Lorentz covaraince of differntial operator

[JDoolin]

Fredrik:

No, the first equality implies the second: $$\frac{\partial}{\partial x'^\nu} = \frac{\partial x^\mu}{\partial x'^\nu} \frac{\partial}{\partial x^\mu}=(\Lambda^{-1})^\mu{}_\nu\frac{\partial}{\partial x^\mu}=\Lambda_\nu{}^\mu\frac{\partial}{\partial x^\mu}$$​

JDoolin:

Ahh, I think that is the Einstein Notation for [STRIKE]divergence[/STRIKE] that I was looking for.

https://www.physicsforums.com/showthread.php?t=511811 Post #8​

Fredrik

I'd go with what Mentz told you in #2.​

My clumsy vocabulary. I meant Einstein Notation for gradient.

$$\nabla u = \begin{pmatrix}<br /> \frac{1}{h_1} \frac{\partial }{\partial x_1}\\ <br /> \frac{1}{h_2} \frac{\partial }{\partial x_2}\\<br /> \frac{1}{h_3} \frac{\partial }{\partial x_3}<br /> \end{pmatrix}u$$

I'll try to work it out in more detail in the other thread.

Edit: Actually, I can put the gist of the question here.

$$\frac{\partial x^\mu}{\partial x'^\nu}\cdot \frac{\partial }{\partial x^\mu} \overset ? = \begin{pmatrix} \frac{\partial ||\vec r||}{\partial ||\vec r`||} \frac{\partial }{\partial r}\\ \frac{\partial ||\vec r||}{\partial ||\vec \theta`||} \frac{\partial }{\partial \theta}\\ \frac{\partial ||\vec r||}{\partial ||\vec \phi`||} \frac{\partial }{\partial \phi} \end{pmatrix} =\begin{pmatrix} \frac{\partial }{\partial r}\\ \frac{1}{r} \frac{\partial }{\partial \theta}\\ \frac{1}{r sin\theta} \frac{\partial }{\partial \phi} \end{pmatrix}$$

 

[Fredrik]

OK, that question doesn't really have anything to do with notation. If you just want a notation with indices for (the components of) $\nabla u$, the answer is of course $\partial_i u$, since $\nabla u=e_i(\nabla u)_i=e_i\partial_i u$. You seem to want the $\nabla$ operator in spherical coordinates. That takes a bit more work. I'm sure this is covered in a lot of books, and probably in some forum posts as well, so I won't do that exercise here. Maybe someone else can link to a proof of this, e.g. at Google Books.

 

[JDoolin]

OK, that question doesn't really have anything to do with notation. If you just want a notation with indices for (the components of) $\nabla u$, the answer is of course $\partial_i u$, since $\nabla u=e_i(\nabla u)_i=e_i\partial_i u$. You seem to want the $\nabla$ operator in spherical coordinates. That takes a bit more work. I'm sure this is covered in a lot of books, and probably in some forum posts as well, so I won't do that exercise here. Maybe someone else can link to a proof of this, e.g. at Google Books.

Okay, you're right. That's a bad example. See if this is a better question.
I find in http://www.mathpages.com/rr/s6-06/6-06.htm the equation:

$$g=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 &0 &0 \\ 0 & 0 & -1 &0 \\ 0 & 0 & 0 & -1 \end{pmatrix} - \frac{2 m}{r}\begin{pmatrix} 1 &0 &0 &0 \\ 0& \kappa x x & \kappa x y &\kappa x z \\ 0& \kappa y x &\kappa y y &\kappa z y \\ 0& \kappa z x &\kappa y z &\kappa z z \end{pmatrix}$$

where

$$\kappa = \frac{1}{r^2(1-2m/r)}$$


I know that
$$g_{11}= \left (\frac{\partial \tau}{\partial t} \right )^2=1-\frac{2m}{r}$$

...but can the other elements of the matrix be expressed in the form of two differentials multiplied together? Or more to the point, what is Einstein notation for g? Of course having finally asked such an unambiguous question, I must try looking it up (pause).

