ok' date=' [math']p=\frac{h}{\lambda}[/math]

 

[math]

h^2f^2=(mc^2)^2+\frac{h^2c^2}{\lambda^2}[/math].

 

[math]\frac{h^2f^2\lambda^2-h^2c^2}{\lambda^2}=m^2[/math]

 

[math]\frac{h(f^2\lambda^2-c^2)}{\lambda^2}=m^2[/math]

 

damn. [math]f^2\lambda^2=c^2[/math]. grrrrrrrrrrrrrrrrrrrrrrrrr.

 

i went back later to finish it for a particle that didn't move at c. i got a formula that says particles UNDER c have imaginary mass. i was under the impression that only tachyons had imaginary mass. what does this mean?

[math]m=\frac{h(\sqrt{v^2-c^2})}{{\lambda}c^2}[/math]

It means you've probably made an error in your mathematics at some point I can't remember the equations, I'm afraid, so I'm probably not the best person to ask at this point.

It would be nice if you could restate your purpose in those manipulations. Actually I don't quite follow them. It is also sometimes a fatal mistake to equate equations simply because they appear to have algebriac equivalance without careful thought. In anycase, it is definitely not appropriate to calculate the "mass" of a photon. A photon may have momentum, energy, wavelength, and frequency, but not mass. You will be able to calculate about everything you need with the use of those parameters instead.

how aren't they equivalent? [math]hf=E\Rightarrowh^2f^2=E^2[/math][math]E^2=(mc^2)^2+(pc)^2[/math] using substitution, we get [math]h^2f^2=(mc)^2+(pc)^2[/math] i was origionally trying to see if i could find a mass, but i fould that mass always cancels out at c. then i decided to not assume c. then i got [math]m=\frac{h(\sqrt{v^2-c^2})}{{\lambda}c^2}[/math].

The math markup isn't working, so it's hard to follow, but E=hf assumes photons. You can't generalize from that to an equation for massive particles with v<c.

it is interesting that photons are the opposite of matter. ftl=normal mass, sublight=imaginary mass, c=0 mass. perhaps that means that imaginary mass does mean something.

You steps from "h²f²=(mc²)²+(pc)²" to "m=[h * sqrt(v²-c²)]/[lambda * c²]" are a bit mysterious and -because of the 2nd equation being wrong- obviously flawed. Note that none of the c's in the 1st equation is the particle´s velocity.

it is interesting that photons are the opposite of matter. ftl=normal mass, sublight=imaginary mass, c=0 mass. perhaps that means that imaginary mass does mean something.

 

How do you conclude this?

athiest, i never said that c was the particles velocity exept when it made m=0.

 

swansont, put speeds in the equation and see for yourself.

athiest' date=' i never said that c was the particles velocity exept when it made m=0.

 

swansont, put speeds in the equation and see for yourself.[/quote']

 

Which equations are you using?

athiest, i never said that c was the particles velocity exept when it made m=0.

I never said you did. It was just an (seemingly wrong) assumption why the v suddenly appears. As I´ve told you before: Please show your calculations step by step (with comments). That´s the only way to find the mistake you made.

frequency times wavelength=speed

yourdadonapogos:

Your algebra is correct, but your interpretation of f\lambda as a particle's velocity is wrong. f\lambda is the wave velocity (which is greater than c) of the particle's wave function. The velocity at which we would interpret the particle to be moving is the group velocity of the wave packet.

This is given by dE/dp, which equals pc^2/E which is less than c.

The problem occurs because you are generalizing the equation to values that are not physically possible. The equation is simply saying that its not possible for the speed to be less than c for a photon.

how can the wave be faster than the particle?

"wave packet" is more accurate than wave - it's often not a single peak, but many oscillations of some form that comprise a wave function. The phase of the peaks within the wave packet can change, and so the peaks can move around faster than c, even though the packet as a whole cannot.

"wave packet" is more accurate than wave - it's often not a single peak, but many oscillations of some form that comprise a wave function. The phase of the peaks within the wave packet can change, and so the peaks can move around faster than c, even though the packet as a whole cannot.

 

Swansont...

 

Is the wave something physical? What is doing the waving, in the case of light?

Swansont...

 

Is the wave something physical? What is doing the waving' date=' in the case of light?[/quote']

 

In the case of light, it is the electric and magnetic fields that are oscillating.

does light also have a probablility wave?

The wave function of a photon is the EM four-potential, A^\mu.

In the case of light, it is the electric and magnetic fields that are oscillating.

 

1. What is an electric field?

2. What is a magnetic field?

3. If |E| denotes the magnitude of an electric field, is |E| a continuous quantity?

4. IF |B| denotes the magnitude of a magnetic field, is |B| a continuous quantity, or a discrete one?

so, the wave-packet can move faster than the photon?

1. What is an electric field?

2. What is a magnetic field?

3. If |E| denotes the magnitude of an electric field' date=' is |E| a continuous quantity?

4. IF |B| denotes the magnitude of a magnetic field, is |B| a continuous quantity, or a discrete one?[/quote']

 

Read a physics book.

so, the wave-packet can move faster than the photon?

 

The peaks and troughs within the wave packet can redistribute faster than c.

Read a physics book.

 

I have, I wanted to know what your mind thinks.

 

Regards

 

My answer's are these:

 

1. Don't know.

2. Don't Know.

3. Don't know, probably discrete.

4. Quantized.