Gravitational time dilation vs velocity time dilation

1. Apr 26, 2010

rhenretta

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:
$$m_{1}$$ = Planet (or black hole) mass
$$m_{2}$$ = Spaceship mass
$$v_{1}$$ = Spaceship velocity rel to $$m_{1}$$
$$T_{1}$$ = Time, as observed on surface of $$m_{1}$$

A spaceship orbiting a super massive black hole (or other body) must have velocity:
$$v_{1}=\sqrt{{Gm_{1}}/r}$$

Using Lorentz transformation
$$T_{1} = T_{2}\sqrt{1-{v^{2}}/{c^{2}}$$

Substituting v with $$v_{1}$$:
$$T_{1} = T_{2}\sqrt{1-{Gm_1}/{r^2}}$$

The problem is, in order to match up with the formula for gravitational time dilation, I need a 2:
$$T_{1} = T_{2}\sqrt{1-{2Gm_1}/{r^2}}$$

2. Apr 26, 2010

George Jones

Staff Emeritus
They are not the same thing, and different methods of "observation" and measurement are used to "see" the two effects.
Unfortunately, things are not this simple.

3. Apr 26, 2010

rhenretta

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. Apr 26, 2010

JesseM

Yes, gravitational time dilation would be experienced even if you weren't orbiting at all, but were just hovering at a constant radius and constant angle using your rockets. In this case if your radius (in Schwarzschild coordinates) was r while the Schwarzschild radius of the black hole (the radius of its event horizon) was r0, your clock would run slow by a factor of $$\sqrt{1 - \frac{r_0}{r}}$$ as seen by an observer at an arbitrarily large distance from the black hole.
Would depend on the mass of the black hole, your radius, and the orbital speed at that radius (not sure what the formula is for orbital speed as a function of radius). As noted by kev in post #35 on this thread, the total time dilation for an orbiting object is just the product of gravitational and velocity-based time dilation.

5. Apr 26, 2010

rhenretta

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:
$$T_{1}=T_{2}\sqrt{1-{Gm_{1}}/r^2} + T_{2}\sqrt{1-{2Gm_{1}}/r^2}$$

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. Apr 26, 2010

JesseM

As I said at the end of my last post (and gave a link to a post by kev with more details), for an orbiting object the total time dilation is the product of gravitational and velocity-based time dilation, not the sum (although as pervect points out in post #19 here, if the time dilation factors are small as would be the case for GPS satellites, taking the sum of the two time dilations is a good approximation)

7. Apr 26, 2010

rhenretta

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. Apr 26, 2010

JesseM

Similarity between which two formulas? Anyway, I don't think you can use the Newtonian circular orbit velocity formula in a situation where spacetime is highly curved like around a black hole, though it will probably work as a decent approximation if you're just talking about an orbit around the Earth where Newtonian gravity is pretty close to accurate. For example, around a black hole the closest possible orbit is at the photon sphere at 3GM/c^2, which is 3/2 times the radius of the event horizon, and at this distance the only possible orbits are those moving at exactly c. That's not what would be predicted by the Newtonian formula for circular orbit velocities!

9. Apr 26, 2010

rhenretta

Very interesting, I did not know this. It introduces a huge question though, if it is not possible to orbit between the event horizon and 3/2 times the radius of the event horizon, wouldn't that make the event horizon 3/2 times itself? Because you would never be able to escape at this new radius, if you are destined to be sucked in.

10. Apr 26, 2010

yuiop

Maybe surprisingly, Kepler's third law:

$$T^2 = R^3\frac{4\pi^2}{GM}$$

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 re-arranged to obtain the coordinate orbital velocity:

$$V = \frac{2\pi R}{T} = \sqrt{\frac{GM}{R}}$$

(So Rhenetta was on the right track). Things get more complicated for non-circular orbits but for now I will stick to circular orbits.

The local velocity is obtained by applying the gravitational time dilation factor so that:

$$v = \frac{2 \pi R}{T\sqrt{1-2GM/(Rc^2)}} = \sqrt{\frac{GMc^2}{(Rc^2-2GM)}}$$

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:

$$T '/T = \sqrt{1-v^2/c^2}\sqrt{1-2GM/(Rc^2)}$$

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):

$$T'/T = \sqrt{\left(1-\frac{GM}{(Rc^2-2GM)}\right)\left(1-\frac{2GM}{Rc^2}\right)}$$

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.

Last edited: Apr 26, 2010
11. Apr 26, 2010

George Jones

Staff Emeritus
No:

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. Apr 26, 2010

yuiop

It just means that there is no natural circular orbit below the photon sphere for an unpowered particle. A rocket with sufficient thrust from its engines can hover below that radius or escape if required (as long as it remains above the event horizon). The other factors mentioned by George are also correct.

13. Apr 26, 2010

JesseM

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, velocity-based or gravitational time dilation:
The two types of time dilation would make an equal contribution when $$\left(1-\frac{GM}{(Rc^2-2GM)}\right) = \left(1-\frac{2GM}{Rc^2}\right)$$, which would mean R=4GM/c^2 (double the Schwarzschild radius). Any closer than that and the velocity-based time dilation makes a greater contribution (with the velocity-based time dilation causing an orbiting clock to approach a ticking rate of 1 - GM/GM = 0 as you approach the photon sphere at R=3GM/c^2, since the local speed of an orbiting object approaches c), and farther than that the gravitational time dilation makes a greater contribution.

