General relativity is such a huge deep subject, but it feels so good when you feel something click and you get some pieces of the puzzle. Your channel was a great way to do that, and it deserves 10 times more subscriptions honestly.
Thank you so much. I was so confused that books do not explicitly say that the derivative of h_\mu u is assumed to be small, yet, they assume it in calculations. I am so relieved that someone explicitly says it.
I'm experiencing some amount of doubt as well... I don't know why so many books are so coy about saying that the derivatives should be small. It's possible we're using some kind of more advanced perturbation theory trick that requires more careful thinking that GR authors just skip over. There's a prof (Valter Moretti) in the comments of this question who says that both assumptions are required: physics.stackexchange.com/questions/593298/linearized-gravity-derivatives-of-the-metric-perturbation
Hi @@eigenchris , I ran into you on the Kathy Loves Physics channel a few days back. You have surely seen the Kerr-Schild metric, suited to all possible forms of Kerr solution (slow Kerr, extreme Kerr and fast Kerr), all of which take that exact same form of a Minkowski metric added to a properly defined perturbation term. That form of metric has some really remarkable properties too and, to add my 2c to this conversation, there is no conceivable way that this perturbation has to be kept small since in the Kerr Schild formulation this perturbation term (added to the Minkowski metric) contains a complete description of the metric for an exact solution of the Einstein equation. The biggest problem that you encounter in the Kerr solution is, of course, the non-removable, enigmatic ring singularity and the branch cut which it heralds.
It is just because of the compatibility between the connection and the metric. A natural assumption in gr. It is usually denoted by covariant derivative of the metric equal to zero but since you can always go in the free falling frame, the covariant reduces to regular partials. Hence your assumption. Cheers, very good videos @@eigenchris
Thanks a bunch ,Chris. Having just finished taking notes on this video, I have to say that it was another amazing epic! I can't thank you enough, you are a star!
Thank you so much for the wonderful work you have been doing! I found your channel just a few weeks ago and I am amazed by how well you explain the topics. In my opinion, everyone should be teaching GR like you do. I'm a seismologist but always wanted to learn GR and now I can! Thanks a lot!
Hei there, first of all thanks, you are gonna save my course! Second, on the notes I got from my professor is shown that we can get rid of the small derivatives of h because they are actually a gauge transformation for h, and we can just not write them then, since h is not uniquely fixed for a given source. For showing this you grab infinitesimal coord transformations and gets rid of a term from the Killing equation assuming that the transformation is small. Hmu if you want it for reference.
Hello! I think that if h is very small but its variations aren't, then h could become not so small and we'd have a contradiction! Other than that thank you so very much for this series of videos! A lot clearer than my GR class, you're saving my life! I particularly appreciate that you give the calculations with all their gory details.
Respectful Sir, Thank you so much your video explains concept in a very nice and simple way...Thank you so much Sir... Thanking All, All Sincerely, Shreyansh
From what I have found about "Lorenz Gauge", it is the divergence of vector potential A in terms of rate of scalar potential V. I am thinking what would be the equations being used in gravitational wave by using Lorenz Gauge. I'm excited on your next video upload of yours sir.
Hi Chris , thank you for your amazing explanations , you make hard concepts look easy . I’ve learned so much from this channel. Can you please do a video about deriving the Einstein field equations non geometrically from the free massless spin 2 field and adding self interaction ? Thank you in advance .
Thanks! Unfortunately I don't know how to do that. I wasn't aware there were any ways to derive the EFE other than generalizing Poisson's Equation or the Einstein-Hilbert action.
Excellent! You seem to have followed Misner Thorne & Wheeler("Gravitation"), but frankly II could never follow the MTW book chapter on linearized gravity. By comparison, you go through all the steps in detail; crystal clear!.. Surprised you did'nt mention it(MTW) here though...maybe you've done so elsewhere
I consulted a number of sources, and most of them do it in a similar way. Sean Carroll has free online GR notes with similar steps. MTW is a fairly old book so maybe many sources use it.
the property that the derivative of h is zero is unnecessary because I think you can rise and lower indices of derivatives of the metric with the metric itself (check me, im not sure)
This is a late comment. When we derived the kappa coefficient of Einstein field equation, kappa = 8*pi*G/c^4. We assume low velocity limit, weak gravity limit and metric tensor is time independent (partial dg/dt). But here in gravitational wave, the metric perturbation h is dependent on time from gravitational wave equation.
