# Measuring the wavelength of a moving light source

Some years ago, I started what is the toughest project I have worked on: I decided to measure the small change in the wavelength of light when emitted from a moving source. Being not a scientist but a simple software programmer, this happens to be a really hard task.

After a few years of trials, today I have a good idea of what I need to complete this project. I have already built some of the necessary assemblies, but honestly, they are pretty bad. I don’t mean they look bad (what they do), but they work badly. The alignment of the beams is complicated. Also, due to what I think is thermal drift I have to re-align the mirrors every 5 minutes to keep the interferometer working. Even worst, the whole setup must lay on the floor for better stability, so all this work should be done in a very uncomfortable position. I will be lucky if I don’t get permanent injuries in the process.

So, I decided to stop, rethink the whole thing, and start again. The idea is to work every component independently and work on them until getting reasonable stability and ease of use. This, in turn, will let me save more data, which means clearer results.

## The experiment

What I need is to measure the small change in the wavelength of a moving source of light. To do this, I use an unbalanced Michelson interferometer. This is a Michelson interferometer where its arms are different in length. This is crucial because that difference acts as a sort of wavelength amplifier. The expected change in wavelength is so small that we will need to amplify it as much as possible, so I made the short arm just a couple centimeters long, and the long arm around 8 meters long (this is why the whole experiment rest on the floor). With this setup, the measured phase shift (after the multiplication effect) is around 0.1 wavelengths. This is easy to see using an oscilloscope.

If you have worked with interferometers, you probably know the floor is not the best place to play with them. Every single vibration is perceived as a change in the interference fringes. This means must of the time you will see moving fringes instead of a stable one (as desired). This is bad, but my design let us get the phase difference between two independent interfering beams, which lets us overcome this limitation.

### The two beams interferometer

On a Michelson interferometer, we usually shine a single laser beam. However in this particular experiment is way better to use two laser beams, in order to get the difference between two interferences, which is useful to get rid of the pesky floor vibrations. Strictly speaking, the vibrations never go away, but you can take your measurements without being affected by them. Given that both beams traverse the same path in the interferometer, they both exhibit the same vibrations, and that means they should be in phase. If there is a difference (which I expect) we should see a phase shift between both signals, and we can measure that difference.

The floor vibrations, however, are too fast to be seen with the naked eye, so we use a photodiode and an oscilloscope to see them. But as soon as you got this signal you will notice that they are way too slow to be useful. As crazy as this could sound, I had to increase the vibrations! A small geared motor hanging from the interferometer do the trick. Now we have sinusoidal signals on the oscilloscope, so we can measure the phase difference.

However, the geared motor has one disadvantage: it vibrates (good) but does not let us know in what direction it is moving at a particular time. And happens this is important, because we should get some phase shift if it is moving in one direction, and get the opposite (negative)  phase shift if it moves in the opposite direction. Also, the velocity of the moving mirror changes direction, what makes signal distinction complicated. If we know in what direction the oscillating mirror is moving we can easily separate one phase shift from the other. Audio exciter (Parts Express)

So, I had to look for a suitable replacement: a voice coil. Parts Express has a range of audio exciters that fit the bill, at very low cost. A simple amplifier feed by the AWG (Arbitrary Wave Generator) port of my Picoscope 2204A run the audio exciter at the desired frequency (120Hz), and with the required amplitude to generate a sinusoidal signal on the oscilloscope. Picoscope (Picotech)

### Saving data

In order to analyze the so obtained waveforms, we must save them. To me, the most friendly and cost-effective way of doing this is by using a Haasoscope. This is an open-source oscilloscope with both excellent capabilities and low cost. A Python user software let me tweak the oscilloscope to save the data as soon as possible, what in the end let me get 120 readings per second (it could go faster, but restrictions apply). Each reading contains 4 waveforms (2 interferometer signals, 1 slider position signal, 1 audio exciter signal). Each waveform is 128 samples wide. Haasoscope (CrowdSupply)

However, due to mechanical vibrations on the slider (due to braking), the data read near zero velocity is too noisy to be useful. Later on, I found the noise could be reduced with better mechanical design of the system, so finally I could save samples also when the slider was near rest.

(Updated on Sept 16, 2019)

(This is a work in progress, last updated on Nov 14, 2019)

# An adjustment to Special Relativity

Let me start saying Albert Einstein was a man truly unique. Who else in the world could ever arrive to the conclusion that time must be relative? That is the idea of a genius. That said, today I have to expose you what, if true, could be a useful adjustment to his Theory of Special Relativity.

