The h Bar

Mr. Feynman

20:03

@imbAF $${a\!\!\!/}{b\!\!\!/}=a_\mu b_\nu\gamma^\mu\gamma^\nu$$

imbAF

I did open it

But I am still researching it

Mr. Feynman

So let's use the linearity of Trace and calculate $$\mathrm{Tr}({\gamma^\mu\gamma^\nu})$$

imbAF

I was about to

calculate that

I assume that a and b can be taken out of the trace right?

Mr. Feynman

That's the linearity I mentioned, yes.

imbAF

Now I need to calculate trace of two matrices

Mr. Feynman

20:06

How would you tackle this problem?

imbAF

Well the tr(AB)=tr(BA)

where A and B are matrices correct?

Mr. Feynman

That's correct and keep it in mind, but you have to do something else before that comes in handy

imbAF

But even if that is the case, I need to show how its the case

I can't simply take it as true

I am thinking

I mean the only thing I can do is the clifford thing

other than that

Mr. Feynman

@imbAF I mean, the cyclic property of trace? That's something you already know from linear algebra

@imbAF Then do it

I will stay 5 minutes more and then retire to my Japanese business

imbAF

tsoto mate

$Tr(2g^{\mu\nu}-\gamma^\nu \gamma^\mu)$

Mr. Feynman

20:11

Right, now?

imbAF

than you have $Tr(\gamma^\mu \gamma^\nu)=-4 - Tr(\gamma^\nu \gamma^\mu)$

Mr. Feynman

It should be 8

imbAF

-8*

Mr. Feynman

Yeah you know what I mean

imbAF

cuz the metric is 1,-1,-1,-1

Mr. Feynman

20:14

Wait, why minus?

No no no no

imbAF

what?

Mr. Feynman

The trace is calculated with one index below and one index above

In other words, you can take the sum of diagonal components only if you have a mixed tensor

imbAF

what?

why?

mixed tensor?

Mr. Feynman

$\mathrm{Tr}{T}:=g_{\mu\nu}T^{\mu\nu}$

Michael

Hey everyone

Mr. Feynman

20:16

$T$ is a tensor with two upper indices

Michael

What is everyones specialty

imbAF

Ok?

Mr. Feynman

So $\mathrm{Tr}(g)=g_{\mu\nu}g^{\mu\nu}=\delta^\mu_\mu=4$

imbAF

I don't understand

the metric tensor is 1 -1 -1 -1

the race is -2 however you look at it

Mr. Feynman

I suggest to search the definition of the trace

3 mins ago, by Mr. Feynman

$\mathrm{Tr}{T}:=g_{\mu\nu}T^{\mu\nu}$

imbAF

20:19

isn't the sum of the main diagonal elements?

Mr. Feynman

No, it only is for tensors with one upper and one lower index

imbAF

This is literally the first time I am being aware of this

Mr. Feynman

You are looking at $g^{\mu\nu}$ now

imbAF

never have I heard this before

Mr. Feynman

To do what you want (i.e. just sum the diagonal elements) you have to consider $g^{\mu}{}_{\nu}$

The notation clearly means that I lowered one index

imbAF

20:21

Yeah it clearly means

but I am dumbfounded from this

Mr. Feynman

But, in case that you don't know, for a metric tensor, it's always true that $g^{\mu}{}_\nu=\delta^\mu_\nu$

imbAF

clearly $g^{\mu\nu}$ if written as a matrix it is 1 -1 -1 -1

And you are telling me that, you cannot sum the elements cuz they don't give you the trace

@Mr.Feynman yeah I am aware of that

but that would be

$gg$

not g itself

Is more than understandable that the trace would change

since you are calculating it for two matrices that multiplied each other

Mr. Feynman

Whenever you are asked to compute the trace of a tensor with two indices up or two indices down, you have to lower/raise one index and then compute the trace that you known

imbAF

but if you would consider just $g^{\mu\nu}$ you get something different, because it is different than $gg$

Mr. Feynman

This is the practical rule

imbAF

20:24

I see

Mr. Feynman

It's just the most general definition of traces

I can explain the reason why if you want to know

imbAF

I mean I don't understand the logic but ok

I want to know why

and at the end tie it with my identity

Mr. Feynman

Okay, so, what's an important property of the trace?

imbAF

because having an 8 doesn't really solve it

Mr. Feynman

Why are traces of matrices interesting, in other words?

@imbAF It does

imbAF

20:25

Idk, maybe because they are conserved or something

Mr. Feynman

@imbAF Yes, invariance.

imbAF

always?

Mr. Feynman

No! That's our problem here!

Michael

Hi @Mr.Feynman, @imbAF

How are you guys

Mr. Feynman

If you sum the diagonal elements of a tensor with both indices above/below you won't get an invariant quantity

imbAF

20:26

Learning dirac algebra

Mr. Feynman

It's not Lorentz invariant

imbAF

the tensor with both indices above/below ?

