New answers tagged

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There is no "point of view of the photons", you can't attach a frame of reference to them. Best if you imagine the photons as waves, and as these waves are propagating with $c$. The classical time dilatation, length contraction formulas (you know, everywhere the $\frac{1}{\sqrt(1-\frac{v^2}{c^2})}$ in them) are defined only for macro-sized objects going ...


1

You have two questions. Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?" The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your ...


4

You're both half right. Your teacher is thinking of the invariant mass or rest mass(see footnote [1]), which is thinking in the more modern and much simpler paradigm of naming properties of a particle-like "thing" in a frame of reference at rest relative to that thing and you're thinking of the transformations of the properties to effective values as seen ...


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This may not be exactly elegant, but it's pretty elementary so it's alright. Label the velocity in the rotating reference frame $v_r$. Let's consider the tangential component and radial component of this velocity, $\frac{dx}{dt}$ and $\frac{dr}{dt}$. From an inertial reference frame, we know that the force required to rotate a body around a point at a ...


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In your question you have proposed the statement: space geometry is changed if and only if there is a kinetic energy difference between them. I need to tidy this up just a bit, so let's consider this related statement: the geometry of space-time is changed if and only if one observer is moving (or even better yet, accelerating) relative to the ...


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Light emitted from the sun is gravitational time dilated, which means it travels slower than the speed of light as it is measured here on earth. I don't know if the difference is as great as the questioner assumes, but it would be significant. Einstein's relativity theories are PRINCIPLE-BASED theories. They are based upon established scientific ...


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There's an exhaustive discussion here.


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I think confusion of the OP is related to the concept of the frame of reference. Frame of reference is a position in the space not a coordinate system. In other word, frame of reference is the position that measurements are evaluated with respect to it and for this evaluation, one can use any kind of coordinate system. So, the book is right when says: ...


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I think inertia doesn't depend on speed, it depends on rate of change in speed, i.e. acceleration. The higher you accelerate the more will be the inertia. It can be understood by taking an example of a motorcycle, in which lower gear gives more traction than the higher one. The higher the acceleration you want the more traction is required due to inertia. ...


2

As is discussed in this answer, the "rest frame" of the cosmic neutrino background would be very similar to that defined by the cosmic microwave background if neutrinos were very light (say $<0.1$ eV). The Sun would be moving with respect to this frame at around 370 km/s. But if neutrinos were more massive(say getting on for 1-2 eV) then they are ...


4

The Earth "compells" an aircraft to rotate with it through the fluid drag of its atmosphere. So a practical answer to your question is then "above the atmosphere", which is at about a $100{\rm km}$ height. This is the von Kármán line, which is often taken as the definition of the edge of space. The definition is made because at this height, a standard ...


1

As soon as you get above the atmosphere (about 100 km off the surface of Earth, give or take), then there's nothing in particular that compels you to follow the Earth's rotation. Of course, once you get there, you will probably already be moving to some degree, depending on which mechanism you use to get yourself up. If you do, however, you can bring along ...


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It doesn't depend on the height. Now you are on the rotating Earth, so you rotate with it (around the axe of Earth's rotation). Its speed is between 0 (on the poles) and around 1.5 Mach on the Equator (1 Mach = the sound of speed). If you want to compensate this rotation, there are many ways, for example, you can simply sit on an airplane capable to go ...


0

If you were at the equator, your ground speed, due to the rotation of the Earth, is around 460 m/s (1 000 mph) so you are really moving. Now say you jump into the air and somehow you delay your fall by 10 secs. Will the earth have rotated a bit under you? No, because you have the same horizontal speed as you had on the ground. This is why rockets are ...


1

Let's try to make the answer as simple as possible. Static weight is written as $w=mg$.(Notice Newtons 2nd law looks similar $F=ma$.) This is $$\mathrm{mass} \times \mathrm{one\ unit\ of\ gravity}$$. On earth, $g=9.8\frac{m}{s^2}$. On a smaller planet, $g$ is less. On a larger planet, $g$ is more. It is important to understand that despite acceleration in ...


