I realized we are both right. At a lower gravitational potential the perception of energy is different. People seldom talk of "proper energy" and "coordinate energy" but that is basically what it is. My expression is correct for a distant observer and your expression is correct for an observer...
I thought that at infinity you have the energy ##E=mc^2## for each of the two particles. After the collision you have ##E=mc^2\sqrt{1-\frac{2GM}{rc^2}}## for each of the two particles. Inserting ##r=3GM/c^2## into the latter expression you get ##E=\frac{mc^2}{\sqrt{3}}## for each of the two...
I think the energy of a nonmoving object in the Swarzschild approximation is:
E=mc^2(1-\sqrt{1-\frac{2GM}{rc^2}})
In that case you get a total energy loss in the collision of \frac{2mc^2}{\sqrt{3}}
In any case you can not get more energy out from the collision than ##2mc^2## which is the...
Well obviously for the Earth and the Sun you can just plug in the known values for the distances and the masses into the known formula for gravitational time dilation and get the results. I was wondering about the next level, how aggregations of stars in our galaxy, our galaxy in total, clusters...
As proven experimentally clocks tick slower deep in a gravitational well and the difference in energy levels between atomic/molecular quantum state also becomes smaller deep in a gravitational well. This is sometimes known as "gravitational time dilation" and "gravitational redshift" I think...
I think that for circular orbits and Schwarzschild metric you always get the same orbital velocity as in classical Newtonian theory if you use "coordinate time"? I mean if you want to calculate the orbits of objects in the solar system you seldom really care about what time the objects...
Okey then. Do note that the paper you have been referring to does just that. Replacing ##m## with ##\gamma{(\mathbf{v})}m## on the left side of ##\frac{d}{dt}(m\mathbf{v})=-\frac{GMm}{r^3}\mathbf{r}## and see where it leads is what the paper is all about.
Actually I do not no if it is...
Note that if you take the classical
##\frac{d}{dt}{(m \mathbf{v})}=-\frac{GMm}{r^3} \mathbf{r}##
and replace ##m## with ##\gamma(\mathbf{v})m##
on both sides of the equation you end up with:
##-\frac{GM}{r^3}\mathbf{r}=\gamma(\mathbf{v})^2 \, \mathbf{a}_{\parallel} + \...
It is easy to run (rewriting sligthly)
##\mathbf{a}=-\frac{GM}{r^3}{\mathbf{r} \cdot \mathbf{v}}\frac{1}{v}(1-\frac{v^2}{c^2}) + \frac{GM}{r^3}\frac{1}{v^2}\mathbf{r}\times\mathbf{v}\times{\mathbf{v}}##
in a numerical integrator and see that you get precisely one third, I do not now how to do...
I am sure you are correct but if you want to calculate how long time it will take for a signal to travel from Earth to a distant spacecraft with and without Jupiter interfering you will get the correct result (I think) if you assume that Fermats principle holds and that the light slows down...
One typical example is of course when you use the Lorentz force as in accelerators that accelerates particles electromagnetically. Then you can use the formula precisely as written by pervect above:
##\frac{q}{m_0}(\mathbf{E}+\mathbf{v}\times\mathbf{B})=\gamma(\mathbf{v})^3 \...
I am sure you are right, yet it is disturbing that the fact that ##E=mc^2## does not hold up when gravity is involved is, as far as I can see, not mentioned at all at for instance the wikipedia page on the relation:
http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence
It is a bit...
Hmm... My interpretation of the Komar mass as described in wikipedia is that, for a stationary object in a spherically symmetric gravitational field, to get the gravitational acceleration right, one must replace:
##\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}##
with...
Let us say that we have small object of mass ##m## at some location far away from the Earth (with zero velocity compared to the earth). The energy of this object is according to relativity ##E=mc^2##.
Now we drop this object and it starts falling towards the earth, transforming potential...
I think Fermats principle holds even in general relativity. If you for instance is sending an electromagnetic signal from Earth to some spacecraft beyond Jupiter, the signal will bend in the gravitational field of Jupiter, but in such a way that the time for the signal to travel from the Earth...