Why is the laboratory frame energy always greater than the center of mass frame energy?

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I have been looking for an answer to 'Why is the laboratory frame energy always greater than the center of mass frame energy during collisions?'.



A lot of resources provided mathematical explanations. I wanted to get some perspectives in terms of physics.



Could you explain this?










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    I have been looking for an answer to 'Why is the laboratory frame energy always greater than the center of mass frame energy during collisions?'.



    A lot of resources provided mathematical explanations. I wanted to get some perspectives in terms of physics.



    Could you explain this?










    share|cite|improve this question









    New contributor




    user7852656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















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      I have been looking for an answer to 'Why is the laboratory frame energy always greater than the center of mass frame energy during collisions?'.



      A lot of resources provided mathematical explanations. I wanted to get some perspectives in terms of physics.



      Could you explain this?










      share|cite|improve this question









      New contributor




      user7852656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I have been looking for an answer to 'Why is the laboratory frame energy always greater than the center of mass frame energy during collisions?'.



      A lot of resources provided mathematical explanations. I wanted to get some perspectives in terms of physics.



      Could you explain this?







      particle-physics inertial-frames collision scattering scattering-cross-section






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      user7852656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited 4 hours ago









      Qmechanic♦

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          3 Answers
          3






          active

          oldest

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          up vote
          3
          down vote













          I should say, in this answer I've set $c=1$.



          In the centre of mass frame, the total momentum is zero, and so the total energy of the system, is just the sum of the masses - e.g. for two particles colliding:



          beginalign
          E=m_1+m_2
          endalign



          In any general frame, the energy takes the form



          beginalign
          E^2=m^2+|vecp|^2
          endalign



          where $m=sum_im_i$ and $vecp=sum_ivecp_i$.



          The masses above are the particle rest masses, and so are frame independent. However $vecp$ is not frame independent, and this is what makes the energies differ between frames.



          Since $|vecp|^2geq0$, one cannot find a frame where $E$ is smaller than when $|vecp|^2=0$ (i.e. $E$ is minimised in the centre of mass frame).




          In words (if you like the "physics", though I think the above is the best way to think about it), in the centre of mass frame, the total energy is just the rest mass energy of all your particles. If you think about the system collectively (e.g. as a "ball" of your particles with momentum given by the net momentum of the system), you will find this "ball" is stationary and has no momentum, and thus does not contribute anymore to the energy.



          If you are in a frame where the total momentum is not zero however, then this "ball" has some net momentum in some direction, and will therefore always add to the energy from the rest masses alone.






          share|cite|improve this answer






















          • Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
            – user7852656
            1 hour ago

















          up vote
          2
          down vote













          We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame).



          If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object is not moving, so it has no kinetic energy. In the lab frame the total object is moving so it does have kinetic energy.






          share|cite|improve this answer




















          • Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
            – user7852656
            1 hour ago


















          up vote
          0
          down vote













          I think you can obtain physics perspective by examining a couple of simple cases. Center of mass (COM) is a mathematical construct so an exact proof does require some algebra which is out-of-scope in this question.



          We can focus on kinetic energies and collisions are irrelevant to the discussion.



          First, what does it mean being in the lab frame or COM frame? We are looking at all the possible inertial frames (frames moving at constant velocity, possibly zero), and asking in which of these, kinetic energy is minimal. The COM frame is well defined, it moves at the same velocity as the COM. The lab frame, can actually be a frame moving at any other velocity.



          The first simple case are two objects fixed in place. Obviously, the optimal inertial frame, minimizing energy, would be the one with no relative velocity to these objects. So you prefer the COM frame to any moving frame.



          Next, consider two objects moving north at equal velocities. Again, it is obvious that your optimal frame should move north and at the same velocity (that's the only frame where energy is zero). So again, you intuitively select the COM frame.



          Finally, assume two objects, one standing and another moving slowly towards the first along the x axis. Obviously, you wouldn't select your optimal frame to be moving very fast towards +X. You would not select it moving very fast towards -X either. So somewhere in between, will be the optimal velocity. The rest is algebra.






          share|cite|improve this answer






















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            I should say, in this answer I've set $c=1$.



