CAN you go faster than light? A thought experiment...

[Phantom1983]

To any astrophysicists out there: a little thought experiment.

Consider a large massive body (a Schwarzchild-type black hole would work best) and a spaceship, positioned a great distance away, which is set gently into motion (or shoved violently; it makes no odds) towards the black hole.

Now, as the ship accelerates toward the black hole (due to gravity) its speed wil increase. However, the relativistic mass increase should have NO effect on the acceleration of the space ship because gravitational field strength INCREASES with mass (indeed, the spaceship's mass would increase to the point where it would attract the black hole TOWARDS it) and no fuel is being expended in the system.

So has the mass increase problem been circumvented or does the logic have a gap in it? Indeed, would there be enough "room" for the ship to reach c before it ran into the event horizon?

(I came up with this a while ago as a way to dodge the old "infinite fuel" issue)

 

[DrPizza]

Related question (I think) but more easily phrased:
Since the minimum velocity (as far as I know) for a object to strike the earth (or rather, it's atmosphere at which point it begins slowing) is equal to the escape velocity, and the escape velocity for a black hole >c,... well, hopefully you see where I'm going with this.

 

[TuxDave]

Maybe space is so warped around the event horizon such that the distance between you and the event horizon become infinite and so you never hit the event horizon.

 

[Phantom1983]

Originally posted by: TuxDave
Maybe space is so warped around the event horizon such that the distance between you and the event horizon become infinite and so you never hit the event horizon.

Very possibly! "Distortion of space" essentially means that the geodesics (shortest paths between two given points) become curves rather than straight lines.

A particle moving at a fixed speed right-to-left along the curve xy=1 will appear, if you are the y-axis, to approach you with an ever decreasing speed, even though we outside the y-axis "know" that its speed is constant. Something very similar may happen at a black hole's EH.

 

[Phantom1983]

Originally posted by: DrPizza
Related question (I think) but more easily phrased:
Since the minimum velocity (as far as I know) for a object to strike the earth (or rather, it's atmosphere at which point it begins slowing) is equal to the escape velocity, and the escape velocity for a black hole >c,... well, hopefully you see where I'm going with this.

Yes, I think I catch your drift. A black hole's escape velocity is equal to c but I can't for the life of me see why a body can't be made even MORE dense so that its escape velocity is greater than c. Then you could just throw a tennis ball at it (which might travel back in time for all we know...).

I suppose it all comes down to if there's some upper limit on how compact matter is allowed to be (there isn't one that I know of but then what do I know...).

 

[Mishkav]

duuuude black!! holes?! i red in orsons book dat if u jump in a boing7fo7 and go speed da light then u come back evedrybody be ded and u be aive! alive! das tight! u know? yea yea yea! u know?

 

[Bobthelost]

If you want to move faster than light all you have to do is shine it into a very dense clear material and hop in a jet.

Light only travels at 1/2 the speed of light (in a vacume, which is the value we all know and love) in fibre optic cable.

It's almost as useful as your suggestion, and physically possible, and theoretically valid.

 

[Zbox]

what happens due to the black hole relativistic effects as you get close though?

 

[Bobthelost]

A black hole has an escape velocity of greater than c. Assuming of course that there's anything inside the event horizon at all. This is the point my physics knowledge starts to fail me.

 

[BEL6772]

A little tangent:

With our way of observing things, we can't observe anything faster than c. I remember a physics test question that asked me to solve for the observed speed of an object that was closing at greater than the speed of light.

The setup was 'object a' is moving towards a stationary object at 0.6c. 'Object b' is moving towards the same stationary object from the opposite direction at 0.5c. If you were on 'object a', what velocity would you observe with respect to 'object b'.
From the stationary object's point of view, 'object a' and 'object b' were closing on each other at a velocity of 1.1c. 'Object a' would observe 'object b' closing at 0.99c, though. I remeber the answer (and that I got it right) but I don't remember all the equations that I used to get there.