Okay, from http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity )

$$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma} = \Lambda^\rho {}_{\bar \mu} \, \Lambda^\sigma {}_{\bar \nu} \, g_{\rho \sigma}$$

...which is probably what I wanted to know. (Once I figure out how to decode the notation, of course)

 

[Fredrik]

Okay, from http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity )

$$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma} = \Lambda^\rho {}_{\bar \mu} \, \Lambda^\sigma {}_{\bar \nu} \, g_{\rho \sigma}$$

...which is probably what I wanted to know. (Once I figure out how to decode the notation, of course)

Posts #3 and #5 in this thread might help. Note by the way that what you've got there is a matrix equation in component form: $$g'=\Lambda^T g\Lambda$$

 

[JDoolin]

Posts #3 and #5 in this thread might help. Note by the way that what you've got there is a matrix equation in component form: $$g'=\Lambda^T g\Lambda$$

Thank you for giving me that extra bit of information Fredrik. I can see now that if

$$\Lambda=\begin{pmatrix}<br /> \frac{\partial t}{\partial \tau} &<br /> \frac{\partial t}{\partial \bar x} & <br /> \frac{\partial t}{\partial \bar y} & <br /> \frac{\partial t}{\partial \bar z} \\<br /> <br /> \frac{\partial x}{\partial \tau} &<br /> \frac{\partial x}{\partial \bar x} &<br /> \frac{\partial x}{\partial \bar y} &<br /> \frac{\partial x}{\partial \bar z} \\<br /> \frac{\partial y}{\partial \tau} &<br /> \frac{\partial y}{\partial \bar x} &<br /> \frac{\partial y}{\partial \bar y} &<br /> \frac{\partial y}{\partial \bar z} \\<br /> \frac{\partial z}{\partial \tau} & <br /> \frac{\partial z}{\partial \bar x} & <br /> \frac{\partial z}{\partial \bar y} & <br /> \frac{\partial z}{\partial \bar z} \\ <br /> \end{pmatrix}$$

then the equations:

$$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma}$$

and
$$g'=\Lambda^T g\Lambda$$

are equivalent.

This is kind of a EUREKA moment for me, because I've never seen the Lorentz Transformation written this way. I have one little thing I want to mention, Wouldn't this technically be the "inverse" lorentz transformation? because:

$$\begin{pmatrix}<br /> dt\\ <br /> dx\\ <br /> dy\\ <br /> dz<br /> \end{pmatrix} = \begin{pmatrix}<br /> \frac{\partial t}{\partial \tau} &<br /> \frac{\partial t}{\partial \bar x} & <br /> \frac{\partial t}{\partial \bar y} & <br /> \frac{\partial t}{\partial \bar z} \\<br /> \frac{\partial x}{\partial \tau} &<br /> \frac{\partial x}{\partial \bar x} &<br /> \frac{\partial x}{\partial \bar y} &<br /> \frac{\partial x}{\partial \bar z} \\<br /> \frac{\partial y}{\partial \tau} &<br /> \frac{\partial y}{\partial \bar x} &<br /> \frac{\partial y}{\partial \bar y} &<br /> \frac{\partial y}{\partial \bar z} \\<br /> \frac{\partial z}{\partial \tau} & <br /> \frac{\partial z}{\partial \bar x} & <br /> \frac{\partial z}{\partial \bar y} & <br /> \frac{\partial z}{\partial \bar z} \\ <br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> d\tau\\ <br /> d \bar x\\ <br /> d \bar y\\ <br /> d \bar z<br /> \end{pmatrix}$$

...or could it be a problem with Wikipedia's equation? Maybe they need to flip their differentials...

 

[Fredrik]

Yes, the two indices on the components of the metric transform "covariantly" (i.e. with the same matrix as the basis vectors of the tangent space) while the index on the components of the coordinates of an event transforms "contravariantly" (with the inverse of the matrix that transforms the basis vectors of the tangent space).

The former transformation matrix is denoted by $\Lambda^{-1}$ and the latter by $\Lambda$. This means that $$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma}$$ isn't equivalent to $g'=\Lambda^Tg\Lambda$ as I said. It's equivalent to $g'=(\Lambda^{-1})^T g\Lambda^{-1}$.

