Summation notation and general relativity derivatives

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  • Thread starter Maniac_XOX
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  • #1
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Does $$\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)$$
mean the same as $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial A^\beta} ?$$
If not could someone explain the differences?
 

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  • #2
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I would have $$\partial^\beta (g_{\alpha \beta} A_\mu A^\mu)=\frac {\partial (g_{\alpha \beta} A_\mu A^\mu)}{\partial (x_\beta)} = \frac {\partial (g_{\alpha \beta} A_\mu A^\mu)} {\partial A^\rho} \frac {\partial A^\rho}{\partial (x_\beta)} .$$
Only when ##\frac {\partial A^\rho}{\partial (x_\beta)}=\delta^{\rho \beta}##, I have ## \frac {\partial (g_{\alpha \beta} A_\mu A^\mu)} {\partial A^\beta}##. Am I right?
 
  • #3
Ibix
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Your up/down indices have got messed up slightly, but that's because the OP has incorrectly defined ##\partial^a=\frac \partial{\partial x^a}## when it should be ##\frac \partial{\partial x_a}##.
 
  • #4
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Hi, @Ibix . Would you point out where I mess up the indices? I have re-check them, and I failed to locate it.
 
  • #5
Ibix
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There's clearly something wrong because, taken as a whole, your last post says (defining ##T_{\alpha\beta}=g_{\alpha\beta}A_\mu A^\mu##) that ##\frac{\partial T_{\alpha\beta}}{\partial x_\beta}=\frac{\partial T_{\alpha\beta}}{\partial A^\beta}## at least in some circumstances, and the indices do not match. I think what you did that was illegitimate was defining the symbol ##\delta^{\alpha\beta}##, since it would imply ##\delta^{\alpha\beta}=g^{\alpha\rho}\delta^\beta_\rho##. We know the right hand side of that is just ##g^{\alpha\beta}##, which doesn't have the properties you want.

If OP had defined ##\partial^\beta=\frac{\partial}{\partial A_\beta}## (which is not generally correct, but at least has consistent indices) then you would have obtained that the definition is correct if ##\frac{\partial A_\rho}{\partial x_\beta}=\delta^\beta_\rho##, and then you are fine.
 
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  • #6
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There's clearly something wrong because, taken as a whole, your last post says (defining ##T_{\alpha\beta}=g_{\alpha\beta}A_\mu A^\mu##) that ##\frac{\partial T_{\alpha\beta}}{\partial x_\beta}=\frac{\partial T_{\alpha\beta}}{\partial A^\beta}## at least in some circumstances, and the indices do not match. I think what you did that was illegitimate was defining the symbol ##\delta^{\alpha\beta}##, since it would imply ##\delta^{\alpha\beta}=g^{\alpha\rho}\delta^\beta_\rho##. We know the right hand side of that is just ##g^{\alpha\beta}##, which doesn't have the properties you want.

If OP had defined ##\partial^\beta=\frac{\partial}{\partial A_\beta}## (which is not generally correct, but at least has consistent indices) then you would have obtained that the definition is correct if ##\frac{\partial A_\rho}{\partial x_\beta}=\delta^\beta_\rho##, and then you are fine.
Hi, @Ibix . Would you point out where I mess up the indices? I have re-check them, and I failed to locate it.
Hello, I am confused here, could we back up and explain what is the correct definition of $$\partial^\beta(g_{\alpha\beta}A_\mu A^\mu) ?$$ You said i messed up the indices but i thought the derivative implied that i was differentiating in terms of the field itself not the 4-position, otherwise the indices differ and I agree. So could you elaborate on how that fraction expands so i understand how Haorong found the answer?
Two more question.
1) Ibix you said "not generally correct" but then said that it would lead to a correct answer, care to explain more?
2) Does it matter if the indices on a kronecker delta are both up, both down or one up and one down? Because throughout my project i've only been doing the third case, please let me know the differences and where they apply.

Sorry for the exhausting reply lol hope you find the time to help out, thank you loads
 
  • #7
anuttarasammyak
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I am not an expert on tensor mathematics so it is a good chance to study.

[tex]A_\mu A^\mu[/tex]
is scalar. May we expect it is as constant in differentiation so the given formula is
[tex]A_\mu A^\mu \partial^\beta g_{\alpha\beta}[/tex]
?
 
  • #8
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I am not an expert on tensor mathematics so it is a good chance to study.

