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paradox 3_14_16 REVIEWED
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Working note by Phil, dated 3.14.16 with a 5.17.16 comment, in the Chapter 10 development files of the Wedge World tensor wedge document. It shows two conflicting results for <Rt us | R tui>, Ris versus δsi, and tests whether φ*T = RT is wrong. It also looks at raising and lowering indices on the basis vectors xei defined by xei = R tui. A later comment says the paradox came from a badly used index and that he no longer works in the φ* = R framework.
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Paradox 3_14_16 involving Section 6.10 PhL 3.14.16
5.17.16: This same paradox keeps popping up where I have a badly used index. But also I am still in the framework of operator φ* = R which I know longer work with, so have lost interest in what is below. This paradox probably appears in 5 docs overall!
1. On the one hand, I know this
en em = g'nm
en em = δnm
en em = g'nm . (7.18.1)' (2.3.2)
and applying this to my x-space of Section 6.10 this says
< xes| xei> = xes xei = δsi
2. On the other hand, I am trying to claim that
(a) φ*T = RT and φ* = R
for the pullback matrix. I believe this from two sources:
I derive it in "is phi star symmetric v2"
It comes out at the end in "Section 10_6 v2".
Now consider my definition of φ*T as stated in Ch 10
(b) φ*T(xei) ≡ Σj=1n Rij (tuj) i = 1..m (10.6.9)
Based on what is stated above (that φ*T = RT), this implies that
(c) RT | xei> = Σj=1n Rij | tuj> .
Can I prove this claim directly? I know from (10.6.6) that
(d) (xei) = R(tui) or |xei> = R| tui> i = 1..n
Inserting this for |xei> in the previous result gives
RT [ R| tui>] = Σj=1n Rij | tuj>
or
(e) RTR | tui> = Σj=1n Rij | tuj>.
Now close this with basis vector <tus| to get
(f) <tus| RTR| tui> = Σj=1n Rij <tus| tui> = Ris
since <tus| tui> = δsi. Now move RT to the left as usual to get
(g) <Rt us | R tui> = Σj=1n Rij <tus| tui> = Ris
which is just a rewrite. Now, as noted above,
|xei> = R| tui> = | R tui>
<xei| = < R tui |
(h) <xes| = < R tus |
Then
(i) <Rt us | R tui> = <xes| xei> = δsi
3. So the paradox is that I have these two conflicting results.
<Rt us | R tui> = Ris
<Rt us | R tui> = δsi
One possible resolution would be this
φ*T = RT is WRONG.
Comment 1. Recall from 10.6 that
[φ*T(xei)]s = Ris . (10.6.11)
How would you write the LHS?
< tus| [ φ*T | xei>] = Ris ?
< tus | φ*T | xei>] = Ris ?
<xei | φ* | tus>] = Ris ? real scalar product
Plan A:
If you define ei the right way, it MUST do ei ej = δij. So now did I define ei in this case?
Go back to this statement from 10_6 v2: which defines xei .
(xei) = R(tui)
or
(xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.4)
How might we raise the label i ? From the second line it would seem that
(xei)j = Rji (xei)j = Rji
and (xei)j = Rji
but the first line is less obvious. Maybe rewrite the second line as
(xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji
Then you could write
(xei) = Rut (tui)
and then we have these two lines:
(xei) = Rdt(tui)
(xei) = Rut(tui)
Summarizing the second case:
(xei) = Rut(tui)
(xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji
and this agrees with what I had.
Plan A:
If you define ei the right way, it MUST do ei ej = δij. So now did I define ei in this case?
I start with this:
(xei) = R(tui)
as definition of the xei . I write this out as
(xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.4)
How might we raise the label i ? From the second line it would seem that
(xei)j = Rji (xei)j = Rji
and (xei)j = Rji
but the first line is less obvious. Maybe rewrite the second line as
(xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji
Then you could write
(xei) = Rut (tui)
and then we have these two lines:
(xei) = Rdt(tui)
(xei) = Rut(tui)
Summarizing the second case:
(xei) = Rut(tui)
(xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji
and this agrees with what I had.