Can Faster-Than-Light Communication Be Achieved with a Long Pole?

Jarfi
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1.
Say there are two Earth's and many light years apart. You have a very long pole which is connected to both Earth's. And one guy on Earth 1 starts pushing/tilting the other end forward. So the guy on Earth 2 feels the pole being pushed forward. guy on Earth pushes pole forward in morse code... Faster than light communication wtf??

2.
say you have a very long pole many light years long pointed out from the earth. It is stuck in a machine that can move it right and left. Let's make the machine move the pole 90 degrees right. Moving the pole takes only 1 second so next to the machine the pole went maybe 5 km p hour i don't know. But at the end of the pole many light years it went much much faster than light... Wtf?

Can anyone explains what happens? and the poles can't break or something like that we assume nothing goes wrong
 
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All mechnical movement - pushing, rotation etc - propagates at a speed of sound in matrial, in metals - about 3-5km/s, about 100000 times slower than c.
 
A ok so the pole would bend and when pushed the movement would travel like a wawe
 
Actually, with modern measuring technology, the measurement of the delay between hitting one end of a 'rigid' ceramic rod and the other end responding can be done over distances like a meter. This is used to study certain bulk properties of ceramics.
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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