Is mass conserved in relativistic collision?

In summary, the invariant mass of a system of particles is conserved in a relativistic collision, but it is not equal to the sum of the invariant masses of the individual particles. This is because the total energy and momentum of the system are conserved, and the invariant mass is a function of these quantities. This concept is important when considering the overall behavior of the universe.
  • #1
Zarif
9
1
For a particle , E2 = (pc)2 + (moc2)2
and for a system of particle , (ΣE)2 = (Σpc)2 + (Σmoc2)2
so in that way before a collision,
(ΣEi)2 = (Σpic)2 + (Σmoic2)2
and after , (ΣEf)2 = (Σpfc)2 + (Σmofc2)2
and as far as i know energy and momentum is conserved . so that means ΣEi=ΣEf
and also Σpi=Σpf
so that leads to Σmoi= Σmof
but as far as i know , rest mass is not conserved in a relativistic collision..
so where am i gettiing it wrong?
 
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  • #2
Zarif said:
For a particle , E2 = (pc)2 + (moc2)2
and for a system of particle , (ΣE)2 = (Σpc)2 + (Σmoc2)2
so in that way before a collision,
(ΣEi)2 = (Σpic)2 + (Σmoic2)2
and after , (ΣEf)2 = (Σpfc)2 + (Σmofc2)2
and as far as i know energy and momentum is conserved . so that means ΣEi=ΣEf
and also Σpi=Σpf
so that leads to Σmoi= Σmof
but as far as i know , rest mass is not conserved in a relativistic collision..
so where am i gettiing it wrong?

Try analysing a two-particle system in its COM frame.
 
  • #3
PeroK said:
Try analysing a two-particle system in its COM frame.
I don't have a good idea on how to approach.. Maybe you could give me some more hint
 
  • #4
Zarif said:
I don't have a good idea on how to approach.. Maybe you could give me some more hint

The simplest system is two particles, ##m_1## and ##m_2## with energies ##E_1## and ##E_2## approaching each other with equal and opposite momenta ##p##.

Imagine they create a single particle, ##M##, which must be at rest.

Do the math, as they say!
 
  • #5
PeroK said:
The simplest system is two particles, ##m_1## and ##m_2## with energies ##E_1## and ##E_2## approaching each other with equal and opposite momenta ##p##.

Imagine they create a single particle, ##M##, which must be at rest.

Do the math, as they say!
I understand now , M comes to be (√((pc)2)+(m1c2)2 + √((pc)2)+(m2c2)2) / c2
so , that is not m1+ m2 .. i understand what i was doing wrong .. i was like writing (a+b)2= a2+b2.
so what do the textbooks actually mean by "for a system of particle , (ΣE)2 = (Σpc)2 + (Σmoc2)2" ?
 
  • #6
Zarif said:
I understand now , M comes to be (√((pc)2)+(m1c2)2 + √((pc)2)+(m2c2)2) / c2
so , that is not m1+ m2 .. i understand what i was doing wrong .. i was like writing (a+b)2= a2+b2.
so what do the textbooks actually mean by "for a system of particle , (ΣE)2 = (Σpc)2 + (Σmoc2)2" ?

It's ##\sum E_i^2 = \sum p_i^2c^2 + \sum m_i^2 c^4##
 
  • #7
PS For a system of particles ##(\sum E_i)^2 - (\sum p_i)^2c^2## is invariant between reference frames. But, it's not equal to ##(\sum m_i)^2 c^4##.
 
  • #8
PeroK said:
PS For a system of particles ##(\sum E_i)^2 - (\sum p_i)^2c^2## is invariant between reference frames. But, it's not equal to ##(\sum m_i)^2 c^4##.
Thanks, I do get the point now . But in the book it says ..let me quote "Notice in particular that the quantity (ΣE)2 = (Σpc)2 + (Σmoc2)2
when applied to a group of particles has two things to
commend it.

 Firstly, it is only a function of total energy and momentum, and
therefore will remain the same before and after the collision.
 Secondly, it is a function of the rest masses (see equation 11) and
therefore will be the same in all reference frames.
"
The first one is the one that is creating all the confusion .. it is said that it will remain same before and after the collision.. i just can't understand the significance of that part
 
  • #9
Zarif said:
Thanks, I do get the point now . But in the book it says ..let me quote "Notice in particular that the quantity (ΣE)2 = (Σpc)2 + (Σmoc2)2
when applied to a group of particles has two things to
commend it.

 Firstly, it is only a function of total energy and momentum, and
therefore will remain the same before and after the collision.
 Secondly, it is a function of the rest masses (see equation 11) and
therefore will be the same in all reference frames.
"
The first one is the one that is creating all the confusion .. it is said that it will remain same before and after the collision.. i just can't understand the significance of that part

What book is this?
 
  • #10
PeroK said:
What book is this?
Upgrade Your Physics by A. C. MACHACEK
 
  • #11
Zarif said:
Upgrade Your Physics by A. C. MACHACEK

The quantity ##(\sum E_i)^2 - (\sum p_i)^2c^2## has those properties. But, as we have shown, it's not equal to ##(\sum m_i)^2 c^4##.

