Needing a solid example of a composite mass value

[lenfromkits]

Hi. I am having a bit of trouble working through all the formulas for calculating the total composite mass of moving particles. If I could just fill in this 'black and white' and very intuitive example then I will be able to use it as a guide to test everything I'm doing.

If we have 2 particles:

A: mass=2Kg, velocity=298,000km/s
B: mass=3Kg, velocity=250,000km/sThen given the kinetic energy of these particles, we know that (based on the idea that the sum of an atom's parts do not add up to the atom's total mass and that the energy involved makes up some of the mass):

mass_Total > mass_A + mass_BIn this example, what exactly would the total mass in Kilograms be?

Thanks 'sooo' much in advance to anyone who can help me with this.

 

[bcrowell]

Your two masses are presumably not interacting (or if they were, you would need to give more information in order to determine what the binding energy was). This is different from an atom, where there are electrical and nuclear binding energies involved.

For the noninteracting particles in your example, you can just add the relativistic mass-energies of the two particles. Each of them has a mass-energy given by m/\sqrt{1-v^2/c^2}, where m is the rest mass (e.g., 2 kg).

 

[lenfromkits]

For the noninteracting particles in your example, you can just add the relativistic mass-energies of the two particles. Each of them has a mass-energy given by m/\sqrt{1-v^2/c^2}, where m is the rest mass (e.g., 2 kg).

So, the total mass would be:

2/\sqrt{1-298000^2/299792.458^2}
=18.3168584Kg

plus

3/\sqrt{1-250000^2/299792.458^2}
=5.43576641Kg

For a total mass of:
23.7526248Kg

Thanks!

 

[lenfromkits]

For the noninteracting particles in your example, you can just add the relativistic mass-energies of the two particles. Each of them has a mass-energy given by m/\sqrt{1-v^2/c^2}, where m is the rest mass (e.g., 2 kg).

Doesn't this kind of suggest that 'relativistic' mass is LESS than 'rest' mass? :)

 

[starthaus]

Hi. I am having a bit of trouble working through all the formulas for calculating the total composite mass of moving particles. If I could just fill in this 'black and white' and very intuitive example then I will be able to use it as a guide to test everything I'm doing.

If we have 2 particles:

A: mass=2Kg, velocity=298,000km/s
B: mass=3Kg, velocity=250,000km/sThen given the kinetic energy of these particles, we know that (based on the idea that the sum of an atom's parts do not add up to the atom's total mass and that the energy involved makes up some of the mass):

mass_Total > mass_A + mass_BIn this example, what exactly would the total mass in Kilograms be?

Thanks 'sooo' much in advance to anyone who can help me with this.

You need to solve:

\gamma(v_1)m_1c^2+\gamma(v_2)m_2c^2=\gamma(V)Mc^2

(conservation of total energy)

and:

\gamma(v_1)m_1v_1+\gamma(v_2)m_2v_2=\gamma(V)MV

(conservation of momentum)

You have two equations (nonlinear) with two unknowns (M and V).

You get a system of equations:

\gamma(v_1)m_1v_1+\gamma(v_2)m_2v_2=\gamma(V)MV
\gamma(v_1)m_1+\gamma(v_2)m_2=\gamma(V)M

Divide the first equation by the second one and you get V=\frac{\gamma(v_1)m_1v_1+\gamma(v_2)m_2v_2}{\gamma(v_1)m_1+\gamma(v_2)m_2}.
From that, you can get

M=\frac{\gamma(v_1)m_1+\gamma(v_2)m_2}{\gamma(V)}.

So:

M<\gamma(v_1)m_1+\gamma(v_2)m_2

 

[lenfromkits]

Divide the first equation by the second one and you get V. From that, you can get M The expressions are very nice.

Hmm. I think I saw all this on the Internet already. Is there any chance you can show me how to plug the numbers into it and get to the mass_total in kilograms?

 

[starthaus]

Hmm. I think I saw all this on the Internet already. Is there any chance you can show me how to plug the numbers into it and get to the mass_total in kilograms?

