CptManering

Length Contraction Just as in the previous post where we found out that time is dependent on the relative speed of an observer it is also true for length. Consider the experiment in post 2 where you are on a space station and a space ship flies past. Now have the experiment on the ship carried out in a way so that the flash of light sent out is in the direction of motion of the passing space ship. So now we have a flash of light sent out and reflected back to the source by a mirror. Let’s define some events The light travels from the source to the mirror a total distance of (L) which we can write as ( C ) the speed of light times the time taken. (CT) On the space station we measure the light travelled a distance of L plus the distance the ship has travelled which is the time taken for the light to reach the mirror times the speed of the spaceship so the distance is (L)+(U)(T), where (U) is the speed of the ship and (T) is the time taken. So we now have (L)+(U)(T)= (CT) with a little bit of maths this can be expressed as (T)= (L)/( C )-(U) (just re-arraging equation) Working out for the return trip of the flash of light in the same way we get (T2) = (L)/( C )+(U) Now the total time taken would be (T) time of light to the mirror plus (T2) the return journey, which is (L)/( C )-(U) plus (L)/( C )+(U) = total time = (2L)/( C )(1-(U2)/( C2 )) If you look closely this looks familiar to 1/square root (1-(U2)/( C2 ) , what we defined to be gamma. So if we multiply both sides we get the change in time divided by gamma = (2L)/( C ) Some more maths of combining previous equations we get an expression for length contraction Where length measured from the reference frame of an observer on the space station is the proper length divided by gamma. Example Superman is a handsome 2 meters tall and towers over me by only 1 inch (that 1 inch makes all the difference). As I am setting my telescope up for a good night viewing I see superman fly past at 0.99( C ). Using the equation from above for length contraction Gamma = 1/square root (1-(0.99C)/( C )) = 7.1 Height of superman 2 meters divided by gamma 7.1 we get 0.282 meters (Hmmm not so big now are we) Summary We have found that time and length are both dependent on the relative speed of the inertial reference frame but are there any quantities that are independent of the reference frame. The answer is yes the speed of light is independent it is constant in every reference frame. After thought Time dilation and length contraction both use the fact that the speed of light is 299 792 458 m / s, that nothing can travel faster than light and that the speed of light is the same in all inertial reference frames. I have only being told these facts so I was wondering if anybody knows of any experiments that have been carried out or observations to validate these claims.

Second Lecture: Special Relativity Time Dilation Say we have a two asteroids heading towards each other and you are observing them from Earth. We see asteroid (A) moving at 0.8C (80% the speed of light) and we see asteroid ( B ) moving at the same speed but in the opposite direction. Now move our reference frame to asteroid (A) using classical velocity rules asteroid (B ) will obviously moving at 1.6C which is asteroid (A) speed minus asteroid (B ) speed (0.8C-(-08C)=1.8C). But hang on nothing can go faster than the speed of light so the standard way to calculate velocities breaks down when speeds approach that of light. So a new way of dealing with this situation is needed. Einstein realized that we needed to think about space and time in a different way to tackle these problems. His idea was not to think of time as absolute but that it can vary; with this in mind different observers will perceive motion differently. So now: "The laws of physics are true in all inertial reference frames and the speed of light in vacuum is the same in all inertial reference frames." This then means that: absolute motion cannot be detected an observer cannot travel at the speed of light events that happen at the same time for one observer may not for another two observers measure distance and time differently Galilean transforms are not valid Example: You are an observer on a space station that for all purposes is stationary. A space ship comes flying past at a velocity (U). As the space ship flies past an experiment is being carried out onboard. A person is sending out a flash of light from the tail end of the ship to the nose a distance (L) where it is hitting a mirror and bouncing back to a detector near the source of the light. We now have to define two inertial reference frames: Frame 1 on the ship: The flash of light travels from the tail end of the ship to the nose and back again a distance (2L) at the speed of light ( C ), so the time taken is just the distance divided travelled by the speed (2L)/( C ). Frame 2 observer on earth: The flash of light travels from the tail end of the ship to the nose but the observer all so sees that the space ship will have also moved so we will have to add this distance to the distance already travelled. Which will be the speed of the ship (U) multiplied by the time taken. So the distance the flash of light travelled in this inertial frame in (using a bit of trig) is 2L = square root of (L)2 + ((D)/2)2 So the distance that the light travels in frame 2 is further than in frame 1. Keeping in mind that now we know the laws of physics are true in all inertial reference frames and that the speed of light is the same we can solve both equations of distance travelled by dividing by the speed of light ( C ). We now have two times for the same event in space, if we carry out a bit of further maths (can attach file if anybodies interested) we get the difference in time for frame 1 is equal to the difference in time in frame 2 multiplied by 1/square root 1/(U)2/( C)2 . We call this gamma ( ϒ ). As you can see gamma is related to how fast you are going so the faster you go the higher value of gamma you will have and so in turn the larger time. Time slows down the faster you go So we have two different times to label a proper time we define it as the time taken between two events in the rest position. Lets see what would happen if we use some real numbers in the above example i've left the maths out, but can upload it as attachment if anybodys interested So let's say the distance from one end of the ship to the other was 20 meters and the speed of the ship was 1000 meters a second The time taken for the experiment would be 0.04 seconds and gamma would be 1+ 0.0000000000005. I f you were to multiply these two together you would still have only 4 seconds as it would be impossible to notice the extremely small change in time (unless it was your boss and you were clocking out early) its some thing like 0.0000 loads of 0000 then maybe a 1 its that small my calculator does not show it. Now let the distance be the same 20 meters but this time the speed to be 0.96 ( C ) you would get gamma to be 3.57 now if you were to multiply this with the 0.04 seconds you would notice a huge difference in time. The new time would be 0.143 seconds Summary: Time is not the same for every observer, at higher speeds close to that of light gamma increases to an amount that is noticble. You should also note that for low velocities gamma is low so does not have to be used in solving equations of such speeds. It would be pointless to work out gamma on a 100 m sprint as the change in time would not register on the clock as it that small.

