Contravariant gradient?


by poophead
Tags: contravariant, gradient
poophead
poophead is offline
#1
Sep11-04, 02:18 PM
P: 10
I'm going to be completely unambiguous on this: the problem I am about to ask is an assigned homework problem so please, do NOT simply just reply with the answer. I have no intention to cheat.

That said, the question I have is with regards to problem 12.55 in Griffith's Intro to Electrodynamics. I'm pretty sure not everyone has the book so I'll sum up the basic points of the question:

Griffiths states that the four-dimension gradient operator d/dx^u (pretend the d's are partials, and pretend the u is superscripted-- sorry, I don't know how some of you manage to get the symbols all nice) functions like a covariant 4-vector, so often times it's written d_u for short. He then states that the corresponding contravarient gradient vector would be d^u = d/dx_u. Now, we're supposed to prove that d^u(phi), that is, the contravarient gradient acting on some scalar function phi is a contravarient 4-vector.

I have no clue as to how to do this. My professor has provided us with the hint of considering the direct Lorentz transformation and how the transformation relates to dx'^u/dx^v, as well as the inverse transformation. We're supposed to use the chain rule somehow to get back to the proof but after laboring on this problem for several hours I'm convinced that I am stuck. Something tells me that this problem shouldn't take more than 10 minutes but all this tensor stuff is all fairly new to me (ie, I didn't learn of any of this stuff till earlier this week!).

Can anyone here provide me with any hints or perhaps a better insight of the problem? From a superficial standpoint, I see that contravarient and covariant vectors differ only in terms of the signs on certain components such as to make their dot product invariant. Is there more to this?

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jcsd
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#2
Sep11-04, 03:39 PM
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PF Gold
P: 2,226
The componets of a pair of dual vectors (i.e. a Lorentz vector and it's one-form) in special relativity are related by: [itex]A^{0} = -A_0[/itex], [itex]A^i = A_i[/itex] (i = 1,2,3) or simply [itex]A^{\alpha} = \eta^{\alpha\beta}A_{\beta}[/itex], [itex]\alpha[/itex] = 0,1,2,3,.

So yes a coontaravrint vecotr and it's covaraint vector only differt by the sign of their time compoments and dot product of a Lorentz vector and a one-form is an invariant scalar.

To prove that it is a contravariant vector you must first set up an equation to show how it's components transforms under a Lorentz transformation, then you must show that the componets in the orginal frame [itex]A^{\beta}[/itex] are related to the compoents in the transformed frame [itex]\bar{A}^{\alpha}[/itex] by:

[tex]\bar{A}^{\alpha} = \frac{\partial\bar{x}^{\alpha}}{\partial x^{\beta}}A^{\beta}[/tex]

summation implied.
poophead
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#3
Sep11-04, 04:38 PM
P: 10
I'm sorry, I'm being dense. I still do not understand exactly how showing the relationship you stated above would show that [tex] \frac{\partial}{\partial x_{\mu}}} \phi [/tex] would give me a contravariant 4-vector.

From:
[tex] x^{\prime\mu} = \Lambda^\mu_{ \ \nu} x^\nu[/tex]

I can sort of see that

[tex] \Lambda^\mu_ {\ \nu} = \frac{\partial x^{\prime\mu}} {\partial x^\nu}[/tex]

Similarily, by multiplying by the inverse, I can see the relationship (albiet hand-wavingly) that:

[tex] (\Lambda^{-1})^{\nu}_{\ \mu} = \frac{\partial x^{\nu}} {\partial x^{\prime\mu}}[/tex]

The hint my professor gives us is to express:
[tex] \frac{\partial}{\partial x^{\prime\mu}} [/tex] in terms of [tex] \frac{\partial x^{\nu}}{\partial x^{\prime\mu}} [/tex] and [tex] \frac{\partial}{\partial x^{\nu}} [/tex]

I have the feeling that he's waved the answer right in front of my face but I just don't understand the problem well enough to reach the final step.

jcsd
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#4
Sep11-04, 04:55 PM
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PF Gold
P: 2,226

Contravariant gradient?


Quote Quote by poophead
I'm sorry, I'm being dense. I still do not understand exactly how showing the relationship you stated above would show that [tex] \frac{\partial}{\partial x_{\mu}}} \phi [/tex] would give me a covariant 4-vector.

