Three Questions About Energy-Mass Equivalence

[NYSportsguy]

I read somewhere that Einstein found out in 1905 that as an object approached the speed of light (c), the object's mass would increase. He also found out that as the object goes faster and approaches "c", it also gets heavier. In fact, at "c" itself, an object's mass and energy should both be infinite.

Now my questions are:

a) How come photons (quantized energy particles of light) can travel the speed of light and yet have no mass? Doesn't that somehow violate what Einstein said in 1905?

b) If photons carry energy, which I am sure they do, shouldn't they have mass considering E= MC²?

c) I read somewhere that in math equations relating to physics, when you get an answer that comes up "undefined" or that approaches an "infinity" vale, you are usually wrong somewhere and have to go back and either add some more missing variables or re-calculate the equation.

Shouldn't the fact that Einstein got a value approaching "infinity" when trying to calculate and object's mass and energy at the speed of light be a red flag of some sort?

 

[ideasrule]

a) How come photons (quantized energy particles of light) can travel the speed of light and yet have no mass? Doesn't that somehow violate what Einstein said in 1905?

Einstein's statement applies for massive objects. Photons have no mass to begin with.

b) If photons carry energy, which I am sure they do, shouldn't they have mass considering E= MC²?

E=mc^2 gives the energy that an object at rest has. A moving object has energy E=sqrt(m^2*c^4 + p^2*c^2). Since m=0, that simplifies to E=pc, which is exactly the energy that a photon has.

Shouldn't the fact that Einstein got a value approaching "infinity" when trying to calculate and object's mass and energy at the speed of light be a red flag of some sort?[/I]

It's a red flag indicating that the object cannot travel at the speed of light in the first place.

 

[NYSportsguy]

Einstein's statement applies for massive objects. Photons have no mass to begin with.



E=mc^2 gives the energy that an object at rest has. A moving object has energy E=sqrt(m^2*c^4 + p^2*c^2). Since m=0, that simplifies to E=pc, which is exactly the energy that a photon has.



It's a red flag indicating that the object cannot travel at the speed of light in the first place.

Ideasrule: Thanks for your replies. But now I have some more inquiries to bring up with some of your responses.

a) If Einstein's statement that as an object increases in mass and heaviness as it approaches "c" or goes faster and faster...how was he able to measure or prove this? And why do photons defy this correlation?

b) Your response to my second question on E=MC² and it not applying to photons was helpful and made sense. I was just curious in the equation E =pc for a moving photon...what does the "p" variable stand for?


c) Your answer to question three seems a bit like circular logic. Once again we have never seen an object with mass go the speed of light, though we can now probably achieve something close to this in the CERN accelerator, so how do we know that it's mass will reach infinity?

 

[Matterwave]

a) Einstein was a theoretical physicist. It wasn't his job to do the experiments. Other people have done the experiments and all (established) experiments have come out in agreement with theory. The statement "as an object increases in speed, it's mass increases" is sloppy language. An object's mass is invariant. What does increase as the speed increases is the kinetic energy. And it appears that as an object obtains more and more kinetic energy, it becomes harder and harder to accelerate it. So, it would appear that the higher the kinetic energy, the higher the "inertia". We call this "relativistic mass". You may want to search for "relativistic mass" on this forum. There are numerous topics discussing it.

b) p stands for the momentum of the photon. Although photons have no mass, they do have momentum.

c) We keep pumping the particles with more and more energy, but they never reach the speed of light. As we pump more energy into the particles, their percent increase in speed gets lower, in accordance with theory. This is a good indication that the theory is correct.

 

[dacruick]

people have always had a hard time explaining to me what inertia means. but i think I've come up with a nifty analogy. Inertia is to movement as specific heat capacity is to temperature change. any holes in that one? it seems pretty crisp

 

[Agent M27]

Dacrulck that would definitely work since inertia is just the amount a given mass resists accelleration (change in direction). I will start using that analogy, provided you don't mind, because it describes it perfectly, granted they know what specific heat is. To the op I had this same misconception of relativistic mass in that I thought the actual object got heavier because I was thinking in terms of classical mechanics taking into account friction. Obviously there is no friction in space so from our frame of reference the object appears to have gained mass via kinetic energy measurements, but it is a relativistic effect such as length contraction and time dilation. If any of that is incorrect please correct me. Hope that helped.

Joe

 

[NYSportsguy]

b) p stands for the momentum of the photon. Although photons have no mass, they do have momentum.