Mr. Feynman

Its trace

imbAF

is not invariant under LT?

Michael

Am I muted?

imbAF

20:27

aha

Mr. Feynman

I mean if you take the naive trace

imbAF

@Michael You are not

Mr. Feynman

@Michael No, Michael. You're not muted. Hello.

Michael

Thanks. I'm looking for a partner and mentor for a human powered aircraft project. I was formerly a software engineer, now I'm studying aeronautics and biomechanics

Mr. Feynman

@imbAF if you're not a little bit familiar with tensor, this discussion will take longer than it should

imbAF

20:29

Depends what you mean

with that

Like I know tensors order etc

But in my lecture

in the homework

no special attention is given to this

at all

Like this is new to me, what you are saying

Mr. Feynman

What you really do in linear algebra is not computing traces of matrices; matrices are just matrices. You really compute traces of linear operators, which are represented by matrices

When we say that the trace is invariant we mean that once we express the operator in another basis i.e. its matrix changes, the trace is preserved

This you already know, don't you?

imbAF

Isn't the linear operator something from physics

Mr. Feynman

Ah

imbAF

but no I didn't know that

Mr. Feynman

Linear operator/linear map

imbAF

20:32

mapping from where to where? curious here

Michael

I guess no one is interested

Mr. Feynman

@Michael It appears that the users currently in the chat are not, we usually focus more on theoretical stuff in this room :P

imbAF

I don't do biomechanics and aeronautics Michael, sorry

Mr. Feynman

@imbAF From a vector space to another

that may be the same vector space

In our case, Minkowski space

imbAF

Ok

gammas would be the linear operators?

Mr. Feynman

20:35

You can think of gammas as operators but in another sense that we don't care about now. For all you care about now they are just matrices. The point is that - very roughly speaking - we have tensor indices and we should carry out the trace appropriately

imbAF

ok

Mr. Feynman

I'm afraid I can't give you the equivalent of a one-semester linear algebra course now, given that I also have to go. For the time being just keep in mind that whenever you have a trace, it means "contract with the metric tensor"

$\mathrm{Tr}{T^{\mu\nu}}:=g_{\mu\nu}T^{\mu\nu}$

imbAF

and what would the result be then?

Mr. Feynman

Of the original identity?

imbAF

$\mathrm{Tr}{T^{\mu\nu}}:=g_{\mu\nu}T^{\mu\nu}$

lhs you have a trace

rhs you don't

what is the result. You want to calculate trace. RHS doesn't have one

Mr. Feynman

20:38

The RHS is a number

It's Einstein convention

imbAF

Ok

Mr. Feynman

It's the definition of the trace

$\mathrm{Tr}{g^{\mu\nu}}:=g_{\mu\nu}g^{\mu\nu}=\delta^{\mu}{}_\mu=1+1+1+1=4$

imbAF

Ok I will take it at face value

but I still can't argue it if I am askd

asked how the 8 comes out

when you have $Tr(2g^{\mu\nu})$

Mr. Feynman

You don't argue definitions

imbAF

Ok. I will just say this

@Mr.Feynman .

Mr. Feynman

20:39

Yes, I'm sure that's understood in your course

It's not something your professor expects you to explain

It's like asking you to explain what's an integral

imbAF

Out of curiousity I will ask anyone in class the next day

see how many are aware of this trivial thing that you just said

Mr. Feynman

That being said, now you have $$\mathrm{Tr}(\gamma^{\mu}\gamma^{\nu})=8-\mathrm{Tr}(\gamma^{\nu}\gamma^{\mu})$$

Use the cyclic property in the RHS

imbAF

Before you go

Mr. Feynman

what

imbAF

Could I also say: $Tr(\gamma^\mu\gamma^\nu)=Tr(\gamma^\nu\gamma^\mu)=Tr(\frac{1}{2}\{\gamma^\mu,\gamma^\nu\})=4g^{\mu\nu}$

Mr. Feynman

20:44

Actually, yes

imbAF

I will ask what we discussed because I am really caught off guard

Mr. Feynman

It's almost the same thing done in reverse order

imbAF

I will ask it tmrw in clas

Ok

I will leave you to your nippon

ciao

Mr. Feynman

@imbAF Bye bye

ciao

imbAF

22:27

When we want to show that $\{\gamma_5,\gamma^\mu\}=0$ should we consider a specific value for $\mu$

or it can be shown without explicit value of it>?

ZeroTheHero

23:46

Interesting discussion on responsibility if AI/chatgpt’s providing incorrect information, albeit in a wholly different context: link.springer.com/article/10.1007/s00146-024-02096-7

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