# Surpassing the speed of light.

1. Jan 30, 2006

### Paindealer

Ok, I am a complete n00b at this stuff, the most I have had is an into to beginning physics and seen a few discovery channel shows, but I was thinking about what I leared and came up with a weird scenario (by the way, I probably have no idea what I am talking about and there is probably a crutial and extremely simple flaw in here so please do correct me).

I heard that it is impossible to surpass the speed of light. Also there is no terminal velocity in a vaccum. SO, if there was a GINORMOUS mass floating afound in space with a vaccum chamber on it that extended gagillions of miles into space, and we dropped something in the vaccum, would it not keep accellerating until it hit the mass? And if the mass was so large and with such strong gravity, that by the time the object dropped reached the ground it would have reached and surpassed the speed of light, then wouldn't that break the law which states that nothing can surpass that speed?

All theoretical of course. I doubt that a mass that large could exist, though one never knows...still, probably not lol.

Last edited: Jan 30, 2006
2. Jan 30, 2006

### dicerandom

You don't need a gigantic hollow mass, you can do something simmilar by dropping an object into a black hole from a very large distance.

The answer, however, is that the object will never exceed the speed of light.

Think of it this way: the speed of light is not special because that's how fast light goes and light is special, it's special because that's as fast as anything can go, and light happens to be one thing which can go that fast.

3. Jan 30, 2006

### franznietzsche

There are a couple of ways to approach this, but all give the same answer. The easiest is probably conservation of energy, which is conserved in flat Minkowksi Spacetime (though its not necessarily conserved in non flat spacetimes, but we'll use Newton's gravity for this, its sufficient for the purpose of example). Potential Energy is:

$$U = G \frac{M}{m}{r}$$

and kinetic energy is

$$K = (\gamma - 1)mc^2$$
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

Where v is the velocity, M is the mass of the large body, m is the mass of the falling body, and c is the speed of light. So the total energy is:

$$E = U + K$$
$$= G \frac{M m}{r} + \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2$$

E is always the same number for this falling object. So, if we solve for v, we get:

$$v = c \sqrt{1 - \left ( \frac{m c^2}{E + mc^2 - G \frac{M m}{r}} \right )^2}$$

If you then take the limit as $$r \rightarrow 0$$ you will get $$v = c$$

This means that yes the object does keep accelerating as it gets closer, but that acceleration becomes smaller and smaller, and the object does not exceed c.

Last edited: Jan 30, 2006
4. Jan 30, 2006

### franznietzsche

The Latex engine has been behaving really problematic for me. I'm trying to get them fixed. Its completely displaying one wrong for me (its displaying a different latex code from elsewhere in my post in two places, when one of those should be something very different).

The first displayed $$r \rightarrow 0$$ should be

$$= G \frac{M m}{r} + \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2$$

I have no idea why its displaying it wrong.

Edit: Fixed it. Its all correct now. Bah, off to the feed back forum on this one.

Last edited: Jan 30, 2006
5. Jan 31, 2006

### pervect

Staff Emeritus
A mass with an escape velocity greater than c is, by defintion, a black hole.

When the mass, falling from infinity, reaches the event horizon of a black hole, its velocity approaches 'c' compared to a hypothetical observer who is stationary with respect to the black hole. So in one sense, you might think that the falling object actually reached the unobtainable, the speed of light.

The problem here turns out to be with the hypothetical observer. In reality, there will be some limit on how close an observer can get to a black hole and "hold station". This is because it requires more and more acceleration to "hold station" - at the event horizon itself, it would require infinite acceleration to hold station, which is not possible.

Thus the falling observer never reaches the speed of light with respect to any real observer.

The net result is the confusing state of affairs where the infalling observer can fall so fast / get so deep that he cannot send light back to the outside observer at infinity.

However, light can always "catch up" to the infalling obsever from outside, there is no problem in that regard.

And the infalling observer will measure the speed of light coming to him from outside the black hole as equal to "c", right up until the last instant when he is finally torn apart by the tidal forces at the heart of the singularity itself.

6. Jan 31, 2006

### pervect

Staff Emeritus
It would be better to use the "effective potential", which is as close as GR comes to having a potential energy under this specific circumstance.

