« on: 02/11/2017 19:32:36 »
How was NASA able to use the relativistic doppler shift equation when the speeds involved are so small compared to the speed of light?
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
I can do the same assuming that the light does travel at c with respect to the Earth.This is true, but there is a subtle difference. If light were traveling at c relative to the observer then the time taken from emission to reception would depend upon the distance from sender to receiver at the time of emission. This would lead to some strange effects. Let's say that Algol is 93.5 light-years from Earth. The Earth is at the maximum distance away from Algol and is approaching Algol. Every time an eclipse occurs the Earth is closer to Algol. This means that the period of time between eclipses will be shorter. 93.5 years later, when the Earth is closest to Algol, the Earth will start receiving those eclipses. So we would have the strange effect of having a shorter period of time between eclipses as the Earth is moving away from Algol.
At the moment the light from one eclipse of Algol occurs, Algol is a certain distance from Earth. If Earth and Algol are moving apart at 66629 mph, then 2.867321 days later, when the light for the next eclipse occurs, Algol will be 4585121.54 miles or 24.6162 light seconds further away from the Earth when that light leaves, This means that the light from this eclipse, traveling at c relative to the Earth, take 24.6162 seconds longer to travel to the Earth than the light from the earlier eclipse did. If the eclipses occurred 2.867321 days apart at Algol, the Earth would see them 24.6162 seconds or 0.00028491 days further apart 2.867321 + 0.00028491 =2.86760591 days apart which is also close to the given time. (quibbling about the small difference here does no good because the Earth's orbital speed itself varies by some 2240.3 mph (.6223 miles per sec) over over the course of an orbit, which would result in a larger difference in answers than what we got.)
The point is that I can get the same type of spacing between observed eclipses by assuming that light does travel at c relative to the Earth.
Because Sue is racing towards the beam of lightI thought it was obvious that I meant from George's point of view.
[G] [A---1 light-second-----B]George (designated by the G) is watching his friends Sue (location A) and Sally (location B) zoom away from him at 0.8c. His friends are located at both ends of a spaceship 1 light-second long. From George's point of view, the ship is only 0.6 light-seconds long. Sue sends a light beam to Sally and it only takes 1 second to get from Sue to Sally. According to George, because Sally is running away from the light, the trip actually takes 0.6/0.2 = 3 seconds. So 1 second for Sue and Sally is equal to 3 seconds for George. Sally decides to send a beam of light to her friend Sue. Once again, it takes 1 second to get from Sally to Sue. Because Sue is racing towards the beam of light, according to George the trip takes 0.6/1.8 = 1/3 seconds. So 1 second for Sue and Sally is equal to 1/3 seconds for George.
T----------F (ship A) (5 light-years in length)So we see that the tail of ship B passes the front of ship A,
<---(0.8c) F------T (ship B) (3 light-years in length)
T------F ----->(0.8c) (ship A) (3 light-years in length)So we see that the front of ship B passes the tail of ship A
F----------T (ship B) (5 light-years in length)
BImagine that there is an observer at the mid-point between A and B. An observer is at the mid-point between B and C.