Recent content by astrodummy

In the slider reference frame you still have the 2m flicker PLUS the time it takes for the slider to clear the 1m gap. At vrel = ##\frac{\sqrt{3}}{2}## it takes ##\frac{2}{\sqrt{3}}~m## of time to clear the gap. So the total flicker time is now ##\frac{2}{\sqrt{3}}+2~m##, the same as in the rail...

You're right, the gradient of the 'current off' and 'current on' world lines will be more timelike so when they intersect with the vertical bulb world line they will have a wider separation.

So in this scenario only the bottom rail in considered 'live' along it's entire length and as the slider bar C breaks contact at A and bar D touches contact B immediately afterwards there will be a 2m delay in the supply to current to the bulb which it will 'see' in 4m time. Assuming electricity...

Actually now that I think about it that cannot be true either. Both the top and bottom rails are always 'live' since they are both connected to the battery at all times. Instantaneously changing the connection from bar A to bar B will not interrupt the current flow when vrel << c. There will be...

Soooo, the electric charge is waiting AT THE SWITCH even through it is open. That makes sense to me, thank you.

What I am getting at is, in the case of the switch being very close to the bulb does this 'signal' move from the switch to the bulb when the switch is thrown, of from the battery to the bulb? If from the battery, what 'signal' traveled from the switch to the battery to get the current moving?

Your 'time evolution' diagrams have a problem. You cannot claim that AB is in 'black out' before bar D gets to B without also acknowledging that the bottom rail between C and D ( the distance between the vertical bars on the slider ) is ALSO in 'black out'. Since there can be no current in...

Suppose you have a battery wired to a light bulb and switch via a long pair of conductors, say 1 light second long. Assume for now that electricity travels at c. The switch is thrown. How long does it take the bulb to light? Does the electricity take 1 second to get from the battery to the...

D'oh! Of course. That 2m section of the rail is "dead" the instant Bar D touches point B. So that 2 meters of 'no power' will get to bulb in due course. So coordinates of Event #3 'power loss reaches bulb' is ##(-4,4)## and coordinates of Event #4 'power restore reached bulb' is ##(-4,6)##. So...

If v_rel << c then ##\gamma## will be 1. So, no length contraction in either reference frame. And since AB = CD the bars will always be in contact with the rails, so the power will never be lost to the bulb. No flicker at all! As Ibix correctly pointed out the quickest way to determine the...

No no no, my bad. ##(-4-2\sqrt{3},4+2\sqrt{3})## equals ##(\frac{2}{\sqrt{3}-2},\frac{-2}{\sqrt{3}-2})## In my defense I'm nearly 60 yo and haven't done this kinda maths for a couple of years. Steve

Thanks Ibix, For the lazy I have attached some screenshots. Steve

Ibix, Thanks for your feedback. I should have known to double-check the coordinates by doing a Lorentz transformation. My coordinates are okay since I mistakenly used 2 metres as the distance between the bulb and point A in the rail frame, instead of 4 metres. And I mislabeled my diagram. D'oh...

Oops ... found an error on the line after: "Find intersection of line [1] and [3] to derive Event #4 coordinates" Next line SHOULD BE: ##-x-\frac{2} {\sqrt{3}}+2 = -\frac{2} {\sqrt{3}}(x+1)## which simplifies to ##x = -6 - 4\sqrt{3}## same as before.

This post builds on a previously closed thread here: https://www.physicsforums.com/threads/lightbulb-paradox.141191/#post-1145674 I will not describe the problem here without the copyright owner's permission. I, like the OP in the original thread, am keen to see if anyone else who has this text...