9

The Sun and Earth attract each other because of gravity. Without that, they would fly apart. Gravity holds them together as they circle. Something else like that would be two people holding on to a rope. They can spin around in circles, held together by the rope. Suppose the people had equal mass. In that case, they both spin around the center of the rope. ...


6

The detection of 2 light pulses by Herman occur at the same time, and at the same place. Let's call it: $$ E_3 = (0, 0)_{\rm Herm} $$ This is one event. An event is a single point in spacetime. I can transform it into any frame (a primed frame, $S'$), and it will be: $$ E_3' = (t', x')_{S'} $$ That is a single event, period. It must be simultaneous and co-...


4

The quadratic term vanishes simply because we're working to first order. In other words, if $\Lambda^\alpha{}_\beta = \delta^\alpha{}_\beta + \omega^\alpha{}_\beta$ exactly, with no infinitesimals, then $\omega$ is not antisymmetric. But if we interpret it as a first order expansion, then the first order term is antisymmetric.


4

You are right. Kepler's third law, as quoted in your question is not exactly true. The situation in the two-body problem (sun and one planet) is more complicated, because both the sun and the planet orbit around their common barycenter (the $\color{red}{+}$ in the animation below). (animated image from Wikipedia: Barycenter - Gallery) The correct form of ...


3

The two-body problem is to predict the motion of two massive objects which are abstractly viewed as point particles that the two objects interact only with one another. In general two planets orbit around the center of mass of the system. In astronomy, the barycenter is the center of mass of two or more bodies that orbit one another and is the point about ...


2

If two matrices are equal, then they are equal component-wise. If you compare the top left entries of the two matrices, then you find that $$\gamma(-v) = \frac{\gamma(v)}{\gamma^2 + v\delta \gamma}$$ Since you've argued that $\gamma(-v)=\gamma(v)$, the result follows directly. If you then compare the top right entries of the two matrices, you find that ...


2

We have the that the standard form of stating the Principle of Equivalence is dependent on providing a definition of inertial frame. My preferred way of dealing with the definition of inertial frame is to go back to the thought demonstration that Galileo offered. Except, rather than using a ship moving along over smooth water we let the protagonists of the ...


2

Some of this depends on how the train applies its brakes. Let's suppose the brake signal is sent from the track, simultaneously in the track frame, from a position on the track to the wheel directly above it. The brakes are perfect and go from $v$ to zero instantly. In this case the short train enters the tunnel, and knowing it will hit the exit door, the ...


2

I don't think it is directly related to Occam's razor, this problem is more related to the Mach's principle, it roughly states that The universe, as represented by the average motion of distant galaxies, does not appear to rotate relative to local inertial frames. Or An isolated body in otherwise space has no inertia. By assuming these we can say whether ...


2

No. Kinetic energy is not a relativistic invariant, meaning that observers in different reference frames will not agree on its value. This is obvious - in my rest frame, the cup I'm holding has zero kinetic energy, but to somebody moving with respect to me, the cup has non-zero kinetic energy.


2

Since simultaneity is relative, how can an observer in a different reference frame explain the outcome of an experiment which depends on simultaneity? Actually, this is essentially the explanation why no experiment’s outcome can depend on the simultaneity of two spatially separated events. a mechanism which cuts a tomato if the rays arrive simultaneously ...


2

This is quite nonstandard approach. Standard approach is to distinguish between coordinates of a point (n-tuple of numbers given by coordinate maps, in this case cartesian) and vectors, that are arrows from one point to another. The moment you say position vector, you already mean arrow from one point to another, the first point being origin of coordinate ...


2

If the net force on the particle is zero, then its momentum is constant, which in particular means that the particle is moving at a constant velocity. Thus, the rest frame of the particle (the frame moving with the particle) is inertial by definition. The rest frame is certainly an important case, but there is nothing special about it in terms of being an ...


1

Suppose that after being rotating with the disk, the static friction force suddenly vanishes. The bead just follows a tangential path, keeping its momentum of the last time of contact. The real situation is similar, except for the existence of a kinetic surface friction. But here, instead of the usual kinetic friction force against velocity, it helps to ...


1

The non inertial frame is not really "bad." Sometimes, using a non inertial frame is quite useful. The important thing is to realize you have fictitious (pseudo) forces in it. Let's consider the free falling elevator. I) Inertial frame. Pretty simple; the net force is simply the force of gravity. $$\Sigma F_y = mg$$ II) Non inertial frame moving ...


