# Question on Dirac Delta function in Griffiths

1. Aug 6, 2013

### yungman

My question is in Griffiths Introduction to Electrodynamics 3rd edition p48. It said

Two expressions involving delta function ( say $D_1(x)\; and \;D_2(x)$) are considered equal if:
$$\int_{-\infty}^{\infty}f(x)D_1(x)dx=\int_{-\infty}^{\infty}f(x)D_2(x)dx\;$$6
for all( ordinary) functions f(x).

In note 6 at the bottom of the page:
This is not as arbitrary as it may sound. The crucial point is that the integral must be equal for any f(x). Suppose
$D_1(x)\; and \; D_2(x)$ actually differed, say, in neighborhood of the point x=17. Then we could pick a function f(x) that was sharply peaked about x=17, and the integrals would not be equal.

I don't understand what the book is talking about. You have two different Dirac Delta functions and can be equal for all f(x).

Say $D_1(x)=\delta (x-2) \; and \; D_2(x)=\delta(x-4)$. Let f(x)=x:
$$\int_{-\infty}^{\infty}x\delta (x-2)dx=f(2) \int_{-\infty}^{\infty}\delta (x-2)dx=2$$
$$\int_{-\infty}^{\infty}x\delta (x-4)dx=f(4) \int_{-\infty}^{\infty}\delta (x-2)dx=4$$

How is these two equal?

Thanks

2. Aug 6, 2013

### king vitamin

They are not equal! Thus,

$$\delta(x-2) \neq \delta(x-4) .$$

I think all that Griffiths is trying to stress here is that these functions really only exist under an integral sign (so really my inequality above is in some sense purely formal). All you should really keep in mind is that when you're trying to prove a property of a delta function, it's almost always better to work with it inside an integral.

3. Aug 6, 2013

### yungman

Thanks for the reply. I understand that delta function only really work inside the integral. The confusion is in the note 6:

The crucial point is that the integral must be equal for any f(x). Suppose $D_1(x)\; and \; D_2(x)$ actually differed, say, in neighborhood of the point x=17. Then we could pick a function f(x) that was sharply peaked about x=17, and the integrals would not be equal.

I read it over and over, it said the two delta function can be different!!! I just don't know what he is trying to say!!! Even if I follow this and set $D_1(x)=\delta(x-16.99)\; and \; D_2(x)=\delta(x-17.01)$, the result is still different!!!

Thanks

4. Aug 6, 2013

### micromass

Staff Emeritus
He's talking about combinations of delta functions. So for example, take $\delta(x)$ and $2\delta(x)$. From the "naive" definition of the delta function, they both are 0 everywhere and infinity in the origin. So there is no real way to distinguish the functions this way.

Furthermore, following integrals are equal

$$\int_{-\infty}^{+\infty} 2x\delta(x) dx = 0 = \int_{-\infty}^{+\infty} x\delta(x)dx$$

So taking the function $f(x) = x$ doesn't work in this case.

But of course, the rule is saying that the two functions are regarded as equal if

$$\int_{-\infty}^{+\infty} 2f(x)\delta(x) dx = \int_{-\infty}^{+\infty} f(x)\delta(x)dx$$

for any (reasonable) function $f(x)$. You can easily find a function for which it fails.

To be honest, his note is pretty confusing, but I think this is what he's trying to say.

5. Aug 6, 2013

### ModusPwnd

That difference manifests only if your function has a non-zero value in the neighborhood. You can pick a particular function for f where your original equality holds. But this does not imply the two delta functions are equal. They are only equal if the original equality holds for all f, not just one cherry picked f.

6. Aug 6, 2013

### yungman

Thanks. I read it over and over, I think he is trying to say $D_1(x)=D_2(x)$ if
$$\int_{-\infty}^{\infty} f(x)D_1(x)dx=\int_{-\infty}^{\infty} f(x)D_2(x)dx$$

7. Aug 6, 2013

### micromass

Staff Emeritus
Yes. And you should take this as the definition of the equality $D_1(x) = D_2(x)$. It's not some consequence that should be proven.

8. Aug 6, 2013

### king vitamin

That's not how I'm reading it. I think the point is, by defining the equality of delta functions by

$$\int_{-\infty}^{\infty}f(x)D_1(x)dx=\int_{-\infty}^{\infty}f(x)D_2(x)dx\ \Longrightarrow D_1(x) = D_2(x)$$

one might worry that there could possibly be several, very different-looking functions which could be deemed "equivalent." However, the fact that $f(x)$ is completely arbitrary actually severely constrains this definition, so that as you can see, $D_1(x)$ and $D_2(x)$ really must take the same "values" everywhere. Read Griffiths' qualifiers: "Suppose they differ... then we could pick a function... so that the integrals would differ." He's doing a proof by contradiction.

EDIT: Sounds like you got this cleared up while I was writing though!