Physics Help Forum The potential energy of born rigid body is a constant ？

 Special and General Relativity Special and General Relativity Physics Help Forum

 Sep 1st 2018, 07:13 PM #1 Junior Member   Join Date: Aug 2016 Posts: 20 The potential energy of born rigid body is a constant ？ $\displaystyle \ \ \ \ \$The sum of rest masses of a born rigid body consists of two parts: the potential energy of a born rigid body, and the actual particles or the material in the form of physical existence. $\displaystyle \ \ \ \ \$ For a born rigid body, there is no conversion between matter and energy. The actual particles or the material in the form of physical existence of a rigid body is constant. So , if the potential energy of born rigid body is a constant, the sum of rest masses of a born rigid body is a constant. $\displaystyle \ \ \ \ \$If rod $\displaystyle AB$ is a Born rigid body, sum of rest masses of every section $\displaystyle DM$ of rod $\displaystyle AB$ is a constant measured in any inertial reference frame. There are $\displaystyle { m_{sum\ 0 }({t_1} ’, DM) } ’ = { m_{sum\ 0 }({t_2} ’, DM) } ’$ (in $\displaystyle K’$),=$\displaystyle { m_{sum\ 0 }({t_3} , DM) }$ （ in $\displaystyle K$ ）. $\displaystyle \ \ \ \ \$So I want to know whether “The potential energy of born rigid body is a constant measured in an inertial reference frame at different time?”, the born rigid body may be in non-inertial motion. $\displaystyle \ \ \ \ \$ We consider a born rigid body model. $\displaystyle \ \ \ \ \$ In $\displaystyle K$ , an isolated rod $\displaystyle AB$ rotates around its mid point $\displaystyle O$ at uniform angular velocity $\displaystyle ω$ , keeps the linear state. For actual rod $\displaystyle AB$, there is micro motion on molecular level. In the process of simplifying and establishing a mathematical model, we do not consider this micro motion. The rod $\displaystyle AB$ is an ideal system without any relative movement between the components relative to $\displaystyle K$ . The distribution of material and potential energy on the rod $\displaystyle AB$ along the length is symmetry $\displaystyle AB$ out mid point $\displaystyle O$ and it does not change with time, measured in $\displaystyle K$ . Rod $\displaystyle AB$ has no thickness along $\displaystyle θ$ direction, if transform $\displaystyle K$ from rectangular coordinate system to polar coordinate system. That is to say consider rod $\displaystyle AB$ as a one-dimensional rod on $\displaystyle x-y$ plane relative to $\displaystyle K$ . $\displaystyle \ \ \ \ \$ For each point on $\displaystyle AB$, the distance from point $\displaystyle O$ to this point measured in $\displaystyle K$ is constant; denote it is $\displaystyle r$ . Such as for point $\displaystyle D$，$\displaystyle r_D$ is a constant. For each point on $\displaystyle OA$, its equation of motion is $\displaystyle x=r\cos(ωt)$ , $\displaystyle y=r\sin(ωt)$ . For each point on $\displaystyle OB$, its equation of motion is $\displaystyle x=r\cos(ωt+π)$ , $\displaystyle y=r\sin(ωt+π)$ . $\displaystyle \ \ \ \ \$The inertial reference frame $\displaystyle K’$ moves relative to $\displaystyle K$ at speed $\displaystyle v$. Last edited by Liuxinhua; Sep 1st 2018 at 07:21 PM.
Sep 1st 2018, 07:21 PM   #2

