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Electromagnetic repulsive force between protons
By Brad Rodgers (P1930) on Sunday, May 27, 2001 - 05:25 am:

Is there a law that represents what acceleration the nuclear force causes at a given radius? I tried to use the fact that a particle travels in a hyperbola when it gets close to another particle to derive what this force would be, but the calculations become too complex to carry out (at least for anyone who doesn't like to simplify 10 lines of r's and thetas). Is there a reasonably elegant way to derive the laws from the hyperbola bit, or is there a way to show the hyperbolic path from another well known law?

Brad

By Sean Hartnoll (Sah40) on Sunday, May 27, 2001 - 10:22 am:

I think you mean the electromagnetic force between protons and protons (the hyperbolic orbits of electrons were used by Faraday to deduce the existence of the nucleus). Anyway the derivation that hyperbolic paths correspond to a repulsive inverse square law, is essentially the same that attractive inverse law gives elliptic orbits, which I think has been covered on NRICH before, but I can't find the link, does anyone know it?

Sean

[I can't find it either. - The Editor]

By Brad Rodgers (P1930) on Sunday, May 27, 2001 - 05:43 pm:

Thanks. I think I know what to do with this now that I know the inverse square part. Who would've guessed that the electromagnetic force would be just the opposite of the gravitational force, though!

Thanks again,

Brad

By Sean Hartnoll (Sah40) on Sunday, May 27, 2001 - 06:34 pm:

Yes, at one level it is a nice coincidence. Note that the sign is the same if the charges are opposite. There are deeper theoretical reasons why it is the sort of law of you might expect, on hindsight. Also, the inverse square law for electromagnetism is just a static approximation (once the charges start moving there are relativistic effects, for instance you cannot get stable orbits in EM like you can with gravity, which is why people need QM to describe the atom). And Newtonian inverse square law is a low speed, low mass approximation to GR, so in a sense an inverse square law is something that comes out naturally as certain limits. But in any case it is certainly a nice thing, and it allows many computational techniques to be imported from one area of physics to another.

Sean

By Brad Rodgers (P1930) on Sunday, May 27, 2001 - 10:27 pm:

When I do the math to find what the electromagnetic orbit should be, I come up with the general form of

r=1/[A×sin(q+D)+(1/k2)]

for k=r2dq/dt

First off, D just determines the angle of view, so we may neglect it. And, that formula seems to produce correct results so long as

either 1/k2 or A is negative (but not both)

and so long as

|1/k2|<|A|.

But the problem comes when I start to think about what k and A actually are. It would seem as though k=r0v0, and then by substitution, A=1/r0-1/(r0v0)2. Clearly the two requirements cannot be met under these conditions. Is there something wrong that I have done (I'd think that if there was, it'd be in the r0v0 part)?

Thanks,

Brad

By Sean Hartnoll (Sah40) on Sunday, May 27, 2001 - 11:17 pm:

Firstly, if you have D'Inverno, this is all done nicely (for the elliptic case) starting on page 192. I think the answer you should get is

r = C/(1+ecos(q - q0))

Which is basically what you got. The criterion that makes it a hyperbaloid is that e (called the eccentricity) is > 1 (this is in fact your second criterion, your first is I think incorrect, in fact they are both positive). C is positive. I'm pretty sure this will work.

Sean

By Brad Rodgers (P1930) on Monday, May 28, 2001 - 12:48 am:

I had the first result because when both are positive or both negative, a parabola or ellipse is produced, so for a hyperbola (one that turns around before the origin), one must be negative and the other positive. Anyway, I think I see the error in my logic. It occurs in the substitutions in the third paragraph.

I'm having a bit of trouble with D'Inverno's notation; what does the r with the ^ on top mean, with the dots? I know that mr^ must = G, but it obviously has more significance than this.

Brad

By Sean Hartnoll (Sah40) on Monday, May 28, 2001 - 09:20 am:

The dot means differentiation. The ^ means unit vector, so for instance r^ is the vector r divided by its modulus. So whilst a vector normally gives you a direction and a magnitude, a vector with a hat on it just gives a direction.

The coefficients are always positive, for ellipses, hyperbolae or parabolae. The point is that if e<1 then the expression is never zero, so the orbit never goes to inifinity and it is closed, whilst if e >= 1 then there is a point at which the denominator goes to zero and so r goes to infinity, because the cosine (or sine) can be negative.

Sean