Anyway, applying that to our ship accelerating towards the black hole... There may be actual closing speeds greater than c, but nobody would be able to observe it directly.

 

[FallenHero]

Originally posted by: Bobthelost
A black hole has an escape velocity of greater than c. Assuming of course that there's anything inside the event horizon at all. This is the point my physics knowledge starts to fail me.

I think physics in general begins to fail at the event horizon.

 

[Matthias99]

Originally posted by: Phantom1983

Originally posted by: DrPizza
Related question (I think) but more easily phrased:
Since the minimum velocity (as far as I know) for a object to strike the earth (or rather, it's atmosphere at which point it begins slowing) is equal to the escape velocity, and the escape velocity for a black hole >c,... well, hopefully you see where I'm going with this.

Yes, I think I catch your drift. A black hole's escape velocity is equal to c but I can't for the life of me see why a body can't be made even MORE dense so that its escape velocity is greater than c.

There's nothing preventing it; in fact, any black hole would have such an area. The escape velocity of a black hole is equal to c at the event horizon. As you get closer to the middle, the escape velocity would become even higher. But since you cannot get over c to begin with, it's sort of irrelevent.

Then you could just throw a tennis ball at it (which might travel back in time for all we know...).

I don't quite follow you. Neither the 'light' nor 'heavy' object will ever reach c, regardless of how massive/dense either one of them is or how fast they are moving to begin with. As you continue accelerating towards c, more and more of the energy gets converted into increased mass rather than velocity, which prevents you from ever reaching c.

There are also time dilation effects (red shift/blue shift) that come into play if you are accelerating relative to something else, but these usually made my head hurt. This is why I was a CS major and not a physicist.

I suppose it all comes down to if there's some upper limit on how compact matter is allowed to be (there isn't one that I know of but then what do I know...).

It doesn't matter in terms of acceleration like this. You can treat the objects like infinitely small point masses and you still get the same results.

 

[MrDudeMan]

I'm not really sure what the real question is here. An observer watching the shuttle approach the black hole would see the events happening at a much different rate once the shuttle approached speeds near the speed of light. The distance from the shuttle to the event horizon, as seen by a stationary observer, is called the proper distance, or Lp. That distance is shorter if measured from inside the moving shuttle. The time it takes to go that distance, however, is called the proper time as measured from the shuttle, and that time will be shorter than what a stationary observer would realize. The difference in time is called time dilation, and the stationary observer records a time that is longer by a factor of gamma, or 1/(sqrt(1-v^2/c^2). This is called special relativity and can be proven with the classic experiment involving mu-mesons and the time it takes for a group of said particles to decay when moving near the speed of light compared to the rest decay time. I know this isn't really related to what you were asking, but I don't see a more 'addressable' question.

A little tangent:

With our way of observing things, we can't observe anything faster than c. I remember a physics test question that asked me to solve for the observed speed of an object that was closing at greater than the speed of light.

The setup was 'object a' is moving towards a stationary object at 0.6c. 'Object b' is moving towards the same stationary object from the opposite direction at 0.5c. If you were on 'object a', what velocity would you observe with respect to 'object b'.
From the stationary object's point of view, 'object a' and 'object b' were closing on each other at a velocity of 1.1c. 'Object a' would observe 'object b' closing at 0.99c, though. I remeber the answer (and that I got it right) but I don't remember all the equations that I used to get there.

Anyway, applying that to our ship accelerating towards the black hole... There may be actual closing speeds greater than c, but nobody would be able to observe it directly.

Actually, it is possible to see things that move faster than C. The shadow of the Earth on the moon, for example, moves faster than c. Ever heard of the Casmir Effect? Photons being exchanged between two uncharged conducting plates can move faster than c (although it is 1 part in 10^20 or less for a very tiny gap, but still faster than c).

Also, two object moving toward each other with a combined apparent speed of greater than c most certainly can be observed at a speed of greater than c, but only by a third party observer. Two rockets at a speed of .6c going directly at each other would have a relativistic velocity of .88c, but to a third party observer the apparent velocity would infact be 1.2c. Based on this, it is possible for a third party observer to see two objects moving away from each other at literally 2c.

