Photons protons and C

[michaellxv (Michael)]

Taking this to a new thread.

Originally Posted by Stu Ward

It is the M part of E=mc2 that makes it all so difficult to accelerate anything but photons to the speed of light. Photons are massless but protons have mass.
The LHC is a particle collider smashing particles with mass at near light speed. If we collide photons then if E=mc2. E=0mc2. E=0.


Stu

I get E=mc2 in relation to nuclear reactions, but how does this relate to accelerating a particle to C?

[renormalised (Carl)]

Originally Posted by michaellxv

Taking this to a new thread.



I get E=mc2 in relation to nuclear reactions, but how does this relate to accelerating a particle to C?

The "m" part of the equation....do the maths with even the slightest amount of mass and the "E" part of the equation goes through the roof. Travel at "c" and the "E" part becomes infinite. In other words, accelerating any object with mass to "c" takes an infinite amount of energy. So, you cannot travel precisely at "c".

[michaellxv (Michael)]

Originally Posted by renormalised

The "m" part of the equation....do the maths with even the slightest amount of mass and the "E" part of the equation goes through the roof. Travel at "c" and the "E" part becomes infinite. In other words, accelerating any object with mass to "c" takes an infinite amount of energy. So, you cannot travel precisely at "c".

There's an assumption in your answer which I was hoping you would state. So I will ask the direct question.

Why does m increase as we accelerate? Is it not this which drives E to infinity?

[renormalised (Carl)]

M increases as we accelerate towards c because we are adding energy (E) to the system by accelerating it. M and E are equivalent, so an increase in E is the same as an increase in M. To achieve c would take an infinite amount of energy. Remember, this is not the rest mass of the object you're accelerating, this is the mass (M) it has relative to the motion it's undergoing. If you were to make an observation of this object as it passed by you, you would see its mass being equal to the object's rest mass multiplied by the acceleration it was undergoing. If you were to observe the object whilst traveling along with it, it would still appear to have the same mass as it had at rest. That's what they mean by Relativity.

Now, we'll look at the case of the photon. A photon is a particle/wave of light, as you would know. It can travel at the speed of light, "c", and only at that velocity (that velocity being dependent on the medium through which the light travels). The moment a photon is created, it is traveling at "c". A photon, as you would know, has a rest mass of zero. So, it has no mass. Therefore, even when a photon is traveling at "c" it gains no net energy therefore no mass....the only energy it has is the energy it was emitted at. So, if the body emitting the photons is producing radiation in the mm range, it produces microwaves. If in the 300-700nm range it's visible light and so on.

To answer the very basis of your question, E=mc^2 applies to your nuclear reactions in exactly the same manner as it applies to accelerating a particle in an accelerator. It's just a matter of different situations.

[michaellxv (Michael)]

I know I'll regret this, but here goes.

If I watch an object accelerate past me this requires additional energy being added to the object to make it accelerate. So the object has more mass. This one more or less makes sense. Except if the object is a photon.

Now if I accelerate myself I must use energy to do that. But you say I don't gain mass in my reference frame. Where in my reference frame does the energy manifest itself and what is stopping me getting to C if I have the same mass?

Some explanations of relativity make perfect sense and it doesn't seem that hard. Others make me think I never really understood it at all.

[renormalised (Carl)]

OK....In your own reference frame, you will appear not to be gaining any mass, if you're accelerating towards c. But to someone watching you from the sidelines, you appear to be getting heavier and heavier as you approach c. The energy which manifests itself in your own reference frame is in the energy you're expending in order to push yourself towards c. The closer to c you get, the more energy you need to expend in order to maintain that acceleration. Think of yourself as flying a rocket. In order to accelerate to and attain c, you'd need to expend an infinite amount of energy to do so. So you'd have to carry an infinite amount of fuel. Remember, I said energy and mass are equivalent. Infinite energy is equivalent to infinite mass....you can never travel at the speed of light. Only photons can do that (and other particles with zero rest mass...e.g. gravitons).

