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Gravitational time dilation vs velocity time dilation 
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#1
Apr2610, 09:06 AM

P: 66

I was watching the new show "Into the Universe with Stephen Hawking", and I found myself a little annoyed by his contrast of gravitational vs velocity time dilation. It was stated that if you took a spaceship, orbiting around a super massive black hole, you'd only get a 2:1 time dilation. However, if you take a spaceship and move fast in a straight line, you'd get an unlimited time dilation ratio.
Now, I'm no physicist, but I am pretty sure the two things were one and the same, so I took it upon myself to prove it. Unfortunately, I found myself missing of all things a digit 2, which must be a mistake on my part. Hopefully someone can show the fault in my math. Definitions: [tex]m_{1}[/tex] = Planet (or black hole) mass [tex]m_{2}[/tex] = Spaceship mass [tex]v_{1}[/tex] = Spaceship velocity rel to [tex]m_{1}[/tex] [tex]T_{1}[/tex] = Time, as observed on surface of [tex]m_{1}[/tex] A spaceship orbiting a super massive black hole (or other body) must have velocity: [tex]v_{1}=\sqrt{{Gm_{1}}/r}[/tex] Using Lorentz transformation [tex]T_{1} = T_{2}\sqrt{1{v^{2}}/{c^{2}}[/tex] Substituting v with [tex]v_{1}[/tex]: [tex]T_{1} = T_{2}\sqrt{1{Gm_1}/{r^2}}[/tex] The problem is, in order to match up with the formula for gravitational time dilation, I need a 2: [tex]T_{1} = T_{2}\sqrt{1{2Gm_1}/{r^2}}[/tex] 


#2
Apr2610, 09:12 AM

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#3
Apr2610, 09:56 AM

P: 66

But, in the context given, he was specifically talking about orbiting a super massive black hole. In order to do this, he would need a velocity approaching the speed of light, and thus time dilation would present itself due to that. Are you saying there is additional time dilation due to gravity? If so, which is greater?



#4
Apr2610, 10:08 AM

Sci Advisor
P: 8,470

Gravitational time dilation vs velocity time dilation



#5
Apr2610, 10:54 AM

P: 66

My question is then:
Does the time dilation due to gravity need to be added to the time dilation due to velocity? IE, GPS satellites  do they have to account for both? IE: [tex]T_{1}=T_{2}\sqrt{1{Gm_{1}}/r^2} + T_{2}\sqrt{1{2Gm_{1}}/r^2}[/tex] Where the first term is the time dilation due to the velocity of the circular orbit, and the 2nd due to gravity. I still think there is a problem in my circular orbit formula, and that the two terms should match. Or am I just way to far off, and I should go RTFM some more 


#6
Apr2610, 11:07 AM

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#7
Apr2610, 11:38 AM

P: 66

sorry, missed that last sentence.
I get that, but I am still lost with my substitutions of my formula. They must be multiplied together, so basically my last formula is a restatement of post 35 you pointed me to, the only difference is I replaced v with the circular orbital velocity formula. My question then, did I perform my replacement correctly? I am still thrown off by the similarity of the two formulas, the only difference being that stupid 2. 


#8
Apr2610, 12:03 PM

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#9
Apr2610, 12:50 PM

P: 66




#10
Apr2610, 12:51 PM

P: 3,967

[tex]T^2 = R^3\frac{4\pi^2}{GM}[/tex] still holds in General Relativity from the point of view of a coordinate observer at infinity in the Schwarzschild metric. The above equation can be rearranged to obtain the coordinate orbital velocity: [tex]V = \frac{2\pi R}{T} = \sqrt{\frac{GM}{R}}[/tex] (So Rhenetta was on the right track). Things get more complicated for noncircular orbits but for now I will stick to circular orbits. The local velocity is obtained by applying the gravitational time dilation factor so that: [tex]v = \frac{2 \pi R}{T\sqrt{12GM/(Rc^2)}} = \sqrt{\frac{GMc^2}{(Rc^22GM)}}[/tex] When R is set to the photon orbit radius 3GM/c^2 the local orbital speed is v=c. The time dilation ratio of a particle with a circular orbit of radius R and local orbital velocity v is: [tex]T '/T = \sqrt{1v^2/c^2}\sqrt{12GM/(Rc^2)}[/tex] The equation obtained earlier for v can be directly inserted into the above expression, to obtain the time dilation ratio of the orbiting particle when the only known variables are the mass of the black hole (M) and the orbital radius (R): [tex]T'/T = \sqrt{\left(1\frac{GM}{(Rc^22GM)}\right)\left(1\frac{2GM}{Rc^2}\right)}[/tex] where T is the time according to an observer at infinity and T' is the time according to the orbiting particle. It is very easy to see that the time dilation of the orbiting particle is unbounded. 


