would see you going $7$ times slower. This is because of the main principle of special relativity. As long as neither of you is accelerating there is nothing to choose between your frame of ...

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You can look at the problem like this: Whether or not one person gets to see all the other persons as connected for some period of time depends on 1) the disk size, say radius R 2) the duration $\tau$ of the connection in the disk frame 3) the speed $v$ of individual components on the rim relative to the disk. Let's denote $T$ the astronauts' period of ...

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In theory this effect is always present, but it may be less present compared to the photoelectric effect. But it is noticeable only if the frequency of the incoming photon has (at least) a magnitude of 10^20-10^21 Hz. If the frequency used is less, the variation of the wavelength is negligible and so I could not identify it.

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The Compton effect is the inelastic scattering of photons by electrons. Compton's initial experiment used electrons in a graphite crystal to act as scatterers. These electrons are not free, they are bound, but the X-ray energies (17 keV) were large compared with the binding energies, so they approximated to free electrons. Photons of lower energies (UV to ...

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Your confusion comes from the fact that you are IMPLICITLY reasoning using the concept "absolute simultaneity", that is: "if two events in different points of the space are simultaneous for one inertial observer then they are simultaneous to all inertial observers". One of the most important features of Special Relativity is precisely that there is NO ...

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You have a good intuition on the possible answer because it does involve oscillations, but you need to visualize the string differently. The origin of the conundrum is not quantum, but relativistic. Here's why: Consider an inertial frame O and set up a string spanning the entire $x$ axis, stretching from $x \rightarrow -\infty$ to $x \rightarrow \infty$. ...

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Simple answer: no. Reason: Let's say you are a photon. A mirror is placed in from of you. If you do not smash in to it then it is traveling as fast as you. there for, the photon will never reach it to be redirected backwards.

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I am tempted to give an experimentalist's point of view, though of course I agree with Lubos and Savanah's answers. Consider a disk of radius R, then circumference is 2πR. Now, make this disk rotate at velocity of the order of c(speed of light). Here it is evident that you are defining a center of mass system for a disk and a person/machinery that will ...

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SR is based upon two postulates: 1)The laws of physics are the same in all inertial frames of reference. 2)Speed of light $c$ is constant for all inertial frames of reference. To arrive at some results in relativity, one only needs to assume one of the postulates, other results are logical conclusions that require assuming the two postulates together. ...

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So Special Relativity states that for all non-accelerating objects of matter the laws of physics are the same. I think the point is just that the constants and the time and space derivatives that appear in a law of physics should not have to change the form of the equation if you measure the time and the space in two frames that move relative to each ...

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When the people see the light they fire thrusters to get up to speed and move in a circle and then after that they fire their thrusters just to maintain uniform circular motion... OK, we have a disk, and light, and a rotating ring of people. No problem. So there are frames where all of the one events happen before all of the other events. And ...

1

It is like to have two numbers each smaller than the other, it is not possible. It's not like that at all. Consider the following true statement: B observes A's clock to run slowly while A observes B's clock to run slowly. This is not a contradiction due to the crucial word observes. Consider an additional true statement: The observer at the ...

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To someone moving on the edge of the disk, the disk is length contracted in the direction of travel. So to them, it is not a circle, no problems with geometry. Euclidean geometry only holds for an inertial plane of simultaneity. By the end of this answer you should know when and how to use Euclidean geometry and what it means (or doesn't). And a collection ...

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Although the answers here are good, in my opinion, they're too mathematical for the OP to get it. Before explaining what Brain Greene intended to convey in his article, I'd like to clear up your misunderstanding. There's no such thing as "time speed" vs "space speed", speed is the magnitude of one's velocity, and velocity is the rate of change of one's ...

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This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way. So I'd like to address the concept of geometry by not having ...

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from the observers perspective information has travelled instantly, the firing of the first guillotine has turned on the light with no delay. Herein lies the flaw in your reasoning. The electrical signal doesn't instantly propagate, it propagates at the speed of light. From the perspective of every observer, the electrical signal from the sensor reaches ...

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No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ...

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What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ...

1

Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$. So we have: \frac{dx^{1}}{dx^{0}} = \frac{d ...

0

If you have two vectors you can project one onto the other. It you orthogonally project a spacetime vector onto your unit tangent then the length of that projection is how much time you observe separating the two events. That might be enough to answer your question right there. So to get an average velocity between two events first compute the vector A ...

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Hint: $T_1=t_A+\dfrac{\text{distance covered by the light beam in the interval} [t_A,t_1] }{c}$ $T_2=t_B+\dfrac{\text{distance covered by the light beam in the interval} [t_B,t_2]}{c}$ You don't get the second terms in these two equations right. This is your only problem. Try to do it again. Your mistake was that you considered the distances traveled by ...

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Imagine a sensor on the front of the train which detects when a guillotine passes infront of it, this then sends an electrical signal down the train to a light at the back of the train which, I think, would turn on at the same time as the back guillotine drops, from the trains perspective That's not possible. Instead of a detector on the front ...

-2

Using this equation for the events in question make the first message received at 32.732seconds Earth time and the second message 360.0595seconds Earth time. The difference between them is 327.3275seconds.

