Photons can't accelerate, can they?

[swansont]

Well where does all the extra energy go? If a gamma-ray goes to a lower gravitational potential, it then becomes a less energetic photon? And if so, where did that extra energy go?

 

What extra energy? If you drop a ball to the ground its gravitational potential energy changes and so does its kinetic energy, but energy is conserved.

[questionposter]

What extra energy? If you drop a ball to the ground its gravitational potential energy changes and so does its kinetic energy, but energy is conserved.

 

But there you go. You raise a ball, so there's more potential energy within the system, then you drop it to the ground and that potential energy becomes kinetic energy and get's released into the ground. The energy is "conserved", but it's location changed. What happens to the location of the energy in a photon when it goes to a lower gravity potential?

[swansont]

But there you go. You raise a ball, so there's more potential energy within the system, then you drop it to the ground and that potential energy becomes kinetic energy and get's released into the ground. The energy is "conserved", but it's location changed. What happens to the location of the energy in a photon when it goes to a lower gravity potential?

 

It's in the photon — it gets blue-shifted. When it hits the ground it has more energy, and that energy gets absorbed by whatever it hits.

[Mystery111]

When a photon is emitted from an electron, is it instantly traveling at c? Doesn't this contradict F=ma*? Or, in other words, doesn't an object have to accelerate if it is to go from one speed (such as rest) to another (such as c)?

 

*I know photons can be considered massless, but then why are they pulled down by gravity?

 

The reason why photons are deflected by gravity is because gravity is the resultant geometry of the curved space. To understand this, you need to learn about GR and how it treats wave equations in a curved region. For instance, let us now concentrate on a specific wave equation, quite a famous one:

 

[math]\frac{\partial^2 \phi}{\partial t^2} = c^2\frac{\partial^2 \phi}{\partial x^2}[/math]

 

This describes two kind of waves, one wave which moves to the left, another to the right. From now on, we will use natural units. To express this equation in three dimensions

 

[math]\frac{\partial^2 \phi}{\partial t^2} = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}[/math]

 

We can rewrite this as

 

[math] \eta^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}}=0[/math]

 

You can derive the relationship for a curved spacetime using the Covariant Derivative. Thus:

 

[math] \frac{\partial}{\partial x^{\mu}} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}}+ \Gamma_{\mu \alpha}^{\mu} g^{\nu \beta} \frac{\partial \phi}{\partial X^{\beta}}=0[/math]

 

[math] \nabla g^{\mu \nu} \frac{\partial \phi}{\partial x^{\nu}}[/math]

 

So to work out the Covariant Derivative is effectively involving Christoffel symbols, and this equation describes the parallel transport between two points as a curve on a manifold and this is the definition of why a wave form like a photon may travel along a certain geodesic - it is purely the curvilinear distortions of spacetime.

Edited September 6, 2011 by Mystery111

[DrRocket]

The speed of light changes because the frame of reference is no longer locally flat.

 

 

 

The local speed of light is c, whether the manifold is flat or not.

 

One relatively (pun intended) way to see this is to note that light follows a null geodesic so that infitesinally [math]\Delta x = c \Delta t[/math].

 

What generates confusion is the use of "coordinate time" (and coordinate space) which is an artifice of the local coordinate system. This can result in a "speed" that appears to be different from c. But coordinate time is not really time and that apparent speed is not real either.

 

All sorts of screwy things can appear to happen due to coordinate effects.

[swansont]

The local speed of light is c, whether the manifold is flat or not.

 

One relatively (pun intended) way to see this is to note that light follows a null geodesic so that infitesinally [math]\Delta x = c \Delta t[/math].

 

What generates confusion is the use of "coordinate time" (and coordinate space) which is an artifice of the local coordinate system. This can result in a "speed" that appears to be different from c. But coordinate time is not really time and that apparent speed is not real either.

 

All sorts of screwy things can appear to happen due to coordinate effects.

 

I was under the impression that the speed of light measured away from the local frame will not be c if you assume that it is flat but it isn't. Which is why there is the Shapiro delay.

[DrRocket]

I was under the impression that the speed of light measured away from the local frame will not be c if you assume that it is flat but it isn't. Which is why there is the Shapiro delay.

 

Shapiro delay is a coordinate effect, which is rather like assuming that things are flat when they are not.

 

Actual measurements of speed can never be truly local. But the local speed of light is always c.

 

There is lots of confusion on this, even misunderstandings of some of Einstein's early writings. In some choices of coordinate systems you get light speeds that differ from c, but that is just a coordinate effect.

 

Another way to think about this is in terms of special relativity. In SR light always travels at c, as a fundamental axiom. But special relativity is the localization of general relativity. So given that velocity is a local concept, the speed of light in general relativity is again always c.