[tex]A_\mu A^\mu[/tex]
is scalar. May we expect it is as constant in differentiation so the given formula is
[tex]A_\mu A^\mu \partial^\beta g_{\alpha\beta}[/tex]
?
I don't think that's correct since ##g_{\alpha\beta}## is also a constant since it's a metric of value ##\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}##
 
  • #9
stevendaryl
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The notation I am familiar with uses ##\partial_\beta## to mean ##\dfrac{\partial}{\partial x^\beta}## and ##\partial^\beta## to mean ##g^{\alpha \beta}\dfrac{\partial}{\partial x^\alpha}##
 
  • #10
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The notation I am familiar with uses ##\partial_\beta## to mean ##\dfrac{\partial}{\partial x^\beta}## and ##\partial^\beta## to mean ##g^{\alpha \beta}\dfrac{\partial}{\partial x^\alpha}##
wouldnt it be ##g^{\alpha \beta}\dfrac{\partial}{\partial x_\alpha}## since the derivative is a contravariant vector ##\partial^\beta## i need to differentiate wrt a covariant vector? ##\partial x_\alpha##
 
  • #11
Ibix
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I am not an expert on tensor mathematics so it is a good chance to study.

[tex]A_\mu A^\mu[/tex]
is scalar.
It's a scalar field, so its partial derivatives are generally non-zero. The assumption here is that the vector field ##A## has a value at every point in spacetime, and hence so does the scalar field ##A_\mu A^\mu##.

Temperature is an everyday example of a scalar field, and its partial derivatives are typically non-zero.
 
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  • #12
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It's a scalar field, so its partial derivatives are generally non-zero. The assumption here is that the vector field ##A## has a value at every point in spacetime, and hence so does the scalar field ##A_\mu A^\mu##.

Temperature is an everyday example of a scalar field, and its partial derivatives are typically non-zero.
That clarified it for me as well actually thanks, what about the previous questions?
 
  • #13
stevendaryl
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wouldnt it be ##g^{\alpha \beta}\dfrac{\partial}{\partial x_\alpha}## since the derivative is a contravariant vector ##\partial^\beta## i need to differentiate wrt a covariant vector? ##\partial x_\alpha##
Coordinates are not vectors, so ##x_a## doesn’t literally make any sense. I don’t know of any meaning of ##\dfrac{\partial}{\partial x_\beta}## except as

##g^{\alpha \beta} \dfrac{\partial}{\partial x_\alpha}##
 
  • #14
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Coordinates are not vectors, so ##x_a## doesn’t literally make any sense.
but isn't ##x_a## a covariant vector with components that correspond to the coordinates? Which actually should be ##x_\beta## since the contravariant derivative is ##\partial^\beta## ?
 
  • #15
anuttarasammyak
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Thanks.   I am familiar with calculation
[tex]\partial_\beta=\frac{\partial}{\partial x^\beta}[/tex]
but how we calculate indecies unside down
[tex]\partial^\beta=\frac{\partial}{\partial x_\beta}[/tex]
What is the relation between the two ?
 
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  • #16
stevendaryl
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but isn't ##x_a## a covariant vector with components that correspond to the coordinates? Which actually should be ##x_\beta## since the contravariant derivative is ##\partial^\beta## ?
As I said, coordinates are not vectors, so there is no such thing as vectors versus covectors with coordinates. ##x^\alpha## is not a vector, it’s just 4 numbers associated with each point in space (or spacetime). For Cartesian coordinates, you can associate the coordinates with a vector (the vector pointing from the origin to the location), but it’s not true for other coordinates.

In the other hand, if you have a parametrized path, or a curve through space (or spacetime) ##x^\alpha(s)##, then the components of the velocity, ##\frac{dx^\alpha}{ds}## is always a vector, even for curvilinear coordinates.
 
  • #17
Ibix
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That clarified it for me as well actually thanks, what about the previous questions?
@stevendaryl is correct, I think, and you should disregard my previous comments to @Haorong Wu about the derivatives.

No, ##x_a## is not a vector or a component of one. There is no such thing as a position vector in non-flat spaces or non-Cartesion coordinates. An infinitesimal change in position, ##\mathrm{d}x^a##, is a vector, so quantities like ##g_{ab}\mathrm{d}x^a\mathrm{d}x^b## can be calculated, but ##g_{ab}x^ax^b## doesn't make sense. Coordinates are a tuple, not a vector.

