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Dale
Dale
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So in my QFT course, my professor said that you can set $c$ and $h$ bar$\hbar$ to 1. And he gave us an example:

$$E = mc^{2}$$

And then set $c = 1$:

$$E = m$$

This seems completely ludicrous to me to do. Doesn't it change the result? Why can this be done and why isn't it wrong?

I mean, $E = mc^{2}$ gives you one answer and $E = m$ gives you another, completely different answer!

So in my QFT course, my professor said that you can set $c$ and $h$ bar to 1. And he gave us an example:

$$E = mc^{2}$$

And then set $c = 1$:

$$E = m$$

This seems completely ludicrous to me to do. Doesn't it change the result? Why can this be done and why isn't it wrong?

I mean, $E = mc^{2}$ gives you one answer and $E = m$ gives you another, completely different answer!

So in my QFT course, my professor said that you can set $c$ and $\hbar$ to 1. And he gave us an example:

$$E = mc^{2}$$

And then set $c = 1$:

$$E = m$$

This seems completely ludicrous to me to do. Doesn't it change the result? Why can this be done and why isn't it wrong?

I mean, $E = mc^{2}$ gives you one answer and $E = m$ gives you another, completely different answer!

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Qmechanic ♦
Qmechanic ♦
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Why can we set c$c$ and h bar$\hbar$ to 1 when it changes the result?

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Tachyon
Tachyon
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Why can we set c and h bar to 1 when it changes the result?

So in my QFT course, my professor said that you can set $c$ and $h$ bar to 1. And he gave us an example:

$$E = mc^{2}$$

And then set $c = 1$:

$$E = m$$

This seems completely ludicrous to me to do. Doesn't it change the result? Why can this be done and why isn't it wrong?

I mean, $E = mc^{2}$ gives you one answer and $E = m$ gives you another, completely different answer!