.Showing Gauss' Law is False with Photon Mass "m"

[saleem]

hi

I have this question, I need your help:

If the photon had mass "m" , show that the Gauss' law would no longer be true.
Note that the electric poential for a point charge would then have a form
V(r) = e/r exp ( -mc/h * r )

Thank you

 

[Meir Achuz]

Just find E=-grad\phi and show that the Gauss ilntegral for that E varies with
the radus of the sphere.

 

[sir-pinski]

for your question. First, it is important to note that the concept of a photon having mass is still a topic of debate and is not universally accepted in the scientific community. However, for the sake of this question, let's assume that the photon does have a mass "m".

Now, let's look at Gauss' Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). Mathematically, this can be written as ∫E⃗⋅dA⃗ = Qenc/ε0.

If we use the electric potential formula given in the question, V(r) = e/r exp (-mc/h * r), and apply it to Gauss' Law, we get:

∫(e/r exp (-mc/h * r) * dA) = Qenc/ε0

Since the electric field, E⃗, is equal to the negative gradient of the electric potential, we can rewrite the left side of the equation as:

-∫(∇V(r) * dA) = Qenc/ε0

Using the definition of electric potential, we know that ∇V(r) is equal to the electric field, E⃗, so we can rewrite the equation as:

-∫(E⃗ * dA) = Qenc/ε0

Now, let's consider the case where the photon has mass "m". This would mean that the electric potential formula for a point charge would no longer be the inverse square law, but rather V(r) = e/r exp (-mc/h * r). This means that the electric field, E⃗, would also have a different form.

If we substitute the new electric field formula into the equation, we get:

-∫(E⃗ * dA) = Qenc/ε0

-∫(e/r^2 exp (-mc/h * r) * dA) = Qenc/ε0

Since the electric field is no longer an inverse square law, the integral on the left side of the equation would no longer be equal to the enclosed charge divided by ε0. This means that Gauss' Law would no longer hold true, as the electric flux through a closed surface would not be equal to the enclosed charge divided by ε0.

In conclusion, if the photon did have