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Equation 24 · The Clock That Comes Back Wrong by Exactly Its Mass

What does this equation mean?

T(b)G(v)T(−b)G(−v)=1T(\mathbf b)G(\mathbf v)T(-\mathbf b)G(-\mathbf v)=\mathbb 1

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Inputs and operationsmathbb 1
Result or conditionT(mathbf b)G(mathbf v)T(-mathbf b)G(-mathbf v)
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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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TT

Symbol T

T is part of the quantity the equation computes from the expression on the right.

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bb

Symbol b

b is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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GG

Symbol G

G is part of the quantity the equation computes from the expression on the right.

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vv

Symbol v

v is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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How to interpret it

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What the article says around this equation

Represent the loop LG\mathcal L_{\rm G} on a Hilbert space. A spatial translation by b\mathbf b is the unitary T(b\mathbf b)=exp⁡(−ib⋅P/ℏ)\exp(-i\mathbf b\cdot\mathbf P/\hbar) , built from the momentum operator P\mathbf P ; a Galilei boost by v\mathbf v is G(v\mathbf v)=exp⁡(−iv⋅K/ℏ)\exp(-i\mathbf v\cdot\mathbf K/\hbar) , built from the boost generator K\mathbf K . The classical loop is T(b\mathbf b)G(v\mathbf v)T(-b\mathbf b)G(-v\mathbf v)=1\mathbb 1 . The operator loop is not automatically the identity, because K\mathbf K and P\mathbf P need not commute as operators even when translations and boosts commute as group elements.

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