Proving d'Alembertian Invariant under Lorentz Transformations

[timman_24]

Homework Statement

Show that (D'Alembertian)^2 is invariant under Lorentz Transformation.

Homework Equations

The book (E/M Griffiths) describes the D'Alembertian as:

$$\square^2=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$

The Attempt at a Solution

I don't really know what it is asking me to do here. Any guidance at where to even start would be very helpful. I am not very good at writing out proofs.

 

[ManyNames]

Homework Statement

Show that (D'Alembertian)^2 is invariant under Lorentz Transformation.

Homework Equations

The book (E/M Griffiths) describes the D'Alembertian as:

$$\square^2=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$

The Attempt at a Solution

I don't really know what it is asking me to do here. Any guidance at where to even start would be very helpful. I am not very good at writing out proofs.

I could be wrong, but it seems that it's asking you to show how the operator manifests itself in the Minkowskian Metric (0,0,0,1); from there, i would assume that any ''rotations'' made in the Cartesian Product always remain invariant... but it's vague, so, i am not entirely sure.

Since the Operator maintains the four dimensions (t, x, y, z), basic rotations in such a ''spacetime'' are considered in special relativity as being invariant. For instance: Take the following Cartesian Distance System

$$s^2=(\Delta x*)+(\Delta y*)+(\Delta z*)+(\Delta t*)$$

Here, the asterisks represent the ''rotations'' i spoke of to you, where the rotations are orientated from the original coordination:

$$s^2=(\Delta x)+(\Delta y)+(\Delta z)+(\Delta t)$$

So in this sense, it is said to be that distance is invariant under any such coordinational changes. Moving on, the covarient expression of the operator is given as $$2=\nabla^{\mu}\nabla_{\mu}$$ which is crucial when taking into account the four functions of $$x^{\mu}$$, but they are merely functions themselves and not exactly vectors. But this is why i first started off with the Cartesian Map, which is a flat spacetime which is identified as a ''harmonic function''.

So there is my bit, and i hope it helps in any way?

 

[ManyNames]

Homework Statement

Show that (D'Alembertian)^2 is invariant under Lorentz Transformation.

Homework Equations

The book (E/M Griffiths) describes the D'Alembertian as:

$$\square^2=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$

The Attempt at a Solution

I don't really know what it is asking me to do here. Any guidance at where to even start would be very helpful. I am not very good at writing out proofs.

This might also give some insight, which maybe more correct towards the answer you are looking for:

http://en.wikipedia.org/wiki/Alternatives_to_general_relativity

''Nordström (1912)

The first approach of Nordström (1912) was to retain the Minkowski metric and a constant value of but to let mass depend on the gravitational field strength . Allowing this field strength to satisfy...

...This theory is Lorentz invariant, satisfies the conservation laws, correctly reduces to the Newtonian limit and satisfies the weak equivalence principle.''

 

[timman_24]

Thanks for the reply, I am looking into the link and will comment once I attempt the problem again. I agree with you though, the question is rather vague.

 

[timman_24]

Okay I think I have this solved. I just went about it the straight forward method by simply crunching the transformations as follows:
$${A}=<br /> \begin{bmatrix}<br /> {\frac{\partial^2}{\partial t^2}\frac{1}{c}t}\\ <br /> {\frac{\partial^2}{\partial x^2}x}\\ <br /> {\frac{\partial^2}{\partial y^2}y}\\ <br /> {\frac{\partial^2}{\partial z^2}z}<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> {\gamma}&{-\beta\gamma}&{0}&{0}\\<br /> {-\gamma\beta}&{\gamma}&{0}&{0}\\ <br /> {0}&{0}&{1}&{0}\\ <br /> {0}&{0}&{0}&{1}<br /> \end{bmatrix}$$

Then I dot:
$${A}\cdot{A'}$$
Using the primed version of matrix A

Which I get:
$$\frac{\partial^4}{\partial x^4}xx' + \frac{\partial^4}{\partial y^4}yy' + \frac{\partial^4}{\partial z^4}zz'$$

After using:
$$\beta= \frac{v}{c},\:\;\: \gamma^2= \frac{c^2}{c^2-v^2}$$
and canceling out differentials that go to zero

This answer equals the answer given by:
$$\begin{bmatrix}<br /> {\frac{\partial^2}{\partial t^2}\frac{1}{c}t}\\ <br /> {\frac{\partial^2}{\partial x^2}x}\\ <br /> {\frac{\partial^2}{\partial y^2}y}\\ <br /> {\frac{\partial^2}{\partial z^2}z}<br /> \end{bmatrix}<br /> \cdot<br /> \begin{bmatrix}<br /> {\frac{\partial^2}{\partial t^2}\frac{1}{c}t'}\\ <br /> {\frac{\partial^2}{\partial x^2}x'}\\ <br /> {\frac{\partial^2}{\partial y^2}y'}\\ <br /> {\frac{\partial^2}{\partial z^2}z'}<br /> \end{bmatrix}$$

Therefore the D'Alembertian is invariant under Lorentz Transformation.

Does this seem like a reasonable way to do this problem?

Thanks

BTW should I prime the partial derivatives in the A' matrices? Then my answer would look like this:

$$\frac{\partial^2}{\partial x^2}x\frac{\partial^2}{\partial x'^2}x' + \frac{\partial^2}{\partial y^2}y\frac{\partial^2}{\partial y'^2}y' + \frac{\partial^2}{\partial z^2}z\frac{\partial^2}{\partial z'^2}z'$$

This wouldn't affect the problem I don't think.