Timeline for Transformation law of vector fields on $\mathbb{R}^n$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2021 at 15:07 | vote | accept | Leonid | ||
| Mar 31, 2021 at 12:36 | answer | added | J. Murray | timeline score: 5 | |
| Mar 31, 2021 at 7:39 | history | edited | Urb | CC BY-SA 4.0 |
added 34 characters in body
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| Mar 31, 2021 at 5:48 | comment | added | QCD_IS_GOOD | ^ (really, perhaps this is not even that true. Temperature and pressure do depend on what reference frame you measure them in, but this is maybe completely besides the point, and not really relevant) | |
| Mar 31, 2021 at 5:46 | comment | added | QCD_IS_GOOD | "If it looks like a duck, swims like a duck, and quacks like a duck, then it probably is a duck." In physics, a vector is precisely something that transforms like a vector As you've observed in your question, (temperature,pressure) does not transform like a vector. Rather, they transform like a pair of scalars. So they are a pair of scalars, not a vector | |
| Mar 31, 2021 at 5:43 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited title; edited tags
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| Mar 31, 2021 at 3:07 | comment | added | Jbag1212 | In physics we look for physical quantities which transform like tensors. If you were just given a function that you don't actually know what it physically represents, then of course you wouldn't be able to proceed. Just because something has multiple indices doesn't mean that it is a tensor. Spinors are one example, And in answer to your question, yes. | |
| Mar 31, 2021 at 2:59 | comment | added | Leonid | What makes this confusing is that suppose I don't actually know what it physically represents, just take it to be a mathematical function, then who am I supposed to trust? The vector transformation law or not? | |
| Mar 31, 2021 at 2:57 | comment | added | Leonid | so just to be clear: If I tell you that the components of F are temperature and pressure it is not a vector field, but if I give you the same function and tell you it is actually supposed to represent the components of a (classical) gravitational field, then suddenly it becomes a vector? | |
| Mar 31, 2021 at 2:52 | comment | added | Jbag1212 | Things obey the vector transformation law if they are vectors which are elements of the tangent space at each point. In $\mathbb{R}^2$ suppose we have a basis for the tangent space, and physically we call it "north-south" and "east-west." Physically speaking, the temperature and pressure do not depend the direction of "north-south" and "east-west." So this vector of temperature and pressure is not in the tangent space. | |
| Mar 31, 2021 at 2:45 | history | edited | Leonid | CC BY-SA 4.0 |
added 29 characters in body
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| Mar 31, 2021 at 2:43 | comment | added | Leonid | @Jbag1212 what do you mean the tangent space needs to be defined? It is defined in the obvious way for $R^2$. And I am aware of all that, what i'm saying is this: I have a function as defined above and it can be easily shown to be a section of the projection map from $TR^2$ to $R^2$, so mathematically it satisfies the definition of a vector field (which is just a section of the projection map), so give me a reason why it shouldn't obey the vector transformation law. | |
| Mar 31, 2021 at 2:37 | comment | added | Mozibur Ullah | The transformation law of a vector field on a manifold is that of a contravariant vector (field). | |
| Mar 31, 2021 at 2:37 | comment | added | Jbag1212 | The tangent space of a point on manifold needs to be defined before you can talk about vectors. This can be done with directional derivatives. This is also why the dimension of the vector space is the dimension of the manifold. The pressure and temperature have no relationship to tangent space. You could define these on a 1D manifold, but the vector space of a 1D manifold must have dimension 1. | |
| Mar 31, 2021 at 2:28 | history | asked | Leonid | CC BY-SA 4.0 |