A question about invariant mass

[Dale]

You have to explain to me why, physically, a hydrogen atom is not a bound system.

Or cite the definition for a bound system and then identify why a hydrogen atom fails to meet the definition. Certainly asking for the definition of an unrelated term is an unresponsive reply

 

[liuxinhua]

We need the definition of invariant mass.
Strict definition

Correct it.
We need the definition of bound system.
Strict definition of bound system.

Under Newton's classical space-time, the bound system is a system that the distance between any components not changes over time. It can also be understood as: there is no relative movement between any components.

Bound system is actually a rigid body. In the special relativity, rigid bodies do not exist.

And The distance between objects needs the coordinates of two positions measuring at the same time. But the simultaneity is no longer absolute. Therefore, we must redefine bound system in the special theory of relativity.There is no transformation between matter and energy in the bound system.
In the two-dimensional plane, defined bound system as: Give a unique markup for every component of the system, such as Ma, Mb…. For every component of the system, such as Ma, it may be non inertial motion and it has its own original timeτ. For every component and its every original time, such as Ma at time τ1, there exists an inertial reference frameK1. And at time t1 of K1 the component of the system Ma is rest in K1. Measured in K1 at time t1, the system has a material distribution and shape. For Ma at timeτ2, there exists another inertial reference frame K2. And at time t2 of K2 the component of the system Ma is rest inK2. Measured in K2 at timet2, the system has another material distribution and shape. If the material distribution and shape of the system don’t change with time τ(τ1,τ2…), the system is a bound system.

 

[liuxinhua]

What does this have to do with whether a hydrogen atom is or is not a bound system? You claimed it is not. I asked why not. You can't answer that question by asking for definitions of terms. You have to explain to me why, physically, a hydrogen atom is not a bound system.

A hydrogen atom is rest. If we consider the structure of hydrogen, the distance between electrons and nuclei is changing.

Sum of 4 momentum of all constituents ( note, radiation has well defined 4 momentum). Norm of the sum is the invariant mass. You choose what constituents to include. This is SR. Invariant mass exists only in an approximate sense in GR.

I think ,Invariant mass is suitable for use in SR.

 

[Ibix]

Under Newton's classical space-time, the bound system is a system that the distance between any components not changes over time. It can also be understood as: there is no relative movement between any components.

This is not correct. It's a system where the components do not separate completely - their separations stay within a finite range. For example, both circular and elliptical orbits are called bound orbits, because the orbiting body does not escape - it is bound to the primary. Parabolic and hyperbolic orbits are unbound, because the orbiting body escapes - it is not bound to the primary. And we've already used the example of a box of gas, where the components are all moving randomly.

 

[Dale]

Bound system is actually a rigid body.

I wouldn’t use that definition. I would call that a rigid body and use “bound” to include systems whose parts stay within some finite distance of each other but not necessarily a fixed distance. This is what we mean when we speak of binding energy.

However, if you use bound in your manner then the Wikipedia article is correct.

 

[liuxinhua]

For example, both circular and elliptical orbits are called bound orbits, because the orbiting body does not escape - it is bound to the primary.

Here are a circular orbit, and a car on the orbit.
The circular orbit and the car form a system.
The circular orbit is rest in K.
Extern force accelerates the car,the speed of the car is getting faster and faster.
The invariant mass of the system, which formed by the circular orbits and the car, is not constant.

 

[liuxinhua]

I wouldn’t use that definition. I would call that a rigid body and use “bound” to include systems whose parts stay within some finite distance of each other but not necessarily a fixed distance. This is what we mean when we speak of binding energy.
However, if you use bound in your manner then the Wikipedia article is correct.

The calculation shows that a rigid body does not rotate and accelerates along the x direction, but its invariant mass will change.

 

[Dale]

The calculation shows that a rigid body does not rotate and accelerates along the x direction, but its invariant mass will change.

The calculation is incorrect.

