# How to calculate relative velocity in SR

1. Oct 8, 2008

### Saw

I assume that, in SR, both observers moving relative to each other calculate the same relative velocity (v). But I wonder how. If they use clocks and rods, I see two possible methods: (a) combination of the measurements of two frames and (b) only "home-made" measurements.

In method (a), Mr A sees Miss B passing by rightwards in a 300,000 km long spacecraft as measured at rest in B's frame. The Front of B's ship passes by at T1 = 0 and the Back at T2 = 1.73205 s. Then if A knew that the relative velocity of B is 0.5c, he would multiply the length of B's spacecraft by sqrt(1-v2/c2), obtain 259,807 km and thus confirm that v is actually 0.5c. But he doesn't know that: v is precisely what he is looking for!

In method (b), instead, A would have an assistant located 519,615 km away to the right and with whom he has synchronised clocks following Einstein's method. He would annotate the time when B passes by (T1) and his assistant would do the same T2. Let us imagine that the time interval is again 1.72305 s. So A would conclude that v is again 0.5c. But what if B does the same operation? I confess I get lost when I try to calculate this. Can anyone help?

But another question: an alternative way to calculate velocities is measuring the Doppler frequency shift of an electromagnetic signal sent to the other frame and reflected back. The relativistic Doppler formula is frequency of reception = frequency of emission * (1-v/c)/(1+v/c). Sorry but I am so bad at algebra I do not manage to solve for v. What is the relative velocity formula based on frequency of reception versus frequency of emission? And are both parties supposed to get the same result by application of such formula? If so, has this been experimentally proved?

2. Oct 8, 2008

### atyy

Use only the "home-made"measurements (you can use other sorts correctly also, but only if you are a masochist). If you want to compare "home-made" measurements of two observers, then they are related by the Lorentz transformations.

B is stationary relative to herself. She should measure A's velocity to be -0.5c if she uses clocks and rods that are stationary relative to her.

It has been experimentally proven, but I am bad at algebra too.
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

3. Oct 8, 2008

### JesseM

Yes, this is the normal method for measuring speed, just finding (change in position)/(change in time) using local measurements. It's easier to just use units of light-seconds in these problems--I guess you're assuming the speed of light is exactly 300,000 km/s (not quite right, but close enough), so that means the assistant is 519,615/300,000 = 1.73205 light-seconds away. This means that if the time interval was 1.72305 seconds, the ship would be moving at the speed of light, not 0.5c. Maybe you meant that the assistant is 0.5*519,615 km away, or 0.5*1.73205 light-seconds = 0.861525 light-seconds away, in which case if B is measured to take 1.72305 seconds to cross this distance, his speed will be 0.861525/1.72305 = 0.5c.
If B places clocks on either end of his ship, which is measured to be 1 light-second long using rulers at rest in his frame, and he synchronizes the clocks using the Einstein synchronization convention, then if he looks at the time T1 that A passes the front clock and the time T2 that B passes the back clock, he'll find that T2 - T1 = 2 seconds, so if A took 2 seconds to cross 1 light-second he must be traveling at 0.5c in B's frame. But to see how this works, let's look at the situation from A's perspective again. In A's frame, the ship is shrunk by a factor of sqrt(1 - 0.5^2) = 0.866025404, and the clocks at either end of the ship are slowed down by the same factor. Also, because of the relativity of simultaneity, if the two clocks are synchronized in B's frame, they are out-of-sync in A's frame; a handy formula to remember is that if two clocks are synchronized and a distance x apart in their rest frame, then in a frame where they are moving at speed v along the axis between them, they will be out-of-sync by vx/c^2 (the leading clock will be behind the trailing clock in terms of its time-reading by exactly this amount). In this case, with x = 1 light-second and v = 0.5c, this means that in A's frame the two clocks are 0.5 seconds out-of-sync. So if the front clock reads T1 = 0 seconds at the moment it passes A, the back clock will read 0.5 seconds at the "same moment" in A's frame. At that moment the back clock is 0.866025404 light-seconds away from A, so traveling at 0.5c it takes a time of 0.866025404/0.5 = 1.732050808 seconds to reach A in A's frame. But the back clock is ticking slow by a factor of 0.866025404 in A's frame, so during this time it only advances forward by 1.732050808*0.866025404 = 1.5 seconds in A's frame; and remember that the back clock already read 0.5 seconds at the time the front was passing A, so by the time the back clock is passing A, it will read T2 = 0.5 + 1.5 = 2 seconds. So, T2 - T1 does indeed work out to 2 seconds when we calculate things in A's frame, just as we predicted from the perspective of B's frame.
If fR = fE * (1-v/c)/(1 + v/c), we can divide both sides by fE to get:

(1 - v/c)/(1 + v/c) = fR/fE

and if we multiply both sides by (1 + v/c) we get:

