# Is this tensor operation valid?

1. Jun 20, 2008

### arroy_0205

Can anybody help me clear my following doubt?
Suppose, I have a relation of the form
$$f(p_{\mu},q_{\mu})=0$$
Then can I multiply the both sides by $$p^{\mu}$$ and then contract?
$$p^{\mu}f(p_{\mu},q_{\mu})=0$$
After this I want to use the identity $$p^{\mu}p_{\mu}=m^2$$ as known in special relativity. So I am first multiplying both sides by a contravariant tensor and then using the summation convention. The question is am I doing it right? I feel this is a valid operation and I have tested one simple example where this is valid but I am not able to prove it in general. Can anybody help? You may refer to some books or website also where I can look up.

2. Jun 20, 2008

### tiny-tim

Hi arroy_0205!

It would have to be fµ, not just f:

$$f_{\mu}(p_{\nu},q_{\nu})=0$$

Then you can contract:

$$p^{\mu}f_{\mu}(p_{\nu},q_{\nu})=0$$

Were you thinking of a particular function f?

3. Jun 20, 2008

### arroy_0205

Hi tiny-tim,
Thanks, and yes you are right it should be $$f_{\mu}$$. In fact I was trying to consider a general form like say
$$a_1p_{\mu}+a_2q_{\mu}=0$$
there is no constant term in the left hand side. But the proof is still not clear to me. This is sort of obvious but without a solid proof I hesitate to accept. May be I am overlooking something obvious and trivial regarding its proof.

4. Jun 20, 2008

### tiny-tim

Hi arroy_0205!

Then, multiplying both sides by $p^{\mu}$ :

$$p^{\mu}(a_1p_{\mu}+a_2q_{\mu})\,=\,0$$

so

$$a_1p^{\mu}p_{\mu}\,+\,a_2p^{\mu}q_{\mu}\,=\,0$$

or

$$a_1m^2\,+\,a_2p\cdot q\,=\,0$$

5. Jun 20, 2008

### arroy_0205

That is true, however by multiplying with $$p^{\mu}$$ I am also introducing summation, this is not like ordinary multiplication by a function but multiplication followed by summation. My doubt is there, whether this kind of operation is appropriate.

6. Jun 20, 2008

### tiny-tim

No, it's fine!

It is like ordinary multiplication by a function …

it's four ordinary multiplications, added together …

which is perfectly valid!

7. Jun 20, 2008

### ismaili

Yes. Of course it is valid.
Just think of this operation component-wise.