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$T_1=\gamma*(t'_1)$ $T_2=\gamma*(t'_2)$ Above both equations are correct. But $T_1 \neq t_A+(0.4c*t_A)/c=1.4t_A$ $T_2 \neq t_B+(300*c*sec-0.4c*t_B)/c=0.6t_B+300$ Check your calculations again.

1

There are two ways to look at the error in this calculation: 1) It's the wrong Lorentz transformation. The Lorentz transformation takes the unprimed coordinates to the primed coordinates, but we need to obtain the unprimed coordinates from the primed coordinates. The correct "inverse" Lorentz transformation is: $$x = (x' + vt')/\gamma$$ $$y = y'$$  z = ...

1

This tells us that your calculations are wrong. Light from the supernova represents the information of the supernova. Information is limited to the speed of light and so the information that the supernova took place cannot outpace the light from the supernova. Even if it is along a direct line-of-sight between the supernova and Earth, the spaceship cannot ...

1

Timaeus's answer could be correct. The $\Lambda$ matrix from your book may have been intended as a passive transformation (one that acts on the coordinate system) and you mistakenly used it as an active transformation (one that acts on the object). Alternatively, the $\Lambda$ matrix from your book may have been intended as the active transformation of a ...

1

Yes, the energy and the increase in energy all depend on your reference frame, but this is NOT special to relativity! The same thing happens in classical mechanics. I wrote a similar answer to the question, "Can you tell your absolute speed in space?" Consider the regular Newtonian mechanics equation, $\mathrm{Ke}=\frac{1}{2}mv^2$. If you weigh 50kg, are ...

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The Theory of Special relativity tells us about the relativistic observations of the observer in an inertial frame (frames moving with constant velocity or experiencing no acceleration). The postulates of relativity drives the theory and make some amazing predictions about the relativistic observations of an inertial observer. Let's talk about a situation ...

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The first thing that bothers me is that for positive ${v}$ I get negative component of the velocity vector If your object was at rest then to someone moving to the right the object at rest is moving to the left in their frame. As for the second question it is same issue. Someone at rest will be moving to the left relative to someone moving faster than ...

2

The answer is YES. It's true that the coordinate system (u′,v′,w′) is related to (x′,y′,z′) also by the orthonormal matrix A, at least under the Lorentz Transformations used in the following. But please, let use other symbols (for example it's custom to use $\;\upsilon\;$ for the algebraic magnitude of the velocity $\:\mathbf{v}=\upsilon\mathbf{n}\:$). ...

0

Is the concept of mass increasing as it approaches the speed of light based on the fact that this is only true relative to the observer? When you say mass increases what you mean is energy increases. You don't want to have a different mass for forces in different directions so the idea of mass changing with speed is now abandoned. You have energy ...

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The famous twins paradox [...] can be resolved with general relativity pretty easily It can be resolved with special relativity. Clocks (and aging people) measure proper time along curves in spacetime. You can compute the length of a curve by breaking it into small pieces and approximating each small piece by a line segment and adding up the proper time ...

2

This is not exactly the twin paradox, but it's close. First, let's make the problem more precise. Let's assume the train is one light year away from a planet, traveling near light-speed, and at the very beginning of the journey, a person on the planet views the clocks as synchronized. Then a person on the train does not view the clocks as synchronized. This ...

2

If both twins undergo inertial motions, parting from one another at point $\mathscr{A}$ and meet again at point $\mathscr{B}$ in spacetime, you are simply describing a situation where the spacetime simply has multiple geodesics joining the two points $\mathscr{A}$ and $\mathscr{B}$. Either twin can age more than the other, depending on which geodesic has ...

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The article you linked to is not about special relativity. There is no problem with special relativity. This is a paper that Einstein started, but did not publish. As the article says, he did not publish it because he saw the mistake in it. The article is disappointing, but not surprising for science reporting. The only reason it is newsworthy is that it ...

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The NPR article you posted discusses Einstein's attempt at solving Lemaitre's theory of and Hubble's discovery of the expanding universe (i.e., the Big Bang) It was trying to solve a big problem of the day. An astronomer named Edwin Hubble had just observed that everything in the universe is moving outward. ... Faced with evidence that the ...

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I think Duffield's answer can be summarized better this way: a putative "worm hole" is our view of a bending of 3-space thru 4 spatial dimensions such that the Euclidean distance from Earth to $DISTANT_PLANET is extremely small. In our 3-D universe, we can't move along this putative 4th dimension, so we're forced to go the "long way" thru space which is, ... -2 How would the twins paradox be affected by wormholes? Not at all. The so-called paradox is that each twin claims time is "passing more slowly" for the other twin. People say Whoa! Paradox! to this, but it isn't a paradox at all. When you and I are separated by distance, I say you look smaller than me and you say I look smaller than you. But we don't ... 0 Let's attack the confusion from the other side. People say light always travels at speed$c.$Light always travels at speed$c$relative to an inertially moving observer. (Or to an inertial frame that is momentarily moving with your observer.) People say that nothing can go faster than$c$Nothing can go faster than speed$c\$ relative to an inertially ...

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