A Momentum calculation of a rotating rod

[liuxinhua]

I suspect you haven't looked at the formal definition of Born rigidity.

Maybe what I call born rigid is different from the expression of the born rigid body in your mind

I have not seen the formal definition of natural Born rigidity, but I think we mean the same thing, Though I'm not sure it.

For any object measured in an inertial frame: $the\ sum \ rest \ masses=\frac {the \ rest \ energy } {c^2}$

 

[PeterDonis]

I have not seen the formal definition of natural Born rigidity

The formal definition of Born rigid motion of an object is that the congruence of worldlines describing the object has zero expansion and zero shear. Since this is an "A" level thread, you should know what those terms mean mathematically.

 

[PAllen]

I have not seen the formal definition of natural Born rigidity, but I think we mean the same thing, Though I'm not sure it.

For any object measured in an inertial frame: $the\ sum \ rest \ masses=\frac {the \ rest \ energy } {c^2}$

This is trivially false. The sum of rest masses for spinning disk does not include kinetic energy contributions to the rest energy.

 

[liuxinhua]

This is trivially false. The sum of rest masses for spinning disk does not include kinetic energy contributions to the rest energy.

The rest energy I understand does not include kinetic energy.

In an inertial reference frame such as $K$, an object is observed.

The total energy （$E$）contains energy includes its rest energy and the kinetic energy relative to $K$;

The invariant mass corresponds to the energy $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$, it contains its rest energy and the kinetic energy relative to MCIF, (the kinetic energy relative to MCIF is not the kinetic energy relative to $K$ ). Maybe physicists call this energy as rest energy. But I think the rest energy of a system should be the following energy.

Rest energy is the amount that add up to all the energy of the system measured from the respective MCIF of each part (infinitely small in length width and height) for each part of the system. The rest energy of a system does not contain any kinetic energy (relative to $K$ or relative to MCIF), corresponding to the sum of the rest masses of the system.For two particles (their rest mass is $m_0$ each, they do not constitute a Born rigid system) moves at velocity $u_1$ and $u_2$ along $x$ direction.

Their total energy is $E=m_0γ(u_1)c^2 + m_0γ(u_2)c^2$.

Their rest energy is $m_0 c^2 + m_0 c^2$.

And $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$=$\sqrt {{(m_0γ(u_1)c^2 + m_0γ(u_2)c^2)}^2- {(m_0γ(u_1) u_1+ m_0γ(u_2) u_2) }^2 c^2}$

$E$>$\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$> $m_0 c^2 + m_0 c^2$

The total energy of a Born rigid body may be changed. The energy $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$ of a Born rigid body may be changed. The rest energy of a Born rigid body is a constant.

For an arbitrary system, its rest energy is not necessarily conserved. Even for isolated systems, the rest energy is not necessarily conserved.

 

[liuxinhua]

Here is one point.
The sum of the rest masses at each location is not the system's invariant mass.
But the sum of the rest energy at each location is the rest energy of the system.

 

[SiennaTheGr8]

This is largely a semantic discussion, but language matters.

Rest energy (or proper energy or invariant energy) and mass are the same thing, just expressed in different units. So the rest energy of a system is indeed the system's total energy as measured in the system's rest frame. It is not the sum of the rest energies of the system's constituents.

As far as I know, there is no special term for the sum of the rest energies (masses) of the constituents of a system. The same is true of kinetic energy: the kinetic energy of a system is not the sum of the kinetic energies of the system's constituents, and there's no special term for the latter.

 

[liuxinhua]

the rest energy of a system is indeed the system's total energy as measured in the system's rest frame.

Accordingly, the rest energy of a born rigid body remains unchanged, that is, the invariant mass remains unchanged. However, as we have discussed, the invariant mass of a born rigid bodies can be changed

 

[PAllen]

The rest energy I understand does not include kinetic energy.

This is false. In fact rest energy is defined as total energy of a system in its COM frame. Invariant mass computed in any frame, times c² equals total energy measured in the COM frame which is defined as rest energy.

