Special Relativity and E=mc² - Part 2 of 5

so each of them alleges that the other has made some kind of error in measuring either distance or time or both and they assign an error which we will call gamma s Prime says that X equals gamma times X prime plus u T Prime whereas s says that X prime is gamma times X minus UT they each use the term gamma because each of them is alleging the other is wrong but neither of them can say that the other is more wrong than they are how are we going to find out what gamma is well we can take these two equations and multiply the left-hand side and multiply the right-hand side and so we get that x times X prime equals gamma squared times X prime plus u T prime times X minus u T and that becomes X X prime equals gamma squared into X prime X plus u T prime X minus X prime u t minus u squared T prime T but now wherever we find T or T Prime we can substitute that T equals x divided by C or T prime equals X prime divided by C and that means we get x x prime equals gamma squared into X prime X plus u X prime over C times X minus X prime u X over C minus u squared X over C X prime over C now we can divide all those terms by X X Prime and that gives us that one equals gamma squared into 1 plus u over C - you over C minus you squared over C squared and that means that gamma equals one over the square root of 1 minus u squared over C squared and that is known as the Lorentz transform and so we see that if we apply this transform X is equal to X prime plus u T prime divided by the square root of 1 minus u squared over C squared and X prime becomes X minus u T divided by the square root of 1 minus u squared over C squared notice on this Lorentz term what happens if u is very small compared to C then determine the square root simply reduces to 1 and you get normal Newtonian mechanics the gamma term is only relevant when u is approaching C and then it is known as becoming relativistic we can also work out formally for time we know that X prime equals X minus u T times gamma that means X prime equals X minus u times X over C because T equals x over C times gamma dividing everything by C you get X prime divided by C equals x over C minus UX over C squared times gamma but of course X prime over C is T primed and x over C is T so T primed equals t minus u over C squared X all times gamma similarly for X x equals x prime plus u T Prime all times gamma x equals x prime plus u X prime over C which is T primed times gamma dividing everything by C X over C equals X prime over C plus UX prime over C squared all times gamma and then of course X over C is T X prime over C is T primed that gives you T equals T primed plus u over C squared X prime all times gamma what is all this telling us it is telling us that when speeds are relativistic in other words getting to the point where the gamma term ceases to be equal to 1 that the distance measured by an observer and the time measured by an observer will depend on the speed of the distance between s and s Prime it may seem odd that time should be different for different people so let's just explore that a little bit more let's imagine that we create a very simple clock the clock is consisting of two mirrors placed one meter apart a light pulse travels between the mirrors and is simply reflected backwards and forwards every time it hits the lower mirror it hits a counter which clicks on by one count we know that the speed of light is three times 10 to the 8 meters per second and we know that the journey from the bottom mirror to the top mirror and back again is a journey of two meters so when that light pulse has completed one point five times ten to the eight cycles one second has gone past and so we can make a clock which simply records the number of times the pulse goes up and down but now let's imagine that that clock is onboard the spacecraft s and we have somebody else observing from a different spacecraft traveling at a different speed and he is looking at that clock relative to that observer where the spacecraft s is moving along at speed or velocity V when the light pulse leaves the bottom mirror the spacecraft will be here but when the light pulse arrives at the top mirror the spacecraft will have moved and so the mirror will have moved and so the light pulse instead of going vertically upwards will have gone at an angle similarly when the light pulse comes back down again the time it takes from getting to thee from the top mirror to the bottom mirror will mean that the spacecraft will have moved again slightly to the right and so the pulse will not come down vertically but at an angle the observer says that the distance that the light pulse has traveled is further than one meter because it has gone up at an angle and since the speed of light is invariant it must have taken longer to get up to that mirror then the person on board the spaceship thinks and that leads to the conclusion which comes from the formula that tells us that T prime equals T minus UX over C squared times gamma what this is saying is that if I observe a moving clock I will say that that clock is running slow every observer observing a moving clock will see that moving clock moving forward in time at a slower rate than their own clock

Channel: DrPhysicsA

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Keywords: Physics

Id: iV96673TZ1g

Length: 8min 26sec (506 seconds)

Published: Sat Jan 21 2012