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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 ...


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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 ...


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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 ...


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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 ...


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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", ...


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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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Have a look at my answer to Time Dilation Effects from simply being on a spinning planet orbiting a star in a rotating galaxy in an expanding universe.. Compared to an observer far from the Earth, time at the Earth's surface runs more slowly by a factor of 0.9999999993. Over a 70 year lifespan this makes a difference of about 1.5 seconds. So the man in ...


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is it possible to consider also the other fundamental forces [...] to be fictitious forces like gravity in the framework of general relativity? No, because the equivalence principle only holds for gravity. If we want a final unification of all fundamental forces, hasn't this feature of gravity to become a feature of the other forces as well? The ...


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The classical theory of electrodynamics can indeed be written as a geometrical theory in a similar way to general relativity. As it happens there is a question and answer addressing just this, but it's in the Maths SE: Electrodynamics in general spacetime. Classical electrodynamics is an example of a class of theories called classical Yang-Mills gauge ...


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On the quantum level, force is not acceleration. The concept of "fictitious force" makes no sense on a QFT level, because forces are interactions between quantum states, not the classical forces you might imagine. Quantum forces are not vector fields in space. The notion of "fictitious force" would mean that, e.g., the strong force is something influencing ...


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I think you may be looking at it wrong - if the vertex of the isosceles triangle is fixed as a hinge for both railings (it seems to be) it'll probably compress the spring. As for conservation of energy, how much energy does it take to get the railings and spring moving? How much energy is being used to compress the spring? (I think but I'm not certain that ...


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This frame is exist. You got wrong result because you ignored that this two photon move in the opposite direction. Set that the first photon move along the z axis and the second photon move against z-axis.$\omega_1$ and $\omega_2$ are the frequency of the first and the second photon correspondingly in the reference frame. In new frame shout be $k'_1=-k'_2$ ...


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In fact, you do not have to find the frame $S'$ because the center of mass is independent of reference frame. It is $\sqrt{(h\nu_1+h\nu_2)^2-(h\nu_1-h\nu_2)^2}=2h\sqrt{\nu_1\nu_2}$.


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The commonly accepted interpretation of Special Relativity is that it's impossible to determine an (inertial) frame of references absolute motion (by performing experiments within that frame). So the default answer (given what we know) would be no, you can't determine your absolute motion relative to the background sea of virtual particles. However this ...


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Kinetic energy is the energy associated with motion. Therefore, we can no more measure kinetic energy free of a reference frame than we can say something is moving free of a reference frame. It must always be moving with respect to something. Energy is conserved because the total energy of a system is the same before and after a process in the same frame of ...


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The exact quantities of kinetic energy (like momentum) depend on your choice of a reference frame. Don't get too worried though; regardless of your choice of a reference frame, you will find that energy (and likewise momentum) is conserved within the reference frame. Therefore, two observers may not agree on the kinetic energy or momentum of an object, but ...


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Because in the frame of reference that is co-rotating, the object doesn't move, and therefore it has no kinetic energy in that frame, which is the frame in which most problems involving objects on earth are looked at. Note that kinetic energy is evidently not a frame-invariant quantity, but it is not required to be.


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Here's a simple demonstration: Consider flat space (i.e. Minkowski), viewed in a rotating frame (in e.g. cylindrical coordinates one just replaces $\phi$ by $\phi'=\phi+\omega t$). One can calculate (without too much trouble) that, in these coordinates, a spatial line element can be expressed in terms of the canonical cylindrical coordinates as $$ ...


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There is some confusion about two points...first, independence of the speed of light on the speed of its source is nothing surprising. It is the same as independence of the speed of water waves on the speed of a boat. Nothing funny there. BUT, independence of the speed of light on the reference frame of two observers in relative motion to the same source, ...


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Consider sound waves. If you have a car with a siren does the car's speed affect the speed of sound? No, instead sound always travels at the same speed (let's ignore for now the fact that the speed of sound can change, e.g. by the density of the medium it is travelling through). Varying the car's speed will cause a Doppler Effect where the sound waves ...


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So, my question is which method is correct and please explain the why one approach works and the other does not. For method one, Liz knows the initial separation and individual rocket velocities so the calculation of the time to impact is straightforward. The most likely problem with method two is due to the relativity of simultaneity. This is a ...


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Why does the angular momentum depends on the position? Angular momentum is always defined relative to a reference point, say $\mathbf r_0$, (which is often, but not necessarily the origin). If the system is invariant under rotation around this reference point the quantity that we call "angular momentum with respect to $\mathbf r_0$" is conserved. (Note, ...


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I physically understand it as the momentum of an object rotating around something given a certain position. However, I can't give a physical explanation to the formula. Why do we multiply the linear momentum by the position? Why does the angular momentum is a function of the position? Angular momentum $L = mv * r$ (This is a late answer, but I ...


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This is an abstract answer, but I find it extremely helpful to the kind of "basic nature" question you seem to be groping for. Think of two things: Noether's theorem and a thought experiment "what if we had evolved as unsighted but clever beings?". As in David Hammen's Answer, it is Noether's theorem that would tell us that if our physical laws are ...


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There are several ways to describe a particle's motion. For example, in 2 dimensions, you could use cartesian $x,y$ coordinates or polar $r,\varphi$ coordinates. To each coordinate, we can associate a 'quantity of motion' or 'generalized momentum'. If a given coordinate corresponds to a symmetry of the system, the corresponding quantity is conserved by ...