I got a bit confused by the right-hand side of $$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma} = \Lambda^\rho {}_{\bar \mu} \, \Lambda^\sigma {}_{\bar \nu} \, g_{\rho \sigma}$$ in your post #33. I would have written this as $$g'_{\mu\nu} = \frac{\partial x^\rho}{\partial x'^{\mu}}\frac{\partial x^\sigma}{\partial x'^{\nu}} g_{\rho\sigma} = (\Lambda^{-1})^\rho{}_\mu (\Lambda^{-1})^\sigma{}_\nu\, g_{\rho\sigma}= \Lambda_\mu{}^\rho\, \Lambda_\nu{}^\sigma \, g_{\rho \sigma}.$$
I don't use the notation that puts primes, bars or whatever on the indices instead of on the variables, so I didn't know that people who use it write row $\mu$ column $\nu$ of $\Lambda^{-1}$ as $\Lambda^\mu{}_{\bar\nu}$.

 

[JDoolin]

I went ahead and posted a question on the wikipedia discussion page to see if it should be changed on Wikipedia.

http://en.wikipedia.org/wiki/Talk:Metric_tensor_(general_relativity))

 

[Fredrik]

I don't think it's necessary to change it. It's just a different notational convention than the one I'm using. They're letting us know that we're dealing with the inverse by moving the bar to lower index, so they don't also have to let us know by changing the horizontal position of the indices.

 

[JDoolin]

Okay, I think I get it.

The equation from the wikipedia article

$$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma} = \Lambda^\rho {}_{\bar \mu} \, \Lambda^\sigma {}_{\bar \nu} \, g_{\rho \sigma}$$


actually implicitly defines

$$\Lambda^\rho { }_{\bar \mu} = \frac{\partial x^\rho}{\partial x^\bar \mu}$$

and

$$\Lambda^\sigma { }_{\bar \nu} = \frac{\partial x^\sigma}{\partial x^\bar \nu}$$

While a Lorentz Transformation from event (t,x,y,z) to $(\tau,\bar x, \bar y, \bar z)$

would be given as

$$\Lambda^{\bar \sigma} { }_\nu = \frac{\partial x^{\bar \sigma}}{\partial x^\nu}$$

I didn't realize the versatility of the $\Lambda$ symbol.

 

[McLaren Rulez]

Hi,

I'm reviving this because I think I get the notation now but the original question still troubles me.

Let's focus on the problem at hand. You got this part right: $$\frac{\partial}{\partial x'^\nu} = \frac{\partial x^\mu}{\partial x'^\nu} \frac{\partial}{\partial x^\mu}$$ The next step is $$=(\Lambda^{-1})^\mu{}_\nu\frac{\partial}{\partial x^\mu}=\Lambda_\nu{}^\mu\frac{\partial}{\partial x^\mu}$$

We have $x'^{\mu}=\Lambda^{\mu}_{\ \ \nu}x^{\nu}\ \ \ (1)$

So $dx'^{\mu}=\Lambda^{\mu}_{\ \ \nu}dx^{\nu}\ \ \ (2)$

which also means that $dx^{\mu}=(\Lambda^{-1})^{\mu}_{\ \ \nu}dx'^{\nu}\ \ \ (3)$

When you did $$\frac{\partial x^\mu}{\partial x'^\nu} \frac{\partial}{\partial x^\mu}=(\Lambda^{-1})^\mu{}_\nu\frac{\partial}{\partial x^\mu}$$ I assumed initially that you used (3) to get $$\frac{\partial x^\mu}{\partial x'^\nu} \frac{\partial}{\partial x^\mu}= \frac{(\Lambda^{-1})^{\mu}_{\ \ \nu}\partial x'^{\nu}}{\partial x'^\nu} \frac{\partial}{\partial x^\mu}$$ and canceled the $\partial x'^{\nu}$ from the numerator and denominator.

But they cannot be canceled out like that since the numerator is a summation over the dummy index $\nu$ whereas the index in the denominator is a fixed one. So what exactly did you do to get that equality?

