# Tag Info

## Hot answers tagged galilean-relativity

16

The answer is simple: Maxwell's equations. Maxwell published his electromagnetic theory in the 1860s. This generated a huge schism in physics. Maxwell's electromagnetism was in direct conflict with Newtonian mechanics. There is no allowance in Maxwell's electrodynamics for the speed of the emitter or the speed of the receiver. The speed of light is constant ...

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, ...

7

Theorem Schmeorem. A Galilean invariant Lagrangian for any number of classical particles interacting with a potential: $$S = \int \sum_k {m_k(\dot{x}_k-u)^2\over 2} + \lambda \dot{u} - U(x_k)\;\;\; dt$$ For any Galilean invariant Lagrangian $L(\dot{x}_k, x_k)$, the Lagrangian $$L'(\dot{x}_k,x_k, \lambda, u) = L(\dot{x}_k-u,x_k) + \lambda \dot{u}$$ ...

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

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

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: $$v' = v(t) - V$$ ...

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

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. ...

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 ...

4

Either I should turn in my medical marijuana card, or the author of your textbook should. The exercise doesn't make any sense. Since we only have one particle, whose mass is fixed, we can set $m=1$. Also, the factor of 1/2 in the equation for KE is purely conventional, so let's drop that as well. We then have the following: Given the definitions $p=v$ and ...

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

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

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, ...

3

Actually given that the first postulate says that all physical laws are the same in all inertial frames, you could replace the second postulate by the postulate: "Maxwell's equations are the physical laws for electromagnetism". From Maxwell's laws you can derive that the speed of light in vacuum has a specific, constant value, in SI units ...

3

I can not answer you question mathematically, but my experience with Burgers equation tells me that there is no such transformation. If you think of Euler equation as Navier-Stokes in the limit where the viscosity vanishes $\nu \to 0$, then time reversal symmetry is simply spontaneously broken. As long as you have a finite viscosity the system is ...

2

Maxwell's theory had predicted that the speed of light varies with the speed of the observer. Initially (prior to Fitzgerald and Lorentz advancing the ad hoc length contraction hypothesis) the Michelson-Morley experiment was compatible with the assumption that the speed of light varies with the speed of the light source (as predicted by Newton's emission ...

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

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

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

There's errors: $E'=\frac{1}{2}m\dot{x}'^2-\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V^2$ should be $$E'=\frac{1}{2}m\dot{x}'^2=\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V+ \frac 1 2 mV^2$$ which is the Galilean transformation of kinetic energy used later. ...

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

The metric doesn't transform at all. The metric stays fixed even in an accelerating coordinate system, with any arbitrary acceleration. This is what it means to say that time is separate from space in Newtonian mechanics. The structure of Newtonian mechanics, however, includes more than the metric, it includes the acceleration law, that the forces are ...

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 ...

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