Consider the field, where is a vector constant. Using suffix notation, show that:

i)

ii)

Using suffix notation, we get

And then the dot product gives us

Just a little unsure where to go from here?

Thanks in advance

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- Mar 8th 2011, 09:49 AMcraigVector Triple Product, Alternating Tensor
Consider the field, where is a vector constant. Using suffix notation, show that:

i)

ii)

Using suffix notation, we get

And then the dot product gives us

Just a little unsure where to go from here?

Thanks in advance - Mar 8th 2011, 09:58 AMAckbeet
It should be

The dot product would then be

[EDIT]: This equation is incorrect. The derivative operator cannot go past the 1/r. See below for correction.

Can you see where this is going?

For the cross product, you'll have a double sum, and then you can reduce using your Levi-Civita to Kronecker Delta identity, that should reduce to the RHS. Make sense? - Mar 8th 2011, 10:16 AMcraig
Ahh sorry a little typo in writing out the question. Not too sure how to evaluate the suffix partial differentiation though :\

Ahh that does make a little sense, I thought that you would end up using the result from the first part so hadn't really given this much of a go.

Will have a play around with it now, thankyou :) - Mar 8th 2011, 10:23 AMAckbeet
Note that is a vector

**constant.**Also note that the Levi-Civita symbol is zero if any of its indices are duplicated. Does that give you any direction? - Mar 8th 2011, 12:27 PMcraig
Because it's a constant, when differentiated it just becomes zero, that's the one bit I knew already haha!

I know the fact about the Levi-Civita symbol being zero if any indices appear more than once, but do we 'sum' over the different s ?

Sorry if I'm not getting this, not had much chance at getting my head around the whole suffix notation thing :\ - Mar 8th 2011, 01:15 PMAckbeet
Hang on a bit, I think I made a mistake earlier. definitely depends on so the partial derivative can't go past that.

Ok, let me make the notation a little clearer:

(using the quotient rule)

Now if the Levi-Civita symbol cancels out the whole expression. So no need to examine that case. Assume Then how does this expression simplify? Recall that

- Mar 8th 2011, 03:33 PMcraig
- Mar 8th 2011, 06:53 PMAckbeet
The reason I'm looking at and not is that the expression is what's in the numerator of the quotient rule. Note that

Why?

Also, if what does the Levi-Civita symbol do? - Mar 9th 2011, 01:37 AMcraig
- Mar 9th 2011, 02:46 AMAckbeet
Right.

Quote:

I've never heard that before, is this just basic knowledge that I should know haha?

Think about it for a second: I'm taking the partial derivative of one*independent variable*with respect to another*independent variable*. If the two variables happen to be the same, then I should get one, right? I mean, you definitely have

and similarly

That you know from Calculus I and Calculus III (or Differential Calculus and Multivariable Calculus, depending on where you took it).

However, let's suppose that and are both independent variables. What should

be?

Suppose I gave you a function and asked you to compute

What would you tell me? And then, if I gave you a function and asked you to compute

what would you tell me? But isn't

So what does that tell you? - Mar 9th 2011, 03:05 AMcraig
Right I'm assuming that this is the obvious 0, treating the as a constant when we take the partial derivative wrt .

We know that both and are both zero, by the same principle as above?

So does this mean that the two partial derivatives in our previous equations are both zero. Hence the dot product is also zero?

Thanks again for the reply. - Mar 9th 2011, 04:41 AMAckbeet
Ok, so it looks like you understand now why

Looking back at post # 6, and looking at the expression

I can simplify down to

But really, the is never going to get a chance to be because if then the Levi-Civita symbol at the beginning of the expression is zero. Thus, we are justified in simply getting rid of that term altogether, and so we now have the expression

Now, given that can you tell me what

is?

If all the indices are throwing you off, just think of

and compute

What do you get? - Mar 9th 2011, 11:01 AMcraig
- Mar 9th 2011, 11:07 AMAckbeet
Your partial derivative of w.r.t. is correct, but your application of it to the expression at hand is incorrect. You need

where I've written the last expression in a (hopefully) suggestive manner. And I think I've answered your question as well.

Do you recognize the expression

Think geometrically here. - Mar 9th 2011, 11:48 AMcraig
Ahh of course it is!!

Right, I don't recognise it exactly, but according to my lecture notes, .

Because of the cyclic property of the Levi-Cevita symbol, and the fact that is a vector constant, this means that it's equal to zero?

Either way, could you explain what the geometric interpretation is haha, could be useful...