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4

0

$begingroup$

To my knowledge, 3 or more indices may not appear in a given term, as I've found in a video produced by "Faculty of Khan":

However, on page 30, Sean Carroll writes:

As obvious, the indices 0 and 1 are repeated 3 times in the RHS of the underlined equation.

I am aware that Carroll keeps an errata for Spacetime and Geometry but I was not able to find information regarding this, so I'm not sure if there is some implied meaning that I'm missing.

tensor-calculus notation

share|cite|improve this question

edited Jul 14 at 3:59

Qmechanic♦

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asked Jul 14 at 3:32

Ken WangKen Wang

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$endgroup$

add a comment | 

4

0

$begingroup$

To my knowledge, 3 or more indices may not appear in a given term, as I've found in a video produced by "Faculty of Khan":

However, on page 30, Sean Carroll writes:

As obvious, the indices 0 and 1 are repeated 3 times in the RHS of the underlined equation.

I am aware that Carroll keeps an errata for Spacetime and Geometry but I was not able to find information regarding this, so I'm not sure if there is some implied meaning that I'm missing.

tensor-calculus notation

share|cite|improve this question

edited Jul 14 at 3:59

Qmechanic♦

111k1212 gold badges214214 silver badges13151315 bronze badges

asked Jul 14 at 3:32

Ken WangKen Wang

10577 bronze badges

$endgroup$

add a comment | 

$begingroup$

To my knowledge, 3 or more indices may not appear in a given term, as I've found in a video produced by "Faculty of Khan":

However, on page 30, Sean Carroll writes:

As obvious, the indices 0 and 1 are repeated 3 times in the RHS of the underlined equation.

I am aware that Carroll keeps an errata for Spacetime and Geometry but I was not able to find information regarding this, so I'm not sure if there is some implied meaning that I'm missing.

tensor-calculus notation

share|cite|improve this question

edited Jul 14 at 3:59

Qmechanic♦

111k1212 gold badges214214 silver badges13151315 bronze badges

asked Jul 14 at 3:32

Ken WangKen Wang

10577 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 14 at 3:59

Qmechanic♦

111k1212 gold badges214214 silver badges13151315 bronze badges

asked Jul 14 at 3:32

Ken WangKen Wang

10577 bronze badges

share|cite|improve this question

edited Jul 14 at 3:59

Qmechanic♦

111k1212 gold badges214214 silver badges13151315 bronze badges

asked Jul 14 at 3:32

Ken WangKen Wang

10577 bronze badges

edited Jul 14 at 3:59

Qmechanic♦

111k1212 gold badges214214 silver badges13151315 bronze badges

edited Jul 14 at 3:59

Qmechanic♦

111k1212 gold badges214214 silver badges13151315 bronze badges

asked Jul 14 at 3:32

Ken WangKen Wang

10577 bronze badges

asked Jul 14 at 3:32

Ken WangKen Wang

10577 bronze badges

1 Answer
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10

$begingroup$

There is nothing wrong with what Carroll wrote, which is why it is not in his errata.

Khan is talking about symbolic contracted indices, which must occur in pairs. You sum over their possible values (typically 0, 1, 2, and 3). In Carroll’s equations the indices already have explicit values and are not being contracted. It doesn’t make sense to “contract over 0 and 1” because you can’t assign values to them.

Carroll’s equation follows from a correct double contraction,

$$F^munu=eta^mualphaeta^nubetaF_alphabeta,$$

when you set $mu=0$ and $nu=1$, write out the 16-term double sum over the contracted indices $alpha$ and $beta$, and use the fact that all the off-diagonal elements of the Minkowski metric and its inverse vanish. After doing this a few times, you can do it in your head.

