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Frames of reference for speed? 
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#19
Feb1113, 07:15 PM

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Car1 > <Car2 v=30 m/s v=30 m/s If the signs of the inputs of a formula, or the result, must be adjusted using intuition, then it's not so good formula. That's just my opinion. But here's a problem: what do you answer when a student asks "what exactly do I plug into this formula"? 


#20
Feb1113, 07:23 PM

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I tell often tell my students... How you would write a computer program [lacking physical intuition] to get the correct answer? One could use a universal formula using signedcomponents.. or else use special cases depending on whether the ships are approaching or receding. 


#21
Feb1113, 07:45 PM

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#22
Feb1213, 05:15 AM

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And then you added post #11: 


#23
Feb1213, 05:26 AM

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But I certainly wouldn't have changed the explanation from one in which the speeds involved were large fractions of the speed of light where the correct sum was significantly different from the incorrect sum to one in which there was no significant difference between doing it the wrong way or doing it the correct way. That was the point he was asking about. Instead, we're off on this tangent that has nothing to do with his question or with the correct answer. 


#24
Feb1313, 03:56 AM

P: 476

Now I will correctly add the velocities of two cars driving on a road at velocities 0.5 c and 0.5 c, relative to the road. I will use the relativistic velocity addition formula.
car1 > < car2 v1=0.5c v2=0.5c We need an observer and an object whose velocity is observed by the observer. Car1 will be an observer and the road will be the observed object, because we know the velocity of the road relative to car1, it's 0.5 c. Then we need another observer and another object whose velocity is observed by the observer. The road will be the observer and car2 will be the observed object, because we know the velocity of car2 relative to the road, it's 0.5 c. Car1 will observe the velocity of car2 to be: [v +u]/[1+ vu/c2] = (The velocity of the road according to car1 + The velocity of car2 according to the road) / (1+(The velocity of the road according to car1 * The velocity of car2 according to the road) / c2) = (0.5c + 0.5c)/(1+ (0.5c*0.5c) / c2) = 0.8c 


#25
Feb1313, 05:00 AM

P: 476

When I just plugged the velocities of the cars into the relativistic velocity addition formula, I made the error of just mechanically plugging the numbers in. Now I have corrected that error in post 24. Wikipedia: A shore observes a ship, which observes a fly. Me in post 24: A car observes a road, which observes another car. Post 24 is correct, isn't it? 


#26
Feb1313, 09:36 AM

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