Angular momentum in different points
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
asked Dec 27 at 21:12
Frogfire
222
add a comment |
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
asked Dec 27 at 21:12
Frogfire
222
add a comment |
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
asked Dec 27 at 21:12
Frogfire
222
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
asked Dec 27 at 21:12
Frogfire
222
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
asked Dec 27 at 21:12
Frogfire
222
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
edited Dec 28 at 12:55
Qmechanic♦
101k121831150
101k121831150
asked Dec 27 at 21:12
Frogfire
222
asked Dec 27 at 21:12
Frogfire
222
asked Dec 27 at 21:12
Frogfire
222
222
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
|
show 7 more comments
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
add a comment |
Angular momentum relative to an origin ${mathcal O_1}$
$$ mathbf{L_{mathcal O_1}} = mathbf{r_{mathcal O_1} times p_{mathcal O_1}}$$
where $mathbf r_{mathcal O_1}$ is the position vector to the particle relative to some origin ${mathcal O_1}$.
Now suppose that angular momentum is conserved in ${mathcal O_1}$. Then
$$ frac{d mathbf L_1}{dt} = mathbf{dot{r_1} times p_1} + mathbf{r_1 times dot{p_1}} = frac{1}{m} mathbf{p_1 times p_1} + mathbf{r_1 times dot{p_1}} =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbf{p_1} = mathbf{p}$). It then follows that
$$ mathbf{r_1 times F_1} =0 . $$
Now, let's look at some other origin $mathcal{O}_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbf{r_2 times F_2} stackrel{?}{=}0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
answered Dec 27 at 21:32
InertialObserver
1,478517
add a comment |
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3 Answers
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3 Answers
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Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
|
show 7 more comments
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
|
show 7 more comments
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
answered Dec 27 at 21:23
Michael Seifert
14.7k22752
14.7k22752
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
|
show 7 more comments
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
1
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 at 23:10
2
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbf{L} = int pmb{tau} , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 at 23:39
1
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 at 0:07
1
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
2 days ago
1
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
2 days ago
|
show 7 more comments
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
add a comment |
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
add a comment |
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
answered Dec 27 at 21:54
Mozibur Ullah
4,64222249
4,64222249
add a comment |
add a comment |
Angular momentum relative to an origin ${mathcal O_1}$
$$ mathbf{L_{mathcal O_1}} = mathbf{r_{mathcal O_1} times p_{mathcal O_1}}$$
where $mathbf r_{mathcal O_1}$ is the position vector to the particle relative to some origin ${mathcal O_1}$.
Now suppose that angular momentum is conserved in ${mathcal O_1}$. Then
$$ frac{d mathbf L_1}{dt} = mathbf{dot{r_1} times p_1} + mathbf{r_1 times dot{p_1}} = frac{1}{m} mathbf{p_1 times p_1} + mathbf{r_1 times dot{p_1}} =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbf{p_1} = mathbf{p}$). It then follows that
$$ mathbf{r_1 times F_1} =0 . $$
Now, let's look at some other origin $mathcal{O}_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbf{r_2 times F_2} stackrel{?}{=}0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
answered Dec 27 at 21:32
InertialObserver
1,478517
add a comment |
Angular momentum relative to an origin ${mathcal O_1}$
$$ mathbf{L_{mathcal O_1}} = mathbf{r_{mathcal O_1} times p_{mathcal O_1}}$$
where $mathbf r_{mathcal O_1}$ is the position vector to the particle relative to some origin ${mathcal O_1}$.
Now suppose that angular momentum is conserved in ${mathcal O_1}$. Then
$$ frac{d mathbf L_1}{dt} = mathbf{dot{r_1} times p_1} + mathbf{r_1 times dot{p_1}} = frac{1}{m} mathbf{p_1 times p_1} + mathbf{r_1 times dot{p_1}} =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbf{p_1} = mathbf{p}$). It then follows that
$$ mathbf{r_1 times F_1} =0 . $$
Now, let's look at some other origin $mathcal{O}_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbf{r_2 times F_2} stackrel{?}{=}0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
answered Dec 27 at 21:32
InertialObserver
1,478517
add a comment |
Angular momentum relative to an origin ${mathcal O_1}$
$$ mathbf{L_{mathcal O_1}} = mathbf{r_{mathcal O_1} times p_{mathcal O_1}}$$
where $mathbf r_{mathcal O_1}$ is the position vector to the particle relative to some origin ${mathcal O_1}$.
Now suppose that angular momentum is conserved in ${mathcal O_1}$. Then
$$ frac{d mathbf L_1}{dt} = mathbf{dot{r_1} times p_1} + mathbf{r_1 times dot{p_1}} = frac{1}{m} mathbf{p_1 times p_1} + mathbf{r_1 times dot{p_1}} =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbf{p_1} = mathbf{p}$). It then follows that
$$ mathbf{r_1 times F_1} =0 . $$
Now, let's look at some other origin $mathcal{O}_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbf{r_2 times F_2} stackrel{?}{=}0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
answered Dec 27 at 21:32
InertialObserver
1,478517
Angular momentum relative to an origin ${mathcal O_1}$
$$ mathbf{L_{mathcal O_1}} = mathbf{r_{mathcal O_1} times p_{mathcal O_1}}$$
where $mathbf r_{mathcal O_1}$ is the position vector to the particle relative to some origin ${mathcal O_1}$.
Now suppose that angular momentum is conserved in ${mathcal O_1}$. Then
$$ frac{d mathbf L_1}{dt} = mathbf{dot{r_1} times p_1} + mathbf{r_1 times dot{p_1}} = frac{1}{m} mathbf{p_1 times p_1} + mathbf{r_1 times dot{p_1}} =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbf{p_1} = mathbf{p}$). It then follows that
$$ mathbf{r_1 times F_1} =0 . $$
Now, let's look at some other origin $mathcal{O}_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbf{r_2 times F_2} stackrel{?}{=}0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
answered Dec 27 at 21:32
InertialObserver
1,478517
edited Dec 27 at 21:46
answered Dec 27 at 21:32
InertialObserver
1,478517
answered Dec 27 at 21:32
InertialObserver
1,478517
answered Dec 27 at 21:32
InertialObserver
1,478517
1,478517
add a comment |
add a comment |
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