# Blueshift at event horizon in rest and accelerated frames

1. Aug 15, 2013

### Stonius

Basic question I'm sure, but here goes;

If the following is correct;

An object may be considered to be 'at rest' when there are no inertial forces acting upon it (ie; it is not accelerating).

A satellite is at rest because it is in freefall. A person standing still on the earth's surface is not truly 'at rest' because their freefall is arrested by the surface of the earth, hence the feeling of 'gravity' (equivalent to inertia).

Bending and red/blueshifting of light has been observed both as a result of speed AND gravity.

Then I get confused with the following;

If a satellite slowly freefalling back to earth is a rest frame, then so is falling into a black hole. Because time passes more slowly in deeper gravitational wells, the light coming in from behind the falling observer (light from infinity) comes from a higher gravitational potential and is blueshifted as a result. Vice versa for light coming from beside and below which is redshifted due to the gravity of the black hole.

Then consider an observer hovering just outside the event horizon. He would have to accelerate very close to the speed of light just to stay put, which is very much *not a rest frame. Aside from a gazillion G forces and tidal nastiness, lets pretend he can stay there quite happily.

Accelerating to C (which is what he will have to do to maintain position), will cause blueshifting of the light in the direction of travel, and redshifting in the opposite direction. Yet there is *also the effect of the blueshifting caused by viewing light from a higher gravitational potential falling in.

Are the effects from gravity and the affects from acceleration cumulative?

Does the hovering observer then see more intense redshift/blueshift?

Will the hovering observer see the falling observer as decelerating *from lightspeed to a stop as he passes the event horizon?

Would the falling observer see the hovering observer at a standstill on the verge of the event horizon, then approaching at him at C?

Many thanks

Markus

2. Aug 15, 2013

### Staff: Mentor

According to one definition of "rest", yes. But there are other possible definitions. "Rest" is not an absolute concept in relativity.

No, this is not correct. There is a perfectly valid sense of "at rest" according to which the person standing on the Earth's surface is at rest and the free-falling satellite is not. Again, "at rest" is not an absolute concept.

Yes. In fact, in the general case, there is no unique way to distinguish the two; there is bending and red/blueshifting of light in a curved spacetime, but no unique way to say how much of it is due to "speed" and how much is due to "gravity".

In the specific case you're discussing, you *can* distinguish the two; but the way that works out is not the way you think.

Consider the case you give, of an observer "hovering" close to the horizon of a black hole, and a second observer free-falling past him. Light is coming in from infinity towards both of them. That light will be blueshifted as it "falls" into the gravity well, yes. But the observer relative to which that blueshift occurs is the *hovering* observer, *not* the free-falling observer. In other words, for the purpose of defining gravitational redshift/blueshift, the correct sense of "at rest" is the sense according to which the hovering observer is at rest. Basically, "at rest" in this sense *means* hovering at a constant altitude (i.e., not orbiting the hole, just hovering, and accelerating in order to maintain the hover).

So from the hovering observer's viewpoint, yes, the incoming light is (greatly) blueshifted. But it is *not* blueshifted to the free-falling observer; in fact, it appears redshifted! From the viewpoint of the hovering observer, this is because the free-falling observer is falling inward at close to the speed of light, close enough so that the blueshift due to falling into a gravity well is more than cancelled out by the Doppler redshift due to the free-faller's inward velocity relative to the hovering observer.

What do things look like from the free-faller's viewpoint? To him, there is no such thing as "gravitational redshift". According to his sense of "at rest", the light sources far away at infinity are moving away from him at close to the speed of light, so of course the light coming from them is heavily redshifted. And the hovering observer that he passes is also moving away from him at close to the speed of light--even *closer* to the speed of light than the light sources at infinity, so the free-faller will expect that the hovering observer will see the incoming light from infinity blueshifted.

3. Aug 15, 2013

### pervect

Staff Emeritus
I usually regard "at rest" as having meaning with respect to a specific set of coordinates, and it means that the position of the body in the chosen set of coordinates is constant.

For this post, I'll call this condition "constant coordinates" just to avoid semantic confusion.

