Would the Pendulum's Shadow Vibrate at PHz on a Moving Wall?

  • Thread starter Thread starter Hippasos
  • Start date Start date
  • Tags Tags
    Shadow
Hippasos
Messages
75
Reaction score
0
Let's experiment with:

A = a very distant source of light

B = pendulum at a given distance from the light source

C = a huge white wall

Let's suppose the pendulum vibrates at PHz.

Let's suppose the white wall is moving in the line of the light source and the pendulum at the very high speed of V.

Would the pendulum's shadow vibrate at the same speed (PHz) on the C regardless of V?

Thanks!
 
Physics news on Phys.org
if the pendulum and light source are stationary and the wall is moving toward the pendulum then the answer is no.
 
Would it be error then if we say:

If the wall is traveling at half light speed towards the light source and the light source is traveling also at half light speed towards the wall the pendulum's (pendulum stationary at the middle) shadow would vibrate at the speed of light?

What would happen to the shadow if both were traveling say like 55% of speed of light?

Thanks again!
 
There is no conflict with relativity, since no actual particles are moving faster than light here, and an FTL shadow cannot be used to transmit information FTL.

Allow me to double check why. If someone is on the earth, and they speak to someone on the moon through a radio, that is approximately light speed communication. Now if I am flashing a laser dot on the moon, I can make the dot on the moon travel faster the light, but I am still not communicating faster then light, because it takes the time for the laser light to get to the moon before the effect is noticed, in this case, also at light speed.
 
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...
Back
Top