Jump to content

Recommended Posts

Posted
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]

Posted

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.

Posted

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.

Posted

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].

Posted

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.

Posted

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.

Posted

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.

Posted
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?

Posted

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.

Posted
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?

Posted
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.

Posted

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.

Posted

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.

Posted

"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.

Posted
"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?

Posted
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.

Posted
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?

Posted
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.

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

 

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

Posted
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.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.