1. Apr 7, 2013

Unicorn.

Hi,
1. The problem statement, all variables and given/known data
The gradient ∇3 can be generalized for spacetime as:
4 =(∇3 ,d/dct)=(d/dx,d/dy,d/dz,d/dct)
Show that ∇4 is a four-vector.

2. Relevant equations

3. The attempt at a solution
I just have to write that :
d/dx'=γ(d/dx-βd/dct)
d/dy'=d/dy
d/dz'=d/dz
d/dct'=γ(d/dct-βd/dx)
And
4 =(d/dx,d/dy,d/dz,d/dct)=(∇3 ,d/dct) ..?

Thanks

2. Apr 7, 2013

haruspex

I don't understand the question. Of course it's a 4-vector. (a, b, c, d) is a 4-vector. Is there some wording missing?

3. Apr 7, 2013

Unicorn.

Hi,
No, this is the complete wording.

4. Apr 7, 2013

WannabeNewton

What? The 4-gradient $\nabla_{a}$ is a derivative operator. It isn't a 4-vector of any kind. I don't think you are phrasing it correctly.

5. Apr 7, 2013

haruspex

A vector operator is still a vector. You can add them, and multiply them by scalar values.

6. Apr 8, 2013

WannabeNewton

An operator acting on a vector isn't an element of the vector space that the vector belongs to. $\nabla_{a}$ is certainly not a 4-vector just because it is linear (it isn't a linear functional - it's just linear). How are you claiming that $\nabla_{a}\in \mathbb{R}^{4}$ (I used $\mathbb{R}^{4}$ because the OP is working in minkowski space-time). That doesn't make any sense.

In fact what is true is that $\frac{\partial }{\partial x}|_{p},\frac{\partial }{\partial y}|_{p},\frac{\partial }{\partial z}|_{p},\frac{\partial }{\partial t}|_{p}$ are 4-vectors in $\mathbb{R}^{4}$.

Last edited: Apr 8, 2013
7. Apr 8, 2013

micromass

The OP is asking about 4-vectors. So saying that ∇ is a 4-vector is the same as saying that ∇ is an element of $\mathbb{R}^4$. This is obviously nonsense.

Of course, we can add ∇ and multiply it by scalars. But that means that it is a vector in a vector space. But since the OP asks about 4-vectors, he's specifically asking about the vector space $\mathbb{R}^4$. At least, that's how I interpret the question.

8. Apr 8, 2013

vela

Staff Emeritus
The OP is supposed to show that $\nabla$ transforms like a four-vector. In other words, he or she needs to show that it makes sense to say that $\partial'_\mu = \Lambda_\mu{}^\nu \partial_\nu$.

Last edited: Apr 8, 2013
9. Apr 8, 2013

Staff: Mentor

Actually, strictly speaking, shouldn't it transform like a 1-form?

10. Apr 8, 2013

WannabeNewton

The coordinate vector fields $\partial_{\mu}$ do indeed transform like one-forms (hence why they are usually written with a lower index) but they are still 4-vectors as they form a basis for the tangent space when evaluated at each point. Also it doesn't make any sense to say $\nabla$ transforms like a 4-vector because it is not a vector in minkowski space-time, it is map on the ring of smooth functions on minkowski space-time. This bad notation is the kind you would see in Srednicki's QFT text.

11. Apr 8, 2013

haruspex

I gather from various posts that, within relativity theory at least, "4-vector" has a more specific meaning than I was aware of. It's not clear to me whether this is a matter of definition or just common usage.

12. Apr 8, 2013

micromass

Ah, maybe you're right. I don't know any relativity. I just looked up 4-vector in wikipedia.

To avoid confusion, perhaps we should let the OP tell us how he defined "4-vector" in his course.