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Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity $\boldsymbol u$ in the inertial and rotating frames: $$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ... 4 To add to David Hammen's answer on the question: When numerically integrating this, together with Euler's equation of rotation, is there a way to ensure that the determinant of R remains equal to one (otherwise \vec{x}(t) will also be scaled)? Method 1 Dumb But Effective Naïve Multiplication Whilst you are getting up to speed with more ... 4 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 ... 2 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 ... 2 I was confused by this too -- pop descriptions of the equivalence principle don't mention the problem where the gravitational field points in different directions in different places. It is true that gravity is equivalent to acceleration, and that as a result, if you are freely falling, you feel like you're in an inertial frame. But this frame is only ... 2 The whole problem is really about knowing what the words mean. An event is a time and a place together as single object. For instance the event where a light sends its first, last, or only pulse. Or the event where a beam or particle touches something and bounces. Anything you can describe with a time and place together is an event. And different people ... 2 Let the equations of motion be expressed in a frame with coordinates q. We now want to switch over to another (arbitrarily moving) frame, whose corresponding coordinates are Q, given by:$$Q = f(q, t)$$For example, if the frame itself is moving with position x(t), we will have:$$Q = q - x(t) (where $x$ is not dynamic, but is completely specified in ...

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That won't work, and here is why. It's subtle. Say that you are on a ship, leaving the solar system with some technology that is able to thrust you in such a way that you experience a constant acceleration of 1g, as measured on the ship. You can measure this by placing a 1kg weight on a scale. From the point of view of the passengers of the ship (where the ...

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The problem with questions like this is that they include many misunderstandings of physics! For example, you say "as one approaches light speed more energy is required to accelerate faster". What you may not be aware of is that in classical mechanics, it's also true that to an observer on the ground, the faster you are going, the more energy you need to ...

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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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You can define energy in an accelerating frame, and you do it every day. The surface of the earth is an accelerating frame. Sometimes you say a frame is close enough to inertial and just treat like it is inertial even though it isn't inertial and hope for the best. Other times you just have to sit down and learn how to do physics in a noninertial frame. ...

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Yash, Imagine S is the Sun sending two photons, P1 and P2, and the object O is represented by two asteroids, O1 and O2, equidistant from each other all the time and from S at $t_0$ - they are moving in the same direction at the same velocity (c/2). So for some time one will be moving towards the Sun and one away from it. So you are right that the speed of ...

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I'm afraid this won't work. If for example you have a rocket motor capable of producing 1g of thrust then it will still produce 1g of thrust no matter how long you accelerate for (assuming you don't run out of fuel). From the perspective of the observer on Earth your acceleration will indeed slow down, but at the same time the Earth observer sees your time ...

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

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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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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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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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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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As you know, dealing with "simultaneous" events in relativity is tricky. If I think two things happen at the same time, you may not. However, it is guaranteed that if if I think two things happen at the same time and the same place, you will agree. That's because these two things are just the same spacetime point. In your frame, we have the simultaneous (in ...

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