Mixing
linear trasnformation
$begingroup$
In Einstein's "The Meaning of Relativity" https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 by putting $λ = 1$, (2b) and (3a) should furnish the conditions:
$ sum limits _{nu }b_{nu alpha }b_{nu beta }=delta _{alpha beta }$
but If I do myself the passages by replacing (3a) in (2b) I get:
$sum_v (sum_alpha b_{nu alpha } Delta x_{alpha})^2 = sum_nu Delta x_{nu }^{2}$
so that the equality condition should be:
$b_{nualpha} = delta_{nualpha}$
What am I doing wrong?
linear-transformations
asked Jan 14 at 9:05
matt fickmatt fick
263
$endgroup$
add a comment |
$begingroup$
In Einstein's "The Meaning of Relativity" https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 by putting $λ = 1$, (2b) and (3a) should furnish the conditions:
$ sum limits _{nu }b_{nu alpha }b_{nu beta }=delta _{alpha beta }$
but If I do myself the passages by replacing (3a) in (2b) I get:
$sum_v (sum_alpha b_{nu alpha } Delta x_{alpha})^2 = sum_nu Delta x_{nu }^{2}$
so that the equality condition should be:
$b_{nualpha} = delta_{nualpha}$
What am I doing wrong?
linear-transformations
asked Jan 14 at 9:05
matt fickmatt fick
263
$endgroup$
$begingroup$
The sum on $alpha$ should be inside the square
$endgroup$
– Ben
Jan 14 at 9:15
add a comment |
$begingroup$
In Einstein's "The Meaning of Relativity" https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 by putting $λ = 1$, (2b) and (3a) should furnish the conditions:
$ sum limits _{nu }b_{nu alpha }b_{nu beta }=delta _{alpha beta }$
but If I do myself the passages by replacing (3a) in (2b) I get:
$sum_v (sum_alpha b_{nu alpha } Delta x_{alpha})^2 = sum_nu Delta x_{nu }^{2}$
so that the equality condition should be:
$b_{nualpha} = delta_{nualpha}$
What am I doing wrong?
linear-transformations
asked Jan 14 at 9:05
matt fickmatt fick
263
$endgroup$
In Einstein's "The Meaning of Relativity" https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 by putting $λ = 1$, (2b) and (3a) should furnish the conditions:
$ sum limits _{nu }b_{nu alpha }b_{nu beta }=delta _{alpha beta }$
but If I do myself the passages by replacing (3a) in (2b) I get:
$sum_v (sum_alpha b_{nu alpha } Delta x_{alpha})^2 = sum_nu Delta x_{nu }^{2}$
so that the equality condition should be:
$b_{nualpha} = delta_{nualpha}$
What am I doing wrong?
linear-transformations
linear-transformations
asked Jan 14 at 9:05
matt fickmatt fick
263
asked Jan 14 at 9:05
matt fickmatt fick
263
edited Jan 14 at 9:47
matt fick
asked Jan 14 at 9:05
matt fickmatt fick
263
asked Jan 14 at 9:05
matt fickmatt fick
263
asked Jan 14 at 9:05
matt fickmatt fick
263
263
$begingroup$
The sum on $alpha$ should be inside the square
$endgroup$
– Ben
Jan 14 at 9:15
add a comment |
$begingroup$
The sum on $alpha$ should be inside the square
$endgroup$
– Ben
Jan 14 at 9:15
$begingroup$
The sum on $alpha$ should be inside the square
$endgroup$
– Ben
Jan 14 at 9:15
$begingroup$
The sum on $alpha$ should be inside the square
$endgroup$
– Ben
Jan 14 at 9:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Plugging $Delta x_nu' = sum_alpha b_{nualpha}Delta x_alpha$
into $sum_nu Delta x_nu'^2 = sum_nu Delta x_nu^2$ gives
$$sum_nu left(sum_alpha b_{nualpha}Delta x_alpharight)^2 = sum_nu Delta x_nu^2$$
Note that the sum on $alpha$ belongs inside the square. Now expand the square on the left side.
