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Anderson Video - Einstein's Velocity Addition Rule

Professor Anderson
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<font color="#ffffff">Okay, one problem that we talked about in class today,</font> <font color="#ffffff">in one of the clicker problems, was the following.</font> <font color="#ffffff">You are sitting here on the earth</font> <font color="#ffffff">and you observe one spaceship that is heading to the right</font> <font color="#ffffff">at speed V, you observe another spaceship that's heading to the left</font> <font color="#ffffff">also at speed V, and we let those V's be very big.</font> <font color="#ffffff">Okay, we said that V</font> <font color="#ffffff">is three quarters of the speed of light and the question was, what is the</font> <font color="#ffffff">observed relative speeds between those two?</font> <font color="#ffffff">Okay, and to do this properly, you need Einstein's velocity addition rule.</font> <font color="#ffffff">And Einstein's velocity addition rule was the following:</font> <font color="#ffffff">Vac equals Vab plus Vbc divided by one plus Vab Vbc, all over C squared.</font> <font color="#ffffff">If these are both three quarter C, which is 0.75 C,</font> <font color="#ffffff">What do we get for the relative speeds as observed by the people on the rocket ships?</font> <font color="#ffffff">Well, we can take this 0.75 and plug it in for each of these</font> <font color="#ffffff">Okay, the C squared cancels with that C because we're going to square it</font> <font color="#ffffff">and we're going to end up with 1.5 on the top,</font> <font color="#ffffff">and then on the bottom we have 1 plus 0.75 squared,</font> <font color="#ffffff">and if you do 0.75 squared, what do you get? Well, that's the same as</font> <font color="#ffffff">9/16ths, which has to be a little bit bigger than a half,</font> <font color="#ffffff">so I think this is probably around 1.5 C over</font> <font color="#ffffff">1.56 C, somebody confirm that in your calculator.</font> <font color="#ffffff">Okay, and if you do this 1.5 over 1.56, that's pretty close to 1,</font> <font color="#ffffff">there's no C on the bottom right? That's gone.</font> <font color="#ffffff">That's pretty close to one but it's a little bit less,</font> <font color="#ffffff">and I know the answer already because we did it this morning, it's 0.96 C.</font> <font color="#ffffff">So, on one of your misconceptual questions in the beginning of your homework,</font> <font color="#ffffff">it is question 26.13, it says this exact problem, right?</font> <font color="#ffffff">Two spaceships heading toward each other at three-quarter C as measured by you on Earth,</font> <font color="#ffffff">what do they measure in the spaceship, right?</font> <font color="#ffffff">It's looking at this other one coming towards it,</font> <font color="#ffffff">we know that anytime you make a measurement</font> <font color="#ffffff">it can't be bigger than C, and so it has to be something less than C</font> <font color="#ffffff">but it's pretty close to C in this case because of the way</font> <font color="#ffffff">those velocities add up.</font> <font color="#ffffff">Okay, any questions about -- about that stuff?</font> <font color="#ffffff">You guys feeling all right?</font>