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Physical Intuition 
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#1
Jun207, 02:16 PM

P: 186

I'm trying to get a physical intuition for length contraction (oxy moron) however is there any type of quantitative way of proving it? time dilation totally makes mathematical and physical sense to me, however length does not other then i ask does this basicly describe it:
If both observers (one stationary one moving w.r.t. the other) agree on the velocity of the moving frame and both agree that the moving observer reaches the same point in space yet they both do no agree on the time of the trip the only way that can be true is if in the moving frame the distance was shorter and thus contracted... mathematically: if O denotes the stationary frame and O' the moving frame then the distance traveled measured in O of the moving frame should be... D = vt while in the moving frame D' = vt', t' = t sqrt(1 v^2/c^2) D' = v t sqrt(1v^2/c^2) > D' = D sqrt(1v^2/c^2) however this is not correct it should be D' = D/sqrt(1v^2/c^2) where might my error occur? 


#2
Jun207, 02:52 PM

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#3
Jun207, 04:19 PM

P: 443

Imagine a meter stick of length D stationary in O, stretching from x=0 to x=D. Think of two events/flashes: event A (O' is passing by begining of stick i.e. by O), event B (O' is passing by end of stick).
From point of view of observers in O frame: O' covered length D with speed V so D = Vt. From point of veiw of observes in O' frame: the meter stick, of stationary length D in O, is Lorentz contracted in O' to length D' = D sqrt(1v^2/c^2) and O covers it moving backwards with speed V for time t' = D' / V = t sqrt(1 v^2/c^2). That is consistent with time dilation since t' is the time between events A and B measured in O' frame. Those two events happen at the SAME PLACE in O', so the time measured in O' must be shortest, and the time measured in any other inertial frame O must be dilated: t' = t sqrt(1 v^2/c^2). The situation is not symmetric between O and O' because using the time dilation formula, you implicitly assumed two events happening at the same place in O' which implies that the meter stick was stationary in O and lorentz contracted in O'. All your formulas were correct except your last statement that D' = D/sqrt(1v^2/c^2). 


#4
Jun207, 06:44 PM

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P: 4,130

Physical Intuition
Learn rapidity (i.e. involving hyperbolic trig functions) from the 1966 Spacetime Physics chapters available here at Ed Taylor's site: http://www.eftaylor.com/download.html
(It's actually is starting to hurt my eyes to see [tex]1/\sqrt{1v^2/c^2}[/tex] when [tex]\cosh\theta[/tex] looks so much nicer.... not to mention that you can use a variant of your Euclidean trigonometric intuition with "Spacetime Trigonometry". You'll see that length contraction boils down to issues involving the "distance between parallel lines".) 


#5
Jun207, 07:31 PM

P: 186




#6
Jun207, 11:20 PM

P: 997

D=D(0)sqrt(1uu/cc) (1) D'=D(0)sqrt(1u'u'/cc) (2) Expressing the right side of (1) as a function of u' via the addition law of velocities we obtain D=D'sqrt(1VV/cc)/(1+Vu'/cc) (3) Equation (3) accounts for the transformation of the wave length in an acoustic wave that propagates with u and u' relative to I and I' respectively. At that point physicists introduce the concept of wave vector k=1/D and k'=1/D' which transforms as k=1/D=k'(1+Vu'/cc)/sqrt(1VV/cc)=(k'+Vf'/cc) (4) f' representing the frequency of the oscillations in the wave. Comments are highly appreciated because it is about a simple (intuitive) way to teach relativity. soft words and hard arguments 


#7
Jun307, 09:46 AM

P: 443

The length of an object measured in its rest inertial frame is always the longest. The measured length is smaller (Lorentz contracted) in any other inertial frame. 


#8
Jun307, 10:13 AM

P: 186

Consider: an observer places a stake a point along the path of a moving platform/ruler what have you (an object with length), the stake is marked point A, he plans to determine the length of the moving platform as it passes by he will mark when the front end passes A and when the back end passes A, thus with his eye he will have "measured the length" not numerically but have an understanding of about how long the object is. As the front of the platform moves past point A the light of this event travels towards the observer so he may say that the front has passed. For very fast objects near the speed of light, the back end will pass point A possibly even before the front event has reached the observers eye, and so the back event light would be propagating very quickly toward the observer who had just noted the front had passed and now has found that the back came across point A very quickly so that the length of the platform must be very short. Does this sound correct? 


#9
Jun407, 07:58 AM

P: 2,954

http://www.geocities.com/physics_wor...ontraction.htm It may be of some help. Best wishes Pete 


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