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The latest about experts, "experts", and talking heads leads me to risk having my head chopped off again, about scientific consensus.
Specifically, I refer to the consensus of scientists working in the subject matter at hand. Consensus means at least 90% agreement -- not mere majority agreement, and especially not "majority rules". 90% agreement, maybe more. (There are always, for various reasons, holdouts.)
It's a fact that the scientific consensus will be accurate most of the time. (Electromagnetism inaccurate? Quantum mechanics inaccurate? None of the electronics in front of your eyes would exist. Our world of electricity wouldn't exist.)
Someone mentioned Einstein's statement about only one person being necessary to overturn relativity. If that one person is going beyond relativity, (for example) arguing that when E/m ~ 10^15, the particle's speed surpasses the speed of light, he is very much part of the consensus. (Compare with E/m ~ 10,000 for protons in the Large Hadron Collider and E/m ~ 100,000 for electrons in the former Stanford Linear Accelerator Center.)
That's just one of many ways one might go beyond relativity.
If we're talking about debunking relativity, that person is going to have to deal with a century's work in relativity (theory, experiment, observation, application). He has to understand the material -- in fact, he probably has to become the best expert in calculating relativity and comparing or contrasting relativity's predictions with (experiment, observation, application) -- if he's going to debunk relativity.
747s circling the world? Low-orbiting satellites and space shuttles losing about 30 microseconds a day? The Global Positioning System (GPS) requiring corrections from General Relativity to get the right answers?
The debunker would have to be able to do the following elementary calculation of relativity, as a prediction of relativity:
A 400-MeV electron is elastically scattered off a free stationary proton. The electron's scattering angle is 30 degrees. Calculate the electron's final energy, and the proton's momentum, kinetic energy, and out-going direction.
Then he would have to compare with experiment. Is it accurate? (My personal experience, with different numbers: yes, it's accurate.) If not, does including the light emitted during the collision make the prediction accurate? ("Radiative corrections" -- this part is enormously harder.)
In principle, this problem is within the scope of a college physics class (freshman, non-calculus) that covers the basics of relativistic mechanics. It only requires relativistic energy and momentum, energy and momentum conservation, algebra, and trigonometry.
In practice, the freshman would find it intractable. Typically, a young graduate student has to be shown the calculation, the way it becomes simple, requiring only a page or so of algebra.
***********
Your relativity debunker would have to calculate that, and so much more, to compare and contrast with experiment. From my own encounters, the would-be debunkers are cranks who don't know the subject. It routinely slips out that they deny (are oblivious to) Newtonian physics as well.
Comments
translation
Some "experts" are talking out of their asses. Scientific consensus is of course based on some assumptions as well as data and a change in our understanding of physics usually comes when those assumptions are shown not to be universally true. They are still true in most normal circumstances on a macro level.
Crackpots
Crackpots don't think that way. They believe that their opinion is worth the same as your years of study, then challenge you to prove yourself.
All too true...
All too true...
-- Daphne Xu
Unfortunately, there are many
Unfortunately, there are many experts who end up in the same situation. "I'm an expert! Why should I have to show evidence?"
It's like elderly people who think they deserve respect just for being old, even from people who don't know them at all. Whereas what they deserve is politeness.
I'll get a life when it's proven and substantiated to be better than what I'm currently experiencing.
They might just be mistaken ...
It is quite easy to mistake as arrogance what in reality is just frustration and resignation - most "members of the public" will be unable to understand proper evidence about scientific issues. After decades of being frustrated that way (you start to explain, get a blank look and/or scepticism, then they start to nod off) you may tend to anticipate the negative reaction and thus skip the "giving evidence" phase.
I'm a computer scientist and just try this as an example:
Try to explain sensible password policies (as opposed to the usual "change on a fixed schedule", "use something cryptic looking" etc. that is often enforced in corporate environments) to someone who has not studied computer science, has no idea of statistics, probability theory or cryptography. Even some of my colleagues were blindly accepting some recipes as a solution for every situation even though those recipes were only appropriate for a very specific niche (e.g. web logins). Laymen try to show you your errors even after you present them with mathematical proof - it runs counter to their perception (and what tey are used to), so it must be false. Argh.
I can understand why scientists reserve presenting evidence for those who can actually understand it.
I don't accept that, no. I
I don't accept that, no. I'm an IT professional as well. It's next to impossible to explain to people like banks that degrees of entropy is more important than complexity for a password. They've been told by so-called experts that an 8-10 character password with a mishmash of letters and number is the BEST thing to have. Those experts never explained it - so you can't refute them.
For those that don't follow, it's simple. The longer your password, the harder it is to guess, no matter how simple it is. 'TheThirdChickenontherightisdinner' is harder to guess than C1cK3N$-
But you can't use a long password at most sites, nor can you bypass the goofy, hard to remember stuff, if your password is long enough.
Anyway - scientists should not have the luxury of refusing to present evidence, even if that is simply "In 30 years of experience in playing with fish, I expect X to happen." (Which, a 40 year fisherman could refute) If people don't understand it, that's a separate problem.
