"Does light move", asked Quantum Quack

anuraganimax said:
Here

λ=h/mv=h/mc
c/ν=h/mc
mc^2=hν
mc^2=E
mc.c=E
p.c=E
Since both relativistic mass and velocity of a photon is fixed the energy should also be fixed. Changing the wavelength or frequency makes no difference as the equations don't change. You'll have to alter the relativistic mass or velocity if you want to want to alter the wavelength or frequency. Impossible as both are constants.
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No. The fully general dispersion relation is $$E^2 = (pc)^2 + (mc^2)^2$$ Since the photon is a massless particle you can't define a rest frame for it, therefore you can't ever have a photon with zero momentum so $$E=mc^2$$ never applies.

Your step where you write $$mc^2 = h \nu$$ is wrong. If instead of $$mc^2$$ you write pc, then the result you get is the de Broglie relationship: $$p = \frac{h}{\lambda}$$: p is frequency / wavelength dependent.
 
prometheus said:
No. The fully general dispersion relation is $$E^2 = (pc)^2 + (mc^2)^2$$ Since the photon is a massless particle you can't define a rest frame for it, therefore you can't ever have a photon with zero momentum so $$E=mc^2$$ never applies.

Your step where you write $$mc^2 = h \nu$$ is wrong. If instead of $$mc^2$$ you write pc, then the result you get is the de Broglie relationship: $$p = \frac{h}{\lambda}$$: p is frequency / wavelength dependent.
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I don't believe this is the case. The equation $$E^2 = (pc)^2 + (mc^2)^2$$ is derived using -
1) E=mc^2
2) p=mv
Therefore if you substitute m0=0 to get E=pc from the above equation the initial equations namely E=mc^2 and p=mv should always apply.
You cannot say that $$E^2 = (pc)^2 + (mc^2)^2$$ is valid for photon but the initial equations are not.
Also the product of wavelength and frequency is constant. If you change the frequency the wavelength must be also be changed in the same proportion. The change in frequency will be canceled out by the change in wavelength anyway and momentum will remain the same.
 
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anuraganimax said:
I don't believe this is the case. The equation $$E^2 = (pc)^2 + (mc^2)^2$$ is derived using -
1) E=mc^2
2) p=mv
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No it's not. You derive $$E^2 = (pc)^2 + (mc^2)^2$$ by taking the norm of the energy momentum 4 - vector:
$$ \underline{p} = \begin{pmatrix} \frac{E}{c} \\ \vec{p} \end{pmatrix}$$
$$ \underline{p}. \underline{p} = \eta^{\mu \nu} p_\mu p_\nu = \frac{E^2}{c^2} - \vec{p}.\vec{p}$$
Dimensional analysis shows this must be equal to $$(mc)^2$$ hence
$$\frac{E^2}{c^2} - \vec{p}.\vec{p}=(mc)^2 \Rightarrow E^2 = \vec{p}.\vec{p}c^2 + (mc^2)^2 $$
All we've assumed in this is that the contraction of a vector with the metric, $$ \eta^{\mu \nu} v_\mu v_\nu$$ is invariant under the symmetries of the metric, in this case the Lorentz group SO(1,3). If you take this equation and assume the particle is at rest, you get $$E= mc^2$$. On the other hand, if you assume the particle is massless you get $$E =pc$$, that since Note $$ \eta^{\mu \nu} p_\mu p_\nu$$ is a constant under Lorentz transformations, so must be $$(mc)^2$$ but since c is a constant, m must also be a constant. Therefore mass is invariant in special relativity.


anuraganimax said:
Therefore if you substitute m0=0 to get E=pc from the above equation the initial equations namely E=mc^2 and p=mv should always apply.
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You cannot say in general that these two equations are true. As I have shown above, $$E= mc^2$$ is a special case of $$E^2 = (pc)^2 + (mc^2)^2$$. Momentum is a little more tricky to show so I wont reproduce it now, but I can if you really want. The definition of momentum comes from Noether's theorem and states momentum is the conserved current that is related to the invariance of the action to spatial translations. For Newtonian mechanics you do the Noether procedure and come out with $$\vec{p} = m \vec{v}$$. For relativistic mechanics when you have a bigger symmetry group than Newtonian mechanics you get $$\vec{p} = \gamma(\vec{v}) m \vec{v}$$. In the non relativistic limit the Newtonian expression for momentum is recovered.

