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Aug 6th 2019 at 4:57:59 PM •••

This is surprisingly not that hard to understand.

Aug 24th 2013 at 5:55:49 AM •••

I think a lot of this article needs to be rewritten. Being unfamiliar with Star Trek, some of the stuff was difficult to understand, but the Momentum section was unreadable. Here's a look through my "frame of reference". :P

Space station and spaceship have identical smaller spaceships. Kira flies from either spaceship or space station to planet. Riker does the same.

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Aug 28th 2013 at 4:04:48 PM •••

Would you perhaps like to do the rewriting of some of the "difficult to understand" bits, so that the rest of us can have a template for how we should rewrite the "unreadable" section?

Sep 26th 2012 at 9:54:03 PM •••

Since the speed of light is a constant, MY hypothesis is that all the light in the universe is standing still and all the SPACE is moving around.

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Aug 24th 2013 at 5:56:25 AM •••

I think you need to read about relativity a little more.

Jun 23rd 2012 at 9:48:32 AM •••

There's some debate about the hottip that says, "Relativistic time dilation will diminish this effect somewhat, but not entirely; O'Brien will still see the the light pulses spaced out, just not as much as Sir Isaac Newton would expect." Specifically, we aren't sure whether time dilation makes it better or worse.

Suppose the Enterprise and Boothby are both at position x=0 at time t=0. The Enterprise moves at speed v. Boothby fires off a light every second. The kth flare catches up to the Enterprise (in Boothby's frame) at time t(k) and position x(k) = v t(k) = c (t(k)-k). Solving, t(k)=k*c/(c-v).

So Isaac Newton thinks O'Brien should see one pulse per c/(c-v) seconds.

Applying the change-of-base formulas, the event "kth pulse passes the Enterprise" happens at (x=kcv/(c-v), t=kc/(c-v)) in Boothby's frame, so it should happen at (x'=0, t'=sqrt(1-v^2/c^2)kc/(c-v)) in O'Brien's frame.

So O'Brien really sees one pulse per sqrt(1-v^2/c^2)c/(c-v) seconds. This is more than 1 second (time if light travels instantaneously) but less than c/(c-v) (time Newton expects).

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Jun 25th 2012 at 9:36:22 AM •••

I agree with your reasoning. However, I still don't think the text in the page is satisfactory. Take the equation for t', straight from the Lorentz transformation (I'm using natural units, I hope you don't mind):

t' = γ (v t_k - v t_k) = γ (1 - v) t_k = γ/γ t_k = t_k/γ

So you see, what is usually called time dilation (the little γ factor on the first couple of equalities) actually does increase the amount of time between pulses in the Enterprise's frame. It's the hyperbolic rotation to a new frame that does most of the work, which is to say that this result is in fact due to length contraction! Time dilation counteracts this, but not enough. It's two effects first order in γ (since length contraction is applied twice) against one.

I will try to fix the text in a way that will be correct yet clear. Please take a look and see what you think.

Edited by piroca
Jun 25th 2012 at 11:25:07 AM •••

Ah, now I see what you're saying. My point in that whole section is that this is all *not relativity*, in the sense that you get the same effects with sound waves. In this case I want to emphasize that the nonrelativistic c/(c-v) is causing a bigger effect than the relativistic effects.

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