From:
[tex] x^{\prime\mu} = \Lambda^\mu_{ \ \nu} x^\nu[/tex]

I can sort of see that

[tex] \Lambda^\mu_ {\ \nu} = \frac{\partial x^{\prime\mu}} {\partial x^\nu}[/tex]

Similarily, by multiplying by the inverse, I can see the relationship (albiet hand-wavingly) that:

[tex] (\Lambda^{-1})^{\nu}_{\ \mu} = \frac{\partial x^{\nu}} {\partial x^{\prime\mu}}[/tex]

The hint my professor gives us is to express:
[tex] \frac{\partial}{\partial x^{\prime\mu}} [/tex] in terms of [tex] \frac{\partial x^{\nu}}{\partial x^{\prime\mu}} [/tex] and [tex] \frac{\partial}{\partial x^{\nu}} [/tex]

I have the feeling that he's waved the answer right in front of my face but I just don't understand the problem well enough to reach the final step.
Whta you need is the chain rule:

[tex]\frac{\partial\bar{\phi}}{\partial\bar{x}^{\mu}} = \frac{\partial\phi}{\partial\bar{x}^{\mu}} = \frac{\partial\phi}{\partial x^{\nu}}\frac{\partial x^{\nu}}{\partial\bar{x}^{\mu}} = \frac{\partial x^{\nu}}{\partial\bar{x}^{\mu}}\frac{\partial\phi}{\partial x^{\nu}}[/tex]

edit oops sorry that's for the actual gradient i.e. the covaraint vector, but you get the picture.
poophead
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#5
Sep11-04, 05:12 PM
P: 10
Thank you jcsd for your speedy replies and I apologize for taking up your time with my inexperience.

I did express it all with the chain rule but ultimately, I'm still confused as to how this end result gives the proof I desire.

After all, I'm concerned about [tex] \frac{\partial}{\partial x_{\mu}}}[/tex] but all our expressions thus far have been with expressions such as [tex] \frac{\partial}{\partial x^{\mu}} [/tex] . How, in the end, does this show that the contravariant gradient acting on a scalar leads to a contravarient 4-vector?

Is the following line of reasoning completely fallacious?

Given the chain rule, we can say: [tex]\frac{\partial\bar{\phi}}{\partial\bar{x}^{\mu}} = \frac{\partial\phi}{\partial\bar{x}^{\mu}} = \frac{\partial\phi}{\partial x^{\nu}}\frac{\partial x^{\nu}}{\partial\bar{x}^{\mu}} = \frac{\partial x^{\nu}}{\partial\bar{x}^{\mu}}\frac{\partial\phi} {\partial x^{\nu}}[/tex]

Given what I wrote about I can say that [tex] (\Lambda^{-1})^{\nu}_{\ \mu} = \frac{\partial x^{\nu}} {\partial x^{\prime\mu}}[/tex] .

If I plug it all in, I get something like [tex]\frac{\partial\bar{\phi}}{\partial\bar{x}^{\mu}} = (\Lambda^{-1})^{\nu}_{\ \mu} \frac{\partial\phi} {\partial x^{\nu}} [/tex]

Am I right so far? And how does this, in the end, all tie back to [tex] \frac{\partial}{\partial x_{\mu}}}[/tex] ?

I realize those are a lot of questions, questions that might seem blindingly obvious yet, it's just not clear to me. Oh, how the raising and lowering of indices in these tensors can cause me so much pain.
jcsd
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#6
Sep11-04, 05:40 PM
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You know that [itex]\phi[/itex] is an invaraint so [itex]\phi = \bar{\phi}[/itex]

[tex]\frac{\partial\bar{\phi}}{\partial\bar{x}_{\mu}} = \frac{\partial\phi}{\partial\bar{x}_{\mu}}[/tex]

Examine the defitnion of a contravarint vector then, simply use the chain rule to derive the corrcet relationship as was done above for the covariant gardient.
poophead
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#7
Sep11-04, 06:11 PM
P: 10
Thanks a lot. I shall ponder on it some more.
mathwonk
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#8
Sep16-04, 04:33 PM
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all you need is definitions of the relevant words. the problem you are asked to prove is essentially the definition of a contravariant vector.


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