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I asked if the rest mass of a photon is still zero in non-vacuum, where it's velocity is less than c. I got the answer that it is indeed still zero, because the length of the 4-vector (E/c,px,py,pz) is zero. And because this is a flat space property it's easily generalised to other coupled fields to describe non-vacuum situations: the result still holds. Now my problem is that I don't know what a flat-space property is, i.o.w. why the four-vector is still zero.

2006-08-15 04:54:50 · 4 answers · asked by helene_thygesen 4 in Science & Mathematics Physics

4 answers

It's so complex, it's simple. The 4 vector is still zero because the vectors are all offsetting, hence the creation of "flat space," which isnt actually flat, but inherently it appears that way because the vectors all cancel each other out.

2006-08-15 05:05:09 · answer #1 · answered by Dunc 1 · 0 0

What an in depth question.

First, a flat space is one that behaves like Euclidean space - i.e., it follows the geometry theorems you studied in high school geometry. In a flat space, straight lines never cross each other and the angles in a triangle add up to 180 degrees, etc, etc,. But all that's important for your answer. I don't know the context of your question, but the relevant information you seek isn't about flat space, it's about the properties of vectors.

Mathematically, a true vector has the property that it's length remains constant under spacial transformations. In other words, you can rotate, translate (move over), or go from a flat space to a curved space, and the length of the vector stays the same.

More generally, vectors belong to a larger class of mathematical objects known as tensors. Tensors all have this property that their length is constant under coordinate (space) transformations. The equations of General Relativity are all tensor-equations. So, very often, when one wants to do a proof of something in GR, one does the proof in flat space, and since the equations are tensor equations, the result holds in curved space. That's probably why you got the answer you got.

I hope that answers your question.

2006-08-15 05:11:55 · answer #2 · answered by Davon 2 · 0 0

Check out:

http://scienceworld.wolfram.com/physics/MomentumOperator.html

Seems to me, as m --> 0, the concept of moment space breaks down. It's been 30 some years since I took advanced physics; so I've forgotten more than I now know.

2006-08-15 05:24:50 · answer #3 · answered by oldprof 7 · 0 0

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2017-01-04 03:26:13 · answer #4 · answered by Anonymous · 0 0

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