What Is the Lightest Weight Measurable by a Sensitive Scales?

[Gokul43201]

Regardless of velocity, a photon will have zero mass.

$$m_r = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

$$m_r$$ is the relativistic mass, which increases with velocity [this is what your trying to say increases]. Now if $$m_0$$ is the rest mass, which is zero for a photon, the quotient would be zero. Therefore the relative mass of a photon is also zero and does not increase with velocity.

But if v=c, which it is, for a photon, then the denominator is also zero! So $m_r$ need not be zero.

 

[Hootenanny]

As I have many times before, I believe the discussion of any mass other than invariant mass is meaningless; it only works to add confusion to any matter in which it is discussed. It is not the 'mass' of a particle which actually increases (i.e. it doesn't get more massive) it is simply the momentum of the particle which increases faster than classically predicted.

 

[disregardthat]

So you mean that p=ma is not correct?
Momentum=mass*velocity

At large scales that is. So when the velocity is rising, the momentum is too, when the mass stays the same, but when you say the momentum is rising because of the high speeds comparable to light do you mean as another factour outside this equation, and not just velocity's effect on momentum?
Becasue if velocity rises, momentum does too, that is clear, but when velocity is c, the momentum is in relativity infinite, so that equation must be wrong.

 

[marlon]

Regardless of velocity, a photon will have zero mass.

$$m_r = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

$$m_r$$ is the relativistic mass, which increases with velocity [this is what your trying to say increases]. Now if $$m_0$$ is the rest mass, which is zero for a photon, the quotient would be zero. Therefore the relative mass of a photon is also zero and does not increase with velocity.

No, that's an incorrect conclusion. We need to v=c for the photon case and you will see that relativistic mass has no physical meaning in the case of a photon.

It is not the 'mass' of a particle which actually increases (i.e. it doesn't get more massive) it is simply the momentum of the particle which increases faster than classically predicted.

Exactly

marlon

 

[ranger]

Hi marlon, I'm kind of confused now. With regards to Hootenanny's reply which you quoted; strictly speaking of Momentum=mass*velocity or $$p = \frac{m_0v} {\sqrt{1 - v^2/c^2}}$$ won't the momentum be zero in both cases? Unless for situations like this, momentum is somehow found by inclusion of the wavelength as the particle has zero mass?

Edit: aw crap, looks like my latex formatting for relativistic momentum is messed up.

 

[Hootenanny]

Unless for situations like this, momentum is somehow found by inclusion of the wavelength as the particle has zero mass?

Exactly! The momentum of a photon is defined as $p = h/\lambda$ and can be derived from general energy expression $E = \sqrt{(pc)^2 +(mc^2)^2}$.

 

[marlon]

Hi marlon, I'm kind of confused now. With regards to Hootenanny's reply which you quoted; strictly speaking of Momentum=mass*velocity or $$p = \frac{m_0v} {\sqrt{1 - v^2/c^2}}$$ won't the momentum be zero in both cases?

Not in the case of a photon because you get 0/0, this formula does not apply to photons !

Unless for situations like this, momentum is somehow found by inclusion of the wavelength as the particle has zero mass?

Edit: aw crap, looks like my latex formatting for relativistic momentum is messed up.

Correctly. Hootenanny answered this question perfectly so i can only quote it to be clear :

Exactly! The momentum of a photon is defined as $p = h/\lambda$ and can be derived from general energy expression $E = \sqrt{(pc)^2 +(mc^2)^2}$.

marlon

 

[vivesdn]

As I have many times before, I believe the discussion of any mass other than invariant mass is meaningless; it only works to add confusion to any matter in which it is discussed. It is not the 'mass' of a particle which actually increases (i.e. it doesn't get more massive) it is simply the momentum of the particle which increases faster than classically predicted.

I've recently read an article that states exactly the opposite. Energy and mass are the same thing, just related by a factor (c^2).
According to this view, the relation of kynetic energy to total energy is a value between 0 an 1, given by 1-(1-v^2/c^2)^1/2. So, if you move at c, all your energy is kynetic energy. So mass at rest has no meaning for a photon as you'll never see a photon not moving at c.