(1 - v/c) = fR/fE + (v/c)*fR/fE

so, by subtracting fR/fE from both sides and adding v/c to both sides, we can rearrange this equation to get all the v/c terms on one side (which is the goal when trying to solve equations like this), giving:

1 - fR/fE = v/c + (v/c)*fR/fE

and the right side can be simplified to (v/c)*(1 + fR/fE), so if we divide both sides by (1 + fR/fE) we get:

(1 - fR/fE)/(1 + fR/fE) = v/c

so then we can just multiply both sides by c and get v = c*(1 - fR/fE)/(1 + fR/fE).
Yes, if both parties are emitting light at the same frequency, both parties will see the light from the other person redshifted (or blueshifted) by the same amount, so they'll both calculate the same speed from the formula above.
Not sure if that specific prediction has been proven but there is lots of evidence that the SR formulas work, see Experimental Basis of Special Relativity.

Last edited: Oct 8, 2008
4. Oct 8, 2008

### granpa

drawing a single spacetime diagram with both coordinate systems on it would probably make it a lot clearer to you

5. Oct 9, 2008

### Saw

Thanks a lot for the prompt and clear answers. Yes, JesseM, I made a mess with my initial numbers. I also forgot to put the sqrt in the relativistic Doppler formula (I suppose that this does not alter the result of the way you solved for v, just we should put (1 - fR/fE)/(1 + fR/fE) within a sqrt. Right?). I will assimilate your replies and revert. Thanks again.

6. Oct 9, 2008

### JesseM

Just as a tip, the numbers often work out more nicely if you use a velocity of 0.6c or 0.8c rather than 0.5c, since sqrt(1 - 0.6^2) = 0.8 and sqrt(1 - 0.8^2) = 0.6.
If (1-v/c)/(1 + v/c) are in a square root, you need to square both sides, so you get (1 - v/c)/(1 + v/c) = (fR/fE)^2 on the first step instead of (1 - v/c)/(1 + v/c) = fR/fE as I wrote it. From then on, wherever I wrote fR/fE, substitute (fR/fE)^2, so the final equation should be v = c*[1 - (fR/fE)^2]/[1 + (fR/fE)^2].

7. Oct 10, 2008

### Saw

I have now reworked my numbers, with the aid of your explanations. I was trying to build an example where a synchronisation operation is carried out and then make the observers calculate relative velocity on the basis of the data obtained through such operation.

This is a better construction of the thought experiment:

STEP 1: A1 and B1 are travelling in long spaceships of unknown length. When they pass by each other, they set their respective clocks to zero. They are standing on the left ends of their ships and their respective assistants (A2 and B2) are standing on the right edges. It just happens that ship A is much longer.

At the time of the meeting, A1 sends a light signal to A2 and B1 sends another one towards B2, precisely in order to synchronise their respective clocks with those of their respective assistants.

This is an approximate picture:

A1 flash >-----------------------------------------------------A2
B1 flash >----------------------B2

STEP 2: When the light signal reaches B2, A2 receives his own light signal and they both happen to be standing by each other:

A1 ----------------------------------------------------- flash >A2

.....................................>B1 ------------------------- flash >B2

STEP 3: B1’s light signal reflects back from B2 and hits B1. She looks at her clock and it marks 2 seconds.

So, in application of Einstein’s convention, she establishes that, at the time of STEP 2, the clock of B2 should have been set to 1/2 of 2 seconds, i.e., 1 second. In other words, the length of car B is 1 light second in B’s frame.

Now we imagine that A ship is crowded with an array of observers who have been constantly watching B1. A1 makes a survey among them with the aim of learning who observed that, when B1 passed by his side, B1’s clock was marking what it should mark at the time of STEP 2, that is to say, 1 second. This guy (A3) happens to be the one who occupies the mid-point of car A, as shown in this reproduction of STEP 2:

A1 ----------------------------A3 --------------------- flash >A2

..........................................>B1 --------------------- flash >B2

This is the clue for A to find out that B is travelling at 0.5 c. Logically, if the light signal has reached A2 at the time of STEP 2 and at such time B1 had traversed half the light path, A must infer that B travels, relative to him, at half c.

Now we should check whether this result is confirmed by transforming B’s figures into frame A’s on the basis of relativistic formulae that use v = 0.5 c.

As to time: B1 has attributed to B2 a time of 1 second for the moment when event STEP 2 took place. In application of the Lorentz Transformation, this must be converted for A2 into (1+0.5)/(0.866025404) = 1.732050808 seconds. In principle, this is the time for A2 when the light signal reached the latter. But time is synchronous for A throughout his own frame, so A3 should be reading the same time when such event STEP 2 took place. And STEP 2 means two things: light reached A2 and B2, but also A3 and B1 coincided in space and time.