It would be instructive to understand why contracting a tensor over a pair of indices produces another tensor, but that is beyond the scope of this particular question.

share|cite|improve this answer

edited Jul 14 at 4:09

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer sir! I did indeed misinterpret 0 as a varying index, rather than referring to a specific entry. With the method that you outlined I was indeed able to verify the equality. (I can't wait to be able to do this in my head as writing out the terms took almost an hour)
  $endgroup$
  – Ken Wang
  Jul 14 at 5:10
  
  

  
  
  
  

add a comment | 

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10

$begingroup$

There is nothing wrong with what Carroll wrote, which is why it is not in his errata.

Khan is talking about symbolic contracted indices, which must occur in pairs. You sum over their possible values (typically 0, 1, 2, and 3). In Carroll’s equations the indices already have explicit values and are not being contracted. It doesn’t make sense to “contract over 0 and 1” because you can’t assign values to them.

Carroll’s equation follows from a correct double contraction,

$$F^munu=eta^mualphaeta^nubetaF_alphabeta,$$

when you set $mu=0$ and $nu=1$, write out the 16-term double sum over the contracted indices $alpha$ and $beta$, and use the fact that all the off-diagonal elements of the Minkowski metric and its inverse vanish. After doing this a few times, you can do it in your head.

It would be instructive to understand why contracting a tensor over a pair of indices produces another tensor, but that is beyond the scope of this particular question.

share|cite|improve this answer

edited Jul 14 at 4:09

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer sir! I did indeed misinterpret 0 as a varying index, rather than referring to a specific entry. With the method that you outlined I was indeed able to verify the equality. (I can't wait to be able to do this in my head as writing out the terms took almost an hour)
  $endgroup$
  – Ken Wang
  Jul 14 at 5:10
  
  

  
  
  
  

add a comment | 

10

$begingroup$

There is nothing wrong with what Carroll wrote, which is why it is not in his errata.

Khan is talking about symbolic contracted indices, which must occur in pairs. You sum over their possible values (typically 0, 1, 2, and 3). In Carroll’s equations the indices already have explicit values and are not being contracted. It doesn’t make sense to “contract over 0 and 1” because you can’t assign values to them.

Carroll’s equation follows from a correct double contraction,

$$F^munu=eta^mualphaeta^nubetaF_alphabeta,$$

when you set $mu=0$ and $nu=1$, write out the 16-term double sum over the contracted indices $alpha$ and $beta$, and use the fact that all the off-diagonal elements of the Minkowski metric and its inverse vanish. After doing this a few times, you can do it in your head.

It would be instructive to understand why contracting a tensor over a pair of indices produces another tensor, but that is beyond the scope of this particular question.

share|cite|improve this answer

edited Jul 14 at 4:09

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer sir! I did indeed misinterpret 0 as a varying index, rather than referring to a specific entry. With the method that you outlined I was indeed able to verify the equality. (I can't wait to be able to do this in my head as writing out the terms took almost an hour)
  $endgroup$
  – Ken Wang
  Jul 14 at 5:10
  
  

  
  
  
  

add a comment | 

$begingroup$

There is nothing wrong with what Carroll wrote, which is why it is not in his errata.

Khan is talking about symbolic contracted indices, which must occur in pairs. You sum over their possible values (typically 0, 1, 2, and 3). In Carroll’s equations the indices already have explicit values and are not being contracted. It doesn’t make sense to “contract over 0 and 1” because you can’t assign values to them.

Carroll’s equation follows from a correct double contraction,

$$F^munu=eta^mualphaeta^nubetaF_alphabeta,$$

when you set $mu=0$ and $nu=1$, write out the 16-term double sum over the contracted indices $alpha$ and $beta$, and use the fact that all the off-diagonal elements of the Minkowski metric and its inverse vanish. After doing this a few times, you can do it in your head.

It would be instructive to understand why contracting a tensor over a pair of indices produces another tensor, but that is beyond the scope of this particular question.

share|cite|improve this answer

edited Jul 14 at 4:09

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges

$endgroup$

share|cite|improve this answer

edited Jul 14 at 4:09

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges

answered Jul 14 at 3:43

G. SmithG. Smith

18.6k11 gold badge3535 silver badges6565 bronze badges