Relativistic optics, w/ acceleration

[ralqs]

The site http://www.anu.edu.au/physics/Savage/TEE/gallery.html tries to show viewers what high-velocity travel would look like, taking into account all sorts of relativistic and optical effects. Very often, it features an object, like a train, accelerating to speeds close to c. They treat acceleration by figuring out what would be seen by an observer at a (constant) speed v, and then let v change continuously.

This doesn't seem valid to me. Wouldn't there be other effects caused by the observer's acceleration?

 

[JesseM]

The site http://www.anu.edu.au/physics/Savage/TEE/gallery.html tries to show viewers what high-velocity travel would look like, taking into account all sorts of relativistic and optical effects. Very often, it features an object, like a train, accelerating to speeds close to c. They treat acceleration by figuring out what would be seen by an observer at a (constant) speed v, and then let v change continuously.

This doesn't seem valid to me. Wouldn't there be other effects caused by the observer's acceleration?

I think it should be valid. Imagine at some moment an accelerating observer is at the same position as an inertial observer, and at that moment they have the same instantaneous velocity. So if we consider a very short time increment (so that we can deal with things like redshift that involve the time between successive peaks of a light beam) where the change in velocity due to acceleration is negligible, shouldn't both see the same thing? Their position as a function of time is nearly identical so the same light rays should be converging on their eyes at each moment during this short increment.

 

[ralqs]

Your logic makes sense, but wouldn't the fact that light has different speeds in the two frames make a difference?

 

[HallsofIvy]

Don't you mean "the fact that light has the same speed in the two frames"?

 

[JesseM]

Your logic makes sense, but wouldn't the fact that light has different speeds in the two frames make a difference?

What two frames? The question of when different light rays hit your eyes (and from what apparent direction, and at what quasi-instantaneous frequency relative to your clock) is frame-independent, you can calculate it in whatever frame you like and get the same answer.

 

[ralqs]

Don't you mean "the fact that light has the same speed in the two frames"?

The two frames I'm referring to are the accelerating frame and its instantaneously co-moving inertial frame. So they won't necessarily measure the speed of light to be the same.

What two frames? The question of when different light rays hit your eyes (and from what apparent direction, and at what quasi-instantaneous frequency relative to your clock) is frame-independent, you can calculate it in whatever frame you like and get the same answer.

The two frames I'm referring to are the accelerating frame and its instantaneously co-moving inertial frame. What do you mean when you say that "[t]he question of when different light rays hit your eyes ... is frame-independent"?

 

[JesseM]

The two frames I'm referring to are the accelerating frame and its instantaneously co-moving inertial frame.

What "accelerating frame"? An accelerating object doesn't have a unique accelerating frame, unlike with inertial frames there is no single "correct" way to construct a coordinate system where an accelerating object is at rest, in fact there are an infinite number of ways of doing this. But as I said, you can calculate what a given object will see visually using any frame you like, there's no need to use a frame where that object is at rest.

What do you mean when you say that "[t]he question of when different light rays hit your eyes ... is frame-independent"?

I mean that if I want to answer a question like "what does my clock read when the light ray from a particular distant event first reaches me eyes" or "at time T on my clock, what event on the worldline of object A am I seeing" or "what is the period according to my clock between seeing successive wave peaks of a given light beam (i.e. the frequency as seen by me)" or "if I am wearing a goldfish bowl on my head with angular markings all over it, what angular coordinates on the globe would the light ray from object A be passing through", I can calculate the answer to this question using any frame I like and get the same answer no matter what. So, all questions about visual appearances (the timing and angle of seeing distant events) have answers which don't depend on what reference frame I use to calculate them. In general, different frames always agree about local facts like what two clocks read at the moment they meet or what one clock reads when the light from a particular event hits it.

 

[ralqs]

Okay fine, but going back to what you said earlier:

I think it should be valid. Imagine at some moment an accelerating observer is at the same position as an inertial observer, and at that moment they have the same instantaneous velocity. So if we consider a very short time increment (so that we can deal with things like redshift that involve the time between successive peaks of a light beam) where the change in velocity due to acceleration is negligible, shouldn't both see the same thing? Their position as a function of time is nearly identical so the same light rays should be converging on their eyes at each moment during this short increment.

