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Frames of reference for speed?

by yoyopizza
Tags: frames, reference, speed
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jartsa
#19
Feb11-13, 07:15 PM
P: 484
Quote Quote by ghwellsjr View Post
You do raise an important point that you have to be careful about signs. However, you still didn't get it correct.
That was not my point, and I don't think I made errors worth mentioning. Here's what I thought, quite simple isn't it?

Car1 ---> <---Car2
v=30 m/s v=-30 m/s


Whenever I use the Velocity Addition formula, I don't worry about the signs of the velocities. It is always obvious whether the final result should be larger or smaller than either of its two inputs.

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"?
robphy
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Feb11-13, 07:23 PM
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Quote Quote by jartsa View Post

Quote Quote by ghwellsjr
Whenever I use the Velocity Addition formula, I don't worry about the signs of the velocities. It is always obvious whether the final result should be larger or smaller than either of its two inputs.
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"?
I would tend to agree.
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 signed-components.. or else use special cases depending on whether the ships are approaching or receding.
Nugatory
#21
Feb11-13, 07:45 PM
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Quote Quote by jartsa View Post
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.
They don't have to be adjusted by intuition, it's just that developing that intuition is both less work and more valuable than memorizing the sign convention rules. The formula represents a model of a physical situation, and you should never be so focused on the model that you lose sight of the actual physical situation. If you do, you aren't understanding physics, you're just crunching numbers by rote, and eventually a computer will replace you.

But here's a problem: what do you answer when a student asks "what exactly do I plug into this formula"?
Ask the student what might be plugged in, try to get him to justify each possible choice as he makes them. Sometimes, it's right to answer a question with a question.
ghwellsjr
#22
Feb12-13, 05:15 AM
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Quote Quote by jartsa View Post
Quote Quote by ghwellsjr View Post
You do raise an important point that you have to be careful about signs. However, you still didn't get it correct.
That was not my point, and I don't think I made errors worth mentioning.
You did make errors worth mentioning. When the OP asked if he should plug the two speeds into the Velocity Addition formula that I linked to, you said in post #9:
Quote Quote by jartsa View Post
No, because observer moving at velocity 30 m/s, observing an object moving at velocity 30 m/s, will obviously observe the velocity of the object to be 0, relative to himself.

But the velocity addition formula gives the result aproximately 60 m/s, if we plug those numbers in.
You stated that the correct answer was obviously 0 and if he plugged the numbers into the formula I gave him, he would get a different answer. Your answer was an error worth mentioning.

And then you added post #11:
Quote Quote by jartsa View Post
I guess there is a formula to calculate what velocity a moving observer observes when observing a moving object.

I guess the formula is:
s = [-1*v +u]/[1+ -1*vu/c2]


This relativistic velocity addition formula here:

s = [v +u]/[1+ vu/c2]

is good for adding speeds, when the observer and the observed object have opposite velocities.

And it's also good for adding the velocity of a ship, observed by a person standing on the shore, and the velocity of a fly, observed by a person sitting inside the ship.
Now you are suggesting that the correct formula was the relative velocity formula and restated that opinion in post #18:
Quote Quote by jartsa View Post
Oh yes, relative velocity formula, it's in Wikipedia too, and it's the right formula for our current task, while the velocity addition formula is not so good.

http://en.wikipedia.org/wiki/Relativ...ativity_theory
But when discussing the relativistic velocity addition formula, you quoted from the wikipedia article concerning the ship and the fly which is exactly the scenario that the OP asked about but with approaching speeds rather than receding speeds. I pointed out your error in post #16 but you reject it. You want to change the OP's question from one similar to the shore, ship and fly orientation where the OP is the shore, the ship is the road and the fly is the other car, to one from the viewpoint on the ship and the OP is the shore traveling away in one direction while the other car is the fly traveling away in the other direction. Of course, the relative velocity formula will also work if you change the sign of one of the speeds but that is totally unnecessary to bring up in the OP's scenario.
Quote Quote by jartsa View Post
Here's what I thought, quite simple isn't it?

Car1 ---> <---Car2
v=30 m/s v=-30 m/s
Looks like another error because you've just set 30 m/s = - 30 m/s.
Quote Quote by jartsa View Post
Quote Quote by ghwellsjr View Post
Whenever I use the Velocity Addition formula, I don't worry about the signs of the velocities. It is always obvious whether the final result should be larger or smaller than either of its two inputs.
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"?
Every formula should include an explanation of what the variables mean as did the wikipedia article on velocity addition that I pointed him to. And the OP had no problem with it until after you told him that he was doing it incorrectly when he wasn't.
ghwellsjr
#23
Feb12-13, 05:26 AM
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Quote Quote by robphy View Post
Quote Quote by jartsa View Post
Quote Quote by ghwellsjr View Post
Whenever I use the Velocity Addition formula, I don't worry about the signs of the velocities. It is always obvious whether the final result should be larger or smaller than either of its two inputs.
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"?
I would tend to agree.
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 signed-components.. or else use special cases depending on whether the ships are approaching or receding.
The OP already knew that he had to add the two speeds, he just didn't know that it took a special formula to add them. His intuition was correct, he didn't need any explanation on what he already understood and he especially didn't need any incorrect explanation.

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.
jartsa
#24
Feb13-13, 03:56 AM
P: 484
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
jartsa
#25
Feb13-13, 05:00 AM
P: 484
Quote Quote by ghwellsjr View Post
Every formula should include an explanation of what the variables mean as did the wikipedia article on velocity addition that I pointed him to. And the OP had no problem with it until after you told him that he was doing it incorrectly when he wasn't.

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?
ghwellsjr
#26
Feb13-13, 09:36 AM
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Quote Quote by jartsa View Post
Post 24 is correct, isn't it?
Yes, and that's exactly what I said in post #16 when I agreed with you that it is important to pay attention to signs.


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