Is g the gravitional constant or a function. If so, what is g(x)=?
Announcement
Collapse
No announcement yet.
Warp - any scientific take on it?
Collapse
X
-
g is the so-called "gamma factor" or "Lorentz factor". I stated that in one of my earlier posts.
g = gamma = gamma(v) = 1/(sqrt(1-v^2/c^2))
The gravitational constant 6.67*10^-11 Nm^2/kg^2 is generally represented as G. g often represents the gravitational field near Earth's surface, but that has nothing to do with problem.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
Comment
-
Originally posted by KrazyHorse
I'm quite calm.
Please post the metric tensor for your postulated wormhole. Or, failing that, the curvature tensor.
As to your question about the metric tensor or curvature tensor used by Thorne in his paper it appears as if it was carefully determined with a view towards being a traversible wormhole.
I certainly don't currently have the background in topology to explain the reasoning here but I'm not sure that qualifies as ridiculous either.
ds^2 = -e^2 redshift(r) dt^2 +dr^2 / [1 - shape(r)/r] + r^2 [d theta^2 + sin^2 theta d phi^2 ]
This should be the "spacetime metric". I apologize if that's not what you're looking for.
You might be best served to see if the paper is electronically available to your institution.
Comment
-
Originally posted by Geronimo
You still haven't posted your "objection". that is to say, what you found most ridiculous about the post. I don't deny there is a problem I simply ask point blank for you to spell out how it was ridiculous.As to your question about the metric tensor or curvature tensor used by Thorne in his paper it appears as if it was carefully determined with a view towards being a traversible wormhole.
Of course it was. And this means something very special.
You're getting much closer to the answer...
ds^2 = -e^2 redshift(r) dt^2 +dr^2 / [1 - shape(r)/r] + r^2 [d theta^2 + sin^2 theta d phi^2 ]
More later on this...
This should be the "spacetime metric". I apologize if that's not what you're looking for.
Again, more on that later...Last edited by KrazyHorse; February 6, 2007, 16:03.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
Comment
-
Hey GP, lemme poke my head in for a sec...
I think you're trying to come into a conceptually difficult problem in the middle (nobody's smart enough). I'm guessing KH is frustrated because you're missing something fundamental. Try this... no math.
You and I are sitting at a table. A light is at the middle of the table. The light goes on, we see it, and we each click our stopwatches. Our stopwatches are now perfectly synchronized.
KH is on a train. He is moving away from us. What he sees, is you click your stopwatch, then me, because the light from you reaches him first (he outruns the light from me for a bit longer). He sees our stopwatches out of synch.
This isn't a trick of the light. That's the key. According to KH, our watches aren't synched. According to us, they are. We're both right. This is why your IMer doesn't solve the problem. You think it's a trick of the light. KH says it's reality (he's right).
Help?
Comment
-
Are you at Fermilab?12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
Comment
-
Look...do you want to wait until I read up on the basics and spend some time thinking about it. Or just have me hack away? Let's say A and B are at rest. They have synchronized clocks. C and D are moving. They have synchronized clocks. Since they are moving they have some time dilation. So their clocks run slower. But they are still at same speed on each ship. And each can be correlated to a time on the clocks of A and B, by undilating. So when A IM's B, D is already past A. Therefore, if C is just hitting B, then C and D are going same speed (and this is all colinear and same direction), so they are maintaining same separation, so when D does an IM (on the C, D clocks), D will still be past A.
Comment
-
Originally posted by TCO
Look...do you want to wait until I read up on the basics and spend some time thinking about it. Or just have me hack away? Let's say A and B are at rest. They have synchronized clocks. C and D are moving. They have synchronized clocks. Since they are moving they have some time dilation. So their clocks run slower.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
Comment
-
HE experiment?12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
Comment
-
If you want to look in on one of my labs, here;
I'll PM the login and password
p.s. feel free to move around the cam.
Comment
-
Cool. I've been doing some LHC phenomenology lately.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
Comment
Comment