 

One difficulty, particularly for experimentalists, is that experiments always deal with coordinate values. For instance you assume that it is fair to compare the time on your wristwatch with the clock on the wall across the room. Strictly speaking, GR does not let you do that. But the approximation that spacetime in the laboratory is flat allows you to resort to special relativity as an approximation and you make that comparison without difficulty. Virtually nothing is lost in this approximation -- and thankfully so or you atomic clock guys would have a hell of a problem comparing clocks. But over large distances, or near black holes, the difference between proper time and coordinate time becomes more important.

 

Let's compare the job of the experimentalist vs the theorist in determining speed: 1) The experimentalist first has to determine what is time and what is space -- he must choose a local coordinate system (nature tends to do this for him). Then he needs a ruler and a stopwatch, and some finite time and spatial arrangement to determine the time required by the moving body to cover a known distance. Then he has to get out a pencil and paper and divide the distance covered by the time recorded on the stopwatch. 2) The theorist just says "the 4-velocity is c", and he is correct because the 4-velocity of anything is c. For light it is easy for both because for light the 4-velocity and the coordinate velocity in Lorentzian coordinates coincide.

 

All of the above depends on the theory being correct. The really subtle problem is translating the abstract stuff into coordinate quantities that are measurable with instrumentation and determining if any disagreement with the theory is due to the approximations made in the coordinates or in solving the difficult equations or if the theory itself is wrong. That takes someone with deeper understanding than mine.

[questionposter]

It's in the photon — it gets blue-shifted. When it hits the ground it has more energy, and that energy gets absorbed by whatever it hits.

 

That doesn't make sense though because the photon should have less potential energy. If I shot an infrared photon into an electron and it raised it up one energy level and then I shot another infra-red photon near a black hole from the same source so that it got blue shifted, it would raise the electron 2 energy levels instead of one. How did energy move from the gravitational field to the photon to allow it to blue shift? Because if I red-shift a photon in the way you describe, it inherently has less energy which is why it won't pump an electron up as many energy levels, so where did that energy go? Were there "mini" photons that got released? How does a gravitational field give and take energy in photons?

Edited September 7, 2011 by questionposter

[swansont]

That doesn't make sense though because the photon should have less potential energy.

Less potential energy. Where did it go? Into the increase in frequency, thus increasing the photon energy.

[questionposter]

Less potential energy. Where did it go? Into the increase in frequency, thus increasing the photon energy.

 

So if I shoot enough photons near a gravitational field to blue shift them, it will eventually siphon out an object's energy?

Edited September 8, 2011 by questionposter

[swansont]

So if I shoot enough photons near a gravitational field to blue shift them, it will eventually siphon out an object's energy?

 

I don't understand your misconception enough to answer this.

 

If you drop a ball and let it hit the ground, does that "siphon off" energy from the object?

[the asinine cretin]

Isn't it possible to "stop" photons with a Bose-Einstein condensate or a Fermionic condensate? What's the story on that?

Edited September 8, 2011 by Ceti Alpha V

[khaled]

Just a curiosity, photon is a boson or a fermion ?

[swansont]

Photons are Bosons. They have spin 1.

 

Isn't it possible to "stop" photons with a Bose-Einstein condensate or a Fermionic condensate? What's the story on that?

That's a press release/journalism cutesy catch-phrase issue. There are stories of "stopped light." What is happening is the light is absorbed and the sample is put in a state where it cannot de-excite on its own, but the sample is prepared in such a way that you can later induce the de-excitation with a laser pulse, and something identical to the original phase-coherent pulse is recreated.

[questionposter]

I don't understand your misconception enough to answer this.

 

If you drop a ball and let it hit the ground, does that "siphon off" energy from the object?

 

You said the energy for the blue shift of a proton comes from the gravitational field itself, so doesn't blue-shifting enough photons use up all the energy from the gravitational field? I mean the Earth can run out of angular momentum form gravity (or the moon), so idk.

Edited September 9, 2011 by questionposter

[swansont]

You said the energy for the blue shift of a proton comes from the gravitational field itself, so doesn't blue-shifting enough photons use up all the energy from the gravitational field? I mean the Earth can run out of angular momentum form gravity (or the moon), so idk.

 

The implication is that photons are affected by gravity just like matter, so it's going to be the same as if it were a ball. A ball dropping adds to its KE. A photon has an increase in frequency, giving a similar increase in E= hv.

[questionposter]

The implication is that photons are affected by gravity just like matter, so it's going to be the same as if it were a ball. A ball dropping adds to its KE. A photon has an increase in frequency, giving a similar increase in E= hv.

 

But if energy can be transferred through a gravitational field, does that mean that the photon actually is taking energy form the object when blue-shifted just as orbit of the moon siphons out Earth's kinetic energy through gravity? I'm kind of not getting any direct answers from you.

