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42

I assume you used the formulae $f_o = fs\sqrt{\frac{1+v/c}{1-v/c}}$ for the clocks ahead of you and $f_o = fs\sqrt{\frac{1-v/c}{1+v/c}}$ for the clocks behind you. Those formulae do imply a singularity for the clock that is closest to you. Which equation to use? The answer is neither. Those expressions assume the travel is along the line of sight to the ...


11

What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ...


10

It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ...


7

Physically, SR can't accomodate observers moving at $c$. Mathematically, the limit is undefined. You have a function $f(v,w)$ defined in the $v-w$ plane, which has a square boundary. You're talking about approaching a corner of the square, but the value of the limit would depend on what path you chose to approach the corner along.


5

As already noted, Special Relativity cannot account for an observer moving at the speed of light. It is also instructive to calculate the proper time for the object ${\bf A}$: as you approach the speed of light the proper time becomes zero. So it is even impossible to define in ${\bf A}$'s frame of reference the moment at which ${\bf B}$ will be emitted.


5

The key is the coriolis force. The coriolis force is $F_c = -2m\Omega \times v $. Here $\Omega$ is the rotation of the frame of reference and $v$ is the linear speed of the satellite. If you do the calculations, left as an exercies for the reader, you'll get the missing force. In case 2 the coriolis force is 0, because the velocity $v$ has to be used in ...


4

I believe the difference comes from the fact that forces can do different amounts of work in different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work ...


4

In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no". There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the ...


4

Your friend is correct. The acceleration of the projectile is determined by the thrust its rocket motor can produce. If the acceleration of the projectile is $2g$ then the thrust of its motor would be $2mg$. But launching the projectile from your space vehicle can't increase it's thrust. You can increase its initial velocity, but once the projectile has ...


4

You're mixing up two things in this question: 1) How we label individual points (Einstein calls them "events") in space-time 2) What results someone would get when looking at their clock when they pass through spacetime points. The first thing is almost completely arbitrary, especially in full general relativity. The second thing is an unambiguous result ...


4

I'm getting the sense that my fears were correct, it's physically a nonsensical situation. Applying a formula outside of the context in which it was derived will likely produce nonsensical results. In the derivation of the relativistic velocity addition formula, and using your notation, there is an object B with uniform velocity $w$ in some inertial ...


4

That is exactly the point: if the field is in a vacuum with respect to observer A, and observer B accelerates uniformly with respect to A, then B will observe a field state with nonzero particle content. It doesn't really matter whether you talk about different observers on different frames of reference, or of a single observer who 'switches' their frame of ...


3

Yes, really. But that doesn't mean that we're floating around loose without a foundation. In Galilean relativity all observer could agree on a number of things about a interaction of process. Things like How long it took (the same for all the bits) How far each part of the system traveled in that time What the mass of bit ... In Einsteinian relativity ...


3

The signal from the clock moving towards you is the Doppler shifted version of the value you "know" it to be - that is, first slow it down by gamma (clock moving relative to your frame of reference), then speed it up by Doppler shift. Ditto, with sign reversed, for clocks behind you. Now the clock moving at right angles shows what you expect and there is no ...


3

Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...


3

Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$ \sum \mathbf p_{i}=\sum\mathbf p_f $$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...


2

Rough impressions can be misleading. The other car really is moving 200 km/h from your point of view. One thing to keep in mind is that you tend to perceive motion more readily when it is closer to you. This is at least partly due to the fact that you really only can see angular speed across your field of view (like degrees per second). To convert this into ...


2

Yes, I can think of two ways to do this and there may be more. In a rotating frame the acceleration is a function of distance from the pivot. If B has sufficiently precise instruments the variation of acceleration with position will be detectable. However if B is confined to a very small space, or doesn't have precise enough instruments the variation of ...


2

It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...


2

To make this easy, I will assume that all speeds are small compared to the speed of light $c$. The motion is in one dimension, so the trajectory of the space vehicle will be the following: $$x(t) = x_0 + v_0 t + \frac12 a t^2$$ We can easily set $x_0 = 0$ if the put the origin of the coordinate system at the start point where the space vehicle started. If ...


2

If you consider a straight-on trajectory, there really will be a discontinuity. That is the same as with the audible Doppler effect. There is a smooth drop in frequency of a fire truck's siren passing you on the street. The reason is, that there is a certain distance between you and the truck at the closest point. If that were not the case, i.e. the siren ...


2

Recalling that the relativistic velocity addition formula in 1+1 dimensions in terms of rapidity $$\beta~:=~\tanh^{-1}\frac{v}{c}$$ is just ordinary addition of rapidities $$\beta_1+ \beta_2,$$ then it becomes clear that OP is essentially asking What is $\infty+(-\infty)$ ? which is not defined mathematically.


2

Some key reading if you want to understand this stuff is chapters two to five of the IERS Technical Note 36, the IERS Conventions (2010). It's not just the J2000/FK5 frame (aka the EME2000 frame) that is associated with some epoch date. Every Earth-centered inertial frame has some epoch date. There are two fundamental reasons why this must be the case: ...


2

For Special Relativity (SR) i think the Michelson-Morley experiment is compatible and provides a verification of SR principle (some other formulations are also compatible with the experiment). Quantum Field Theory and especially the Dirac prediction and verification of positron is also a verification of SR (and many other expreriments in this context) For ...


1

Where's the flaw? Here's one: Let γ be the squareroot of 1−v²/c² but, in fact, $$\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ But, there is a flaw in your reasoning too so correcting the error in the formula for $\gamma$ will not get you to the correct answer. Let's work it out with coordinates to see explicitly what's going on. Let the ...


1

The friend rotating and experiencing the centrifugal force may observe several effects that his linearly accelerating friend doesn't: the acceleration at different points of the box is slightly different i.e. the apparent gravitational field is non-uniform there is the extra Coriolis force acting on objects that are moving relatively to the rotating frame ...


1

Torque is defined as the cross product of two vectors: r and F. r points from the pivot point to the point where the force F is applied. Both vectors are not bound and stay the same vectors no matter where you put them. However, if you combine them via mathematical operations, you have to stick to their rules (here: T1=r2*F3-r3*F2 etc...). There are no ...


1

You're focusing on the centrifugal force. You forgot about the coriolis effect. Neither has much of an effect for an object dropped from the top of a tree, even a very, very tall tree. The first thing you need to realize is that "down" (the direction in which a plumb line points) is generally not toward the center of the Earth. That is only true at the ...


1

A more particle-based view of zakk and Moonraker's observation that photons experience zero time between emission and absorption is this: The postulated event that a massless particle (say a photon) could emit another massless particle (say another photon) in the retrograde direction — or in any direction for that matter — is always zero. A photon from its ...


1

Actually it's not that difficult (but a neat problem), there's only one crucial step in the development that I will show you, but let's start from a bit earlier. Let's first write out the two forces on interest here (in terms of magnitudes, as we know already they're on parallel trajectories): The coulomb repulsion between the charges: $$F_C = ...



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