We're assuming that the EFE is the correct equation for general relativity and works in all cases. The low-velocity limit is what we need to check to confirm the equations align with Newtonian gravity.
13:31 Why can’t the Alpha and Beta summation indices in the Einstein Tensor not be relabeled consistently with the other summation indices? I seem to be missing something here
Great video!! Just a question, why do we raise/ lower indices with the minkowski metric? I was under the impression that to do that you needed to use the "full" metric tensor, so for example in calculating the ricci tensor and scalar why do we multiply with the minkowski metric instead of the whole g_mn?
Thanks, was just doing these calculations myself last week, along my reading of Sean Carroll's "Spacetime and Geometry". Will you also derive the non-covariant form of these equations after decomposing the metric perturbation h into phi, w_i, psi and s_ij? (This is similar to decomposing the field strengh tensor F_mu,nu into electric and magnetic fields in EM, you lose the covariance but you gain more insight into the physical degrees of freedom.) Thanks for mentioning explicitly the assumptions on h, especially that derivatives are also small. This was exactly one of my questions as Carroll didn't write it explicitly and just because a function is small, its derivative is not necessarily.
For your first question, I admit I'm not familiar with that. I'll have to read up on Sean Carroll's notes. For the derivatives of h being small, I'm uncomfortable that so few sources mention it. It makes me feel like I'm doing something wrong... but they all seem to use the assumption implicitly, so I thought I'd just say it out loud.
@@eigenchris I _think_ this is a feature of perturbation theories. If I understood correctly, we don't actually take a "small h". We take h with a small coefficient in the formal power series. And we use the same coefficient in the "formal derivative" so the contribution to the derivative of g should be small. I far from understood everything I read about this(and it wasn't much to begin with), but maybe this can give you a hint on where to find your answer, if nothing else.
@@narfwhals7843 I don't really think that you gain anything from pulling out an explicit epsilon. If you replace the decomposition g = eta + h by g = eta + epsilon * h, then the combination |epsilon*h|h (where ' denotes any derivative). Unless all h'h', h'h and h''h terms would cancel out themselves, h' and h" being small is an assumption you have to make as well. And when I did this calculation I checked that they don't cancel which you can already see on the level of Christoffel symbols. I tried to think physically about it, what it would mean for h to be small but h' or h" being large and if that scenario makes sense. Probably not, because in the Einstein equations the energy momentum tensor couples to the g, g' and g", so any of these quantities to be large means T_mu,nu has to be large, which is not a small deviation from Minkowski spacetime then where T=0. So I think that any derivative of the metric being large is just not in the spirit of perturbation theory. But I'm not sure, this is my own argument, not something I read.
Hi @eigenchris. I wish you let us to see the 109 c. d & e, videos. I'll really apreciatte it, I'm preparing mi final test about RG. Thanks a lot. (5 unavailable videos are hidden :( )
Those are actually drafts of the 109 a and b videos that I deleted... I'm working on videos c, d, and e now. 109c might be out on Sunday. The other two might take a week each. If this isn't enough time you can try reading Chapter 6 of Sean Carrol's GR notes: www.preposterousuniverse.com/grnotes/
Yes, more general relativity! Hey Chris, I've been meaning to ask, are you familiar with geometric algebra and the way it interprets certain equations?
@@eigenchris Not very specific. I learned about this theory a while ago and I'm frankly surprised it's so obscure. It gets to the very heart of the matter--what exactly is a wave function, anyway? What does the imaginary unit in the Schrödinger equation mean? Just what is spin and how does it factor into quantum theory? GA (at least what I've read up on it) provides a lot of insight into these almost-metaphysical questions, so I wondered if you'd heard of it (since you explain everything with great clarity) and would perhaps give the topic some exposure in the future. In any case keep up the good work!
@@orktv4673 I wasn't aware it gave any insight into the Schrodinger Equation or the imaginary unit. Can I ask how that works? But yes, GA is pretty important for understanding spinors, I believe. I plan to do videos on that aspect later this year. Have you seen the video "A Swift Introduction to Geometric Algebra"?
