# Tag Info

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

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 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 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, ...

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 ...

2

Let's look to your own statements. First, time derivative after transformations isn't equal to an "old" derivative: for $\mathbf r' = \mathbf r - \mathbf u t = \mathbf r - \mathbf u t' \Rightarrow \mathbf r = \mathbf r' + \mathbf u t'$ $$\partial_{t'} = (\partial_{t'}\mathbf r )\partial_{\mathbf r} + (\partial_{t'}t) \partial_{\mathbf t} = (\mathbf u \cdot ... 1 In the S' frame, your variables are x' = x - t\cdot u \cos\theta  and y' = y - t\cdot u \sin\theta. If you do the change of variable, you get that the motion now is described by$$x' = 0y' = -\frac{g}{2}t^2$$So in your new frame of reference you have vertical free fall from rest. This is not very helpful in finding out when or where does the ... 1 Here I would like to expand some of the arguments given in Ron Maimon's inspiring answer. Consider N point particles with positions {\bf r}_1, \ldots, {\bf r}_N. The Galilean transformation group is, e.g., explained here. The only transformation, which we will bother to mention explicitly from now on, is the shear transformation$$ t \longrightarrow t, ...

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