Hi all would just like to mention that the following blogs on special relativity are from my notes taken at uni. I have tried putting the ideas that were taught to me by my professor into my own words but really all the work is his and i've just stolen it ha ha har. I will not name my professor but all credit should go to him and thanks for letting me post this online. Hopefully it will help me in understanding the subject spark a few discussions and some people might find it an interesting read. Lecture 1 Special Relativity First we must define relativity in the classical sense: Relativity is events that happen with respect to a certain view point (reference frame). So are a set of coordinates in the x,y,z direction with an observer located at the origin. An internal reference frame is one that is not accelerating. This does not mean there is no motion just that the motion is at a constant rate. This then implies that any internal reference frame must travel at a constant velocity with respect to any other internal reference frame. "There is no unique internal reference frame and no absolute velocity can be measured." Example A train goes by and someone throws a ball from one end to the other. A person on the train would see it travel, from their internal reference frame at say 12 meters a second. A person on a bank watching the train go by would see from their inertial reference frame the ball travel at 12 meters a second plus the speed of the train. Who is correct? Well both are which means there is no unique reference frame and no absolute velocity. Galilean Transformations Coordinates in one frame are related to coordinates in another frame. Newton’s 1st and 2nd laws are true in all inertial reference frames. Newton’s 1st law If no external force acts then an object in motion stays in motion of if at rest will remain at rest Newton’s 2nd law If an external force acts on an object then it will accelerate directionally proportional to the force, the constant of proportionality being mass. In an inertial reference frame (A) a car is moving at constant velocity (U) now let another inertial reference frame ( B ) move along the horizontal x axis at a velocity (V) then the velocity of the car in this new frame is simply the velocity of the new frame (V) minus the velocity (U) of the car. Position transforms If you have in inertial reference frame (A) a particle moving along the x axis then an internal reference frame ( B ) moving along the x axis with a difference of velocity (U) multiplied by time (T), then the position of the particle in inertial reference ( B ) will be related to that in internal reference (A) by. Position of (A) = Position of ( B ) + (U) x (T) Note: acceleration does not change between reference frames. Hope this is straight forward to understand any questions more than happy to talk about

What this all about? A few years ago while having a few drinks round a friend’s house, towards the end of the night we got talking about space and time. Being a sci-fi fan and educated to BBQ level on the subject I had some small ideas of how the universe works, anyhow we all had a great night talking about time slowing down and worm holes etc. The following day since I enjoyed the subject so much I decided to quit work and go back to school and learn physics (not the only reason but it started me thinking about it). I am now at university and just about to start modern physics section of the course. The first part is on special relativity and I thought some people might find it interesting to read my notes from the lectures if I were to write them up in a blog. I will try and keep the math’s to a minimum and just deliver the main concepts. I am learning this subject as I go along so please don't expect the answers to everything (42). I just thought some people would like to know what gets taught on such a course and I’ll have some notes to review when I come to revise for exams. Hopefully I will be able to add the first entry tomorrow thanks for reading.