 

[liuxinhua]

The calculation is incorrect.

A rigid body.
Before t0, the rigid body is rest in K.
At t0, the rigid begin to accelerate along x-axis.
At time t1’, the rigid body is rest relative to K’.
At time t1, different part of the rigid body has different velocity relative to K.
By definition of invariant mass, calculate in K, the invariant mass of the rigid body at time t1 is not equal to it at time t0.

 

[vanhees71]

We need the definition of invariant mass.
Strict definition

The invariant mass (squared) is one of the Casimir operators of the proper orthochronous Poincare group, defined as $P_{\mu} P^{\mu}=m^2$ (taking natural units with $c=1$). The other Casimir operator is the square of the Pauli-Lubanski vector.

 

[vanhees71]

A rigid body.
Before t0, the rigid body is rest in K.
At t0, the rigid begin to accelerate along x-axis.
At time t1’, the rigid body is rest relative to K’.
At time t1, different part of the rigid body has different velocity relative to K.
By definition of invariant mass, calculate in K, the invariant mass of the rigid body at time t1 is not equal to it at time t0.

A rigid body in the usual sense does not exist within relativistic physics. It's easy to understand why: A rigid body has by definition an infinite sound velocity, which contradicts Einstein causality within relativistic theory. For more details search this forum for "Born rigid body".

 

[liuxinhua]

A rigid body in the usual sense does not exist within relativistic physics. It's easy to understand why: A rigid body has by definition an infinite sound velocity, which contradicts Einstein causality within relativistic theory. For more details search this forum for "Born rigid body".

On the thirty-seventh floor, I define the bound system.
Dale said he used to call it as rigid body.
So I used the word here.

Of course, rigid body do not exist. But an object can behave like a rigid body through external action, which means this.

 

[Dale]

By definition of invariant mass, calculate in K, the invariant mass of the rigid body at time t1 is not equal to it at time t0.

This is not a calculation. You should actually work through the calculation here. The conclusion is incorrect

 

[Dale]

A rigid body in the usual sense does not exist within relativistic physics. It's easy to understand why: A rigid body has by definition an infinite sound velocity, which contradicts Einstein causality within relativistic theory. For more details search this forum for "Born rigid body".

This is true, but an object can accelerate in a Born rigid manner due to external forces. So the discussion is OK with some small caveats about wording, as you mention.

 

[liuxinhua]

This is not a calculation. You should actually work through the calculation here. The conclusion is incorrect

To facilitate explanation, the system is simplified into two particles. For these two particles, the static mass of a single particle is m₀. Under the action of external forces, the distance between two particles is invariable as a rigid body.

In K, at time t₀, the two particles are rest in K, and the system's invariant mass is 2m₀.
In K, at time t₁, the velocity of the two particles is different, one is u₁, another is u₂. and the system's invariant mass is greater than 2m₀.

Is that enough to explain:calculate in K, the invariant mass of the rigid body at time t₁is not equal to it at time t₀

 

[Dale]

To facilitate explanation, the system is simplified into two particles. For these two particles, the static mass of a single particle is m0. Under the action of external forces, the distance between two particles is invariable as a rigid body.

Ok, with the understanding that the distance is the distance in the sense of Born rigid motion.

In K, at time t0, the two particles are rest in K, and the system's invariant mass is 2m0.

Yes

In K, at time t1, the velocity of the two particles is different, one is u1, another is u2.

Yes.

and the system's invariant mass is greater than 2m0.

No. This is not a calculation. Please show your calculation. Exactly how much is the invariant mass? When you calculate it you will find that your claim is wrong.

