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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?
particle-physics inertial-frames collision scattering scattering-cross-section
edited 4 hours ago
Qmechanic♦
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up vote
1
down vote
favorite
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
edited 4 hours ago
Qmechanic♦
98.1k121701061
asked 4 hours ago
user7852656
61
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.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited 4 hours ago
Qmechanic♦
98.1k121701061
asked 4 hours ago
user7852656
61
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
particle-physics inertial-frames collision scattering scattering-cross-section
edited 4 hours ago
Qmechanic♦
98.1k121701061
asked 4 hours ago
user7852656
61
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.
edited 4 hours ago
Qmechanic♦
98.1k121701061
asked 4 hours ago
user7852656
61
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.
edited 4 hours ago
Qmechanic♦
98.1k121701061
edited 4 hours ago
Qmechanic♦
98.1k121701061
edited 4 hours ago
Qmechanic♦
98.1k121701061
98.1k121701061
asked 4 hours ago
user7852656
61
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.
asked 4 hours ago
user7852656
61
asked 4 hours ago
user7852656
61
61
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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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New contributor
user7852656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3 Answers
3
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.
answered 4 hours ago
Garf
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
add a comment |Â
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.
answered 3 hours ago
Luke Pritchett
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
add a comment |Â
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.
answered 1 hour ago
npojo
1,332118
add a comment |Â
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.
answered 4 hours ago
Garf
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
add a comment |Â
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.
answered 4 hours ago
Garf
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
add a comment |Â
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.
answered 4 hours ago
Garf
826112
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.
answered 4 hours ago
Garf
826112
edited 3 hours ago
answered 4 hours ago
Garf
826112
answered 4 hours ago
Garf
826112
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
add a comment |Â
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
add a comment |Â
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.
answered 3 hours ago
Luke Pritchett
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
add a comment |Â
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.
answered 3 hours ago
Luke Pritchett
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
add a comment |Â
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.
answered 3 hours ago
Luke Pritchett
2,05769
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.
answered 3 hours ago
Luke Pritchett
2,05769
answered 3 hours ago
Luke Pritchett
2,05769
answered 3 hours ago
Luke Pritchett
2,05769
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
add a comment |Â
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
add a comment |Â
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.
answered 1 hour ago
npojo
1,332118
add a comment |Â
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.
answered 1 hour ago
npojo
1,332118
add a comment |Â
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.
answered 1 hour ago
npojo
1,332118
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.
answered 1 hour ago
npojo
1,332118
edited 1 hour ago
answered 1 hour ago
npojo
1,332118
answered 1 hour ago
npojo
1,332118
answered 1 hour ago
npojo
1,332118
1,332118
add a comment |Â
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