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Karozo
Karozo
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At the end of spontaneous symmetry breaking I get these mass terms:

$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$$$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp i W_{\mu}^{2} \bigr )$$

$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

So I have $$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$

Is it right? Or there are too many terms and it is enough:

$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$

At the end of spontaneous symmetry breaking I get these mass terms:

$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$$

$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

So I have $$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$

Is it right? Or there are too many terms and it is enough:

$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$

At the end of spontaneous symmetry breaking I get these mass terms:

$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp i W_{\mu}^{2} \bigr )$$

$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

So I have $$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$

Is it right? Or there are too many terms and it is enough:

$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$

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Qmechanic ♦
Qmechanic ♦
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At the end of spontaneous symmetry breaking I get these mass terms:

$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$$

$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

So I have $M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$

Is it right? Or there are too many terms and it is enough:

$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$

At the end of spontaneous symmetry breaking I get these mass terms:

$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$

$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$

So I have $M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $

Is it right? Or there are too many terms and it is enough:

$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $

At the end of spontaneous symmetry breaking I get these mass terms:

$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$$

$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

So I have $$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$

Is it right? Or there are too many terms and it is enough:

$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$

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Karozo
Karozo
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Glashow-Weinberg-Salam mass terms

At the end of spontaneous symmetry breaking I get these mass terms:

$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )$

$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$

So I have $M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $

Is it right? Or there are too many terms and it is enough:

$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $