E=pc question

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username970964
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I know m=0 because it's 0 for a photon. And I know how to write its 3-momentum. But I don't know how to work out its momentum value in GeV/c. I know I have to use the equation E=pc. Do I need to use p=mc as well? The lecturer said p=1 GeV/c but I don't know how she got that.

For p=E/c, p=1/3x10^8 and that doesn't come out as 1.

For p=mc, m=0 so p=0 which is wrong as well.
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Eimmanuel
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(Original post by Airess3)
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I know m=0 because it's 0 for a photon. And I know how to write its 3-momentum. But I don't know how to work out its momentum value in GeV/c. I know I have to use the equation E=pc. Do I need to use p=mc as well? The lecturer said p=1 GeV/c but I don't know how she got that.

For p=E/c, p=1/3x10^8 and that doesn't come out as 1.

For p=mc, m=0 so p=0 which is wrong as well.
This is just simple algebraic manipulation which you have gotten.

 E = pc => p = \dfrac{E}{c}

 p = \dfrac{1 \text{GeV}}{c} =1 \dfrac{\text{GeV}}{\text {c}}

When the unit includes c, you don't times or divides the numerical values with the speed of light.
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alow
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(Original post by Eimmanuel)
This is just simple algebraic manipulation which you have gotten.

 E = pc => p = \dfrac{E}{c}

 p = \dfrac{1 \text{GeV}}{c} =1 \dfrac{\text{GeV}}{\text {c}}

When the unit includes c, you don't times or divides the numerical values with the speed of light.
The "implies that" symbol can be printed in \LaTeX by using the \implies command: \implies.
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