mctai Posted January 1, 2006 Report Share Posted January 1, 2006 this may seem a naive question but how does the dimension of a cube shrink as its velocity becomes a sizable fraction of speed of light say a cube with dimension 1 x 1 x 1 . how does its dimensions get affected as it travels at 0.9c 2nd question . how does travelling at 0.9c constant velocity to galaxy then return afterwards make someone younger than someone who was in earth the whole time? this book said they just are but didn't explain it enough by how much (question asked 40 years in total) THX ... Quote Link to comment Share on other sites More sharing options...
kamil Posted January 2, 2006 Report Share Posted January 2, 2006 this may seem a naive question but how does the dimension of a cube shrink as its velocity becomes a sizable fraction of speed of light say a cube with dimension 1 x 1 x 1 . how does its dimensions get affected as it travels at 0.9c This is because of length contraction, one of the conseqeunces of Special Relativity. The dimension that will shrink wil only be the one that is parralel to the direction of motion. The other dimensions wil remain the same. Quote Link to comment Share on other sites More sharing options...
TheBigDog Posted January 2, 2006 Report Share Posted January 2, 2006 This is because of length contraction, one of the conseqeunces of Special Relativity. The dimension that will shrink wil only be the one that is parralel to the direction of motion. The other dimensions wil remain the same.kamil, If I was riding the cube @.9 C, how would this shift be evident to me, if at all? Bill Quote Link to comment Share on other sites More sharing options...
kamil Posted January 2, 2006 Report Share Posted January 2, 2006 kamil, If I was riding the cube @.9 C, how would this shift be evident to me, if at all? Bill If you were riding the cube, you would be moving at the same speed as the cube. This would mean you would oserve the cube as bieng stationary, so it wouldnt affect the dimensions of it. Quote Link to comment Share on other sites More sharing options...
TheBigDog Posted January 2, 2006 Report Share Posted January 2, 2006 If you were riding the cube, you would be moving at the same speed as the cube. This would mean you would oserve the cube as bieng stationary, so it wouldnt affect the dimensions of it.Thanks, I had another question, but I will try and figure it out myself. Bill Quote Link to comment Share on other sites More sharing options...
kamil Posted January 2, 2006 Report Share Posted January 2, 2006 Glad to help:) Quote Link to comment Share on other sites More sharing options...
mctai Posted January 2, 2006 Author Report Share Posted January 2, 2006 This is because of length contraction, one of the conseqeunces of Special Relativity. The dimension that will shrink wil only be the one that is parralel to the direction of motion. The other dimensions wil remain the same. but by how much will that 1 x 1 x 1 cube shrink? or is there no definite answer . i dont think my test question ask that but just curious say someone who was 20 went to a journey in 2000, travelling at 0.96c, returns in 2040, what will be her age? THX . Quote Link to comment Share on other sites More sharing options...
kamil Posted January 2, 2006 Report Share Posted January 2, 2006 You can determin by how much length is contracted and by how much time is dilated through these formula's: t'=t*sqrt[1-(v/c)^2]L'=L*sqrt[1-(v/c)^2] where t is the duration of time meausured in the stationary frame, t' is the amount of time measured in the moving frame(which moves at speed v). L is the length of the cube measured by someone moving at the same speed as the cube, L' would be the length of the cube as measured by someone moving alongside the cube(at speed v) c is the speed of light. The derivation of these formula's come form the two postulates of relativity (if your interested ill show you how) CraigD 1 Quote Link to comment Share on other sites More sharing options...
mctai Posted January 12, 2006 Author Report Share Posted January 12, 2006 then is it possible to calculate how someone should age in a trip . Quote Link to comment Share on other sites More sharing options...
kamil Posted January 12, 2006 Report Share Posted January 12, 2006 then is it possible to calculate how someone should age in a trip . Yes:) Quote Link to comment Share on other sites More sharing options...
sanctus Posted January 12, 2006 Report Share Posted January 12, 2006 t'=t*sqrt[1-(v/c)^2]L'=L*sqrt[1-(v/c)^2] ¨Kamil shouldn't it be t'=t/sqrt[1-(v/c)^2]L'=L*sqrt[1-(v/c)^2] because otherwise (your formula) you would also have contraction of time... Quote Link to comment Share on other sites More sharing options...