But by definition, doesn't p= mv? (Momentum = mass* velocity). If a photon has no mass, how can it have momentum?

 

[Matterwave]

p=mv is the classical definition of momentum. In relativity, $$p=\gamma mv; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}$$

As a photon's velocity is c, the gamma factor goes to infinity while the mass goes to 0 (and 0 times infinity is an indeterminate form). So, for a photon even this equation doesn't tell us much. We can get the momentum of the photon from the equation ideasrule gave: E=pc or p=E/c.

 

[NYSportsguy]

p=mv is the classical definition of momentum. In relativity, $$p=\gamma mv; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}$$

As a photon's velocity is c, the gamma factor goes to infinity while the mass goes to 0 (and 0 times infinity is an indeterminate form). So, for a photon even this equation doesn't tell us much. We can get the momentum of the photon from the equation ideasrule gave: E=pc or p=E/c.

You state "0 times infinity is an indeterminate form"...

But isn't zero x anything (including infiniti) still equal to ZERO? Once again, how can we get the momentum of a photon from the equation E = pc when p = zero?? You still didn't answer the question sufficiently.

 

[Matterwave]

0 times anything is 0 except for when you try to multiply it with infinity (or negative infinity). Doing this is kind of like trying to divide by 0. It's not well defined. That's why I suggested you use the equation E=pc. You can see from that equation that p=E/c so since photons have energy E, they have momentum E/c.

The equation p=mv doesn't apply for high speeds, and it doesn't apply for photons.

If you are to work within the model of relativity, you can't use p=mv. The momentum is derived differently in special relativity.

If you want a little more background. In quantum mechanics we find that the energy a photon has is: $$E=hf$$, so that if we use the equation p=E/c $$p=E/c=\frac{hf}{c}=\frac{h}{\lambda}$$

So since the photon has some wavelength lambda, it has a corresponding momentum. If you're asking where the equation p=E/c comes from, then I could only say that you'd have to study special relativity. A full derivation of such an equation is not within my scope of expertise.

 

[russ_watters]

But isn't zero x anything (including infiniti) still equal to ZERO? Once again, how can we get the momentum of a photon from the equation E = pc when p = zero?? You still didn't answer the question sufficiently.

P isn't equal to zero. You just quoted the equation that shows that!

 

[NYSportsguy]

P isn't equal to zero. You just quoted the equation that shows that!

Well duh, I wasn't the one that came up with that equation...I was questioning it. Someone else brought it up as a possible answer to my original question above, I just want to know why that would be correct considering photons usually don't have mass, so how can momentum (the "p" part of the equation) be possible?

 

[NYSportsguy]

If you want a little more background. In quantum mechanics we find that the energy a photon has is: $$E=hf$$, so that if we use the equation p=E/c $$p=E/c=\frac{hf}{c}=\frac{h}{\lambda}$$

So since the photon has some wavelength lambda, it has a corresponding momentum. If you're asking where the equation p=E/c comes from, then I could only say that you'd have to study special relativity. A full derivation of such an equation is not within my scope of expertise.

I actually have seen this formula before, $$E=hf$$, in quantum mechanics. I believe it describes the Energy of a Photon = the Photon's Frequency * Planck's constant.

And I realize that a Photon's Frequency can be re-written as [Frequency = speed of light/ wavelength.] But again, where is momentum in all of this?

 

[Matterwave]

If you don't accept the equation E=pc or p=E/c, I'm not sure you have an easy way of finding the momentum of the photon independently.

Would it satisfy you to say that experiments have shown the momentum of the photon to be equal to E/c?

 

[NYSportsguy]

Well it's not that I don't accept it, it's just I want to know WHY this is the formula. I believe the question I posed was a valid one and if the formula holds true there should be an answer for my question.

I plan on doing some research online and seeing why "photons have momentum". If I can get a logical answer then I can accept that equation.

 

[Matterwave]

Ah, well we can derive the Energy equation, which is not too complicated.

In relativistic mechanics the total energy and momentum of a particle are given by:

$$E=\gamma mc^2$$

This might look weird since the equation looks independent of velocity; however, the velocity of the particle is encoded in the gamma term.

$$p=\gamma mv$$

So, if you put the energy in terms of momentum, you get the equation:

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

If you don't believe that, then you can expand the terms yourself and see that its equal.

From that equation we can see that if m=0 as is the case for photons, then E=pc drops out.

