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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 ...


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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 ...


3

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 ...


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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 ...


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You seem to be neglecting the change in the moment of inertia of the planet in the definition of angular momentum: $$ \mathbf L=I\boldsymbol\omega $$ Since $\boldsymbol\omega$ increases, then $I$ must decrease to satisfy angular momentum conservation. There isn't any need for angular momentum to somehow "reach out" and take it from a nearby planet.


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You say: But for the observer on the planet, since the total angular momentum of the star about its axis is zero it should remain zero. But the observer on the planet does not occupy an inertial frame. An observer in a rotating frame measures fictitious forces. So there is no reason why angular momentum should be conserved.


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There are a number of different frames of references. For the velocities of celestial objects we use: (i) The geocentric frame: This is a velocity measured with respect to the Earth's centre. Obviously this is quite useful for artificial satellites, but also for things like meteors. (ii) The heliocentric frame: this is the velocity as seen from the centre ...


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There are two separate questions there. The easiest one to answer is how we measure the vleocity of the Earth, Milky Way etc, because we measure it relative to the cosmic microwave background (or CMB). If you measure the CMB in all directions and find it's the same in all directions then you are stationary in comoving coordinates. However if you find the ...


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Speed is a distance (separation between two well defined points in space) traveled over a time. The speed of Earth you quote is the orbital velocity. We know how far away the Sun is and we know the shape of the orbit, so we know how far the Earth travels relative to the Sun (distance) per year (time). Likewise the speed of the Milky Way is also given ...


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The only outright requirement is that you compute all the angular momenta in your problem around the same center (modulo applying the parallel axis theorem to break the angular momentum of extended bodies into of-and-around the CoM parts). So you can freely chose any single point to use Now, as with most such "free" choices in physics there are generally ...


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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 ...


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This is the so called "principle of uniformity". Basically, it stipulates that the laws of physics are the same everywhere in space and time. Now, why should we believe in such principle ? I have thought a bit about that, and here is my reasoning. Let's define two "types" (in the sense of logic) of "objects". Call pobject, any physical object you can see, ...


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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!


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The discussion is mostly semantic. They are both calculated relative to a point, in the case of the torque the point has the additional meaning that if you put an axle trough the point, the object will start to rotatte around it if the net torque is not zero. It happens also that the torque will be the same if you chose any other point along the axis ...


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Assume that the origin of frame $r$ is moving with acceleration $\newcommand{\a}{\mathbf{a}}\a$, with respect to a fixed frame $f$, but that the coordinate axes of $r$ are aligned with $f$'s. Then if an object has an apparent acceleration $\a_r$ in the frame $r$, then it's actual acceleration $\a_f$ in the fixed frame is $\a_r + \a$. Now when your allow the ...


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I see what you're trying to ask. Let me try to rephrase it: Does an observer at the bottom of a massive gravity well perceive that the clocks of actors outside of the gravity well move faster? The answer is yes, but the intensity depends on the depth of the gravitational well. For anyplace far away from exotic things like event horizons and perhaps neutron ...


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Suppose you're standing at the edge of an enormous merry-go-round. You'll feel a "force" that seems to be pushing you outwards. This is the centrifugal force. The centrifugal force is the force you feel because you're at a certain distance away from the axis of rotation. It is written as $\vec{F}_{\rm centrifugal} = m~\vec{\omega} \times (\vec{r} \times ...


2

Look in Wikipedia http://en.wikipedia.org/wiki/Coriolis_effect. For understanding intuitively the Coriolis force effect, assume an object moving according to a static (inertial) frame of reference, in the plane perpendicular to the rotation axis, and along the radius, In the rotating frame, see the animation in Wikipedia, the Coriolis force imposes an ...


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The Coriolis force $\vec F_{\text{coriolis}} = -2m \, \vec \omega \times \vec v$ only depends on velocity. The centrifugal force $\vec F_{\text{centrifugal}} = -m \, \vec \omega \times (\vec \omega \times \vec r)$ only depends on position. Finally, if the object is not rotating uniformly ($\dot {\vec \omega} \ne 0$), then yet another fictitious force comes ...


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I was doing a question about if a train fits in a tunnel. Did the question assignment include a specific consistent definition of what's meant there by "to fit", in the first place? Presumably, in the setup which is typically considered, the ends of the tunnel (say participants $A$ and $B$) are supposed to be at rest to each other, the ends of the ...