Thank you

 

[Fredrik]

(1) is the four components of the matrix equation $x'=\Lambda x$. Multiply with $\Lambda^{-1}$ from the left, and you get $\Lambda^{-1}x'=x$. The $\mu$th component of this equation is $$x^\mu=(\Lambda^{-1})^\mu{}_\nu x'^\nu=\Lambda_\nu{}^\mu x'^\nu.$$ The expression $$\frac{\partial x^\mu}{\partial x'^\nu}$$ denotes the $\nu$th partial derivative of the function $x^\mu:\mathbb R^4\rightarrow\mathbb R$ defined by $$x^\mu(x'^0, x'^1, x'^2, x'^3)=\Lambda_\nu{}^\mu x'^\nu=\Lambda_0{}^\mu x'^0+\Lambda_1{}^\mu x'^1+\Lambda_2{}^\mu x'^2+\Lambda_3{}^\mu x'^3.$$ All you have to do now is to compute the partial derivatives.

Edit: Regarding the "cancelling out" issue, note that the chain rule says that $$\frac{\partial f}{\partial x'^\nu}=\frac{\partial x^\mu}{\partial x'^\nu}\frac{\partial f}{\partial x^\mu}.$$

 

[McLaren Rulez]

Sorry but I still cannot see how it all cancels out.

Let us consider $$\frac{\partial}{\partial x'^{2}}$$

The chain rule says $$\frac{\partial}{\partial x'^{2}}=\frac{\partial x^{0}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}}+\frac{\partial x^{1}}{\partial x'^{2}}\frac{\partial}{\partial x^{1}}+\frac{\partial x^{2}}{\partial x'^{2}}\frac{\partial}{\partial x^{2}}+\frac{\partial x^{3}}{\partial x'^{2}}\frac{\partial}{\partial x^{3}}$$

Now, we express the first term on the RHS as follows:$$\frac{\partial x^{0}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}} = \frac{\Lambda_{0}^{\ \ 0}\partial x'^{0}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}}+\frac{\Lambda_{1}^{\ \ 0}\partial x'^{1}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}}+\frac{\Lambda_{2}^{\ \ 0}\partial x'^{2}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}}+\frac{\Lambda_{3}^{\ \ 0}\partial x'^{3}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}}$$

And we get similar expressions for the three other terms. But the expression we should get is only the third term on the RHS (where we can "cancel out" the $\partial x'^{2}$ term) i.e. $$\frac{\partial x^{0}}{\partial x'^{2}}\frac{\partial}{\partial x^{0}}=\Lambda_{2}^{\ \ 0}\frac{\partial}{\partial x^{0}}$$

This correctly satisfies $$\frac{\partial}{\partial x'^{\nu}}=\Lambda_{\nu}^{\ \ \mu}\frac{\partial}{\partial x^{\mu}}$$

Where do the remaining terms in the earlier expression go?

Thank you

EDIT: Oh dear! Of course, they're all zero! Mega brain fart there. Sorry about that, and thank you so much.

 

[McLaren Rulez]

I think something that needs to be mentioned is that there are two "kinds" of 4-vector.
Ones that transform like the coordinates: $x' = \Lambda x$, in physics we
call these contravariant vectors or just vectors. we generally write them in index notation
as a beast with an upper index like $x^i$. in this notation we have
$x'{}^{i} = \Lambda^i_{\;j} x^j$.

the second kind are like the derivatives and transform with the inverse matrix:
$p' = \Lambda^{-1} p$. in physics we call these covariant vectors
or co-vectors. we generally write them in index notation as something with a lower
index like $p_i$. in this notation we have
$p'_{i} = \left( \Lambda^{-1} \right)_i^{\;j} p_j$

Ok one more thing based on qbert's post.

The four vector $$(\frac{\partial}{\partial x^{0}},\frac{\partial}{\partial x^{1}},\frac{\partial}{\partial x^{2}},\frac{\partial}{\partial x^{3}})$$ is a covariant one because it transforms the "opposite" way i.e. with the inverse matrix. So should it be represented as $\partial_{\mu}$ since covariant vectors have the lower index?

Also, does $$(\frac{\partial}{\partial x_{0}},\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial x_{2}},\frac{\partial}{\partial x_{3}})$$ then become the contravariant four vector i.e. $\partial^{\mu}$?

Lastly, would it make sense to think of the covariant vectors as row vectors and contravariant vectors as column vectors? I read in another thread that this is usually ok (with the caveat that it is not always correct) but it seems to be a bit inconsistent if applied here.

Thank you and sorry to drag the thread on for so long.

 

[Fredrik]

Short answer: yes, yes and yes.

 

[McLaren Rulez]

Great, thank you!