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jby

Any books recommended for dummies? All books that I've found starts with contraviant and covariant tensor, which seems misleading to me.

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jby

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- #2

marcus

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Let me guess. You looked at "Shaum College Outline Series" inexpensive good book with lots of worked problems----nothing wrong with book, you were just put off by words "contravariant and covariant". Problem is not with book or with you but with ice-cold nomenclature.Originally posted by jby

cant help with other book-recommendation---surely others will

So I will take a bizarre attitude and attempt to persuade you that the words covariant and contravariant are actually warm and fuzzy. they are your friends and the really bad terminology in that bunch is "tensor"

my private sentiment is that the word "tensor" is misleading

(probably "dingus" would do as well and it is just some

historical accident that they started saying tensor-----but covar

and contravar are descriptive names, therefore could be helpful to understand what they mean.

are you all right with the idea of a smooth mapping from some place to another place?

(how you think of the "places" depends on the problem, or on mathematical habit and convention, they might be the surfaces of two different potatos or even two copies of the same potato)

a covariant dingus is something that the mapping carries along

a contravariant dingus is something that a mapping from X to Y "pulls back" from Y to X, so it gets translated opposite or

Let's write f: X --> Y to mean a map from X to Y

some gizmos (functions, calculus tools, mathematical contraptions) that you can define on Y are of a sort that f pulls back in a natural way to wherever it is defined, namely to X.

other bits of hardware are of a sort that if they are defined on Y the map will not pull them back but if they are defined on X the map will carry them over and make versions of them defined on Y,

directional derivatives work that way.

for example a plain old real valued function G(y) defined on Y will pull back, using the map f,

to become a new function G(f(x)) defined on X

but a real-valued function G(x) defined on X will not push forward to a function defined on Y----the mapping f:X-->Y is going the "wrong way" to accomplish that, you would need to have an inverse map of f and that is not always defined or available.

however a trajectory or path in X is COvariant because the map f will transform it into an image path in Y. And taking the derivative of some real-valued function defined on X at a point along that path translates into taking the derivative of a function defined on Y along the image path.

so the solemn ritual of taking a derivative of whatever and in whatever direction goes along WITH the map, getting carried along from X to Y in the same direction as the map goes.

in the end one just has to draw pictures. the two potatoes and the arrow going from one to another, and think about the chain rule of ordinary freshman calculus (nothing ever gets done without applying the chain rule at some point)

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Geroch, Robert; General Relativity from A to B;The Unversity of Chicago Press(1978)

https://www.amazon.com/exec/obidos/...90-0933546?v=glance&s=books&tag=pfamazon01-20

Lilley, Sam;Discovering Relativity for yourself;Cambridge University Press(1981)

https://www.amazon.com/exec/obidos/...andallA/104-8288190-0933546&tag=pfamazon01-20

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jcsd

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http://www.geocities.com/zcphysicsms/

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Chi Meson

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Dingus it is then! I've always hated the term "tensor." Great explanation too. I'll be using it later this year if you don't mind. (And even if you did, how would you stop me?)Originally posted by marcus

my private sentiment is that the word "tensor" is misleading

(probably "dingus" would do as well and it is just some

historical accident that they started saying tensor-----but covar

and contravar are descriptive names, therefore could be helpful to understand what they mean.

- #6

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Can I take a dingus at Y and carry it back to X and compare it with the dingus at X and take the derivative? That'd be like going CONTRARY to the mapping and hence CONTRAvariant, isn't it?Originally posted by marcus

however a trajectory or path in X is COvariant because the map f will transform it into an image path in Y. And taking the derivative of some real-valued function defined on X at a point along that path translates into taking the derivative of a function defined on Y along the image path.

so the solemn ritual of taking a derivative of whatever and in whatever direction goes along WITH the map, getting carried along from X to Y in the same direction as the map goes.

- #7

pmb

Use caution when using these notes. They were written by a well known crackpot. Most of it is okay I guess (simple stuff copied from texts) but other parts are very wrong.Originally posted by jcsd

http://www.geocities.com/zcphysicsms/

The author used to post here for a short time. He came here and imediately started flaming me when he couldn't convince me that a scalar was not defined in modern physics/math as a tensor of rank zero. He was banished when he started to flame the moderator. He was warned to cease flaming but continued and was tossed out.

Pete

- #8

pmb

What do you find misleading about this? Covariant/contravariant refer to the geometric object in most cases (sometimes it refers to a vector and its dual)Originally posted by jby

If you'd like more on covariant vs. contravariant see

http://www.geocities.com/physics_world/co_vs_contra.htm

Pete

- #9

jcsd

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I haven't looked over all of them, but they do seem okay to me, I've the feeling he might of lifted them from a textbook.Originally posted by pmb

Use caution when using these notes. They were written by a well known crackpot. Most of it is okay I guess (simple stuff copied from texts) but other parts are very wrong.

The author used to post here for a short time. He came here and imediately started flaming me when he couldn't convince me that a scalar was not defined in modern physics/math as a tensor of rank zero. He was banished when he started to flame the moderator. He was warned to cease flaming but continued and was tossed out.

Pete

- #10

pmb

Some but not all.Originally posted by jcsd

I haven't looked over all of them, but they do seem okay to me, I've the feeling he might of lifted them from a textbook.

For instance: The author keeps trying to define force as mass times acceleration. That has always gotten him into trouble in the past and I'm sure it will in the future too. In this particular case he tried to derive an expression for the Coriolis force. So he derived an expression for the Coriolis acceleration and then multiplied it times rest mass and calls that the "Coriolis force." His result is incorrect for that reason. His bogus derivation of the Coriolis force is here

http://www.geocities.com/zcphysicsms/chap6.htm#BM80

For a correct derivation see

http://arcturus.mit.edu/8.962/notes/gr6.pdf

Force is defined as the time rate of change of momentum - not as mass times acceleration. Bertschinger gets it right. The author of that web site you referanced was shown this and all he could do was his usual backpeddling - It was too silly of response to repeat here. For those interested go to sci.physics.relativity and read the thread "Centrifugal etc fict-force relativistic derivation"

Then there is the subject regarding tidal forces. He botches that up pretty bad. He does something bizzare - he defines tidal force differently than the rest of the world and then proceeds to show that there is no such thing as a gravitational tidal force. He tries to define tidal force in terms of 4-forces which is quite unlike the term is defined in General Relativity. His definition is that the tidal force is the gradient of a 4-force of a particle in free-fall which is total nonsense since the gravitational force is an inertial force and not a 4-force. Any decent GRist knows that as fact.

Then there is the time where he was totally unable to comphrend what a tidal force tensor was or how it is related to the Riemann tensor.

See - http://www.geocities.com/physics_world/tidal_force_tensor.htm

for the Newtonian definition. The GR definition is that the tidal force tensor is the Riemann tensor. They're related.

Then there is the part where he claims that by "curvature" cosmologists mean "Gaussian curvature" which is bogus as well.

Then there is his claim that a uniform gravitational field has spacetime curvature when its really defined as having no tidal forces.

Pete

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- #11

Phobos

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Do me a favor and let's stop that line of discussion here and get back to the original question. Thanks.Originally posted by pmb

It goes on and on and on. He's just a crackpot - plain and simple.

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