# Frames of reference for speed?

1. Feb 8, 2013

### yoyopizza

So sense speed is relative, could I observe something traveling faster than light? Like in a car, when I see oncoming traffic it appears much faster, so could I not be traveling at, say 1/2c and observe somebody approaching at 1/2c therefore moving at 1c relative to me?

2. Feb 8, 2013

### ghwellsjr

No, look up Velocity Addition in wikipedia.

3. Feb 9, 2013

### ghwellsjr

I decided to draw some spacetime diagrams to illustrate this scenario. We start with diagrams for the Inertial Reference Frame (IRF) of someone stopped on the road like a radar cop who is shown as a thick black line with dots every second of the cop's time. I am using scales such that the speed of light (and radar signals) is 1 to make it easy to do calculations. This makes the radar signals travel along lines on a 45 degree diagonal.

You are shown in going to the right at one half the speed of light. You should be able to confirm that for every two seconds of Coordinate Time, you travel one light-second of Coordinate Distance. But since you are traveling at such a high speed, time for you is dilated by the factor gamma which is 1.1547 at 0.5c and your one-second blue dots are spaced farther apart than the coordinate time. The dots always show the Proper Time for each observer and are only the same as the Coordinate Time if the observer is not moving in the IRF. In that case the observer's path will be shown as a vertical line.

Coming towards you from the right is a green car also traveling at -0.5c and with one-second green dots spaced farther apart, just like yours.

In this first diagram, the cop is using radar signals shown in red to measure your speed:

The cop sends two red radar signals and receives two echoes. The first signal is sent at Coordinate Time of -6 and received at -2. (Remember, since the cop is not moving, the Proper Time on his clock matches the Coordinate Time.) From this data, the cop can conclude that halfway between these two points, at his Proper Time of -4, your distance was one-half of the difference between his two Proper Times which is (-2+6)/2 = 4/2 = 2 light-secs. You can see that at Coordinate Time -4, your distance from the cop was 2 light-secs.

The second red radar signal is sent at the cop's Proper Time -3 and received at -1. So applying the same process we conclude that at Coordinate Time -2, your distance was (-1+3)/2 = 2/2 = 1 light-sec which you can also confirm.

Now to determine your speed, the cop has to take the difference between your two distances that he measured and divide it by the difference in the two times that he placed those measurements at. This calculates as (2-1)/(-4+2) = 1/-2 = -0.5c. This means that the cop measures that you are coming towards him at 0.5c.

Now we look at how you in the blue car can do the same thing with your radar gun aimed at the cop as depicted in this diagram:

Since you are traveling with respect to the Coordinate Time, you need to use your own Proper Time as indicated by the blue dots. If you follow what the cop did with his Proper Time, you can see that your measurement is exactly the same as what the cop did and you will conclude that the cop is moving towards you at 0.5c.

However, if you look at the diagram, the distances don't appear to be correct as depicted on the drawing. But you can't tell that, all you can tell is what the data from your measurements tell you.

But if use the Lorentz Transformation process to create a new diagram for an IRF in which you are stationary, we can see how the distances would come out consistent with your measurements. Since you are traveling to the right at 0.5c we use an IRF that is traveling to the left of the first one at -0.5. This is what we get:

Now it is easy to see that the distances you measured match the Coordinate Distances but remember, you can't tell the difference as all your observations are dependent on radar signals traveling at c.

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4. Feb 9, 2013

### ghwellsjr

Now that we know how to measure the speed of another object, we are ready for you to measure the speed of the green car that is traveling on the road at -0.5c toward you while you are traveling on the road at 0.5c. You do the same thing that you did to measure the speed of the cop relative to you but of course you will get different numbers. Here is the diagram for the original IRF with the measurements you make:

Now the distance calculation for the first radar signal is (-2+18)/2 = 16/2 = 8 light-seconds at a time that is at the midpoint between when you sent and received the signals, at -10 seconds. For the second radar signal these calculations are (-1+9)/2 = 8/2 = 4 light-seconds and the midpoint is at -5 seconds. Taking the differences as we did before we get (8-4)/(-10+5) = 4/-5 = -0.8c which means you see the green car approaching you at 0.8c.

You will note that if you followed the Velocity Addition formula linked to in post #2, you would have calculated:

(0.5+0.5)/(1+0.52) = 1/(1+0.25) = 1/1.25 = 0.8

Now one last diagram to show the IRF in which you are at rest:

You should be able to confirm that both of the IRF's in this post present exactly the same information.

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5. Feb 9, 2013

### Naty1

no. velocities do not add EXACTLY as s = v + u....that is a low speed approximation.

Under SPECIAL RELATIVITY in the above link from ghwells, use the first formula for all speeds, even those close to c.

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

and substitute for v,c = 1/2c and u = 1/2c

and figure out what answer you get...it will be less than c.

6. Feb 9, 2013

### yoyopizza

How, then, do you know if you are using special relativity or normal relativity? Also, v and u just mean the speed of the observer and the speed of the object observed, correct?

7. Feb 9, 2013

### Staff: Mentor

There's no such thing as "normal" relativity. There's "special relativity", called that because it is a special case of a more general theory, and there's that more general theory, which is naturally called "general relativity". They don't conflict. Any problem that can be solved using SR can also be solved using GR, and you'll get the same answer; but there are some problems (any in which the effect of gravity is significant) that can only be solved using GR.

We're using special relativity in this thread because there are no gravitational effects involved.

8. Feb 9, 2013

### Naty1

yes....my post was supposed to read 'substitute 1/2c for u, and for v.

9. Feb 9, 2013

### jartsa

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.

10. Feb 10, 2013

### Naty1

My post #8 is relative to the originally posed question regarding my interpretation of the '1/2c' velocities used in the example.