Another common expression of faster than light (FTL) travel is through virtual photons. Since they are virtual, and not bounded by the Heisenberg uncertainty principle, virtual photons can exceed c.

Galaxies can move away from each other faster than c. I don't know much about this, but it has something to do with Hubble's constant and how far apart the galaxies are.


There are many more common examples of things that can move FTL, but those are most of the classical arguements I am aware of. I am sure you can find more in depth articles online about everything I just mentioned, plus a whole lot more. This is not my area of expertise, but it is very interesting so I thought I would share.

 

[AluminumStudios]

I don't think what you're saying would work. Black holes have a massive gravitational gradient near them due to their compaction, but as you get a few AUs away from the event horizon their pull is the same as the pull of a star of the same mass.

In order to have a runway long enough to hit the speed of light, you would be far enough away that the pull you felt would be the same as the pull of a large star. Or if you were dealing with a supermassive black hole like you find in the center of galaxies the pull would be the same as any object in that galaxy.

So basically I don't believe that the gravitational pull could ever get you to the speed of light before the event horizon. As far as your mass vs. your speed playing a role - you'd be subject to the same physics that currently govern moving bodies in solar systems and the galaxy - and we don't see everything making a bee-line for the nearest black hole. So I don't suspect that this line of reasoning works.

Bigsm00th: "but to a third party observer the apparent velocity would infact be 1.2c. Based on this, it is possible for a third party observer to see two objects moving away from each other at literally 2c. "

Bigsm00th, I think you might be misundertanding relativity. Wouldn't a third party observer, observe each object moving away from HIMSELF at 1c? I don't think 2c would be observed anywhere. I don't fully recall how it was explained but the book Elegent Universe by Brain Greene has an example of how nothing can ever be observed going faster than c.

 

[MrDudeMan]

Originally posted by: AluminumStudios
snip


Bigsm00th: "but to a third party observer the apparent velocity would infact be 1.2c. Based on this, it is possible for a third party observer to see two objects moving away from each other at literally 2c. "

Bigsm00th, I think you might be misundertanding relativity. Wouldn't a third party observer, observe each object moving away from HIMSELF at 1c? I don't think 2c would be observed anywhere. I don't fully recall how it was explained but the book Elegent Universe by Brain Greene has an example of how nothing can ever be observed going faster than c.

You, as a third party observer, see the relativistic doppler shift of the objects approaching each other from an angle. Johann Doppler actually intended his work to explain the shift in frequency or wavelength of the light emitted by moving atoms and astronomical objects. This is commonly referred to as 'red shift' and 'blue shift', which, interestingly enough, don't use visible or even infrared waves; instead, microwaves emitted by galaxies or other distant objects are captured and the resulting absorption lines then give a 'shift'.

Anyway, two objects approaching each other give a third party observer an apparent relative speed of greater than c. This should intuitively make sense.



A fixed light source in frame S emits waves separated in space by the wavelength L (Lambda) and moving outward at a speed of c as seen from frame S, right? An observer in frame S' does not see L; he/she will see L', or the wavelength emitted from the source in frame S as seen from frame S'. Note: this is assuming frame S and S' have coincident x-axes and they are approaching each other with a speed V. This means an observer outside of both S and S' will see the apparent velocity of the waves from the source in S at c and also add the velocity V that relates how fast S' is approaching S.


If you really want to get into relativity, lets talk about relativistic rocket travel. If a ship sets out from Earth toward a distant galaxy at a speed of lets say .5c, time dilation would slow down the clocks inside the ship by a factor of about .15 compared to a clock on Earth. This means the astronauts would be travelling further in their lifetime than what would be normal to an observer on Earth, and if you could go fast enough, that speed would be greater than c as seen by someone on Earth. This isn't truly FTL travel, but it can be observed from someone in a different inertial frame as being FTL. The big advantage here is the ship will cover a great distance in the lifetime of the astronaut, enabling the people on-board to get further and live longer than if time dilation didn't have an effect on them.