            In the centre of mass frame, the total momentum is zero, and so the total energy of the system, is just the sum of the masses - e.g. for two particles colliding:



            beginalign
            E=m_1+m_2
            endalign



            In any general frame, the energy takes the form



            beginalign
            E^2=m^2+|vecp|^2
            endalign



            where $m=sum_im_i$ and $vecp=sum_ivecp_i$.



            The masses above are the particle rest masses, and so are frame independent. However $vecp$ is not frame independent, and this is what makes the energies differ between frames.



            Since $|vecp|^2geq0$, one cannot find a frame where $E$ is smaller than when $|vecp|^2=0$ (i.e. $E$ is minimised in the centre of mass frame).




            In words (if you like the "physics", though I think the above is the best way to think about it), in the centre of mass frame, the total energy is just the rest mass energy of all your particles. If you think about the system collectively (e.g. as a "ball" of your particles with momentum given by the net momentum of the system), you will find this "ball" is stationary and has no momentum, and thus does not contribute anymore to the energy.



            If you are in a frame where the total momentum is not zero however, then this "ball" has some net momentum in some direction, and will therefore always add to the energy from the rest masses alone.






            share|cite|improve this answer






















            • Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
              – user7852656
              1 hour ago














            up vote
            3
            down vote













            I should say, in this answer I've set $c=1$.



            In the centre of mass frame, the total momentum is zero, and so the total energy of the system, is just the sum of the masses - e.g. for two particles colliding:



            beginalign
            E=m_1+m_2
            endalign



            In any general frame, the energy takes the form



            beginalign
            E^2=m^2+|vecp|^2
            endalign



            where $m=sum_im_i$ and $vecp=sum_ivecp_i$.



            The masses above are the particle rest masses, and so are frame independent. However $vecp$ is not frame independent, and this is what makes the energies differ between frames.



            Since $|vecp|^2geq0$, one cannot find a frame where $E$ is smaller than when $|vecp|^2=0$ (i.e. $E$ is minimised in the centre of mass frame).




            In words (if you like the "physics", though I think the above is the best way to think about it), in the centre of mass frame, the total energy is just the rest mass energy of all your particles. If you think about the system collectively (e.g. as a "ball" of your particles with momentum given by the net momentum of the system), you will find this "ball" is stationary and has no momentum, and thus does not contribute anymore to the energy.



            If you are in a frame where the total momentum is not zero however, then this "ball" has some net momentum in some direction, and will therefore always add to the energy from the rest masses alone.






            share|cite|improve this answer






















            • Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
              – user7852656
              1 hour ago












            up vote
            3
            down vote










            up vote
            3
            down vote









            I should say, in this answer I've set $c=1$.



            In the centre of mass frame, the total momentum is zero, and so the total energy of the system, is just the sum of the masses - e.g. for two particles colliding:



            beginalign
            E=m_1+m_2
            endalign



            In any general frame, the energy takes the form



            beginalign
            E^2=m^2+|vecp|^2
            endalign



            where $m=sum_im_i$ and $vecp=sum_ivecp_i$.



            The masses above are the particle rest masses, and so are frame independent. However $vecp$ is not frame independent, and this is what makes the energies differ between frames.



            Since $|vecp|^2geq0$, one cannot find a frame where $E$ is smaller than when $|vecp|^2=0$ (i.e. $E$ is minimised in the centre of mass frame).




            In words (if you like the "physics", though I think the above is the best way to think about it), in the centre of mass frame, the total energy is just the rest mass energy of all your particles. If you think about the system collectively (e.g. as a "ball" of your particles with momentum given by the net momentum of the system), you will find this "ball" is stationary and has no momentum, and thus does not contribute anymore to the energy.



            If you are in a frame where the total momentum is not zero however, then this "ball" has some net momentum in some direction, and will therefore always add to the energy from the rest masses alone.






            share|cite|improve this answer














            I should say, in this answer I've set $c=1$.