Okay, now suppose there were two observers, A and B, with A at rest w.r.t. a city and B moving at a speed v w.r.t. A. Suppose A and B both had cameras, manufactured such that they can pass through each other. Further suppose that at a certain moment both cameras do overlap, and they both take pictures. So by the same logic, wouldn't the same light rays converge on both cameras? So wouldn't they both see the same thing?

 

[JesseM]

Okay, now suppose there were two observers, A and B, with A at rest w.r.t. a city and B moving at a speed v w.r.t. A. Suppose A and B both had cameras, manufactured such that they can pass through each other. Further suppose that at a certain moment both cameras do overlap, and they both take pictures. So by the same logic, wouldn't the same light rays converge on both cameras? So wouldn't they both see the same thing?

The set of events they're seeing, and the visual angle they seem them at, are identical at that moment, but the redshift/blueshift would be different because frequency deals with the time between successive incoming wave peaks according to your clock, so you need at least a small (perhaps infinitesimal) interval of time to define it, and their different velocities cause them to measure different values for frequencies of incoming rays in that time.

edit: actually come to think of it the visual angles (and thus the angular size of various objects) won't be the same for both of them, because in one's rest frame the other is length-contracted, so if they each define angles on something like a goldfish bowl helmet with angular markings on it like I suggested above, in either one's rest frame the other's goldfish bowl is distorted in shape from a sphere to an ellipsoid.

 

[ralqs]

The set of events they're seeing, and the visual angle they seem them at, are identical at that moment, but the redshift/blueshift would be different because frequency deals with the time between successive incoming wave peaks according to your clock, so you need at least a small (perhaps infinitesimal) interval of time to define it, and their different velocities cause them to measure different values for frequencies of incoming rays in that time.

So everything would be the same, except for colour? I don't think that's right; wouldn't there be aberration effects and such for one observer and not the other?

 

[ralqs]

edit: actually come to think of it the visual angles (and thus the angular size of various objects) won't be the same for both of them, because in one's rest frame the other is length-contracted, so if they each define angles on something like a goldfish bowl helmet with angular markings on it like I suggested above, in either one's rest frame the other's goldfish bowl is distorted in shape from a sphere to an ellipsoid.

Just noticed your edit...but if they're seeing different things, then the way light hits the photographic plate has to be different for both cameras. So how do you explain that?

 

[JesseM]

Just noticed your edit...but if they're seeing different things, then the way light hits the photographic plate has to be different for both cameras. So how do you explain that?

In any frame where they have different speeds, their shapes should be distorted by different degrees of length contraction. Even if you pick a frame where they have equal and opposite speeds so their shapes look identical, if you are imagining light being registered on an extended surface like a photographic plate or the fishbowls I was imagining, you have to consider that each surface will define simultaneity differently in its own rest frame, and therefore the set of photon registrations at different positions that are considered to have happened at "the same time" (and thus give a snapshot image at a single instant) will be different.

 

[ralqs]

But if the photographic plates are perpendicular to their motion, then length contraction and relativity of simultaneity shouldn't matter, no?

 

[JesseM]

But if the photographic plates are perpendicular to their motion, then length contraction and relativity of simultaneity shouldn't matter, no?

In that case I do think they should register the same image, even if colors due to redshift/blueshift are different. Though my guess is that calculations in relativistic optics would typically involve not simultaneous registration of photons on an extended flat surface like a photographic plate, but rather simultaneous convergence of photons at a single point, with the optical angles determined by something akin to my "comoving spherical fishbowl with angular markings on it" picture. In this case, the light rays converging at the center of the fishbowl would be the ones that all strike the surface of the fishbowl simultaneously in the fishbowl's rest frame, and in any frame where the fishbowl is moving, the point at the center is moving too so light rays that hit the surface at at the same time and same distance from the center in this frame won't generally hit the center at the same moment.

 

[ralqs]

Okay, that makes sense. Thanks for your patience.