Transforming co- and contravariant 4-vectors

[EricTheWizard]

I'm slightly confused by the difference between covariant and contravariant 4-vectors and how they transform under Lorentz boosts. I'm aware that $x_{\mu} = (-x^0 ,x^1, x^2, x^3) = (x_0 ,x_1, x_2, x_3)$, but when I do a Lorentz transform of the covariant vector, it seems to transform exactly like a contravariant vector would:

$$x_{\mu} \Lambda^{\mu}_{\nu} = \pmatrix{\gamma & -\gamma\beta & 0 & 0\\-\gamma\beta & \gamma & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1} \pmatrix{x_0\\x_1\\x_2\\x_3} = \pmatrix{\gamma(x_0 -\beta x_1)\\\gamma(x_1 -\beta x_0)\\x_2\\x_3}$$

But I've heard people say that they transform differently; so am I doing this wrong?
I was also hoping someone would explain how differential operators behave/transform as well (does $\frac{\partial}{\partial x^\mu} = \partial_\mu$ transform like a covariant vector? What would $\frac{\partial}{\partial x_\mu}$ mean?)

 

[PeterDonis]

Some observations that might help for the first of your questions:

(1) When you write a contravariant vector, you need to write the vector itself with an "upstairs" abstract index, not a "downstairs" one; also, you shouldn't include the signs if you are writing the components out with indexes--the signs come in in a different way, as we'll see in a moment:

$$x^{\mu} = (x^{0}, x^{1}, x^{2}, x^{3})$$

$$x_{\mu} = (x_{0}, x_{1}, x_{2}, x_{3})$$

(2) You use the metric to raise and lower indices, i.e., to convert between contravariant and covariant representations of the same vector, which is implicitly what you were trying to do when you wrote your formula for $x_{\mu}$. That means that:

$$x_{\mu} = \eta_{\mu \nu} x^{\nu}$$

So if we have specific numerical values for $x^{\nu}$, say $x^{\nu} = (t, x, y, z)$ , then we can use the above formula to find that $x_{\mu} = (-t, x, y, z)$ .

(3) The Lorentz transformation matrix has one upper and one lower index, so you can use the same matrix to transform either kind of vector, by just switching which index you sum over (lower vs. upper):

$$x^{\nu} = \Lambda^{\nu}_{\rho} x^{\rho}$$

$$x_{\mu} = \Lambda^{\sigma}_{\mu} x_{\sigma}$$

If you write this out in matrix form, you will see that the first equation represents multiplying the Lorentz transformation matrix by a column vector, whereas the second represents multiplying it by a row vector--i.e., covariant vectors are row vectors, not column vectors. However, since the Lorentz transformation matric is symmetric, the two types of multiplication work out the same way as far as how they manipulate the components of the vector.

 

[PeterDonis]

(does $\frac{\partial}{\partial x^\mu} = \partial_\mu$ transform like a covariant vector?

Yes.

What would $\frac{\partial}{\partial x_\mu}$ mean?)

It's equivalent to raising the index of $\partial_{\mu}$ using the metric, as described in my last post. In other words, the partial derivative with respect to a contravariant vector field is a covariant vector field, and vice versa. (Note that now we're talking about "vector fields", not just "vectors", since we're talking about derivatives; the derivative with respect to a single vector makes no sense, since the whole point of the derivative is to describe how something else changes as the vector changes.)

 

[EricTheWizard]

Ahh thank you for your post. So both vectors transform the same way, then.
And just to make sure I have this right, you're saying that $\frac{\partial}{\partial x_\mu} = \partial^\mu = \eta^{\mu\nu}\partial_\nu$?

So if I were to construct a relativistically-correct 4-momentum operator, would it be of the form $\hat{p}_\mu = -i\hbar \partial_\mu = (-\frac{1}{c} i\hbar\frac{\partial}{\partial t}, -i\hbar \nabla) = (-\frac{1}{c} \hat{E},\hat{p})$? (I ask because this kind of ties in with a homework assignment)

 

[PeterDonis]

you're saying that $\frac{\partial}{\partial x_\mu} = \partial^\mu = \eta^{\mu\nu}\partial_\nu$?

Yes.

(I ask because this kind of ties in with a homework assignment)

We're not supposed to directly give answers to homework assignments, but it looks like you're basically on the right track.