            In the centre of mass frame, the total momentum is zero, and so the total energy of the system, is just the sum of the masses - e.g. for two particles colliding:



            beginalign
            E=m_1+m_2
            endalign



            In any general frame, the energy takes the form



            beginalign
            E^2=m^2+|vecp|^2
            endalign



            where $m=sum_im_i$ and $vecp=sum_ivecp_i$.



            The masses above are the particle rest masses, and so are frame independent. However $vecp$ is not frame independent, and this is what makes the energies differ between frames.



            Since $|vecp|^2geq0$, one cannot find a frame where $E$ is smaller than when $|vecp|^2=0$ (i.e. $E$ is minimised in the centre of mass frame).




            In words (if you like the "physics", though I think the above is the best way to think about it), in the centre of mass frame, the total energy is just the rest mass energy of all your particles. If you think about the system collectively (e.g. as a "ball" of your particles with momentum given by the net momentum of the system), you will find this "ball" is stationary and has no momentum, and thus does not contribute anymore to the energy.



            If you are in a frame where the total momentum is not zero however, then this "ball" has some net momentum in some direction, and will therefore always add to the energy from the rest masses alone.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 4 hours ago









            Garf

            826112




            826112











            • Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
              – user7852656
              1 hour ago
















            • Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
              – user7852656
              1 hour ago















            Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
            – user7852656
            1 hour ago




            Hi @Garf, a really great insight. So the answer to the question is because the total momentum transfer for collisions in laboratory frame energy is not zero, whereas it is zero in center of mass frame. But does the difference in total momentum transfer arise from assuming inelastic collisions?
            – user7852656
            1 hour ago










            up vote
            2
            down vote













            We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame).



            If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object is not moving, so it has no kinetic energy. In the lab frame the total object is moving so it does have kinetic energy.






            share|cite|improve this answer




















            • Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
              – user7852656
              1 hour ago















            up vote
            2
            down vote













            We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame).



            If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object is not moving, so it has no kinetic energy. In the lab frame the total object is moving so it does have kinetic energy.






            share|cite|improve this answer




















            • Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
              – user7852656
              1 hour ago













            up vote
            2
            down vote










            up vote
            2
            down vote









            We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame).



            If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object is not moving, so it has no kinetic energy. In the lab frame the total object is moving so it does have kinetic energy.






            share|cite|improve this answer












            We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame).



            If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object is not moving, so it has no kinetic energy. In the lab frame the total object is moving so it does have kinetic energy.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Luke Pritchett

            2,05769




            2,05769











            • Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
              – user7852656
              1 hour ago

















            • Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
              – user7852656
              1 hour ago
















            Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
            – user7852656
            1 hour ago





            Hi @LukePritchett, thanks for your insight. I wanted to ask for some clarifications for "not moving". So once collisions in the center of mass frame occurs, the positions (on a coordinate in a general sense) of the particles are not the same as prior to the collisions. Could you provide more explanations for what you mean by "not moving"? or the reasons behind why they would not be moving?
            – user7852656
            1 hour ago











            up vote
            0
            down vote













            I think you can obtain physics perspective by examining a couple of simple cases. Center of mass (COM) is a mathematical construct so an exact proof does require some algebra which is out-of-scope in this question.



            We can focus on kinetic energies and collisions are irrelevant to the discussion.



            First, what does it mean being in the lab frame or COM frame? We are looking at all the possible inertial frames (frames moving at constant velocity, possibly zero), and asking in which of these, kinetic energy is minimal. The COM frame is well defined, it moves at the same velocity as the COM. The lab frame, can actually be a frame moving at any other velocity.



            The first simple case are two objects fixed in place. Obviously, the optimal inertial frame, minimizing energy, would be the one with no relative velocity to these objects. So you prefer the COM frame to any moving frame.



            Next, consider two objects moving north at equal velocities. Again, it is obvious that your optimal frame should move north and at the same velocity (that's the only frame where energy is zero). So again, you intuitively select the COM frame.



