Deriving parallel transport equations
$begingroup$
From Wald's General Relativity page 37,
It is easiest to compute the change in $v^a$ when we return to $p$ by letting $omega_a$ be an arbitray dual vector field and finding the change in the scalar $v^aomega_a$ as we traverse the loop.
1) Where is the vector $v^a$ in the picture above?
2) Why do they consider $omega_a$?
For small $Delta t$ the change, $delta_1$, in $v^aomega_a$ on the first leg of the path,
$$delta_1 = Delta t frac{partial}{partial t}(v^aomega_a)|_{(Delta t/2,0)}$$
where, by evaluating the derivative at the midpoint, we have ensured that this expression is accurate to second order in $Delta t$. We may rewrite $delta_1$ as
$$ delta_1 = Delta t T^b triangledown_b (v^aomega_a)|_{(Delta t/2,0)}$$
3) How can we introduce the $T^b triangledown_b$?? I do not see where this came from?
geometry differential-geometry vector-spaces tensors general-relativity
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
$endgroup$
add a comment |
$begingroup$
From Wald's General Relativity page 37,
It is easiest to compute the change in $v^a$ when we return to $p$ by letting $omega_a$ be an arbitray dual vector field and finding the change in the scalar $v^aomega_a$ as we traverse the loop.
1) Where is the vector $v^a$ in the picture above?
2) Why do they consider $omega_a$?
For small $Delta t$ the change, $delta_1$, in $v^aomega_a$ on the first leg of the path,
$$delta_1 = Delta t frac{partial}{partial t}(v^aomega_a)|_{(Delta t/2,0)}$$
where, by evaluating the derivative at the midpoint, we have ensured that this expression is accurate to second order in $Delta t$. We may rewrite $delta_1$ as
$$ delta_1 = Delta t T^b triangledown_b (v^aomega_a)|_{(Delta t/2,0)}$$
3) How can we introduce the $T^b triangledown_b$?? I do not see where this came from?
geometry differential-geometry vector-spaces tensors general-relativity
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
$endgroup$
add a comment |
$begingroup$
From Wald's General Relativity page 37,
It is easiest to compute the change in $v^a$ when we return to $p$ by letting $omega_a$ be an arbitray dual vector field and finding the change in the scalar $v^aomega_a$ as we traverse the loop.
1) Where is the vector $v^a$ in the picture above?
2) Why do they consider $omega_a$?
For small $Delta t$ the change, $delta_1$, in $v^aomega_a$ on the first leg of the path,
$$delta_1 = Delta t frac{partial}{partial t}(v^aomega_a)|_{(Delta t/2,0)}$$
where, by evaluating the derivative at the midpoint, we have ensured that this expression is accurate to second order in $Delta t$. We may rewrite $delta_1$ as
$$ delta_1 = Delta t T^b triangledown_b (v^aomega_a)|_{(Delta t/2,0)}$$
3) How can we introduce the $T^b triangledown_b$?? I do not see where this came from?
geometry differential-geometry vector-spaces tensors general-relativity
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
$endgroup$
From Wald's General Relativity page 37,
It is easiest to compute the change in $v^a$ when we return to $p$ by letting $omega_a$ be an arbitray dual vector field and finding the change in the scalar $v^aomega_a$ as we traverse the loop.
1) Where is the vector $v^a$ in the picture above?
2) Why do they consider $omega_a$?
For small $Delta t$ the change, $delta_1$, in $v^aomega_a$ on the first leg of the path,
$$delta_1 = Delta t frac{partial}{partial t}(v^aomega_a)|_{(Delta t/2,0)}$$
where, by evaluating the derivative at the midpoint, we have ensured that this expression is accurate to second order in $Delta t$. We may rewrite $delta_1$ as
$$ delta_1 = Delta t T^b triangledown_b (v^aomega_a)|_{(Delta t/2,0)}$$
3) How can we introduce the $T^b triangledown_b$?? I do not see where this came from?