As to length: Based on what is said above, the length of car A is 1.732050808 light seconds in A’s frame. Thus A3 would conclude that, when B1 was passing by, it had traversed half the length of car A (which is 0.866025404 light seconds) in precisely 1.732050808 seconds. This gives back a speed for B of 0.5 c.

We can reach the same conclusion by transforming B's measurement of length. B claims that her car B is 1 light second long, but (as measured in A’s frame) B suffers length contraction: for A, that car is only 0.866025404 light seconds long. So the distance that B1 has traversed is 1.732050808 light seconds minus the length of car B, which (as noted) is for A 0.866025404 light seconds. This difference is also 0.866025404 light seconds. Therefore, we confirm again that B1’s velocity is 0.866025404 light seconds/1.732050808 seconds = 0.5 c.

And now let us take the perspective of B’s frame: A claims that his car is 1.732050808 light seconds long, but as measured in B’s frame A suffers length contraction: for B, that car is only 1.5 light seconds long. Hence the distance that A1 has traversed to the left is equal to the length of car A as measured by B (1.5 light seconds) minus the length of car B also as per B’s standard (1 light second). This difference is 0.5 light seconds. Conclusion: A1 has traversed 0.5 light seconds at the time of STEP 2 (1 second) and therefore A’s velocity is 0.5 light seconds/1 second = 0.5 c.

Thus it is confirmed, as you pointed out, that both parties get the same relative velocity for each other. But a funny situation arises. The distance A1-B1 in the above picture will be:

- For A, 0.866025404 light seconds.
- For B, 0.5 light seconds.

Likewise, time will be:

- For A3: 1.732050808 seconds.
- For B1: 1 second.

What is funny is that, both with regard to length and time, in order to shift from one number to the other, you cannot multiply by the gamma factor or 1/gamma. The ratio between the two of them is precisely the factor 1.732050808. This is not strange since the whole calculation was based on the Lorentz Transformation, which is based on gamma but also on the factor vx/c^2.

What is the technical explanation for this? Does it have to do with the relativity of simultaneity?

I am aware that I have included in my reasoning two coincidences of space-time in two distant places (A2-B2 and A3-B1). But that is no my fault: if A2 and B2 happen to be together when the light flashes reach them, that is not my fault; if A3 happens to meet B1 when the latter’s clock reads 1s, that is not my fault, either; and the rest is pure application of the relativistic formulae… or maybe not?

8. Oct 10, 2008

### JesseM

OK, makes sense so far.
But if B1 read 1 second, this will be simultaneous with the event of B2 receiving the flash in the B frame, but not in the A frame; if the clocks at B1 and B2 are synchronized in the B frame, then in the the A frame the clock at B1 is actually ahead of the clock at B2 by some constant amount. So B2 won't read 1 second until some time after B1 read 1 second in the A frame, and if B1 reads 1 second when it passed A3, and B2 reads 1 second at the time the flash reaches it and it's next to A2, that means in the A frame B2 has yet to reach A2 at the moment that B1 is passing A3. Your diagram above only shows what's happening in the B frame, not the A frame.

If the diagram above represents the B frame, then that must mean that in the B frame, the distance from A3 to A2 is 1 light-second, and since A3 was the midpoint that means that in the B frame the distance from A1 to A2 is 2 light-seconds. But the distance from A1 to A2 must be longer than this in the A frame (since it's Lorentz-contracted in the B frame), it must be 2/sqrt(1 - v^2/c^2), whatever the value of v in this problem.

We also know that in the A frame, the distance from B1 to B2 is 1*sqrt(1 - v^2/c^2) due to Lorentz contraction, so if B1 is at position x=0 at t=0 when the light is emitted from B1 (with A1 also at x=0), then the position of B2 as a function of time is x(t) = vt + sqrt(1 - v^2/c^2). Meanwhile, the position of the light beam emitted from x=0 at t=0 is of course x(t) = ct. Thus, in the A frame the light will catch up to B2 when these are equal, i.e.:

vt + sqrt(1 - v^2/c^2) = ct

And, we also know that when the light catches up with B2, B2 is at the same position as A2, which as I said earlier must be at position x = 2/sqrt(1 - v^2/c^2). So, since the light's position as a function of time is x(t) = ct, we know it will reach A2 (and B2) at the value of t such that ct = 2/sqrt(1 - v^2/c^2), or t = 2/c*sqrt(1 - v^2/c^2). So we can substitute this value of t into the previous equation:

[2v/c*sqrt(1 - v^2/c^2)] + sqrt(1 - v^2/c^2) = 2/sqrt(1 - v^2/c^2)

Now we can multiply both sides by sqrt(1 - v^2/c^2) which gives:

2v/c + 1 - v^2/c^2 = 2

Subtract 2 from both sides, then multiply both sides by -c^2 to get:

v^2 - 2vc + c^2 = 0

This simplifies to:

