2 deleted 1 characters in body edited Jul 30 '12 at 12:02 Grisha Kirilin 1,9921010 silver badges1717 bronze badges There is no gravitational waves for a rotating uniformly axiallyrotating axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fieldsThe classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the field of an axially symmetric body rotating uniformly about its axis still also be constant. However in this case the time directions are no longer equivalent by any means -- if the sign of time is changed, the sign of the angular velocity is changed. Therefore in such constant gravitational fields (we shell call them stationary fields) the components $$g_{0i}$$ of the metric tensor are in general different from zero. The reason is very simple. For an axially symmetric body, the distribution of mass in the lab frame coincides with that in the rotating system, thus the solution of Einstein equation can be found in the rotating system where the body and metric are static and then in the lab frame by means of $$r'=r$$, $$z' = z$$, $$\phi' = \phi + \Omega t$$ coordinate transformation ($$r$$, $$\phi$$, $$z$$ are cylindrical coordinates). Therefore all derivatives $$\partial x_{\alpha}/\partial x_{\beta}$$ do not depend on time. Hence the metric of an rotatinga uniformly rotating axially symmetric body is time-independent. For example, the component $$g_{0i}$$ outside of the slow rotating body ($$M\ll c m r_{g}$$) has the form: $$g_{0i}=-\frac{2G}{c^{3}}\,M_{ij} \frac{n_{j}}{r^{2}},$$ where $$M_{ij}$$ is the total angular momentum antisymmetric tensor. UPD. Concerning the answer of David Bar Moshe. The conclusions presented above are valid only for axial symmetric body. The total power radiated by an non axial symmetric body is proportional to the third power of difference of inertia tensor eigen values for axes transverse to the rotation axis, i.e., $$(I_{1}-I_{2})^{3}$$. There is no gravitational waves for a rotating uniformly axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the field of an axially symmetric body rotating uniformly about its axis still also be constant. However in this case the time directions are no longer equivalent by any means -- if the sign of time is changed, the sign of the angular velocity is changed. Therefore in such constant gravitational fields (we shell call them stationary fields) the components $$g_{0i}$$ of the metric tensor are in general different from zero. The reason is very simple. For an axially symmetric body, the distribution of mass in the lab frame coincides with that in the rotating system, thus the solution of Einstein equation can be found in the rotating system where the body and metric are static and then in the lab frame by means of $$r'=r$$, $$z' = z$$, $$\phi' = \phi + \Omega t$$ coordinate transformation ($$r$$, $$\phi$$, $$z$$ are cylindrical coordinates). Therefore all derivatives $$\partial x_{\alpha}/\partial x_{\beta}$$ do not depend on time. Hence the metric of an rotating uniformly axially symmetric body is time-independent. For example, the component $$g_{0i}$$ outside of the slow rotating body ($$M\ll c m r_{g}$$) has the form: $$g_{0i}=-\frac{2G}{c^{3}}\,M_{ij} \frac{n_{j}}{r^{2}},$$ where $$M_{ij}$$ is the total angular momentum antisymmetric tensor. UPD. Concerning the answer of David Bar Moshe. The conclusions presented above are valid only for axial symmetric body. The total power radiated by an non axial symmetric body is proportional to the third power of difference of inertia tensor eigen values for axes transverse to the rotation axis, i.e., $$(I_{1}-I_{2})^{3}$$. There is no gravitational waves for a uniformly rotating axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the field of an axially symmetric body rotating uniformly about its axis still also be constant. However in this case the time directions are no longer equivalent by any means -- if the sign of time is changed, the sign of the angular velocity is changed. Therefore in such constant gravitational fields (we shell call them stationary fields) the components $$g_{0i}$$ of the metric tensor are in general different from zero. The reason is very simple. For an axially symmetric body, the distribution of mass in the lab frame coincides with that in the rotating system, thus the solution of Einstein equation can be found in the rotating system where the body and metric are static and then in the lab frame by means of $$r'=r$$, $$z' = z$$, $$\phi' = \phi + \Omega t$$ coordinate transformation ($$r$$, $$\phi$$, $$z$$ are cylindrical coordinates). Therefore all derivatives $$\partial x_{\alpha}/\partial x_{\beta}$$ do not depend on time. Hence the metric of a uniformly rotating axially symmetric body is time-independent. For example, the component $$g_{0i}$$ outside of the slow rotating body ($$M\ll c m r_{g}$$) has the form: $$g_{0i}=-\frac{2G}{c^{3}}\,M_{ij} \frac{n_{j}}{r^{2}},$$ where $$M_{ij}$$ is the total angular momentum antisymmetric tensor. UPD. Concerning the answer of David Bar Moshe. The conclusions presented above are valid only for axial symmetric body. The total power radiated by an non axial symmetric body is proportional to the third power of difference of inertia tensor eigen values for axes transverse to the rotation axis, i.e., $$(I_{1}-I_{2})^{3}$$. 1 answered Jul 30 '12 at 11:55 Grisha Kirilin 1,9921010 silver badges1717 bronze badges There is no gravitational waves for a rotating uniformly axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the field of an axially symmetric body rotating uniformly about its axis still also be constant. However in this case the time directions are no longer equivalent by any means -- if the sign of time is changed, the sign of the angular velocity is changed. Therefore in such constant gravitational fields (we shell call them stationary fields) the components $$g_{0i}$$ of the metric tensor are in general different from zero. The reason is very simple. For an axially symmetric body, the distribution of mass in the lab frame coincides with that in the rotating system, thus the solution of Einstein equation can be found in the rotating system where the body and metric are static and then in the lab frame by means of $$r'=r$$, $$z' = z$$, $$\phi' = \phi + \Omega t$$ coordinate transformation ($$r$$, $$\phi$$, $$z$$ are cylindrical coordinates). Therefore all derivatives $$\partial x_{\alpha}/\partial x_{\beta}$$ do not depend on time. Hence the metric of an rotating uniformly axially symmetric body is time-independent. For example, the component $$g_{0i}$$ outside of the slow rotating body ($$M\ll c m r_{g}$$) has the form: $$g_{0i}=-\frac{2G}{c^{3}}\,M_{ij} \frac{n_{j}}{r^{2}},$$ where $$M_{ij}$$ is the total angular momentum antisymmetric tensor. UPD. Concerning the answer of David Bar Moshe. The conclusions presented above are valid only for axial symmetric body. The total power radiated by an non axial symmetric body is proportional to the third power of difference of inertia tensor eigen values for axes transverse to the rotation axis, i.e., $$(I_{1}-I_{2})^{3}$$.