Is Special Relativity Incomplete Without Considering Intrinsic Momentum?

[Goccam]

A little problem with special relativity.

The equation for general relativity E^2= M^2c^4 + P^2c^2 is usually reduced in special relativity using the argument:
If the mass is at rest, momentum is zero, therefore
E^2 = 0 + M^2c^4
Therefore
E = Mc2.
Similarly if M is zero,
E^2 = p^2c^2 + 0
And therefore E = pc

The problem with this is that pc is only = to mc^2 if m is at rest.

If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
Comments please.

 

[malawi_glenn]

??!

1) Why are you posting this 3 times??

2) Why are you posting this in "High energy, Nuclear, Particle physics"-forum?

pc can NEVER be equal to mc^2 if the particle is at rest, if the particle is at rest, then pc=0.

Also kinetic energy is not equal to 1/2Mv^2 in special relativity..

 

[Doc Al]

The equation for general relativity E^2= M^2c^4 + P^2c^2 is usually reduced in special relativity using the argument:

That equation is a result of special relativity.

If the mass is at rest, momentum is zero, therefore
E^2 = 0 + M^2c^4
Therefore
E = Mc2.

OK. That's the energy of a particle with mass M at rest (its rest energy). Meaningless for photons.

Similarly if M is zero,
E^2 = p^2c^2 + 0
And therefore E = pc

OK. This applies for photons (and other massless particles).

The problem with this is that pc is only = to mc^2 if m is at rest.

Huh? If you're talking about a massive particle at rest, pc = 0, not mc^2. (Again, meaningless for photons.)

If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.

Huh? A true statement would be that for a massive particle in motion, the total energy is greater than mc^2 (the rest energy) by a kinetic energy term. That kinetic energy term is not simply 1/2mv^2 (that's only true nonrelativistically).

As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.

Sorry. No idea what you're talking about here.

Edit: I merged the multiple threads you created. Once is enough!

 

[malawi_glenn]

What is:

"intrinsic momentum" and "inertial momentum" ?

 

[yuiop]

A little problem with special relativity.

The equation for general relativity E^2= M^2c^4 + P^2c^2 is usually reduced in special relativity using the argument:
If the mass is at rest, momentum is zero, therefore
E^2 = 0 + M^2c^4
Therefore
E = Mc2.
Similarly if M is zero,
E^2 = p^2c^2 + 0
And therefore E = pc

The problem with this is that pc is only = to mc^2 if m is at rest.

If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
Comments please.

The total energy of a particle is $$E^2= p^2c^2 + m_0^2c^2$$ where m0 is the rest mass. A photon has a rest mass of zero so for a photon E=pc. You might argue that since p=mv and the mass of a photon is zero then p=0 and E=0 but that would be confusing rest mass with inertial mass. The momentum in terms of rest mass is $$p=\frac{m_0c}{\sqrt{(1-v^2/c^2)}}$$

so for a photon p=0/0=inderminate. That means there is insufficient information to determine momentum for a photon using that equation. The required information is that the energy of a photon is E= hf where h = Planck's constant and f=frequency. So for a photon hf=pc and momentum p=hf/c=mc and m=hf/c^2 and E=mc^2=pc=hfc^2/c^2=hf. In fact it is a mute point whether it is valid to ask what the rest mass of a photon is, when there is no valid reference frame where the photon is at rest.

For a particle the rest mass is given by:

$$m0 = \frac{E}{\sqrt{c^4+\frac{v^2}{(1-v^2/c^2)}}}$$

and when v=c it can be seen that m0=0. In short a particle like a photon without rest mass still has inertial mass, momentum and kinetic energy. In relativity, Total Energy = Kinetic Energy + Rest Energy and in the case of a photon the Total Energy is all due to Kinetic Energy. In short a photon has inertial mass, momentum and Kinetic Energy despite having zero rest mass.

 

[Goccam]

First, apologies with the multiple posting, I had finger trouble with the module.
I obviously did not ask the question correctly , by pc = to Mc^2 I was talking about the equivalence,

In relativity, Total Energy = Kinetic Energy + Rest Energy and in the case of a photon the Total Energy is all due to Kinetic Energy. In short a photon has inertial mass, momentum and Kinetic Energy despite having zero rest mass.

And of course if the mass is at rest, the total energy is the rest mass energy, and if p = mc,
Then (mc)c is the same as mc^2. And of course the other equivalent forms of writing energy, hf, frequency, wavelength, electron volts, are all related to rest mass
However if the mass is not at rest, the total energy = kinetic energy + rest energy, and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.

 

[Doc Al]

However if the mass is not at rest, the total energy = kinetic energy + rest energy,

Right.

and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.

Not right. Those expressions for momentum and kinetic energy are only valid non-relativistically: for speeds much lower that the speed of light. (In his post, kev gives the relativistic expression for the momentum of a massive particle.)

 

[tiny-tim]

Welcome to PF!

If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
Comments please.

Hi Goccam! Welcome to PF!

No, the "additional kinetic energy component" is:

E = 1/2mv^2 - 1/24mv^4 + 1/720mv^6 - …

"intrinsic momentum" is not a vector or a tensor or anything else geometric.

Nor is "inertial momentum".

Only by adding the two do we get a vector!

We add the two only because the result fits into a mathematical category (of vectors).

It is virtually impossible to talk about anything which fits into no mathematical category … and why would one want to?

Does that help?

 

[yuiop]

...
However if the mass is not at rest, the total energy = kinetic energy + rest energy, and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.

In relativity Kinetic Energy = Total energy -Rest energy = $$\frac{m_0c^2}{\sqrt{(1-v^2/c^2)}}-m_oc^2$$

which is not the same as 1/2mv^2 except at very low velocities.

 

[malawi_glenn]

However if the mass is not at rest, the total energy = kinetic energy + rest energy, and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.

no, the momentum is:

$$p = \frac{m_0v}{\sqrt{1-(v/c)^2}}$$

And kinetic energy is:
$$E_k = (\gamma - 1)m_0c^2$$

Where: $$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$$

Now to your "question", can you defice intrinsic vs. inertial momentum for us?

You seems to be stuck in the old Newtonian system and its definition of linear momentum...

 

[Goccam]

OK
Thanks very much