##\partial_af## is the change in a function ##f## as you move an infinitesimal distance in the direction in which only the ##a##th coordinate changes. Yes, this is ##\frac{\partial}{\partial x^a}f## (Carroll notes that, very roughly speaking, "##x^\mu## has an upper index, but when it is in the denominator of a derivative it implies a lower index on the resulting object."). ##\partial^af##, then, cannot mean ##\frac{\partial}{\partial x_a}f##, because ##x_a## doesn't mean anything. Instead, it is an accepted shorthand for ##g^{ab}\partial_bf##.

So to return to your original question, you should interpret ##\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)## as $$\begin{eqnarray*}
&&\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)\\
&=&g^{\beta\sigma}\partial_\sigma(g_{\alpha\beta}A_\mu A^\mu)\\
&=&g^{\beta\sigma}\frac{\partial}{\partial x^\sigma}\left(g_{\alpha\beta}A_\mu A^\mu\right)
\end{eqnarray*}$$
 
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  • #18
anuttarasammyak
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Instead, it is an accepted shorthand for gab∂bf.
D'Alembert opertor
[tex]\square=\partial_\mu \partial^\mu =g^{\mu\beta}\partial_\mu \partial_\beta[/tex]
is another example of this way.
 
  • #19
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@stevendaryl is correct, I think, and you should disregard my previous comments to @Haorong Wu about the derivatives.

No, ##x_a## is not a vector or a component of one. There is no such thing as a position vector in non-flat spaces or non-Cartesion coordinates. An infinitesimal change in position, ##\mathrm{d}x^a##, is a vector, so quantities like ##g_{ab}\mathrm{d}x^a\mathrm{d}x^b## can be calculated, but ##g_{ab}x^ax^b## doesn't make sense. Coordinates are a tuple, not a vector.

##\partial_af## is the change in a function ##f## as you move an infinitesimal distance in the direction in which only the ##a##th coordinate changes. Yes, this is ##\frac{\partial}{\partial x^a}f## (Carroll notes that, very roughly speaking, "##x^\mu## has an upper index, but when it is in the denominator of a derivative it implies a lower index on the resulting object."). ##\partial^af##, then, cannot mean ##\frac{\partial}{\partial x_a}f##, because ##x_a## doesn't mean anything. Instead, it is an accepted shorthand for ##g^{ab}\partial_bf##.

So to return to your original question, you should interpret ##\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)## as $$\begin{eqnarray*}
&&\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)\\
&=&g^{\beta\sigma}\partial_\sigma(g_{\alpha\beta}A_\mu A^\mu)\\
&=&g^{\beta\sigma}\frac{\partial}{\partial x^\sigma}\left(g_{\alpha\beta}A_\mu A^\mu\right)
\end{eqnarray*}$$
Right, that makes it clear with Carrol's quote thank you very much. basically whenever i'm in the situation where i have ##\partial^\mu## i cannot use ##\frac{\partial}{\partial x_\mu}## as ##x_\mu## represents coordinates in flat spacetime, which in this Minkowski spacetime don't exist. Therefore i must lower the indices of that partial derivative from ##\partial^\beta## to ##g^{\beta\alpha} \partial_\alpha## and operate ##g^{\beta\alpha} \frac{\partial}{\partial x^\alpha}##. Is this correct?
 
  • #20
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@stevendaryl is correct, I think, and you should disregard my previous comments to @Haorong Wu about the derivatives.

No, ##x_a## is not a vector or a component of one. There is no such thing as a position vector in non-flat spaces or non-Cartesion coordinates. An infinitesimal change in position, ##\mathrm{d}x^a##, is a vector, so quantities like ##g_{ab}\mathrm{d}x^a\mathrm{d}x^b## can be calculated, but ##g_{ab}x^ax^b## doesn't make sense. Coordinates are a tuple, not a vector.

##\partial_af## is the change in a function ##f## as you move an infinitesimal distance in the direction in which only the ##a##th coordinate changes. Yes, this is ##\frac{\partial}{\partial x^a}f## (Carroll notes that, very roughly speaking, "##x^\mu## has an upper index, but when it is in the denominator of a derivative it implies a lower index on the resulting object."). ##\partial^af##, then, cannot mean ##\frac{\partial}{\partial x_a}f##, because ##x_a## doesn't mean anything. Instead, it is an accepted shorthand for ##g^{ab}\partial_bf##.

