Tag Info

16

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, ...

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 ...

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 ...

6

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. ...

5

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, ...

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

Your prof is using slightly wrong words: the group is a Lie group of dimension 10, not order 10. A group's order is the number of its elements, which is here uncountably infinite. A group of order 10 is a finite group (and indeed there are only two possible groups with 10 elements). I'm not sure how much continuous group theory (Lie group) theory you have ...

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

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 ...

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 ...

4

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 ...

3

The basic assumptions on the space-time structure in classical mechanics are: (1) Time intervals beetween events are absolute. (2) Space intervals beetween contemporary events are absolute. We may refere to this two properties as to the "galilean space-time structure". In the first chapter of Arnold's "Mathematical Methods of Classical Mechanics" you can ...

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

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

The Galilean spacetime is indeed the affine space $\mathbb{A}^4$. Affine space can be considered as a 'space with no origin', which makes intuitively sense because why would some point (the origin) be special. For example a trivial Galilean space is $\mathbb{E}\times \mathbb{E}^3$ where $\mathbb{E}$ is Euclidean space. The $\mathbb{R}\times \mathbb{R}^3$ ...

3

Unlike light, sound can only travel through a medium - in most situations, air. The velocities in your equations are relative to a fixed reference frame - that of the body of air in which the sound is travelling (which in a typical physics problem is the same as the ground's reference frame). So there really is a tangible difference between the case where ...

2

An example of a law that is not invariant: $F = -\mu mv$. That is, some kind of universal friction slows down all moving objects. This requires a point of reference they are slowing down compared to, so it is not invariant. Any law that can be written in the form of a tensor is invariant, but this law cannot be written in that form. Not unless you have some ...

2

In short, usually I see people saying that Galilean Relativity is bound to one certain structure of spacetime where space is relative and time is absolute. What really is the relation between such structure and the principle stated? Another possibility is that the implication is the other way around: absolute time implies Galilean Relativity. But I also ...

2

In your derivation you've implicitly assumed that the wavefunction does not change its values when you go to the Galilean boosted frame. In other words, you've assumed $\psi'(x', t') = \psi \bigl(x(x',t'), t(x',t') \bigr)$. However, this isn't right. The wavefunction encodes information about a particle's momentum, so when you go to a different frame the ...

2

If an object has no acceleration in one inertial frame of reference that means no real forces acting on it. now suppose you observe the same object from a different inertial frame,its not possible that just because you are observing the same object from a different inertial frame somehow a real force will start acting on the object. But if you observe now ...

2

My argument would be something along the following lines: assume that your spacetime is $\mathbb E^4$ and determine its isometry group. This turns out to be $O(4)\ltimes\mathbb R^4$, which doesn't coincide with the Galilei group (indeed in special relativity $O(1,3)\ltimes\mathbb R^4$ is the full Poincaré group, which is also the isometry group of flat ...

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible