Measuring the speed of light in a straight line

[MikeandSuch]

So recently I watched a video detailing how it is impossible to measure the speed of light in a straight line because it's not possible to synchronize two-time measuring devices without first knowing the speed of light.



But I was thinking if light can orbit a black hole in the photon sphere could you not shine a laser in one direction then measure how long it takes for it to orbit the black hole and reach you? That way the light travels in a straight line but doesn't require two separate synchronized measuring devices.
Obviously, this information would be worthless since it could never be transmitted out of the black holes orbit so whoever does the measurement will be the only human to ever know the true speed of light before being turned to spaghetti.

 

[phyzguy]

How is this different from simply using a mirror to reflect the light back at you?

 

[Ibix]

An experiment in curved spacetime like that can be made to return more or less any result you like because what you mean by "distance travelled" is open to interpretation. Certainly you'll get different results with different orbits. Look up "Shapiro delay".

It doesn't matter how clever your experimental design, you cannot get around the fact that the one way speed of light is conventional.

 

[Baluncore]

So recently I watched a video detailing how it is impossible to measure the speed of light in a straight line because it's not possible to synchronize two-time measuring devices without first knowing the speed of light.

Since the speed of light has been defined as 299 792 458 metres/second, are you actually measuring the length of one metre or the time of one second ?

 

[sophiecentaur]

Summary:: How to measure the speed of light in a straight line using a black hole and a laser.

the speed of light in a straight line

An experiment in curved spacetime like that can be made to return more or less any result you like because what you mean by "distance travelled" is open to interpretation.

When we observe Gravitational Lensing, the light on a whole range of path lengths arrives here at the same time (is why an image can be seen). The speed of the light at all points is the same (for someone measuring it there) so the times / distances involved on each path must change to allow this.
The trip round a back hole is just an extreme case of this. The time for the round trip would be the same because of the profile of the gravity field and you could see a similar lensing effect.

 

[Ibix]

When we observe Gravitational Lensing, the light on a whole range of path lengths arrives here at the same time (is why an image can be seen). The speed of the light at all points is the same (for someone measuring it there) so the times / distances involved on each path must change to allow this.

Not sure what you are saying here. Gravitationally lensed images can certainly appear different ages - in principle you could see a supernova in a gravitationally lensed galaxy at two (or more) times, once in each image.

 

[Vanadium 50]

in principle you could see a supernova in a gravitationally lensed galaxy at two (or more) times, once in each image.

Already happened. https://en.wikipedia.org/wiki/SN_Refsdal

 

[sophiecentaur]

Not sure what you are saying here. Gravitationally lensed images can certainly appear different ages - in principle you could see a supernova in a gravitationally lensed galaxy at two (or more) times, once in each image.

Doesn't it ever appear during one exposure?

 

[Ibix]

Doesn't it ever appear during one exposure?

Seeing a supernova at exactly the same time in two images could only happen if you, the lens, and the supernova were colinear, I think.

 

[sophiecentaur]

Seeing a supernova at exactly the same time in two images could only happen if you, the lens, and the supernova were colinear, I think.

I would have thought that seeing a ring in the same image (just hours of exposure) would imply that the paths from the explosion have all the same propagation times. Otherwise any image would be of just one section of the ring. Different images would have different sectors of the lensed image.
?

 

[phyzguy]

I would have thought that seeing a ring in the same image (just hours of exposure) would imply that the paths from the explosion have all the same propagation times. Otherwise any image would be of just one section of the ring. Different images would have different sectors of the lensed image.
?

You only get a ring when the lensing galaxy is spherically symmetric. With real galaxies, which are lumpy and not spherically symmetric, you get individual images. In this image of SN Refsdal, there were four distinct images of the SN with time delays of a few days, and a fifth image further away that re-appeared about a year later:

https://arxiv.org/pdf/1512.04654.pdf

 

[sophiecentaur]

Seeing a supernova at exactly the same time in two images could only happen if you, the lens, and the supernova were colinear, I think.

