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- From: franjion@spot.Colorado.EDU (John Franjione)
- Subject: Re: tensors: How about 3rd, 4th rank?
- Message-ID: <1992Aug21.230104.15702@ucsu.Colorado.EDU>
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- Organization: University of Colorado, Boulder
- References: <1992Aug20.190041.6215@pellns.alleg.edu> <israel.714359708@unixg.ubc.ca> <5130@tuegate.tue.nl> <5134@tuegate.tue.nl> <3djygrk@rpi.edu>
- Date: Fri, 21 Aug 1992 23:01:04 GMT
- Lines: 29
-
- Tom Pierce <pierct@rpi.edu> writes:
-
- >What would you use a third or highr order tensor for? Any PHYSICAL
- >applications?
-
- First, keep in mind that tensors are basically a notational convenince
- for expressing linear relationships (e.g. a second order tensor is a
- linear transformation from one first order vector space to another).
-
- I do research in fluid mechanics of solids emmersed in very viscous
- fluids. Since the governing partial differential equations are
- linear, tensors come in handy. For example, the force on a particle
- is linearly related to the velocity of the particle. This is
- conveniently expressed as:
- f = A . u
- where f is the force vector, u is the velocity vector, and A is a
- second order tensor.
-
- Now say the particle is immersed in an ambient flow field, which is
- characterized by a velocity gradient, grad(u). The force on the
- particle is linearly related to the velocity gradient:
-
- f = G . grad(u)
-
- Since grad(u) is itself a second order tensor, G is a third order
- tensor. In this case, the "dot" product is actually a contraction
- over 2 indices, rather than 1.
-
- John Franjione
-