(v - c)^2 = 0

But of course this only works for v = c, which means there's no sublight velocity which will make your scenario work. You can see this if you play around with the numbers a bit too. For example, if we assume v=0.6c, that means sqrt(1 - v^2/c^2) = 0.8, so in the A frame the distance from A1 to A2 is 2/0.8 = 2.5, whereas the distance from B1 to B2 is 1*0.8 = 0.8. So if B2 is given by x(t) = 0.6*t + 0.8, B2 will reach A2 when 0.6*t + 0.8 = 2.5, which means t = 2.8333..., but by that time the light will have reached position 2.8333..., which is past the position of A2. If we assume v=0.8c, the distance from A1 to A2 is 2/0.6 = 3.333..., and the distance from B1 to B2 is 0.6, so B2 will reach A2 when 0.8*t + 0.6 = 3.333... which means t = 3.41666..., but by that time the light will be at position 3.41666..., which is still past A2 (though by a smaller amount than in the case of v=0.6c). If we assume v=0.999c, the distance from A1 to A2 is 2/0.044710178 = 44.73254408, and the distance from B1 to B2 is 0.044710178, so B2 will reach A2 when 0.999*t + 0.044710178 = 44.73254408, which gives t = 44.723566, but by that time the light will be at position 44.732566, which is still past A2, though by a very tiny amount. You can see from these numbers that it's plausible that the light would only be at the same position as A2 when B2 reaches it in the limit as v approaches c.

9. Oct 10, 2008

### Saw

OK. This means that, what is one single STEP 2 (containing two simultaneous events) for B, is a sequence of two events in A's frame:

- 2.1, when B1 meets A3 and B1's clock reads 1 s, while B2 has not yet reached A2, has not been yet reached by the light signal and B2's clock reads 1 s minus something.

- 2.2, when B2 (whose clock does read "now" 1 s) meets A2 and they are both hit by their light signals, while B1 has already left A3 behind and B1's clock reads 1s plus something.

But then you say...

It is clear that, following your correction, the diagram only represents B frame, where the distance from A3 to A2 is 1 light-second. But then you deduce that "since A3 was the midpoint" (of A's car) "that means that in the B frame the distance from A1 to A2 is 2 light-seconds". Then on the basis of that assumption you conclude that the length of A car is longer in A frame, being equal to 2/sqrt(1-v^2/c^2). However, I have a doubt about the assumption. A3 is the midpoint of car A in A's frame. But in B's frame car A is, as it is usually said, "foreshortened in the direction of motion": since, as viewed by B, A is moving to the left, car A is to be shortened by its left side. So could I not hold that in B's frame A3 is not the midpoint of car A, but is shifted to some extent to the left, as if the car had been squeezed like a spring from the left side? Or, on the contrary, does the Lorentz contraction act proportionally on all parts of the body in question?

Finally

Does this mean that I should give up with my search for meaningful numbers for the example of which you said, at the beginning, that it made sense, so far?

I mean: what I seek is a display where both parties get the same relative velocity for each other and, additionally, they do it in the context of light synchronization signals flying around. Why? Because only that way can I feel comfortable that the measurement of relative velocity with clocks synchronized in the prescribed way does give the expected result: agreement on relative velocity.

So, if we choose any other relative velocity that may be more comfortable, if we place observer A3, instead of in the midpoint of car A, in any other location, if we assume that in STEP 2 B1 measures any other time and length values that happen to be more appropriate, can´t the above example be worked out...?

10. Oct 10, 2008

### JesseM

Right.
Yes, Lorentz contraction is just a uniform squashing along the axis of motion, it doesn't distort proportional distances along this axis. If that wasn't true, if in B's frame A3 was closer to A2, for example, then the Lorentz formula would not correctly predict the distance between A3 and A2.
No, it just means you'd need to change the assumption that A3 was at the midpoint between A1 and A2. Suppose instead that A3 was 1/3 of the way from A1 to A2, which means in the B frame A is 1.5 light-seconds long, so it must be 1.5/sqrt(1 - v^2/c^2) long in the A frame. Then the distance from B1 to B2 is still 1*sqrt(1 - v^2/c^2) in the A frame, so if B1 is at position x=0 at t=0 when the light is emitted from B1 (with A1 also at x=0), then the position of B2 as a function of time is still x(t) = vt + sqrt(1 - v^2/c^2). So again we find that in the A frame the light will catch up to B2 when t satisfies the following equation:

vt + sqrt(1 - v^2/c^2) = ct

And as before, when the light catches up with B2, B2 is at the same position as A2, which is now at position x = 1.5/sqrt(1 - v^2/c^2). Since the light's position as a function of time is x(t) = ct, we know it will reach A2 (and B2) at the value of t such that ct = 1.5/sqrt(1 - v^2/c^2), or t = 1.5/c*sqrt(1 - v^2/c^2). So we can substitute this value of t into the previous equation:

[1.5*v/c*sqrt(1 - v^2/c^2)] + sqrt(1 - v^2/c^2) = 1.5/sqrt(1 - v^2/c^2)

Now we can multiply both sides by sqrt(1 - v^2/c^2) which gives:

1.5*v/c + 1 - v^2/c^2 = 1.5

Subtract 1.5 from both sides, then multiply both sides by -c^2 to get:

v^2 - 1.5*vc + 0.5*c^2 = 0

You could use the quadratic formula here, but I'll take a shortcut and note that if you add 0.0625*c^2 to both sides you get:

v^2 - 1.5vc + 0.5625c^2 = 0.0625c^2

and the left side simplifies to:

(v - 0.75c)^2 = 0.0625c^2

So, take the square root of both sides, which gives two possible solutions:

v - 0.75c = +0.25c or -0.25c

If we use v - 0.75c = +0.25c, that again leaves us with v=c which we don't want. But if we use the second solution, v - 0.75c = -0.25c, that gives v = 0.5c.

Just to check that works, note that in the A frame, A is 1.5/0.866025404 = 1.732050808 light-seconds long, and B is 1*0.866025404 light-seconds long. So, since x(t) for B2 is given by x(t) = 0.5*t + 0.866025404 (because it's moving at 0.5c and it started at x=0.866025404 light-seconds), we want the value of t such that 0.5*t + 0.866025404 = 1.732050808 (when B2 has reached A2 at x=1.732050808 light-seconds), which means t = 1.732050808. And of course, if x(t) for the light is given by x(t) = 1*t, then of course at t = 1.732050808 seconds the light is also at position x = 1.732050808 light-seconds.

11. Oct 11, 2008

### Saw

I am afraid of boring you. This is what I have understood (hope it is self-explanatory):

#### Attached Files:

• ###### Forum.jpg
File size:
42.2 KB
Views:
98
12. Oct 11, 2008

### JesseM

Don't worry about boring me, I find these exercises satisfying somehow...and yes, everything in your diagram looks exactly right.

13. Oct 11, 2008

### morrobay

14. Oct 11, 2008

### JesseM

There isn't any realistic situation where a ship would be measured to move at the speed of light, I was just pointing out that there was a mistake in the numbers Saw posted, he had wanted the ship to be moving at 0.5c but the numbers he gave would mean it was moving at 1c. This was because Saw had two people at different locations measuring the time the ship passed them, using clocks which were synchronized in their rest frame--Saw said the first person would see the ship pass him when his clock read 0 seconds, and the second person would see the ship pass him when his clock read 1.73205 seconds, and the distance between the two people was 519,615 km. In general if two people are a distance of x apart, and the time between each one measuring an object passing them is t (using synchronized clocks), then that object must be moving at a velocity of x/t, since speed is just defined as change in position over change in time. Here x=519,615 km and t=1.73205 seconds, so x/t = 300,000 km/second, the speed of light. Again, this was just because Saw made a mistake with the numbers, he really wanted x/t to be 150,000 km/second (half the speed of light), so the distance between the observers should have been 259,808.5 km.

15. Oct 12, 2008

### Saw

Coming back to the point (how to calculate RV), I see that the system of home-made measurements works, but I have questions as to the combination of measurements.

I know that only the first is practical. The second leads to a vicious circle, because you cannot combine the foreign measurement with your own without first transforming the former and you cannot transform without knowing v. But once v is known, the method should work…

Let us take the example of the displacements of:

- In frame A: B1 as it travels from A1 to (A1-A2)/2.
- In frame B. A3 as it travels from (B1-B2)/2 to B1.

As to home-made measurements, it is clear:

- In frame A: B1 travels 0.866025404 light-seconds in 1.732050808 s, so v = 0.5c.
- In frame B: A3 travels 0.5 light seconds in 1 s, so v= 0.5 c.

As to combined measurements:

In frame A [B1 travels from A1 to (A1-A2)/2]:

(a) My length – your time:

- My length: B1 has travelled in A frame 0.866025404 light-seconds.
- Your time: The observer in the midpoint of A1-A2 looks at B1’s watch and reads 1.5 s. He realises that this is time dilatated (too few long seconds) and enlarges it by 1/sqrt(1-v^2/c^2) into 1.732050808 s.
- The division is 0.5 c, which is OK.

(b) My time – your length:

- Your length: B1 is asked which distance the observer at the midpoint of A has travelled in her B frame. Her answer is 0.75 light-seconds. If the A observer then assumes that this distance is length contracted (too many short light-seconds) and shrinks it by sqrt(1-v^2/c^2), he will get 0.64951905 light-seconds.
- My time: B1 has travelled in A frame during 1.732050808 s.
- The division is 0.375 c, which is not OK.