Is that enough to explain:calculate in K, the invariant mass of the rigid body at time t1is not equal to it at time t0

No. You need to actually do this calculation. This is not calculating, this is “hand waving”

 

[liuxinhua]

For these two particles, the static mass of a single particle is m₀.
“In K, at time t₁, the velocity of the two particles is different, one is u₁, another is u₂. The system's invariant mass is greater than 2m₀.” This is a ready-made conclusion.If necessary, I can list its calculations

Also can see, https://en.wikipedia.org/wiki/Invariant_mass
Because the invariant mass includes the mass of any kinetic and potential energies which remain in the center of momentum frame, the invariant mass of a system can be greater than sum of rest masses of its separate constituents.

 

[Dale]

This is a ready-made conclusion.If necessary, I can list its calculations

I believe this is the fourth time I have asked you to provide those calculations. Your ready made conclusion is wrong, and at this point your unwillingness to post the calculations is exceptionally irritating.

If you are correct, then why did you simply not post the calculations after the first request? Now, if you are wrong, instead of me being willing to patiently help you learn I am going to already be exasperated and unwilling to help, and if you are right I am going to be exasperated and unwilling to learn. Either way you have already exhausted my goodwill in this conversation. That is not a good strategy. This is now a more antagonistic conversation than it needed to be.

 

[liuxinhua]

I'm sorry. I hope Dale has not left because of anger.
I do not know how to issue formulas in the forum. So I want to use character to explain directly.
But now it seems that words can not explain this problem.
I'll just use a picture. Due to the limitation of the size of the file in the forum, it may not be clear enough.

For these two particles, the static mass of a single particle is m₀.
In K, at time t₁, the velocity of the two particles is different, one is u₁, another is u₂. The system's invariant mass is greater than 2m₀.

 

[PeterDonis]

A hydrogen atom is rest. If we consider the structure of hydrogen, the distance between electrons and nuclei is changing.

Yes, but that doesn't make it not a bound system. As others have already commented, your definition of "bound system" is much too strict. To give a Newtonian example, the Sun and its planets constitute a bound system, but the distances between the Sun and planets are changing.

 

[PeterDonis]

I'm sorry. I hope Dale has not left because of anger.

No, he's just telling you that you need to actually do the calculations and show your work.

I do not know how to issue formulas in the forum.

Use the PF LaTeX feature:

https://www.physicsforums.com/help/latexhelp/

We can't comment on equations in images.

 

[liuxinhua]

For these two particles, the static mass of a single particle is m₀.
In K, at time t₁, the velocity of the two particles is different, one is u₁, another is u₂. The system's invariant mass is greater than 2m₀.

Set $β_1=\frac 1 {\sqrt {1 - \frac {u_1^2} {c^2}}}$ $β_2=\frac 1 {\sqrt {1 - \frac {u_2^2} {c^2}}}$

$E=\frac {{m_0} {c^2}}{\sqrt {1 - \frac {u_1^2} {c^2}}}+\frac {{m_0} {c^2}}{\sqrt {1 - \frac {u_2^2} {c^2}}}$

$E=β_1 {m_0} {c^2} +β_2 {m_0} {c^2}$

$P=β_1 {m_0} { u_1} +β_2 {m_0} { u_2}$

${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| p \right\|}^2$

$=(β_1+β_2)^2{m_0^2}{c^2}-(β_1 m_0 u_1 +β_2 m_0 u_2) ^2$

$\frac {m_{i0}^2c^2} {m_0^2}= (β_1+β_2)^2{c^2}-(β_1 u_1 +β_2 u_2) ^2$

$={β_1}^2c^2+{β_2}^2c^2+2β_1β_2c^2-{β_1}^2{u_1}^2-{β_2}^2{u_2}^2-2β_1β_2 u_1 u_2$

$={β_1}^2(c^2-{u_1}^2)+{β_2}^2(c^2-{u_2}^2)+2{β_1}{β_2}(c^2-{u_1}{u_2})$

$>4 c^2$ ， So， $m_{i0}>2 m_0$

Because ${β_1}^2(c^2-{u_1}^2)= c^2$. ${β_2}^2(c^2-{u_2}^2)= c^2$. $2{β_1}{β_2}(c^2-{u_1}{u_2})>2 c^2$The following proof: $2{β_1}{β_2}(c^2-{u_1}{u_2})>2 c^2$ at ${β_1}≠{β_2}$

