# Measurement is theory-laden part 2

In my previous post in this series I discussed the idea that all measurement is theory laden and gave an example of a bad explanation of a measurement. In this post, I will give a good example, from the special theory of relativity. The point is to illustrate  that a good explanation consists of taking an idea seriously and working out its consequences, not of sweeping apparently strange implications under the rug. Doing that is an error since either those new implications are false and your idea is a dud or they are true and you have discovered something new and important. You can’t find out which of those two possibilities is true without working out the consequences of the idea.

First, two definitions that will be relevant. An frame of reference is a set of physical systems used to measure the motion of other systems. An inertial reference frame (IRF) is a frame of reference in which objects that do not suffer some external force move in straight lines. Empty space far away from any masses is a good approximation to an inertial frame. If you measure everything relative to your car while it is accelerating that won’t be a good approximation to an inertial frame because you will see things accelerating when no forces act on them.

Special relativity uses two assumptions.

(1) All IRFs are the same with respect to the laws of physics. For example, if two people use different IRFs, they will both see that objects obey conservation of momentum.

(2) The speed of light in empty space has the same value in all IRFs, denoted by c.

The second assumption looks a bit strange. If you were to get in your car and drive down the road, the cars coming in the opposite direction would be moving at a different velocity relative to you if you speed up or slow down. If you’re driving at 30mph with respect to the pavement and the chap driving in the opposite direction is also travelling at 30mph then you will see him coming at you at 60mph. But you can consistently work out the consequences of these assumptions and that’s part of what matters when it comes to doing experimental tests. (It is not the only thing that matters since some consistent and well worked out ideas don’t have empirically testable consequences.)

I will illustrate two consequences of these assumptions to illustrate what I’m talking about. The consequences in question are that if an object is moving at constant velocity (moving at a constant speed and in a constant direction) with respect to an IRF will act as if it is shorter along the direction in which you are moving than in an IRF in which it isn’t moving and any clocks attached to that object will run slower. This happens in such a way as to make measurements of the speed of light come out the same in both IRFs.

Suppose that you have a clock X that is at rest in an IRF S. This clock consists of two mirrors A and B, a distance L apart with a pulse of light bouncing between them. Each time the light pulse bounces off one of the mirrors it causes a clock attached to mirror A to tick. In the frame S the time T between ticks is 2L/c.

Now suppose there is another IRF S‘ that is moving at a speed v with respect to S at a right angle to the pulse of light bouncing between the mirrors: The distance between the mirrors in S isn’t going to change because that distance is at right angles to the direction in which the clock is moving. The time it takes the clock to tick in S‘ is T‘. The light takes T‘/2 to hit the mirror B in S‘ and during that time the mirror has moved vT‘/2. The light then takes another T‘/2 to hit the first mirror again at position C. So using Pythagoras theorem the total distance the light travels in S‘ is $AB + BC = 2\sqrt{L^2+(vT'/2)^2}$.

The total distance travelled by the light pulse must be cT‘ because the speed of light doesn’t change between the two frames, so $cT' = 2\sqrt{L^2+(vT'/2)^2}$

and a bit of rearrangement gives: $T' = \frac{2L}{\sqrt{c^2-v^2}}$.

Since L = cT/2 we can write $T' = \frac{T}{\sqrt{1-v^2/c^2}}$

Now, $1 > \sqrt{1-v^2/c^2}$

because v < c so the amount of time it takes for the clock to tick increases.

Now suppose that we consider another frame S” moving  at speed parallel to the direction in which the light travels in the clock: Since the far end of the clock is moving away from the light pulse it has to travel farther than the length of the clock in S” to get to the mirror at the far end of the clock: $L'' + vt'' = ct''$.

When the light pulse is travelling back from the far end of the clock to where it started, it travels a shorter distance: $L'' - vu'' = cu''$.

The total time is $T'' = u'' + v'' = \frac{2L''c}{c^2-v^2}=\frac{2L''/c}{1-v^2/c^2}$.

The time measured in S‘ and S” will be the same so: $T'' = \frac{T}{\sqrt{1-v^2/c^2}}$ $L'' = L\sqrt{1-v^2/c^2}$

and so moving objects will act as if they are shorter.

On the scales of speed and distance we use in everyday life these effects are very small, but they can be measured for particles travelling near the speed of light and they have been found. About conjecturesandrefutations
My name is Alan Forrester. I am interested in science and philosophy: especially David Deutsch, Ayn Rand, Karl Popper and William Godwin.

