I Differential Forms in GR: Higher Order Derivatives

kent davidge
Messages
931
Reaction score
56
The differential form of a function is
\partial{f(x^1,...,x^n)}=\frac{\partial{f(x^1,...,x^n)}}{\partial{x^1}}dx^1+...+\frac{\partial{f(x^1,...,x^n)}}{\partial{x^n}}dx^nIs there (especially in General Relativity) differential of higher orders, like \partial^2{f(x^1,...,x^n)}? If so, how is it computed?
 
Physics news on Phys.org
jedishrfu said:
This paper discusses Differential forms, Tensors and uses in General Relativity so I would say yes higher order forms are used and are useful in General Relativity.

http://www.aei.mpg.de/~gielen/diffgeo.pdf
But at the same time no, there are no forms that are the second derivative of a function (as the OP suggests) as the exterior derivative applied twice gives zero (the any-dimensional equivalent of curl(grad(f))=0).

Of course there are other forms of higher order.
 
Orodruin said:
But at the same time no, there are no forms that are the second derivative of a function (as the OP suggests) as the exterior derivative applied twice gives zero (the any-dimensional equivalent of curl(grad(f))=0).

Of course there are other forms of higher order.

Yes, d^2 always produces zero. However, you can get something sort of conceptually similar to d^2:
  1. Operate on F with d to produce dF.
  2. Take the Hodge dual, *dF.
  3. Operate on THAT with d, to produce d * dF
This isn't necessarily zero, and is sort of like a higher-order derivative. In 3-D, if F is a scalar function, then * d * d F = \nabla^2 F.
 
Orodruin said:
But at the same time no, there are no forms that are the second derivative of a function (as the OP suggests) as the exterior derivative applied twice gives zero (the any-dimensional equivalent of curl(grad(f))=0).

Of course there are other forms of higher order.
stevendaryl said:
Yes, d^2 always produces zero. However, you can get something sort of conceptually similar to d^2:
  1. Operate on F with d to produce dF.
  2. Take the Hodge dual, *dF.
  3. Operate on THAT with d, to produce d * dF
This isn't necessarily zero, and is sort of like a higher-order derivative. In 3-D, if F is a scalar function, then * d * d F = \nabla^2 F.

In General Relativity one frequently deal with differential forms, say W. What actually is it? I know it has to be a completely antisymmetric (0,p) tensor. But what are its components Wμ1...μp? Would it be some array of derivatives of a function?
 
The components don't have to be derivatives. They are just antisymmetric, which makes threm taylor-fit for integration over manifolds.
 
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

Replies
1
Views
989
Replies
2
Views
1K
Replies
10
Views
2K
Replies
4
Views
2K
Replies
14
Views
1K
Replies
38
Views
6K
Replies
9
Views
2K
Back
Top