Tweaking nature's laws

Dennett's main point in Freedom Evolves is that assuming we live in a deterministic universe, then the interesting question is: how different would things have to be in order to produce a different outcome? I know that I can't run a 2 minute mile--that would require deep physiological changes. I know that I can't speak Mandarin right now--that would require deep neural changes. But I might be able to calculate the differential cross-section for some scattering process. That's within the bounds of my capability, though I may fall short.

I think this approach is also valuable in thinking about science. It's useful to ask the question: "How different would the universe have to be in order for X to be the case?" In biology, we could imagine primates evolving to have 6 limbs. It's harder to imagine them evolving to have 2 on one side and 3 on the other, but not impossible.

In physics, we could imagine the value of G being different (even if it would create a universe that could not support life). It's harder to imagine a universe in which the small oscillations around a point of equilibrium is described by a triangle wave rather than a sinusoidal wave. I think that's about like trying to imagine a universe in which 2 + 2 = 5. You sort of can, but it's very hard to do.

Well, here's my rumination on a question my brother asked me, about why E=mc^2 and not E=m(kc)^2. I imagine that this falls closer to the 2+2=5 case than the different G case, but i'm not fully convinced by my explanation. I haven't seen a more complete explanation anywhere else though.


Hey Stu,

Let me try to do a better job answering your question about why E=mc^2 and not E=m(kc)^2.

I think it really centers on frame-invariant quantities. If you want something to be frame-invariant, it needs to be represented in an appropriate 4-vector. An example is the position 4-vector (t, x, y, z), for which s^2 = t^2 - x^2 - y^2 - z^2 is invariant. If you're on a train and I'm on a platform, we may disagree about the time between 2 events or the space between 2 events, but we'll always agree on the s^2 of the 2 events.

Well, rest-mass is something that is also invariant. You and I will always agree on the rest-mass of an object, even if we disagree on its momentum and energy. We can make a 4-vector that obeys these
properties [is this the only way to get a frame-invariant rest-mass?]: (A/c, px, py, pz), where
m^2*c^2 = A^2/c^2 - px^2 - py^2 - pz^2
= A^2/c^2 - p^2.

What is A? Well, note that p = gamma * m * v, with gamma = (1-v^2/c^2)^(-1/2). [You need the gamma if you want conservation of momentum to hold in different reference frames.] Now, let's take an enlightened guess about what A is:

A = gamma * m * (kc)^2.

Then, we can expand the gamma with a Taylor series (trust me if you forget how to do this):

A = m*(kc)^2 * (1 + v^2/2c^2 + 3v^4/8c^4 + ...)
= m*(kc)^2 + (1/2)m(kv)^2 + (3/8)mv^4(k/c)^2 + ...

The third and higher terms are negligibly small, so let's drop those.
Then if we set k=1, we get

A = mc^2 + (1/2)mv^2

All that we can measure are changes in energy. So what we will measure are changes in (1/2)mv^2, which just happens to be the same as the expression for kinetic energy that we're familiar with. If, however, we had left k not equal to 1 then changes in A would not correspond to changes in Newtonian kinetic energy. And we would not be able to set A = E.

Not sure if that helps. It's a good question, one that hits at some subtle issues in relativity. Most of the time we just use it because we know that it works! :)