We will make use of three well proven tools:
1) The time dilation equation
2) The Hafele-Keating experiment (ok, not really a tool, but a reference to a real world evidence)
3) A thought experiment to put it altogether.

Let’s suppose you are the pilot of a spaceship traveling the deep space, far away from any gravitational source. You just wake up from a long hibernation state, so you don’t know what your speed is. All you know is your spaceship is capable of accelerate and decelerate on a straight line, and that the maximum speed reached is far from the speed of light (these assumptions are appropriate to simplify our calculations).

You are bored, so you prepare an experiment to verify (one more time) the validity of the theory of special relativity. Given the spaceship can only travel longitudinally, you plan to go to the middle of the ship and from there throw two high precision clocks, one to fore-end, and one to aft-end. A pile of pillows will catch the clocks at the end, for they don’t get damaged. Of course, both clocks are already synchronized to the spaceship clock. Being on a spaceship, you throw 2 clocks on opposite directions to measure time dilation.

While doing the calculations, you find there is missing information. You can’t make any prediction about the output of this experiment, because  the time dilation depends not only on the speed of the clocks, but also on the spaceship speed, which is unknown. Anyway, you throw both clocks with the same speed, and then you compare them both with the spaceship clock.

Let me stop here. Have you noticed something rare on the text above? No? We have the equations but we can’t predict the output (so the readings of both clocks will be undefined for us). What will the clocks show? Will they show an undefined value? I really doubt it. They will delay exactly in the appropriate amount predicted by special relativity.

1) Both clocks show exactly the same time dilation (that is, the spaceship is at rest)
or
2) One clock shows more time dilation than the other, that is, one clock speed up and the other one slow down (what means the spaceship has some speed > 0 in the direction of the slower clock)

Please note you now have one of 2 possible outputs, but most important, you already know what is your state of movement, that is, when you read the clocks you will know for sure if you are at movement or at rest. Nothing new here right? Everybody knows that.

Now, let’s check what we have. By the simple fact of reading the clocks, your unknown speed “collapses” (a Quantum Physics term) to a known state. That is, you certainly know if you are at movement or at rest. But moving relative to what? There is not a planet involved, there is not another spaceship either. It’s just your spaceship, and anyway you know what your speed is. If you know your speed, then you know your speed relative to something, maybe relative to rest. But what is rest? We have no point of reference! It doesn’t matter. We don’t know what is it but it is there anyway, as an absolute reference frame for time dilation. And clocks someway “perceive” that frame of reference. That is what experiments tell us (Hafele-Keating experiment, among others).

So what? If this is true, we may need to do a little “conceptual” adjustment to the theory of special relativity to account for this observation. According to Einstein, no frame of reference is absolute. No experiment can let you know what your own speed is. However, clocks disagree, as we have seen above. Please notice this experiment could be done in a fully closed spaceship, with no windows at all, and it will work anyway.

Now we are talking about an absolute frame of reference I must mention that this absolute frame is the real solution to the Einstein Twins Paradox. Previous solutions talking about the asymmetry of the movements of both twins don’t apply, sorry. We will talk about that in a later article.

# The centripetal force work problem – solved!

A few weeks ago, I wrote the article “Does the centripetal force do some work?“. Since then, I keep myself searching for an answer. Last week, the right answer finally arrived to my inbox. And the answer came from Henry Reich, the creator of MinutePhysics. To be honest, when I wrote him the email my chances of get an answer were really low. However, to my surprise, Henry was patient enough to show me where my error was. Here is the transcription (simplified) of the problem I exposed to Henry, and his very clear response.

First, I explain him why I thought the centripetal force does work:

Henry,
Let me start saying I have some formal physics instruction so I am aware of the energy conservation laws. Actually I don’t think this law can be easily broken. I also know that there is no work done when making a circular path, because of the Work=F*d*Cos(angle) formula.

Now, what takes a man aware of the physics laws to think that it requires work to make a circle? The problem of deflecting an asteroid.

The “deflect an asteroid” problem:
Imagine an asteroid, 1 million tons weight, traveling the flat spacetime with velocity Vx (lets think of it as traveling on a x-y plane). Please note there is no string attached to the asteroid. Just an asteroid happily traveling in a straight line.

You can exert a force Fy over the asteroid, and that force will deflect its trajectory, so Fy will do a work Wy=Fy*dy .
or
You can exert a Fx over the asteroid, opposed to its velocity, so you can slow down the asteroid, and the work done will be Wx=-Fy*dx (negative work, by convention).
To this point every physicist agree there is work done in each of these cases. Perfect!

Now, what if we apply both forces simultaneously? For most physics, work done disappears. Why? Because a circle does not require energy to be created. I know that. The force is always perpendicular to the trajectory, so no work is done.