In an inertial reference frame such as $K$, an object is observed.

The total energy （$E$）contains energy includes its rest energy and the kinetic energy relative to $K$;

You are defining your own nonstandard terminology, which you will find others have no interest in trying to understand. In most cases, there will be no frame in which all parts of a system are at rest, even momentarily, but there will be a frame in which total momentum of all constituents (including fields) sums to zero.

A spinning body is an example of a system where KE is irremovable and is part of rest energy. An even simpler one is s box of gas, with the box having no overall motion in some frame. Then all the KE of the molecules is part of the rest energy of the system.

The invariant mass corresponds to the energy $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$, it contains its rest energy and the kinetic energy relative to MCIF, (the kinetic energy relative to MCIF is not the kinetic energy relative to $K$ ). Maybe physicists call this energy as rest energy. But I think the rest energy of a system should be the following energy.

See above.

Rest energy is the amount that add up to all the energy of the system measured from the respective MCIF of each part (infinitely small in length width and height) for each part of the system. The rest energy of a system does not contain any kinetic energy (relative to $K$ or relative to MCIF), corresponding to the sum of the rest masses of the system.

Physicists do not define such a thing because they have found no utility in it.

For two particles (their rest mass is $m_0$ each, they do not constitute a Born rigid system) moves at velocity $u_1$ and $u_2$ along $x$ direction.

Their total energy is $E=m_0γ(u_1)c^2 + m_0γ(u_2)c^2$.

Their rest energy is $m_0 c^2 + m_0 c^2$.

And $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$=$\sqrt {{(m_0γ(u_1)c^2 + m_0γ(u_2)c^2)}^2- {(m_0γ(u_1) u_1+ m_0γ(u_2) u_2) }^2 c^2}$

$E$>$\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$> $m_0 c^2 + m_0 c^2$

The total energy of a Born rigid body may be changed. The energy $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$ of a Born rigid body may be changed. The rest energy of a Born rigid body is a constant.

For an arbitrary system, its rest energy is not necessarily conserved. Even for isolated systems, the rest energy is not necessarily conserved.

One major reason physicists do not define any such quantity as you describe is that adding this up as computed in many different MCIFs does produce any useful quantity for analysis. In your case, it leads to persistence of several conceptual errors.

 

[PAllen]

Accordingly, the rest energy of a born rigid body remains unchanged, that is, the invariant mass remains unchanged. However, as we have discussed, the invariant mass of a born rigid bodies can be changed

No, what we found is that born rigid bodies undergoing acceleration or rotation are not closed systems, and speaking of conservation laws without including all elements is not meaningful. Rotation is far more subtle, because what is missing is in equilbrium in the COM frame, but not in another inertial frame. However, what is missing to provide local momentum conservation, and a 'cause' for maintenance of non-inertial motion of consitutents, is present irrespective of frames, and must be included to treat the system as closed.

 

[liuxinhua]

PAllen is right. I can't define a system by myself. I must obey everyone's rules.

The rest energy I said on post#30 and post#36 is a concept that no one else defines and uses.
In fact, $\sqrt {(E^2-\left\|{p}^2\right\|c^2)}$ is equal to rest energy of a system what everybody define
.
So what I post on #30 is not mean “the rest energy of born rigid body is a relativistic scalar”. I only want to insist “the potential energy of born rigid body is a relativistic scalar.”

I said on post#37 : “Accordingly, the rest energy of a born rigid body remains unchanged” is refer to : “Accordingly SiennaTheGr8 said on post #36 “the rest energy of a system is indeed the system's total energy as measured in the system's rest frame”, the rest energy of a born rigid body remains unchanged” , that is, the invariant mass remains unchanged. But the invariant mass of a born rigid bodies can be changed

 

[liuxinhua]

“Accordingly, the rest mass of the spinning disk acquires an additional term representing potential energy of this deformation.” in https://arxiv.org/abs/0712.3891 is referring to “When the rotational speed of the ring increases, its potential energy will increase”? This is consistent with my view on post# 17.
What's different is that he also considered the contribution of KE to rest mass.