The effective potential is covered in, for instance, http://www.fourmilab.ch/gravitation/orbits/

You can easily find from this web page that

dr/dtau, though, is not really a velocity, but a rapidity. Working out velocities is harder. I've done it in some other post, but since the search engine seems to be stalled finding it, I can't post a link to it. I recall that George independently worked out the same results as I did for this case.

7. Jan 31, 2006

### rbj

??

is not the Newtonian PE this:

$$U = -G \frac{Mm}{r}$$ ?

(assuming $m$ is invariant mass, which i better get used to assuming so we don't have semantic confusion, i guess.)

ahhh! now it's almost right. (needs a minus sign with the G.)

in fact, i think we can give E a name (assuming you drop the ball with zero velocity). ($E = U+K = 0$)

i get (assuming i haven't dropped a minus sign somewhere):

$$v = c \sqrt{1 - \frac{1}{\left(\frac{G M}{c^2 r} + 1 \right)^2} }$$ .

Last edited: Jan 31, 2006
8. Jan 31, 2006

### franznietzsche

Bah, my Latex error.

I always assume G carries the negative. Maybe not the best convention though, but I'm consistent on it (at least, I mean to be). But no, not an error, just unclear convetion.

Yes, but there is no need to assume that. And no reason to, better to leave it more general.

Yes, I was just trying to keep it simple for the poster whose level of experience I'm unaware of.

It would be, if I knew enough GR to do that off-hand. But thats why I like having you around.
This is why I wanted to use Newtonian gravity and SR. Its much simpler, and therefore (hopefully) more clear and understandable to the OP. Even if it is a flawed combination.

9. Jan 31, 2006

### leandros_p

Paindealer,

you asked: "And if the mass was so large and with such strong gravity, that by the time the object dropped reached the ground it would have reached and surpassed the speed of light, then wouldn't that break the law which states that nothing can surpass that speed?"

The key word in your question is the phrase "by the time". When a physical object is accelerating towards the speed of light, the dimension of time becomes smaller and smaller (time contracts), and just before reaching the speed of time the dimension of time becomes so "small" that the object does not have time to accelerate any further.

You can use the calculations of "the Lorentz transformations of space and time" found at the Wikipedia article about "Special relativity".

So even if enormous forces are implemented on an accelerating object, the dimension of time is decelerated when the speed of the object approaches the speed of light.

Another way to understand the phenomenon of time contraction is to understand that in a fast accelerating object there is an increase its "mass" (this is called "relativistic mass"). This does not mean that the object becomes actually more massive in itself. It means that for an observer in a different frame the object’s mass is increasing as the velocity of the object increases. Therefore, it requires even more and more force/power in order to achieve further acceleration. When the speed of the object is closing to the speed of light, the object becomes so (relativistic) massive that its (relativistic) mass inertia becomes infinite, so that the amount of power/force that is needed to accelerate it further becomes infinite too – this is an impossible task.

So, the key word in your question is "time". Time is "running" more slowly when an object is moving relative to another frame of reference. For a constantly accelerating object (in relation to another frame of reference), time is becoming a physical term in shortage.

Leandros

10. Jan 31, 2006

### pervect

Staff Emeritus
This reminds me of something I forgot to mention that came up in previous discussions. The velocity of an observer infalling into a black hole actually depends on where the observer is that is measuring it. (Black holes are relevant, again, because only a black hole has an escape velocity of greater than 'c', thus an infalling object could be naievely be expected to reach 'c' only for a black hole).

An observer at infinity will see the velocity of an object infalling into a black hole as approaching zero, due to the immense gravitational time dilation effects. According to the observer at infinity, the infalling observer will never actually reach the event horizion.

An observer co-located with the infalling observer but stationary with respect to the black hole will, however, read a velocity that approaches 'c' as the co-located observer is placed closer and closer to the event horizion, as I mentioned previously.

When the search engine gets fixed, I'll try and post some links to the past discussion, if I can find them.

Last edited: Jan 31, 2006
11. Jan 31, 2006

### DaveC426913

All (or at least, most): These formulae are great and everything, but are they really likely to answer the OP's question as a self-described n00b?