@@eigenchris I have not, I will check it out! As for the imaginary unit, this is identified as a volume form σ1σ2σ3 which indeed has the property of squaring to -1. The Pauli algebra, which we tend to understand as a "quantum" phenomenon, turns out to have a quite natural, classical-geometric interpretation, as does the whole Clifford algebra substructure. (These sigmas are indeed Pauli matrices reinterpreted as GA paravectors, and the imaginary unit that pops up when taking products becomes a lot less mystical when you look at it that way.) In fact the very notion of a wave function becomes that of a rotor, a kind of rotational operator, which makes the "half spin" property of spinors manifest because when the rotor has an angle π, it will rotate the object it's working on by 2π; and Hestenes goes so far as to propose that spin was actually part of the Schrödinger/Dirac equation all along, and not something which should or can be tacked on after the fact. I can't give a lot of details, but if you read Hestenes's Oersted Medal lecture you'll quickly get the gist. GA also has some neat consequences for how we understand electrodynamics, as it allows for the electric and magnetic field to be considered a single complex vector F = E + i B, where again the i is actually a volume element. From this the plane wave solution e^i(ωt - kx) starts looking more sensible, and the notion of the magnetic field being a "pseudovector" makes sense, as i B is in fact an oriented area element (by pulling out the i, the B becomes a proper vector). Also with this notation all of Maxwell's equations can be merged into a single equation, ∇F = J, though I'm not yet fully convinced this is just as elegant as the conventional picture of tensors and four-vectors.
So Linearized Gravity is a "halfway" approximation between the full field Equations and Riemann Normal Coordinates at a point (where all Christoffel Symbols are zero)? Also, what is the physical interpretation of partial derivative operators with a raised index? Is it just to sum with the Einstein Notation? At first glance I thought they were antiderivatives.
If you like you can think of the metric tensor being expressed as an infinite taylor series (with "zero-order" term, a linear term, quadratic term, cubic term, etc). Linearized gravity ignores all the terms except the "zero order term" (which is the flat metric) and the linear term (the h-part). I'm not sure I have a good physical interpretation of partial derivatives with a raised index. To me it's just a notational shortcut.
Chris what does lowering or raising index of a Tensor by raing or lowering index of del really mean? Is the tensor changing or is it like changing matrix to its transpose? For instance del sq can be thought as del row operating on del column? Raising or lowering of index by metric or Minkowski tensor changes the tensor. What does lowering or raising index of Tensor by del means exactly?
I talk about this in more detail in my Tensors for Beginners Video 16: ruclips.net/video/_z9R7OMpxhY/видео.html The original idea is that every vector V has a covector partner "g(V, __) = V · __". Raising and lowering indices with the metric lets you change betwen the vector components of V and the covector components of its covector partner "V · __". The concept of raising and lowering indices can be extended to any tensor with upper/lower indices. Most of the time in this video it's just handy notation or a quick algebraic shortcut.
Thank you for the video. Though I have question about the lower and upper indicies; does it change how we do the calculations or is it just used for convention and just notation purposes?
It's mostly just a convention for notational purposes. Vector components normally have upper indices, and covector components normally have lower indices, but you can swap between them using the metric. You can search for my "Tensors for Beginners 16" video if you want to learn more. I spend around 15 minutes explaining it.
At 5:59 you told that the derivative of "h" is zero then wouldn't be the connection coefficient become zero because it contains only derivatives of "h".?
@@eigenchris At 9:11 you raised the indices such that "nu" in all partial derivative becomes "sigma" but at 8:07 you raised the indices such that some "alpha" becomes "mu" but at the same time you raises the one final index "mu" of "h" instead of raising "alpha" present in partial derivative why is that.?
@@keshavshrestha1688 In the particular case of 8:07 I wanted that term to have 2 lowered partial derivatives... so that when it comes time to get the formula for the Ricci tensor, the first two terms are obviously the same. I basically knew the answer I wanted ahead of time and followed the steps necessary to get the correct result. The metric can raise or lower any indices as needed.
@@eigenchris oh so at 8:07 if i raise the index "alpha" of partial derivative instead of "mu" of "h" then that process is also right but i may not get the correct and concluding result, right.?
First of all, thank you very much Chris for another class. Secondly, I wanted to ask you what exactly is linearized gravity? I didn't quite understand this part
@@eigenchris So does it mean that for small perturbations in space-time the quantities related to field equations such as the Ricci tensor and the Christoffel symbols remain in their original form?
