Calculations i cant understand and make me crazy

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In summary, Q2 asks what factors increase the refractive index. Q3 has questions about why different waves have different frequencies and how the physicists deal with it. These questions are not answered in the content.
  • #1
katatonia911
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hi,

im not involved into physics but i read sometimes about it and have very curious questions:

Q1-

if E=mc2 then as i know that electromagnetic waves have mass of zero

so this mean E=0xc2 which means that c2=E/0

here i get that the E=0 , and c=infinite..which i guess is impossible

so what is wrong in my understanding of the equation?Q2-

since the refractive index of glass is typically around 1.5, meaning that light in glass travels at c / 1.5 ≈ 200,000 km/s; the refractive index of air is about 1.0003, so the speed of light in air is very close to c.

what is the factor that increase the refractive index??

and if we supposed that we have a material of a very high refractive index which its value ≈ 300,000,000 what will will happen to the light? will it become somthing with mass that we can see?

Q3-

i always read when people talking about light travel speed as if light is something fixed but as i know its composed of different waves with different frequencies (rainbow colors, infra-red,UV,..) ..its getting me crazy

i don't get it.when scientists put their calculations do they put these fact's in their mind?
how they deal with it?thank you all
 
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  • #2
Hello kat, welcome to PF :)

katatonia911 said:
Q1-

if E=mc2 then as i know that electromagnetic waves have mass of zero

so this mean E=0xc2 which means that c2=E/0

here i get that the E=0 , and c=infinite..which i guess is impossible

so what is wrong in my understanding of the equation?

The m in the equation E=mc2 is in simple terms the total mass which comprises the rest mass and the kinetic mass. For a stationary particle the total mass is simply the rest mass but for a moving object there is a kinetic component. For something moving at the speed of light all the mass is kinetic mass, so m=0 does not apply to a photon.

E = mc2 = moc2/sqrt(1-v2/c2)

where mo is the rest mass. For a photon mo=0 and sqrt(1-v2/c2)=0 so E = 0/0*c2 which is not the same as zero or infinite but is indeterminate. All that means is that the answer for this situation can not be calculated using this method. The equation that can be used for a photon (or any particle) is:

E = sqrt(mo2c4 + p2c2)

where p is the relatistic momentum of the particle. For a particle with rest mass, the relativistic momentum is equal to:

p = mov/sqrt(1-v2/c2)

and for a photon the momentum is:

p = hf

where f is the frequency of the photon and h is Planck's constant, so for a photon:

E = mc2 = sqrt(mo2c4 + p2c2) = pc

In short, the rest mass of a photon is zero, but its total mass is not and in the famous E=mc2 equation, the m does not refer just to rest mass.

katatonia911 said:
Q3-
i always read when people talking about light travel speed as if light is something fixed but as i know its composed of different waves with different frequencies (rainbow colors, infra-red,UV,..) ..its getting me crazy

The wavelength and frequency of electromagnetic waves is related according to:

c = f*w where w is the wavelength. A photon with a large frequency has a small wavelength and vice versa, but the product is always a constant speed equal to c.

Hope that helps.
 
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  • #3
The answers for 1 and 3 given above by kev are good, so I will just do this one
katatonia911 said:
Q2-

since the refractive index of glass is typically around 1.5, meaning that light in glass travels at c / 1.5 ≈ 200,000 km/s; the refractive index of air is about 1.0003, so the speed of light in air is very close to c.

what is the factor that increase the refractive index??

and if we supposed that we have a material of a very high refractive index which its value ≈ 300,000,000 what will will happen to the light? will it become somthing with mass that we can see?
The refractive index is the speed of light in vacuum divided by the speed of light in the material, and the speed of light in any material is given by
[tex]\frac{1}{\sqrt{\mu \epsilon}}[/tex]
where [itex]\mu[/itex] is the permeability and [itex]\epsilon[/itex] is the permittivity of the material. The permeability of most materials is not terribly far off from that of vacuum, but dielectric materials have permittivity values that can be quite far from vacuum. That is why things like glass, plastic, diamond, and crystal all have high refractive indexes and all are good electrical insulators, the two properties are both a result of the high permittivity.
 