It is invariant between reference frames because total energy-momentum is a four-vector (being the sum of four-vectors).

It is the same before and after collisions for the reason given: total energy and momentum are conserved.

Its square root is equal to is the total energy in the COM frame. This is only equal to the sum of the rest energies if all particles are at rest. This is always true for a single particle (it's at rest in its COM frame), but it's not in general true for a system of particles.
 
  • #12
Zarif said:
as far as i know , rest mass is not conserved in a relativistic collision.
The invariant mass of the system is conserved. It is just not generally equal to the sum of the invariant masses of the parts of the system.
 
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  • #13
PeroK said:
The quantity ##(\sum E_i)^2 - (\sum p_i)^2c^2## has those properties. But, as we have shown, it's not equal to ##(\sum m_i)^2 c^4##.

It is invariant between reference frames because total energy-momentum is a four-vector (being the sum of four-vectors).

It is the same before and after collisions for the reason given: total energy and momentum are conserved.

Its square root is equal to is the total energy in the COM frame. This is only equal to the sum of the rest energies if all particles are at rest. This is always true for a single particle (it's at rest in its COM frame), but it's not in general true for a system of particles.
Thanks @PeroK .. Everything is quite clear now
 
  • #14
Dale said:
The invariant mass of the system is conserved. It is just not generally equal to the sum of the invariant masses of the parts of the system.
Considering the universe as a system comprised of massive constituents, is it accurate to say dark energy is its "binding energy"?
 
  • #15
stoomart said:
Considering the universe as a system comprised of massive constituents, is it accurate to say dark energy is its "binding energy"?

No. The universe as a whole is not a bound system, and the concept of "binding energy" is not well defined for it. Neither is the concept of "invariant mass", for that matter. Those concepts only are well defined for isolated systems--systems of finite spatial extent surrounded by empty space. The universe is not such a system.
 
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  • #16
PeterDonis said:
No. The universe as a whole is not a bound system, and the concept of "binding energy" is not well defined for it. Neither is the concept of "invariant mass", for that matter. Those concepts only are well defined for isolated systems--systems of finite spatial extent surrounded by empty space. The universe is not such a system.
Which of these systems best describes the universe: open, closed, isolated, other, none?

https://en.m.wikipedia.org/wiki/File:Diagram_Systems.png
 
  • #17
A closed system in a thermodynamic context makes the most sense to me, since it would be compatible with LQC, which also makes the most sense to me.
 
  • #18
stoomart said:
Which of these systems best describes the universe: open, closed, isolated, other, none?

None of them. All of those definitions assume a system with a boundary that separates it from something else. The universe has no such boundary.
 
  • #19
Dale said:
The invariant mass of the system is conserved. It is just not generally equal to the sum of the invariant masses of the parts of the system.
Could you please elaborate on what that means?


 
  • #20
Battlemage! said:
Dale said:
The invariant mass of the system is conserved. It is just not generally equal to the sum of the invariant masses of the parts of the system.
Could you please elaborate on what that means?

The invariant mass of a system is $$\frac{\sqrt{(\sum E_i)^2 - (\sum p_i)^2c^2}}{c^2}$$i.e. (total energy in the COM frame*) divided by ##c^2##.

Then see:
PeroK said:
The quantity ##(\sum E_i)^2 - (\sum p_i)^2c^2## has those properties. But, as we have shown, it's not equal to ##(\sum m_i)^2 c^4##.

It is invariant between reference frames because total energy-momentum is a four-vector (being the sum of four-vectors).

It is the same before and after collisions for the reason given: total energy and momentum are conserved.

Its square root is equal to is the total energy in the COM frame. This is only equal to the sum of the rest energies if all particles are at rest. This is always true for a single particle (it's at rest in its COM frame), but it's not in general true for a system of particles.

*COM frame = Centre Of Momentum frame = the frame in which total momentum is zero = the frame in which the system as a whole may be considered to be at rest.
 
  • #21
DrGreg said:
The invariant mass of a system is $$\frac{\sqrt{(\sum E_i)^2 - (\sum p_i)^2c^2}}{c^2}$$i.e. (total energy in the COM frame*) divided by ##c^2##.

Then see:*COM frame = Centre Of Momentum frame = the frame in which total momentum is zero = the frame in which the system as a whole may be considered to be at rest.
Thanks. So basically, the particles might be moving? If so then that is pretty obvious, since velocity depends on reference frame.
 
  • #22
Battlemage! said:
So basically, the particles might be moving?
If you mean "relative to each other", then, yes. The original question was about collisions, so in that case there must be relative motion.
 
  • #23
Battlemage! said:
Could you please elaborate on what that means?
In addition to what @DrGreg said above, let's consider a concrete example. I will use the four momentum ##\mathbf{P}=(E,p_x,p_y,p_z)## and I will use units where c=1, so that the invariant mass is the Minkowski norm of the four momentum: ##m^2=|\mathbf{P}|^2=E^2-p_x^2-p_y^2-p_z^2##. For convenience I will choose units such that the mass and energy are scaled such that the rest mass or rest energy of an electron is 1.