My try at an explanation: B1’s “length” measurement in reality hides a time measurement. She has done one of these two things:

• She has recorded with her clock how much time it has elapsed between the passage of A1 and the arrival of (A1-A2)/2. This time interval is 1.5 s. (This would be a new picture not shown in the diagram). Then she has multiplied it by 0.5 and obtained 0.75 light-seconds. But her measurement of time is dilatated and converts for A into 1.732050808 s, which multiplied by 0.5 gives 0.866025404 light-seconds.
• An assistant of hers, with whom she has synchronized clocks, annotates the time when (A1-A2)/2 passes by and she does the same. The difference is again 1.5 s, which multiplied by 0.5 gives 0.75 s. But this time difference is time dilatated…

In frame B [A3 travels from (B1-B2)/2 to B1]:

(a) My length – your time:

- My length: A3 has travelled in B 0.5 light-seconds.
- Your time: For A3 the travel time is 1.154700538 s. If B thought that this is time dilatated, she would enlarge it and convert it into 1.333… But this would not work. Instead, she accepts that it is her own B clock that slows down and shrinks it to 1s, which is what her clock actually reads.

(b) My time – your length:

- Your length: If asked, A3 would say it is 0.577350269 light-seconds. If B thinks that this distance is length contracted and shrinks it by sqrt(1-v^2/c^2), she will obtain 0.5 light-seconds.
- My time: 1s.
- Division is 0.5 c.

Questions:

Are the numbers right by chance? (I am totally dizzy now and I agree with Atyy that his exercise was masochist… In fact, in a previous calculation which got deleted, there was also a problem with foreign measurement in frame B...).

It seems as if there is lack of reciprocity.

Would the concept of “proper” measurement play a role here?

16. Oct 15, 2008

### JesseM

Yes. Note, though, that this only works because at t=0 in the A frame, the B1 clock also read t'=0, which ensures that at t=T in the A frame the B1 clock will always read t'=T*sqrt(1 - v^2/c^2). If you had tried something similar with B2 without taking into account simultaneity issues it wouldn't have worked.
But that would be the wrong procedure for figuring out how far B1 traveled between when it was lined up with A1 and when it was lined up with the midpoint of A (call this midpoint Am for short), in the frame of A. After all, suppose you have a third rod at rest relative to B and 0.75 light-seconds long in B's frame, placed so that the right end lines up with Am at t'=0 in B's frame (at the same time B1 is lined up with A1 in this frame), and the left end lines up with Am at the moment it reaches the same position as B1 (and since this rod is at rest relative to B, that means the left end is always lined up with B1). The two ends of this rod show the distance Am travels between the moment A1 is lined up with B1 and the moment Am is lined up with B1.

Now if we switch to A's frame, it's true that this third rod is shrunk by sqrt(1 - v^2/c^2), so its length is 0.64951905 l.s. However, in this frame it is not true that this rod has its right end lined up with Am at the moment that B1 is lined up with A1, because of simultaneity issues (in B's frame the event of B1 and A1 lining up was simultaneous with the event of the right end of the third rod lining up with Am, so these events must be non-simultaneous in A's frame). Therefore you can't use the length of this rod to judge the distance Am traveled between the moment A1 lined up with B1 and the moment Am lined up with B1.
OK, that works.
With observers at different positions you also have to take into account that if their clocks are synchronized in B's frame, they are out-of-sync in A's frame--in this case, in A's frame the clock at B1 will be ahead of the clock at the assistant's position (which is 0.75 l.s. away from B1 in the B frame) by (0.75 l.s.)*(0.5c)/c^2 = 0.375 seconds. So at t=0 in the A frame, B1's clock reads t'=0 s, and the clock of the assistant reads t'=-0.375 s. If B1 is at x=0 in the A frame at this moment, then the assistant is at x = (0.75 l.s.)*(sqrt(1 - 0.5^2)) = 0.64951905 l.s. at this moment in the A frame, and the midpoint of A must be at x=0.8660254 l.s. at this moment in the A frame since this is half the rest length of A. So the distance from the midpoint to the assistant at t=0 in the A frame is 0.216506 l.s., meaning the midpoint will line up with the assistant at t = 0.216506/0.5 = 0.43301 seconds in the A frame. The assistant's clock will have elapsed 0.43301*0.8660254 = 0.375 seconds in this time, and since it started off reading -0.375 s at t=0 in this frame, it will read exactly 0 when the midpoint passes it.