There is a force Fy applied simultaneously with a Fx force, the asteroid deflects (more than when the Fy force was applied alone), and still there is no work done. Don’t you think this sounds odd, just a little?

If this is true, we can actually deflect such a big asteroid in a circular path, and it will use no energy to do that, because no work has been done!
We can even literally stop the asteroid on a 1 meter circle without any energy expended! I don´t think that could happen in a real world.

Of course, many argue that in fact we will require lots of energy to generate the forces, and that can be true. However, the energy required to generate the force is not considered in my calculation, just the net work done over the asteroid by the force applied. If it takes energy to generate the force, that will add to the net energy required to deflect the asteroid. And that means even more energy would be required.

Does this makes sense?

Please note this goes against what we have learned (It took me months to start to believe this could be true). Also note I am aware that there are lots of things rotating forever with zero energy required (I will explain this later, if we find the asteroid case makes sense).
All we have learned to this day say a circle requires no energy, but please, don’t discard this possibility too fast. Let’s imagine the asteroid and ask your self -no equations involved- Can we really deflect an asteroid without work?

Henry’s answer to this problem was, of course, brilliant. This is the only response that has taken two apparently different realities and generated a coherent response:

Stalin,

I think you are putting too much faith in your intuition and not enough in your logical reasoning.

Also, I think you are confusing what is meant by “work”: you’re totally correct that in certain cases it can take the expenditure of a LOT of energy to keep an object rotating in a circle – moving an asteroid with rockets is one example – but that is not what work is. “Work” in physics parlance, means specifically energy ADDED to the system in question, and while the rockets may expend a lot of energy changing the asteroid’s direction of motion so that it travels in a circle, if they do so entirely by firing perpendicularly then they do not change the asteroid’s speed or kinetic energy – just the direction of the velocity. The point is, if they stopped firing and let the asteroid continue on its path, it would have the exact same energy it started with (though be traveling in a new direction). And in physics parlance, this means that zero work has been done on the asteroid. This doesn’t mean that nothing happened, or that no one expended any energy! It just means that the energy OF THE ASTEROID has not changed.

I think that you are kind of intuitively using a different definition for “work” than what is meant by physicists. All physicists mean by “work” is the change of total energy of an object over time. If an asteroid in free space starts with velocity V and that velocity changes direction but not magnitude (and nothing else changes), then the total energy of the asteroid is still 1/2*m*V^2 and so the change of energy is zero. If you want “work” to mean something different, that’s fine, but you can’t then go argue with physicists that they’re wrong saying “no work happens”, because by their definition, they are right.

This is why I hate the use of the word “work” by physicists. If I could abolish that word forever and just have everyone say “change of energy” then I think all sorts of confusion would be avoided, because the word “work” has many colloquial meanings, many of which are closer to “power” or “force” or other things than “change of energy”.

Henry

# Bell’s spaceship thread does not breaks

Is interesting the kind of ideas that arise from relativity. One of the widely accepted ones are about some kind of “relativistic stresses”. As wikipedia says:

In general, it was concluded by Dewan & Beran and Bell, that relativistic stresses arise when all parts of an object are accelerated the same way with respect to an inertial frame, and that length contraction has real physical consequences.

This assertion is correct, as far as this condition is met: ” when all parts of an object are accelerated the same way with respect to an inertial frame”

However, this is a highly unnatural situation, far from the original “spirit” of the Bell’s spaceship paradox wich initial conditions say:

“Consider two identically constructed rockets at rest in an inertial frame S. Let them face the same direction and be situated one behind the other”. It is obvious that, under this conditions, both spaceships will have exactly the same speed for every instant of time. Moreover, having the same speed they will share the same frame of reference, so B will stay at rest relative to C, and this yields to no broken thread. We can not say the same about A. A will stay at rest, so for A, B and C velocities will be different. Let’s check the facts:

Initial conditions:
$latex V_a=0\\ V_b=0\\ V_c=0\\$

Now, let’s analyze the situation by dividing the flight time on very little intervals $latex \Delta t$, and we will assume that $latex \Delta t$ is so small that the initial condition holds for the entire interval.

Step 1 (B and C start motors). The initial conditions of this interval say $latex V_a = V_b = V_c = 0$, so A, B and C share a common reference frame, so me assume they share the reference frame for the whole $latex \Delta t$. This means $latex \Delta t_a = \Delta t_b = \Delta t_c = \Delta t$, because if A, B and C share the reference frame, time flows equally for all of them.
At the end of this step, we will have:
$latex V_a=0\\ V_b=a\Delta t\\ V_c=a\Delta t\\ V_b=V_c=V_1=a\Delta t$

Step 2. A is a rest, while B and C has a velocity $latex V_1$. This makes B and C to share a common reference frame, while A keeps its original reference frame. This means time flows now equal to B and C, but different for A (by time dilation) $latex \Delta t_b = \Delta t_c = \Delta t$. $latex \Delta t_a \neq \Delta t$. B and C now share the same reference frame, so for all practical purposes they are at rest relative to each other. This of course means the thread doesn’t break during this interval.