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The site rules forbid us from giving the answers to homework problems, but this problem illustrates a fundamental issue in relativity so I think it's worth some general comments. Incidentally you may be interested to read the Wikipedia article on the ladder/barn paradox, though in it's efforts to be comprehensive I think the article gets a bit confusing. ...


2

Your first sentence is not true. There is a whole family of freefall frames that are co-incident at any spacetime event. They correspond to different "initial velocities" as they diverge from that event: in geometric language, their origins follow the many different geodesics defined by different tangent vectors of the tangent space at that spacetime event, ...


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Because: $$\vec p = \vec r_2 - \vec r_1 -r_3,$$ where $\vec r_3$ is the unlabelled side of the triangle that is parallel to $\vec H$. $$ \vec r_2 - \vec r_1 = \vec p + \vec r_3$$ $$ (\vec r_2 - \vec r_1) \times \vec H = \vec p \times \vec H + \vec r_3 \times \vec H$$ but $\vec H$ and $\vec r_3$ are parallel so $\vec r_3 \times \vec H = 0$, and $$ (\vec r_2 ...


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I guess there is no single definition, since it varies according to author. Personally, I think the term should just be abandoned. It's too ambiguous. It mixes up two ideas that need to be cleanly separated, but often aren't: the frame of reference as a grid of sensors, and a single sensor that receives incoming information, like a camera. Mixing up these ...


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The "reference" for acceleration, is its own previous "state". What this means is that, for linear acceleration, it is the initial point prior to the start of linear acceleration (at t = 0). And for angular acceleration, it is the imaginary line defined by the center of rotation and the initial position at t = 0 ($theta$ = 0).


2

My guess would be that $\mathbb E^n$ denotes Euclidean space. In addition to having geometric structure (angles and distances) and motions (rotations, translations, reflections) - not all of it terribly useful in the 1-dimensional case - it is an affine space. Affine spaces have no notion of distinguished origin or zero point. We can use a vector space like ...


-1

I have yet to find a physics book that doesn't make this really confusing. If one has a vector fixed in inertial space, its components as viewed in a moving frame are obtained by the dot product of the vector with the moving unit triad fixed to the body but moving relative to inertial space. While the inertial frame would measure its components as constants ...


0

What this refers to is the Rotation Reversal Theorem - rotating first about axis z with angle az , and then about the rotated y axis by angle ay , followed by rotation by the now twice rotated z axis by angle bz is the same as rotating first about the original z axis by bz, followed by rotation about the ORIGINAL y axis by ay and then finally about the ...


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I don't see the difference between your "coordinate time" and just time. Yes, time is relative. Since time is relative, it is incorrect to find a reason (or cause). A cause is, by definition, some local event in the past, which affects some other event (consequence) in the future. Time relativity is not an event. It is not local, and is not in the past. ...


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At the research of the answer, I would recommend to consider the example of an astronaut going very far in a lifetime (example: moving near speed of light, he reaches a star 5000 light years away in only 50 years). I would ask the reversal of "Why is coordinate time frame dependent?": why proper time is passing slower than the observer's dilated time? It ...


1

Tilting an object in space changes its apparent dimension (think of trying to get furniture through a door: the width of an object depends on its orientation). Objects in relative motion are tilted in space and time (or rather, spacetime), and different observers will see things unfold under different perspectives. Personally, I find relativity of ...


0

One reason for the difference is time dilation--a given coordinate clock at rest in one frame will be running slow as measured by coordinate clocks at rest in another frame. It can be demonstrated that this follows logically from the two postulates of special relativity, using the "light-clock" thought-experiment detailed here. The second postulate says that ...


0

People who are studying something something generally use a frame of reference which is reasonably close to the things of interest. While it might in theory be possible to measure the stature of a man by very accurately determining the distance from the center of the Earth to the bottoms of his feet, as well as the distance from the center of the earth to ...


0

A couple of things first... 1) Time dilation is not a consequence of high speeds, but of ANY speed - it just the effects grow large rapidly within about 10% the speed of light. Low speeds can have measurable consequences as is the case with magnetic fields for example. 2) All identical clocks "tick" away at the same rate under all circumstances,** metering ...