### 10 Responses to Measurement is theory-laden part 2

1. Maciek Markiewicz says:

Alan, I found your page through physics.stackexchange and your answer to this question: http://physics.stackexchange.com/questions/103572/local-epr-experiments-with-photons-in-vacuum (to which I left a comment as bright magus). It’s great to meet once in a while a person concerned with real physics (i.e. reality) insead of just mathematics! As to your post here. Notice one thing: The derivation of the equations for time dilatation and length contraction you show (and which is commonly accepted by mainstream physics) is completely false. It’s like with lots of other things that seem obvious so they go unnoticed, and yet are completely wrong. So why is it false? Because it is physically imposiible for a single ray of light (a photon) to be seen by two different observers. One can see (measure) only the photon that comes directly to his eyes (measurement device). One cannot see light that runs tangentially. It’s not part of his data, so there is no way to empirically conduct the experiment you are showing on your diagram. It’s impossible that two persons located in two different reference frames would compare the routes of some photon as seen by both of them, because only one of them would see it. That’s one of the funny and unphysical assumptions in Relativity that make the theory so riddled with paradoxes (i.e. signs of error).

• conjecturesandrefutations says:

Special and general relativity are entirely consistent and have stood up to experimental testing, see:
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
http://relativity.livingreviews.org/Articles/lrr-2006-3/.

• Maciek Markiewicz says:

Is this all you have to say? Just links and no comment? I’m disappointed that you avoid the problem (you are not the only one, but from what you said at physics.stockexchange you seemed to have the courage to follow your own intelligence and not the crowd). I’m not going to bother you anymore then. I will just leave you a puzzle concerning one of your “proofs” as a goodbye. These guys here: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html (and many others) claim (in the tables included in the 4th and 5th picture) that muon undergoes both time and distance dilatation. However, Special Relativity claims time dilatation and length contraction (as you must know, right?). And yet this experiment is considered a proof of SR. So, what do you think of such “consistency”, as you called it? (Also, how can you prove that clock A runs slower than clock B, and at the same time clock B runs slower than clock A? Because that’s what SR means when it says that no frame of reference is preferred. How can you prove that the two clocks are both slow with respect to each other? Any proof that clock A is slow with respect to clock B, or that clock B is slow with respect to clock A means SR is just plain wrong. And if you intoduce accelerations to show the reason, you will be cheating, because in SR time dilatation is a concequence of the difference in uniform velocity only; that’s what the equation shows – the bigger the difference in uniform velocity, the bigger the dilatation; there is no acceleration in there). So are you going to give it a moment of thought or will you just shrug your shoulders and call me a crackpot?

• conjecturesandrefutations says:

I assumed when you made your comment that you were interested in how SR could be tested, so I linked you information about that.

The page you linked went to a webpage with no descriptions of experiments with muons. I found this page
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html
at that site using Google, but it doesn’t fit the description you gave: it doesn’t claim that the muons dilate in both time and space. If you think SR is wrong you should make a more specific argument.

You write:

“Also, how can you prove that clock A runs slower than clock B, and at the same time clock B runs slower than clock A? Because that’s what SR means when it says that no frame of reference is preferred. How can you prove that the two clocks are both slow with respect to each other? Any proof that clock A is slow with respect to clock B, or that clock B is slow with respect to clock A means SR is just plain wrong.”

If you measure the rate at which clock A runs in a frame in which B is stationary then you find that A runs slower than B. If you ,measure the rate at which B runs in the frame in which A is stationary then you find that B runs slower. Those two measurements are not measuring the same thing and so there is no particular reason to expect them to have the same result. Nor is there any problem of consistency if they have different results.

You ask how anybody can prove this is happening. The answer is simple: they can’t. Proof is impossible.Scientific theories (and all other knowledge) are conjectures controlled by criticism., see
http://fallibleliving.com/thinkers/popper.
In science, experimental tests are used to criticise theories, not to prove them.

• Maciek Markiewicz says:

By the way, I just noticed that the website you liked is by John Baez. Perhaps (out of pure scientific curiosity) you want to check my reply here: http://physics.stackexchange.com/a/111471/43402, (look under the EDIT) where I show that Baez simply cheats when he proves accelerations can by handled by Special Relativity. (If you care, you can also see another answer of mine (again, the EDIT) here: http://physics.stackexchange.com/a/113863/43402, where I comment on the answer by prof Orzel, he gave to this question; it’s all very “interesting”, I would say. And telling as regards SR.)

2. Maciek Markiewicz says:

I provided the wrong link, sorry. Look at the tables by the 4th and 5th figure. They say Muon: distance 2 km, time 6.8 µs, Earth: distance 10 km, time 34 µs. So for earth both variables are larger. Now check your equations for time dilatation and length (distance) contraction – gamma is in the denominator of one and in the nominator of the other. This means that if for Earth time is larger than for Muon, then distance cannot also be larger for Earth than for Muon. This is simple maths. Time and distance are in inverse proportions according to SR, and yet this page shows they are in direct proportions.

