# Tag Info

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Analyzing one moving clock from the perspective of one stationary person will be inadequate to derive special relativity from. With just that set-up, you aren't actually using the key fact that the speed of light is the same for all observers – all you're actually using is just the fact that the speed of light is finite. With just taking into account that ...

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When you're trying to understand the mechanics of a system it's usually convenient to choose coordinates that reflect the symmetry of the system. The solar system is roughly centrally symmetric because the Sun is by far the largest mass in it, and the coordinates that reflect this symmetry are polar coordinates with the Sun at the centre. For example in ...

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It's all about the context in which you want to analyze particular issue. If you are studying the solar system, the most suitable, would be to consider the sun as the center of the system. If you are studying the Milky Way, the sun is not a good reference point, you should take the center of the galaxy. Similarly, to locate the stars from an observer on ...

9

Relativity is not needed: If you replace the clock with a strong laser which fires a ray of light every second with an atomic clock, you know that the ticks will not slow down because every tick will be followed by another after a second from the laser's perspecive . But how can you arrange that with the fact that the light move further and further away ? ...

8

why not relative to the Earth? Scientists do express things relative to the Earth, where that makes sense. I couldn't imagine trying to forecast the weather or model the global circulation of the Earth's atmosphere from the perspective of a non-rotating frame with it's origin at the solar system barycenter. Astronomers, at least those dealing with ...

4

A reference frame at rest with the Sun is, with a good approximation, an inertial system (much better than one at rest with our planet or other bodies in the Solar system, essentially in view of the hugely larger mass of the Sun). Physics in inertial reference frames has the simplest form. For instance the motion of planets around the Sun is described along ...

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 ...

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 ...

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

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 ...

3

Your written text says "t = 1/sqrt[1-(v^2/c^2)]". If you used that equation, it's no wonder you got a nonsensical value. In the equation in the scanned image, that's a letter $t$ in the numerator, not the digit $1$. Also, although it works for this problem, the scanned image should really say something like $$\Delta t' = \frac{\Delta ... 3 Passive Lorentz transformations are what everybody learns first. There is one physical reality, and you're just describing that one physical reality using two different coordinate systems, where one coordinate system uses a set of axes that are rotated and/or boosted relative to the axes used by the other coordinate system. You're just using a different ... 2 If the curve is a geodesic then in the coordinate system of an observer moving along the geodesic coordinate time and proper time are the same. That's because in the freely falling observer's coordinates dx = dy = dz = 0 and therefore ds^2 = -c^2dt^2 = -c^2d\tau^2. This makes proper time a natural way of parameterising the curve because it's just the ... 2 In special relativity there is a distinction between 'experiencing events' and the concept of an observer: Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from In ... 2 A closed system can not speed itself up, that's the momentum conservation law which is also the key to your problem. As far as I can see you are implicitly supposing the following three equalities to hold$$E_i+A=E_f\\p_i=p_f\\m_i=m_f where subscripts $i$ and $f$ stays for the initial (before acceleration) for the final (after acceleration) states ...

2

As a simple example, suppose we translate a function $f(x)$ by $a$ in the $x$-direction and by $b$ in the $y$-direction. We take the active viewpoint, viz., if $f(x)$ had a peak of height $f_{0}$ centered at $x=x_{0}$, it would be translated to be a peak of height $f_{0} +b$ located at $x=x_{0}+a$. Surprisingly, however, what we end up doing is a ...

2

The problem here isn't a simple algebra error, but rather an issue with the physics. A medium which at rest is isotropic no longer behaves as an isotropic medium when it is moving relativistically. Instead, it behaves as a nonreciprocal bianisotropic material. In particular, the phase velocity of light at a particular frequency in a medium is no longer ...

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 ...

1

Linear momentum will be conserved when the Lagrangian that describes your system is unchanged by translations in space. This is a consequence of Noether's theorem, and it's as close as you're going to get to a fundamental explanation for the conservation of momentum. In general a Lagrangian written in the coordinate system of an accelerating observer is not ...

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1)Definition: An inertial frame of reference is a frame of reference where Newton's first law applies (uniform motion if without external force). Now if we have other frame of references that are moving relative to this inertial frame with uniform relative velocities, then all the others are also called inertial frame of references. 2)Transformation between ...

1

I've heard that inertial frames are frames in which Newton's laws hold. The modern view of Newton's first law is that it defines the concept of an inertial frame. It also, at least conceptually, provides a mechanism for testing whether a frame of reference is an inertial frame. Suppose you know that no forces act on some particle. If that particle ...

1

Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$ With $x= h \cot \theta$, ...

1

Nothing happens to the initial angular momentum. It is simply irrelevant. Let's imagine that the pendulum is placed on a non-rotating body (or at the earth's equator). When the string is cut, the pendulum falls straight to the center. The path it traces is a straight line. In the non-equatorial case, the pendulum structure has some rotation. At the ...

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Your understanding is correct. And switching between different perspectives (what physicists would call different inertial frames of reference) like that is a very useful tool in physics, because it turns out that the laws of physics have the same form no matter which inertial frame of reference a problem is described in. For more information, see ...

1

The block is accelerating, but that's due to the pseudo force, not the frictional force, which is zero. You can't just look at the horizontal forces on the block to determine whether or not static friction will hold. You have to consider what's going on with the other surface as well. Here's a simpler situation to consider first in an inertial frame. ...

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In this context, the least ambiguous reference frame is the comoving rest frame. In essence, this is the local frame moving along with the local Hubble expansion. We can accurately determine earth's velocity with respect to this comoving frame (and thereby obtain our so-called peculiar velolocity) by subtracting out the dipole anisotropy from the Doppler ...

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When you're in a vehicle turning on a road, you'll feel a centrifugal force on the vehicle. This will be in the horizontally outward direction. The centrifugal force and weight add up vectorially. \begin{align} \vec{F_w} &= m\vec{g} \\ \vec{F_c} &= m |\vec{\omega}|^2 \vec{r} \\ \therefore \vec{F_{net}} &= \vec{F_w} + \vec{F_c} \end{align} Now, ...

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It is really valid for distributions too, but you need to use the integral form. To see why it is only "valid" for point charges, take a look at the equation (for a point particle, the one you have probably been looking at). It refers to the distance from the source, which is only defined for a point, not a distribution. It is however not valid for moving ...

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Coulomb's law does not apply to two charged bodies of finite sizes, say two charged spheres. It is because, the distribution of charge does not remain uniform, when the two bodies are bought together.$_1$ Credits: $_1$ Modern's abc of Physics-Satish K Gupta, 23rd edition, pg.14.

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No matter in which direction the clock moves (away or towards you) , the time will slow down in the clock. Time will always dilate as the clock moves faster and faster, but will be apparent to a human eye only once it reaches speeds close to the light speed. This is not taking into effect the doppler shift which is merely the increase/decrease in the ...

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