Raised Indices

Raised Indices - Сообщения

#1 Опубликовано: 22.03.2012 23:23:46
Steve Swinbanks

Steve Swinbanks

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Hi,

I'm sorry if this is a dumb question, but is there a way to have raised indices, just as there is with lowered indices?



Also is there a way to sum over two or more indices?

Thanks

Steve
#2 Опубликовано: 23.03.2012 03:27:53
Radovan Omorjan

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Hello Steve,

I can not see your image, sorry.
Wrote

I'm sorry if this is a dumb question, but is there a way to have raised indices, just as there is with lowered indices?

The question is, why do you need raised indices? If this is like literal lower index (obtained with dot) I think that raised indices will make confusion with power operator.
Wrote

Also is there a way to sum over two or more indices?

I hope I understood your question and yes, there is. You can use summation operator from the palette or loops like "for" or "while". Visit the Wiki, there you will find Tutorials and Examples about it.

Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
#3 Опубликовано: 23.03.2012 04:19:12
Steve Swinbanks

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Radovan,

Sorry about that, should have read the Wiki more carefully on inserting images. Anyway what I am trying to do is manipulate tensors in General Relativity and need to combine contra and covariant tensors, such as the metric tensor.
Hopefully you can see what I mean in the image below.



Steve
#4 Опубликовано: 23.03.2012 05:41:24
mr_raf

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Hi there.
I guess that for B You should use:

B:sum((el(y,i)^i),i,0,n)

As below


As You see C value is different than You wrote.
Best regards. Rafal
#5 Опубликовано: 23.03.2012 07:36:55
Steve Swinbanks

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Rafel, thanks, but that is not exactly what I want.

[/img]

As you can see I am after something that expands like contravariant tensors. Not sure if it possible or not.

Regards

Steve
#6 Опубликовано: 23.03.2012 07:42:14
mr_raf

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Ok. I thought a little bit, and i think that this should do the trick



Risen_indices - wiki_upload_file

Although still does not have a "nice" (transparent) appearance. Perform calculations correctly.

Something simpler. I went to far with "n" variable.


Risen_indices_2 - wiki_upload_file
Best regards. Rafal
#7 Опубликовано: 23.03.2012 12:08:38
Radovan Omorjan

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Hello,

By following this discussion it crossed my mind something. It is quite questionable that this will work in SMath
[MATH=eng]A:sum(el(x,i),i,0,3)[/MATH]
[MATH=eng]A=el(x,0)+el(x,1)+el(x,2)+el(x,3)[/MATH]
But, on the other hand, there is no way that the vector element of x with zero index can get a value, neither numerical nor symbolical. In SMath the minimal matrix index is one but in these cases SMath will not complain if you put any index value.
Do not know how to explain this.

Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
#8 Опубликовано: 23.03.2012 15:00:39
mr_raf

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Hello Radovan.
From what I know matrixes and vectors elements position variable needs to be defined as a positive integers. It is questionable if zero is an integer. In my opinion zero is zero, a strange structure which does not fit into any other sets of numbers.
In Smath as I noticed matrixes and vectors start from number 1. Thats why i used i:=1.
Using i:=0 generates an error, that there is no such element. Which is understandable, because in our vector there is no element [MATH]x,0[/MATH].

Radovan said:
Цитата

But, on the other hand, there is no way that the vector element of x with zero index can get a value, neither numerical nor symbolical. In SMath the minimal matrix index is one but in these cases SMath will not complain if you put any index value.
Do not know how to explain this.



Exacly, using summation of matrix/vector elements alows to put i:=0, but creating matrixes/vectors requires usage of elements number i≥1.
Best regards. Rafal
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Radovan Omorjan 23.03.2012 15:24:00
#9 Опубликовано: 23.03.2012 15:35:36
Radovan Omorjan

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Hello Rafal,

Thanks for the reply. On my second thoughts, I think that nothing is strange if we look at this like a symbolic expression. For instance:
[MATH=eng]sum(el(x,i),i,-3,2)=el(x,-3)+el(x,-2)+el(x,-1)+el(x,0)+el(x,1)+el(x,2)[/MATH]
Therefore, we can accept that symbolic engine accepts any index number, but not numerical. As you sad, only integers greater or equal to 1.
Hope I am right about all of this.

Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
#10 Опубликовано: 23.03.2012 16:05:23
mr_raf

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Wrote

Hello Rafal,

Thanks for the reply. On my second thoughts, I think that nothing is strange if we look at this like a symbolic expression. For instance:
[MATH=eng]sum(el(x,i),i,-3,2)=el(x,-3)+el(x,-2)+el(x,-1)+el(x,0)+el(x,1)+el(x,2)[/MATH]
Therefore, we can accept that symbolic engine accepts any index number, but not numerical. As you sad, only integers greater or equal to 1.
Hope I am right about all of this.

Regards,
Radovan



If we assume that the indicator of the element is a number lying on the axis of integers, and we treat it as an ordinal it seems to me that, yes, ther`s nothing strange.
On the other hand, assuming that it is an element in the first quadrant of the Cartesian, that is part of the real numbers possible to the physical, "touchable" definition, than it is not.
The real thing is to have (n>0) or not have (n=0) an apple, it isn`t physically possible to have negative number of apples (n<0), this statement is a virtual, contractual. While looking at the table You can only say if ther`s an apple (or more apples), or ther`s none. Therefore, while building a vector/matrix, an element can either exist or not, you can not build a matrix for which the indicator of elements will not be possible for defining, that is they will lay in negative part of Cartesian.

I hope that my understanding isn`t wrong.
Best regards. Rafal
#11 Опубликовано: 24.03.2012 20:19:16
Steve Swinbanks

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Thanks guys for your comments. I tend to agree with Omorr, the indices are only markers and can take any integer form including zero. Anyway it doesn't get me any closer to raised indices so I think I will give that up for the time being.

Steve
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