 

 

 

[davey2222]

Interesting.

 

But even if the earth's rotational (or translational) kinetic energy is used up in blue-shifting a photon, its gravitational field is still there. Then further blue-shifting through the g-field will result in what? Perhaps you meant to ask if the g-field will become weaker instead?

[csmyth3025]

You said the energy for the blue shift of a proton comes from the gravitational field itself, so doesn't blue-shifting enough photons use up all the energy from the gravitational field? I mean the Earth can run out of angular momentum form gravity (or the moon), so idk.

A photon traveling through space and a ball traveling through space will both gain energy relative to their energy at the point of origin as they move towards a source of gravity. This isn't because the source of gravity is "giving up" its energy, it's because the source of gravity is, by definition in general relativity, the source of curvature of space-time. The photon and the ball are also sources of curvature of space-time, but their effect is immeasurably small.

 

You're right that the Earth can lose angular momentum to the Moon:

 

...some of the Earth's rotational momentum is gradually being transferred to the Moon's orbital momentum, and this causes the Moon to slowly recede from Earth at the rate of approximately 38 millimetres per year...

(ref. http://en.wikipedia....Tidal_evolution )

 

However, the Earth can't "run out of angular momentum" by transferring it to the Moon:

 

The Moon is gradually receding from the Earth into a higher orbit, and calculations suggest that this would continue for about fifty billion years. By that time, the Earth and Moon would become caught up in what is called a "spin–orbit resonance" in which the Moon will circle the Earth in about 47 days (currently 27 days), and both Moon and Earth would rotate around their axes in the same time, always facing each other with the same side. However, the slowdown of the Earth's rotation is not occurring fast enough for the rotation to lengthen to a month before other effects change the situation: about 2.1 billion years from now, the increase of the Sun's radiation will have caused the Earth's oceans to vaporize, removing the bulk of the tidal friction and acceleration.

 

Note that even if the Earth-Moon system remains undisturbed for fifty billion years, the Earth will still retain angular momentum (it will be rotating around its axis once every 47 days). The Earth's angular momentum won't decrease beyond that because the Earth and the Moon will be tidally locked.

 

More to your point, though, is the fact that the Earth will still have just as much gravity fifty billion years from now as it does today. The transfer of rotational energy from the Earth to the Moon doesn't change the gravitational field of the Earth any more than a photon or a ball falling toward the Earth does.

 

The gravitational field of the Earth is dictated by its mass. Things like photons and balls "falling" to Earth can (minutely) increase the Earth's mass and thus (minutely) increase the strength if its gravitational field. They can't "use up" the energy of Earth's gravitational field.

 

Chris

[swansont]

But if energy can be transferred through a gravitational field, does that mean that the photon actually is taking energy form the object when blue-shifted just as orbit of the moon siphons out Earth's kinetic energy through gravity? I'm kind of not getting any direct answers from you.

 

If you feel you are not getting any direct answers it's because I don't understand the questions. The moon "siphons" rotational KE from the earth via gravity but that's because of tidal interactions (it also give a context for the question you asked earlier that made no sense to me). I don't know if there is a mechanism in GR that would let you do something along those lines.

[questionposter]

If you feel you are not getting any direct answers it's because I don't understand the questions. The moon "siphons" rotational KE from the earth via gravity but that's because of tidal interactions (it also give a context for the question you asked earlier that made no sense to me). I don't know if there is a mechanism in GR that would let you do something along those lines.

 

So then how does a photon actually get energy by going into a high gravitational potential?

[swansont]

So then how does a photon actually get energy by going into a high gravitational potential?

 

How does a ball get/give energy by doing the same thing?

 

Do you get why I don't understand how to answer your questions? Because they are physics 101 questions with the added part about gravity affecting photons. Are you looking for the physics 101 answer, or are you trying to find out something else?

[davey2222]

I think the issue questionposter brought up was about conservation of energy. If the photon gets blue-shifted due to gravity, where did the energy to blue-shift the photon come from?

[questionposter]

I think the issue questionposter brought up was about conservation of energy. If the photon gets blue-shifted due to gravity, where did the energy to blue-shift the photon come from?

 

That's actually what I was asking, and then I was also asking the details, such as how gravity itself conserves that energy.

 

How does a ball get/give energy by doing the same thing?

 

Do you get why I don't understand how to answer your questions? Because they are physics 101 questions with the added part about gravity affecting photons. Are you looking for the physics 101 answer, or are you trying to find out something else?

 

If your a physics major how can't you answer physics 101 questions? That's like saying Isaac Newton could't do 15+7.

Edited September 10, 2011 by questionposter

[davey2222]

In physics there is the problem in unifying gravity with EM and Quantum mechanics. Perhaps this is the result of our unsuccessful attempt to unify gravity with them, that we are unable to answer such a question satisfactorily?