Intuition on integral term in D'Alembert's formula

[bosque]

If $$\phi(t,x)$$ is a solution to the one dimensional wave equation and if the initial conditions $$\phi(0,x) , \phi_t(0,x)$$ are given, D'Alembert's Formula gives
$$\phi(t,x)= \frac 12[ \phi(0,x-ct)+ \phi(0,x+ct) ]+ \frac1{2c} \int_{x-ct}^{x+ct} \phi_t(0,y)dy . \tag{1}$$
which is commonly written (letting c = 1 and defining g and h)
$$\phi(t,x)= \frac 12[ g(x-t)+ g(x+t) ]+ \frac12 \int_{x-t}^{x+t} h(y)dy . \tag{2}$$
My question is:
What is the physically intuitive meaning of the integral term? I think I understand the displacement IC. It is the velocity IC that confuses me. I know h is the vertical displacement velocity (the time derivative of the vertical displacement) and the integral sums up the effects on the displacement of the velocity from each point on the x-axis between x-ct to x+ct. But, for just one example of my confusion, it doesn't seem that the points on the x-axis between x-ct and x+ct (excluding x-ct and x+ct) should contribute since they would reach the point (t,x) too early. For another example, velocity h is being integrated wrt distance not time so how does displacement result? I know the 1/c factor straightens out the dimensions but I am still confused. Does writing the differential as dy/c (in eq. 2) yield any insight? Maybe the x/t diagrams are confusing me. A diagram or drawing might help.

 

[Orodruin]

But, for just one example of my confusion, it doesn't seem that the points on the x-axis between x-ct and x+ct (excluding x-ct and x+ct) should contribute since they would reach the point (t,x) too early.

This is only true in three dimensions. In one dimension the effects linger after the wave front has passed, which can be seen as the Green's function of the one-dimensional wave equation is a step function whereas the one in three dimensions is a delta function at radius $r = ct$.

You can also study the one-dimensional problem using Hadamard's method of descent where a source of the one-dimensional wave equation would correspond to a plane source in three dimensions. Given a point in three dimensions, for all times when $ct$ is greater than the distance to the plane source there will be a contribution from somewhere in the plane arriving. It is rather straight forward to convince yourself that this results in the step function Green's function. (The signal arrives from a growing circle, but the distance to that circle increases and exactly cancels that growth.)

Edit: Also the two-dimensional wave equation has a Green's function that remains non-zero after the wave front passes. However, in the two-dimensional case, the wave front is singular and the lingering effect goes to zero as $t \to \infty$.

 

[bosque]

This is only true in three dimensions. In one dimension the effects linger after the wave front has passed, which can be seen as the Green's function of the one-dimensional wave equation is a step function whereas the one in three dimensions is a delta function at radius $r = ct$.

Looking at the integral term and its limits $x-ct,x+ct$, it seems to me that the effect of an initial velocity seen at x would also occur BEFORE $x=ct$.

 

[Orodruin]

Looking at the integral term and its limits $x-ct,x+ct$, it seems to me that the effect of an initial velocity seen at x would also occur BEFORE $x=ct$.

What do you mean with this? Of course there will be an effect of the initial velocity regardless of the position depending on how the initial velocity depends on position. The point is that the integral covers the region $x-ct$ to $x + ct$ so only the initial velocity within a distance $ct$ from the point you are considering are relevant at time $t$.

 

[bosque]

This is only true in three dimensions. In one dimension the effects linger after the wave front has passed, which can be seen as the Green's function of the one-dimensional wave equation is a step function whereas the one in three dimensions is a delta function at radius $r = ct$.

Looking at the integral term and its limits $x-ct,x+ct$, it seems to me that the effect of an initial velocity seen at x would also occur BEFORE $x=ct$.

What do you mean with this? Of course there will be an effect of the initial velocity regardless of the position depending on how the initial velocity depends on position. The point is that the integral covers the region $x-ct$ to $x + ct$ so only the initial velocity within a distance $ct$ from the point you are considering are relevant at time $t$.

For example, for the displacement IC only the values of the initial displacement at $x-ct$ and $x+ct$ effect the results (the first two terms in eq. 1 or 2). But for the velocity IC all the values of the initial velocity between and including $x-ct$ and $x+ct$ effect the results. There is a basic difference there that I don't understand. I do understand how it derives from the equations but don't understand it intuitively (or even physically).