I could be mistaken, and I admit that, but I am pretty sure the preceding information and examples are correct. I even cracked a textbook to verfiy some of it.

 

[DrPizza]

Originally posted by: Bobthelost
If you want to move faster than light all you have to do is shine it into a very dense clear material and hop in a jet.

Light only travels at 1/2 the speed of light (in a vacume, which is the value we all know and love) in fibre optic cable.

It's almost as useful as your suggestion, and physically possible, and theoretically valid.

Are you suggesting we hop in a jet and fly 93,000 miles per second (or slightly faster, so we are going faster than the light in your fiber optic cable.)
Incidentally, is the index of refraction for fiber optic cable that high? (2)?

 

[MrDudeMan]

Ok, I thought more about this, and I know for sure you can observe two objects with a combined apparent velocity faster than c. Assume for a moment that this scenario is possible. Relate all measurements to the x-axis:

We observe ship 1 moving in the positive-x direction: u1 = +c;
We observe ship 2 moving in the negative-x direction: u2 = -c;

If we then ask the question, ?What is the value (vector) u1 ? u2?? Obviously, u1 ? u2 = 2c.

If we ask the question, ?What is the velocity of ship 1 relative to ship 2??, then the answer is different because the question implies that we want to measure the speed of ship 1 in the f.o.r of ship 2. To answer that question, we must transform from our f.o.r to the f.o.r. of ship 2. That involves the Lorentz transformation. However, the transformation is impossible because the factor gamma is infinite. No two f.o.r.?s may have a relative speed of c.

An equivalent question that avoids the infinity problem is this (use b for beta, and e for epsilon):
We observe ship 1 moving in the positive-x direction: u1 = +bc;
We observe ship 2 moving in the negative-x direction: u2 = -bc;
You can let b = 1 ? e, where e as small as you want, as long as it is non-zero; for example, take e = 1E-10.
Still, the quantity u1 ? u2 = 2bc = 2(1 ? e)c = 2c ? 2ec, which is almost 2c.

Now, what is the velocity of ship 1 relative to ship 2? Transform from our f.o.r. to the f.o.r. of ship 2. The relative velocity is v = u2 = -bc. Then the speed of ship 1 transforms as:

u1? = (u1 ? v)/( 1 ? u1 v/c^2 ) = (2bc)/( 1 + b^2 ) = 2(1-e)c/(1+1-2e+e^2)
or u1? = c( 2(1 ? e)/( 2 ( 1 ? e + 0.5e^2) ) ); the best final form for the result is:
u1? = c ( 1 / ( 1 + (0.5e^2)/(1-e) )
which, I hope you can see, is less than c, even if e = 1e-10.

Finally, I will point out that no material object can travel at a speed equal to c.

 

[Matthias99]

Originally posted by: DrPizza
Incidentally, is the index of refraction for fiber optic cable that high? (2)?

I believe so (at least for some cables; according to some quick Google searching, 'average' ones are more like 1.5).

Fiber optics rely on total internal reflection, which (to maintain any kind of reasonable minimum bend radius) requires a relatively high IOR for the core of the cable.

 

[Velk]

Originally posted by: DrPizza

Originally posted by: Bobthelost
If you want to move faster than light all you have to do is shine it into a very dense clear material and hop in a jet.

Light only travels at 1/2 the speed of light (in a vacume, which is the value we all know and love) in fibre optic cable.

It's almost as useful as your suggestion, and physically possible, and theoretically valid.

Are you suggesting we hop in a jet and fly 93,000 miles per second (or slightly faster, so we are going faster than the light in your fiber optic cable.)
Incidentally, is the index of refraction for fiber optic cable that high? (2)?

I'd imagine the reasoning he was using was :
If optical fiber cable, which is not very dense, causes light to travel at half the speed, then a really dense transparent substance will cause it to travel much more slowly than that.

