Is this tensor operation valid?

  • Thread starter arroy_0205
  • Start date
Can anybody help me clear my following doubt?
Suppose, I have a relation of the form
[tex]
f(p_{\mu},q_{\mu})=0
[/tex]
Then can I multiply the both sides by [tex]p^{\mu}[/tex] and then contract?
[tex]
p^{\mu}f(p_{\mu},q_{\mu})=0
[/tex]
After this I want to use the identity [tex]p^{\mu}p_{\mu}=m^2[/tex] 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

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Homework Helper
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Can anybody help me clear my following doubt?
Suppose, I have a relation of the form
[tex]
f(p_{\mu},q_{\mu})=0
[/tex]
Then can I multiply the both sides by [tex]p^{\mu}[/tex] and then contract?
[tex]
p^{\mu}f(p_{\mu},q_{\mu})=0
[/tex]
After this I want to use the identity [tex]p^{\mu}p_{\mu}=m^2[/tex] 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! :smile:

It would have to be fµ, not just f:

[tex]f_{\mu}(p_{\nu},q_{\nu})=0[/tex]

Then you can contract:

[tex]p^{\mu}f_{\mu}(p_{\nu},q_{\nu})=0[/tex]

Were you thinking of a particular function f? :smile:
 
Hi tiny-tim,
Thanks, and yes you are right it should be [tex]f_{\mu}[/tex]. In fact I was trying to consider a general form like say
[tex]
a_1p_{\mu}+a_2q_{\mu}=0
[/tex]
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

Science Advisor
Homework Helper
25,790
242
[tex]a_1p_{\mu}+a_2q_{\mu}=0[/tex]
Hi arroy_0205! :smile:

Then, multiplying both sides by [itex]p^{\mu}[/itex] :

[tex]p^{\mu}(a_1p_{\mu}+a_2q_{\mu})\,=\,0[/tex]

so

[tex]a_1p^{\mu}p_{\mu}\,+\,a_2p^{\mu}q_{\mu}\,=\,0[/tex]

or

[tex]a_1m^2\,+\,a_2p\cdot q\,=\,0[/tex] :smile:
 
That is true, however by multiplying with [tex]p^{\mu}[/tex] 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

Science Advisor
Homework Helper
25,790
242
No, it's fine! :smile:

It is like ordinary multiplication by a function …

it's four ordinary multiplications, added together …

which is perfectly valid! :biggrin:
 
161
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That is true, however by multiplying with [tex]p^{\mu}[/tex] 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.
 

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