So if we have an electron and a positron at rest then the four-momentum of the electron is ##(1,0,0,0)## which has invariant mass 1, and the four-momentum of the positron is also ##(1,0,0,0)## which is also an invariant mass of 1. So the system as a whole has four-momentum ##(1,0,0,0)+(1,0,0,0)=(2,0,0,0)## and so the system invariant mass is 2.

After anhillation, assume that the photons are aligned along the x axis. Then one has four-momentum ##(1,1,0,0)## and the other has four-momentum ##(1,-1,0,0)##, so each has invariant mass 0. But the system as a whole has four-momentum ##(1,1,0,0)+(1,-1,0,0)=(2,0,0,0)## and so the system invariant mass is still 2. Thus we now have a system with invariant mass remains 2, even though the sum of the invariant masses of the parts of the system is 0.
 
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  • #24
So for a system, momentum and energy are included as part of invariant mass?

FYI I'm at the B level (or lower).
 
  • #25
The momentum and energy of the components of the system as measured in the frame where the system, on average, is not moving, yes.
 
  • #26
Imager said:
So for a system, momentum and energy are included as part of invariant mass? )

Momentum and energy are the components of a 4-vector. Mass is the magnitude of that vector.
 
  • #27
Ibix said:
The momentum and energy of the components of the system as measured in the frame where the system, on average, is not moving, yes.

So it is "kinda" like counting the energies inside a proton as part of the proton's rest mass?
 
  • #28
Imager said:
So it is "kinda" like counting the energies inside a proton as part of the proton's rest mass?
Yes. Any internal energy (as opposed to kinetic energy of the system as a whole) counts, as far as I know.
 
  • #29
Thank you @Ibix and @Mister T! That really helps. One misunderstanding down, 114.7 Duodecillion to go!
 
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  • #30
Dale said:
In addition to what @DrGreg said above, let's consider a concrete example. I will use the four momentum ##\mathbf{P}=(E,p_x,p_y,p_z)## and I will use units where c=1, so that the invariant mass is the Minkowski norm of the four momentum: ##m^2=|\mathbf{P}|^2=E^2-p_x^2-p_y^2-p_z^2##. For convenience I will choose units such that the mass and energy are scaled such that the rest mass or rest energy of an electron is 1.

So if we have an electron and a positron at rest then the four-momentum of the electron is ##(1,0,0,0)## which has invariant mass 1, and the four-momentum of the positron is also ##(1,0,0,0)## which is also an invariant mass of 1. So the system as a whole has four-momentum ##(1,0,0,0)+(1,0,0,0)=(2,0,0,0)## and so the system invariant mass is 2.

After anhillation, assume that the photons are aligned along the x axis. Then one has four-momentum ##(1,1,0,0)## and the other has four-momentum ##(1,-1,0,0)##, so each has invariant mass 0. But the system as a whole has four-momentum ##(1,1,0,0)+(1,-1,0,0)=(2,0,0,0)## and so the system invariant mass is still 2. Thus we now have a system with invariant mass remains 2, even though the sum of the invariant masses of the parts of the system is 0.
Dale, I am assuming you can add these just like regular vectors? But magnitudes would be done by squaring each term and adding them with the +,-,-,- or -, +,+,+ signatures? (basically a dot product)

And is the invariant mass such a magnitude?
 
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  • #31
Battlemage! said:
Dale, I am assuming you can add these just like regular vectors? But magnitudes would be done by squaring each term and adding them with the +,-,-,- or -, +,+,+ signatures? (basically a dot product)

And is the invariant mass such a magnitude?
Yes to all 3 questions.

But for the 2nd question, you also need to square-root the absolute value of the final answer (as I think you know but forgot to say!).
 
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Related to Is mass conserved in relativistic collision?

1. Is mass conserved in relativistic collision?

According to the theory of relativity, mass is not conserved in relativistic collisions. This means that the total mass of the particles involved in the collision before and after the collision may not be the same.

2. How is mass-energy equivalence related to relativistic collision?

Mass-energy equivalence, as described by Einstein's famous equation E=mc^2, is a fundamental concept in relativity that explains how mass and energy are interchangeable. In relativistic collisions, mass can be converted into energy and vice versa.

3. Can mass be created or destroyed in relativistic collisions?

No, mass cannot be created or destroyed in relativistic collisions. It can only be converted into energy or redistributed among the particles involved in the collision.

4. What is the role of momentum in relativistic collision?

Momentum is conserved in relativistic collisions, meaning that the total momentum of the particles before and after the collision remains the same. This is a crucial concept in understanding the behavior of particles in relativistic collisions.

5. How does the speed of the particles affect mass conservation in relativistic collisions?

The speed of the particles is a key factor in determining whether mass is conserved in relativistic collisions. As particles approach the speed of light, their mass increases and the conservation of mass becomes less apparent. However, the conservation of energy and momentum still hold true in these situations.

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