$(c {u_1}- c {u_2})^2>0$

$-2c^2 {u_1} {u_2}> -c^2 {u_1}^2 - c^2{u_2}^2$

$c^4-2c^2 {u_1} {u_2}+{u_1}^2{u_2}^2> c^4- c^2 {u_1}^2 - c^2{u_2}^2+{u_1}^2{u_2}^2$

$(c^2-{u_1}{u_2})^2> (c^2-{u_1}^2) (c^2-{u_2}^2)$

$(c^2-{u_1}{u_2})^2> (1-\frac {{u_1}^2} {c^2} )(1-\frac {{u_2}^2} {c^2}) c^4$

$\frac {(c^2-{u_1}{u_2})^2} {(1-\frac {{u_1}^2} {c^2} )(1-\frac {{u_2}^2} {c^2})}> c^4$

$\frac {(c^2-{u_1}{u_2})} {\sqrt{{(1-\frac {{u_1}^2} {c^2} )(1-\frac {{u_2}^2} {c^2})}}}> c^2$

$2{β_1}{β_2} {(c^2-{u_1}{u_2})}> 2 c^2$

 

[PeterDonis]

In K, at time t1, the velocity of the two particles is different, one is u₁, another is u₂.

Correct.

The system's invariant mass is greater than 2m₀.

Incorrect.

What your calculation shows is that the energy of the system composed of the two particles, in K, is greater when the two particles are moving. But if the particles are moving, energy divided by $c^2$ is not equal to invariant mass. The particles also have nonzero momentum in K, and you have to take that into account in calculating the invariant mass.

Now go look up the correct formula for invariant mass, including both energy and momentum, and do a correct calculation of the invariant mass of the system composed of the two particles. You will see that it is unchanged.

(I am assuming that the two particles in question are being accelerated in a straight line in a Born rigid manner, so that the distance between them, in either of their instantaneous rest frames, is constant. As far as I can tell, that is the case you are calculating, but you have not been very clear about that.)

 

[PAllen]

In any given inertial frame, an accelerating Born rigid body has different proper acceleration at different positions, therefore velocities of constituent elements don't remain that same over time, therefore invariant mass changes. This is no issue because a Born rigid object is not a closed system - there is differential force exerted on different parts to maintain such rigidity. To have system including an accelerating born rigid object that conserves momentum, for example, you must include something else that applies force to each part of the body. The invariant mass of this overall system will remain constant, but the invariant mass of the born rigid object will not be constant.

What is true is that the definition of Born rigidity requires that in the MCIF (momentarily comoving inertial frame) of any constituent at any time, all parts of the body are momentarily at rest, thus the invariant mass at this moment in this inertial frame is the sum of rest masses. However, for any given such inertial frame, this is true only for one moment.

 

[liuxinhua]

I use
${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| p \right\|}^2$
not
${m_{i0}^2}{c^2}=(\frac E {c})^2$
I have take the nonzero momentum into account in calculating the invariant mass.

But if the particles are moving, energy divided by $c^2$ is not equal to invariant mass. The particles also have nonzero momentum in K, and you have to take that into account in calculating the invariant mass.

$\sum E=\frac {{m_0} {c^2}}{\sqrt {1 - \frac {u_1^2} {c^2}}}+\frac {{m_0} {c^2}}{\sqrt {1 - \frac {u_2^2} {c^2}}}$
$\sum P=β_1 {m_0} { u_1} +β_2 {m_0} { u_2}$

 

[PAllen]

Note, it is true that you cannot have two particles of different velocities in some inertial frame have invariant mass equal to sum of rest masses. This is easy to show with 4-vectors. You desire that norm (m1 U1 +m2 U2) = m1 + m2. This requires that U1 dot U2 = 1. This requires that the vectors are equal.