geometry differential-geometry vector-spaces tensors general-relativity
geometry differential-geometry vector-spaces tensors general-relativity
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
edited Dec 3 '18 at 21:15
Permian
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
asked Dec 3 '18 at 20:12
PermianPermian
2,2061135
2,2061135
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- The vector $v^a$ is just an arbitrary vector at $p$
It is just a convenient step, you will see how it pays off later in the development. At this point, just note that by introducing $omega_a$, we can build the quantity $v^aomega_a$, which is a scalar, and easier to work with: one less index to worry about
The chain rule
$$
frac{partial }{partial t} = left(frac{partial x^b(t, s)}{partial t}right)_sfrac{partial}{partial x^b} = T^b partial_b
$$
where
$$
T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s
$$
is the tangent to the curves of constant $s$
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
$endgroup$
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
|
show 1 more comment
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- The vector $v^a$ is just an arbitrary vector at $p$
It is just a convenient step, you will see how it pays off later in the development. At this point, just note that by introducing $omega_a$, we can build the quantity $v^aomega_a$, which is a scalar, and easier to work with: one less index to worry about
The chain rule
$$
frac{partial }{partial t} = left(frac{partial x^b(t, s)}{partial t}right)_sfrac{partial}{partial x^b} = T^b partial_b
$$
where
$$
T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s
$$
is the tangent to the curves of constant $s$
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
$endgroup$
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
|
show 1 more comment
$begingroup$
- The vector $v^a$ is just an arbitrary vector at $p$
It is just a convenient step, you will see how it pays off later in the development. At this point, just note that by introducing $omega_a$, we can build the quantity $v^aomega_a$, which is a scalar, and easier to work with: one less index to worry about
The chain rule
$$
frac{partial }{partial t} = left(frac{partial x^b(t, s)}{partial t}right)_sfrac{partial}{partial x^b} = T^b partial_b
$$
where
$$
T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s
$$
is the tangent to the curves of constant $s$
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
$endgroup$
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
|
show 1 more comment
$begingroup$
- The vector $v^a$ is just an arbitrary vector at $p$
It is just a convenient step, you will see how it pays off later in the development. At this point, just note that by introducing $omega_a$, we can build the quantity $v^aomega_a$, which is a scalar, and easier to work with: one less index to worry about
The chain rule
$$
frac{partial }{partial t} = left(frac{partial x^b(t, s)}{partial t}right)_sfrac{partial}{partial x^b} = T^b partial_b
$$
where
$$
T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s
$$
is the tangent to the curves of constant $s$
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
$endgroup$
- The vector $v^a$ is just an arbitrary vector at $p$
It is just a convenient step, you will see how it pays off later in the development. At this point, just note that by introducing $omega_a$, we can build the quantity $v^aomega_a$, which is a scalar, and easier to work with: one less index to worry about
The chain rule
$$
frac{partial }{partial t} = left(frac{partial x^b(t, s)}{partial t}right)_sfrac{partial}{partial x^b} = T^b partial_b
$$
where
$$
T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s
$$
is the tangent to the curves of constant $s$
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
answered Dec 5 '18 at 14:06
caveraccaverac
14.5k31130
14.5k31130
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
|
show 1 more comment
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
3) I still cant see how to apply the chain rule
$endgroup$
– Permian
Dec 5 '18 at 14:13
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
@Permian Forget the arguments $$ frac{partial f}{partial t} = frac{partial x}{partial t}frac{partial f}{partial x} $$
$endgroup$
– caverac
Dec 5 '18 at 14:14
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
I dont uderstand how you can just say $T^b stackrel{.}{=} left(frac{partial x^b(t, s)}{partial t}right)_s$?. I understand the reasoning for the RHS but I cannot see how we can call this $T$ and say its just a tensor
$endgroup$
– Permian
Dec 7 '18 at 20:54
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
@Permian I never said it was a tensor, you can call it whatever convenient way you prefer
$endgroup$
– caverac
Dec 7 '18 at 21:02
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
$begingroup$
Ill give it to you anyway
$endgroup$
– Permian
Dec 7 '18 at 22:01
|
show 1 more comment
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