Added:
In general to expand something like this, you do:
$$(sum_alpha c_{alpha})^2 = sum_{alpha,beta} c_{alpha}c_{beta}$$
See if you can get it from there.
answered Jan 14 at 9:21
BenBen
3,861616
$endgroup$
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
add a comment |
Your Answer
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1 Answer
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$begingroup$
Plugging $Delta x_nu' = sum_alpha b_{nualpha}Delta x_alpha$
into $sum_nu Delta x_nu'^2 = sum_nu Delta x_nu^2$ gives
$$sum_nu left(sum_alpha b_{nualpha}Delta x_alpharight)^2 = sum_nu Delta x_nu^2$$
Note that the sum on $alpha$ belongs inside the square. Now expand the square on the left side.
Added:
In general to expand something like this, you do:
$$(sum_alpha c_{alpha})^2 = sum_{alpha,beta} c_{alpha}c_{beta}$$
See if you can get it from there.
answered Jan 14 at 9:21
BenBen
3,861616
$endgroup$
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
add a comment |
$begingroup$
Plugging $Delta x_nu' = sum_alpha b_{nualpha}Delta x_alpha$
into $sum_nu Delta x_nu'^2 = sum_nu Delta x_nu^2$ gives
$$sum_nu left(sum_alpha b_{nualpha}Delta x_alpharight)^2 = sum_nu Delta x_nu^2$$
Note that the sum on $alpha$ belongs inside the square. Now expand the square on the left side.
Added:
In general to expand something like this, you do:
$$(sum_alpha c_{alpha})^2 = sum_{alpha,beta} c_{alpha}c_{beta}$$
See if you can get it from there.
answered Jan 14 at 9:21
BenBen
3,861616
$endgroup$
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
add a comment |
$begingroup$
Plugging $Delta x_nu' = sum_alpha b_{nualpha}Delta x_alpha$
into $sum_nu Delta x_nu'^2 = sum_nu Delta x_nu^2$ gives
$$sum_nu left(sum_alpha b_{nualpha}Delta x_alpharight)^2 = sum_nu Delta x_nu^2$$
Note that the sum on $alpha$ belongs inside the square. Now expand the square on the left side.
Added:
In general to expand something like this, you do:
$$(sum_alpha c_{alpha})^2 = sum_{alpha,beta} c_{alpha}c_{beta}$$
See if you can get it from there.
answered Jan 14 at 9:21
BenBen
3,861616
$endgroup$
Plugging $Delta x_nu' = sum_alpha b_{nualpha}Delta x_alpha$
into $sum_nu Delta x_nu'^2 = sum_nu Delta x_nu^2$ gives
$$sum_nu left(sum_alpha b_{nualpha}Delta x_alpharight)^2 = sum_nu Delta x_nu^2$$
Note that the sum on $alpha$ belongs inside the square. Now expand the square on the left side.
Added:
In general to expand something like this, you do:
$$(sum_alpha c_{alpha})^2 = sum_{alpha,beta} c_{alpha}c_{beta}$$
See if you can get it from there.
answered Jan 14 at 9:21
BenBen
3,861616
edited Jan 15 at 11:37
answered Jan 14 at 9:21
BenBen
3,861616
answered Jan 14 at 9:21
BenBen
3,861616
answered Jan 14 at 9:21
BenBen
3,861616
3,861616
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
add a comment |
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
Can you show me, how to expand the left side?
$endgroup$
– matt fick
Jan 15 at 11:31
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
$begingroup$
@mattfick I added a suggestion, see if you can figure it out from there!
$endgroup$
– Ben
Jan 15 at 11:38
add a comment |
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$begingroup$
The sum on $alpha$ should be inside the square
$endgroup$
– Ben
Jan 14 at 9:15