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The angular momentum is a concept analogous with the linear momentum p = mv, in which m is the mass of the body and v its velocity. Now, see where the angular momentum comes from. Consider for simplicity a body moving on a circle around some axis, and let ω be the angular velocity, i.e. the angle by which the object rotates, in a unit time. The ...


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Ultimately, what's special about angular momentum is this: Look up in the sky. A certain set of physical laws pertain in that direction. Look to the north. A certain set of physical laws pertain in that direction. Look to the west. A certain set of physical laws pertain in that direction. Those physical laws: They're the same in all directions. There's ...


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Consider something like a door. A piece of wood with a hinge on one edge. Maybe it is one meter tall and three meters long. Now say that you're trying to hold the door in place, at the position half a meter from the hinge, while someone else throws a baseball at the other side of the door. If the baseball hits the hinge, you don't have to push at all. If ...


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Maybe you can see it this way: The modulus of a vector multiplication is like this: $$|\mathbf{L}|=|\mathbf{r}\times\mathbf{p}|=rp\sin{\hat{rp}}$$ where you can see the main feature of angular momentum: position and linear momentum of the matter considered need to be both proportional to $L$ and inversely related to each other. That is how you guarantee ...


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Have a look at the video here I hope that when you see the man spinning around and moving the weights (changing $r$) you can see that $r$ is important. Remember $r$ is the distance from each part of a rotating object to the axis of rotation, (which is not exactly the same as the position)


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Let's consider a simple experiment in which a stone tied to a string is moving in a uniform circular motion in a horizontal plane. We can analyze this experiment from inertial and non-inertial frames. An observer in an inertial frame sees the stone having a radial acceleration and concludes that there must be a radial force causing it. He observes the taut ...


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In an inertial frame the only force that causes a particle to move in a circular motion is the centripetal force. The reason that a particle does not "fall" into the center is because it has some tangential velocity, so it moves away from the center tangentially as it is falling towards it. The relationship between the centripetal acceleration and tangential ...


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Law of Special Relativity: The speed of light remains constant for all observers no matter their state of motions. $L$ and therefore $t$ would therefore remain constant irrelevant to your state of motion.


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First off, you state If the time it took light is, $t=\frac{L}{c}$ then I know that the train was stationary. I must ask, stationary with respect to what? It may not seem so to you but answering this question with most likely highlight the error in your reasoning/understanding of the principles involved. Equivalently, a correct answer to this ...


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An intuitive way to see why this method would not work (or any other method inside a truly sealed train) is that the speed of the train is always relative to something else. There is nothing stopping me from describing the situation as rails moving under a stationary train at a certain speed. Now, you propose to measure the speed at which the rails move ...


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Update and note: In the answer below, I do assume the OP and reader are aware of the Galilean relativity of motion but wonder why the invariance of the speed of light cannot be used to find an absolute rest frame. If this isn't he case, then Rod Vance's excellent answer is more appropriate. I switch on the torch and measure the amount of time it takes ...


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By your assumptions, the train will always be stationary. Light will always take time $\frac{L}{c}$ to traverse the distance simply because $c$ is a constant. Additionally, even if your train was moving at $c$, it would still not matter because you are still at zero velocity with respect to the coach. Relativistic measurements would come into picture if you ...


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The fundamental postulate of special relativity, indeed of Galilean relativity, is that there is no experiment that determine the state of motion of any inertial frame relative to the outside world unless the measurement uses data gleaned from outside the frame. Read Galileo's wonderful and very famous allegory of Salviati's Ship for a poetic and rock ...


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Virtual particles aren't real. It's in the name. The reason people say there are "vacuum fluctuations" in form of virtual particle-pairs forming and annihilating again is because they misunderstood Feynman diagrams. In Feynman diagrams, internal lines are called virtual particles, since the external lines (i.e. the open-ended lines) correspond to real ...


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For a start, we need to remember that local distance does not change for a person moving along with the object. Such a person (riding one of your objects) would not see any of the objects change length, but also - and which is important here - he would not see the gap between the objects become shorter or larger. Special Relativity states that a person ...


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I don't think it's good to consider length contraction a law of special relativity. The law of special relativity is that light is measured the same for all inertial observers, or that the interval $(ct)^2-x^2$ is conserved from reference frame to reference frame. That's THE law. Length contraction, on the other hand, is a special case deduced when you look ...


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Acceleration If your exact need is, as you say, to determine "if the vehicle is moving at a constant speed" then there is a wide range of accelometers available on the market with various degrees of accuracy. If the accelometer shows 9.8 m/s2 downwards (assuming you install it in a fixed orientation relative to the vehicle), then it is moving at a constant ...


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Actually special relativity states that all inertial reference frames are equivalent, so there is no such thing as absolute speed. You can't ask "What is my speed", this question is just not well formed. You can ask: "What is my speed in reference to this object", and in fact all "real life" examples of asking "What is my speed" actually have some implied ...


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Are you traveling in a vacuum? If not, use a pitot tube


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Does humidity and similar cues count in the answer or not? For example if moving continously after a while certain environmental parameters will change. Which cannot be screened in the vehicle. For example humidity, temperature etc.. Assuming the vehicle can actually screen all those parameters. Then one can use non-inertial cues: For example rotations, ...



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