            Finally, assume two objects, one standing and another moving slowly towards the first along the x axis. Obviously, you wouldn't select your optimal frame to be moving very fast towards +X. You would not select it moving very fast towards -X either. So somewhere in between, will be the optimal velocity. The rest is algebra.






            share|cite|improve this answer


























              up vote
              0
              down vote













              I think you can obtain physics perspective by examining a couple of simple cases. Center of mass (COM) is a mathematical construct so an exact proof does require some algebra which is out-of-scope in this question.



              We can focus on kinetic energies and collisions are irrelevant to the discussion.



              First, what does it mean being in the lab frame or COM frame? We are looking at all the possible inertial frames (frames moving at constant velocity, possibly zero), and asking in which of these, kinetic energy is minimal. The COM frame is well defined, it moves at the same velocity as the COM. The lab frame, can actually be a frame moving at any other velocity.



              The first simple case are two objects fixed in place. Obviously, the optimal inertial frame, minimizing energy, would be the one with no relative velocity to these objects. So you prefer the COM frame to any moving frame.



              Next, consider two objects moving north at equal velocities. Again, it is obvious that your optimal frame should move north and at the same velocity (that's the only frame where energy is zero). So again, you intuitively select the COM frame.



              Finally, assume two objects, one standing and another moving slowly towards the first along the x axis. Obviously, you wouldn't select your optimal frame to be moving very fast towards +X. You would not select it moving very fast towards -X either. So somewhere in between, will be the optimal velocity. The rest is algebra.






              share|cite|improve this answer
























                up vote
                0
                down vote










                up vote
                0
                down vote









                I think you can obtain physics perspective by examining a couple of simple cases. Center of mass (COM) is a mathematical construct so an exact proof does require some algebra which is out-of-scope in this question.



                We can focus on kinetic energies and collisions are irrelevant to the discussion.



                First, what does it mean being in the lab frame or COM frame? We are looking at all the possible inertial frames (frames moving at constant velocity, possibly zero), and asking in which of these, kinetic energy is minimal. The COM frame is well defined, it moves at the same velocity as the COM. The lab frame, can actually be a frame moving at any other velocity.



                The first simple case are two objects fixed in place. Obviously, the optimal inertial frame, minimizing energy, would be the one with no relative velocity to these objects. So you prefer the COM frame to any moving frame.



                Next, consider two objects moving north at equal velocities. Again, it is obvious that your optimal frame should move north and at the same velocity (that's the only frame where energy is zero). So again, you intuitively select the COM frame.



                Finally, assume two objects, one standing and another moving slowly towards the first along the x axis. Obviously, you wouldn't select your optimal frame to be moving very fast towards +X. You would not select it moving very fast towards -X either. So somewhere in between, will be the optimal velocity. The rest is algebra.






                share|cite|improve this answer














                I think you can obtain physics perspective by examining a couple of simple cases. Center of mass (COM) is a mathematical construct so an exact proof does require some algebra which is out-of-scope in this question.



                We can focus on kinetic energies and collisions are irrelevant to the discussion.



                First, what does it mean being in the lab frame or COM frame? We are looking at all the possible inertial frames (frames moving at constant velocity, possibly zero), and asking in which of these, kinetic energy is minimal. The COM frame is well defined, it moves at the same velocity as the COM. The lab frame, can actually be a frame moving at any other velocity.



                The first simple case are two objects fixed in place. Obviously, the optimal inertial frame, minimizing energy, would be the one with no relative velocity to these objects. So you prefer the COM frame to any moving frame.



                Next, consider two objects moving north at equal velocities. Again, it is obvious that your optimal frame should move north and at the same velocity (that's the only frame where energy is zero). So again, you intuitively select the COM frame.



                Finally, assume two objects, one standing and another moving slowly towards the first along the x axis. Obviously, you wouldn't select your optimal frame to be moving very fast towards +X. You would not select it moving very fast towards -X either. So somewhere in between, will be the optimal velocity. The rest is algebra.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                npojo

                1,332118




                1,332118




















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