 

[liuxinhua]

Note, it is true that you cannot have two particles of different velocities in some inertial frame have invariant mass equal to sum of rest masses. This is easy to show with 4-vectors. You have that norm (m1 U1 +m2 U2) = m1 + m2. This requires that U1 dot U2 = 1. This requires that the vectors are equal.

At floor 42

The calculation shows that a rigid body does not rotate and accelerates along the x direction, but its invariant mass will change.

U₁ and U₂ is along the x direction.
The same positive and negative
U1 dot U2 = 1

 

[PAllen]

U₁ and U₂ is along the x direction.
The same positive and negative
U1 dot U2 = 1

You are thinking of 3 vectors and the Euclidean dot product. I was referring to 4-vectors and the Minkowski dot product. The comment was meant for someone with the appropriate background. Please read about 4-vectors before commenting, as your comment above indicates you have not.

Very briefly, the notation U means a unit 4-vector. The dot product of two unit timelike 4-vectors will result in gamma of their relative velocity, noting that this is as determined by relativistic velocity addition/subtraction. The result is always positive as long as the two vectors are both future pointing or both past pointing. The result is 1 only if the the unit vectors are the same, otherwise it is greater than 1. Note that use of 4 vectors unifies energy and momentum into one vector quantitiy.

 

[liuxinhua]

Thank PAllen very much for pointing out my mistakes and carelessness. I will recalculate my process.

$E_1=γ_1 {m_0} {c^2}$

$\frac {i E_1} c =iγ_1 {m_0} c$

$u_{1x}= u_1$$P_1= \begin{bmatrix} γ_1 m_0 u_{1} \\ 0 \\ 0 \\ iγ_1 m_0c \end{bmatrix}$$P= \begin{bmatrix} γ_1 m_0 u_{1}+γ_2 m_0 u_{2} \\ 0 \\ 0 \\ iγ_1 m_0c+ iγ_2 m_0c \end{bmatrix}$

${\left\| P \right\|}^2 =\left| (γ_1 m_0 u_1 +γ_2 m_0 u_2) ^2– (γ_1 +γ_2) ^2 {m_0}^2c^2 \right|$

${\left\| P \right\|}^2 ≠(γ_1 m_0 u_1 +γ_2 m_0 u_2) ^2$

I have change βto γ

 

[PAllen]

Don’t you mean gamma where you have beta?

 

[liuxinhua]

I have a new puzzle .

A particle is rest in $K$. Its rest mass is $m_0$

How much is its invariant mass$m_{i0}$?

$E= {m_0} {c^2}$

$P= \begin{bmatrix} 0 \\ 0 \\ 0 \\ i m_0c \end{bmatrix}$

${\left\| P \right\|}^2 = \left|-{m_0} ^2{c^2}\right|={m_0} ^2{c^2}$

${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| P \right\|}^2$

$= {m_0}^2 {c^2}-{m_0} ^2{c^2}$

$=0$If $m_{i0}=0\ ?$

 

[PeterDonis]

A particle is rest in $K$. Its rest mass is $m_0$

How much is its invariant mass $m_{i0}$?

It's $m_0$.

You wrote $P$ as a 4-vector, not a 3-vector, so the invariant mass in this formalism is not $E^2 - P^2$, it's just the norm of $P$, which, in natural units, is just $m_0$. (If you insist on using units in which $c$ is not 1, you need to divide the norm by $c$.)

 

[liuxinhua]

It's $m_0$.

You wrote $P$ as a 4-vector, not a 3-vector, so the invariant mass in this formalism is not $E^2 - P^2$, it's just the norm of $P$, which, in natural units, is just $m_0$. (If you insist on using units in which $c$ is not 1, you need to divide the norm by $c$.)

We confirm a thing first .

$P$ as a 4-vector, $p$ as a 3-vector.

${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| P \right\|}^2$ is right .

Or ${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| p\right\|}^2$ is right.

