Speed of Light Affecting Objects' Appearances?

[connorp]

So this has been really bugging me over the past few days (and forgive me if the answer is so simple). Let's say we're observing the Sombrero galaxy. It is about 29 million ly away and 50 thousand ly in diameter. So we should be observing the "front" of it at what it looked like 29 million years ago, and the "back" of it 29.05 million years ago. Why doesn't this extra distance change the galaxy's shape? If, for example, the galaxy was moving directly away from us in a straight line (not that it is), wouldn't the galaxy be compressed? Hope this makes sense.

 

[Jonathan Scott]

So this has been really bugging me over the past few days (and forgive me if the answer is so simple). Let's say we're observing the Sombrero galaxy. It is about 29 million ly away and 50 thousand ly in diameter. So we should be observing the "front" of it at what it looked like 29 million years ago, and the "back" of it 29.05 million years ago. Why doesn't this extra distance change the galaxy's shape? If, for example, the galaxy was moving directly away from us in a straight line (not that it is), wouldn't the galaxy be compressed? Hope this makes sense.

Try a calculation!

It's true that there's a time difference between the back and the front. However, for this to affect the shape by some fraction, say 1%, you would need the difference in speed of the back and front to be that fraction of the speed of light. Most astronomical objects are moving many orders of magnitude more slowly than that.

 

[connorp]

I guessed something like this, but still am slightly confused. Anyway you could elaborate a bit more? Thank you.

 

[Jonathan Scott]

I must admit that referring to "difference in speed" was too restrictive, but I was thinking of what might distort the shape in a visible way.

Overall motion towards or away from the observer would have little effect, as for galactic distances we normally have no direct way to determine accurate relative distances of stars within a galaxy along the same line of sight. What we would typically do is match up the appearance (including such aspects as relative velocity towards or away from the observer at the edges, deduced from redshift) with a common galaxy shape and deduce the approximate orientation from that appearance.

I think the most visible distortion would be if the galaxy as a whole was moving sideways relative to the line of sight from the observer, as that overall motion could be much faster than any relative motion (which would tend to make the galaxy come apart).

If the galaxy was for example overall moving sideways at speed v and was of diameter D then the time delay for light from the far side would be t = D/c so it would appear to be lagging behind by distance vt = vD/c = (v/c) D. This means that the fraction of the diameter by which the far side would be lagging behind the near side is v/c. For the far side to appear 1% behind its true position, the galaxy would have to be moving overall sideways at about 1% of the speed of light, that is at about 3000km/s (which is probably possible) but I do not think the effect of a 1% lag on the visible shape would be detectable.

For very distant galaxies, the effective recession velocity for a high redshift could in theory cause significant flattening in the direction away from the observer, but as we don't have any means of measuring the distance to individual stars in such galaxies, this would not affect the visible shape.

 

[connorp]

Great explanation. Thanks!

 

[CWatters]

I calculate light from the edge of the sun takes about 2 seconds longer to get to us than light from the centre of the disc because the centre is nearer. So the position of the edge is 2 seconds "older" and should lag the position of the centre as it appears to us. It wouldn't make it look a different shape but a sun spot that is geometrically in the centre would appear very slightly out of position relative to the edges.

The sun takes 24 hours to "move" 360 degrees around the Earth so a central sun spot would appear (360*2)/(24*60*60) = 0.008 degrees out of position. Please check my reasoning and my sums!