 

[Velk]

Originally posted by: Bigsm00th
Ok, I thought more about this, and I know for sure you can observe two objects with a combined apparent velocity faster than c. Assume for a moment that this scenario is possible. Relate all measurements to the x-axis:

We observe ship 1 moving in the positive-x direction: u1 = +c;
We observe ship 2 moving in the negative-x direction: u2 = -c;

If we then ask the question, ?What is the value (vector) u1 ? u2?? Obviously, u1 ? u2 = 2c.

If we ask the question, ?What is the velocity of ship 1 relative to ship 2??, then the answer is different because the question implies that we want to measure the speed of ship 1 in the f.o.r of ship 2. To answer that question, we must transform from our f.o.r to the f.o.r. of ship 2. That involves the Lorentz transformation. However, the transformation is impossible because the factor gamma is infinite. No two f.o.r.?s may have a relative speed of c.

An equivalent question that avoids the infinity problem is this (use b for beta, and e for epsilon):
We observe ship 1 moving in the positive-x direction: u1 = +bc;
We observe ship 2 moving in the negative-x direction: u2 = -bc;
You can let b = 1 ? e, where e as small as you want, as long as it is non-zero; for example, take e = 1E-10.
Still, the quantity u1 ? u2 = 2bc = 2(1 ? e)c = 2c ? 2ec, which is almost 2c.

Now, what is the velocity of ship 1 relative to ship 2? Transform from our f.o.r. to the f.o.r. of ship 2. The relative velocity is v = u2 = -bc. Then the speed of ship 1 transforms as:

u1? = (u1 ? v)/( 1 ? u1 v/c^2 ) = (2bc)/( 1 + b^2 ) = 2(1-e)c/(1+1-2e+e^2)
or u1? = c( 2(1 ? e)/( 2 ( 1 ? e + 0.5e^2) ) ); the best final form for the result is:
u1? = c ( 1 / ( 1 + (0.5e^2)/(1-e) )
which, I hope you can see, is less than c, even if e = 1e-10.

Finally, I will point out that no material object can travel at a speed equal to c.

I freely admit that while I can accept this as a concept, the side effects give me a headache.

For example, if you have two ships one light year apart, travelling toward each other at just under C, and your third party observer halfway between, at what time do they run into each other on each one's individual reckoning ?

If their closing velocity relative to each other is less than C, then it should take them slightly more than a year to impact with each other, whereas the stationary observer can see each of them approach at C from different directions, so they should collide in half a year, having travelled 1/2 of a light year to reach the observer.

It just seems to be very counterintuitive.

 

[Matthias99]

Originally posted by: Velk
I freely admit that while I can accept this as a concept, the side effects give me a headache.

For example, if you have two ships one light year apart, travelling toward each other at just under C, and your third party observer halfway between, at what time do they run into each other on each one's individual reckoning ?

There is no 'absolute' right answer; the amount of time that passes is different depending on where the observer is. They would all agree on when the collision occurs, but they would disagree on the amount of time that passed before the collision.

If their closing velocity relative to each other is less than C, then it should take them slightly more than a year to impact with each other, whereas the stationary observer can see each of them approach at C from different directions, so they should collide in half a year, having travelled 1/2 of a light year to reach the observer.

That's more or less correct. A clock attached to one of the moving objects would seem to run 'more slowly' than one that is 'stationary'.

Keep in mind that 'stationary' is relative as well! Someone on the 'moving' ship would see it as the clock on the 'stationary' one running faster than normal.

 

[MrDudeMan]

Originally posted by: Velk

I freely admit that while I can accept this as a concept, the side effects give me a headache.

For example, if you have two ships one light year apart, travelling toward each other at just under C, and your third party observer halfway between, at what time do they run into each other on each one's individual reckoning ?

If their closing velocity relative to each other is less than C, then it should take them slightly more than a year to impact with each other, whereas the stationary observer can see each of them approach at C from different directions, so they should collide in half a year, having travelled 1/2 of a light year to reach the observer.

