Don’t Marry Your Arbitrary

This fall, I’m TAing a course on General Relativity. I haven’t taught in a while, so it’s been a good opportunity to reconnect with how students think.

This week, one problem left several students confused. The problem involved Christoffel symbols, the bane of many a physics grad student, but the trick that they had to use was in the end quite simple. It’s an example of a broader trick, a way of thinking about problems that comes up all across physics.

To see a simplified version of the problem, imagine you start with this sum:

g(j)=\Sigma_{i=0}^n ( f(i,j)-f(j,i) )

Now, imagine you want to sum the function g(j) over j. You can write:

\Sigma_{j=0}^n g(j) = \Sigma_{j=0}^n \Sigma_{i=0}^n ( f(i,j)-f(j,i) )

Let’s break this up into two terms, for later convenience:

\Sigma_{j=0}^n g(j) = \Sigma_{j=0}^n \Sigma_{i=0}^n f(i,j) - \Sigma_{j=0}^n \Sigma_{i=0}^n f(j,i)

Without telling you anything about f(i,j), what do you know about this sum?

Well, one thing you know is that i and j are arbitrary.

i and j are letters you happened to use. You could have used different letters, x and y, or \alpha and \beta. You could even use different letters in each term, if you wanted to. You could even just pick one term, and swap i and j.

\Sigma_{j=0}^n g(j) = \Sigma_{j=0}^n \Sigma_{i=0}^n f(i,j) - \Sigma_{i=0}^n \Sigma_{j=0}^n f(i,j) = 0

And now, without knowing anything about f(i,j), you know that \Sigma_{j=0}^n g(j) is zero.

In physics, it’s extremely important to keep track of what could be really physical, and what is merely your arbitrary choice. In general relativity, your choice of polar versus spherical coordinates shouldn’t affect your calculation. In quantum field theory, your choice of gauge shouldn’t matter, and neither should your scheme for regularizing divergences.

Ideally, you’d do your calculation without making any of those arbitrary choices: no coordinates, no choice of gauge, no regularization scheme. In practice, sometimes you can do this, sometimes you can’t. When you can’t, you need to keep that arbitrariness in the back of your mind, and not get stuck assuming your choice was the only one. If you’re careful with arbitrariness, it can be one of the most powerful tools in physics. If you’re not, you can stare at a mess of Christoffel symbols for hours, and nobody wants that.

3 thoughts on “Don’t Marry Your Arbitrary

  1. Mr. Eldritch

    This … doesn’t seem right to me. It’s specifically because f(i, j) is an arbitrary function that we can’t just swap its arguments in one of the terms, because it is not true that for functions in general f(i, j) = f(j, i). (for instance, f(i,j) = ie^j will not actually yield the same values if you swap it for je^i)


      1. 4gravitonsandagradstudent Post author

        To clarify for readers who might be confused about something similar: it’s important that i and j are summed over. When the other side of the equation was just g(j), you couldn’t swap i and j because a specific choice of j was part of the “question” you were answering. When you take the sum, though, j becomes arbitrary, it isn’t part of your “output”.



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