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I am trying to understand simple examples of space-time curvature.

Assume for the moment that $c$ is infinite (classical curvature due to Newton's laws). Also, I will only consider 1+1-dimensional space for simplicity: $(x,t)$.

Case I. I will consider uniform acceleration due to gravity g along the positive x-direction pervading the entire space. Then all particles will have $$x-x_0 = \frac{1}{2}g(t-t_0)^2.$$ For this case, we can use the transformation \begin{align} x' &= x - \frac{1}{2}g(t-t_0)^2 \\ t' &= t\end{align} to get a transformed space-time $(x',t')$ such that any path of the form $$x-x_0 = \frac{1}{2}g(t-t_0)^2$$ in the original co-ordinate system is equivalent to the form $$x' = vt' + x_0'$$ representing uniform motion in the new $(x',t')$ coordinate system. Thus, if space-time is warped as described by the transformation from $(x,t)$ to $(x',t')$ system, all objects just follow a straight line in the transformed system.

Case II. Now I will consider a slightly more complex scenario. Here, $$g(x) = -\omega^2 x.$$ Can we obtain transformation $x'=x'(x,t)$ and $t'=t'(x,t)$ such that uniform motion in $(x',t')$ is equivalent to Simple Harmonic motion in $(x,t)$?

Case III. Now assume that acceleration due to gravity is $g$ for $-1 \le x \le 1$ and is 0 everywhere else. What about this field?

I don't have any experience in differential geometry or any obscure mathematics. My current goal is to see how much we can understand relativity just with simple mathematics.

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  • $\begingroup$ This may not be what you're looking for, but graph the function $r = \sec \theta$ in polar coordinates, and see what that gives you. $\endgroup$ – HDE 226868 Nov 25 '14 at 21:37
  • $\begingroup$ $r=\sec\theta$ is the same as $t=1$ in the $(x,t)$, am I correct? I assumed you meant $r=\sqrt{t^2+x^2}$ and $\theta=\tan^{-1}(x/t)$. So how do I proceed from there? Is your intuition trying to solve case II or case III? $\endgroup$ – Gowtham Nov 26 '14 at 7:26
  • $\begingroup$ Is it not this the essence of the Hamilton-Jacobi formulation of classical mechanics? To make non-linear problems into linear ones via a suitable transformation (canonical)? $\endgroup$ – Marion Mar 20 '15 at 10:11
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If you are allowing non-linear transformations of space-time (as you have done in your Case I), then I don't see why you can't write any motion of an object as x' = 0. i.e. Suppose the motion of the object in the (x,t) coordinates can be written as f(x,t) = 0 for some non-linear function f. Then, by setting x' = f(x,t), t' = t, the motion in (x',t') can be simply written as x' = 0. For example, the answer to your second case can be written as: Transform (x,t) to (x' = A*sin(wt+d),t'=t), then the motion is simply x'=0.

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    $\begingroup$ The issue with your answer is that the transformation of space time depends on the values of A and d. I want the transformation to be independent of the parameters A and d. $\endgroup$ – Gowtham Nov 26 '14 at 17:35
  • $\begingroup$ The A and d are just coming from initial conditions. Even in your case I, you have $x_o$ and $t_o$ showing up in the transformation. It is the same as that. $\endgroup$ – MasterShake Nov 26 '14 at 23:31
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The Lorentz transformation simplifies to the classical transformation for all your cases since you took $c$ as infinity. There is no time dilation. The transform is just an identity multiplication for all 3 cases with:

$$x' = x - vt$$ and $$t' = t$$

Now that you are in the classical world in one dimension, your first case is merely a tilted line with objects rolling down the slope, no matter from where on the line you look. Like standing on a hill watching a ball roll down toward you, or away from you.

In second case: $$x' = (-w^2)x' = (-w^2)(x - vt)$$ and $$t' = t$$

In the third case: acceleration due to gravity will be g also for moving observer but just seen at coordinates shifted from $-1 <= x <= 1$ becomes $-1 <= x'+vt' <= 1$ or $-1-vt' <= x' <= 1-vt'$ and $0$ everywhere else

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Ok I solved case 2. Let $t'=\tan^{-1}(\omega t)$ and $x'=x\sqrt{1+t'^2}$. Also observe that $x=A\cos \omega t + B\sin \omega t$ to represent simple harmonic motion. We get $x'=A+Bt'$. So only case 3 remains open.

Also as u can see, I just did guesswork, without proper techniques for how to solve this in general.

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protected by Qmechanic Jun 25 '15 at 13:55

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