Also, at t=0 the midpoint was at a distance of 0.8660254 from B1 in this frame, and B1's clock read 0 at that moment. So, the midpoint will catch up with B1 at t = 0.8660254/0.5 = 1.732051 seconds, and since B1's clock was running slow by a factor of 0.8660254, B1's clock will read 1.732051*0.8660254 = 1.5 seconds at the moment the midpoint catches up with it.
No, there's no need for B to conclude her own clock slows down. Again, she can explain these numbers using the relativity of simultaneity--in the A frame the clocks at A1 and A3 are synchronized and a distance of 0.577350269 l.s. apart, so in the B frame the clock at A3 must be ahead of the clock at A1 by a time of (0.577350269 l.s.)*(0.5c)/c^2 = 0.28867513 seconds. So at t'=0 in the B frame, the clock at A1 reads 0 seconds, and the clock at A3 reads 0.28867513 seconds. This means that if A3 reads 1.154700538 seconds when it passes B1, it has elapsed a time of 1.154700538 - 0.28867513 = 0.866025 seconds. And if B assumes that A3 was running slower than her clock (not the other way around) by a factor of 0.866025, she will conclude that in her frame 0.866025/0.866025 = 1 second has passed between the moment A1 passed her and the moment A3 passed her. So if A3 travelled 0.5 light-seconds in this time, it must have moved at 0.5c.
Yes, B will say the distance from A1 to A3 is 0.5 light-seconds.
That's right.
If you take into account the relativity of simultaneity along with time dilation and length contraction, you do have reciprocity--each observer says that points on the other rod which are a distance L apart in the other rod's rest frame are a distance of L*sqrt(1 - v^2/c^2) apart in the observer's own frame, each observer says that any individual clock moving along with the other rod is slowed down by a factor of sqrt(1 - v^2/c^2) in the observer's own frame, and each observer says that if there are two clocks attached to points on the other rod which are a distance L apart in the rod's rest frame, and the two clocks are synchronized in the rod's rest frame, then in the observer's own frame the trailing clock will be ahead of the leading clock by vL/c^2.

17. Oct 20, 2008

### Saw

Thanks. Yes, I had complicated myself by considering too many situations (in fact, it suffices to analyse just one from different viewpoints) and with a bad approach to the issue of combined measurements.

Just in case anyone is interested, I post here a “clean” version of how my question gets answered:

Velocity is distance divided by a time interval. Any measurement of velocity is the “story” of two events: two points of one frame (L1 and L2) meet successively with one single point of another frame. The length is the difference L2 – L1. As to time, it may happen that:

* At L1 and L2 there are synchronised clocks C1 and C2, which register the time (T1 and T2) when the single point of the other frame passes by each of them. The time interval is the difference between the two readings = T2 – T1. Thus in this case the measurement is constructed with the instruments of one single frame. We could all it a purely “local or home-made measurement”.

* Time is registered, instead, by a single observer, who annotates the readings of his single clock when it meets L1 and L2 (T1 and T2). Time is again the difference between the two readings = T2 – T1. Eventually, the two frames may swap roles: the single observer gets a space reference in his frame to measure distance; one of the two observers steps out and the remaining one measures time. As this method mixes measurements from two different frames, we can call it a “combined measurement”.

In classical relativity, due to the homogeneity of time and length intervals, as well as of simultaneity, everything is easier.

For example, these two stories:

- in A frame, the transit of B1 from A1 to A3 and
- in B frame, the transit of A3 from Bm to B1

are identical, because the length A1-A3 = the length B1-Bm, the transit time for B1 to go from A1 to A3 = the transit time for A3 from Bm to B1 and the two stories start and end at the same time.

That is why it is expected that local measurements will give the same result not only in terms of speed: also the denominator (length) and the numerator (time interval) will be identical. Likewise, in combined measurements, each frame can rely on the other frame’s measurement as if it were his or her own.

In SR, things are more complicated. Both frames do obtain in the end the same speed for each other, but the “stories” of the measurements are different.

The story may be in B frame, for example, like in the picture of classical relativity = A3 meets Bm (at the same time as A1 and B1 meet) and later A3 meets Bm. Thus B frame would make these “local measurements”:

When A3 passes by Bm, the latter annotates the time (TB1) = 0 s.
When A3 passes by B1, the latter annotates the time (TB2) = 1 s.
The difference = dTB = TB2 – TB1 = 1 s.

The distance between B1 and Bm in B frame is 0.5 l.s.

Hence v = division of B length by B time = 0.5 l.s. / 1 s = 0.5 c.

However, this is STORY B. But A would measure a different story. If the pairs A1-B1 and A3-Bm are aligned simultaneously in B frame, they are forcefully not in A frame. In A frame, the rod B1-Bm is shorter and so, at TB1 = 0 s, Bm is not yet lined up with A3 and this will only happen a little later, after Bm traverses a little additional path.

Thus if A frame wants to make his local measurement of velocity, he has to choose between two stories:

STORY A: B1 travels from A1 to A3.
STORY A bis: B1 travels to A3, although not from A1, but from the place that B3 occupies when (as per A’s line of simultaneity) Bm reaches A3 = B1 travels from A.1.1 to A3.

