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13

In an ideal situation (no air resistance) there will be absolutely no difference in the place where the coin lands! Whether you toss the coin up from inside the train or while standing on the roof, the coin will land back in your hand (provided you've tossed it perfectly vertically). However, in practice, while standing on a fairly fast train's roof, ...


6

We have to be careful in stating exactly what we're going to allow ourselves to assume here. We need some sort of principle of relativity -- that the laws are the same for both observers. But we don't want to assume anything else a priori, right? For instance, we don't want to assume at first that rulers have the same length for both observers -- we need to ...


5

I suspect your confusion is because you're holding two conflicting notions in your head, something like: "Motion is relative, so physics works exactly the same inside a moving train car as inside a stationary one." and: "Directions are absolute, so if I throw the coin straight up from the roof of the train, it goes the same way whether the train is ...


5

how Galilean transformations which are wrong (are approximately correct) give the correct answer for k? The Lorentz prediction and the Galilean prediction must agree in the limit that $v \to 0$ (or in the limit that $c \to \infty$). This is because $v=0$ corresponds to no transformation at all, so they had better both agree there. So if you take ...


5

If I remember correctly, assuming only a homogeneous and isotropic spacetime, on top of an arbitrary group structure, the only 4D spacetime symmetries that are allowed are either galileo group SO(3,1) (that is Lorentz group) SO(4) (that is euclidean 4D rotations). The relevant references where this was shown are (according to link below) J-M. ...


5

Take the unprimed frame to be your and my rest frame. For some body we measure $v(t)$ and by differentiating it we get $a(t)$. Now consider another observer in the primed frame moving at constant velocity $V$ relative to us. because of the law of addition of velocities we know that the other observer measures the velocity of body to be: $$ v' = v(t) - V $$ ...


4

The answer is negative. There is no action of the free particle invariant under the Galilean group. In the following, a heuristic explanation will be given and in addition a reference where a more detailed proof is provided. The basic reason is that the Galilean group cannot be realized on the Poisson algebra of functions on the phase space of the free ...


3

I'll point out the more detailed differences below, but a nice rule of thumb to follow for these is that since the Galilean transformation gets it's name from a man who lived several centuries ago, the physics formulation for them is more basic than the Lorentz transformation, which is a more modern interpretation of physics. That way you can remember that ...


3

In special relativity, proper acceleration is defined as $$ a = \frac{du}{dt}, $$ where $$ u = \frac{dx}{d\tau} = v\frac{dt}{d\tau} $$ is the proper velocity, and $$ d\tau = dt\sqrt{1-v^2/c^2} $$ is the proper time. So $$ \frac{d}{dt}\left(\frac{v}{\sqrt{1-v^2/c^2}}\right) = a. $$ If we integrate this over a time interval $[0,t]$, we get, if $a$ is constant, ...


3

In the context of Special Relativity, you do need to be careful about constraints such as "assume constant acceleration" without further qualification because, just as there is a need to distinguish between proper time and coordinate time, one must distinguish between proper acceleration (acceleration measured by an accelerometer) and coordinate ...


3

In general, it won't drop to the place right above where you dropped it in the roof-of-the-train case, but its motion will be different. The situation is different in that generally there are other forces on the coin: the air is moving relative to the coin not only vertically, but also along the direction of the train's motion. In other words, when on the ...


3

You seem to be totally skimming by conservation of momentum. Let's look at your situation initially, if you look from there frame it does not matter whether they are at rest or moving with constant velocity. Let's say the mass of small robot(blue circle) is $m$ and mass of big spaceship(large rectangle) is $M$ ($m \lt M$) If you just see in there frame, ...


2

Time translations are generated by $H$, as expected. The generator for Gallilean boosts $\vec{K}$ is related to the CM operator $\vec{R}=1/M\sum_i m_i\vec{x}_i$, $\vec{K}=M\vec{R}$. See http://en.wikipedia.org/wiki/Schr%C3%B6dinger_group for some references.


2

Not the technical answer you're looking for, but perhaps a helpful thought that occurred to me when first read your question. It's a well-known feature of (special) relativity that you can recover Galilean relativity (and thereby Newtonian mechanics) by taking the limit as c goes to infinity of any result in SR (for example, look at the velocity addition ...


2

Galileo proposed that all inertial systems are equal. That means there is no absolute observer. He stated that if you sit in the belly of a ship which is moving on a calm sea, you cannot know or measure its velocity with respect to the ground of the ocean. Special relativity makes this statement a little different as it introduces the maximum speed $c$, but ...


2

There is no contradiction between quantum mechanics and the equivalence principle. There has never been any such contradiction. String theory makes the compatibility of the principles explicit but to explain the compatibility at the level of the question, we don't really need any characteristically stringy arguments. So the answers to the numbered questions ...


2

Let's look to your own statements. First, time derivative after transformations isn't equal to an "old" derivative: for $\mathbf r' = \mathbf r - \mathbf u t = \mathbf r - \mathbf u t' \Rightarrow \mathbf r = \mathbf r' + \mathbf u t'$ $$ \partial_{t'} = (\partial_{t'}\mathbf r )\partial_{\mathbf r} + (\partial_{t'}t) \partial_{\mathbf t} = (\mathbf u \cdot ...


2

Here I would like to expand some of the arguments given in Ron Maimon's inspiring answer. Consider $N$ point particles with positions ${\bf r}_1, \ldots, {\bf r}_N$. The Galilean transformation group is, e.g., explained here. The only transformation, which we will bother to mention explicitly from now on, is the shear transformation $$ t \longrightarrow t, ...


2

The flight time from the USA to China may be different to the return trip, but if so I'd gues this is due to the jet stream rather than the rotation of the Earth. Anyhow, your intuition about the train is correct. The two passengers A and B see the A to B and B to A speeds of the ball to be the same, while an external observer sees them to differ by twice ...


1

You have said: If,for instance,the relative motion observed between two frames of reference is that of uniform acceleration, how can we determine which frame is the unaccelerated system? It is obviously not possible. and Another part of this very question is also: How can we call the occupied frame of reference as being inertial regardless of whether other ...


1

In the $S'$ frame, your variables are $x' = x - t\cdot u \cos\theta $ and $y' = y - t\cdot u \sin\theta$. If you do the change of variable, you get that the motion now is described by $$x' = 0$$ $$y' = -\frac{g}{2}t^2$$ So in your new frame of reference you have vertical free fall from rest. This is not very helpful in finding out when or where does the ...


1

If the ship is at rest and you apply a force to the fly, it will crash against the wall. If you make the fly stop, you have to apply a force in the opposite direction. In the moving frame, this would mean the have the same speed again.


1

Godparticle, It's difficult to refer to your drawings because they are blurred and pale. However, I think that your confusion might come from the fact that you do not specify (and seem not to take into account) whether there are other forces (than its own engine controlled by you) acting on the fly, like the ones present on earth. So is there any ...



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