Unraveling the Mystery of the Massless Photon

[Quarlep]

We know the equation E=pc or P=mc both of them are avaliable for photon.But photon is a massless particle.How can that be possible ?
Thanks

 

[ZapperZ]

What exactly is the problem here?

And have you read the FAQ in the Relativity forum?

Zz.

 

[jtbell]

P=mc

I guess you got this from p = mv which is non-relativistic. Photons are relativistic objects and therefore follow the relativistic equation $E^2 = (mc^2)^2 + (pc)^2$ where m is the invariant mass a.k.a. "rest mass" (yes, it's a silly name for the mass of a photon which is never at rest), which gives you your E = pc.

The usual equation for relativistic momentum in terms of velocity is
$$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$
but this doesn't work for photons because it gives p = 0/0 which is undefined mathematically.

 

[Quarlep]

There happened math error can you wrote it again please ?

 

[Dale]

Please read: https://www.physicsforums.com/showthread.php?t=511175

 

[xox]

We know the equation E=pc

Yes.

or P=mc both of them are avaliable for photon.

False, in the case of photon $$p=\frac{h}{ \lambda}$$

But photon is a massless particle.How can that be possible ?
Thanks

It isn't , you are making a mistake, see above.

 

[PAllen]

Is it possible OP means:

P = E/c = (E/c²)c = mc ?

Thus a gross abuse of relativistic mass, and another reason it has fallen out of favor.

 

[jtbell]

There happened math error can you wrote it again please ?

Are you referring to my equations? I see them OK. Are you using the PF app on a smartphone or other mobile device? I think LaTeX equations are properly visible only in a web browser. I can see them in Safari on my wife's iPad, as well as in Firefox on my main computer.

However, using the PF app on the iPad, I see only the LaTeX source code.

 

[DrStupid]

The usual equation for relativistic momentum in terms of velocity is
$$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$
but this doesn't work for photons because it gives p = 0/0 which is undefined mathematically.

It is not undefined. After substitution with
$$E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$
the result for the removable discontinuity is
$$\left| p \right| = \mathop {\lim }\limits_{\left| v \right| \to c} \frac{{E \cdot \left| v \right|}}{{c^2 }} \cdot \frac{{\sqrt {c^2 - v^2 } }}{{\sqrt {c^2 - v^2 } }} = \frac{E}{c}$$

 

[xox]

It is not undefined. After substitution with
$$E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$
the result for the removable discontinuity is
$$\left| p \right| = \mathop {\lim }\limits_{\left| v \right| \to c} \frac{{E \cdot \left| v \right|}}{{c^2 }} \cdot \frac{{\sqrt {c^2 - v^2 } }}{{\sqrt {c^2 - v^2 } }} = \frac{E}{c}$$

You cannot do this types of manipulations since both $$E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$ and $$p=\frac{mv}{\sqrt {1 - (v/c)^2}}$$ are valid ONLY for $$m \ne 0$$.

As such, the formulas cannot be applied to the photon, as you are trying to do in your calculations.

 

[DrStupid]

You cannot do this types of manipulations since both $$E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$ and $$p=\frac{mv}{\sqrt {1 - (v/c)^2}}$$ are valid ONLY for $$m \ne 0$$

Did you noticed the limes?

 

[xox]

Did you noticed the limes?

Yes, I noticed, your STARTING point is wrong, i.e. your error occurs BEFORE you take the limits.

 

[pervect]

We know the equation E=pc or P=mc both of them are avaliable for photon.But photon is a massless particle.How can that be possible ?
Thanks

Hopefully you've read the FAQ by now.

When we say that a photon has no mass, we mean that it has no invariant mass or rest mass. When you write P=mc, you are using relativistic mass, which is equal to E/c^2.

Thus the term "mass" is ambiguous. Modern textbooks and papers tend to use invariant mass, while older textbooks and popularizations still often use relativistic mass.

Given that we know that the "mass" in your second equation is the old-fashioned "relativistic mass", the second equation is just a restatement of the first

P = m_relativistic * c is the same as P = (E/c^2)*c = E/c which is the same as E = pc

As a side note, the general relativistic relationship between energy, momentum, and invariant mass is given by:

E^2 = (p*c)^2 + (m*c^2)^2

 

[DrStupid]

Yes, I noticed, your STARTING point is wrong, i.e. your error occurs BEFORE you take the limits.

What exactly is wrong with my starting point?

 

[xox]

What exactly is wrong with my starting point?

I thought this is quite clear.

 

[DrStupid]

I thought this is quite clear.

No, it's not. As I do not set m=0 it does not refer to my calculation. And as you noticed the limes you should be aware of that. So what's your point?

 

[xox]

It is not undefined. After substitution with
$$E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$
the result for the removable discontinuity is
$$\left| p \right| = \mathop {\lim }\limits_{\left| v \right| \to c} \frac{{E \cdot \left| v \right|}}{{c^2 }} \cdot \frac{{\sqrt {c^2 - v^2 } }}{{\sqrt {c^2 - v^2 } }} = \frac{E}{c}$$

You CANNOT start with $E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$ since it DOES NOT apply to photons. One day you'll get it.

 

[DrStupid]

You CANNOT start with $E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$ since it DOES NOT apply to photons. One day you'll get it.

I demonstrated how to do it. It's quite simple math. One day you'll get it.

 

[Dale]

Closed pending moderation