0

Imagine, for example, that two different inertial observers, one sitting on a train moving through a station with uniform velocity $v$ with respect to ground. The experiment will consist of turning on a flashlight aimed at a mirror directly above on the ceiling and measuring the time it takes the light to travel up and be reflected back on its starting ...


0

This is to be expected from the relativistic velocity addition formula, which has a way of "compressing" differences in speed. Consider the simple case of two stars with velocities $u_1$ and $u_2$ in the galaxy's rest frame. (For definiteness let's say both $u_1$ and $u_2$ are positive.) Their speeds $s_1$ and $s_2$ relative to an observer moving with ...


0

No, the velocity dispersion cannot change because you are always using a diference of velocities to compute the dispersion, and the difference does not change when you change to another intertial frame because all velocities are added the same speed and cancels. If you want you can post the code to see where is the error. Assuming you use ...


9

In general relativity, angular motion actually does have some "relativity" to it as well. When you're in close proximity to a spinning object, you'll actually be dragged along with it. This is known as the Lense-Thirring effect, or just "frame-dragging". The most dramatic example is the ergosphere of a spinning black hole, a region where no object can remain ...


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Special relativity deals with "inertial" or "non-accelerating" frames. Physics in inertial frames are equivalent independent of their velocity and the velocity of inertial frames are relative. You are free to assume any inertial frame is stationary and all other frames are moving relative to it. Rotating frames are not inertial, they are accelerating ...


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Velocity is relative. There is no special reference frame that would be "at rest". But acceleration is not and was never claimed to be. Reference frames in free fall are special and reference frames that are accelerating relative to the ones in free fall contain inertial forces (circular motion involves acceleration towards the centre; the corresponding ...


3

Easy way to distinguish between gravity and rotating space station: Throw a ball straight up in the air. If it comes straight down, gravity. If it moves away from you (behind your tangential velocity), it's a rotating space station.


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General covariance applies only to freely falling observers -- once you invoke non-gravitational forces, like the inward pressure of the wall, the observer is no longer freely falling.


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If the occupants of the space station were not aware of its design and could not look out a window then there is no way to tell if it is rotating or they are near a earth size planet that causes the gravity. Orbiting around another space station will causes a sensation of gravity, and it seems you are contradicting yourself. If there is any rotational ...


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Two questions and two errors: The (hypothetical) point of view of the photon is not "frozen" because you need time to perceive something frozen. But the photon has proper time zero, everything is reduced to one instant, thus nothing can be frozen. "Dodge" a photon: Information is transmitted with light speed. As the photon is moving with light speed it is ...


2

A photon cannot be said to have its own inertial reference frame, because inertial reference are defined to be a family of coordinate systems that satisfy the two fundamental postulates of SR, one of which is that light moves at c in all frames. You could construct a coordinate system where the photon was at rest, but since this coordinate system wouldn't be ...


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How to understand non-associative composition of velocities in STR? Special relativity introduces a weirdness about how your axes can be related to other observers' axes: if your axes are aligned with observer A's axes and theirs are aligned with observer B then special relativity (i.e. the Lorentz transformations) say that B's axes will be rotated with ...


1

As pointed out you can't travel at the speed of light but you can look at the limits we are tending towards as we approach it. So, if I were to travel in a spacecraft at the speed of light, would I freeze and stop moving? From the perspective of a stationary observer if your spacecraft was traveling at close to the speed of light, time on the ...


0

Let us assume as you move (velocity $v$) you will see the relative kinetic energy of the Universe increases, therefore energy of $KE_0$ we measure is different to the kinetic energy in perspective of the Universe of us $KE_1$, and since the mass of universe is greater than you we can assume the kinetic energy will also be greater, therefore we will model ...


3

Are Lorentz transformations more adequate representations of motion, than the more intuitive velocities? Yes. The non-associativity that bothers you simply arises because there is no group of three dimensional boosts. Confined to one dimension, boosts form a rather lovely one parameter subgroup of the Lorentz group $SO^+(1,3)$. So everything "works", ...


1

There are two types of time dilation: dilation caused by being near a large body, and dilation caused by traveling very fast relative to another observer. Relativistic time dilation plays a bigger role for astronauts aboard a space station similar to the ISS. Even though velocity and gravity produce opposite time dilation, in this scenario, time dilation due ...



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