Contravariant Four-gradient ESN in Wikipedia appears wrong

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SUMMARY

The discussion centers on the contravariant four-gradient representation in relativity as presented in Wikipedia. Participants assert that the expression E0∂0 - E1∂1 - E2∂2 - E3∂3 = Eα∂α contains incorrect negative signs. They clarify that the correct form should be Eα∂α = E0∂0 + E1∂1 + E2∂2 + E3∂3, emphasizing the proper application of Einstein summation notation without negative signs in the initial terms.

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  • Understanding of Einstein summation notation
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  • Basic knowledge of partial derivatives in the context of relativity
  • Concept of four-gradient in physics
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SpecialEd
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Homework Statement


I am self studying relativity. In Wikipedia under the four-gradient section, the contravariant four-vector looks wrong from an Einstein summation notation point of view.

https://en.wikipedia.org/wiki/Four-vector

Homework Equations



It states:

E
00-E11-E22-E33 = Eαα

The Attempt at a Solution


As it is it looks wrong to me. Is it wrong or is it NOT ESN or something else?
 
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Welcome to PF!
I agree with you that it doesn't look right. For the contravariant form of the 4-gradient they have
upload_2016-10-11_22-11-0.png

I don't think there should be any negative signs in the first four lines. The last four lines look correct to me. So, I believe the Einstein summation works as follows $$\partial = E_\alpha \partial^\alpha = E_0 \partial ^0 + E_1 \partial ^1 +E_2 \partial ^2 + E_3 \partial ^3$$ where $$\partial^1 \equiv \frac{\partial}{\partial x_ {_1}} = -\frac{\partial}{\partial x^1}$$ etc.
 
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TSny said:
Welcome to PF!
I agree with you that it doesn't look right. For the contravariant form of the 4-gradient they have
View attachment 107315
I don't think there should be any negative signs in the first four lines. The last four lines look correct to me. So, I believe the Einstein summation works as follows $$\partial = E_\alpha \partial^\alpha = E_0 \partial ^0 + E_1 \partial ^1 +E_2 \partial ^2 + E_3 \partial ^3$$ where $$\partial^1 \equiv \frac{\partial}{\partial x_ {_1}} = -\frac{\partial}{\partial x^1}$$ etc.

Thanks so much for your comment and help!
 

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