@@victormarchant217 What do you mean by "their original form"? My goal for the video was to show the new equations for all the tensors/symbols under the linearized gravity assumption.
@@eigenchris In my view, some of the equations maintain a form that was already known only with the metric tensor gμv, as if you only replaced gμv with hμv in the equations, as if it were an invariant transformation, but you said that you get new equations. maybe I have to review the concepts of linearity and invariance, I believe I'm confusing these two things
@@eigenchris Just keep producing this content. It is above my level but that doesn't mean I didn't enjoy it. I watch/enjoy the PreMath channel RUclips videos but there is something you may notice about how he explains his problems/content. I am not saying that you should do this or change, it is just that every step in working a problem/concept he goes back to fundamentals. Thanks for allowing me feedback.
@@prbmax I agree with Justin that TC 21-26 might help with this. But for my 108-110 videos, I'm assuming viewers are pretty comfortable with the content from Relativity 107. I'd be curious to know if you've seen those too? Most of these videos aren't really meant to be watched on their own.
General relativity is such a huge deep subject, but it feels so good when you feel something click and you get some pieces of the puzzle. Your channel was a great way to do that, and it deserves 10 times more subscriptions honestly.
Thank you so much. I was so confused that books do not explicitly say that the derivative of h_\mu
u is assumed to be small, yet, they assume it in calculations. I am so relieved that someone explicitly says it.
I'm experiencing some amount of doubt as well... I don't know why so many books are so coy about saying that the derivatives should be small. It's possible we're using some kind of more advanced perturbation theory trick that requires more careful thinking that GR authors just skip over. There's a prof (Valter Moretti) in the comments of this question who says that both assumptions are required: physics.stackexchange.com/questions/593298/linearized-gravity-derivatives-of-the-metric-perturbation
Hi @@eigenchris , I ran into you on the Kathy Loves Physics channel a few days back.
You have surely seen the Kerr-Schild metric, suited to all possible forms of Kerr solution (slow Kerr, extreme Kerr and fast Kerr), all of which take that exact same form of a Minkowski metric added to a properly defined perturbation term.
That form of metric has some really remarkable properties too and, to add my 2c to this conversation, there is no conceivable way that this perturbation has to be kept small since in the Kerr Schild formulation this perturbation term (added to the Minkowski metric) contains a complete description of the metric for an exact solution of the Einstein equation.
The biggest problem that you encounter in the Kerr solution is, of course, the non-removable, enigmatic ring singularity and the branch cut which it heralds.
It is just because of the compatibility between the connection and the metric. A natural assumption in gr. It is usually denoted by covariant derivative of the metric equal to zero but since you can always go in the free falling frame, the covariant reduces to regular partials. Hence your assumption. Cheers, very good videos @@eigenchris
Thanks a bunch ,Chris. Having just finished taking notes on this video, I have to say that it was another amazing epic! I can't thank you enough, you are a star!
Thank you so much for the wonderful work you have been doing! I found your channel just a few weeks ago and I am amazed by how well you explain the topics. In my opinion, everyone should be teaching GR like you do. I'm a seismologist but always wanted to learn GR and now I can! Thanks a lot!
Thanks! That means a lot. I can be a bit slow with uploading but I hope you enjoy the next ones.
Thanks for the video. I am currently in a GR class and the only things that are helping me are your videos and a DG class I took last semester
You always come with an amazing video. I sincerely appreciate it!
>Dark mode gone
>1 AM
>Sed
You're so good man, i love your videos.
Hei there, first of all thanks, you are gonna save my course! Second, on the notes I got from my professor is shown that we can get rid of the small derivatives of h because they are actually a gauge transformation for h, and we can just not write them then, since h is not uniquely fixed for a given source. For showing this you grab infinitesimal coord transformations and gets rid of a term from the Killing equation assuming that the transformation is small. Hmu if you want it for reference.
Hello! I think that if h is very small but its variations aren't, then h could become not so small and we'd have a contradiction!
Other than that thank you so very much for this series of videos! A lot clearer than my GR class, you're saving my life! I particularly appreciate that you give the calculations with all their gory details.
Respectful Sir,
Thank you so much your video explains concept in a very nice and simple way...Thank you so much Sir...
