Can you prove a theorem from Bergmann chapter 8 using Fig. 8?

  • Thread starter Thread starter supakorn
  • Start date Start date
  • Tags Tags
    Theorem
supakorn
Messages
11
Reaction score
0
From Bergmann chapter 8.Please show that (see at Fig. below)...thank you.
 

Attachments

  • B1.GIF
    B1.GIF
    1.1 KB · Views: 443
  • B2.GIF
    B2.GIF
    995 bytes · Views: 464
Physics news on Phys.org
Sounds like a homework problem.
What have you tried?
What is \varphi? Does it have any special properties?
 
it not HW but i read and try to prove.For PHI, it is a skewsymmetric.
 
I'm Done (but Not Sure).phi Defined On The Flat Space,i Will Post My Answer Next Time.now I'm Happy.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
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...

Similar threads

Back
Top