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There are two very similar question today itself. I am pasting my answer from the other one here - It is shown in movie "The contact" where it passes just a second or so on earth, but during the same time, the astronaut records many hours of static. What people on earth saw was that the space ship crashed before even taking off, but astronaut experienced ...


1

In special relativity, you can't do it. Much like a straight line is the shortest distance between two points in Euclidean geometry, a straight (nonaccelerating) worldline is the longest distance between two points in Minkowskian geometry (the geometry of spacetime). If one twin accelerates and the other doesn't, the one who doesn't accelerate will be older, ...


1

Consider two clocks $A$ and $B$ at the two ends of your pipe, and two identical clocks $A'$ and $B'$ right next to $A$ and $B$. An observer stationary with respect to the pipe then synchronizes $A$ with $A'$, then (all at once) synchronizes $A$ with $B$ using light and $A'$ with $B'$ using sound. Now, according to that observer, all four clocks are ...


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First, the authors positioned their reference frame at the center of mass, if you calculate the momentums that way, the total momentum will be zero, at least instantaneously. But that is not the point. In addition to the effects of other objects in our own solar system, mentioned in the other answer, remember that the Sun travels around the milky way ...


1

The conclusion would be the same, but quantitative results would be harder to derive. Why the conclusion would be the same: The following relies on the assumption that the speed of sound in the train frame is known to be lower than the speed of light. Now imagine the train observer, in the middle of the train, sending out both light and sound waves ...


1

I don't know what numbers you are using, or how much precision you are expecting, but the problem could be your assumption that the earth - sun system is an isolated system. First there is earth's moon that (if not taken into account) might induce errors. Then there is the effect of other planets, especially jupiter. All of these must be considered if you ...


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I think the best way to think of it is as follows.(It's not too different from what everyone has said, but may be put into better perspective). Choosing a frame of reference is a completely different job than setting up of coordinates. To observe an event in spacetime you must belong to some frame of reference(or equivalently, you create a frame of ...


1

To test Lorentz invariance rigorously, one has to consider theoretical models where Lorentz invariance is violated that are not already ruled out. One can do that by considering the Standard Model and then adding terms that violate Lorentz invariance and studying the most general such model that is physically plausible. This has been done in this article ...


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Regarding the experiment mentioned with Francois Arago in 1810 measuring the speed of light when it hit the telescope, we are only measuring the speed of light once it hits earth's atmosphere. This does not tell us the speed of light out in space.


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You are absolutely right; the value of the momentum of an object depends on the reference frame in which it is measured. The interesting question that might be bothering you about this is then: How can momentum possibly be useful if its value can be anything? Am I right? If so, here is the simple answer: The laws of physics pertaining to momentum are ...


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You are correct that the momentum of an object depends on the frame of reference of the observer. However, in a collision, the change in momentum of each object in the collision is found to be the same by all observers in all inertial frames of reference.


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Yes, momentum does change from one reference frame to another. In Newtonian mechanics it is a 3 vector. Also velocity. In special relativity it is part of a 4 vector, with the fourth component (or more often the zeroth) the energy. It is the energy momentum vector. It transforms in Minkowski spacetime as a 4 vector. In general relativity is is part of a ...


0

Your logic is very good, indeed. And it makes sense, what you describe. I agree. But then we try to measure it. And here comes the big problem: what we measure in our world doesn't follow this logic! It seems very logic, but the world just doesn't behave that way. Weird, yes. But apparently it is true. A famous example: Put a light measurement device ...


1

Lets imagine for a moment that for some reason or other only one object was left existing within the universe, and it was a spaceship. It can accelerate and decelerate, thus movement is in effect here, movement across space. However, whether it is alone in the universe or not, it has a maximum speed of which it can move across the vacuum of space. That being ...


1

The Lorentz transformation may shed some light on this... $$\gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}$$ Assume one body is, dare I say, "stationary" and the other is traveling away at velocity v. If the relative velocity between these two bodies moving apart is equal to the speed of light, then the denominator in the Lorentz transformation ...


