The reason for all this science-speak? These days, biotech breakthrough make headlines, theory-of-everything books lead best-seller lists, and robots clean our rugs. We may not know more science, but we expect fancier terminology.
The Fantastic Four filmmakers' motto: If you're going to butcher the science, do it with flair. In one early scene in the screenplay I read, the group's leader, Reed Richards, wonders if he'll be able to use his immersive visualization technology in combination with submolecular string theory to more precisely fix the time and place of the big bang. (Why didn't you think of that, Hawking?)
He also uses Spider-Man 2 as his favorite example:
. . . my favorite part is when Dr. Otto Octavius, a nuclear physicist who turns evil after one of his experiments goes awry, explains that he needs giant, artificially intelligent mechanical arms fused to his spine with nanowires and held in check by an inhibitor chip to control the elements of his homebuilt fusion reactor.
In another section of the article, he brings up the point:
This doesn't make the story more plausible--it makes it funnier. Maybe I need to get out more, but I get bigger laughs from the psuedoscientific gibberish spouted in these films than I do from the tired one-liners of standard summer comedies. It's a new genre, a kind of geek comedy.
"A kind of geek comedy." This is funny to me because it exists on lots of levels. Sure, there are the "Wil Wheaton" style geeks who are cute and funny but also utterly useless, but then there are the real geeks, the original geeks, who laugh at things that even the "geeks" let slip by.
My favorite example of this is also from also from Spider-Man 2. I went to see this in the theatre with a few friends of mine. Only three or four of us were engineers, and only two or three of us were ECEs. There were two of us in particular, who now go to grad school together, who had a particular problem when Dr. Curt Conners turned to his class and asked, "What are the eigenvalues?" Peter Parker, who had recently stopped being Spider-Man in order to catch up on studies, work, and relationships, eagerly shot up his hand, was called on, and excitedly responded something like ".213 electron volts."
Now, I don't have a problem with Dr. Conners asking for eigenvalues and Parker only giving a singular response. After all, the linear operator in question could have had repeated eigenvalues. What I do have a problem is that his response had UNITS. Eigenvalues don't have units. Eigenvalues are scalars. They tell you how much a linear operator scales a vector along a particular eigenvector (which could be a number of different things, including a function).
Now, when I heard him give this quantity, I thought I was probably being too harsh. However, my friend, who is noticeably cooler than I am, looked at me during the movie and had the SAME COMPLAINT!
[ I also have a problem with Peter Parker being portrayed like a smart undergrad who could anticipate major problems with Dr. Octavius' work; undergrads are miles away from knowing enough to do that, regardless of what subject they study and where they go to school. However, this really is a different sort of complaint. ]
I've also heard astronomers complain about skies in movies not looking at all like the sky from earth. They really get upset about stars not being in the right place. Would anyone else notice?
So, you see, I think there are different levels of geek humor and geek complaints about these things, and I think it's great that screenwriters can manage to touch all levels of them. Right or wrong, novel or cliche, they do their best to do their worst, and I think that's pretty nifty.
I think this relates to that new book, Everything Bad Is Good for You: How Today's Popular Culture Is Actually Making Us Smarter. Even if it gets my blood boiling in the theatre, in the end, I'm happier it's there. In the end, it gets more people sending me e-mails or coming to my office to ask, "So what is an eigenvalue, anyway?" In the LONG run, how could that be a bad thing?
4 comments:
So. Apparently you goggled "spiderman 2 eigenvalue" and proceeded to post a snarky comment about this on my blog. So, in turn I must post something snarky on yours.
There are many problems that can be reduced to an eigenvalue problem, discrete (eigenvectors) and otherwise (eigenfunctions). As long as the problem has the form Ax = lambda x where lambda is unknown, it qualifies. To claim that eigenvalues cannot have physical units is incorrect. It has the units of the physical problem it is associated with. By that logic, the action of any linear operator is unitless in general as well.
I don't think I posted anything snarky, and because you didn't post anything that can return people to your blog, anyone else who reads this will just have to take my word over yours.
I was just looking for the exact quote. I was hoping to hit one of the many screenplay sites out there, but instead I got all blogs. Your quote seemed the most like the one I remembered, so I used it.
With regard to eigenvalues, these scalars are unitless. The units are in the basis vector. In the case of coordinate eigenvectors, it's appropriate to assume the units are in the vector (x, in your example). However, in general the units are elsewhere in the problem.
And you're incorrect that every Ax=\lambda x qualifies. In that case, x must be nontrivial. More specifically, x must be an eigenvector corresponding to the eigenvalue lambda.
Eigenvalues are scalars. I'm not saying this to be snarky. I'm saying this because this is how they are.
If you really know what you're talking about, then you'll setup an example where the eigenvalue of a linear operator has units. Then I will tell you what you did wrong, and you'll see that the eigenvalues are unitless.
And if that doesn't satisfy you, then maybe we can meet and have a pissing contest.
Okay. Say you have the velocity of a particle encoded as a matrix. The eigenvalues represent the special cases where the particle translation reduces to multiplication by a scalar. (ie the translation occurs in the direction of an eigenvector.)
You would hardly argue that this multiplication is 'unitless', since it's still the application of a velocity. Equating 'scalar' with 'unitless' is odd because it implies all 1D physics is unitless.
And yes, your comment was snarky. It took a clearly absurd story and added a pedantic "I know math! And I know more than you!" And if your audience wants to read it, it's the second hit when you google "spiderman 2 eigenvalue".
The velocity of a particle would be encoded as a column vector. Each element of the vector would represent the component of velocity in a particular direction with respect to a set of basis vectors. The basis vectors would have the units.
However, if you like, you can simplify this to say that the state vector (the velocity of the particle) would carry the units. That's fine too.
Either way, the eigenvalue is unitless. It scales quantities that have units.
Let's say you go from "ft/s" to "m/s". Your explanation somehow suggests that this changes the eigenvalue. However, the eigenvalue stays the same. The operator still scales the vector in the same way... The vector just happens to be a different size to begin with.
And I'm sorry you took the comment that way. Even if that was the tone, I think it was warranted given the tone of your post was, "I know math! And I know more than the screenwriters!"
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