4 ms·
> it’s the only length that can be derived from the constants c, G, and h without adding some arbitrary constant—so it may retain the same value in all referenc
by ajuc 10y ago
> it’s the only length that can be derived from the constants c, G, and h without adding some arbitrary constant—so it may retain the same value in all reference frames, not subject to any Lorentz contraction. But the Planck length is derived from universal constants, so it must have the same value in all reference frames; it can’t change according to a Lorentz contraction.
The arbitrary constant is 1. How is 1 less arbitrary than 2, 42, or 3.14?
I don't get it - is it trying to say objects of Planck length wouldn't contract due to relativity? If so - why?
- chriswarbo 10y ago1 is the multiplicative identity, i.e. x * 1 = x, hence we can always leave out a "* 1" factor without changing things. A different factor, like 2, is more arbitrary, since "* 2" can't be left out without changing the value. You can make the claim that a "* 1" factor is lurking in there, but even from that perspective it's less arbitrary than "* 2", since we could claim that "* 2" also has a "* 1" lurking in it, and so on :)
- ajuc 10y agoI mostly meant - how is this proving that object of length 1 * planck length won't contract, and object of 2 * planck length will. And if it's not - how is the length special?