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I'm pretty sure this question has been asked before, Here and Here and a non stack exchange explanation can be found Here and another, perhaps the easiest read of the bunch: Here Now, I've read (not word for word), but I've read a chunk of those answers and I still find it a little confusing without a clear and concise "yes" or "no" - which would be nice, ...


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Firstly particles can reach event horizon in finite time in the frame of an observer at infinitely far away (This is the frame of reference for describing black hole radiation). But the above phenomenon of particle reaching to the event horizon in finite time has nothing to do with black hole radiation. Hawking radiation is a quantum effect. The pair of ...


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Yes so-called pseudo forces do work and if they were to be describable as a conservative force, then yes the corresponding mechanical energy would be conserved. The best example I can find is the gravitational pull we feel at the surface of the Earth. It is in fact the sum of the "true" gravitational force owing to Newton's law of gravitation and the, ...


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You are calculating an unreal force by using Newton's second law. Remember that Newton's laws are valid in inertial frames of reference. And Newton defined inertial frames as those frames where an object continues to be at rest or in constant motion unless acted upon by a real physical force.In your example the train or you is not an inertial frame. Hence, ...


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It's not a force that would break the glass, it's an uneven force (i.e., a baseball smacking the center of the glass). A great illustration of this is air pressure. Air pressure has a force of about 100,000 $\frac{N}{m^2}$ - so a normal glass case big enough to hold a fire extinguisher or something would have about 10,000$N$ of force evenly distributed ...


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"Rest, in physics, refers to an object being stationary relative to a particular frame of reference or another object." - Wikipedia (emphasis mine) While on Earth, the planet is often treated as the default frame of reference. It is not a perfect frame of reference, but for many purposes it is good enough. Since there is no absolute frame of reference, ...


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"A state of rest" is a relative term. Relative means - measured in comparison to the things around it. When you sit in a train and sip from a cup of coffee, you can do so because the cup is still relative to you even though both of you might be hurtling through the countryside at 200 km/h. For most experiments, objects can be considered "at rest" if they ...


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Revolving around the sun is equivalent to free fall around the sun, so the revolution allows you not to 'feel' the sun's gravity. The rotation of the earth is something that can be measured: (i) a centrifugal force which is a small offset on gravity, and (ii) causes the coriolis force. Both these are small effects, so can often be ignored for laboratory ...


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When you're in a moving vehicle and see trees or buildings, who is moving? Are you moving forward, or are the trees and buildings moving backward? Its counter-intuitive for beginners but both these views are absolutely correct. We can only describe the motion of an object from a reference frame. A reference frame is a specific configuration from where you ...


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Solving lots of physics problems involved you choosing a frame of reference. As long as all your formulations are from a particular frame of reference, for example the energy of objects then the laws will work. Since you are also travelling with the object, (as the earth rotates) it is easier to say you are at rest. If you are travelling at constant velocity ...


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You must remember that the 'body-frame' (which is the frame 'inside' the rotating body) is a frame and that OTHER bodies (which may, for example, be rotating about the same point with a different speed) will appear in the 'body-frame' to have an angular speed less than if observed in an inertial (stationary) frame. To help visualise, if you stick your arm ...


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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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There is no reason why we could not be viewing the BB in our frame. We can only view anything from our position in space-time, and in our frame. We don't see the whole BB 3-sphere, just that 2-sphere section of it which is located at just the right distance that light can reach us in the time that has since elapsed. That's why what we see is spread over a ...


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The acceleration they experience locally is different due to the different masses, whilst the acceleration they experience one to the other is of course equal (or at least synchronizing during any approach), otherwise A would hit B (when the planets crash together) later or sooner than B would hit A.


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Your misconception has nothing to do with gravity - you're just getting a little mixed up about acceleration vs. relative acceleration. Let's dispense with gravity, since it's a red herring here. Say there are two cars. Car A accelerates at $+3 ~\rm m/s/s$ (to the right). Car B accelerates at $-5 ~\rm m/s/s$ (that is, to the left). So far, so good, right? ...