@PAllen, Two identical objects $A$ and $B$ rest in $K$, before time $t_0$:
View attachment 229839
They have the same sum rest masses.

One of them ($B$) is deformed under external force, from time $t_0$ to $t_9$:
View attachment 229840
After time $t_9$, they don’t have the same sum rest masses.

In fact, we consider the internal tension or internal potential energy as rest mass.

It does not mean that the potential energy of a rotating ring (its angle velocity is constant measured in an inertial reference frame relative to which the rotating ring center is static) is different when the potential energy of the rotating ring is measured in a certain inertial reference frame at different time.

 

[liuxinhua]

In the front, there are some misunderstandings because of my confusion of the definition of rest energy. Now let's get back to the subject.

Is this correct? “The potential energy of born rigid body is a constant measured in an inertial reference frame at different time.”

In a confirmed inertial reference frame，When a born rigid body is subjected to external action, the change is its kinetic energy and its momentum. However, the relative position of each component does not change, which does not depend on the reference frame. This is the characteristic of born rigid body.

 

[vanhees71]

No, it's the opposite, and that's why everybody insists to use invariant mass only, because invariant mass is a Lorentz scalar, while energy is the spatial component of a four-vector, i.e.,
$$m^2 c^2=p_{\mu} p^{\mu}=E^2/c^2-\vec{p}^2$$
is independent of the reference frame, while $E$ and $\vec{p}$ change according to the transformation law under a Lorentz transformation
$$p^{\prime \mu}={\Lambda^{\mu}}_{\nu} p^{\nu},$$
i.e., the four-momentum components transform as a Lorentz four-vector (guess why it's called the four-momentum vector to begin with ;-)).

 

[liuxinhua]

In general, the internal potential energy of an ordinary object will change. but for a born rigid body, I feel that its internal potential energy will not change, but there is no reliable reason.

Similarly, PAllen , and vanhees71, say that the invariant mass is a Lorentz scalar, and the potential energy is not a Lorentz scalar. But it can not be deduced to that the internal potential energy of born rigid body will change.

Whether a quantity is conserved is not necessarily related to whether or not it is Lorenz scalar.

In addition, I posted errors on post#30, #31，#34 ，#35 and #37. You can skip it and ignore it.

 

[PAllen]

Let's get back to some points early in this discussion made by Pervect and myself, and try to sharpen them up.

A consistent dynamical model in SR requires vanishing divergence of the stress energy tensor (SET). It is only this property that leads to global conservation laws for general closed systems in SR. Your model of a rotating rod whose elements consist of nothing but rest energy in their local MCIF cannot be a valid description because you have a net rate of change of 4-momentum of the 'rod particles' with no local compensation to make the SET divergence zero. To achieve this you have to add tension, and you cannot bundle this into rest energy or you recreate same divergence failure. Then, when transforming the SET (from the element MCIF) to another frame to express it in some overall inertial coordinates, the tension components of the SET mix with the energy density, totally separately from the affect of the boost on the energy. In your K' frame, the result is that the tension and energy density of a rod element expressed in these coordinates are both time varying.

Your approach of claiming only rest energy exists in an element's MCIF will never work consistently for these reasons.

 

[liuxinhua]

If the divergence of a physical quantity is vanishing, the physical quantity is conserved.

For the example at post#26, the divergence of potential energy is vanishing can be used to get the potential energy in $K'$ (remains unchanged) at the moments before and after the acceleration.

We can also get the potential energy unchanged at the moments before and after the acceleration by the distance between points and points unchanged.
Of course, this is not enough to show that the potential energy of rods (in my model ) remains unchanged in $K'$.

But I haven’t effective method to calculate the divergence of the rod’s potential energy.
Perhaps this problem has to be shelved temporarily.