So to return to your original question, you should interpret ##\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)## as $$\begin{eqnarray*}
&&\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)\\
&=&g^{\beta\sigma}\partial_\sigma(g_{\alpha\beta}A_\mu A^\mu)\\
&=&g^{\beta\sigma}\frac{\partial}{\partial x^\sigma}\left(g_{\alpha\beta}A_\mu A^\mu\right)
\end{eqnarray*}$$
The most i have managed to simplify so far is to this point:
$$\partial^\beta F_{\beta\alpha} -\partial_\alpha (A_\mu A^\mu)\beta + \mu^2 A_\alpha =-2A_\alpha (\partial_\rho A^\rho) \beta + \frac {4\pi}{c} J_\alpha$$
therefore $$\partial^\beta F_{\beta\alpha} + 2A_\alpha (\partial_\rho A^\rho) \beta + \mu^2 A_\alpha =+ \partial_\alpha (A_\mu A^\mu)\beta +\frac {4\pi}{c} J_\alpha$$

but it could also become: $$\partial^\beta F_{\beta\alpha}+\mu^2 A_\alpha =+ \beta[A_\nu (\partial_\alpha A^\nu) + A_\mu (\partial_\alpha A^\mu) -2A_\alpha (\partial_\rho A^\rho)] +\frac {4\pi}{c} J_\alpha$$

Any further tips to simplify? or is this alright :)
 
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  • #21
robphy
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$$\partial^\beta F_{\beta\alpha}+\mu^2 A_\alpha =+ \beta(A_\nu \partial_\alpha A^\nu + A_\mu \partial_\alpha A^\mu -2A_\alpha \partial_\rho A^\rho) +\frac {4\pi}{c} J_\alpha$$
Since
[itex] A_\nu \partial_\alpha A^\nu = A_\mu \partial_\alpha A^\mu[/itex] ,
it follows that
[itex] A_\nu \partial_\alpha A^\nu + A_\mu \partial_\alpha A^\mu=2 A_\nu \partial_\alpha A^\nu [/itex]
 
  • #22
PeterDonis
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Any further tips to simplify?
@Maniac_XOX please note that there is no need to open a new thread to ask this same question. I have merged the separate thread you started on the same question with this one.
 
  • #23
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Thanks for the clarification, @Ibix . So in Minkowski spacetime, could I write ##\partial^\alpha=\frac {\partial}{\partial x_\alpha} ##? Or is that also wrong? I remember I read somewhere that I could use this kind of euqtions. Maybe in QFT? The QFT I learnt is in Minkowski spacetime, so maybe this equation holds.
 
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  • #24
anuttarasammyak
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From derivative law
[tex]\frac{\partial}{\partial x_\alpha}=\frac{\partial}{\partial x^\beta}\frac{\partial x^\beta}{\partial x_\alpha}[/tex]
I am not still sure if we say
[tex]\frac{\partial x^\beta}{\partial x_\alpha}=g^{\alpha\beta}[/tex]
I observe
[tex]x^\beta=g^{\gamma\beta}x_\gamma[/tex]
So in Minkowsky spacetime, where ##g^{\alpha\beta}## are constants, it seems alright.

I would like to know such derivative law calculation also applies in GR with or without taking ##\partial ## as covariant derivative.

[EDIT]
[tex]\frac{\partial}{\partial x_\alpha}=\frac{\partial x^\beta}{\partial x_\alpha}\frac{\partial}{\partial x^\beta}[/tex]

Derivation does not apply to ##\frac{\partial x^\beta}{\partial x_\alpha}##

[EDIT2]
Taking \delta as covariant derivative
[tex]\frac{\partial x^\beta}{\partial x_\alpha}=g^{\alpha\beta}+\frac{\partial g^{\gamma\beta}}{\partial x^\alpha}x_\gamma[/tex]
where
[tex]\frac{\partial g^{\gamma\beta}}{\partial x^\alpha}=\frac{g^{\gamma\beta}}{\partial x^\delta}\frac{\partial x^\delta}{\partial x_\alpha}=0\cdot\frac{\partial x^\delta}{\partial x_\alpha}=0[/tex]

So we observe ##\partial^\alpha,\partial_\alpha## behaves as usual vector whose index goes up or down by applying metric tensor g.
 
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  • #25
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Since
[itex] A_\nu \partial_\alpha A^\nu = A_\mu \partial_\alpha A^\mu[/itex] ,
it follows that
[itex] A_\nu \partial_\alpha A^\nu + A_\mu \partial_\alpha A^\mu=2 A_\nu \partial_\alpha A^\nu [/itex]
So even though the index has changed from the original by lowering it this still remains the same right?
 

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