It just seems to be very counterintuitive.

and there you have it...relativity is counterintuitive all the way around! I'm not kidding either. That is exactly what I said in class when I first started learning about relativity. My professor laughed and said "You are starting to understand it."


Matthias99, actually as long as the third party observer is within the light cone, or the future region of the collision, then the passed time will not vary through observation regardless of the position. The only thing that matters as far as the third party observer is concerned is where the ships are in the third party frame. Assuming they start on the x-axis at a point equadistant from the origin, they will collide at the origin in exactly half a year as far as the third party observer is concerned. From inside the ship, the journey will take considerably less. Here is the math:

From Velk's example, lets say each ship is going .995c (close enough to c that the result should be comprehendable). From an outside observer, the time to the collision will take 1/2 of a light year divided by .995, or .503 years...so basically half a year earth time.

Now, from inside the spaceship (either one since they are both going the same speed), the result is MUCH different. The formula to figure out the proper time (the proper time is always measured by the clock inside the moving intertial frame) is delta-T = gamma*delta-Tp where delta-T is the time measured by the observer, gamma is the conversion factor, and delta-Tp is the proper time.

gamma = 1/sqrt(1-v^2/c^2)

plugging in from above...
gamma = 1/sqrt(1-(.995*3*10^8)^2/(3*10^8(^2) = 10.01

so the proper time (the time inside the ship) is...
delta-T/gamma = .503/10.01 = .05 years

.05 years * 365.26 days/year = 18.3 days.

hope this helps...

 

[Matthias99]

Originally posted by: Bigsm00th
Matthias99, actually as long as the third party observer is within the light cone, or the future region of the collision, then the passed time will not vary through observation regardless of the position. The only thing that matters as far as the third party observer is concerned is where the ships are in the third party frame. Assuming they start on the x-axis at a point equadistant from the origin, they will collide at the origin in exactly half a year as far as the third party observer is concerned.

Rereading my last post, it was a little unclear. I didn't mean to imply that the elapsed time would vary depending on the position of the 'stationary' third party, just that it varies between the 'stationary' and 'moving' observers.

It is, indeed, EXTREMELY counterintuitive that the observed speed of light would remain constant and the rate of time passage would be relative. Standard Newtonian mechanics gives a totally different answer here than relativistic physics. But the relativistic answer appears to be how things work in reality, which is hard to argue with.

 

[MrDudeMan]

Originally posted by: Matthias99

Originally posted by: Bigsm00th
Matthias99, actually as long as the third party observer is within the light cone, or the future region of the collision, then the passed time will not vary through observation regardless of the position. The only thing that matters as far as the third party observer is concerned is where the ships are in the third party frame. Assuming they start on the x-axis at a point equadistant from the origin, they will collide at the origin in exactly half a year as far as the third party observer is concerned.

Rereading my last post, it was a little unclear. I didn't mean to imply that the elapsed time would vary depending on the position of the 'stationary' third party, just that it varies between the 'stationary' and 'moving' observers.

It is, indeed, EXTREMELY counterintuitive that the observed speed of light would remain constant and the rate of time passage would be relative. Standard Newtonian mechanics gives a totally different answer here than relativistic physics. But the relativistic answer appears to be how things work in reality, which is hard to argue with.

Ah, ok. I see what you meant now :beer:

I don't mean to beat a dead horse but I have to say this again - counterintuitive is the best word to describe Relativity

Newtonian Relativity, based off of Newtonian mechanics, actually makes sense. It doesn't take into account time dilation for different inertial frames though, which even though the answers obtained with his formulas make sense, they are incorrect. It just can't be that easy...nothing ever is.

For anyone that doesn't know, basically what I just said is the factor gamma doesn't exist in Newtonian Relativity. The proper time and the stationary clock time are the same, which is why the answers turn up wrong every time (although the difference is negligible at speeds << c).