Thanking All,
All Sincerely,
Shreyansh
From what I have found about "Lorenz Gauge", it is the divergence of vector potential A in terms of rate of scalar potential V. I am thinking what would be the equations being used in gravitational wave by using Lorenz Gauge. I'm excited on your next video upload of yours sir.
Thank you very much, this was a great help
Hi Chris , thank you for your amazing explanations , you make hard concepts look easy . I’ve learned so much from this channel. Can you please do a video about deriving the Einstein field equations non geometrically from the free massless spin 2 field and adding self interaction ? Thank you in advance .
Thanks! Unfortunately I don't know how to do that. I wasn't aware there were any ways to derive the EFE other than generalizing Poisson's Equation or the Einstein-Hilbert action.
Excellent! You seem to have followed Misner Thorne & Wheeler("Gravitation"), but frankly II could never follow the MTW book chapter on linearized gravity.
By comparison, you go through all the steps in detail; crystal clear!..
Surprised you did'nt mention it(MTW) here though...maybe you've done so elsewhere
I consulted a number of sources, and most of them do it in a similar way. Sean Carroll has free online GR notes with similar steps. MTW is a fairly old book so maybe many sources use it.
Nice video. You got yourself a subscriber.
13:32 Wondering if LIGO backs up ‘ small’ assumptions?
"At 9:12, the index is raised for \( h^{\alpha}_{\sigma} \) and not on the partial derivative \( \partial_{
u} \) in the second term."
the property that the derivative of h is zero is unnecessary because I think you can rise and lower indices of derivatives of the metric with the metric itself (check me, im not sure)
This is a late comment.
When we derived the kappa coefficient of Einstein field equation, kappa = 8*pi*G/c^4. We assume low velocity limit, weak gravity limit and metric tensor is time independent (partial dg/dt). But here in gravitational wave, the metric perturbation h is dependent on time from gravitational wave equation.
We're assuming that the EFE is the correct equation for general relativity and works in all cases. The low-velocity limit is what we need to check to confirm the equations align with Newtonian gravity.
13:31 Why can’t the Alpha and Beta summation indices in the Einstein Tensor not be relabeled consistently with the other summation indices? I seem to be missing something here
Great video!! Just a question, why do we raise/ lower indices with the minkowski metric? I was under the impression that to do that you needed to use the "full" metric tensor, so for example in calculating the ricci tensor and scalar why do we multiply with the minkowski metric instead of the whole g_mn?
You can use the full metric, but the "h" part will always get sent to zero when it multiplies with another really small term.
@@eigenchris oh makes sense. Thanks!
Thanks, was just doing these calculations myself last week, along my reading of Sean Carroll's "Spacetime and Geometry". Will you also derive the non-covariant form of these equations after decomposing the metric perturbation h into phi, w_i, psi and s_ij? (This is similar to decomposing the field strengh tensor F_mu,nu into electric and magnetic fields in EM, you lose the covariance but you gain more insight into the physical degrees of freedom.)
Thanks for mentioning explicitly the assumptions on h, especially that derivatives are also small. This was exactly one of my questions as Carroll didn't write it explicitly and just because a function is small, its derivative is not necessarily.
For your first question, I admit I'm not familiar with that. I'll have to read up on Sean Carroll's notes.
For the derivatives of h being small, I'm uncomfortable that so few sources mention it. It makes me feel like I'm doing something wrong... but they all seem to use the assumption implicitly, so I thought I'd just say it out loud.
@@eigenchris I _think_ this is a feature of perturbation theories. If I understood correctly, we don't actually take a "small h". We take h with a small coefficient in the formal power series. And we use the same coefficient in the "formal derivative" so the contribution to the derivative of g should be small.
I far from understood everything I read about this(and it wasn't much to begin with), but maybe this can give you a hint on where to find your answer, if nothing else.
@@narfwhals7843 I don't really think that you gain anything from pulling out an explicit epsilon. If you replace the decomposition g = eta + h by g = eta + epsilon * h, then the combination |epsilon*h|h (where ' denotes any derivative).
Unless all h'h', h'h and h''h terms would cancel out themselves, h' and h" being small is an assumption you have to make as well. And when I did this calculation I checked that they don't cancel which you can already see on the level of Christoffel symbols.
