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Can Light Actually Slow Down?


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Keeping it short and simple,

If light is traveling (at the speed of light of course) in the exact opposite direction to the pull of a black hole, will it not HAVE to slow down?

Light is effected by the gravitational pull of black holes, and usually spirals into them, but if a photon is traveling in -exactly- the opposite direction to the pull of the black hole, then it will have only one way to accelerate: backwards. It seems to me that physicists are always saying light can't slow down in a constant medium...but if light didn't slow down here what would happen? I can't see another possible solution.

 

Please reply with your thoughts, I have tried talking with person after person about this, but no one seems to care or understand. If I'm wrong I want to be SHOWN why I'm wrong, I'd love that! It's just that no one has ever given me an answer to this and it's driving me nuts! :-(

Edited by CameronFriday
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To keep it simple, light always travels in a straight-line path through space-time. Gravitational bodies warp and bend space, causing the shortest-distance-path to seem curved, the way the line connected two points on the surface of a sphere is curved.

 

So, let's think about your situation: where would the photon be coming from if its direction were exactly orthogonal to the surface of the sphere (with radius the schwarzchild radius of the black hole)? The only way this could happen is if the photon came out from the black hole (from the other side of the event horizon). This is clearly impossible.

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The photon is moving towards the black hole, and hits an object which is also moving towards the black hole. This object will be going slower than the photon, of course, because nothing can reach the speed of light. The photon hits the object just before it enters the event horizon so that it reflects directly opposite the pull of the black hole.

Edited by CameronFriday
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To answer the original question, the photon always moves at c.

 

Kyrisch question is a bit deeper.

So, let's think about your situation: where would the photon be coming from if its direction were exactly orthogonal to the surface of the sphere (with radius the schwarzchild radius of the black hole)?

Suppose some spacefaring species finds a massive, non-rotating black hole. They outfit a small, robust probe with a transmitter and null the transverse component of the probe's velocity vector with respect to the black hole. The probe will fall toward the black hole. The combination of a massive black hole and a small, robust probe means the probe will withstand the tidal forces as it nears the event horizon.

 

From the perspective of the observers safely ensconced far from the black hole, they will see the transmissions from the probe redshift and become fainter and fainter. Suppose the transmitter is quite powerful and the receivers are quite sensitive, both in frequency and signal strength. It is as if the probe comes to a stop at the event horizon. (This motivated one of the earlier names for black holes, frozen stars.)

 

The photons are nonetheless moving at c. They instead lose energy via a change in frequency.

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Ah. So you're saying the photon would not lose velocity, but instead constantly lose energy in different ways like reduced wavelength etc. until it either escaped the black hole at a lower energy level or was depleted, being absorbed into the black hole completely?

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If light is traveling (at the speed of light of course) in the exact opposite direction to the pull of a black hole, will it not HAVE to slow down?
No. It slows down when it approaches the black hole. Of course, you can't measure this locally, but the coordinate speed as measured by a distant observer reduces.

 

Light is effected by the gravitational pull of black holes, and usually spirals into them, but if a photon is traveling in -exactly- the opposite direction to the pull of the black hole, then it will have only one way to accelerate: backwards.
Gravity isn't actually a "pull". Yes, the path of a photon is affected by the gravity of a black hole, and a photon can spiral in, but it's better to think of it as a "veer". The black hole alters the space around it so that the photon doesn't travel straight any more.

 

It seems to me that physicists are always saying light can't slow down in a constant medium...but if light didn't slow down here what would happen? I can't see another possible solution.
It isn't a constant medium. If it was, a passing photon would go straight as a die and would shoot right on past the black hole.


Merged post follows:

Consecutive posts merged
Ah. So you're saying the photon would not lose velocity, but instead constantly lose energy in different ways like reduced wavelength etc. until it either escaped the black hole at a lower energy level or was depleted, being absorbed into the black hole completely?
A photon escaping a black hole doesn't actually lose any energy. Conservation of energy applies. The frequency doesn't actually change.

 

Imagine you've got some emitter, and you can measure frequency down near a black hole, and then again way up in space. You'd measure the frequency to be different, but it wasn't the photons that changed. It was you and your measuring devices. You were time dilated down near the black hole, so you measured the frequency as high. Up in space you aren't, so you measure the frequency as lower. So the light looks like it's been redshifted and has lost energy climbing out of the gravitational field, but it hasn't.

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No. It slows down when it approaches the black hole.

No and yes. The no answer first: Using local rulers and local meter sticks, photons travel at c. Always. For the yes answer, see post #4.

 

 

Light is effected by the gravitational pull of black holes, and usually spirals into them, but if a photon is traveling in -exactly- the opposite direction to the pull of the black hole, then it will have only one way to accelerate: backwards.[/quote']Gravity isn't actually a "pull". Yes, the path of a photon is affected by the gravity of a black hole, and a photon can spiral in, but it's better to think of it as a "veer". The black hole alters the space around it so that the photon doesn't travel straight any more.

Terrible answer! What is the difference between a "pull" and a "veer"? (Hint: None.)

 

 

The black hole alters the space around it so that the photon doesn't travel straight any more.

Photons always travel in straight lines. The problem is that our concept of a straight line is straightjacketed by our Euclidean mindset.

 

 

A photon escaping a black hole doesn't actually lose any energy. Conservation of energy applies. The frequency doesn't actually change.