Anyway, in Schwarzschild coordinates, the body with constant coordinates experiences a blueshift due to gravity. If you imagine a body with constant coordinates at the same location (but different velocity) as a body that's free-falling into the black hole, if we assume that the free falling body is moving towards the black hole, it will experience an additional redshift due to its motion.

in fact, one can write the following expression

f_infinity = frequency of signal emitted at infinity
f_constant = frequency of signal received by constant coordinate body
f_falling = frequency of signal received by infalling body

Then

f_infinity / f_falling = (f_infinity / f_constant) * (f_constant / f_falling)

So you can see that the total frequency ratio aka blueshift is the product of two terms.

This redshift or doppler shift due to the motion of the infallig body may and often is larger than the blueshift due to gravity. The details depend on the speed of the infalling body, but if the infalling body started out with no velocity at infinity, then the redshift will exceed the blueshift. This requires a non-trivial calculation, which ahs been posted, but I won't get into the detials unless there is some interest.

The received frequency of an infalling light beam is a coordinate independent concept, and defining doppler shift as the ratio of the received frequency to the transmitted frequency at infinity is also coordinate independent. However, the "partioning" or explanation of what parts of the blueshift are "due to gravity" and what parts are "due to motion" depends on the details of the coordinates one chooses.

4. Aug 15, 2013

### Stonius

Thanks for the replies.

I suppose my presumptions of what is 'at rest' stem from the solution to the twin paradox and the equivalence principle.

I thought that the equivalence principle meant that inertia from acceleration and G forces from gravity were equivalent, which is why it is the travelling twin's rapid deceleration and acceleration at the half-way point that allows both twins to be the same age when they are reunited.

If being 'at rest' is arbitrary, even in spite of acceleration, then surely both twins can expect each other to be younger at the reunion. It is during the acceleration that the travelling twin ages more quickly and 'catches up'. From this I figured that a rest frame is intrinsically different from an accelerating frame, hence my conclusions that freefalling and hovering must provide different experiences of black holes.

So if I read you correctly, a freefalling observer sees no gravitational time dilation, only doppler shifting because the distance between the falling observer and the outside universe is increasing. Is the light in the direction of the black hole also red-shifted? Is there no direction from which light is blueshifted? Does the universe simply grow dark around the freefalling observer?

In truth, my curiosity comes from the modern assertion that you could easily cross the event horizon of a large black hole without even noticing it. I find it hard to imagine the blueshifting and bending of light wouldn't be a bit of a giveaway, given that we can see gravitational lensing from earth.

Thanks
Markus

5. Aug 15, 2013

### Staff: Mentor

That's one way of putting it, yes.

That's one way of analyzing the scenario, yes. The Usenet Physics FAQ goes into this:

The entire FAQ, of which the link here is just one section, is worth reading.

Being at rest is arbitrary, but that doesn't mean acceleration is arbitrary. Being in free fall is still physically different from being accelerated. But the difference is not usefully viewed as "being at rest" vs. "being in motion".

If you rephrase this as "freely falling is intrinsically different from being accelerated", it would be fine. Plus, it would still justify this:

That conclusion only requires freely falling to be different from accelerating (hovering); it doesn't require that either one corresponds to "rest".

More precisely, the freefalling observer's most natural way of interpreting what he sees is to attribute all of the redshift he observes in light coming in from infinity to Doppler shift, and none of it to gravitational time dilation. As I noted in my previous post, in general there is no unique way to draw the distinction between "Doppler shift" and "gravitational redshift/blueshift".

You have to be careful thinking about "light in the direction of the black hole". Light can't escape from the black hole, so the only light a free-falling observer would see while still above the horizon would be light emitted closer to the horizon than he is, but still outside it. Whether he would see that light to be redshifted or blueshifted would depend on what the light source was doing when the light was emitted. If the light source was "hovering" closer to the horizon than the free-faller is when he sees the light, the light would appear blueshifted to him; but if the light source was free-falling into the hole ahead of him, and emitted the light just before it fell below the horizon, it would appear redshifted to him.

Inside the horizon, it's impossible to "hover"; everything must fall inward towards the singularity. So any light coming from ahead of the free-faller would appear more and more redshifted.

However, there's another wrinkle here too: everything I said above applies to light that is only moving radially. Light can also move tangentially, and that considerably complicates the picture.

Yes, that assertion is often made sloppily. The more precise way to say it is that you could easily cross the horizon of a large black hole without noticing anything *locally*. Observations of light coming from the rest of the universe are not local observations. When those are taken into account, yes, there are definitely optical effects that happen around the horizon. There are some good animations here:

6. Aug 16, 2013

### Stonius

Thanks Peter, I appreciate it. All makes sense now.