 

[vanhees71]

I've no clue what you mean by "divergence of the potential energy". The divergence is an operation on a vector field. In component it reads $\partial_{\mu} j^{\mu}$. If (and only if!) you have
$$\partial_{\mu} j^{\mu}=0,$$
you can show that the quantity
$$Q(t)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} j^0(t,\vec{x})$$
is a Lorentz scalar and that it is conserved
$$\dot{Q}(t)=0.$$

 

[liuxinhua]

Thank PAllen and vanhees71 providing two methods to determine whether the potential energy of an object is conservation.

What I want to emphasize is that born rigid body have its own characteristics. The distance between any two particles remains unchanged.

 

[pervect]

I've no clue what you mean by "divergence of the potential energy". The divergence is an operation on a vector field. In component it reads $\partial_{\mu} j^{\mu}$. If (and only if!) you have
$$\partial_{\mu} j^{\mu}=0,$$
you can show that the quantity
$$Q(t)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} j^0(t,\vec{x})$$
is a Lorentz scalar and that it is conserved
$$\dot{Q}(t)=0.$$

I believe what liuxinhua is actually after is expressing the relativistic version of the conservation of energy. Which becomes an expression of the conservation of energy-momentum in the covariant formulation.

I've suggested a few times now that the vanishing of the divergence of the stress-energy tensor is the conservation law that he needs. But he doesn't seem to be listening, it's hard to say why exactly.

 

[liuxinhua]

We can use the method mentioned by PAllen and vanhees71 to determine whether the energy (potential energy of a rigid body) is conserved.
Using methods similar to the method mentioned by PAllen and vanhees71, we can get the potential energy of rod AB at $t_1’$ is less than it at $t_3’$。

What PAllen said is a necessary and sufficient condition for the conservation of energy.
What vanhees71 said is another necessary and sufficient condition for the conservation of energy.
Is there another necessary and sufficient condition for the conservation of the potential energy of a rigid body?

What I express is that the potential energy of a rigid body will not change directly from the definition of potential energy.

It is like "we have always thought that when a proton is accelerating, its rest mass remains unchanged".
But when a proton is accelerating, it can not be considered that the proton is a particle. When a proton accelerating, its invariant mass will change.

But I still think that the proton potential energy does not change when a proton accelerates. It is the relative velocity of each component which is changing, and then the kinetic energy of each component that causes the change of its invariant mass, when a proton accelerating. And the sum rest masses of a proton's entire component will not change when a proton accelerating.

 

[liuxinhua]

If a proton is considered a particle, its rest mass will not change.

When a proton is considered to be a rigid body with a volume, and at acceleration, the velocities of each component is different, measured in an inertial reference frame. Its invariant mass will change.
But the distance between the components will remain unchanged.

 

[vanhees71]

A proton is a complicated bound state of quarks and gluons and as such can be excited, and these excited states manifest themselves as many baryon resonances.

 

[PeterDonis]

When a proton is considered to be a rigid body with a volume, and at acceleration, the velocities of each component is different, measured in an inertial reference frame. Its invariant mass will change.
But the distance between the components will remain unchanged.

A proton is not composed of a fixed number of components at rest relative to each other. You need to stick to classical objects.

 

[liuxinhua]

A proton is not composed of a fixed number of components at rest relative to each other. You need to stick to classical objects.

I just want to say that proton is not a particle. It is made up of smaller components. Of course, this composition can not be described by classical physics.

 

[PeterDonis]

I just want to say that proton is not a particle. It is made up of smaller components. Of course, this composition can not be described by classical physics.

You're just repeating what I've already said. But you failed to notice the key additional thing I said: a proton is not made up of a fixed number of smaller components. So the "distance between components" for a proton is not even well-defined.

 

[PeterDonis]

At this point everything that could be said relevant to the thread topic has been said. Thread closed.

 

[PeterDonis]

@PAllen has provided a good reference that is relevant to this thread topic:

https://arxiv.org/abs/gr-qc/0411145