#11
Apr2610, 12:54 PM

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P: 6,248

1) a rocket can escape form this region; 2) light directed upwards can escape from region; 3) a baseball thrown upwards can escape from this region. 


#12
Apr2610, 01:08 PM

P: 3,967




#13
Apr2610, 01:26 PM

Sci Advisor
P: 8,470

Thanks kev! That last equation also can be used to answer the question about which makes a greater contribution to the total time dilation of an orbiting object, velocitybased or gravitational time dilation:



#14
Apr2610, 01:35 PM

P: 3,967

[tex]v_e = \sqrt\frac{2GM}{R}}[/tex] See http://en.wikipedia.org/wiki/Escape_velocity This is the velocity attained by a particle initially at rest at infinity (loosely speaking) when it falls to a radius R as its potential energy is converted to kinetic energy. Inserting this velocity into the SR time dilation equation gives: [tex]T '/ T = \sqrt{1\frac{v_e^2}{c^2}} = \sqrt{1\frac{2GM}{Rc^2}}[/tex] This is the time dilation ratio of a particle hovering at R. For a particle orbiting at R you have to multiply gravitational time dilation at R by the time dilation due the local orbital velocity of the particle. 


#15
Apr2610, 02:31 PM

HW Helper
P: 1,495

If an observer moves along the r direction the Schwarzschild metric reduces to (spherical symmetry): [tex] ds^2=(1\frac{r_s}{r})c^2dt^2\frac{dr^2}{1\frac{r_s}{r}} [/tex] which would yield: [tex] d\tau=\sqrt{\left(1\frac{r_s}{r}\right)\left(1\frac{r_s}{r}\right)^{1} \left(\frac{v}{c}\right)^2}dt [/tex] 


#16
Apr2610, 02:36 PM

P: 100

Between the photon sphere and the event horizon it is possible to escape, but even at light speeds you would need to be headed within a cone outward from the radial axis that tightens to a spike near the event horizon. If you are between the photon sphere and the event horizon you could accelerate to where your velocity is within the escapre cone and leave the region. If you free fell into this region you would need to accelerate to avoid hitting the event horizon. In my space travels I like to stay more than 10 Rs from the event horizon, just to be safe :) 


#17
Apr2610, 05:05 PM

P: 66

Thank you all for the explanations. While I still have a lot of fundamentals to learn, it is nice to know I am getting closer to understanding this  not bad for a software programmer with nothing more than a high school diploma from a sub par ghetto school.



#18
Apr2610, 10:05 PM

P: 3,967

[tex]d\tau = dt \sqrt{1v_L^2/c^2}\sqrt{1R_s/R}[/tex] (Eq1) Now I was using the equation in context of horizontal orbital velocity, but lets see if it work in the radial direction too. If we want to compare it to the Schwarzschild metric we need to convert the local velocity [itex]v_L[/itex] as measured by a stationary observer at R to a coordinate velocity v as measured by the Schwarzschild observer at infinity using the relation: [tex]v = v_L (1R_s/R)[/tex] (Eq2) Substituting this into Eq1 gives: [tex]d\tau = dt\sqrt{1\frac{v^2}{c^2(1R_s/R)^{2}}} \sqrt{1\frac{R_s}{R}}[/tex] (Eq3) which simplifies to the equation you gave: [tex]d\tau = dt \sqrt{\left(1\frac{r_s}{r}\right)\left(1\frac{r_s}{r}\right)^{1} \left(\frac{v}{c}\right)^2}[/tex] (Eq4) Now if the locally measured radial velocity of a particle initially at rest falling from infinity is the Newtonian escape velocity [itex]\sqrt{(2GM/R)} = R_s c^2[/itex], then the coordinate time dilation of the free falling particle using Eq1 is: [tex]d\tau = dt \sqrt{1R_s/R}\sqrt{1R_s/R} = dt(1R_s/R)[/tex] (Eq5) It can be seen that the magnitude of the velocity time dilation of the free falling particle is equal to the magnitude of the gravitational time dilation of the free falling particle (but it is subject to both effects). To a local observe at R, the time dilation of the free falling particle is only due to velocity and numerically equal to [itex]\sqrt{(1v^2/c^2)}[/itex] or [itex]\sqrt{(1R_s/R)}[/itex]. Supporting evidence: According to mathpages http://www.mathpages.com/rr/s607/607.htm the coordinate velocity of a falling particle using G = c = 1 is: [tex]\frac{dr}{dt} = \left(1\frac{2M}{R} \right) \sqrt{1\left(1\frac{2M}{R}\right)k}[/tex] From mathpages it can be seen that the parameter K is unity when the particle is initially at rest at infinity. The local velocity using my Eq2 is then: [tex]\frac{dr '}{dt '} = \sqrt{1\left(1\frac{2M}{R}\right)} = \sqrt{\frac{2M}{r}}[/tex] which is the Newtonian escape velocity. 


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