Or just ${m_{i0}^2}{c^2}={\left\| P\right\|}^2$ is right.Of course in natural units where c = 1,

${m_{i0}^2}= E ^2-{\left\| P \right\|}^2$ is right.

Or ${m_{i0}^2}= E^2-{\left\| p\right\|}^2$ is right.

Or just ${m_{i0}^2}={\left\| P\right\|}^2$ is right. In fact $E^2-{\left\| p\right\|}^2={\left\| P\right\|}^2$I think ：

${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| p\right\|}^2$ is right.
In natural units where c = 1,
${m_{i0}^2}= E^2-{\left\| p\right\|}^2$ is right.I don’t know whether PAllen and PeterDonis is the same means with me .

My previous calculation on the 57 floor use ${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| p\right\|}^2$

（${m_{i0}^2}= E^2-{\left\| p\right\|}^2$）.

On the 57 floor, there are two clerical errors. $β$ should be $γ$

$P=γ_1 {m_0} { u_1} +γ_2 {m_0} { u_2}$ should be $p=γ_1 {m_0} { u_1} +γ_2 {m_0} { u_2}$

 

[PeterDonis]

$P$ as a 4-vector, $p$ as a 3-vector.

Ok.

${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| P \right\|}^2$ is right .

No.

Or ${m_{i0}^2}{c^2}=(\frac E {c})^2-{\left\| p\right\|}^2$ is right.

Yes.

Or just ${m_{i0}^2}{c^2}={\left\| P\right\|}^2$ is right.

Yes, since ${\left\| P\right\|}^2 = (\frac E {c})^2-{\left\| p\right\|}^2$.

 

[pervect]

I'd also like to point out some of the usual sign conventions that are different than what the OP used

Modifying wiki's sign conventions to use "i c t", https://en.wikipedia.org/wiki/Four-momentum, we'd write the following for the 4-momentum, which we'll calll $\vec{P}$, where E is the energy, p_x, p_y, and p_z are the x, y, and z components of the 3-momentum.

We've basically just added the "i" to wiki's approach, wiki doesn't use the i.

$$
\vec{P}= \begin{bmatrix}
i E / c \\
p_x \\

p_y \\
p_z
\end{bmatrix}
$$

Here $p_x, p_y, p_z$ of the 3-momentum, and $\vec{P}$ is the energy- momentum 4 vector.

Then the squared magnitude of the 4-vector, $\vec{P} \cdot \vec{P}$, is just the sum of the squares of it's components, giving

$$\vec{P} \cdot \vec{P} = -E^2/c^2 + p_x^2 + p_y^2 + p_z^2$$

given that $i^2 = -1$

With this sign convention $\vec{P} \cdot \vec{P} = -m_0^2 c^2$, if I haven't made an error with the factors of c, which I usually set to 1, personally. The simplest way to see this is to set all the momentum components to zero. Then $\vec{P} \cdot \vec{P} = -(m_0^2 c^4)/c^2$

This is equivalent to Wiki's relationship

$$
E^2/c^2 = p_x^2 + p_y^2 + p_z^2 + m_0^2 c^2
$$

with a re-arrangement of terms. As far as the factors of c goes, $m_0 c$ has units of momentum, mass times velocity, so $m_0^2 c^2$ has units of momentum squared, which is what we want. The minus sign comes from the -+++ metric signature, which is convenient as we only have one factor of "i", in the 4-vector.

Note that we can write $E = \gamma m_0 c^2$, $p_x = \gamma m_0 v_x$, $p_y = \gamma m_0 v_y$, $p_z = \gamma m_0 v_z$, where $\gamma$ is the relativistic gamma factor, $1/\sqrt{1- (v_x/c)^2 - (v_y/c)^2 - (v_z/c)^2}$.

The relativistic gamma factor was apparently omitted in the OP's original formulation I believe. Wiki incoroporates it in their dicusssion in a different way.