The question: The relativistic generalisation of the ▾ symbol is the four-derivative a aμ = == მუს 1 a cǝt' (4) where the upper index in the numerator of the derivative has become a lower index in . I.e., the partial derivative with respect to a contravariant coordinate, itself transforms as a covariant 4-vector. We will now go through the steps to show that, and then investigate a few expressions involving four-derivatives. g) Find expressions for the components of "f, (x) for a four-vector-valued function f". My answer: g) and fu(x) диам= бри S. and fulas st+f+f да tofa = (c² 30² ² + √² ) ( f° + f³) o-component== = de f° + ð²f° + i-component> ½ f² + 2 Fi The answer: gl Express and" fu(x). The contraction here is between derivatives, so this is really the two a d² operator acting on each component of f separately So, amam fulxl is 0-component: (covariant) 4-vector with: 202* fol+) + 2; 2² f(x) i-component = - - ·½ de fo(x) ¯♡² fo(x) ¿½ filx) ²fi(x) + < either of these کے would be a good answer.
The question: The relativistic generalisation of the ▾ symbol is the four-derivative a aμ = == მუს 1 a cǝt' (4) where the upper index in the numerator of the derivative has become a lower index in . I.e., the partial derivative with respect to a contravariant coordinate, itself transforms as a covariant 4-vector. We will now go through the steps to show that, and then investigate a few expressions involving four-derivatives. g) Find expressions for the components of "f, (x) for a four-vector-valued function f". My answer: g) and fu(x) диам= бри S. and fulas st+f+f да tofa = (c² 30² ² + √² ) ( f° + f³) o-component== = de f° + ð²f° + i-component> ½ f² + 2 Fi The answer: gl Express and" fu(x). The contraction here is between derivatives, so this is really the two a d² operator acting on each component of f separately So, amam fulxl is 0-component: (covariant) 4-vector with: 202* fol+) + 2; 2² f(x) i-component = - - ·½ de fo(x) ¯♡² fo(x) ¿½ filx) ²fi(x) + < either of these کے would be a good answer.
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Transcribed Image Text:The question:
The relativistic generalisation of the ▾ symbol is the four-derivative
a
aμ
=
==
მუს
1 a
cǝt'
(4)
where the upper index in the numerator of the derivative has become a lower index in .
I.e., the partial derivative with respect to a contravariant coordinate, itself transforms as a
covariant 4-vector. We will now go through the steps to show that, and then investigate a
few expressions involving four-derivatives.
g) Find expressions for the components of "f, (x) for a four-vector-valued function f".
My answer:
g) and fu(x)
диам= бри
S.
and fulas st+f+f
да
tofa
= (c² 30² ² + √² ) ( f° + f³)
o-component== = de f° + ð²f°
+
i-component>
½ f² +
2 Fi
The answer:
gl Express and" fu(x).
The contraction here is between
derivatives, so
this is really
the two
a d² operator
acting on each component of f separately
So, amam fulxl is
0-component:
(covariant) 4-vector with:
202* fol+) + 2; 2² f(x)
i-component
=
-
-
·½ de fo(x) ¯♡² fo(x)
¿½ filx) ²fi(x)
+
< either of these
کے
would be
a good
answer.
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