I tried to think physically about it, what it would mean for h to be small but h' or h" being large and if that scenario makes sense. Probably not, because in the Einstein equations the energy momentum tensor couples to the g, g' and g", so any of these quantities to be large means T_mu,nu has to be large, which is not a small deviation from Minkowski spacetime then where T=0. So I think that any derivative of the metric being large is just not in the spirit of perturbation theory. But I'm not sure, this is my own argument, not something I read.
If we use an epsilon in front of h_{ij} in the definition of g_{ij} and claim that epsilon
in 3:32 the eta is before the k but in the next step the order is changed. why is that possible?
Hi @eigenchris. I wish you let us to see the 109 c. d & e, videos. I'll really apreciatte it, I'm preparing mi final test about RG. Thanks a lot. (5 unavailable videos are hidden :( )
Those are actually drafts of the 109 a and b videos that I deleted... I'm working on videos c, d, and e now. 109c might be out on Sunday. The other two might take a week each. If this isn't enough time you can try reading Chapter 6 of Sean Carrol's GR notes: www.preposterousuniverse.com/grnotes/
Yes, more general relativity! Hey Chris, I've been meaning to ask, are you familiar with geometric algebra and the way it interprets certain equations?
I'm familiar with the basics. Do you have a specific question?
@@eigenchris Not very specific. I learned about this theory a while ago and I'm frankly surprised it's so obscure. It gets to the very heart of the matter--what exactly is a wave function, anyway? What does the imaginary unit in the Schrödinger equation mean? Just what is spin and how does it factor into quantum theory? GA (at least what I've read up on it) provides a lot of insight into these almost-metaphysical questions, so I wondered if you'd heard of it (since you explain everything with great clarity) and would perhaps give the topic some exposure in the future. In any case keep up the good work!
@@orktv4673 I wasn't aware it gave any insight into the Schrodinger Equation or the imaginary unit. Can I ask how that works? But yes, GA is pretty important for understanding spinors, I believe. I plan to do videos on that aspect later this year. Have you seen the video "A Swift Introduction to Geometric Algebra"?
@@eigenchris I have not, I will check it out! As for the imaginary unit, this is identified as a volume form σ1σ2σ3 which indeed has the property of squaring to -1. The Pauli algebra, which we tend to understand as a "quantum" phenomenon, turns out to have a quite natural, classical-geometric interpretation, as does the whole Clifford algebra substructure. (These sigmas are indeed Pauli matrices reinterpreted as GA paravectors, and the imaginary unit that pops up when taking products becomes a lot less mystical when you look at it that way.) In fact the very notion of a wave function becomes that of a rotor, a kind of rotational operator, which makes the "half spin" property of spinors manifest because when the rotor has an angle π, it will rotate the object it's working on by 2π; and Hestenes goes so far as to propose that spin was actually part of the Schrödinger/Dirac equation all along, and not something which should or can be tacked on after the fact. I can't give a lot of details, but if you read Hestenes's Oersted Medal lecture you'll quickly get the gist.
GA also has some neat consequences for how we understand electrodynamics, as it allows for the electric and magnetic field to be considered a single complex vector F = E + i B, where again the i is actually a volume element. From this the plane wave solution e^i(ωt - kx) starts looking more sensible, and the notion of the magnetic field being a "pseudovector" makes sense, as i B is in fact an oriented area element (by pulling out the i, the B becomes a proper vector). Also with this notation all of Maxwell's equations can be merged into a single equation, ∇F = J, though I'm not yet fully convinced this is just as elegant as the conventional picture of tensors and four-vectors.
So Linearized Gravity is a "halfway" approximation between the full field Equations and Riemann Normal Coordinates at a point (where all Christoffel Symbols are zero)? Also, what is the physical interpretation of partial derivative operators with a raised index? Is it just to sum with the Einstein Notation? At first glance I thought they were antiderivatives.
If you like you can think of the metric tensor being expressed as an infinite taylor series (with "zero-order" term, a linear term, quadratic term, cubic term, etc). Linearized gravity ignores all the terms except the "zero order term" (which is the flat metric) and the linear term (the h-part). I'm not sure I have a good physical interpretation of partial derivatives with a raised index. To me it's just a notational shortcut.
@@eigenchris It wasn't my question, but thanks for explaining that!