I'll get a life when it's proven and substantiated to be better than what I'm currently experiencing.
On the other hand...
Scientists don't need to spend all their time explaining basic concepts to people who never bothered to learn them in the first place. I rarely bother debating with flat earthers.
I didn't say they should
I didn't say they should explain the concepts. Just that they should present the evidence. It's up to the audience to research the concepts. Now, if his _peers_ demand he explain his concepts, that's a different story, and everyone else should stay out of weapons reach of the ensuing argument.
I'll get a life when it's proven and substantiated to be better than what I'm currently experiencing.
ScienceSplaining
I once had someone try to prove some wacky flat earth style theory by asking why the air gets colder when you go up a mountain. I forgot the reasons and justifications, but everyone knows that hot air rises, so it should get hotter as you go higher.
And, indeed, you see that in bodies of water. Drop a temperature probe (or dive) in a lake or ocean, and it gets colder as you go down.
How much time do I spend talking to him about Boyle's Law? P1*V1/T1=P2*V2/T2 As the air goes up, the lower pressure causes it to expand. As it expands, it cools.
This guy isn't going to say, "Gee, you're right. I guess I need to throw out my pet theory."
(By the way, water doesn't cool when the pressure goes down because it doesn't expand. The volume remains constant. Mostly.)
Oh, I say it anyhow, just so that bystanders are less likely to be deceived. But I'm sure that he's still running around preaching his gospel about them ivory tower scientists don't know nuthin.
Interesting thought
In your experiment with an electron scattering off a proton your formula assumes the proton is stationary. While in reality the earth is moving therefore the test chamber the proton is in along with the proton are both moving. So in truth the proton must be moving at the same speed and direction as the observer for the formula to be correct.
Which leads to that interesting thought I had while reading this blog;
What effect would we see if the observer was moving at different speeds and directions relative to the experiment?
We the willing, led by the unsure. Have been doing so much with so little for so long,
We are now qualified to do anything with nothing.
Stationary proton
First, we're talking about the frame in which the lab is at rest. (Part of relativity is understanding that notion.) The initial electron energy and scattering angle are the values in that frame.
The proton is only approximately stationary, of course. I imagine the proton as embedded in a material such as polyethylene -- specifically, a hydrogen atom in the molecule. Its kinetic energy is a fraction of an eV, and its momentum is tens of thousands of eV/c. An MeV is a million eV. The proton's final momentum would be in the hundreds of MeV/c, and its final KE in the tens of MeV. In other words, any initial proton motion is completely negligible if our answer is limited to a few significant figures.
Go to a different frame? How fast are you imagining? 1% the speed of light? 10%? 90%? Also, which direction are you going?
In any case, one would have to Lorentz-transform the initial electron energy and the scattering angle. Alternatively, one could solve the problem in the lab frame, and Lorentz-transform all the answers. Your one person attempting to debunk relativity would be able to figure out how to do it.
And also, to be complete, there may also the problem of Lorentz-transforming the measuring device.
-- Daphne Xu
The measuring device.
Regarding Lorentz-transforming the measuring device, here is a bit of a trick question. A spaceship's length is 100 meters. Joe measures the length with a meter stick, laying it down and picking up, and counting the number of times he does it. How many times does he lay down and pick up the meter stick? 100 times. (An obvious answer.)
The spacecraft is flying fast enough so that in the frame of the earth, it's now only 50 meters long. (Lorentz-Fitzgerald contraction.) How many times (in the earth's frame) does Joe lay down and pick up the meter stick? Answer: 100 times. The meter stick has also shrunk 50%.
-- Daphne Xu
Hm... in that case, he may
Hm... in that case, he may not have the time to actually lay the stick down 100 times - or at least, everyone on Earth interested in the answer will have died of old age.
I'll get a life when it's proven and substantiated to be better than what I'm currently experiencing.
Not at all.
Not at all. Gamma is two in this situation. If it takes him (say) ten minutes to do the job, those on the earth would view it as taking 20 minutes. (Actually watching it, it would be doppler-effected further, depending on whether he was approaching or receding, or moving transverse.)
-- Daphne Xu
All true, but...
If in your example Joe is not onboard the ship and uses the same measuring stick to measure the apparent length of the ship, now he would only pick up the stick 50 times. Thus my statement of it being relative to the observer. it's an interesting exercise in thought :)
We the willing, led by the unsure. Have been doing so much with so little for so long,
We are now qualified to do anything with nothing.
Doing It
If he's not on board the spaceship, or comoving with the spaceship, he couldn't do it. With gamma equal to two, the spaceship's speed is sqrt(3)/2 times the speed of light (0.87c). If he's stationary but manages to hold the meter stick next to the spaceship once, the second time, the spaceship is close to a million kilometers away.
Not to mention, Joe doesn't want to be anywhere near a moving spacecraft.
-- Daphne Xu
What if they then strike an
What if they then strike an object weighing approximately 1/10th of a gram?
(If a tree falls on a forester, and there's no one around to hear him scream, does he make a noise?)
I'll get a life when it's proven and substantiated to be better than what I'm currently experiencing.