I would have thought that seeing a ring in the same image (just hours of exposure) would imply that the paths from the explosion have all the same propagation times. Otherwise any image would be of just one section of the ring. Different images would have different sectors of the lensed image.
?

You only get a ring when the lensing galaxy is spherically symmetric. With real galaxies, which are lumpy and not spherically symmetric, you get individual images. In this image of SN Refsdal, there were four distinct images of the SN with time delays of a few days, and a fifth image further away that re-appeared about a year later:

https://arxiv.org/pdf/1512.04654.pdf

View attachment 283789

ok. I was quoting an exceptional situation but isn’t it all an example of Fermat’s principle?

 

[phyzguy]

ok. I was quoting an exceptional situation but isn’t it all an example of Fermat’s principle?

Fermat's principle just says the the time of travel is a local extremum. There can be multiple local minima. What about this case, where you are simultaneously seeing an object directly and in a mirror. You would not expect those light travel times to be equal, would you?

 

[sophiecentaur]

Fermat's principle just says the the time of travel is a local extremum. There can be multiple local minima. What about this case, where you are simultaneously seeing an object directly and in a mirror. You would not expect those light travel times to be equal, would you?View attachment 283794

I didn't think I was discussing that. My point is that lensing increases the apparent brightness of an object as the images of the object appear in the same place. When this happens, say in a paraboloid reflector or in a convex lens, the same transit time is involved for all paths then the image is focussed. I thought that lensing was more than just refracting the light as it goes past a massive object. Is that wrong?

 

[Ibix]

I would have thought that seeing a ring in the same image (just hours of exposure) would imply that the paths from the explosion have all the same propagation times. Otherwise any image would be of just one section of the ring. Different images would have different sectors of the lensed image.
?

But the rings we see are galaxies, which have been emitting light for a long time. SN Refsdal must have been more or less exactly behind the lensing object to get multiple copies of the supernova on the same day - presumably there's something wildly asymmetric in the lens that produces a fifth copy.

I thought that lensing was more than just refracting the light as it goes past a massive object. Is that wrong?

I think so. Gravitational lenses are absolutely dreadful, optically speaking. That's why you get a point source smeared into a ring, worst case. You may get more light, total, than you would if there were no lens (as you would with @phyzguy's mirror setup) but it's not a crisp image. Have a look at a couple of orbiting black holes: https://svs.gsfc.nasa.gov/13831

 

[sophiecentaur]

The location of the image in that plane mirror diagram is due to one light path. There is no 'enhancement' of an image. The equivalent to lensing (even just rubbish quality) is surely a concave reflector, producing a coherent image from light via many paths. Better would be a convex lens where the light paths are still the same and that relies on the different path lengths in the glass of the lens. The gravitational gradient on the path past the black hole produces the same effect by causing light, initially leaving the object in different directions to follow paths with the same delay. We use that explanation for basic optics so why not with space-time 'optics'?
I guess all I am saying is that a gravity gradient will produce refraction effects (however poor quality). Its works the 'same way' for both - the light goes slower whilst near the centre of the hole or lens. So why not allow Fermat's principle there?
I can't see any relevance in saying that the galaxies have been running for a long time. If your lab object is a femtosecond flash or a DC LED, the image will be in the same place. Non-ideal dispersion effects will blur the DC image and change the time profile of the short pulse but Fermat's still at work, imo.

 

[phyzguy]

So I think you are saying all of the light in a given image arrives at about the same time, and I would agree. But look at the picture. There are multiple images of the same supernova. Each image follows a different path. Think of it as the warped gravitational field of the highly complex galaxy corresponds to multiple lenses, each with a slightly different curvature.

 

[sophiecentaur]

But look at the picture.

It would make a rubbish camera, I agree but what do we call the process - is my question?

 

[Vanadium 50]

presumably there's something wildly asymmetric in the lens that produces a fifth copy.

I believe the lens is a cluster, not a single galaxy.

 

[sophiecentaur]

I believe the lens is a cluster, not a single galaxy.

A really really crumby camera then!