The first and last sentences are wrong. The middle sentence is kinda right, kinda wrong; it depends on what you mean by "energy" and by "conserved". The conservation principles we know and love are consequences of Noether's Theorem applied to Euclidean space/universal time. General relativity does not have a global time coordinate, so conservation laws are a bit problematic in general relativity. That said, there are analogs to conservation of energy in general relativity.

 

Generalizing from Newtonian mechanics, one would expect the energy of a photon to be a combination of the energy due to the photon's frequency and some analog of gravitational potential energy. That is indeed the case. The analog to gravitational potential energy is the metric tensor. The conserved quantity is the energy due to a photon's frequency and the equivalent of gravitational potential energy. The end result: photons redshift as they climb away from a black hole (or any massive object).

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No and yes. The no answer first: Using local rulers and local meter sticks, photons travel at c. Always. For the yes answer, see post #4.
Photons always travel at 299,792,458 m/s because we define the second and the metre using the local motion of light. See the NIST fountain clock for this, wherein we count 9,192,631,770 incoming microwave peaks and then say a second has elapsed. Hence the frequency is 9,192,631,770 Hertz by definition, regardless of time dilation. The metre doesn't change because the slower light and the larger second cancel each other out. This is why the locally measured speed of light is always the same.

 

Terrible answer! What is the difference between a "pull" and a "veer"? (Hint: None.)
The difference is there's no action at a distance. The earth doesn't pull a photon towards it, it "conditions" the surrounding space so that it's inhomogeneous, and the photon then moves in a curvilinear fashion. Hence it veers. See Einstein's 1920 Leyden address re inhomogeneous space.

 

Photons always travel in straight lines. The problem is that our concept of a straight line is straightjacketed by our Euclidean mindset.
No, the motion is curvilinear. It's curved, and we can see that it's curved. See the Foundation of the General Theory of Relativity for curvilinear motion. We call this curved spacetime, and we talk of geodesics, but those lines are curved, not straight.

 

The first and last sentences are wrong. The middle sentence is kinda right, kinda wrong; it depends on what you mean by "energy" and by "conserved". The conservation principles we know and love are consequences of Noether's Theorem applied to Euclidean space/universal time. General relativity does not have a global time coordinate, so conservation laws are a bit problematic in general relativity. That said, there are analogs to conservation of energy in general relativity.
Those sentences are not wrong. You've got an inhomogeneous region of space, and a photon moving out away from it. There are no other particles. There is no mechanism by which the photon loses energy. The initial photon energy is the same as the final photon energy. The measured difference is due to the change in the environment. When you're time dilated along with your clocks, the photon frequency appears to be high. Move to a higher altitude where you're not, and the frequency looks lower.

 

Generalizing from Newtonian mechanics, one would expect the energy of a photon to be a combination of the energy due to the photon's frequency and some analog of gravitational potential energy. That is indeed the case. The analog to gravitational potential energy is the metric tensor. The conserved quantity is the energy due to a photon's frequency and the equivalent of gravitational potential energy. The end result: photons redshift as they climb away from a black hole (or any massive object).
Newton mechanics has been superseded by general relativity. The climbing photon doesn't lose energy, and the descending photon doesn't gain energy. In similar fashion the gravitational potential energy of a body at altitude is in the body. As it falls, the body loses this potential energy and acquires kinetic energy instead. But the sum total of kinetic+potential energy of that body doesn't increase. If it did, the gravity of a collapsing spherical shell would increase, because energy causes gravity. It doesn't increase. There is no additional energy added. Conservation of energy applies. The "force" of gravity does not add energy to the falling body.
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Conservation of energy applies.

 

The change in frequency does not violate conservation of energy, because conservation of energy applies within a frame of reference. Energy is not invariant between frames, and you are comparing it between frames.

Edited by swansont
punctuation
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Photons always travel at 299,792,458 m/s because we define the second and the metre using the local motion of light.

This is a fallacious argument, and you know that it is.

 

For everyone else reading this thread, what Farsight said is correct -- now, that is. The current definition of a meter is the distance traveled by light in vacuum for 1⁄299,792,458 of a second. What Farsight has omitted saying, and the reason this is a fallacious argument, is that this is a very recent definition (October 1983). For the 23 years before that, the meter was defined in terms of the wavelength of a specific frequency of light. For 161 years before that, the meter was defined in terms of various prototype meter-long bars.

 

In other words, the standards committee gave the scientific community more than 100 years to disprove that notion that the speed of light is the same to all observers. This idea did not start with Einstein. It started with James Maxwell. In fact, a large portion of Einstein's 1905 paper is on Maxwell's equations.

 

The Michelson-Morley experiment, the Kennedy-Thorndike experiment, and a slew of others showed that this mind-bending concept is correct -- and all of those pre-1983 experiments were based on definitions of length and time that did not depend on the speed of light.

 

In the minds of the standards committee, a century's worth of cumulated experimentation was more than enough to justify the switch to making the speed of light fundamental.

 

No, the motion is curvilinear.

What is straight? The motion is curvilinear only from our insistence on describing things in terms of Euclidean space. To many physicists, what is bent is our stubborn insistence on using a 2300 year old concept. A photon follows a straight view.

 

Those sentences are not wrong.

Yes, they are.

 

There are no other particles.

There's a flippin' big black hole.

 

There is no mechanism by which the photon loses energy.

You seem to be forgetting that there's a flippin' big black hole again.

 

You also should do some reading on the metric tensor.

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