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It doesn't have to be modified, it's fine as it is. There's no paradox at all. The force that attracts both to each other is indeed $F=G\frac{M_A M_B}{r^2}$. But the acceleration they experience is not the same (at least provided $M_A \neq M_B$) because their inertias (masses) are not the same. For one $F=M_A a_A$, for the other $F=M_B a_B$. There's no ...


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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 ...


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How do we know the speed of light is constant and spacetime dilates rather than vice versa? We know that the speed of light is not constant. I'm afraid it's a popscience myth that the speed of light is constant. See Irwin Shapiro talking about it here: Some conspiracy nut was telling me that Einstein was BS and there's a giant conspiracy that he's ...


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You might be confusing some issues. In special relativity, space and time do not stretch or compress. It really comes down to measurements with clocks and rulers made by people that are moving uniformly with respect to each other. One option that is consistent with observations for SR is that there is one family whose clocks and rulers are right and ...


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Yes, though the effect is greater on earth-bound or terrestrial tracking stations than on spaceborne or orbiting satellites. See Table 2.9 ("Perturbing accelerations acting on a GNSS satellite") and sec. 10.1.2 ("Site Displacement Modeling: Solid Earth Tides, Pole Tides, and Permanent Tides") in the BERNESE Software Manual


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If the rigid body is rotating then in general the primed axes will be changing with time. An easier way to see this is to look at the Euler angles themselves as in this diagram. If, for instance, $\alpha$ is changing, then both the line of nodes (the $N$-axis in the diagram) and the $z'$-axis (the $Z$-axis in the diagram) are changing.


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The clocks are all in sync in the ground frame, but they are not in sync in the train's frame. An observer on the train would think that the clocks in the front of the train are ahead, while those to the rear are behind. Measuring the forward traveling beam against the nearby clocks will show a long time difference, while measuring the rearward traveling ...


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\begin{equation} t_{\beta}^{\prime} = \gamma t_{\alpha} + \frac{v\gamma t_{\alpha}}{c} \tag{01} \end{equation} Your equation (01) is right, but your equation (02) is wrong : \begin{equation} t_{\beta} \ne \frac{1}{\gamma}t_{\beta}^{\prime} \tag{02} \end{equation} The right (02) equation is : \begin{equation} t_{\beta} =\gamma t_{\beta}^{\prime} ...


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Absolute four-momentum is not an observable. Relative four-momentum is. We cannot find the four-momentum of the lab itself, but we can (and do, regularly) measure the four-momentum of particles relative to a given lab, which then allows us to calculate the four-momentum of said particles relative to any frame you care to name. Whoever said that wasn't ...


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"a stationary rocket ship in close proximity to ours, stationary relative to our reference frame" If the two ships are stationary with respect to one another, then their clocks will run identically (in fact, all physics will be identical in both ships) unless there are actual sources of gravity with non-negligible tidal forces across the distance separating ...


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Q: How reasonable it it to conclude that, from a remote observer’s frame, matter falling towards a black hole never crosses the event horizon, because ∆ t → 0 as v → c (according to the Lorentz transform)? This is not reasonable at all because the property of the equivalence principle indeed does say that the infalling object falls into the black hole. The ...


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Considering light travels at relativistic speeds much faster than you can in your car, the exposure is the same. The same logic does not apply as for rain as rain doesn't fall at relativistic speeds.


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The Earth as a rest frame is arbitrary. So saying that two cars have velocities toward each other is just a choice of reference frame. You are right that you can consider yourself to be at rest while the car coming toward you has some speed which is twice what the Earth frame would say it has. As it comes toward you, you observe it to have a contracted ...


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What you have to realize is, both the man and the ground beneath him are travelling at the same speed. What that means is, yes, the ground underneath the man is travelling east very quickly, but the man is also travelling east just as quickly. The gravitational force causes the man to come back down, but it has nothing to do with making sure he stays over ...



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