1

But now what if you have cameras installed at the two sets of doors, and you have the cameras take a picture as soon as the doors close? [...] they can't see the ladder partially sticking out of the barn (the doors are closed!) The spacetime events of the two doors closing are spacelike separated. This means that at the time either door closes, the light ...


1

One is the time as measured by the stationary observer, $t$. The other is the time as measured by the moving observer, $t'$. For the stationary observer the travelled distance is longer, $d$, than for the moving observer, $d'$. But the speed of light is same, $c$, for both. With a different observed distance but same speed the observed time must be different ...


1

In time dilation formula, $t$ is time between two events measured in frame S and $t'$ is time between the same two events measured in frame $S'.$ In the derivation of dilation formula, one usually takes time $t$ to be the time as measured by the observer B in your picture. This is the time of one tick of B's clocks in the rest frame of the clock. Then you ...


1

The measure $d^3p$ is unaffected by rotations, neither is $p^0$. For boosts can restrict to 1+1 dimensions as the transverse momentum components are not affected by a boost. Now parameterise the mass-shell by the rapidity $s$, so that $$ p^0=m\cosh s\\ p^1=m\sinh s $$ then $ dp^1= m\cosh s \,ds$ and $$ \frac{dp^1}{p^0}=ds. $$ Under a Lorentz boost with ...


1

To a mathematician, it is only a matter of how you choose your coordinate system. You can let the sun orbit the earth, but the physics make heliocentric easier.


1

The two body problem tells us that two bodies rotate(because of gravitational pull of each other) about their centre of mass as sun is very heavy so the centre of mass of sun, earth system is very close to the sun. :) It rotates but very unobservable.(but just for fun a fact that our solar system also revolves about the centre of Milky way) So I hope that ...


1

Here are alternative approaches to @Philip's answer (where the masses of your particles are assumed equal). Although Galilean-velocities add linearly, [real-world Minkowskian-]velocities don't...as you found. In fact, Euclidean-slopes also don't add linearly. (Linearity of Galilean-velocities is more the exception, rather than the rule.) However, angles (...


1

Your intuition is correct, in the zero momentum frame you'd assume both objects to be moving towards you at the same speed. (I'm assuming, of course, that the two objects are identical.) So the first question to ask yourself is "What should this speed be?" The assumption you made ($v=u/2$) is incorrect, as while $q$ will move towards you with a ...


1

The components of the acceleration are: $$\begin{bmatrix} a_x \\ a_y \\ a_z \\ \end{bmatrix}=\left[ \begin {array}{c} {\omega}^{2}x+\dot\omega \,y+2\,\omega\,{\dot y }\\{\omega}^{2}y-\dot\omega \,x-2\,\omega\,{\dot x} \\ 0\end {array} \right]$$ with : $$\dot x=v\,\cos(\theta)~,\dot y=v\,\sin(\theta)$$ $$\begin{bmatrix} a_x \\ a_y \\ a_z \\ ...


1

There is no disagreement about the simultaneity of two events that take place at the same location. You can convince yourself of this using the Lorentz transformation. In the Einstein train thought experiment, there is disagreement as to whether the lightning bolts struck the front and the rear of the train car at the same time. But both observers agree that ...


1

Yes. Geocentric models of the solar system such as the Ptolemaic system required very complicated and ad-hoc geometric mechanisms to explain the observed motions of the planets. Kepler's laws of planetary motion, which introduced elliptical orbits with the sun at one of the foci, were an enormous simplification. Newton's law of universal gravitation then ...


1

There is no photon frame as that is in direct contradiction to the fact that photon moves with the same speed in every inertial frame on which whole relativity is build.


1

Actually, your matrix can be greatly simplified as $$ M = \begin{bmatrix} \frac{1}{\sqrt{a^2 - p^2}} a & \frac{p}{\sqrt{a^2 - p^2}} & 0 & 0 \\ \frac{p}{\sqrt{a^2 - p^2}} & \frac{a}{\sqrt{a^2 - p^2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$ since $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ forms ...


1

I think your terminology is confusing you - the correct term is pseudo-force, not pseudo-acceleration. Introducing a horizontal pseudo-force in the reference frame of an accelerating cart allows us to pretend that objects that are stationary in the reference frame of the cart are in equilibrium, even though we know they are really not. If the cart is moving ...


1

Copy/pasted from https://physics.stackexchange.com/questions/591368/galilean-invariance-of-schrödinger-equation as a reaction on @MikeStone's answer. It is too long for a comment. The Schrödinger equation is not Galilei covariant, even in the absence of electromagnetic fields. There is a symmetry group, the Schrödinger group, that describes the symmetry of ...


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