Nevertheless, both STORY A and STORY A bis lead to the same measurement of relative velocity:

In STORY A, A frame measures that B1 travels a longer length in longer time:

The distance between A1 and A3 in A frame (LA) is 0.577350269 l.s.

When B1 passes by A1, the latter annotates the time (TA1) = 0 s.
When B1 passes by A3, the latter annotates the time (TA2) = 1.154700538 s.
The difference = dTA = TA2 – TA1 = 1.154700538 s.

Hence v = division of LA by dTA = 0.5 c.

In STORY A bis, A frame measures that B1 travels a shorter length in shorter time:

The distance between A1.1 and A3 in A frame (LA bis) is 0.43301270 l.s., which is the same as the distance B1-Bm in A frame = half the length of B car in A frame.

When B1 passes by A1.1, the latter annotates the time (TA1bis), which can be guessed as follows: The distance that B1 must traverse in A frame so that Bm lines up with A3 is equal to the difference between A1-A3 (0.55577350269 l.s.) and B1-Bm as measured in A frame (0.43301270 l.s.), which is 0.14433757 l.s. Travelling at 0.5 c, B1 will need 0.28867513 s to traverse such distance. Besides, we can also obtain the same result reasoning with the sync factor: B1 will judge that, when she meets A1 and Bm meets A3, A3 (which is moving away) will be ahead by the factor LAv/c^2, LA being the distance that separates A1 and A3 in A frame (0.55577350269 l.s.) = 0.28867513 s. So TA1 bis = 0.28867513 s.
When B1 passes by A3, the latter annotates the time (TA2bis) = 1.154700538 s.
The difference = dTA bis = TA2 bis – TA1 bis = 0.866025 s.

Hence v = division of LA bis by dTA bis = 0.5 c.

If we put together all these numbers, we would realise that A measures, depending on the A story we choose, either more length and more time than B or less length and less time than B, but the two things always go together (if you increase time, you increase length as well; if you reduce length, you reduce time) and the proportion for either increasing or reducing B numbers is always the same [(sqrt(1-v^2/c^2)]. Otherwise it would be impossible that both frames measured the same relative velocity…

If we now consider “combined measurements”, we find that the two frames also measure the same relative velocity.

Let us analyse, first, the hypothesis where the two ends of the rod B1-Bm pass by A3:

- Length is measured in B frame as the length B1-Bm = LB = 0.5 l.s.

- Time is measured by A3’s clock:

TA1 = the reading of A3’s clock when he meets Bm, as noted above, is = 0.28867513 s.

TA2 = the reading of A3’s clock when he meets B1, as noted above is = 1.15470053 s

dTA = 0.86602540 s.

- If B wants to combine her length LB with A3’s dTA, she must assume the latter is too short and enlarge it through multiplication by 1/sqrt(1-v^2/c^2) = 1 s. Thus division of 0.5 l.s. by 1 s = 0.5 c.

- If A3 wants to combine his time dTA with LB, he must assume the latter is too long and shrink it trough multiplication by sqrt(1-v^2/c^2) = 0.43301270 l.s. Thus division of 0.43301270 l.s. by 0.86602540 s is again 0.5 c.

Now the case where the two ends of the rod A1-A3 pass by B1:

- Length is measured in A frame as the length of the rod A1-A3 = LA = 0.577350269 l.s.

- Time is measured by B1’s clock:

TB1 = the reading of B1’s clock when A1 arrives, as noted above, is = 0 s
TB2 = the reading of B1’s clock when A3 arrives, as noted above, is = 1 s
dTB = 1 s

- If A wants to combine her length LA with dTB, she must assume the latter is too short and enlarge it through multiplication by 1/sqrt(1-v^2/c^2) = 1.15470053 s. Thus division of 0.577350269 l.s. by 1.15470053 s = 0.5 c.

- If B1 wants to combine his time dTB with LA, he must assume the latter is too long and shrink it trough multiplication by sqrt(1-v^2/c^2) = 0.5 l.s. Thus division of 0.5 l.s. by 1s is again 0.5 c.

Conclusion: it is true, thanks again. If you consider the three effects together (TD, LC and RS) there is reciprocity in each of them and, in spite of that, the two frames obtain the same relative velocity for each other, no matter whether they use local or combined measurements.

One question: I understand “speed” does not include direction, while “velocity” does. Given this, technically speaking, which term should I be using throughout this text? Since we talk about a common concept (each frame may deem itself to be stationary for measurement purposes, but they agree that in fact there is relative motion between them), I feel that direction plays no clear role here. But what exists, instead, is the difference between approach and recession, which sometimes are referred to conventionally as positive and negative velocity, respectively. So would “velocity”, in the end, be the right choice?

Finally, to close the circle, I just need to understand the frequency method… But that is for another day.

Last edited: Oct 21, 2008