Angular momentum in different points












3














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?










share|cite|improve this question





























    3














    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?










    share|cite|improve this question



























      3












      3








      3







      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?










      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 28 at 12:55









      Qmechanic

      101k121831150




      101k121831150










      asked Dec 27 at 21:12









      Frogfire

      222




      222






















          3 Answers
          3






          active

          oldest

          votes


















          5














          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.






          share|cite|improve this answer

















          • 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



















          4















          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.






          share|cite|improve this answer





























            3














            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.






            share|cite|improve this answer























              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "151"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f450722%2fangular-momentum-in-different-points%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5














              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.






              share|cite|improve this answer

















              • 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
















              5














              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.






              share|cite|improve this answer

















              • 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














              5












              5








              5






              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.






              share|cite|improve this answer












              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.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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














              • 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











              4















              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.






              share|cite|improve this answer


























                4















                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.






                share|cite|improve this answer
























                  4












                  4








                  4







                  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.






                  share|cite|improve this answer













                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 27 at 21:54









                  Mozibur Ullah

                  4,64222249




                  4,64222249























                      3














                      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.






                      share|cite|improve this answer




























                        3














                        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.






                        share|cite|improve this answer


























                          3












                          3








                          3






                          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.






                          share|cite|improve this answer














                          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.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 27 at 21:46

























                          answered Dec 27 at 21:32









                          InertialObserver

                          1,478517




                          1,478517






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Physics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f450722%2fangular-momentum-in-different-points%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Human spaceflight

                              Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                              張江高科駅