pervect said:
I am not quite sure I follow your derivation, but does knowing that ~E = 1 for a stationary object at r=infinity help?
(1-2GM/rc^2) * (dt/dtau)
is the correct expression for the "energy at infinity" which is conserved. Actually I probably should have called it the "specific energy at infinty", i.e. the energy at infinity per unit mass. That's what the tilde is for.
Yep that's what I got. From the previous post:
kmarinas86 said:
pervect said:
Take a look at
http://www.fourmilab.ch/gravitation/orbits/ for the orbital equations of a test particle orbiting a black hole. (Or consult a GR textbook, the equations on this webpage are pretty much taken directly from MTW's "Gravitation").
A conserved energy parameter for the orbit is written in this webpage as
~E. it is sometimes called the "energy at infinity". Similarly ~L is the conserved angular momentum. So any orbit will have constant values for ~E and ~L (if we ignore loses due to gravitational radiation).
How is this?
Givens:
\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}
dt=\frac{d\tau}{\sqrt{1-2M/r}} (is this still applicable here?)
Results:
\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}
\left(\sqrt{1-2M/r}\right)\frac{dt}{d\tau}=1
\sqrt{1-2M/r}=\tilde{E}
Is it really that simple...
The part that agrees is:
\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}
Converting from natural units in regular units, we know that M becomes M*G/c^2.
Notice the lack of a square root.
pervect said:
When I say mass above in "energy per unit mass", I mean invariant mass. The invariant mass of an object does not depend on its state of motion. If you think of a 1 kg test mass like they use at the bureau of standards, a platinium-iridium bar, you can imagine the 1kg mass I mean above as the mass of a physical copy of that platinium-iridium bar.
Note that at r=infinity, for a stationary object dt/dtau = 1, hence the energy at infinity per unit mass is 1 as I said before. So if we have a 1kg platinum iridium standard bar motionless at r=infinity, this equation says that its energy is equal to its mass in geometric units, or mass*c^2 in non-geometric units.
For a rapidly moving object, dt/dtau > 1 and hence the energy at infinity per unit mass is greater than 1. In fact, it is equal to the factor gamma, so that if our bar were moving with a time dilation factor gamma , it's energy at infinity would be gamma times its rest mass (i.e. gamma times its invariant mass).
This where my "results" i had ealier disagreed with the standard results. This is why:
kmarinas86 said:
Givens:
\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}
This agrees what with you said, but notice how the RHS (right hand side) is not written gamma. If \tilde{E} was gamma, this would make:
\frac{dt}{d\tau}=\frac{\gamma}{1-2M/r} ?
kmarinas86 said:
dt=\frac{d\tau}{\sqrt{1-2M/r}} (is this still applicable here?)
Results:
\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}
\left(\sqrt{1-2M/r}\right)\frac{dt}{d\tau}=1
\sqrt{1-2M/r}=\tilde{E}
Is it really that simple...
Notice that the third-to-last equation (variant of the first given) is divided by the second to last equation (variant of the second given) to get the first-to-last equation. So either the second given is incorrect for some reason or the results are right.
I'm thinking that I've been mixing "stationary" with "moving", so I'm not sure if what I did was correct.
pervect said:
The correct relationship between dt, dtau, dr, etc. is given by the Schwarzschild metric - I think this may have been what you were trying to express in your second equation, but I'm not sure, i.e:
dtau^2 = (1-2M/r) dtau^2 - dr^2 / (1-2M/r) - r^2 dtheta^2 - r^2 sin^2(theta) dphi^2
Actually
its:
http://upload.wikimedia.org/math/d/0/4/d04cde2566a8e949e0e88b2bd98815e7.png
I know. I know =P.
Still, I think we agree that:
\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r} is correct when the energy is
at infinity.
and
\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E} is correct when the energy is
at infinity.
What I just now realized is that (when the object is at r=\infty):
\left(1-2M/r\right)=1
Making:
\tilde{E}=\frac{dt}{d\tau}
But is there a case where both time:
\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}
from (http://www.fourmilab.ch/gravitation/orbits/) .
and, Gravitational time Dilation
dt=\frac{d\tau}{\sqrt{1-2M/r}}
...can be satisfied (exactly)... or is this impossible?
Are their constraints contradictory?
But apparently(!) this can only hold at infinity if \tilde{E}=1.
As for the first equation, why have r, if the equation only applies for infinity?
The weird thing is this, looking at the time part of the first equation, it is the time as perceived
from infinity. With the \tilde{E} part, it is energy
at infinity.
And if \tilde{E} is equal to \gamma, then I'm afraid that the two equations cannot co-exist with each other unless if \gamma=1 and r=\infty.
Once more, if both equations are correct then \tilde{E} must equal \sqrt{1-2M/r}. If scientists are right about these equations, and IF they can apply for the same test particle, what are we to make of \tilde{E}=\gamma?
Mr. Hillman, are you there? I believe you will know plenty about this... help us resolve this...