Can you explain why gravity spelled with a V and not a B, would make more sense.
Chris what does lowering or raising index of a Tensor by raing or lowering index of del really mean? Is the tensor changing or is it like changing matrix to its transpose? For instance del sq can be thought as del row operating on del column? Raising or lowering of index by metric or Minkowski tensor changes the tensor. What does lowering or raising index of Tensor by del means exactly?
I talk about this in more detail in my Tensors for Beginners Video 16: ruclips.net/video/_z9R7OMpxhY/видео.html
The original idea is that every vector V has a covector partner "g(V, __) = V · __". Raising and lowering indices with the metric lets you change betwen the vector components of V and the covector components of its covector partner "V · __". The concept of raising and lowering indices can be extended to any tensor with upper/lower indices. Most of the time in this video it's just handy notation or a quick algebraic shortcut.
Ok I will review.
I send you love from Bangladesh 🇧🇩
Sir,can you please make a video on derivation of warp drive metric
Sorry but I haven't stuied that metric.
Thank you for the video. Though I have question about the lower and upper indicies; does it change how we do the calculations or is it just used for convention and just notation purposes?
It's mostly just a convention for notational purposes. Vector components normally have upper indices, and covector components normally have lower indices, but you can swap between them using the metric. You can search for my "Tensors for Beginners 16" video if you want to learn more. I spend around 15 minutes explaining it.
@@eigenchris Thank you for clearing things up. Love your videos btw.
At 5:59 you told that the derivative of "h" is zero then wouldn't be the connection coefficient become zero because it contains only derivatives of "h".?
I was trying to say that "h*(∂h)" goes to zero because both "h" and "∂h" are small. "∂h" on its own is not small enough to go to zero.
@@eigenchris At 9:11 you raised the indices such that "nu" in all partial derivative becomes "sigma" but at 8:07 you raised the indices such that some "alpha" becomes "mu" but at the same time you raises the one final index "mu" of "h" instead of raising "alpha" present in partial derivative why is that.?
@@keshavshrestha1688 In the particular case of 8:07 I wanted that term to have 2 lowered partial derivatives... so that when it comes time to get the formula for the Ricci tensor, the first two terms are obviously the same. I basically knew the answer I wanted ahead of time and followed the steps necessary to get the correct result. The metric can raise or lower any indices as needed.
@@eigenchris oh so at 8:07 if i raise the index "alpha" of partial derivative instead of "mu" of "h" then that process is also right but i may not get the correct and concluding result, right.?
@@keshavshrestha1688 Yes... it's a "legal"/"allowed" algebraic operation. But it might make the formula messier instead of cleaner.
First of all, thank you very much Chris for another class. Secondly, I wanted to ask you what exactly is linearized gravity? I didn't quite understand this part
I should have explain it better. It's an approximation for weak gravity where we only consider small changes away from the flat metric.
@@eigenchris So does it mean that for small perturbations in space-time the quantities related to field equations such as the Ricci tensor and the Christoffel symbols remain in their original form?
@@victormarchant217 What do you mean by "their original form"? My goal for the video was to show the new equations for all the tensors/symbols under the linearized gravity assumption.
@@eigenchris In my view, some of the equations maintain a form that was already known only with the metric tensor gμv, as if you only replaced gμv with hμv in the equations, as if it were an invariant transformation, but you said that you get new equations.
maybe I have to review the concepts of linearity and invariance, I believe I'm confusing these two things
❤️
cool
Yeah, I'm watching this but truthfully, he lost me at the one minute mark but I continued to the end.
Sorry to hear that. It's a very algebra-heavy video. Is there something I can do to help?
I'd recommend watching TC 21 and up for the Riemann Curvature Tensor and Ricci Tensor. They are excellent videos
@@eigenchris Just keep producing this content. It is above my level but that doesn't mean I didn't enjoy it. I watch/enjoy the PreMath channel RUclips videos but there is something you may notice about how he explains his problems/content. I am not saying that you should do this or change, it is just that every step in working a problem/concept he goes back to fundamentals. Thanks for allowing me feedback.
@@prbmax I agree with Justin that TC 21-26 might help with this. But for my 108-110 videos, I'm assuming viewers are pretty comfortable with the content from Relativity 107. I'd be curious to know if you've seen those too? Most of these videos aren't really meant to be watched on their own.