Is this tensor operation valid?

[arroy_0205]

Can anybody help me clear my following doubt?
Suppose, I have a relation of the form
<br /> f(p_{\mu},q_{\mu})=0<br />
Then can I multiply the both sides by p^{\mu} and then contract?
<br /> p^{\mu}f(p_{\mu},q_{\mu})=0<br />
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.

 

[tiny-tim]

Can anybody help me clear my following doubt?
Suppose, I have a relation of the form
<br /> f(p_{\mu},q_{\mu})=0<br />
Then can I multiply the both sides by p^{\mu} and then contract?
<br /> p^{\mu}f(p_{\mu},q_{\mu})=0<br />
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.

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?

 

[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
<br /> a_1p_{\mu}+a_2q_{\mu}=0<br />
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.

 

[tiny-tim]

a_1p_{\mu}+a_2q_{\mu}=0

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

 

[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.

 

[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!

 

[ismaili]

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.

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