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.