- #1
the one
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Hi
assume that a Cylinder with radius \[<br /> R<br /> \] , proper mass \[<br /> M_0 <br /> \] and height \[<br /> h<br /> \] which is rotating at a constant angular speed \[<br /> \omega <br /> \]
In order to calculate the relativistic mass we use the proper mass element to calculate the relativistic mass element , so :
\[<br /> dM = \frac{{dM_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}<br /> \]
But \[<br /> dM_0 = \rho _0 dV_0 <br /> \] where \[<br /> \rho _0 <br /> \] is the proper mass density and \[<br /> dV_0 <br /> \] is the proper volume element . so :
\[<br /> \begin{array}{l}<br /> dM = \frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\ <br /> M = \int\limits_V {\frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} = \int\limits_0^R {\int\limits_0^{2\pi } {\int\limits_0^h {\frac{{\rho _0 rdrd\phi dz}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} } } = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} \\ <br /> but:v = \omega r \\ <br /> M = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{\omega ^2 }}{{c^2 }}r^2 } }}} \\ <br /> \end{array}<br /> \]
Now , make the substitution
\[<br /> u = 1 - \frac{{\omega ^2 }}{{c^2 }}r^2 \Rightarrow du = - 2\frac{{\omega ^2 }}{{c^2 }}rdr \Rightarrow 2rdr = - \frac{{c^2 }}{{\omega ^2 }}du<br /> \]
so :
\[<br /> \begin{array}{l}<br /> M = - \frac{{\pi h\rho _0 c^2 }}{{\omega ^2 }}\int\limits_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } {\frac{{du}}{{\sqrt u }}} = - \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left[ {\sqrt u } \right]_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } = \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> but:M_0 = \rho _0 V_0 = \pi R^2 h\rho _0 \\ <br /> M = \frac{{2M_0 c^2 }}{{R^2 \omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> \end{array}<br /> \]
now , there is something make me confused in this equation . If we put \[<br /> \omega R = c<br /> \] we find that the relativistic mass is \[<br /> M = 2M_0 <br /> \] . How it can be ?
I know that any thing has a v = c it's mass goes to infinity .
Again , How it can be ?
Thanks
assume that a Cylinder with radius \[<br /> R<br /> \] , proper mass \[<br /> M_0 <br /> \] and height \[<br /> h<br /> \] which is rotating at a constant angular speed \[<br /> \omega <br /> \]
In order to calculate the relativistic mass we use the proper mass element to calculate the relativistic mass element , so :
\[<br /> dM = \frac{{dM_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}<br /> \]
But \[<br /> dM_0 = \rho _0 dV_0 <br /> \] where \[<br /> \rho _0 <br /> \] is the proper mass density and \[<br /> dV_0 <br /> \] is the proper volume element . so :
\[<br /> \begin{array}{l}<br /> dM = \frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\ <br /> M = \int\limits_V {\frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} = \int\limits_0^R {\int\limits_0^{2\pi } {\int\limits_0^h {\frac{{\rho _0 rdrd\phi dz}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} } } = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} \\ <br /> but:v = \omega r \\ <br /> M = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{\omega ^2 }}{{c^2 }}r^2 } }}} \\ <br /> \end{array}<br /> \]
Now , make the substitution
\[<br /> u = 1 - \frac{{\omega ^2 }}{{c^2 }}r^2 \Rightarrow du = - 2\frac{{\omega ^2 }}{{c^2 }}rdr \Rightarrow 2rdr = - \frac{{c^2 }}{{\omega ^2 }}du<br /> \]
so :
\[<br /> \begin{array}{l}<br /> M = - \frac{{\pi h\rho _0 c^2 }}{{\omega ^2 }}\int\limits_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } {\frac{{du}}{{\sqrt u }}} = - \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left[ {\sqrt u } \right]_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } = \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> but:M_0 = \rho _0 V_0 = \pi R^2 h\rho _0 \\ <br /> M = \frac{{2M_0 c^2 }}{{R^2 \omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> \end{array}<br /> \]
now , there is something make me confused in this equation . If we put \[<br /> \omega R = c<br /> \] we find that the relativistic mass is \[<br /> M = 2M_0 <br /> \] . How it can be ?
I know that any thing has a v = c it's mass goes to infinity .
Again , How it can be ?
Thanks