14. Apr 26, 2010

yuiop

There is a sort of relationship between velocity and gravitational time dilation, but it is the escape velocity at a given radius that is required. The Newtonian escape velocity is:

$$v_e = \sqrt\frac{2GM}{R}}$$

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:

$$T '/ T = \sqrt{1-\frac{v_e^2}{c^2}} = \sqrt{1-\frac{2GM}{Rc^2}}$$

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.

Last edited: Apr 26, 2010
15. Apr 26, 2010

Cyosis

I don't see how we can relate velocity and gravitational time dilation of an observer moving along the r direction in a gravitational field by using the definition of proper time from special relativity.

If an observer moves along the r direction the Schwarzschild metric reduces to (spherical symmetry):

$$ds^2=(1-\frac{r_s}{r})c^2dt^2-\frac{dr^2}{1-\frac{r_s}{r}}$$

which would yield:

$$d\tau=\sqrt{\left(1-\frac{r_s}{r}\right)-\left(1-\frac{r_s}{r}\right)^{-1} \left(\frac{v}{c}\right)^2}dt$$

Last edited: Apr 26, 2010
16. Apr 26, 2010

utesfan100

The photon sphere is the closest one could ever hope to orbit a black hole.

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. Apr 26, 2010

rhenretta

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. Apr 26, 2010

yuiop

In #10 I gave an equation that indicates that the coordinate time dilation of a particle moving in a gravitational field is product of the time dilation due its local velocity $v_L$ and time dilation due to gravity, i.e:

$$d\tau = dt \sqrt{1-v_L^2/c^2}\sqrt{1-R_s/R}$$ (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 $v_L$ as measured by a stationary observer at R to a coordinate velocity v as measured by the Schwarzschild observer at infinity using the relation:

$$v = v_L (1-R_s/R)$$ (Eq2)

Substituting this into Eq1 gives:

$$d\tau = dt\sqrt{1-\frac{v^2}{c^2(1-R_s/R)^{2}}} \sqrt{1-\frac{R_s}{R}}$$ (Eq3)

which simplifies to the equation you gave:

$$d\tau = dt \sqrt{\left(1-\frac{r_s}{r}\right)-\left(1-\frac{r_s}{r}\right)^{-1} \left(\frac{v}{c}\right)^2}$$ (Eq4)

Now if the locally measured radial velocity of a particle initially at rest falling from infinity is the Newtonian escape velocity $\sqrt{(2GM/R)} = R_s c^2$, then the coordinate time dilation of the free falling particle using Eq1 is:

$$d\tau = dt \sqrt{1-R_s/R}\sqrt{1-R_s/R} = dt(1-R_s/R)$$ (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 $\sqrt{(1-v^2/c^2)}$ or $\sqrt{(1-R_s/R)}$.

Supporting evidence:

According to mathpages http://www.mathpages.com/rr/s6-07/6-07.htm the coordinate velocity of a falling particle using G = c = 1 is:

$$\frac{dr}{dt} = \left(1-\frac{2M}{R} \right) \sqrt{1-\left(1-\frac{2M}{R}\right)k}$$

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:

$$\frac{dr '}{dt '} = \sqrt{1-\left(1-\frac{2M}{R}\right)} = \sqrt{\frac{2M}{r}}$$

which is the Newtonian escape velocity.

Last edited: Apr 26, 2010
19. Apr 26, 2010

starthaus

I asked him the same question, long ago. He does it through a simple algebraic trick, by factoring out $$\sqrt {1-\frac {r_s}{r}}$$ and by relabelling $$\frac {v}{1-\frac {r_s}{r}}$$ as $$v_L$$.

20. Apr 27, 2010

rhenretta

All this discussion re Schwarzschild brings up another question I had recently. The Schwarzschild radius refers to the radius of volume a mass must be compressed to in order to become a black hole. The question is, once it becomes a black hole, is the Schwarzschild radius and the event horizon the same thing?

21. Apr 27, 2010

JesseM

For an ideal black hole that isn't spinning (no angular momentum) and has no electric charge, yes. A realistic black hole would probably have some angular momentum (but negligible charge), and the event horizon of a rotating black hole is an ovoid rather than a perfect sphere.

22. May 20, 2010

Passionflower

Comparing apples with apples is good, comparing apples with pears is asking for trouble but comparing apples with nothing is always wrong.

Time dilation is always a comparison between clock rates. The gravitational time dilation component between two objects is caused by the difference of proper acceleration between them. The velocity time dilation component is nothing more than the observed speed, both uniform and accelerated, between them.

23. Aug 27, 2010

starthaus

True for "natural" satellites only. Not true for self-propelled satellites (like the GPS) that can have any period they wish, based on their speeds. Also, definitely false for the airplanes used in the Haefele-Keating experiment. (for the same reason, they can have any speed they want).

The first equality is true (trivially), the second one is true only for "natural" satellites (see above). This is not the biggest problen with your proof, the biggest problem can be seen below.

Err, no. This is where things go South in your derivation.

Not.

No, again.

Last edited: Aug 27, 2010