You have all the information, you need to learn how to do things by yourself, it does not help you if I do all your homework.

 

[lenfromkits]

You have all the information, you need to learn how to do things by yourself, it does not help you if I do all your homework.

Why did you bother responding? I asked a specific question.

 

[starthaus]

Why did you bother responding? I asked a specific question.

You have the complete answer, you are just too lazy to finish the calculations.

 

[starthaus]

So, the total mass would be:

2/\sqrt{1-298000^2/299792.458^2}
=18.3168584Kg

plus

3/\sqrt{1-250000^2/299792.458^2}
=5.43576641Kg

For a total mass of:
23.7526248Kg

Nope, this is incorrect.

 

[Passionflower]

So:

M<\gamma(v_1)m_1+\gamma(v_2)m_2

You mean:

M>\gamma(v_1)m_1+\gamma(v_2)m_2

?

 

[starthaus]

You mean:

M>\gamma(v_1)m_1+\gamma(v_2)m_2

?

no, exactly what i wrote, since \gamma(V)>1

 

[Passionflower]

no, exactly what i wrote, since \gamma(V)>1

Oh yes indeed, we have to calculate the Lorentz factor from V we get in the first formula.

This has been a very long time ago I worked with these things, a numerical refresher would be good..

Ok, so M would represent the total mass of the system right?

Let's see if I get this right, in steps:

TotalEnergy = KineticEnergy + FieldEnergy = Total Mass - ParticleMass = 17.15728546 - 5 = 6.59533939

But isn't the total energy = 23.75262485?

Something does not add up (for me of course).

 

[starthaus]

Oh yes indeed, we have to calculate the Lorentz factor from V we get in the first formula.

yes

Ok, so that would represent the total mass of the system right?

yes

Let's see if I get this right, in steps:

KineticEnergy + FieldEnergy = Total Mass - ParticleMass = 17.15728546 - 5 = 6.59533939

I don't think that this is correct, mass is not additive/subtractive in SR.

What is correct is:

E^2-(Pc)^2=(Mc^2)^2

where

P=\gamma(V)MV=\gamma(v_1)m_1v_1+\gamma(v_2)m_2v_2
E=\gamma(V)Mc^2=\gamma(v_1)m_1c^2+\gamma(v_2)m_2c^2

 

[Passionflower]

Total Rest Mass of the System: = 17.15728546
Total Momentum of the System: 22.74028327
Total Energy of the System := 23.75262485
Total Particle Rest Mass := 5
Kinetic Energy of the system:= 23.75262485 - 5 = 18.75262485
Field Energy of the system: = -6.59533939

No?

Edited: Please ignore these numbers, they are all wrong!

 

[starthaus]

Total Rest Mass of the System: = 17.15728546
Total Momentum of the System: 22.74028327
Total Energy of the System := 23.75262485
Total Particle Rest Mass := 5
Kinetic Energy of the system:= 23.75262485 - 5 = 18.75262485
Field Energy of the system: = -6.59533939

No?

I have no idea what you are doing, try writing down symbolic formulas and I'll tell you what you are doing wrong, it is impossible for me to tell you looking at the numbers.
I can tell you for a fact that this one is wrong:

"Kinetic Energy of the system:= 23.75262485 - 5 = 18.75262485 "

since :

KE=E-Mc^2

You are doing:

KE=E-(m_1+m_2)c^2

and that is definitely wrong.

 

[Passionflower]

I have no idea what you are doing, try writing down symbolic formulas and I'll tell you what you are doing wrong, it is impossible for me to tell you looking at the numbers.

Oops, I apologize, please forget those numbers above as I used proper velocities which messed up things big time.

