- #1
the one
- 13
- 0
Hi
assume that a Cylinder with radius [tex]\[
R
\][/tex] , proper mass [tex]\[
M_0
\][/tex] and height [tex]\[
h
\][/tex] which is rotating at a constant angular speed [tex]\[
\omega
\][/tex]
In order to calculate the relativistic mass we use the proper mass element to calculate the relativistic mass element , so :
[tex]\[
dM = \frac{{dM_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}
\][/tex]
But [tex]\[
dM_0 = \rho _0 dV_0
\][/tex] where [tex]\[
\rho _0
\][/tex] is the proper mass density and [tex]\[
dV_0
\][/tex] is the proper volume element . so :
[tex]\[
\begin{array}{l}
dM = \frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\
M = \int\limits_V {\frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} = \int\limits_0^R {\int\limits_0^{2\pi } {\int\limits_0^h {\frac{{\rho _0 rdrd\phi dz}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} } } = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} \\
but:v = \omega r \\
M = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{\omega ^2 }}{{c^2 }}r^2 } }}} \\
\end{array}
\][/tex]
Now , make the substitution
[tex]\[
u = 1 - \frac{{\omega ^2 }}{{c^2 }}r^2 \Rightarrow du = - 2\frac{{\omega ^2 }}{{c^2 }}rdr \Rightarrow 2rdr = - \frac{{c^2 }}{{\omega ^2 }}du
\][/tex]
so :
[tex]\[
\begin{array}{l}
M = - \frac{{\pi h\rho _0 c^2 }}{{\omega ^2 }}\int\limits_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } {\frac{{du}}{{\sqrt u }}} = - \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left[ {\sqrt u } \right]_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } = \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\
but:M_0 = \rho _0 V_0 = \pi R^2 h\rho _0 \\
M = \frac{{2M_0 c^2 }}{{R^2 \omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\
\end{array}
\][/tex]
now , there is something make me confused in this equation . If we put [tex]\[
\omega R = c
\][/tex] we find that the relativistic mass is [tex]\[
M = 2M_0
\][/tex] . 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 [tex]\[
R
\][/tex] , proper mass [tex]\[
M_0
\][/tex] and height [tex]\[
h
\][/tex] which is rotating at a constant angular speed [tex]\[
\omega
\][/tex]
In order to calculate the relativistic mass we use the proper mass element to calculate the relativistic mass element , so :
[tex]\[
dM = \frac{{dM_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}
\][/tex]
But [tex]\[
dM_0 = \rho _0 dV_0
\][/tex] where [tex]\[
\rho _0
\][/tex] is the proper mass density and [tex]\[
dV_0
\][/tex] is the proper volume element . so :
[tex]\[
\begin{array}{l}
dM = \frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\
M = \int\limits_V {\frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} = \int\limits_0^R {\int\limits_0^{2\pi } {\int\limits_0^h {\frac{{\rho _0 rdrd\phi dz}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} } } = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} \\
but:v = \omega r \\
M = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{\omega ^2 }}{{c^2 }}r^2 } }}} \\
\end{array}
\][/tex]
Now , make the substitution
[tex]\[
u = 1 - \frac{{\omega ^2 }}{{c^2 }}r^2 \Rightarrow du = - 2\frac{{\omega ^2 }}{{c^2 }}rdr \Rightarrow 2rdr = - \frac{{c^2 }}{{\omega ^2 }}du
\][/tex]
so :
[tex]\[
\begin{array}{l}
M = - \frac{{\pi h\rho _0 c^2 }}{{\omega ^2 }}\int\limits_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } {\frac{{du}}{{\sqrt u }}} = - \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left[ {\sqrt u } \right]_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } = \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\
but:M_0 = \rho _0 V_0 = \pi R^2 h\rho _0 \\
M = \frac{{2M_0 c^2 }}{{R^2 \omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\
\end{array}
\][/tex]
now , there is something make me confused in this equation . If we put [tex]\[
\omega R = c
\][/tex] we find that the relativistic mass is [tex]\[
M = 2M_0
\][/tex] . 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