# Zwiebach page 183

1. Oct 10, 2007

### ehrenfest

Zwiebach page 182

1. The problem statement, all variables and given/known data
Zwiebach says equation 10.107 means h_II is traceless when p^2 equals 0. But it seems to me like p^2 = 0 would mean that every element in the diagonal would need be zero as opposed to only their sum? There is no implied summation in 10.107 is there?

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 10, 2007
2. Oct 10, 2007

### nrqed

That's not what he says. The fact that h is traceless comes from Eq 10.100 (see also 10.101).

Then, when he gets to 10.107, he says that if p^2=0, there is no condition at all on h^(IJ) arising from 10.107. But then he reminds the reader that H^(IJ) is not completely free since it must obey the equation 10.101, i.e. be traceless.

So the traceless condition does not arise from 10.107.

But if p^2 =0, it means that we know nothing at all about the elements h^(IJ)!! I don't see why you say that it tells us that all the elements are zero!

3. Oct 11, 2007

### ehrenfest

Okay, so it seems like there is an implied summation in 10.101 since apparently h^II really means the h_11 + h_22 + h_33 + h_44 ... h_dd = 0.

4. Oct 11, 2007

### nrqed

Zwiebach uses Einstein's summation convention: whenever two indices are repeated, a summation over those that index is implied (there *are* some equations where an index is repeated but the summation is not implied. In those cases the author usually says explicitly that a summation is not implied. But if the author says nothing and an index is repeated, summation is implied).

5. Oct 11, 2007

### ehrenfest

In this equation, the index I appears twice as a superscript. I thought the Einstein summation convention only summed over indices that appeared once up and once down?

6. Oct 11, 2007

### nrqed

You are right if one uses a covariant formalism. But if one uses a noncovariant formalism like the light-cone coordinates, then one usually extends the summation convention to include indices that are both upstairs or downstairs.

7. Oct 11, 2007

### ehrenfest

I see. Thanks.