anuraganimax said:
You cannot say that $$E^2 = (pc)^2 + (mc^2)^2$$ is valid for photon but the initial equations are not.
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You're going the wrong way. $$E^2 = (pc)^2 + (mc^2)^2$$ is more general than the other equations you're using, not less.

anuraganimax said:
Also the product of wavelength and frequency is constant. If you change the frequency the wavelength must be also be changed in the same proportion. The change in frequency will be canceled out by the change in wavelength anyway and momentum will remain the same.
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It's true that the product of frequency and wavelength is a constant, but individually they are not invariant. You can only ever derive (correctly) a formula for p that includes either frequency or wavelength which should tell you that momentum is dependent on them. It would be independent it p depended only on the product $$\nu \lambda$$ but it does not.
 
Well according to my book

E=mc^2=m0c^2/sqrt(1-v^2/c^2).......1
P=mv=m0v/sqrt(1-v^2/c^2).......2

Square 1 and 2

E^2=m0^2c^4/sqrt(1-v^2/c^2).........3
p^2=m0^2v^2/sqrt(1-v^2/c^2)

p^2c^2=m0^2v^2c^2/sqrt(1-v^2/c^2)............4

Subtract 4 from 3

E^2 - p^2c^2=m0^2c^4(1-v^2/c^2)/(1-v^2/c^2)=m0^2c^4

E^2=p^2c^2+m0^2c^4

If you look at this way then E=mc^2 and p=mv are the basic equations
 
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prometheus said:
No it's not. You derive $$E^2 = (pc)^2 + (mc^2)^2$$ by taking the norm of the energy momentum 4 - vector:
$$ \underline{p} = \begin{pmatrix} \frac{E}{c} \\ \vec{p} \end{pmatrix}$$
$$ \underline{p}. \underline{p} = \eta^{\mu \nu} p_\mu p_\nu = \frac{E^2}{c^2} - \vec{p}.\vec{p}$$
Dimensional analysis shows this must be equal to $$(mc)^2$$ hence
$$\frac{E^2}{c^2} - \vec{p}.\vec{p}=(mc)^2 \Rightarrow E^2 = \vec{p}.\vec{p}c^2 + (mc^2)^2 $$
All we've assumed in this is that the contraction of a vector with the metric, $$ \eta^{\mu \nu} v_\mu v_\nu$$ is invariant under the symmetries of the metric, in this case the Lorentz group SO(1,3). If you take this equation and assume the particle is at rest, you get $$E= mc^2$$. On the other hand, if you assume the particle is massless you get $$E =pc$$, that since Note $$ \eta^{\mu \nu} p_\mu p_\nu$$ is a constant under Lorentz transformations, so must be $$(mc)^2$$ but since c is a constant, m must also be a constant. Therefore mass is invariant in special relativity.
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Hmmm... I did not know of this way to derive it I will look into this if I can find this derivation.
 
prometheus said:
It's true that the product of frequency and wavelength is a constant, but individually they are not invariant. You can only ever derive (correctly) a formula for p that includes either frequency or wavelength which should tell you that momentum is dependent on them. It would be independent it p depended only on the product $$\nu \lambda$$ but it does not.
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Even if you do alter frequency to alter momentum you have automatically altered the wavelength as well which will shift momentum back to its original value as c=fλ=constant
P depends on both frequency as well as wavelength
p=hv/c
and
p=h/λ
 
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anuraganimax said:
Well according to my book

E=mc^2=m0c^2/sqrt(1-v^2/c^2).......1
P=mv=m0v/sqrt(1-v^2/c^2).......2

Square 1 and 2

E^2=m0^2c^4/sqrt(1-v^2/c^2).........3
p^2=m0^2v^2/sqrt(1-v^2/c^2)

p^2c^2=m0^2v^2c^2/sqrt(1-v^2/c^2)............4

Subtract 4 from 3

E^2 - p^2c^2=m0^2c^4(1-v^2/c^2)/(1-v^2/c^2)=m0^2c^4

E^2=p^2c^2+m0^2c^4

If you look at this way then E=mc^2 and p=mv are the basic equations
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What book is this?

