Warp - any scientific take on it?

[Jon Miller]

Yeah, I had it in my first semester, nonphysics class.

Jon Miller
(We have first term seminars which fulfilled a writing requirement and one other requirement. I dicided to not be smart and to take one that also fulfilled a science/math requirement (Which I would have way more then enough of) and took science and science fiction. We talked about newtonian mechanics (Simple stuff) and special relativity and read science fiction novels (Timeline?) and watched some science fiction movies and wrote a science fiction short story. Was a cool class. But it was for general students, although I think that they struggled a bit on the math parts.)

[Perfection]

Originally posted by TCO
But you have failed to make me understand it.

TOUGH! It's not the role of technical folks to explain everything in detail to nontechnical folks or folks of other technicality especially when they aren't willing to take the time to look at the sources available to them (like wikipedia).

[self biased]

answer my question, dammit!

[TCO]

On the example we were talking about (with the 4 corner info transfer), how much earlier will A get his message before he sent it. How many minutes earlier?

[KrazyHorse]

I already told you that. vd/c^2 where v is the velocity of the C&D wrt A&B, d is the distance between them and c is the speed of light.

[TCO]

I don't see how the message will move back in time at all, since we IMd everything.

60mph*100LY/(3x10^8)sq

Hmmm. Looks like I need to convert some units.

I need to get the perpindicular thing out of my mind also. BTW, does it matter if D crosses A headed towards B ("closing direction") or away from B "opening direction"?

[KrazyHorse]

Originally posted by TCO
I don't see how the message will move back in time at all, since we IMd everything.

Again, that is because you do not understand relativity. At all. Your statements imply a galilean separation of space and time. The Lorentz transforms describe boosts as a hyperbolic rotation of the space and time coordinates. There is an off-diagonal term in the matrix. A mixing term. You have yet to accept that.

I need to get the perpindicular thing out of my mind also. BTW, does it matter if D crosses A headed towards B ("closing direction") or away from B "opening direction"?

I assumed that the velocity of C and D was in the same direction as the displacement vector from A to B. I we make the v negative...let me run through the calculation.

[KrazyHorse]

I don't know how better to explain it than this:

In relativity, what is, in one frame, an "IM", that is, a spacelike displacement vector (t, d, 0, 0) (where t is the time of transmission...which I assumed to be 0, but which will still work as long as it is less than d/c) is, in another frame (moving in x direction), some linear combination (pt+qd, rt+sd,0,0). The coefficients p,q,r and s are dependent on the velocity between the two frames of reference.

Now, I called that displacement vector "spacelike" because the spatial separation is larger than the time separation (using a factor of c to match units). When we have "timelike" displacement vectors (the time separation is larger than the spatial separation), it is easy to show that no possible p,q,r and s for a boost of velocity v < c will end up making the displacement have a negative time component. THE SAME IS NOT TRUE WHEN YOU HAVE A SPACELIKE DISPLACEMENT. There will be a number of different values for v which will lead to p,q,r and s so that changing frames ("boosting" by v) will get a negative time component to the displacement in the new frame.

So, if the two endpoints of a spacelike displacement (A and B or C and D) are placed in causal contact (i.e. can send IMs) then it is possible to place A at t = 0 into two-way causal contact with itself at an earlier time. In other words, you've allowed a backwards time IM.

[KrazyHorse]

Making v point in the opposite direction does not cause backwards time transmission, by the way.

But what it does do is cause A to see his own message being received by D at vd/c^2 after it sent the message...even though the messages were instantaneous in the reference frame of sender and receiver and there was no turnaround time. So where, in your galilean understanding of space-time does that vd/c^2 delay come from?

[TCO]

What leg of the journey did it creep in? As far as I see, we stake out a cartesian grid on the universe. We can now consider each step of the "round the block" transmission and for each event (msg initiated or recievied) write down exactly what point each ABCD is at in terms of the grid. We can also write down their specific time on their ship. Given that the IMs are instantanoues, the ships never move (even the moving ones), during the 4 instantaneous events (I'm ignoring the minute delay thing from before). The only displacement of the ships, or time elapsing for the stationary guys occurs from delays in forwarding the message. I don't see how that bakes down to getting a message before I sent it.

I'll go read what you wrote now.

[Kuciwalker]

There's such a thing as instantaneous rate of change.

[TCO]

The derivative?

[Kuciwalker]

The point being that the Lorentz tranformation applies even though everything is "isntantaneous" and nothing moves.

[TCO]

I think if you plot the whole thing out as a table and list each person's location and apparent time, as each event occurs, you will see that there is not the problem.

[Kuciwalker]

I think you're obviously wrong.