# Index question

Question
In index notation, can you have more than two occurances of the same index in the same term? Let me provide and example:

Let's say I have a two index tensor, $M{\alpha \beta}$, and I contract it with itself:

$$M_{\alpha \beta} M^{\alpha \beta}$$

Then let's say I wish to operate on this product with some sort of two index operator, $\chi_{\alpha \beta}$. Is it "legal" to write,

$$\chi_{\alpha \beta}[M_{\alpha \beta} M^{\alpha \beta}]$$

or must I introduce new indices,

$$\chi_{\gamma \sigma}[M_{\alpha \beta} M^{\alpha \beta}]$$

and then use the metric tensor and delta function to clean things up? None of the primers I've read are particularly clear on this point.

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Hmm, I'm not sure that my example makes sense, so I'll give another one.

Let's say I've got some vector $M^{\alpha}$. Then,

$$M_{\alpha} M^{\alpha} = M_{0} M^{0} + \vec{M}\cdot\vec{M} = M_{0} M^{0} + M^{2}$$

Is it "proper" to write:

$$\frac{\partial}{\partial M^{\alpha}} \left( M_{\alpha} M^{\alpha} \right)$$?

If so, what is it? If $M^{\alpha}$ were just plain old variable in one dimension, like $x$, then I would just be thinking:

$$\frac{\partial}{\partial x} x^{2} = \frac{1}{2} x$$

but I'm not exactly sure what $\frac{\partial}{\partial M^{\alpha}}$ is here. From stuff I've read, I'm tempted to write it as,

$$\frac{\partial}{\partial M^{\alpha}} \equiv \partial_{\alpha}$$

where it is simply implied that it only acts on $M$. Then I might write:

$$\partial_{\alpha} \left( M_{\alpha} M^{\alpha} \right) = \left( \partial_{\alpha} M_{\alpha} \right) M^{\alpha} + M_{\alpha} \left( \partial_{\alpha} M^{\alpha} \right)$$

Dick
Homework Helper
No, having an repeated index both upper and lower is the Einstein convention indicating that the index is summed over. After you sum over it there is no grammatical sense to introduce another index with the same name.

No, having an repeated index both upper and lower is the Einstein convention indicating that the index is summed over. After you sum over it there is no grammatical sense to introduce another index with the same name.
Thanks.

(On a side note, I just noticed that a typo in my second post is that:
$$\frac{\partial}{\partial x} x^{2} = \frac{1}{2} x \rightarrow \frac{\partial}{\partial x} x^{2} = 2 x$$
...of course.)

So I suppose the correct way to write it would be:

$$\frac{\partial}{\partial M^{\beta}} (M_{\alpha} M^{\alpha})$$

I'm still not quite sure if it's alright to do this bit:

$$\Rightarrow \partial_{\beta} \left( M_{\alpha} M^{\alpha} \right)$$

which would then give:

$$\partial_{\beta} \left( M_{\alpha} M^{\alpha} \right) = ( \partial_{\beta} M_{\alpha} ) M^{\alpha} + M_{\alpha} ( \partial_{\beta} M^{\alpha})$$

which itself isn't very enlightening, but I'll stare at it a bit more.

Okay, I've answered my question. (But I would appreciate anyone who'd like to look it over and tell me that it's the correct way to go about it.)

I'd like to solve:

$$\frac{\partial}{\partial M^{\beta}} ( M_{\alpha} M^{\alpha} )$$

I first apply the product rule:

$$\left( \frac{\partial}{\partial M^{\beta}} M_{\alpha} \right) M^{\alpha} + M_{\alpha} \left( \frac{\partial}{\partial M^{\beta}} M^{\alpha} \right)$$

Then I use a clever trick that took me forever to realize, but was staring me straight in the face:

$$M^{\alpha} = \delta_{\beta}^{\alpha} M^{\beta} \Rightarrow \partial M^{\alpha} = \delta_{\beta}^{\alpha} \partial M^{\beta} \Rightarrow \frac{\partial}{\partial M^{\beta}} M^{\alpha} = \delta_{\beta}^{\alpha}$$

Using this for the second term, we get:

$$\left( \frac{\partial}{\partial M^{\beta}} M_{\alpha} \right) M^{\alpha} + M_{\alpha} \delta_{\beta}^{\alpha}$$

or

$$\left( \frac{\partial}{\partial M^{\beta}} M_{\alpha} \right) M^{\alpha} + M_{\beta}$$

For the first term, I raise the index using the metric tensor, so that we have:

$$\left( \frac{\partial}{\partial M^{\beta}} g_{\alpha \gamma} M^{\gamma} \right) M^{\alpha} + M_{\beta}$$

Then pulling out the metric tensor, and using the same trick, that is:

$$M^{\gamma} = \delta_{\beta}^{\gamma} M^{\beta} \Rightarrow \frac{\partial}{\partial M^{\beta}} M^{\gamma} = \delta_{\beta}^{\gamma}$$

we can change the first term, and the whole thing is:

$$g_{\alpha \gamma} \delta_{\beta}^{\gamma} M^{\alpha} + M_{\beta}$$

or

$$\delta_{\beta}^{\gamma} M^{\alpha} + M_{\beta}$$

which is

$$M_{\beta} + M_{\beta} = 2 M_{\beta}$$

...the anticipated result!

Dick
Homework Helper
You've got it. What you are doing there is taking the functional derivative with respect to the components of the function M. As opposed to taking the partial derivative with respect to the coordinate x.

Thanks Dick. The math mostly makes sense, but I think I'm still only barely comfortable with the concepts. I saw the functional derivative earlier, but in a different context (where $q$ and $\dot{q}$ were the variables). Let me try to take it a step further to make sure I understand.

(Let me just mention that when I preview my posts, it inserts all of the TeX from the above posts instead of what I've actually written, so this post might just look like jibberish when I submit it.... if not, then just ignore this comment.)

Let's say I have some functions (or functionals?):

$$f_{i} = f (\dot{q} (t) , q (t), t)$$

and that

$$f_{1} = q^{2}$$

then

$$\frac{\partial}{\partial q} f_{1} = 2 q$$

and

$$\frac{\partial}{\partial \dot{q}} f_{1} = 0$$.

But then if,

$$f_{2} = \dot{q}^{2}$$

then

$$\frac{\partial}{\partial q} f_{2} = 0$$

and

$$\frac{\partial}{\partial \dot{q}} f_{1} = 2 \dot{q}$$? ?

Dick
Homework Helper
The TeX is clear enough. Yes, they are 'functionals' (functions of other functions). And, yes, your derivatives look correct. It's the same as in Euler-Lagrange equations.

The TeX is clear enough. Yes, they are 'functionals' (functions of other functions). And, yes, your derivatives look correct. It's the same as in Euler-Lagrange equations.
Excellent, thanks! Then it would appear that

$$\frac{\partial}{\partial (\partial^{\sigma} M^{\beta})} M_{\alpha} M^{\alpha} = 0$$

while

$$\frac{\partial}{\partial (\partial^{\sigma} M^{\beta})} \partial_{\alpha} M^{\alpha} = 1$$