You misunderstood .... it red shifts ... but does not slow down ...Originally Posted by Digix
the speed of light is the same in every frame and is a constant. Space and time can vary but their ratio needs to be constant with respect to c.
Order of Kilopi
You misunderstood .... it red shifts ... but does not slow down ...
the speed of light is the same in every frame and is a constant. Space and time can vary but their ratio needs to be constant with respect to c.
Order of Kilopi
Light will not slow down when escaping a massive body. It will redshift though, and when escaping from the event horizon of a black hole, the redshift will be infinite (in other words, it will have zero energy and infinite wavelength). When coming from farther down (below the event horizon), the redshift will remove more energy than the photon has, in essence preventing it from escaping (but never slowing it down - it either escapes at c or it doesn't escape).
Established Member
Ok so that was misunderstanding.
so if we go back to topic
according to these formulas
http://en.wikipedia.org/wiki/Gravitational_redshift
Infinite redshift happens if we calculate path of photon which travels from event horizon to infinity.
But if photon goes smaller distance it still has some remaining energy left. If we use energy of another particle to refill loss It can go a little further. So it can climb up this lader until escapes completely
Also I found another problem with this formula of gravitational redshift:
redshift between radius and infinity is infinite at horizon level, but below it becomes imaginary.
That is quite an absurd, because infinity is more or less valid but imaginary absolutely is unacceptable in physic in any form.
So I suspect something must be wrong here.
Order of Kilopi
No, it cant. There is not a path out. Once inside the EH there is no path out, no ladder, no taking a climb up then getting colliding with another photon etc ...
There is no path out.
Established Member
Let me put it this way, if you send two beams over an equal distance, and send one through an area of higher gravity then the other, it will take longer to reach the distance then the unaffected beam.
Both beams will appear to move at C when hitting the sensors, though, and if you could set up a system that measured the speed of part of the beam inside the time dilated area, that too would show the beam to be traveling at C, since it measure it in the dilated frame of referance.
It is important to realise that this is not a slowing of the light itself, it is a change in time flow, things just happen slower inside the time dilated area than outside.
One example of this effect is the result of the Shapiro experiment, where a radar signal was sent from the Earth towards Venus so that it passed by the Sun. The return signal showed the predicted time delay(that is, the predicted compound delay of gravitational curvature of the signal and the gravitational time dilation) within accuracy limits.
Several space probes have done similar experiments, and the effect must be taken into account when designing and using probe radio sytems. Satellite navigation systems(and other satelite systems) where changes in travel time of the signal may cause problems in the results will have to compensate for the effect too.
Order of Kilopi
Huh? I disagree with this. Light travels at C from all frames ... if we see light near a black hole it still travels at c from our vantage point as well as from a local observers vantage point.
The only thing that could make light take longer to get to a target would be if it curved so that the path was longer.
Light doesnt travel slower in a gravitational well.
Established Member
So in that case,
If I understand it well then when you move towards horizon redshift between 2 vertically arranged points increases until it reaches infinity in zero distance.
resulting infinity divided by zero redshift at horizon level.
that is the only reasonable say to prevent everything from escaping.
because in other way we can use external refueling to climb up.
In this way everything that reaches horizon level should be ripped apart instantly, no matter how big is black hole, because acceleration gradient between 2 points is always infinite.
Established Member
The curvature of light is a different effect, as it is gravity affecting space, not time.
Light always travel at C in local frames of referance, you can not measure the speed of light in another referance frame(like the observer close to the black hole), only your own, and there the time dilation factor would be equal for you and the light. Therefore, regardless of where you are, light always move at C.
Or put another way, I would say that the concept that light always moves at C to an observer requires light to be subject to gravitational time dilation, if not, light would appear to move faster to an observer in a high gravity field than one in a low gravity environment, since his/her time would move slower.
Established Member
seems to be valid point.
if we assume 2 receivers inside of gravity well they should notice that beam goes faster than light.
But that will conflict with acceleration-gravity equivalence.
No acceleration is felt what means no time dilation can occur.
Established Member
the formation of black-holes is not possible in the first place
because , black-holes inorder to form , can only do so one dimensionally
inotherwords
the focus of the rotation can be from the N or S but not both at the same time
Order of Kilopi
North, you will keep your own ATM out of this Q&A thread.
Get up, a get-get, get down.
Established Member
Yes, if light was not subject to time dilation, detectors inside the gravity well would measure a higher speed of light than outside the well.
I would expect that in an accelerating frame of referance, the effect would be identical.
I am not sure what you mean by no acceleration is felt, what is it that does not feel acceleration?
Order of Kilopi
The collapse of a core into a BH isn't from pressure. It is simply from gravity overcoming neutron degeneracy
Established Member
Solar mass black holes take about one second to form an event horizon, so some photons do escape, but the last photons reach the event horizon from the inside, too late to escape. If the hypothesis is correct, these photons will travel inside the event horizon for trillions plus of years, relative to both inside and outside observers. Likely some of what we think we know about black holes and relativity is wrong, but that probably does not mean we need to reject the existence of black holes, nor relativity. Neil
Order of Kilopi
My bold for reference. Can you demonstrate, in appropriate mathematical detail, why you think this assertion has any validity?
Established Member
Ok here:
first we need to calculate redshift amount between close alder steps.
v=sqrt(2g(h1-h2) where g= GM/r
this velocity change between lader steps can be used to calculate dopler shift.
g increases linearly while we move towards mass center.
so assuming that distance between lader steps is dh we get this:
dv=sqrt(2GM/r *(dh) )
this proves that Doppler shift between closely spaced lader steps is never infinite and no matter how close to mass center, It is always possible to climb up some small distance.
