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Light as a particle wrong?

When E=hc/λ and E=mc^2........eq. them we get (m1+m0)c=h/λ
can we take m=0 if its a photon and convert eq. into h=0 or there is some mass in a photon when its moving

Reply 1

username1560589

A photon has no mass.
Converting all of a photon's energy into mass(or vice versa), you would equate these equations, giving hc/lambda = mc^2
m = h/(lambda * c)
The particle produced has mass, but is not a photon.(or the mass gets destroyed to produce a photon)

Reply 2

Phichi

Original post by morgan8002

The photon cannot just transform it's energy into mass unfortunately, but could create a pair of particles when interacting with another particle, of which could be unchanged in the interaction.

Original post by 1DA

When E=hc/λ and E=mc^2........eq. them we get (m1+m0)c=h/λ
can we take m=0 if its a photon and convert eq. into h=0 or there is some mass in a photon when its moving

The equation you're referring to here

E=mc^2

has many forms, and this strictly isn't the full equation. A photon has no mass, but has energy, due to the fact it has momentum.

This equation may shed more light for you:

E^2 = (pc)^2 + (m_0c^2)^2

However this is usually represented by

E=\gamma m_0c^2

Which links both rest mass and momentum to energy. For a massive particle with zero momentum however, this would reduce down to the equation you suggested

E=mc^2

(edited 9 years ago)

Reply 3

tomtjl

Original post by Phichi

The photon cannot just transform it's energy into mass unfortunately, but could create a pair of particles when interacting with another particle, of which could be unchanged in the interaction.

The equation you're referring to here

E=mc^2

has many forms, and this strictly isn't the full equation. A photon has no mass, but has energy, due to the fact it has momentum.

This equation may shed more light for you:

E^2 = (pc)^2 + (m_0c^2)^2

However this is usually represented by

E=\gamma m_0c^2

Which links both rest mass and momentum to energy. For a massive particle with zero momentum however, this would reduce down to the equation you suggested

E=mc^2

I don't think he was saying that all of a photon's energy could be converted to mass, but theoretically if it were translated to mass then that's the equation that you would use, just to shed a little light on the matter.

@OP - Phici has it down to be honest. The E=MC^2 equation is only for objects at rest, since photons are moving their momentum has to be included as well.

Also, as a general rule, plugging "x=0" (where x can be any variable) into an equation usually yields few results. If we take your example we get that 0*lambda = 6.63x10-34 which just doesn't make sense. Even if we took h as 0, we'd get that 0*lambda = 0 meaning that lambda could take on any size it wants and the equation would hold. None of this makes sense or yields useful information and inevitably leads to 0/0 = lambda, which again makes no sense.

Reply 4

1DA

When eq. is represented by

and we know that γ=0 when the P=0 by eq. P=γm0*v so by putting it in eq.1 we would get E=0.

to prove this when we put γ=P/m0c for photons in eq.1 then we get result to be E=pc that's right for m=0
.......

Reply 5

Phichi

Original post by 1DA

When eq. is represented by

You're doing exactly what someone else in this thread said not to do, you can't just stick

p=0

into

E=\gamma m_0v

and then say that

\gamma=0

as a consequence. Secondly in the case of a photon, it has both momentum and velocity. When you stick zero into these equations of multiple variables, you are destroying solutions to your equations. E.g, if I told you;

x=0, y=2, z=0

With the equation

xy=z

you could say that

y=0

, when clearly it isn't.

Also with these wild assumptions, you are mistaking what

\gamma

is, perhaps you should google the lorentz factor. As

\gamma \geq 1

in every case.

If you let

m_0=0

E^2 = (pc)^2 + (m_0c^2)^2

you'll get

E=pc

You've derived the equation on the assumption

m_0=0

which would cause your fundamental step to fail.

\gamma=\frac{p}{0}

Also

p=\gamma m_0v

is an indeterminate form, and can only be evaluated properly as a limit.

(edited 9 years ago)

Reply 6

Kallisto

Entertainment Forum Helper

Original post by 1DA

When E=hc/λ and E=mc^2 (...)

Could it be that you confuse an issue with one another? Einstein's well known formula means that mass of material in form of particles can be changed to Energy and vice versa.

That is to say Enstein's formula does not mean the mass of light. So it makes no sense to equal the Energy of light with the theory of relativity.

Reply 7

Phichi

Original post by Kallisto

The full Einsteinian equation where the total momentum is non-zero leads is the energy-momentum relation. Which in turn can reduce down to the energy-mass equivalence. It makes sense to equal the 'energy of light with the theory of relativity', however the case where

E=mc^2

is purely a reduction relying on the fact momentum is zero, in the case of light, it is not.

Reply 8

Kallisto

Entertainment Forum Helper

Original post by Phichi

E=mc^2

is purely a reduction relying on the fact momentum is zero, in the case of light, it is not.

I see. The Einstenian equation could really have be used for all kinds of materials, if the fact what you have written above would not apply. But I wonder why the momentum is zero. The photon is always in motion with an incredible velocity, the fastest one anyway, if I am not mistaken. So what is the reason for zero momentum?

(edited 9 years ago)

Reply 9

Phichi

Original post by Kallisto

I think you've misunderstood what I said. I mentioned the fact that the energy-momentum relation will boil down to

E=mc^2

if the momentum is zero, however in the case of a photon, this is not true, hence you must use the energy-momentum relation, which is technically the full version of the Einsteinian equation;

E^2=(pc)^2+(m_0c^2)^2

Original post by Phichi

however the case where

E=mc^2

is purely a reduction relying on the fact momentum is zero, in the case of light, it is not.

(edited 9 years ago)

Reply 10

Kallisto

Entertainment Forum Helper

Original post by Phichi

I think you've misunderstood what I said. I mentioned the fact that the energy-momentum relation will boil down to

E=mc^2

if the momentum is zero (...)

Yeah, I have misunderstood you but this statement made it clearer. Thank you.

Reply 11

Phichi

Original post by Kallisto

Yeah, I have misunderstood you but this statement made it clearer. Thank you.

No problem :smile:

Reply 12

WishingChaff

I posted about this in a previous forum. This is a common mistake amongst physicists. I have copied my previous post from before.

"This is not necessarily true. Most people make the same mistake here.

The key point is to distinguish between light-like particles and time-like particles. In the first case,

E = \gamma m c^2

, this holds for particles travelling along time-like trajectories. The latter formula can be used to describe both light-like particles and time-like particles. It is this confusion that often leads to questions regarding massless particles having no energy i.e.

E = mc^2

, and

m = 0

then surely this implies

E=0

.

An interesting question here is the following. What energy-momentum relation would space-like particles be described by? You will find these types of particles have some very strange properties."

The point that people miss is the geometry of special relativity (and to some extent general relativity). In the common construction of special relativity (as people are using here) we have different regions (geometries) in which these equations are valid. Another key example, in which these types of questions arise, is when considering physics near event horizons. Sometimes the problem can be inherently physical, or can be due to a poor choice of coordinates. i.e. Eddington-Finkelstein coordinates.

This is one of the key points of special and general relativity, namely that the physics should not depend on the choice of coordinate system (essentially a statement of general covariance). This is where it is much better to recast the language of general relativity into that of differential geometry and primarily that of differential forms.

These problems also arise when you wish to consider general Lagrangians for particle motions in background independent space-times (i.e. Manifolds with non-dynamical metric tensor fields.)

I realise this may be a bit more detailed than the initial poster wanted, however I added more as it seems there are a few undergraduates here who may benefit from this information.