11. Feb 10, 2013

### jartsa

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.

12. Feb 10, 2013

### yoyopizza

Unless the object is moving towards the observer, and the observer towards the object. Then the velocity of the object would be 60m/s relative to the observer(assuming they are both moving 30m/s towards each other) right?

13. Feb 10, 2013

### Staff: Mentor

No. As you've written it, this is an impossible situation because an observer is never moving relative to himself, so the observer cannot be moving at 30 m/s towards the object. As far as the observer is concerned, he's at rest and the object (and perhaps everything else nearby, even the ground under his feet) is moving towards and past him.

However, I think that you may really be asking about a different but similar situation: There's a second observer, he's at rest relative to himself (of course!), he sees the object approaching from the left at 30 m/s, and he sees the vehicle carrying the first observer approaching from the right at 30 m/s. This is basically the same situation as if the second observer were standing at the side of a road, watching two cars approach from opposite directions at 30 m/s.

This second observer will see the object and the first observer both approaching him at a speed of 30 m/s each, and therefore moving towards each other at 60 m/s.

The first observer will consider himself to be at rest while the second observer is moving towards him at 30 m/s. However, he will not see the object approaching him at 60 m/s; it will be something less and you'll have to use the relativistic velocity addition to calculate what it is. Notice especially that the two observers do not agree about the speed of the object relative to each other.

14. Feb 11, 2013

### ghwellsjr

That's exactly how I interpreted the situation except I made the second observer be a radar cop in a patrol car parked at the side of the road and, of course the speeds are 0.5c, not 30 m/s. Where'd that come from?

Note that I showed the calculation of the resultant speed between the two diagrams in post #4 as 0.8c.

15. Feb 11, 2013

### jartsa

According to the observer the velocity of the object is very close to 60 m/s.

But how do we calculate that result?

Let's try relativistic velocity addition formula: s = [v +u]/[1+ vu/c2]
v=30 m/s, u=-30 m/s
result of calculation is 0 m/s

Here's a warning for all newbies: If you ask this question about approaching cars in any forum, very likely you will be told to use the relativistic velocity addition formula, which will give wrong results, if you plug in the velocities of the cars.

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16. Feb 11, 2013

### ghwellsjr

You do raise an important point that you have to be careful about signs. However, you still didn't get it correct.

The example in the wikipedia article quoted Galileo's example of v being the speed of a ship moving away from the shore and u being the speed of a fly moving away from the ship. So if you want to define the speed of an object relative to an observer as being positive when the object is moving away from the observer, then you would say that when traveling on a road, a radar cop parked on the road is moving at a negative speed relative to the driver in the car and an approaching car is also moving at a negative speed relative to the cop and the resultant speed is also negative.

So you should have said:

v=-30 m/s, u=-30 m/s
result of calculation is very close to -60 m/s

And if you look at my derivations in posts #3 and #4, you will see that I was consistent in this regard and I got -0.8c as the speed of the approaching car relative to the OP's car. I could have been more consistent when I plugged numbers into the formula by making both of the being negative but I was just concerned about the magnitude of the answer.

The OP's question had to do with two very large speeds, 0.5c, which when added are not close to their algebraic sum. They add to 0.8c, not 1.0c or even close to 1.0c. By using very slow speeds, you obfuscate the issue that the OP was asking about because with your example the correct answer is so close to the algebraic answer as to be indistinguishable.

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.

17. Feb 11, 2013

### robphy

It's helpful to use descriptive subscripts.
(All quantities are signed... think "x-components of velocities" (not speeds).)

Relative-velocity formula
$$v_{C\ wrt\ B}=\frac{v_{C\ wrt\ A}-v_{B\ wrt\ A}}{1-v_{C\ wrt\ A}v_{B\ wrt\ A}}$$
(You can drop the "wrt"="with respect to" when the notation is clear.)

Since $v_{A\ wrt\ B} = -v_{B\ wrt\ A}$, you can write
$$v_{C\ wrt\ B}=\frac{v_{C\ wrt\ A}+v_{A\ wrt\ B}}{1+v_{C\ wrt\ A}v_{A\ wrt\ B}}$$

The Galilean versions are:
Relative-velocity formula
$$v_{C\ wrt\ B}=v_{C\ wrt\ A}-v_{B\ wrt\ A}=(v_{C\ wrt\ Z}-v_{A\ wrt\ Z})-(v_{B\ wrt\ Z}-v_{A\ wrt\ Z})=(v_{C\ wrt\ Z}-v_{B\ wrt\ Z})$$

$$v_{C\ wrt\ B}=v_{C\ wrt\ A}+v_{A\ wrt\ B}=(v_{C\ wrt\ Z}-v_{A\ wrt\ Z})+(v_{A\ wrt\ Z}-v_{B\ wrt\ Z})=(v_{C\ wrt\ Z}-v_{B\ wrt\ Z})$$

To do the analogue of this for special relativity, use rapidities $\theta_{C\ wrt\ B}$ (which are additive like the Galilean velocities are), where velocity=tanh(rapidity). You'll make use of the identity for the tanh of a difference.

18. Feb 11, 2013

### jartsa

19. Feb 11, 2013

### jartsa

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

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. Feb 11, 2013

### robphy

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.

21. Feb 11, 2013

### Staff: Mentor

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.

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.

22. Feb 12, 2013

### ghwellsjr

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:
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:
Now you are suggesting that the correct formula was the relative velocity formula and restated that opinion in post #18:
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.
Looks like another error because you've just set 30 m/s = - 30 m/s.
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.

23. Feb 12, 2013

### ghwellsjr

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.

24. Feb 13, 2013

### jartsa

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. Feb 13, 2013

### jartsa

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?