In contrast escape velocity is calculated assuming that dh is infnite.
however my calculations now depend on fact if gravity can slow down light speed itself.
Established Member
This deserves additional discussion since I always expected that light speed is not affected by gravity but you note seems valid.
However how will you explain this:
According to your theory if we send light beam through massive tube it will require more time to reach destination.
But if we slowly drop some ball through same tube it will reach destination faster than ball going same distance at same initial speed in free space.
Something must be wrong here
Established Member
The special theory of relativity is called such because it is a special case. Inertial motion only, in flat space, and measured locally, will find that c=c.
How can gravity bend light if there is no change in 'velocity'?
Special relativity does not override general relativity. It's the other way around.
c=c for special cases, ok?
Order of Kilopi
Huh? gravity curves space and time.
Order of Kilopi
But what i think you are forgetting is that BOTH time and space are dialated in such a way that c is kept constant. As time dilates, space is also curved to keep a ratio of c. If A is in a deep gravitational well, B is a distance away in no gravitational well and C is distant from both A and B.
BH--A----------B------------C
And B releases a series of photons. C, B and A will measure the speed of the photons at the speed of light. If A releases the photons C,B, and A will agree that those photons are moving at hte speed of light.
Established Member
A ball? Hmmm... Well, a ball is subject to acceleration by gravity, if that is what you mean... Light would not speed up, since it already moves at the maximum speed in that reference frame, it just gains energy as an increase in the frequency of its constituant waves.
Order of Kilopi
Is this small distance curved? Please dont forget that space-time is very curved around the EH. It seems that your small step may actually not be in the direction of the EH at all, but curved back to the singualrtiy.
There is no future other than the singularity.
Order of Kilopi
Gravity does not slow down light speed.
Order of Kilopi
Huh? c=c period. Gravity bends light by changing the geodesic. Light always travels in a straight line accross its geodesic. When space time is curved, light takes a different path.
Think of the moon. The moon is travelling in a straight line along its geodesic.
Order of Kilopi
I think you are really confused. Light gains energy as it falls into a gravity well and spends energy when it leaves the well, similar to the ball.
the speed of light stays constant.
Established Member
Ok, lets leave blackholes alone for some time and discuss this problem.
can you explain somehow that fact about light speed.
T----->R1------>R2----->R3
R1 and R2 are in gravitational well while T and R3 are in free space
If light speed is constant beam travel time between R1 and R2 will be less than 1/3 of total time. Because R1-R2 clocks tick slower so for them light will appear to propagate in speed greater than C
How do you solve this?
I also expect that light speed should be constant , but something seems wrong here.
Established Member
Not so simple. gravitational acceleration is independent to mass, so if you use electric particle accelerator particle cannot surpass speed of light because in the end it just gains mass while speed remains nearly constant. Light speed is not some "sonic limit", but inability to accelerate object of infinite mass.
However in gravity field this is not the case, because acceleration does not depend on particle mass and it is theoretically seems possible to exceed c.
In gravitational field even particle with infinite mass will continue to accelerate at same rate like particle with low mass.
Established Member
I am not sure there is any way for A B and C to know how the light's travel time was affected by gravity in the experiment you show here, since the sender and recievers are in different non-inertial frames, any system they use for syncronisation of the experiment would be affected by the time dilation.
In the Shapiro experiment things were a bit different:
Venus ♀ <========== radar signal =====> ⊕ Earth
Sun--------------->☉
The radar signal was bounced of Venus past the Sun. So therefore the sender and reciever was in the same reference frame, and their time flow would be syncronous. So when the result showed that the traveltime was longer than it should have been, and that matched within the accuracy limits of the experiment the predicted compound delay of curved space and dilated time in this system, that shows that light is affected by both. if only one effect affected the light the result would be different
Yes, exactly, so you have to take in to account both the effects of curved space and of curved time.
You said "The only thing that could make light take longer to get to a target would be if it curved so that the path was longer." I interpreted that to mean that you were claiming that the light would be affected by curved space, but not by curved/dilated time.
So what I meant with it being a different effect was that light takes longer to pass a massive object than curvature of space alone accounts for, so you have to take gravitational time dilation(or gravitational time curvature) in to account too.
Light gains energy in the form of blueshift, as I said, while the ball would gain energy as an increase in velocity relative to the mass, the effect of leaving the gravity fields would be opposite, of course.
I am not confused really, but I suppose I should have mentioned that the effect of going up the well would take away the energy gained.
Order of Kilopi
You might want to look up momentum and exactly how mass, momentum, and energy interact to correct your misunderstanding.
It does? Care to show exactly how the relativistic equations don't apply? Hint, again, you need to stop concentrating on mass and look up momentum.
Established Member
Honestly, there is just so much one can practicaly write in a discussion, if every post were to mention every aspect of physics and their effect on something, they would be more like books than posts on a forum. So as the balls wouldn't be likely to reach relativistic speeds, I didn't mention it.
Gravity will not accelerate anything past the speed of light in the local reference frame, though there are some theoretical things like worm holes that connect different parts of space time and warp drives that could cause a global speed higher than c by using warping of space-time. The thing with these is that no mass ever reaches c in a local reference frame, so they do not really violate the constancy of c. Anyway, this is realy a subject for another discussion.
An infinite mass wouldn't accelerate at all, though you may find that it would pull the universe it was in into itself in a big crunch event.
Posting Permissions