 

[Orodruin]

Looking at the integral term and its limits $x-ct,x+ct$, it seems to me that the effect of an initial velocity seen at x would also occur BEFORE $x=ct$.

Yes, you said that already, but that statement does not make sense. Do you mean to say that it depends on the initial condition at all $x'$ such that $x-x' < ct$? In that case, yes, it does. I already covered this in post #2.

For example, for the displacement IC only the values of the initial displacement at $x-ct$ and $x+ct$ effect the results (the first two terms in eq. 1 or 2). But for the velocity IC all the values of the initial velocity between and including $x-ct$ and $x+ct$ effect the results. There is a basic difference there that I don't understand. I do understand how it derives from the equations but don't understand it intuitively (or even physically).

Let us consider the wave equation for the displacement of an infinite string for definiteness in terminology. A non-zero initial condition in the initial displacement carries no transverse momentum. A non-zero initial condition on the velocity does carry momentum. The result is that the initial condition on the displacement does not give any effect on the integral
$$
U(t) = \int_{-\infty}^\infty [u(x,t) - u(x,0)] dx
$$
as the total transverse momentum is conserved (as you can easily show either directly from the wave equation or using Noether's theorem) and equal to zero. Note that $U(0) = 0$ and that
$$
U'(t) = \int_{-\infty}^\infty u_t(x,t) dx,
$$
which essentially just the total transverse momentum (up to multiplication by the linear density). However, for an initial condition on the velocity, the total momentum is non-zero and that initial condition will generally lead to an overall displacement of the string. This is why the effect has to linger after the wave front has passed.

I also suggest you consider the case of a single delta impulse for the velocity, i.e., the Green's function, with the above in mind.

 

[vanhees71]

I'm a bit puzzled, what is argued about in this thread. The d'Alembert solution is simply the complete solution of the (1+1)-dimensional wave equation,
$$(\partial_0^2-\partial_1^2) \phi(x^0,x^1)=0.$$
I use the usual relativistic notation with $x^0=c t$ and $\partial_1=\partial/\partial x^1$ etc.

Now to see how d'Alembert's solution comes about we use a bit of operator algebra. The partial derivatives commute (if defined on a sufficiently well-behaved class of functions, which I assume in the typical physicist's way without specifying what "well-behaved" exactly means), and thus we can write
$$\partial_0^2-\partial_1^2=(\partial_0 - \partial_1)(\partial_0+\partial_1).$$
Thus we get for sure a solution of the wave equation, if the function is made vanishing by one of the factors in the operator product), i.e.,
$$(\partial_0 + \partial_1) \phi_+=0, \quad (\partial_0-\partial_1) \phi_-=0.$$
These equations are very simply solved. Let's look at $\phi_-$. It obviously obeys
$$\partial_0 \phi_-=\partial_1 \phi_- \; \Rightarrow \; \phi_-=f_-(x^0+x^1),$$
where $f_-$ is an arbitrary twice-differentiable function. In the same way you find
$$\phi_+=f_+(x^0-x^1).$$
The general solution then is the superposition of these solutions
$$\phi=f_+(x^0-x^1)+f_-(x^0+x^1).$$
The functions $f_{\pm}$ have to be determined by initial (and also often boundary) values.

 

[Orodruin]

I'm a bit puzzled, what is argued about in this thread.

The OP's problem is (or hopefuly was) that if you assume the initial conditions $\phi(x,0) = g(x)$ and $\phi_t(x,0) = h(x)$, then adapting to these initial conditions gives d'Alembert's formula (as quoted in Eq (2) of the OP). Whereas the solution at $\phi(x,t)$ only depends on the initial condition $g(x')$ at the points a distance $ct$ away from $x$, i.e., $x' = x\pm ct$, it depends on the initial condition $h(x')$ in the entire interval between those points. The OP was asking for a physical explanation why this was so and as far as I understood did not have any issues with the actual derivation of the solution.