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This is the fundamental postulate of special relativity: Light (in vacuum) moves at the same speed no matter what you measure it relative to. Pretty much everything in SR is just a mater of figuring out the deductive consequences of this basic fact. It is an experimental fact that it is so, and it was so even before Einstein -- in particular, light had ...


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The simple answer is 'with respect to anything'. For instance if I am standing somewhere and you are in a spaceship then we will always measure our relative speeds to be less than $c$. Equally, if I am standing somewhere and two spacecraft are passing me in opposite directions, then I will always measure the speeds of the spacecraft to be less than $c$, ...


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When you talk about the speed of light, regardless of what you say the speed of light is relative to, the speed of light remains the speed of light which is 1,86,000 miles/s. We just cannot travel at the speed of light because we humans are made up of particles which have mass, and according to Einstein's theory of relativity, the closer an object gets to ...


-4

The moon is dynamically and almost rigidly (but for librations) tied to the earth by means of an invisible solid rod, as it were. That is why we inhabitants of earth do not get to see the far side of the moon from our base. Solar eclipse occurs when the invisible rod places the moon between earth and sun casting a tiny shadow on earth. Lunar eclipse ...


44

The Moon's orbit must be concave toward the Sun. The Moon's orbit with respect to the Sun is always convex. This is easily proven by comparing the minimum possible gravitational acceleration of the Moon toward the Sun (5.7 mm/s2) and the maximum possible gravitational acceleration of the Moon toward the Earth (3.1 mm/s2). The acceleration vector, and ...


45

Incorrect Path I'm curious as to what does the moon's orbit around the sun looks like? One might think the orbit (in the sun's rest frame) follows the path of an epitrochoid. A (very) over exaggerated view of this motion (for unrealistic parameters, thus, not an accurate representation) can be seen in the following animation: Note that if you change ...


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Before answering let me mention that there is a terrific free applet showing the orbits, including the velocity vectors of the system Sun/Earth/Moon: https://phet.colorado.edu/en/simulation/gravity-and-orbits It is in java so pretty easy to download and use. The moon's orbit must be concave toward the sun. The Moon' orbit around the Sun is a ...


10

The orbital speed of the earth around the sun is about 30 km/s, whereas the orbital speed of the moon around the earth is about 1 km/s. From this it follows that at no point of its path around the sun the moon will ever show a backwards motion. The path is similar to the trajectory of a point (moon) on the perimeter of a (somewhat sliding) wheel rolling ...


2

Use a Focault Pendulum, measure the rotation period of the plane of oscillation and compare it with the theoretical value $$T=\frac{24\, h}{|\sin\lambda|},$$ where $\lambda$ is the latitude. If the precession is clockwise you are in the northern hemisphere, otherwise you are in the southern hemisphere. No precession means you are at the equator line.


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The basic assumption in the Big Bang cosmolocigal model is that the CMB departures from a uniform black body radiation observed in the map All-sky map of the CMB, created from 9 years of WMAP data. are a relic of the density of matter at the time of the decoupling of photons, about 380000 years after the BB. In contrast, neutrino decoupling, if ...


2

Here is what I understand: if you have a particle at state $|x \rangle$, active translating it by $a$ means moving the particle to state $ | x + a \rangle$. Passive transformation means you keep the particle in the same place, and change the coordinate by new variable $x = x' + a$ (note that the coordinate system is translated backwards $-a$). I am not very ...


3

Newton would not say that an individual standing on the Earth represents an inertial frame of reference. An inertial frame is one in which Newton first law applies i.e. an object moves in a straight line at constant speed. Since the observer dropping the apple observers the apple to accelerate that observer's frame is non-inertial. However you are correct, ...


2

The drone flies by its propellers exercising force against the air inside the airplane. It flies with respect to the air inside the airplane. Since the air is being carried by the airplane, the drone will fly with respect to the airplane. It'll fly forward or in whichever direction you point it, with respect to the X. Assuming the airplane is airtight.


0

There is another issue that must be clarified before making any definite conclusion. The question is:"Is rigid body in free state or is it constrained"? This question must be asked when dealing with rotors constrained by gravity in the bearings, constituting a rotor/ bearing system. Then next question arises:"Does this system has a resonance (natural motion) ...



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