[AGarvin]

Hi all,

I could be wrong here so feel free to correct. The equation E=mc^2 is actually a shortened version of the full equation and is only used to demonstrate the relationship between mass & energy. It does not take into consideration momentum. Also, the m is actually rest mass. The full equation is :

E^2 - (pc)^2 = (mc^2)^2

where p is momentum and m is rest mass.

This translates to E^2 = (pc)^2 + (mc^2)^2

As p increases so does the energy, however rest mass actually stays the same. This is why an object with zero rest mass (ie, photons) must travel at c .... if m is zero then (mc^2)^2 would also be zero. A photon travelling at less than c would require m to be greater than zero.

Hope that's correct .

Andrew.

[CraigS]

Putting aside the equations (and the maths) for a minute, the constraints bounding all of this are:

- the Conservation of Momentum;
- the Conservation of Energy;
- the constant speed of light in a vacuum.

Momentum, (a measure of the quantity motion), is actually boundless (in classical Physics), and can be made to increase without limit. If velocity is limited by c, then the only variable left is m. Which is why m has to increase with increasing motion.

This all then changes, when viewed from a non-comoving frame of reference (which then reconciles it with Relativity).

… (my 2 cents worth, anyway … and how it makes sense to me, in Classical Physical terminology).

Cheers

[michaellxv (Michael)]

Coming as I do from a maths background the maths actually makes more sense to me. And almost looks obvious when I look at the equations.

It is the real world understanding that I struggle with, and why is it so!

So for photons m=0, therefore p=0 so the whole equation is 0. That's why they are free to travel at C.

[renormalised (Carl)]

If the maths is easy for you, then I can't see why the real world understanding should be such a problem. The maths applies here just as it does in any abstract rendition of the equations. You seem to be having trouble with reconciling the pure logic and applied sides of the equations.

[sjastro]

Originally Posted by AGarvin

Hi all,

I could be wrong here so feel free to correct. The equation E=mc^2 is actually a shortened version of the full equation and is only used to demonstrate the relationship between mass & energy. It does not take into consideration momentum. Also, the m is actually rest mass. The full equation is :

E^2 - (pc)^2 = (mc^2)^2

where p is momentum and m is rest mass.

This translates to E^2 = (pc)^2 + (mc^2)^2

As p increases so does the energy, however rest mass actually stays the same. This is why an object with zero rest mass (ie, photons) must travel at c .... if m is zero then (mc^2)^2 would also be zero. A photon travelling at less than c would require m to be greater than zero.

Hope that's correct .

Andrew.

Hello Andrew,

In the equation E^2 = (pc)^2 + (mc^2)^2

E is the total energy which is the sum of the particles rest energy and its kinetic energy.
It can be written as E= mc^2 + K where K is the kinetic energy.

Hence (mc^2 +K)^2 = (pc)^2+ (mc^2)^2

For a photon m=0 and E=K.
Hence E = pc or p=E/c

While K=0.5m(v^2) for Newtonian theory, in special relativity
K= mc^2(1/(1-(v^2/c^2))^.5-1)
If a photon has mass, K becomes infinitely large when v=c, hence m=0 for a photon to travel at c.

Regards

Steven

[AGarvin]

Originally Posted by sjastro

Hello Andrew,

In the equation E^2 = (pc)^2 + (mc^2)^2

E is the total energy which is the sum of the particles rest energy and its kinetic energy.
It can be written as E= mc^2 + K where K is the kinetic energy.

Hence (mc^2 +K)^2 = (pc)^2+ (mc^2)^2

For a photon m=0 and E=K.
Hence E = pc or p=E/c

While K=0.5m(v^2) for Newtonian theory, in special relativity
K= mc^2(1/(1-(v^2/c^2))^.5-1)
If a photon has mass, K becomes infinitely large when v=c, hence m=0 for a photon to travel at c.

Hi Steve, it's been a long time since I played with this stuff, but yep, that makes sense.

Originally Posted by CraigS

Which is why m has to increase with increasing motion.

The idea of mass increasing with motion is actually refering to the redundant concept of relativistic mass, which is rest mass plus kinetic energy (as per Steves post). From memory in both Special & General relativity m is always rest mass.

Cheers,
Andrew.