Killing vectors - 1

This video deals with the process of how the Killing equation arises from the Lie derivative of the metric for some manifold. It interprets the solutions to the Killing equation as being isometries or symmetry directions on the manifold.

[@lengooi6125]

Thank you for taking the time to clearly explain this rather confusing subject. The detail analysis really helps and is something rarely found in most text on Gen Relativity.This really helps.

[@TensorCalculusRobertDavie]

Thank you Leng, it is nice to hear such words of generous praise.

[@lengooi6125]

Is there a place where your lecture notes are kept and is it accessible.? Thanks

[@TensorCalculusRobertDavie]

@@lengooi6125 Not just yet but I will at some future stage.

[@andrewzerwekh6855]

This really cleared up some details for me! Thanks for the guidance. Big fun!

[@TensorCalculusRobertDavie]

Hello Andrew and thank you for that. Glad you liked it.

[@szazorkan]

Why is the killing equation ∇iUj+∇jUi = 0 and not simply ∇iUj = 0 ? As i and j are dummy indices.... isn't it equivalent anyways ?

[@TensorCalculusRobertDavie]

In the context of Riemannian geometry, the Killing equation is typically stated as ∇iUj + ∇jUi = 0, where ∇ denotes the covariant derivative operator and U is a vector field. Regarding your question about whether ∇iUj = 0 is equivalent, it's important to understand the distinction between the two formulations. The equation ∇iUj = 0 indeed implies that the covariant derivative of the vector field U with respect to the index i is zero, but it doesn't necessarily capture the full symmetry of a Killing vector field. The Killing equation in its more general form ∇iUj + ∇jUi = 0 ensures that the derivative of the vector field U behaves symmetrically with respect to the indices i and j. This symmetry is crucial because it reflects the property of a Killing vector field to preserve the geometry of a manifold along certain directions. While i and j are indeed dummy indices, the equation ∇iUj + ∇jUi = 0 captures the covariant derivative's symmetry explicitly, which is essential for characterizing Killing vector fields and their associated isometries. So, while ∇iUj = 0 ensures certain properties of the vector field U, the full Killing equation ∇iUj + ∇jUi = 0 provides deeper insights into the geometric symmetries induced by Killing vector fields on a Riemannian manifold. I hope this helps clarify the rationale behind the specific form of the Killing equation!

[@TensorCalculusRobertDavie]

Imagine you have a vector field, which is like a bunch of arrows attached to each point on a surface. These arrows can represent things like velocity or force at each point. Now, in geometry, there are special types of vector fields called "Killing vector fields." These are like special patterns of arrows that, when you move them around on the surface, they don't change the shape of the surface. It's like if you have a rubber sheet and you stretch it or twist it, but certain patterns of arrows stay the same no matter what. The equation ∇iUj+ ∇jUi = 0 is a fancy way of saying that these special arrow patterns, when you take their "derivative" (which is like measuring how they change) in two different directions, they always balance out perfectly. It's like saying if you measure how the arrows change when you move them to the left and then to the right, the total change is zero. Now, someone might ask, "Hey, isn't it enough to just say that when you move the arrows in one direction, they don't change?" That's what ∇iUj = 0 means. But the full equation, ∇iUj + ∇jUi = 0, tells us even more. It tells us that no matter which two directions you choose, as long as they're different, the change in these special arrow patterns always balances out perfectly. This extra bit of symmetry is really important for understanding how these special patterns affect the shape of the surface. So, in simpler terms, the full equation captures more about the special way these arrow patterns behave on the surface, making it really useful for understanding the geometry of the space.

[@szazorkan]

@@TensorCalculusRobertDavie Thank you for the time spent explaining, but I am still not sure I've understood correctly. I agree about the conceptual interest of formulating like this, but, from a computational standpoint, is it false that, from the Killing equation, we can get : ∇iUj = - ∇jUi thus ∇iUj = -∇iUj thus ∇iUj = 0 ? Meaning that both formulations are equivalent.

[@TensorCalculusRobertDavie]

@@szazorkan Your second last line is the same as saying 2 = -2 which is clearly not correct. The reasoning presented in the question contains an error. Let's break it down: The Killing equation is ∇iUj + ∇jUi =0, which expresses the symmetry of the covariant derivative of a vector field U. The question attempts to derive ∇iUj = 0 from this equation, which is incorrect. Here's the breakdown of the reasoning and why it's flawed: Starting point: ∇iUj + ∇jUi = 0 (Killing equation) Incorrect step 1: Trying to isolate ∇iUj on one side: ∇iUj + ∇jUi = 0 ∇iUj = −∇jUi (Incorrect) Incorrect step 2: Attempting to simplify further: ∇iUj = −∇iUj (Incorrect) Incorrect conclusion: Concluding that ∇iUj = 0: This conclusion is incorrect because the step above (∇iUj = −∇iUj) is invalid. The error lies in assuming that ∇iUj = −∇jUi implies ∇iUj = 0, which is not true. The correct interpretation of the Killing equation is that it provides a relationship between the covariant derivatives of a vector field that ensures certain symmetries, but it does not imply that the individual covariant derivatives are zero. In summary, the two formulations of the Killing equation, ∇iUj + ∇jUi = 0 and ∇iUj =0, are not equivalent. The former expresses a specific symmetry property of the covariant derivative, while the latter would imply that the covariant derivative of U with respect to i is zero for all indices, which is not generally the case.

[@FunctionalIntegral]

Very nice and excellent explanation!

[@TensorCalculusRobertDavie]

Thank you for that. Glad it was helpful to you.

[@rickandelon9374]

Extremely helpful, thank you!

[@TensorCalculusRobertDavie]

Hello Sakun and thank you for your comment, it is much appreciated. I am glad to hear that it was useful to you.

[@askil2574]

At 5:20, in the fourth line evaluating L_u(w_iv_j), isn't it right that the second term is T_kj, not T_jk?

[@TensorCalculusRobertDavie]

Thanks for that. It is okay if T is symmetric, w_iv_j = w_jv_i.

[@askil2574]

@@TensorCalculusRobertDavie Yes, thanks

[@emrahboke983]

İ have a question.do you have a idea for integral of X(f)dx here X is a killing vector field on M ( M is a n dimensional manifold

[@TensorCalculusRobertDavie]

Hello Emrah. I made use of Killing vectors when looking at the question of the conservation of energy and momentum on the space-time manifold. Have a look at the following videos. kzbin.info/www/bejne/boPFaYmertF4ndU kzbin.info/www/bejne/jJrPd4uJbs17aNE kzbin.info/www/bejne/mH21f42pedN0rs0

[@bennyribeiroguedes8369]

Hello, Professor Robert! Would you indicate any reference (book) for studying vectors and Killing equation? Especially the part that deals with generators and how we got to them. Big hug, Benny.

[@TensorCalculusRobertDavie]

Thank you Benny. In the GR context there are many good books and free material on the web; www.physics.usu.edu/Wheeler/GenRel2013/Notes/GRKilling.pdf for more depth, core.ac.uk/download/pdf/32549495.pdf

[@bennyribeiroguedes8369]

@@TensorCalculusRobertDavie Hello good day! Thank you very much for your attention and for the directions! Big hug!!

[@asenhikov3122]

At 5:45 isn't the term in the {Luvj}, {vkdju^k} with a negative sign in front of it? and also why is the u^k with an upper indice, shouldnt it be with a lower?

[@TensorCalculusRobertDavie]

You should follow the pattern shown at the top of the page which in turn was derived on the previous page.

[@1.4142]

killed it

[@TensorCalculusRobertDavie]

Thank you!