Summation convention with multiple terms

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  • #1
olgerm
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I got another basic question: should the summation in einstein notation start from first occurance of index or in beginning of equation?
For eampledoes this equation ##R_{\alpha \beta }={R^{\rho }}_{\alpha \rho \beta }=\partial _{\rho }{\Gamma ^{\rho }}_{\beta\alpha }-\partial _{\beta }{\Gamma ^{\rho }}_{\rho \alpha }+{\Gamma ^{\rho }}_{\rho \lambda }{\Gamma ^{\lambda }}_{\beta \alpha }-{\Gamma ^{\rho }}_{\beta \lambda }{\Gamma ^{\lambda }}_{\rho \alpha }## from wikipedia mean:
## \sum_{j_1=0}^D(\sum_{j_2=0}^D(\frac{\partial{\Gamma ^{j_1}}_{\beta \alpha }}{\partial x^{j_1}}-\frac{{\Gamma^{j_1}}_{j_1 \alpha}}{\partial x^{\beta}}+{\Gamma^{j_1}}_{j_1 j_2}{\Gamma ^{j_2}}_{\beta \alpha }-{\Gamma ^{j_1}}_{\beta j_2 }{\Gamma^{j_2}}_{j_1 \alpha }))##
or
##\sum_{j_1=0}^D(\frac{\partial{\Gamma ^{j_1}}_{\beta \alpha }}{\partial x^{j_1}}-\frac{{\Gamma^{j_1}}_{j_1 \alpha}}{\partial x^{\beta}}+\sum_{j_2=0}^D({\Gamma^{j_1}}_{j_1 j_2}{\Gamma ^{j_2}}_{\beta \alpha }-{\Gamma ^{j_1}}_{\beta j_2 }{\Gamma^{j_2}}_{j_1 \alpha }))##?
 
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  • #2
Nugatory
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I got another basic question: should the summation in einstein notation start from first occurance of index or in beginning of equation?
Neither. Each individual term is summed by itself: ##A^iB_i+P^kQ_k## is to be read as##\sum_iA^iB_i+\sum_kP^kQ_k##
 
  • #3
olgerm
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Neither. Each individual term is summed by itself: ##A^iB_i+P^kQ_k## is to be read as##\sum_iA^iB_i+\sum_kP^kQ_k##
so it is same as last one beacuse ##\sum_i(A^iB_i)+\sum_k(P^kQ_k)=\sum_i(A^iB_i+P^iQ_i)##?
 
  • #4
Nugatory
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so it is same as last one beacuse ##\sum_i(A^iB_i)+\sum_k(P^kQ_k)=\sum_i(A^iB_i+P^iQ_i)##?
In this particular case, yes, that just happens to work. In more complex expressions you won't be able to combine terms under one summation the way you're trying to do. Consider, for example, ##A^iB_i+P^{ij}Q_{ij}## - the first term is a four-element summation and the second is a sixteen-element summation.
 
  • #5
olgerm
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Consider, for example, ##A^iB_i+P^{ij}Q_{ij}##
if every term has its own summation it is ##\sum_{i=0}^D(A^iB_i)+\sum_{i=0}^D(\sum_{j=0}^D(P^{ij}Q_{ij}))=\sum_{i=0}^D(A^iB_i+\sum_{j=0}^D(P^{ij}Q_{ij}))##?

Can you say clearly whether it is true that the equation ##R_{\alpha \beta }={R^{\rho }}_{\alpha \rho \beta }=\partial _{\rho }{\Gamma ^{\rho }}_{\beta\alpha }-\partial _{\beta }{\Gamma ^{\rho }}_{\rho \alpha }+{\Gamma ^{\rho }}_{\rho \lambda }{\Gamma ^{\lambda }}_{\beta \alpha }-{\Gamma ^{\rho }}_{\beta \lambda }{\Gamma ^{\lambda }}_{\rho \alpha }## from wikipedia means:
##\sum_{j_1=0}^D(\frac{\partial{\Gamma ^{j_1}}_{\beta \alpha }}{\partial x^{j_1}}-\frac{{\Gamma^{j_1}}_{j_1 \alpha}}{\partial x^{\beta}}+\sum_{j_2=0}^D({\Gamma^{j_1}}_{j_1 j_2}{\Gamma ^{j_2}}_{\beta \alpha }-{\Gamma ^{j_1}}_{\beta j_2 }{\Gamma^{j_2}}_{j_1 \alpha }))##?
 
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  • #6
Nugatory
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Can you say clearly whether it is true that the equation ##R_{\alpha \beta }={R^{\rho }}_{\alpha \rho \beta }=\partial _{\rho }{\Gamma ^{\rho }}_{\beta\alpha }-\partial _{\beta }{\Gamma ^{\rho }}_{\rho \alpha }+{\Gamma ^{\rho }}_{\rho \lambda }{\Gamma ^{\lambda }}_{\beta \alpha }-{\Gamma ^{\rho }}_{\beta \lambda }{\Gamma ^{\lambda }}_{\rho \alpha }## from wikipedia means:
##\sum_{j_1=0}^D(\frac{\partial{\Gamma ^{j_1}}_{\beta \alpha }}{\partial x^{j_1}}-\frac{{\Gamma^{j_1}}_{j_1 \alpha}}{\partial x^{\beta}}+\sum_{j_2=0}^D({\Gamma^{j_1}}_{j_1 j_2}{\Gamma ^{j_2}}_{\beta \alpha }-{\Gamma ^{j_1}}_{\beta j_2 }{\Gamma^{j_2}}_{j_1 \alpha }))##?
I can say clearly that it means$$R_{\alpha \beta }={R^{\rho }}_{\alpha \rho \beta }=\sum_\rho\partial _{\rho }{\Gamma ^{\rho }}_{\beta\alpha }-\sum_\rho\partial _{\beta }{\Gamma ^{\rho }}_{\rho \alpha }+\sum_{\rho,\lambda}{\Gamma ^{\rho }}_{\rho \lambda }{\Gamma ^{\lambda }}_{\beta \alpha }-\sum_{\lambda,\rho}{\Gamma ^{\rho }}_{\beta \lambda }{\Gamma ^{\lambda }}_{\rho \alpha }$$I haven't checked your algebra, but it is plausible that you've found a valid way of obscuring manipulating the formula.
 
  • #7
olgerm
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obscuring
more easily readable for me. waiting foranswer whether it's correct or not.
 
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  • #8
PeterDonis
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more easily readable for me. waiting foranswer whether it's correct or not.
If you insist on writing things differently from the way everybody else writes them, you can't expect everybody else to check your work. We have standard ways of writing things in physics for a reason.

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