Hopefully it is right now:
v1 = 0.9940210037
v2 = 0.8339102380

Calculate the gammas using (1/sqrt(1-v²)):
y1 = 9.158429219
y2 = 1.811922137

Get the momenta using (p = myv):
p1 = 18.20734201
p2 = 4.532941263

Get the total energy using (m1 y1 + m2 y2):
E_t = 4.532941263

Get the total momentum using (p1 + p2):
P = 22.74028327

Get V using (P/E_t)
V = 0.9573798018

Get yV
yV = 3.462220442

Get the total rest mass of the system using (E_t/yV)
M = 6.860517765

Get the kinetic energy of the particles using (E_k = E_t - (m1+m2))

E_k = 18.75262485

Get the field energy using (E_t - M - (m1+m2))

E_f = -16.89210708

So then:
M = (m1+m2) + E_k + E_f

I looked at your update about kinetic energy. If I want the kinetic energy of the particle I take the relativistic mass (energy) and subtract the rest mass of each particle?

 

[starthaus]

Oops, I apologize, please forget those numbers above as I used proper velocities which messed up things big time.

Hopefully it is right now:
v1 = 0.9940210037
v2 = 0.8339102380

Calculate the gammas using (1/sqrt(1-v²)):
y1 = 9.158429219
y2 = 1.811922137

Get the momenta using (p = myv):
p1 = 18.20734201
p2 = 4.532941263

Get the total energy using (m1 y1 + m2 y2):
E_t = 4.532941263

Get the total momentum using (p1 + p2):
P = 22.74028327

Get V using (P/E_t)
V = 0.9573798018

Get yV
yV = 3.462220442

Get the total rest mass of the system using (E_t/yV)
M = 6.860517765

Get the kinetic energy of the particles using (E_k = E_t - (m1+m2))

E_k = 18.75262485

OK, so far.
You could also write:

KE=(\gamma(V)-1)Mc^2

Get the field energy using (E_t - M - (m1+m2))

What is this? What physical meaning do you attach to this?

E_f = -16.89210708

So then:
M = (m1+m2) + E_k + E_f

If

E_f=E-M-(m_1+m_2)

then, it follows that:

M=E-E_f-(m_1+m_2)

I looked at your update about kinetic energy. If I want the kinetic energy of the particle I take the relativistic mass (energy) and subtract the rest mass of each particle?

Yes:

ke=(\gamma-1)mc^2

By definition.

 

[Passionflower]

What is this? What physical meaning do you attach to this?

Well the total rest mass of the system is not equal to the rest mass of the constituents + their kinetic energies, we should add the energy of the field as well. No?

 

[starthaus]

Get the field energy using (E_t - M - (m1+m2))

E_f = -16.89210708

So then:
M = (m1+m2) + E_k + E_f

I looked at your update about kinetic energy. If I want the kinetic energy of the particle I take the relativistic mass (energy) and subtract the rest mass of each particle?

I think I know what you are trying to do.If the collision were lossless, then the kinetic energy after the collision would be:

KE=E-Mc^2

But the collision is not lossless, some energy is dissipated in the form of radiative energy (what you call "field energy"):

E_f=E-Mc^2-KE

In the above, you no longer have

KE=(\gamma-1)Mc^2

but:

KE<(\gamma-1)Mc^2

for reasons of dissipation through radiation.

 

[Vanadium 50]

I have removed a number of messages involving an overly speculative discussion.

I have reopened the thread, but if it goes anywhere other than answering the specific question asked, it will be closed again.

 

[Vanadium 50]

This is the sort of mess you get when you don't think about invariants.

The mass is given by m² = E² - p² (units where c = 1).

E = \gamma m and p=\beta\gamma m, so we can work it out: the total energy is 22.7 kg, the total momentum is +12.7 kg, so the total mass is 18.9 kg.

 

[starthaus]

This is the sort of mess you get when you don't think about invariants.

The mass is given by m² = E² - p² (units where c = 1).

E = \gamma m and p=\beta\gamma m, so we can work it out: the total energy is 22.7 kg, the total momentum is +12.7 kg, so the total mass is 18.9 kg.

..which is exactly the same result you get from post 5. The same calculation (based on invariants) as yours is shown already in post 14. In addition, post 5 gives you the speed after collision.