  • #4
kev said:
E = sqrt(mo2c4 + p2c2)

Yeah, E=mc^2 is just plain wrong unless you define the 'mass' of the particle in a very weird way. I guess the physicists think that it's nice that the general public knows atleast one physics formula, even if the formula is completely wrong. :-)



About the refractive index: http://www.scientificamerican.com/article.cfm?id=what-visions-in-the-dark&sc=I100322"might be of some interest to you
 
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  • #5
clamtrox said:
Yeah, E=mc^2 is just plain wrong unless you define the 'mass' of the particle in a very weird way. I guess the physicists think that it's nice that the general public knows atleast one physics formula, even if the formula is completely wrong. :-)

Actually, it isn't wrong, because it tells you exactly how much energy one gets when an amount of mass is converted. It is just that it isn't the whole story.

It is very similar to using F=ma. You wouldn't call that wrong, would you? Structural engineers use it all the time to build houses and buildings. Yet, it is only valid for constant "m", i.e. it is not the full story. So when you have varying m, then using that equation is not appropriate. This is exactly what is going on with the relativistic energy equation. People who haven't learned enough are using the wrong equations for the wrong situation. It is not the equation that is wrong.

Zz.
 
  • #6
Thank you all, clear answers and to the point:smile:
 
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  • #7
ZapperZ said:
It is very similar to using F=ma. You wouldn't call that wrong, would you? Structural engineers use it all the time to build houses and buildings. Yet, it is only valid for constant "m", i.e. it is not the full story. So when you have varying m, then using that equation is not appropriate. This is exactly what is going on with the relativistic energy equation. People who haven't learned enough are using the wrong equations for the wrong situation. It is not the equation that is wrong.
Zz.

... and if 90% of the time everyone used F=ma to systems where mass wasn't constant, would you say that equation was still right? Or if structural engineers tried to build houses from photons and they'd say "these guys can't feel any forces 'cause their mass is zero".
 
  • #8
clamtrox said:
... and if 90% of the time everyone used F=ma to systems where mass wasn't constant, would you say that equation was still right? Or if structural engineers tried to build houses from photons and they'd say "these guys can't feel any forces 'cause their mass is zero".

I'm not sure what your point is here, since you are actually supporting what I just said. It really doesn't matter how many percent of the time people use F=ma incorrectly. It is still incorrect.

E=mc^2 IS correct for what it is conveying. When you want to calculate the energy conversion from an amount of mass, what else do you use? For that specific usage, it is correct. So what are we arguing about here? I'm hoping this is not about "covariant mass" versus "rest mass" versus "relativistic mass", because that has been discussed ad nauseum.

Zz.
 

FAQ: Calculations i cant understand and make me crazy

What is the purpose of calculations in scientific research?

Calculations are used in scientific research to quantify and analyze data, make predictions and test hypotheses. They help scientists to draw conclusions and make informed decisions based on the results of their experiments.

Why do some calculations in science seem so complex and difficult to understand?

Some calculations in science can be complex because they involve a combination of different mathematical concepts and formulas. Additionally, they may require a thorough understanding of the underlying scientific principles and a certain level of mathematical proficiency.

What can I do if I am struggling to understand and solve a particular calculation?

If you are having trouble understanding a calculation, it is important to ask for help from a teacher, colleague, or mentor. They can provide additional explanations and guidance to help you better understand the concepts and solve the calculation.

Is it necessary to memorize all the formulas and equations for different calculations?

While it is important to have a basic understanding of common formulas and equations, it is not necessary to memorize them all. In most cases, you can refer to a textbook or online resources to find the relevant formulas and equations for a specific calculation.

How can I improve my skills in performing calculations in science?

To improve your skills in performing calculations, it is important to practice regularly. You can also attend workshops or courses that focus on mathematical and scientific methods, or seek guidance from experienced scientists or tutors. Additionally, using technology and software tools can also help to simplify and streamline complex calculations.

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