Re those animations; I don't suppose there's an easily digestible account of those Schwarzschild bubbles anywhere?

Cheers
Markus

7. Aug 16, 2013

### yuiop

I thought the whole point of the twin paradox was that the twins would have different ages when they re-united?

This is not true. The time dilation of a free falling observer clock relative to a clock at infinity can be obtained from the Schwarzschil metric as:

$$\frac{d tau}{d t} = \sqrt{ (1-2GM/r) - \frac{1}{(1-2GM/r)}\left(\frac{dr}{dt}\right) ^2- r^2 \left(\frac{d \omega }{dt} \right)^2}$$

where I am using units of c=1. The first term (1-2GM/r) inside the square root on the right is the time dilation due to gravity. The second term is the time dilation due to the vertical velocity (squared) and the third term is the time dilation due to angular or orbital velocity (squared).For an object falling from rest at infinity the velocity reached at height r is:

$$v = \sqrt{\frac{2GM}{r}}$$

This is the velocity as measured by a stationary observer that remains at r as the object passes. The velocity as seen by the observer at infinity is:

$$\frac{dr}{dt} = \sqrt{\frac{2GM}{r}} (1-2GM/r)^2$$

We can now substitute this into the first equation to obtain:

$$\frac{d tau}{d t} = \sqrt{ (1-2GM/r) - (1-2GM/r)\frac{2GM}{r} - r^2 \left(\frac{d \omega }{dt}\right )^2}$$

Assuming no angular motion so that $\omega =0$ and simplifying we obtain:

$$\frac{d tau}{d t} = (1-2GM/r)$$

Note that the time dilation of the free falling clock is the square of the time dilation of stationary clock in the gravitational field so the proper time of the free falling clock remains real even when the clock falls below the event horizon.

The above expression can be rewritten as:

$$\frac{d tau}{d t} = \sqrt{ 1-2GM/r} *\sqrt {1-2GM/r}$$

$$\frac{d tau}{d t} = \sqrt{ 1-2GM/r} *\sqrt {1-v^2}$$

where v is the locally measured velocity (vertical or horizontal) of the object as measured by an observer that is stationary in the gravitational field at r. This equation can be viewed as the product:

(Total time dilation) = (Gravitational time dilation)*(Special Relativity time dilation).

For observed red or blue shift you have to factor in the Doppler shift due to relative velocity into the above equations.

Last edited: Aug 16, 2013
8. Aug 16, 2013

### Staff: Mentor

This is the "viewpoint" of an observer at infinity, not the viewpoint of the free falling observer. The statement of mine that Stonius was responding to was about how things look from the viewpoint of the free falling observer--more precisely, how things look in his local inertial frame. Your derivation is all fine, but it's done in global Schwarzschild coordinates, and the split you give between what is "time dilation due to gravity" and "time dilation due to velocity" is only valid if you adopt those specific coordinates; it does not hold if you adopt other coordinates.

9. Aug 16, 2013

### yuiop

I accept your observation that this split between gravitational time dilation and velocity time dilation is unique to Schwarzschild coordinates. I question if a free falling observer can even ask how much of his time dilation is gravity or velocity induced because generally speaking any observer is not aware of their own time dilation. Time dilation is always the point of view of another observer at a different potential or with motion relative to the clock under consideration.

10. Aug 16, 2013

### Staff: Mentor

True, but "time dilation" here is really just an indirect reference to the observed redshift, and "time dilation due to gravity" vs. "time dilation due to velocity" are just two different ways of explaining the redshift. The free-falling observer can ask whether the redshift he observes in incoming light is due to gravity or velocity--and the most natural answer for him, based on his local inertial frame, is that it's all due to velocity.

11. Aug 16, 2013

### Stonius

Yes, of course you're right. I was looking at a minkowski diagram where the planes of simultaneity seem an first glance to show same age at reunion. But yes, the travelling twin ages less in relation to the stationary one. Thanks.

12. Aug 16, 2013

### WannabeNewton

Stonius, with regards to your initial statements in post #1, I think what you meant to say instead of 'rest' was 'locally inertial'. Certainly freely falling observers are locally inertial, because locally they resemble inertial observers in flat space-time. So freely falling observers are special in that they constitute locally inertial frames; rest frames however can be attributed to any arbitrary observer.

13. Aug 16, 2013

### Stonius

Thanks, yes I see that now. Rest is arbitrary, but inertia is not. :-)