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49

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 ...


18

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 ...


11

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 ...


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 ...


3

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

In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent. Question: why is this true? Actually this isn't quite true. General relativity doesn't have frames of reference (except locally, which is trivially true because GR is the same as SR locally). A better way of ...


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 our frame, the light beam approaches the rear end of the train at speed $c+v$, because both are moving and in opposite directions. Note that this does not mean anything is traveling faster than light. Similarly, the light beam approaches the front of the train at speed $c-v$. We want to choose the position of emission so that they arrive at the same ...


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

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 ...


1

Spin is an intrinsic property of quantum objects that, unlike a particle's orbital momentum, does not depend on the frame of reference you are considering. Another intrinsic quantity of that kind would be charge, which is also just a fundamental number you assign to a particle, no matter its state of motion. One possible source of confusion when talking ...


1

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, ...


1

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 ...


1

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.


1

In relativity the notion of simultaneity is relative to the obeserver. While one observer (the one "standing") see all the rays of light arrive at the same time, another observer (the one going near the speed of light) will see one ray arriving before another. The paradox here is you think simultaneity can be defined in an absolute way independent of the ...


1

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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