kamil Posted January 12, 2006 Report Share Posted January 12, 2006 :rolleyes: No, because gamma is always bigger than 1 when v<c (which is basically always). Therefore it would mean that time would move faster when traveling at a faster speed and this is certainly not the case. http://www.drphysics.com/syllabus/time/time.html This is a simple derivation of the formula. BTW, if a very fast moving car drives past a road then that driver should measure the speed of the road to be the same as the road measures the speed of the car. v=l/t v'=l'/t' l/t=l'/t' Substitute t'=t*sqrt(1-[v/c]^2) l/t=l'/t*sqrt(1-[v/c]^2) t cancels out on both sides. l=l'/sqrt(1-[v/c]^2) l*sqrt(1-[v/c]^2)=l' So the time dilation formula must be the one i stated in order for it to also be consistent with length conrtaction. Quote Link to comment Share on other sites More sharing options...
sanctus Posted January 12, 2006 Report Share Posted January 12, 2006 Yes you are right my previous post is wrong (I mixed up between I and * but i knew that once you have a * and once a /) but yours still is wrong as well. To speak the same language gamma=1/sqrt(1-(v/c)^2), so gamma tends to ininitiy when v tends to c and tends to 1 when v tends to 0. So if we want time dilatation we have to have t'=t/ gammawith t' the time in the moving frame (where the system is at rest) and t the time in the lab. So we see that the time in the frame where an object moves (t) is much bigger then where it is at rest (t') ergo time dilatation To have contraction we have L'=gamma*L with L' the length in the frame where the body is at rest and L the length measured in the lab. So as gamma > 1 the length where the object is at rest (L') is bigger than the length (L) in the frame where the object moves ergo length contraction. This as well exactly what your site says.... Quote Link to comment Share on other sites More sharing options...
kamil Posted January 12, 2006 Report Share Posted January 12, 2006 So if we want time dilatation we have to have t'=t/ gammawith t' the time in the moving frame (where the system is at rest) and t the time in the lab. So we see that the time in the frame where an object moves (t) is much bigger then where it is at rest (t') ergo time dilatation Asume theres a space station 1 light year away. A space ship goes there and back at a speed of 0.8c. Therefore the clock on earth would measure 2/0.8=2.5 years, but the clock on the space ship would measure 1.5 years at the end of the trip. But if we assume the space ship is stationary then the distace between the earth and the space station is moving at 0.8c. Therefore the distance would be contracted by 0.6. so the time would be 1.2/0.8=1.5 From what I understood you claim that the time dilation formula must be changed from t'=t/gamma to t'=t*gamma when we consider the moving frame ro be stationary. But as you see length contraction is the reason why the space station clock measures slower time from his own perspective. Rather than adjusting the time dilation formula. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted January 12, 2006 Report Share Posted January 12, 2006 Hold it a sec, gamma is at the denominator for both t and x in the the coordinate transformations, but remember that the two observers aren't seeing the same pair of space-time points as being the ends of the rod at a given instant. It's a bit more complicated than applying the Lorentz transform to a pair of points having delta x = L. You need to consider the two world lines, one for each of the rod's ends. Sorry, I'm running short on time!:rolleyes: Quote Link to comment Share on other sites More sharing options...
kamil Posted January 13, 2006 Report Share Posted January 13, 2006 http://cmtw.harvard.edu/Courses/Phys16/l1_latex/l1_latex.html Look under lorentz transformations. You will see that the trnasformation equations are derived from 5 equations. One of them bieng: t=t'/sqrt(1-v^2/c^2) which is basically t'=t*sqrt(1-v^2/c^2) Quote Link to comment Share on other sites More sharing options...
kamil Posted January 14, 2006 Report Share Posted January 14, 2006 BTW, i found that t'=(t-vx/c^2)*gamma can be used to solve the twin paradox better than t'=t/gamma , though take a look at this: t'=(t-vx/c^2)*gammat'=(t-v(vt)/c^2)*gammat'=t(1-v^2/c^2)*gamma (1-v^2/c^2) is the denominator of gamma sqauared. THerefore: t'=t/gamma :lol: Quote Link to comment Share on other sites More sharing options...
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