 

[vanhees71]

Ah sorry, I overread this. That's indeed an interesting question. Of course Eq. (2) in #1 follows from my general solution. In my notation you have the initial conditions
$$\phi(0,x)=g(x), \quad \partial_0 \phi(0,x)=h(x).$$
This translates into
$$g(x)=f_+(-x)+f_-(x), \quad f'_{-}(x)+f'_{+}(-x)=h(x).$$
The 2nd equation translates into
$$-f_+(-x)+f_-(x)=H(x)+C,$$
where
$$H(x)=\int_0^x \mathrm{d} x' h(x').$$
From this you get
$$f_+(-x)=\frac{1}{2} [g(x)-H(x)+C], \quad f_-(x)=\frac{1}{2} [g(x)+H(x)-C].$$
So finally we have
$$\phi(x^0,x^1)=\frac{1}{2} [g(x^1-x^0)+g(x^1+x^0)]+\frac{1}{2} [H(x^1+x^0)-H(x^1-x^0)],$$
which is what's given in Eq. (2) of #1.

This doesn't explain the physics behind this. The point is a very interesting one indeed. Our intuition for the wave propagation in the source-free room is somehow biased by "Hugens's principle", which however is only true for the wave equation in (1+d) space-time dimensions, where $d \geq 3$ is odd. For even-dimensional spatial manifolds you have wake fields for $d \geq 2$, and $d=1$ is a special case.

An alternative derivation, which also works for any $d$, is to use Green's theorem in (1+d) dimensions using a Green's function of the corresponding D'Alembert operator. The most intuitive one is the retarded propagator. For $d \geq 3$ odd it's proportional to (derivatives) of the $\delta(x^0-|\vec{x}|)$ distribution, leading to a generalized version Huygens's principle of optics or acustics (for $d=3$ it's $\propto \delta(x^0-|\vec{x}|)$, for which you get the literal version of Huygens's principle). If you want, I can try to derive it (at least for $d=1,2,3$ :-)). You find it discussed in great detail in the great book

S. Hassani, Mathematical Physics, 2nd edition, Springer Verlag (2013) p. 685ff

 

[Orodruin]

If you want, I can try to derive it (at least for d=1,2,3d=1,2,3d=1,2,3 :-)).

I believe you already covered the $d = 1$ case ... Just let $h(x) = \delta(x)$ (generally the retarded GF can be written $\theta(t) f(x,t)$, inserted into the wave equation gives $f_t(x,0) = \delta(x)$ and $f(x,0) = 0$).

The $d = 3$ case is just a matter of a rather straight forward Fourier transform (and the GF of the harmonic oscillator) and the easiest way I know for finding the $d = 2$ GF is using Hadamard's method on the $d = 3$ one. Perhaps a good exercise for OP.

 

[vanhees71]

What's Hadamard's method? I think you can just use the 3D Gf. and integrate out one of the spatial components to get the 2d Gf, but I have to think about that again. It's too long ago, I've done this exercise, but it's a very nice exercise indeed.

 

[Orodruin]

What's Hadamard's method?

This is:

I think you can just use the 3D Gf. and integrate out one of the spatial components to get the 2d Gf

Essentially you use convolute the 3D GF with a line inhomogeneity $\delta(t) \delta(x) \delta(y)$.

 

[bosque]

However, for an initial condition on the velocity, the total momentum is non-zero and that initial condition will generally lead to an overall displacement of the string. This is why the effect has to linger after the wave front has passed.

Great Explanation--Thanks! Is this explanation to be found anywhere else?

What about the idealized infinite string which has no mass (ie. no momentum)--it seems like many of the online pdf's derivations of D'Alembert's formula use that sort of string.

Also, in the online pdf's which derive D'Alembert's formula, mass is not involved--but it seems like it should be involved if the string's momentum is key to having the velocity's effects persist.

 

[Orodruin]

What about the idealized infinite string which has no mass (ie. no momentum)--it seems like many of the online pdf's derivations of D'Alembert's formula use that sort of string.

Such a string would have an infinite wave speed. You might be thinking of a case where units are chosen such that $\rho$ and the string tension are equal to one. Can you provide a link to such a pdf?

 

[bosque]

Such a string would have an infinite wave speed. You might be thinking of a case where units are chosen such that $\rho$ and the string tension are equal to one. Can you provide a link to such a pdf?

Sorry, I reviewed my pdfs and the ones that did mention the physical properties of the string give it a mass density. The others start with the 1D wave equation and do not mention physical properties (mostly math pdfs).