$latex V_a=0\\ V_b=v_1+a\Delta t=2a\Delta t\\ V_c=v_1+a\Delta t=2a\Delta t\\ V_b=V_c=V_2=2a\Delta t$

Step 3. A keeps at rest, while B and C has a velocity $latex V_2$. B and C share the same reference frame, so they both keep ar rest relative to each other, so no thread breaks.

As you can see we can keep doing this analysis for a long time, and the result will not change. B and C keeps always at rest relative to each other, so the thread never breaks. Now the question is: why this was not seen before? Simple: we are humans, and by starting with A, B and C from the same reference frame and by being acceleration the same from the begining, we easily tend to think B and C will keep synchronized to A. By the way, A keeps synchronized to B and C? No. Why? By the relativity of simultaneity. When B and C get a velocity V the line of simultaneity for them get a little inclination relative to A rest frame, so simultaneous events for B and C are now out of phase for A. The phase difference is given by: $latex \Delta t=Vx/c^2$, been $latex x$ the distance from A to B or C, and V the velocity of B or C. This means (folowing our analysis) that B and C get velocity $latex V_2$ at the end of step 2 simultaneously for its own frame of reference, but B get that velocity slightly after C from the A frame of reference. This means the minkowski diagrams for this paradox are wrong too, because they assume B and C get velocity $latex V_2$ simultaneously as seen from A frame of reference. The right diagram shows below.

Do you see of any error in this article? Let me know! I’ll really appreciate it.

After seing the article Does the centripetal force do some work? some questions arise: why we could’n see it before? Why are there so many respectable books claiming this? I think I have a culprit: a well known figure that clearly explains why there is no work done when a force is perpendicular to displacement. Yes, it’s all about a simple image. However I must admit it is a very convincing one. Haven’t I found accidentally the asteroid problem we still stay under its hypnotic power.

Let me show, your highness, the innocent gilty of 200 years of conceptual darkness:

Is a really convincing one, don’t you think? However I must say, to its favor, that the image is clear and has ever tried to trick us. This figure is only guilty of being misunderstood.

First, let’s check what the image is showing us: it is a simple block, moving over the floor, and an arrow showing us the force applied. Of.course, there is the already classical $latex \Theta$ angle, pointing out that the force is applied at an angle. The block moves only horizontally (what other direction could it take for christ sake?), so it’s obvious that only the horizontal component of the applied force will do work over the mass. That is what we always have seen. Is there any else?

Yes! There is a normal force, pointing upwards, equal to the vertical component of the applied force. Don’t you see it? Don’t worry, nobody draws it. Can this missing arrow change things? Of course! When this arrow appears is obvious that the vertical component of the applied force is exactly cancelled by the normal force, so the net force will be zero, and zero force do zero work. There is my friend, the source of this error.

Now think for a minute: will the mass move horizontally if no floor is there to stop it? I rest my case, Your Honor.

# Does the centripetal force do some work?

Would you like to show to your physics teacher something he doesn’t know about? Here I leave you the demonstration that the centripetal force, contrary to our beliefs, do work.

Before proceeding, let’s remember that although a circle looks like a “natural” trajectory (after all we see lots of things moving on circles), it happen that bodies don’t move ‘spontaneously’ on circular paths, but on rectilinear paths. Making a circle requires a force that changes the mass trajectory. Even more, it requires a continuous application of this force to keep it circular. Once this force disappears, the object stops making circles and start again a rectilinear motion.

What do we need to create a circle? From the cinematic point of view, all we need is a centripetal acceleration. From the dynamics point of view, all we need is a centripetal force. Now the question is: From the energy point of view, does it requires energy to make a circle? Does this centripetal force do any work over the mass moving on a circle? Or is this just a static force, making no displacement, so the total work will be zero?

To analyze the possibility of any work done over the mass we are going to decompose the centripetal acceleration on its $latex x-y$ components, and calculate a $latex \Delta x$ and $latex \Delta y$ displacements over a $latex \Delta \Theta$ angle.
$latex \Delta x = R\Delta\Theta Cos \Theta$     (1)
$latex \Delta y = -R\Delta\Theta Sin \Theta$     (2)