Re: Ye Old One Is A Tit!
- From: josephus <dogbird@xxxxxxxxxxxxx>
- Date: Mon, 01 Oct 2007 08:29:25 -0500
spintronic wrote:
On 30 Sep, 16:02, josephus <dogb...@xxxxxxxxxxxxx> wrote:you are not the only one that makes a mistake. i meant to say T/(1-V^2/c^2) the larger the velocity the slower the clock. by the definitions we use here. T/(1 - (gBH/r)/c^2) ) collecting terms
spintronic wrote:
On 29 Sep, 17:57, josephus <dogb...@xxxxxxxxxxxxx> wrote:
spintronic wrote:
Hell look at that! Another Bob. You have so much in common.
The other Bob has tourettes and is repetitive as well!
I spent some time looking at the equation (1- E^2*V^2/c^2) = t1
Corr, another piece of chewing gum, how the hell did you track me
down?
I'll save you some time, that equation makes no sense!
If E = energy you are saying,
(1- m^2C^4V^2/C^2) = (1- m^2C^2V^2)
the point of my eqution was to associate a know mass relation with
velocity. then I said (1-V^2/c^2) = (1- (G*M/r)/c^2) if you want to
pretend <snip>
No you said "(1- E^2*V^2/c^2) = t1 "
t'= T/(1- GBH/(r*C^2))
GBH/r = V^2 then T/(1 - GBH/(c^2*r))
Now that in no way = t.
If E = escape velocity, you are saying
(1- V^4/C^2) which again in no way is equat to t'
snip> some gibberish! Then this.
3*10^10/ 299799285^2 = 333.78
Nope!
Ive already stated
I dont read minds and I dont read all of any other posts you make. some times you do stupid thing like say-mg a quadratic equation has more than 2 solutions.
please act sensibly.
SR = 2GM/C^2 = 4.45455458*10^3m
point (m) in most calculations is GM/c^2 and is the term for relativistic energy V^2 +1/r^2 = U*(2/r-1/A) + 2*m*h^2/r
for our sun 1.3271244E20/C^2 == 1476.6 km this is 'm'
you need to justify this as energy, because RV^2 is energy. and in orbital terms R*V^2 + m/r
r* GBH + GBH/c^2*r
GBH(R + 1/c^2*r) this becomes GBH*R to accuracy 1/c^2
GBH*r = V^2 * r this is the energy of a circular orbit.
E = GBH*r the object is in an orbit. it can be parabolic, hyperbolic, elliptical, circular or rectangular a degenerate form. but no matter what the matter is doing it must be a ballistic trace.
H is defined if e > 0 -> A*(1-e^2) = p h^2= aPH = 1/25m = 4*10^-2m
R = H + SR = 4.45459458*10^3m
Escape Velocity at R = sqrt(2GM/R) = 2.99791112004869*10^8m/s
this doe not relate to the orbit.
ether use gm/r = v^2 or U(2/r -1/A) +)1/r^-3m/r^3)
t' = t/ sqrt(1-(EV^2)/(C^2)) = 333.7s
a little thought and I recognized it is the ratio of seconds
between the observer and the object in a circular orbit. the object is
that much slower than the observers clock 1 to 3333.78 but the mass in
question is 10^10 solar masses.
No, no, no.
An observer in circular orbit has an additional time dilation thus
t'_extra = 1+ [(t'-t) + (t/sqrt(1- (v_orbital^2)/(C^2))-t)]
t'_extra, is in no way proportional to t'
as the Orbital V is equal to Escape velocity / sqrt(2).
Your Time dilation due to Escape velocity =
3.3271374*10^2s
Your Joint Time Dilation is only
3.3412795*10^2s
Do you see the problem?
Your maximum orbital velocity if it were allowed is C/sqrt(2)
t'_Orbilal max = sqrt(2)
light speed limitations are arbitrary and not associated with orbital
motions. it simply means you cannot go that fast for ANY REASON.
What, did I say they was? Rhetorical, don't bother!
you are sort of right but you seem confused.
lol, thats fresh. Mr C^C.
you get excited and leave out how the eqaution was predicated. where did all of your powers come from.
Thats due to my hard work, if you want to know how the equation was
derived, do your own work!
the obsever is not in orbit
Never said he was.
the object is and the observer sees a time
"extension" the object has a slower clock than the observer. which are
in different frames. in each frame 1 second is 1 second. but due to
the black hole the objects clock is slower than the observer. in
addition the limit for any velocity is C. this means an object cannot
be faster and will fall into the black hole.
And??
the eqaution V^2= gm/r posts a limit. when sqrt(gm/r) tries to exceed
the speed of light then all objects will fall into the black holr
the value for sqrt(GBH/r) = c is the inner most radius that any matter can exist and not fall in. the event horison is here.
sqrt(Gm/r) Never gets anywhere neer the speed of light.
The closest it gets is C/sqrt(2).
that is false. you are saying that an object orbiting a BH will never go faster than C/sqrt(2) ? if the orbit is elliptical it will vary between two values. the problem is you think there are limits to matter velocity in an orbit. since the circular speed is sqrt(GBH/r) = C if it tries to accelerate, it will be too slow and will curve into the BH. that is why an acceleration ring exists around a BH. You dont even need to have advance of perihelion to fail to orbit and fall in.
the question is the time differential. two different frames.
Already posted!
please post it again because that is not what you said. humor me and clarify this issue.
It might clarify things if you explain what you beleive is true and why. and then you can show us how you think matter behaves near a BH and how time is bent.
josephus
--
I go sailing in the Summer and
look at STARS in the Winter.
"Everybody is igernant, jist on differt subjects"
Will Rogers Jr.
"it aint what you know that gets you in trouble
it is what you know that aint so"
Josh Billings.
.
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