For a little more background on how the relativistic total energy was obtained, go to the relativistic kinetic energy section of the wiki article: http://en.wikipedia.org/wiki/Kinetic_energy

It has a nice derivation of the kinetic energy. Next just remember that the total energy is the kinetic energy plus the rest energy.

 

[NYSportsguy]

Ideasrule already brought up this equation if you read his response. The problem ONCE AGAIN is how come "p" in LaTeX Code: E=\\sqrt{p^2c^2+m^2c^4} has a value?

Remember "p" stands for momentum and no matter what equation you use to define momentum there is always mass involved again...an we know in this yet again, that a photon don't have a "mass". Why aren't both the "m" and the "p" values equal to zero?

Therefore my question (again) is why does a photon have momentum when there is no mass??

I already understand that a photon is total kinetic energy cause it never stops moving but the equation doesn't account for it having momentum due to the fact that it has no mass. Both the "m" and "p" values in that equation should be 0.

 

[russ_watters]

Well duh, I wasn't the one that came up with that equation...I was questioning it. Someone else brought it up as a possible answer to my original question above, I just want to know why that would be correct considering photons usually don't have mass, so how can momentum (the "p" part of the equation) be possible?

Because mass isn't in the equation!

Remember "p" stands for momentum and no matter what equation you use to define momentum there is always mass involved again...an we know in this yet again, that a photon don't have a "mass". Why aren't both the "m" and the "p" values equal to zero?

Again, you quoted the equation so I know you saw it: it doesn't have mass in it.

 

[Matterwave]

Ideasrule already brought up this equation if you read his response. The problem ONCE AGAIN is how come "p" in LaTeX Code: E=\\sqrt{p^2c^2+m^2c^4} has a value?

Remember "p" stands for momentum and no matter what equation you use to define momentum there is always mass involved again...an we know in this yet again, that a photon don't have a "mass". Why aren't both the "m" and the "p" values equal to zero?

Therefore my question (again) is why does a photon have momentum when there is no mass??

I already understand that a photon is total kinetic energy cause it never stops moving but the equation doesn't account for it having momentum due to the fact that it has no mass. Both the "m" and "p" values in that equation should be 0.

Your assumption that "no matter what equation you use to define momentum there is always mass involved" is incorrect, as I and Ideasrule have just shown you the equation WITHOUT a mass in it!

 

[dacruick]

i don't clearly understand this concept either NY. But through reading these posts, it seems that there is a distinct different between momentum and relativistic momentum. Its not that physicists decided that the momentum was a certain value. they did tests and they said, what the heck, photons have a momentum equal to this. then they tried to figure out what the relationship was and came to see that the equation given above is quite accurate. it might be something that you either have to just accept, or read about in depth.

 

[Matterwave]

i don't clearly understand this concept either NY. But through reading these posts, it seems that there is a distinct different between momentum and relativistic momentum. Its not that physicists decided that the momentum was a certain value. they did tests and they said, what the heck, photons have a momentum equal to this. then they tried to figure out what the relationship was and came to see that the equation given above is quite accurate. it might be something that you either have to just accept, or read about in depth.

Well, the designation is not arbitrary. Especially during experiment. We can use conservation of momentum to figure out the momentum of photons emitted by decay processes, or by scattering photons off of electrons like in Compton scattering. .

 

[dacruick]

Well, the designation is not arbitrary. Especially during experiment. We can use conservation of momentum to figure out the momentum of photons emitted by decay processes, or by scattering photons off of electrons like in Compton scattering. .

i didnt mean to say that it is arbitrary. i meant to say that it is experimentally proven to be that.

 

[NYSportsguy]

Yes, but the word and formula definitons for the concept of "momentum" still contain a mass unit in it. You can't calculate any type of momentum unless you have mass...it's part of what makes up momentum! Just look at the two formulas posted already for Relativisitic and Classical momentum by Ideasrule and Matterwave.

Once again, how can you calculate momentum if mass is part of the equation (an not just mass by itself as in the Kinetic Energy formula but also as part of the "p" in momentum?!?

 

[Matterwave]

lol, this discussion is getting nowhere. We keep telling you how to calculate the momentum without invoking a mass, and all you say is that we can't...

Then I said then you can just look at experimental evidence, and you said that's not satisfying to you either.

I don't know what you want.

Do you want us to tell you..."Yea you're right! All of us physicists have been wrong, special relativity is wrong, our experiments are faulty! In fact photons have no momentum!"

If that's the case, I can't help you any further.