Firstly, the m you have in your formulae are relativistic masses. $$m_{\text{rel}} = \gamma m$$. Relativistic mass is a nice way to think about the effects of relativity for beginners, because you have Newtonian intuition you can still work with Newtonian thinking by allowing mass to depend on speed. Unfortunately, that's really the only good thing about it. Relativistic mass is not actually a mass at all because, as I showed above, mass is invariant in SR. Relativistic mass is really another way to look at the energy of the system.

I guess this is a (dodgy) way to do it. It's dodgy because, for massless particles $$\gamma m = \infty \times 0$$ which is undefined. With my way you make less assumptions and the result is stronger.

PS In your equations 3 and 4 there should be $$1-\frac{v^2}{c^2}$$ in the denominator each time, not $$\sqrt{1-\frac{v^2}{c^2}}$$
 
anuraganimax said:
Even if you do alter frequency to alter momentum you have automatically altered the wavelength as well which will shift momentum back to its original value as c=fλ=constant
P depends on both frequency as well as wavelength
p=hv/c
and
p=h/λ
Click to expand...

No it doesn't. Plug some numbers in and see for yourself - if you change the frequency, and also change the wavelength accordingly, the momentum will change - it is not invariant.
 
prometheus said:
What book is this?
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An old one when I was in school maybe 8 years or so years old edition.

prometheus said:
Firstly, the m you have in your formulae are relativistic masses. $$m_{\text{rel}} = \gamma m$$. Relativistic mass is a nice way to think about the effects of relativity for beginners, because you have Newtonian intuition you can still work with Newtonian thinking by allowing mass to depend on speed. Unfortunately, that's really the only good thing about it. Relativistic mass is not actually a mass at all because, as I showed above, mass is invariant in SR. Relativistic mass is really another way to look at the energy of the system.

I guess this is a (dodgy) way to do it. It's dodgy because, for massless particles $$\gamma m = \infty \times 0$$ which is undefined. With my way you make less assumptions and the result is stronger.
Click to expand...
I will check your way as well if I can find the means.

prometheus said:
PS In your equations 3 and 4 there should be $$1-\frac{v^2}{c^2}$$ in the denominator each time, not $$\sqrt{1-\frac{v^2}{c^2}}$$
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Yes thanks for pointing that out.
 
You just mixed a non-relativistic quantum expression with a relativistic approach so your deduction is wrong.

The actual expression for a de Broglie wavelength is $$\lambda = \frac{h}{p}$$. For a slow moving object (if its not moving at light speed then it must have non-zero mass) you have the Newtonian approximation of p = mv. You implicitly assumed that, which is not true for things moving near or at the speed of light.

Yet another example of how your ignorance leads to flawed conclusions.
 
AlphaNumeric said:
You just mixed a non-relativistic quantum expression with a relativistic approach so your deduction is wrong.

The actual expression for a de Broglie wavelength is $$\lambda = \frac{h}{p}$$. For a slow moving object (if its not moving at light speed then it must have non-zero mass) you have the Newtonian approximation of p = mv. You implicitly assumed that, which is not true for things moving near or at the speed of light.

Yet another example of how your ignorance leads to flawed conclusions.
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Shut your trap . I gotta work with what I have at the moment. And by the way hotshot its not me who has to be competent in this field it is you. I study physics for fun and nothing more.
 
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anuraganimax said:
May I know the reason as to why you are so interested in knowing about the author's name?
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Because I'm interested. I have a fairly large collection of books and I also have access to a physics library at work so the chances are I have access to it.
 
prometheus said:
Because I'm interested. I have a fairly large collection of books and I also have access to a physics library at work so the chances are I have access to it.
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Sorry to disappoint you but my book is quite old and torn and I can't make out the name of the author as both cover and front pages are missing.
 
I've got to say I don't believe it either. If you bought a book 6 years ago and tore the cover off you should surely still be able to remember who wrote it?
 
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