Paindealer: whether by gravitational acceleration or by rocket propulsion, it does not matter - an object with mass will not exceed the speed of light. As it approached c, its acceleration (due to gravity) would slow.

12. Jan 31, 2006

### Physics_Kid

13. Jan 31, 2006

### pervect

Staff Emeritus
Not so interesting - Van Flandern doesn't have a lot of credibility. See for instance

http://math.ucr.edu/home/baez/RelWWW/wrong.html#speed [Broken]

also the sci.physics.faq on the speed of gravity

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

According to GR, gravity waves should travel at a speed of 'c' through a vacuum. There may be very small changes in the propagation speed through a media that lower the speed - I don't know the exact magnitude of this effect offhand, but I'd expect it to be very tiny.

Last edited by a moderator: May 2, 2017
14. Feb 1, 2006

### Oxymoron

A hypothetical observer at rest w.r.t a black hole right on the Schwarzchild surface watches his companion falling into the black hole from infinity. Theoretically the falling companion's speed approaches c relative to the observer. At the moment the observer calculates the companion's speed to hit exactly c the companion has unfortunately crossed the Schwarzchild radius. The observer was really hoping his companion could talk to him, describe what travelling faster than light was like, maybe even to tell him the winning lottery numbers before they were drawn. But unfortunately the companion, although theoretically now travelling faster than c relative to the observer sitting on the Schwarzchild surface, cannot relay any information to the observer due to the light cones of the companion tipping over. So, is this how a Black hole hides any breach of causality that would result from something travelling faster than light? Is it like nature realised that black holes could violate causality so it simply threw a blanket over the whole contraption blocking communication from inside the surface to the rest of the universe?

15. Feb 1, 2006

### DaveC426913

Well, no. He doesn't necessarily reach c - and needn't.

What happens is that, as the falling companion approaches the Schwarzchild radius, he begins to slow down in time. He will never actually reach the S.R. (from the external viewer's POV) , rather, he will spend all eternity just above it, getting closer to it.

16. Feb 2, 2006

### Longstreet

Q related to a thread I started recently. If there is an observer near the event horizon, will they not possible see objects further away from the SR than they are moving much faster than next to them. And possibly even faster than light?

17. Feb 2, 2006

### pervect

Staff Emeritus
Yes, but there is not a lot of physical significance to this. What's physically significant is the velocity when the speed is measured with local clocks and rulers, i.e. when an observer measures the speed of another object at the same place the observer is (or close enough to the observer that the metric has not changed much).

18. Feb 2, 2006

### Longstreet

I think it is physically significant to the question of whether things can apear to exceed the speed of light. From equivilance we know the c is the same in every local frame. But really how can you say there *is* a local frame in GR at all; maybe a single 1D line representing a potential. But between potentials you can have the example I just gave. I don't see how that's not physically significant to this question. Or is it just significant when objects conveniently apear to travel slower than c?

19. Jul 31, 2008

### stragz

hi, i'm most likely the dumbest person around here
but as i was making a report on black holes/white holes for my teacher
i've stumbled across this forum
and could not help from noticing that some people still use the natural laws of physics about talking beyond the event horizon
and i've read that space and time become distorted as you are closing in on a black hole
once you've crossed the horizon that space/time even switch places
going on from this point,i get the fact that at the even horizon:
gravity pull= speed of light, this is the "frozen effect"
now as for the claim that time and space become extremly distorted approaching a black hole,
and they shift places , could it not be that the gravitational pull is HIGHER then the speed of light beyond the event horizon?

i'm not trying to be a know-it-all
its is rather disturbing and hard to grasp how mucht a dimension as time is being influcent by mass, if someone could explain this to me, i would find it easier to grasp,
if you want to anwser with formulas, be my guest, i had enough maths in school and i know how to use a search button to understand the formulas myself
but please consider that spacetime inside a black hole becomes timespace :p
or am i wrong about that? i got those ideas after reading it a couple of times on different sites

20. Jul 31, 2008

### HallsofIvy

No one here is going to find a question like this annoying but many might be annoyed by the fact that you asked in someone else's thread!

How about starting a thread for your question specifically. Oh, and it would help if you asked as specific a question as you could.