The Spacecraft, I assume?
It's a standard collision problem in relativity. The moving spacecraft hitting a stationary object (in the earth's frame)? Does the tiny object embed itself in the spacecraft? Does it bounce off the spacecraft? In what direction? If kinetic energy is conserved, the collision is elastic. If the object embeds itself in the spacecraft, the collision isn't elastic, but the spacecraft gains (a notch under) 0.2 grams of mass, and slows down a tiny notch.
Other bad things can happen as well.
The problem can be done in either frame (any frame).
The problem I posed in this original post is such a collision problem -- specifically, elastic.
I think that one should do Newtonian collision problems, before attempting Einsteinian collision problems.
-- Daphne Xu
A 0.1-gram rock
... moving at 10^6 m/s, hits a stationary 10^10-gram asteroid, and they stick together. How fast does the resulting asteroid go?
Or a 10^10-gram asteroid moving at 10^6 m/s hits a stationary 0.1-gram rock. How fast does the resulting asteroid go?
In both cases, what's the change in velocity of the asteroid?
Standard freshman physics problem.
-- Daphne Xu
The general practice
... is to do all calculations in one frame, then (if necessary) Lorentz-transform to a different frame. In this case, the calculation is done in the lab frame, so no further transformation is needed.
Comparing theory with experiment: The electron detector is placed at 30 degrees. A proton detector is placed at the calculated angle. Both detectors are also set to detect a range of momentum, and that range is (approximately) centered on the calculated answers. Each particle's momentum is measured. The final energies are calculated from the momentum. There are other data analysis issues that I don't wish to go into.
The questions for a particular electron scattering: does the electron go into its detector within the range of momentum? Does the proton go into its detector within the range of momentum? If the proton's angle is wrong, then the proton won't go into the detector. If the calculated final momenta (plural of momentum) are way off, then the detectors won't detect them. If the particles are detected, then the question is, do the particles actually have the final momenta calculated?
In electron accelerator nuclear physics experiments, this experiment is done so often (with different electron energies and angles), it's no longer a real experiment. Instead, it's done as part of calibrating a nuclear physics experiment. It's been done probably thousands, perhaps tens of thousands, of times.
Your would-be relativity debunker would have to deal with these successes.
Your would-be relativity debunker has to understand the subject.
I hope that you (Nuuan) noticed that when you mentioned Einstein regarding the scientific consensus, you blundered head-on into a subject I was very familiar with.
-- Daphne Xu
But you are still looking at it
But you are still looking at it from the view point of the measuring device (observer) within the same space/time frame as the experiment. Of course the formula would work as predicted when the measuring device is moving at the same relative speed.
I'm asking what happens when the experiment is measured from a perspective outside of the same space/time. When the experiment and the measuring device are moving at different speeds than each other. I am actually curious how someone with your level of knowledge on the subject, which is higher than mine, answers this?
We the willing, led by the unsure. Have been doing so much with so little for so long,
We are now qualified to do anything with nothing.
No.
"Of course the formula would work as predicted when the measuring device is moving at the same relative speed." If relativity were wrong, the final electron energy would be wrong. And once the final electron energy (and momentum) are wrong, then the final proton angle, momentum, (and energy) are wrong. The whole set of calculations fail..
So you imagine that the situation is identical, but the detectors are moving. If they're both moving at the same velocity, transform to their rest frame. You would have a different initial electron energy. The one difference it makes to the calculation is that now, the protons might have a (non-negligible) initial momentum.
If you're thinking of speeds such as the earth's orbital speed around the sun, v/c = 10^-4, and gamma = 1 + 0.5*10^-8. Try again.
If the two detectors are moving at different velocities, try imagining setting up such an experiment. Perhaps out in space. Also, two detectors moving at different velocities would have a huge trouble measuring the appropriate scattering angles.
I hope that you realize that space-time issues of relativity lead to mass-energy-momentum issues -- different relations between mass, speed, energy, and momentum. This experiment tests the mass-energy-momentum issues.
BTW, the calculation with the initial electron energy and scattering angle that I gave, gives the final proton momentum of 200 MeV/c. Newtonian mechanics is adequate to approximate its kinetic energy, and I hope that you can do it. The proton mass is 938 MeV/c^2.
Also, I hope that you can compare the Newtonian and Einsteinian results for the momentum of the initial electron, given the energy of 400 MeV and the mass of 0.511 MeV/c^2. They are completely different. (For the Newtonian calculation, the electron energy is its kinetic energy. For the Einsteinian kinetic energy, subtract off mc^2, but that's lost in the insignificant figures.) You might also compare the speed calculations (Newtonian vs. Einsteinian). Again, they're completely different.
-- Daphne Xu
Responxe to Measuring Spaceship example
Actually the meter stick would be laid down 100 times BUT picked up 99 times!! LOL
Good one Diana
But you know his mother would see that he left his meter stick lying about and make him pick it up and put it away, so it would be 100 times :)
We the willing, led by the unsure. Have been doing so much with so little for so long,
We are now qualified to do anything with nothing.