0

I haven't found a satisfactory answer to this question.

In special theory of relativity

$$E=\sqrt{m_{0}^2c^4 + p^2c^2}.$$

When we consider photons where $m_{0}=0$ then $E=pc$ but we also know that

$$E= m_{0}c^2\left(1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\dots\right.\,.$$

If we put $m_{0}=0$ in this equation then $E=0$.

It immediately follows from $E=pc$ that $p=0$. It means that photons neither have energy nor momentum.

It seems that I'm going wrong somewhere because moving photons definitely have kinetic energy.

Ruslan
  • 28,862
Shreyansh Pathak
  • 3,339

3 Answers3

5

The second conclusion doesn't quite work because the second equation comes from the Taylor expansion of $\gamma$, i.e. $\frac{1}{\sqrt{1-v^2/c^2}}$. Since photons move at the speed of light, this factor is infinite, and infinity multiplied by zero is undefined.

Alternatively, you can write out more terms in the Taylor expansion, and you should see that when you add them all up with $v=c$ you get infinity.

Allure
  • 20,501
  • Doubt: then how would we be able to calculate energy? – Shreyansh Pathak Apr 29 '20 at 06:17
  • 1
    @Sarcasm for a massles particle such as a photon you can't derive energy or momentum from speed (because it's always $c$). You have to use the quantum expression $E=h\nu$ and calculate them from frequency. – Ruslan Apr 29 '20 at 08:09
2

Two ways :

$$E = \sqrt{ m_0^2c^4 + p^2c^2 }$$

First try $m_0=0$ and you get :

$$E = \sqrt{ p^2c^2 } = pc$$

which is perfectly correct.

Second try not dividing by zero (as you did when you expanded) and instead try dividing by $pc$ :

$$E = \sqrt{ m_0^2c^4 + p^2c^2 } = pc \sqrt{1 + \frac{m_0^2c^2}{p^2}}$$

Expanding that square root gets you an expansion with terms of powers of $m_0$ except for the constant term $1$ at the start and that results in $E=pc$ as expected.

Your error is dividing by something you plan to set to zero, which is undefined in this context. What you tried was (in effect) :

$$\sqrt{1+\frac 1 0 } = 1 + \frac 1 {2\times 0} + ...$$

which of course makes no sense.

StephenG - Help Ukraine
  • 9,034
2

How one takes limits is very important...

Consider the energy-momentum diagram showing the 4-momentum of a particle with nonzero mass.

The pink arrow represents the relativistic kinetic energy $T$,
which is the relativistic energy $E$ (the tip of the temporal leg)
minus
the rest-energy $mc^2$ (the intersection of the mass-shell with the temporal leg).

  • With fixed $E$, taking the limit of decreasing $m$ increases $T$. With zero $m$, you get $T\rightarrow E_{initial}$.
  • With fixed $p$, taking the limit of decreasing $m$ increases $T$. With zero $m$, you get $T\rightarrow p_{initial}c$.
  • With fixed $v$, taking the limit of decreasing $m$ decreases $T$. With zero $m$, you get $T\rightarrow 0$.

robphy-energyMomentum-limiting-KineticEnergy

robphy
  • 11,748