# In a mass spectrometer what mediates the acceleration?

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1. Apr 30, 2013

### mesa

R=mv/qB describes the radius of a charged particle moving through a static Bf with the acceleration directed radially inward. The cross product is used to calculate the Force that results in this acceleration. Why does the cross product work?

2. May 1, 2013

### Staff: Mentor

Physics cannot answer "why" the laws of physics are like that. We can just observe them, and describe them. The cross product for the magnetic force is one of those descriptions.

3. May 1, 2013

### mesa

That was the answer I was expecting. At least it works although it is unfortunate these interactions have not been figured out yet.

4. May 1, 2013

### sophiecentaur

It's nothing to lose any sleep over, afaiac. There will always be 'something more' that we haven't sorted out yet. It just happens that the "cross product" works for describing and predicting a lot of physical processes. You could, perhaps, say that in itself is a bit 'magical' - but the same can be said about the way a lot of Maths applies across the whole of Science.
In any case. What IS Maths??

5. May 1, 2013

### mesa

We know so much today and it seems there is still much left to discover, how exciting! I should have started Physics when I was younger... lol

6. May 1, 2013

### ModusPwnd

I wouldn't say it just happened, I would say that it was specifically designed for that purpose.

There is a paragraph about it in wikipedia.

http://en.wikipedia.org/wiki/Cross_product#History

7. May 1, 2013

### sophiecentaur

The cross product is just a part of the whole field of vector manipulations and I would say that it just happens (not "just happened") to be a subset of Maths that describes many Scientific relationships. Not surprisingly, in that Wiki article, the Cross Product appears as a paragraph within a much larger body of information about vectors in general.
Maths is (I'd contend) a bigger set than Science because not all of Maths relates to Science. In a separate Universe, somewhere, with an entirely different Physics, there could be Mathematicians who could be coming up with the same Maths theorems as they do here - but a different set might apply the their 'reality'.