Register to reply 
Wanting to make sure I understand length contraction and time dilation 
Share this thread: 
#1
Oct2605, 12:34 AM

P: 145

I made my own problem and tried to answer it to see if I understand relativity well. Can you please let me know if I got the answers right?
A spaceship moving at .866c compared to a pair of stars. The stars are located at points A and B which appear to be 2 light seconds apart from an observer stationary to the stars at point M (the midpoint of the two stars). The spaceship is flying strait from one star to the other. Therefore the spaceship will see the two stars separated by a distance of: L = L0 (1 – (v^2 / c^2))^.5 L = 2ls (1 – ((.866c)^2 / c^2))^.5 L = 2ls * .5 L = 1 light second Now, at the instant the spaceship is at point M Star A and Star B go nova simultaneously. When will: (a) the spaceship see the stars going nova? (b) the man at point M see the stars go nova? (a) Since the space ship is at point M when this occurs Star A .5 light seconds away from his position, by definition of midpoint. Now because it is a known fact that light travels at the velocity of c in all reference frames it must be concluded that the spaceship will see the nova from Star A .5 seconds later. The exact same logic and results can be achieved for Star B which also yields a result of .5 seconds later. (b) By similar logic the stationary man at point M will see the novas of Star A and Star B 1 second after the spaceship passes point M. Time dilation can also be used to solve this problem. It is possible to find how long it will take the spaceship to see the novas by using the time dilation equation from the stationary mans frame. This can make more since because it is more intuitive that the stationary man will see the novas after 1 second. t0 = t (1 – (v^2 / c^2 )).^5 t0 = 1s (1 – ((.866c)^2 / c^2))^.5 t0 = 1s * .5 t0 = .5s As expected the solutions match those found when using length contraction. 


#2
Oct2605, 01:12 AM

Sci Advisor
HW Helper
PF Gold
P: 562

When you say Stars A and B go nova "simultaneously", I assume you mean simultaneously according to a stationary observer at M.
The same events will NOT be simultaneous for the spaceship observer, and your calculation for the times the spaceship sees the novas are incorrect. In those calculations, you have ignored the distance the spaceship moves in the time it takes light to travel from each star to the spaceship. You've also ignored the fact that the novas didn't happen simultanously in the spaceship reference frame. 


#3
Oct2605, 06:59 AM

P: 145

I guess this points out where I am getting confused. Can you by chance help me to understand the flaw of this logic?
First Arguments Argument 1. De Sitter observed that no matter if a star is moving towards, away, or is stationary to you if the star is X light years away then the light from that star will take X light years to reach us. Argument 2. It makes no difference if the star is moving towards us or we are moving towards the star with constant velocity. Conclusion. If we are moving towards, away, or are stationary to a star X light years away the light from the star will reach us in X years. Second Argument Argument 1. The First Argument is independent of any distance traveled by either us or the star. Argument 2. If we are moving towards, away, or are stationary to a star X light years away the light from the star will reach us in X years. Conclusion. It is irrelevant if we have moved when calculating how long light takes to reach us. Third Argument Argument 1. Light travels at a constant velocity for all reference frames. Argument 2. Reference frames travel with different velocities compared to each other. Conclusion. The velocity at which light propagates is not equal as viewed from one reference frame to another, i.e. to reference frames are traveling at v1 and v2 where v1 and v2 are not equal. Both reference frames will perceive light to travel at C (argument 1) and both have different velocities (argument 2). Therefore, the speed of light can be calculated in the first reference frame for both frames by: (v1  v1) + c = c And (v1  v2) + c <> c The first and second arguments are what I used to solve my original problem and the third argument is just a logical statement that seems to come from the first two arguments. I’m sure there is a flaw in the logic somewhere I’m just not sure where. Thank you for the help. 


#4
Oct2605, 08:48 AM

Mentor
P: 11,749

Wanting to make sure I understand length contraction and time dilation
[tex]u' = \frac {u  v} {1(uv/c^2)}[/tex] (assuming for simplicity that the velocities all lie along the same line) Let u = c (that is, let "the object" be a bit of light), solve for u' and see what you get... 


#5
Oct2605, 08:36 PM

Sci Advisor
HW Helper
PF Gold
P: 562

Wizardsblade:
It's all a matter of choosing one reference frame and working with that. It is probably simpler to look at things in the rest frame of the star, in fact, since the star can be assumed to be "fixed" in space, if you like. 


#6
Oct2705, 03:49 PM

P: 145

Ok I think I understand better now. I'm going to try to rewrite these so they work.
Revised First Argument Statement 1. De Sitter observed that no matter if a star is moving towards, away, or is stationary to you, if the star is X light years away then the light from that star will take X light years to reach us. Statement 2. We always view from our reference frame and from our reference frame the star is the only one moving at a constant velocity. Conclusion. If a star is moving towards or away from us that we measure X light years away (while we are moving at a constant velocity) the light from the star will reach us in X years. Second Argument Resolved Third Argument Statement 1. Light travels at a constant velocity for all inertial reference frames. Statement 2. Reference frames travel with different velocities compared to each other. Conclusion. If I am in moving with velocity V to your reference frame and I tell you how fast I see light. From your reference frame it will appear I am seeing light travel faster then what you see light to travel. Therefore, the speed of light you see (as an inertial reference frame) and the speed of light you perceive me to see (i.e. not using relativity to find how fast I actually see light) can be calculated from your reference frame for both frames by: (v1  v1) + c = c And (v1  v2) + c <> c Thanks again I think this is really helping. 


#7
Oct2805, 01:06 AM

Sci Advisor
HW Helper
PF Gold
P: 562

But if I use the theory of relativity, for example, then my prediction of what you see for the speed of light (c) is the same as what you actually see, which shows that relativity is a good theory. 


#8
Oct2805, 05:08 AM

P: 145

Ok cool , very good point, I think I understand what you’re saying about the third argument. Did I get the revised first argument right? Thanks so much.



#9
Oct3005, 05:58 PM

Sci Advisor
HW Helper
PF Gold
P: 562

Your first argument is ok.
Light emitted from an object always travels at the same speed c, regardless of the speed of the observer or emitter. Thus, in ANY reference frame, if the light travels a distance measured to be x, the time measured for it to cover that distance will be x/c, provided that the time and distance are both measured in the same reference frame. There's nothing relativistic about that, except for the constancy of the speed of light. The theory of relativity itself is most important when we want to deduce what one observer measures for time, distance or some other physical quantity, based on what a different observer measures for the same quantity. 


Register to reply 
Related Discussions  
Time Dilation to Length Contraction  Advanced Physics Homework  22  
Time dilation, length contraction  Special & General Relativity  6  
Length contraction & time dilation  Introductory Physics Homework  10  
Length contraction and time dilation  Special & General Relativity  7  
Time Dilation and Length Contraction  Special & General Relativity  8 