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 Originally Posted by Liuxinhua $\displaystyle \ \ \ \ \$The sum of rest masses of a born rigid body consists of two parts: the potential energy of a born rigid body, and the actual particles or the material in the form of physical existence. $\displaystyle \ \ \ \ \$ For a born rigid body, there is no conversion between matter and energy. The actual particles or the material in the form of physical existence of a rigid body is constant. So , if the potential energy of born rigid body is a constant, the sum of rest masses of a born rigid body is a constant. $\displaystyle \ \ \ \ \$If rod $\displaystyle AB$ is a Born rigid body, sum of rest masses of every section $\displaystyle DM$ of rod $\displaystyle AB$ is a constant measured in any inertial reference frame. There are $\displaystyle { m_{sum\ 0 }({t_1} ’, DM) } ’ = { m_{sum\ 0 }({t_2} ’, DM) } ’$ (in $\displaystyle K’$),=$\displaystyle { m_{sum\ 0 }({t_3} , DM) }$ （ in $\displaystyle K$ ）. $\displaystyle \ \ \ \ \$So I want to know whether “The potential energy of born rigid body is a constant measured in an inertial reference frame at different time?” the born rigid body may be in non-inertial motion. $\displaystyle \ \ \ \ \$ We consider a born rigid body model. $\displaystyle \ \ \ \ \$ In $\displaystyle K$ , an isolated rod $\displaystyle AB$ rotates around its mid point $\displaystyle O$ at uniform angular velocity $\displaystyle ω$ , keeps the linear state. For actual rod $\displaystyle AB$, there is micro motion on molecular level. In the process of simplifying and establishing a mathematical model, we do not consider this micro motion. The rod $\displaystyle AB$ is an ideal system without any relative movement between the components relative to $\displaystyle K$ . The distribution of material and potential energy on the rod $\displaystyle AB$ along the length is symmetry $\displaystyle AB$ out mid point $\displaystyle O$ and it does not change with time, measured in $\displaystyle K$ . Rod $\displaystyle AB$ has no thickness along $\displaystyle θ$ direction, if transform $\displaystyle K$ from rectangular coordinate system to polar coordinate system. That is to say consider rod $\displaystyle AB$ as a one-dimensional rod on $\displaystyle x-y$ plane relative to $\displaystyle K$ . $\displaystyle \ \ \ \ \$ For each point on $\displaystyle AB$, the distance from point $\displaystyle O$ to this point measured in $\displaystyle K$ is constant; denote it is $\displaystyle r$ . Such as for point $\displaystyle D$，$\displaystyle r_D$ is a constant. For each point on $\displaystyle OA$, its equation of motion is $\displaystyle x=r\cos(ωt)$ , $\displaystyle y=r\sin(ωt)$ . For each point on $\displaystyle OB$, its equation of motion is $\displaystyle x=r\cos(ωt+π)$ , $\displaystyle y=r\sin(ωt+π)$ . $\displaystyle \ \ \ \ \$The inertial reference frame $\displaystyle K’$ moves relative to $\displaystyle K$ at speed $\displaystyle v$.
Ummmmm... Okay.

What is a "born" mass? And what does $\displaystyle m_{sum\ 0 }({t_2} ’, DM) ’$ mean?

-Dan
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 Sep 1st 2018, 07:48 PM #3 Junior Member   Join Date: Aug 2016 Posts: 20 It's my model on the post#1, which satisfies the definition of born rigid body. Let's consider the model on the post#1. Whether the potential energy of it is constant in any inertial reference frame. $\displaystyle { m_{sum\ 0 }({t_2} ’, DM) } ’$ is the sum of rest masses of section $\displaystyle DM$ on rod $\displaystyle AB$ measured in $\displaystyle K’$ at time $\displaystyle {t_2} ’$.
Sep 1st 2018, 09:52 PM   #4

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 Originally Posted by Liuxinhua It's my model on the post#1, which satisfies the definition of born rigid body. Let's consider the model on the post#1. Whether the potential energy of it is constant in any inertial reference frame. $\displaystyle { m_{sum\ 0 }({t_2} ’, DM) } ’$ is the sum of rest masses of section $\displaystyle DM$ on rod $\displaystyle AB$ measured in $\displaystyle K’$ at time $\displaystyle {t_2} ’$.
And these are? I don't understand how a rigid body can be "born." Perhaps this translates badly to English?

Let's try the other one this way: How do you sum over rest masses if you only have one mass?

-Dan
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 Sep 2nd 2018, 12:43 AM #5 Junior Member   Join Date: Aug 2016 Posts: 20 $\displaystyle \ \ \ \ \$ We consider a model. $\displaystyle \ \ \ \ \$ In $\displaystyle K$ , an isolated rod $\displaystyle AB$ rotates around its mid point $\displaystyle O$ at uniform angular velocity $\displaystyle ω$ , keeps the linear state. For actual rod $\displaystyle AB$, there is micro motion on molecular level. In the process of simplifying and establishing a mathematical model, we do not consider this micro motion. The rod $\displaystyle AB$ is an ideal system without any relative movement between the components relative to $\displaystyle K$ . The distribution of material and potential energy on the rod $\displaystyle AB$ along the length is symmetry $\displaystyle AB$ out mid point $\displaystyle O$ and it does not change with time, measured in $\displaystyle K$. Rod $\displaystyle AB$ has no thickness along $\displaystyle θ$ direction, if transform $\displaystyle K$ from rectangular coordinate system to polar coordinate system. That is to say consider rod $\displaystyle AB$ as a one-dimensional rod on $\displaystyle x-y$ plane relative to $\displaystyle K$ . $\displaystyle \ \ \ \ \$ For each point on $\displaystyle AB$, the distance from point $\displaystyle O$ to this point measured in $\displaystyle K$ is constant; denote it is $\displaystyle r$ . Such as for point $\displaystyle D$，$\displaystyle r_D$ is a constant. For each point on $\displaystyle OA$, its equation of motion is $\displaystyle x=r\cos(ωt)$ , $\displaystyle y=r\sin(ωt)$ . For each point on $\displaystyle OB$, its equation of motion is $\displaystyle x=r\cos(ωt+π)$ , $\displaystyle y=r\sin(ωt+π)$ . $\displaystyle \ \ \ \ \$The inertial reference frame $\displaystyle K’$ moves relative to $\displaystyle K$ at speed $\displaystyle v$. $\displaystyle \ \ \ \ \$ Is its potential energy a constant measured in an inertial reference frame ($\displaystyle K$ or $\displaystyle K’$) at different time.
Sep 2nd 2018, 02:08 AM   #6

