Can the equation E = pc be applied to particles?

[Mr.somebody]

Homework Statement

Can the equation E = pc be applied to particles? Why or why not?

Homework Equations

The Attempt at a Solution

It can be applied to particles that DONT have a rest mass (photons, which are particles). It can not be applied to particles that have a rest mass (almost everything).

 

[Dr. Courtney]

OK start, but it would be more complete to consider the equations that express the relative energy of particles that do have rest mass.

 

[andrevdh]

So its only applicable to photons (and gluons)?
But you only make a stament and do not explain why it is so?

 

[gleem]

Try this. Evaluate the quantity

m²c⁴ - m₀²c⁴
where of course mc² = E

 

[PWiz]

where of course mc2 = E

Just a clarification of what gleem is saying:
E over here is the total energy of the object. So the actual formula for this will be $E= \gamma m_0 c^2$, where $$\gamma = \frac {1}{\sqrt{1-\frac{v^2}{c^2}}}$$. By m, gleem means the relativistic mass (an orthodox concept, really) $m=\gamma m_0$.
(I'm providing this clarification in case the OP accidentally uses the rest mass for the formula for E over here)

 

[andrevdh]

I get up to the point where it evaluates to E²(v²/c²)?

 

[PWiz]

I get up to the point where it evaluates to E²(v²/c²)?

OK. Now substitute $E=\gamma m_0 c^2$ for E. What does this reduce to? Do you know the relativistic momentum expression?

 

[andrevdh]

I think it comes to (pc)²?
Which approaches the energy-momentum relation from the other side.
How is this relevant to the original question?
We evaluated a difference between two terms and found
that they are related to the relativistic momentum of the entity?

 

[PWiz]

I think it comes to (pc)²?
Which approaches the energy-momentum relation from the other side.
How is this relevant to the original question?
We evaluated a difference between two terms and found
that they are related to the relativistic momentum of the entity?

It is relevant because you just derived the forumula $pc=\sqrt{{\gamma}^2m_0^2 c^4 - m_0 ^2c^4} = \sqrt{E^2-m_0^2 c^4}$, proving the fact that $E≠pc$ if $m_0 ≠0$, which I believe was your original question.

 

[andrevdh]

I'll think about it.
It seems to make sense.
Thank you.