“Those two measurements are not measuring the same thing.” Sure they are, because you can conduct two parallel experiments with only two clocks. Imagine there are two rockets (A and B) moving in opposite directions with one clock in each of them. And in each rocket there are two experimenters (using the same clock). In each rocket one experimenter assumes he is stationary with respect to the other rocket and the other experimenter assumes he is moving. This way, if you compare the clocks, one experimenter in rocket A (let’s call him A1) should be able to see that his clock is faster than the clock in rocket B, and the other experimenter (A2) should be able to see that his clock (the same clock A) is slower than the clock in rocket B. No way this can happen.

OK, let’s have it your way: In science theories are falsifiable. The above experiment is the falsification of SR as it currently stands.

• conjecturesandrefutations says:

You have not explained what those numbers referred to and as a result you have made an accusation of inconsistency where none exists. The distance the muon travels in the Earth frame is 10km. What distance does this correspond to in the muon frame? It is 0.2*(the earth frame distance) = 2km.

The time in the muon frame for the muon to get to the ground refers to the time that would be recorded on a clock travelling with a muon: a clock in the muon’s frame. So then if you consider what rate that clock will tick at in the Earth frame the answer is that it ticks slower in the Earth frame and so the time in the Earth frame = time in muon frame/0.2 = 34 microseconds.

Your rocket example is very unclear. You start by saying
(1) “Imagine there are two rockets (A and B) moving in opposite directions with one clock in each of them.”

Then you say

(2) “In each rocket one experimenter assumes he is stationary with respect to the other rocket and the other experimenter assumes he is moving. This way, if you compare the clocks, one experimenter in rocket A (let’s call him A1) should be able to see that his clock is faster than the clock in rocket B, and the other experimenter (A2) should be able to see that his clock (the same clock A) is slower than the clock in rocket B.”

Either the rockets are moving wrt one another or they are not. In either case, one of the experimenters on each rocket is correct (the experimenter who says the rockets are moving wrt one another) and their observations will be consistent with the correct assumption of that experimenter.

Or do you mean that one of the experimenters in each rocket is floating in space in a different frame from the rest of the rocket? In that case he will see clocks on both rockets running slower.

You have not explained any inconsistency and your descriptions of what is happening are unclear.

“OK, let’s have it your way: In science theories are falsifiable.”

Do you think Popper is right or not?

• Maciek Markiewicz says:

Excuse me, Alef, I did not realize you needed an explanation. Still, I see you got around somehow. Now, I didn’t ask you for an explanation. I know what these numbers mean. They mean that what the muon experiment shows (if it is correct) is that Special Relativity is false. You need explanation? No problem. Earth is the stationary frame of reference and muon is the moving one. And we can see that the time numbers are consistent – there is time dilatation shown, and muon’s clock is slower than Earth’s clock. However, the problem is, there is length (distance) dilatation instead of contraction. Because distance as measured from stationary frame (Earth) is again larger than as measured from the moving frame (Muon). All is clear now?

“Either the rockets are moving wrt one another or they are not. In either case, one of the experimenters on each rocket is correct.” Say what?! Are you telling me that you don’t know the foundations of the Special Relativity (for which you show the derivation of basic equations)? That you don’t know that there is no absolute rest, and no absolute movement? That – according to Special Relativity – you are absolutely free to choose, which frame of reference is moving and which is at rest? Is this some kind of joke to brake the ice?

You still need more explanation? Be my guest.
2 spaceships – A and B – moving at v with respect to each other. 4 experimenters, 2 in each rocket (A1, A2, B1, B2). Now, according to SR, each of these experimenters can claim that he is at rest, and that the other rocket is moving. Let’s say that experimenters work in pairs: (1) A1 with B1 and (2) A2 with B2. The first (1) pair agree that A is moving and B is at rest, while the other pair (2) agree that A is stationary and B is moving. So they proceed to comparing their clocks. In pair (1) they should see that clock A is slower than clock B, while pair (2) should see that clock B is slower than clock A. You see the contradiction now? We already have proved it is impossible, but to make it more obvious I provided only one clock in each rocket, so that A1 and A2 must usie the same clock A, while B1 and B2 must use the same clock B. Can you see it now? Can you see that Special Relativity understood as real changes in time is incorrect? Clock A cannot be both slow and fast with respect to clock B. Even David Copperfield could not do that.

Instead of discussing Popper (what’s the purpose?) let’s go back to my first post here, where I showed that the derivation you conducted (following the logic of most textbooks on the subject) is incorrect. Again, two different observers (in two different reference frames) cannot see and measure the same photon. It’s physically impossible. (More on that you will find in my first post, which you apparently didn’t care to read thoroughly, as I didn’t even mention SR test – I didn’t even say a single word about it).