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 Originally Posted by Liuxinhua $\displaystyle \ \ \ \ \$ We consider a model. $\displaystyle \ \ \ \ \$ In $\displaystyle K$ , an isolated rod $\displaystyle AB$ rotates around its mid point $\displaystyle O$ at uniform angular velocity $\displaystyle ω$ , keeps the linear state. For actual rod $\displaystyle AB$, there is micro motion on molecular level. In the process of simplifying and establishing a mathematical model, we do not consider this micro motion. The rod $\displaystyle AB$ is an ideal system without any relative movement between the components relative to $\displaystyle K$ . The distribution of material and potential energy on the rod $\displaystyle AB$ along the length is symmetry $\displaystyle AB$ out mid point $\displaystyle O$ and it does not change with time, measured in $\displaystyle K$. Rod $\displaystyle AB$ has no thickness along $\displaystyle θ$ direction, if transform $\displaystyle K$ from rectangular coordinate system to polar coordinate system. That is to say consider rod $\displaystyle AB$ as a one-dimensional rod on $\displaystyle x-y$ plane relative to $\displaystyle K$ . $\displaystyle \ \ \ \ \$ For each point on $\displaystyle AB$, the distance from point $\displaystyle O$ to this point measured in $\displaystyle K$ is constant; denote it is $\displaystyle r$ . Such as for point $\displaystyle D$，$\displaystyle r_D$ is a constant. For each point on $\displaystyle OA$, its equation of motion is $\displaystyle x=r\cos(ωt)$ , $\displaystyle y=r\sin(ωt)$ . For each point on $\displaystyle OB$, its equation of motion is $\displaystyle x=r\cos(ωt+π)$ , $\displaystyle y=r\sin(ωt+π)$ . $\displaystyle \ \ \ \ \$The inertial reference frame $\displaystyle K’$ moves relative to $\displaystyle K$ at speed $\displaystyle v$. $\displaystyle \ \ \ \ \$ Is its potential energy a constant measured in an inertial reference frame ($\displaystyle K$ or $\displaystyle K’$) at different time.
I don't see that any potential energy the rod has would vary. You haven't introduced anything that could act as such potential. All the rod has is total energy E and this is due to it's rotational kinetic energy (and, of course, its rest energy.)

Now if you were doing this in a gravitational field then, yes, the potential energy could be affected. On the other hand, that gets us into General Relativity territory and that's going to make things messy.

-Dan
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 Sep 2nd 2018, 04:10 AM #7 Junior Member   Join Date: Aug 2016 Posts: 20 Don’t put it in a gravitational field. The influence of gravity is not considered. I mean the internal potential energy of the rod at the molecular level or the smaller space. The rod has is total energy E (constant) and its rest energy (will change). The rod will bent observed in $\displaystyle K’$.The shape of the rod in $\displaystyle K$ and $\displaystyle K’$ is like a line in the picture below. Attached Thumbnails
 Sep 2nd 2018, 07:45 AM #8 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 987 I think you are making things very difficult with unusual terminology. Is a Born rigid body anything to do with the contributions of Max Born to elasticity? https://en.wikipedia.org/wiki/Max_Born As I understand your discussion you are saying that the tangential acceleration of the particles of a whirling rod increases along its length away from the pivot to the rim. This is equivalent to a force in Newtonian mechanics which is resisted by the flexural rigidity of the rod. There are formulae that were developed by Elasticity pioneers Prescott, Southwell and Inglis for this effect in rotating elastic disks, when bursting of turbines and shafts became an issue in the early 1900s. topsquark likes this.
 Sep 3rd 2018, 07:17 AM #9 Junior Member   Join Date: Aug 2016 Posts: 20 Its potential energy is a constant measured in inertial reference frame $\displaystyle K$ at different time. I consider this problem in special relativity. I want to calculate the momentum of the rod in $\displaystyle K’$. So I want to know whether the potential energy of the rod in $\displaystyle K’$ at different time is a constant.. About born rigid body, can see : https://en.wikipedia.org/wiki/Born_r...metric_motions The sum rest masses is “cut the rod into numerous segment, then sum each segment’s rest mass”.
 Sep 3rd 2018, 11:24 AM #10 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 987 Surely if a subject is worth considering and discussing, it is worth writing the few more lines needed to make your posts intelligible. topsquark likes this.

 Tags body, born, born rigid, constant, energy, potential, potential energy, rigid

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