Order of indices using index notation
I'm trying to solve a problem in Lagrangian mechanics involving index notation. I'm wondering if the expression
$$ A_{ij}dot{q^i}q^j = A_{ji}dot{q^j}q^i $$
is true. Our professor skipped over this notation convention and I think the problem is made much easier if it is true.
Thanks in advance for any help.
classical-mechanics index-notation
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
add a comment |
I'm trying to solve a problem in Lagrangian mechanics involving index notation. I'm wondering if the expression
$$ A_{ij}dot{q^i}q^j = A_{ji}dot{q^j}q^i $$
is true. Our professor skipped over this notation convention and I think the problem is made much easier if it is true.
Thanks in advance for any help.
classical-mechanics index-notation
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
add a comment |
I'm trying to solve a problem in Lagrangian mechanics involving index notation. I'm wondering if the expression
$$ A_{ij}dot{q^i}q^j = A_{ji}dot{q^j}q^i $$
is true. Our professor skipped over this notation convention and I think the problem is made much easier if it is true.
Thanks in advance for any help.
classical-mechanics index-notation
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
I'm trying to solve a problem in Lagrangian mechanics involving index notation. I'm wondering if the expression
$$ A_{ij}dot{q^i}q^j = A_{ji}dot{q^j}q^i $$
is true. Our professor skipped over this notation convention and I think the problem is made much easier if it is true.
Thanks in advance for any help.
classical-mechanics index-notation
classical-mechanics index-notation
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
asked Nov 29 at 15:12
Aaron Fitzpatrick
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If you have trouble thinking about this sort of thing, take it in stages:
$$A_{ij}dot{q}^iq^j=A_{kl}dot{q}^kq^l=A_{ji}dot{q}^jq^i,$$where at the first equals I impose $i,,jmapsto k,,l$ while at the second I impose $k,,lmapsto j,,i$.
answered Nov 29 at 15:15
J.G.
22.3k22034
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
add a comment |
Yes, it is the same. Indices don't carry any meaning, they are just labels. As an example
$$
sum_i a_i b^i = sum_p a_p b^p
$$
or in Einstein's notation
$$
a_i b^i = a_p b^p
$$
So, coming back to your example
$$
A_{ij}dot{q}^i q^j = A_{m n}dot{q}^m q^n = A_{j i}dot{q}^j q^i
$$
answered Nov 29 at 15:16
caverac
13.1k21029
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you have trouble thinking about this sort of thing, take it in stages:
$$A_{ij}dot{q}^iq^j=A_{kl}dot{q}^kq^l=A_{ji}dot{q}^jq^i,$$where at the first equals I impose $i,,jmapsto k,,l$ while at the second I impose $k,,lmapsto j,,i$.
answered Nov 29 at 15:15
J.G.
22.3k22034
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
add a comment |
If you have trouble thinking about this sort of thing, take it in stages:
$$A_{ij}dot{q}^iq^j=A_{kl}dot{q}^kq^l=A_{ji}dot{q}^jq^i,$$where at the first equals I impose $i,,jmapsto k,,l$ while at the second I impose $k,,lmapsto j,,i$.
answered Nov 29 at 15:15
J.G.
22.3k22034
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
add a comment |
If you have trouble thinking about this sort of thing, take it in stages:
$$A_{ij}dot{q}^iq^j=A_{kl}dot{q}^kq^l=A_{ji}dot{q}^jq^i,$$where at the first equals I impose $i,,jmapsto k,,l$ while at the second I impose $k,,lmapsto j,,i$.
answered Nov 29 at 15:15
J.G.
22.3k22034
If you have trouble thinking about this sort of thing, take it in stages:
$$A_{ij}dot{q}^iq^j=A_{kl}dot{q}^kq^l=A_{ji}dot{q}^jq^i,$$where at the first equals I impose $i,,jmapsto k,,l$ while at the second I impose $k,,lmapsto j,,i$.
answered Nov 29 at 15:15
J.G.
22.3k22034
answered Nov 29 at 15:15
J.G.
22.3k22034
answered Nov 29 at 15:15
J.G.
22.3k22034
answered Nov 29 at 15:15
J.G.
22.3k22034
22.3k22034
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
add a comment |
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
Cool thank you! Now that you say it it seems nearly obvious. I think the fact that it's assumed we should already be familiar with the convention despite having never seen it before has caught me off-guard.
– Aaron Fitzpatrick
Nov 29 at 15:18
add a comment |
Yes, it is the same. Indices don't carry any meaning, they are just labels. As an example
$$
sum_i a_i b^i = sum_p a_p b^p
$$
or in Einstein's notation
$$
a_i b^i = a_p b^p
$$
So, coming back to your example
$$
A_{ij}dot{q}^i q^j = A_{m n}dot{q}^m q^n = A_{j i}dot{q}^j q^i
$$
answered Nov 29 at 15:16
caverac
13.1k21029
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
add a comment |
Yes, it is the same. Indices don't carry any meaning, they are just labels. As an example
$$
sum_i a_i b^i = sum_p a_p b^p
$$
or in Einstein's notation
$$
a_i b^i = a_p b^p
$$
So, coming back to your example
$$
A_{ij}dot{q}^i q^j = A_{m n}dot{q}^m q^n = A_{j i}dot{q}^j q^i
$$
answered Nov 29 at 15:16
caverac
13.1k21029
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
add a comment |
Yes, it is the same. Indices don't carry any meaning, they are just labels. As an example
$$
sum_i a_i b^i = sum_p a_p b^p
$$
or in Einstein's notation
$$
a_i b^i = a_p b^p
$$
So, coming back to your example
$$
A_{ij}dot{q}^i q^j = A_{m n}dot{q}^m q^n = A_{j i}dot{q}^j q^i
$$
answered Nov 29 at 15:16
caverac
13.1k21029
Yes, it is the same. Indices don't carry any meaning, they are just labels. As an example
$$
sum_i a_i b^i = sum_p a_p b^p
$$
or in Einstein's notation
$$
a_i b^i = a_p b^p
$$
So, coming back to your example
$$
A_{ij}dot{q}^i q^j = A_{m n}dot{q}^m q^n = A_{j i}dot{q}^j q^i
$$
answered Nov 29 at 15:16
caverac
13.1k21029
answered Nov 29 at 15:16
caverac
13.1k21029
answered Nov 29 at 15:16
caverac
13.1k21029
answered Nov 29 at 15:16
caverac
13.1k21029
13.1k21029
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
add a comment |
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
Thank you for the help!
– Aaron Fitzpatrick
Nov 29 at 15:20
add a comment |
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