Deriving Ampere's law from the Biot-Savart law

[fayled]

If we write the Biot Savart law as
B(r)=μ₀/4π∫(J(r')xn/n²)dV'
where B is the magnetic field which depends on r=(x,y,z), a fixed point, J is the volume current density depending on r'=(x',y',z'), and n=r-r', a vector from the volume element dV' at r' to the point r. Note we integrate over the primed coordinates as J, the source of the current varies with these.

Then take the curl, making use of curl(AxB)=A(∇.B)+(B.∇)A-(A.∇)B-B(∇.A), and noting ∇.J=0 (it depends on the primed coordinates) and that (n/n².∇)J=0 for the same reason, we get
∇xB=μ₀/4π∫J(∇.n/n²)dV'-μ₀/4π∫(J.∇)n/n²dV'.

Now term two can be shown to integrate to zero, which I understand (incidentally, the book says this second term is integrated over a volume enclosing all current, as I suspected - this is related to my problem below), but I have a problem with term 1. We get
∇xB=μ₀/4π∫J(∇.n/n²)dV'
and ∇.n/n²=4πδ³(r-r'). So
∇xB=μ₀/4π∫4πJ(r')δ³(r-r')dV'
∇xB=μ₀∫J(r')δ³(r-r')dV'
Now according to my book, this reduces to
∇xB=μ₀J(r), which makes sense in a way because we can say
∇xB=μ₀∫J(r)δ³(r-r')dV'
because only the value of J at the 'spike' is actually useful anyway. This is constant so
∇xB=μ₀J(r)∫δ³(r-r')dV' and then this integral is simply one giving the desired result.

However, we're integrating over (x',y',z') and so varying r', keeping r fixed. As our integral only need cover all of our current, and r could be outside of our current distribution, (according to my brain) we need not even integrate over the 'spike' of the delta function at r'=r, which makes the integral zero.

What is it I am misunderstanding? Thanks.

Edit: If the answer is something along the lines of 'we may as well integrate over all space because there's no current anywhere else anyway', I ask, if there's no current anywhere else anyway, why do we get two different answers by integrating/not integrating over all space.

 

[jambaugh]

You are not wrong!

As our integral only need cover all of our current, and r could be outside of our current distribution, (according to my brain) we need not even integrate over the 'spike' of the delta function at r'=r, which makes the integral zero.

You are not wrong... just considering a special case. IF you pick a point outside the current distribution the curl of B is zero as your mathematical intuition indicates. You are deriving Ampere's law -after all- which states that the curl of B at a point is proportional to the current density at that point. But r need not be outside the current distribution in which case you DO need to integrate over the spike and wherein B and J are not zero but still proportional.

All is right with the world.

 

[fayled]

You are not wrong... just considering a special case. IF you pick a point outside the current distribution the curl of B is zero as your mathematical intuition indicates. You are deriving Ampere's law -after all- which states that the curl of B at a point is proportional to the current density at that point. But r need not be outside the current distribution in which case you DO need to integrate over the spike and wherein B and J are not zero but still proportional.

All is right with the world.

Oh dear, that is pretty obvious now Thankyou :)

 

[vanhees71]

Well, and if $\vec{r}$ is at a point outside the region where an electric current is flowing, your equation is correct too (for stationary currents only, of course!), because there $\vec{j}(\vec{r})=0$. The local Maxwell Equations are always right!

 

[paranormal]

It seems that you are misunderstanding the concept of a delta function. A delta function is not a traditional function, but rather a mathematical tool used to represent a spike or impulse at a specific point. In this case, the delta function δ3(r-r') represents a spike at the point r=r'. This means that the integral over all space is necessary in order to capture the contribution of the current at that specific point.

In other words, the integral over all space is not just a convenient choice, but rather a necessary one in order to properly account for the contribution of the current at that specific point. If you do not integrate over all space, you are essentially ignoring the contribution of the current at that point, which is not a valid approach.

Additionally, the integral over all space does not necessarily mean that there is current everywhere in space. It simply means that we are accounting for the contribution of the current at all points in space, including the point where the delta function is located.

In summary, the integral over all space is necessary in order to properly account for the contribution of the current at the specific point represented by the delta function. Ignoring this integral would lead to incorrect results.