Question about total energy in quantum mechanics

[5614]

Special relativity says that speed of light is constant... this includes being constant in all frames.

[Johnny5]

Special relativity says that speed of light is constant... this includes being constant in all frames.

 

Special relativity does not say that the speed of light is constant. Einstien got his idea that the speed of light is constant from classical Maxwellian electrodynamics. He then wrote that the speed of light is c, in all reference frames in which the laws of classical electrodynamics are true. He and others presumed that those frames were inertial frames. The actual answer is that they are true in the frames in which the experiments that measured [math] \epsilon_0 [/math] and [math] \mu_0 [/math] were carried out, namely the gravitational rest frame of the earth.

 

Recall:

 

[math] c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} [/math]

 

 

[math] \epsilon_0 [/math] is the permittivity of free space

[math] \mu_0 [/math] is the permeability of free space

 

The units of c are that of a speed, specifically the speed of a wave in Maxwell's theory.

[swansont]

[math] c = \sqrt \frac{1}{\epsilon_0 \mu_0} [/math]

[swansont]

The second postulate of special relativity is that c is the same for all inertial observers. I can't prove it. But if you can disprove it, you disprove all of relativity. I'm pretty confident that relativity is correct.

[Johnny5]

The second postulate of special relativity is that c is the same for all inertial observers. I can't prove it. But if you can disprove it, you disprove all of relativity. I'm pretty confident that relativity is correct.

 

Being pretty confident, isn't being certain. The simplest way to prove that the second postulate leads to a contradiction, is to prove that simultaneity is absolute not relative. The barn and ladder problem does this perfectly.

 

Regards

[Tom Mattson]

swansont: An inertial observer can never be in a frame where the photon is at rest.

 

Johnny5: Prove your statement please.

 

It's a direct consequence of the 2nd postulate. If the speed of a light pulse is c in any inertial frame, then it cannot be 0 in any frame.

[5614]

The e represented in [math]e = hf[/math] and the e in [math]e^2 = (pc)^2 + (m_0 c^2 )^2[/math] represent different types of energy.

 

Reason:

if it was the same type of energy, which i thought it was then:

[math]e^2 = (pc)^2 + (m_0 c^2 )^2 = (hf)^2[/math]

and if m=0 then

[math]e^2 = (pc)^2 = (hf)^2[/math]

[math]e = pc = hf[/math]

but how can pc = hf when f is a variant depending on the photon's frequency and pc is a constant? The answer is that pc doesn't equal hf because they're referring to different types of energy... maybe that was obvious, i just realised it though!

 

=========

 

moving onto the photons don't have rest frames thing... if you are going to disagree with relativety then fine, but in the universe as we know it photons do not have rest frames.

[Johnny5]

The e represented in [math]e = hf[/math] and the e in [math]e^2 = (pc)^2 + (m_0 c^2 )^2[/math] represent different types of energy.

 

Reason:

if it was the same type of energy' date=' which i thought it was then:

[math']e^2 = (pc)^2 + (m_0 c^2 )^2 = (hf)^2[/math]

and if m=0 then

[math]e^2 = (pc)^2 = (hf)^2[/math]

[math]e = pc = hf[/math]

but how can pc = hf when f is a variant depending on the photon's frequency and pc is a constant? The answer is that pc doesn't equal hf because they're referring to different types of energy... maybe that was obvious, i just realised it though!

 

=========

 

moving onto the photons don't have rest frames thing... if you are going to disagree with relativety then fine, but in the universe as we know it photons do not have rest frames.

 

What on earth are you trying to show me? It would help me follow you, if you actually derived the formula we are talking about from your starting point.

 

Regards

[5614]

It's just you said that:

 

 

in the very first post but im saying you can't do that because you cant combine e=hf and [math] e^2 = (pc)^2 + (m_0 c^2)^2[/math] together as they are for different types of energy.

[Johnny5]

It's just you said that:

 

[math]e = hf + m_0 c^2[/math]

 

in the very first post but im saying you can't do that because you cant combine e=hf and [math]e=(pc)^2 + (m_0 p^2)^2[/math] together as they are for different types of energy.

 

This is getting all screwed up' date=' because no effort is being made to derive the formula

 

[math'] E^2 = (pc)^2 + (m_0 c^2)^2 [/math]

 

The derivation can be made to come from only one assumption, namely:

 

[math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

 

Your formula is

 

[math]E=(pc)^2 + (m_0 p^2)^2[/math]

 

Where I have taken the liberty of writing e=E=total energy.

 

Your formula is not...

 

[math]E=(pc)^2 + (m_0 c^2)^2[/math]

 

Why are these two formulas different?

 

Regards

[5614]

OMG, sorry...

 

I meant [math]e^2 = (pc)^2 + (m_0 c^2 )^2[/math] my mistake, it was meant to be a c at the end and not a p....

 

the whole reason this came up from my post was that in your very first post it combined the above mentioned forumula and combined it with the e=hf formula.... im saying you can't do that because they are different types of e.

[Johnny5]

OMG' date=' sorry...

 

I meant [math']e^2 = (pc)^2 + (m_0 c^2 )^2[/math] my mistake, it was meant to be a c at the end and not a p....

 

the whole reason this came up from my post was that in your very first post it combined the above mentioned forumula and combined it with the e=hf formula.... im saying you can't do that because they are different types of e.

 

Answer me a question please. In the formula you have right here;

 

[math]e^2 = (pc)^2 + (m_0 c^2 )^2[/math]

 

Is it the case that:

 

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

 

?

 

And my second question is, what does e denote?

 

My third question is, how did you derive the formula?

 

Thank you

[Tom Mattson]

Of course photons have rest frames' date=' and if m0=0 then for a photon you will get E=pc, but as I said, photons do have rest frames.

[/quote']

 

There is no inertial frame in which the photon is at rest. An intertial frame is defined by a coordinate system that is moving without acceleration. When v=c, the t-axis of this "frame" is undefined. Since you can't have a coordinate system when one or more of the coordinate axes are undefined, you can't have an inertial frame moving at v=c in SR.

[5614]

How did I derive [math]e^2 = (pc)^2 + (m_0 c^2)^2[/math] well it's a well known equation that I've learnt.

 

I'm not sure which type of energy is given by the above equation and e=hf but they cannot be the same (read post #32) and so I asked my physics teacher and he said they weren't the same type of energy.

[Johnny5]

How did I derive [math]e^2 = (pc)^2 + (m_0 c^2)^2[/math] well it's a well known equation that I've learnt.

 

I'm not sure which type of energy is given by the above equation and e=hf but they cannot be the same (read post #32) and so I asked my physics teacher and he said they weren't the same type of energy.

 

Post 32 where?

 

Regards

[5614]

each post is surrounded in a thin blue line with a thicker blue line at the top of it, above the name of the poster it says the date in that thick blue line... look on the right side of that line it says #41 or any number, that is the number of the post.

[Johnny5]

How did I derive [math]e^2 = (pc)^2 + (m_0 c^2)^2[/math] well it's a well known equation that I've learnt.

 

I'm not sure which type of energy is given by the above equation and e=hf but they cannot be the same (read post #32) and so I asked my physics teacher and he said they weren't the same type of energy.

 

Telling me that it is a well known equation which you have learned doesn't answer the question. I clearly see that you say you aren't sure what type of energy is given by the above equation. Why don't you try deriving it so you do know? And then after you know what 'kind' of energy it is, then you will be able to decide whether or not it is equivalent to (hf)^2?

 

Regards

[5614]

Sadly this is when physics is still a bit too much for me, I am young, I am learning, I do not know all the answers to all of your questions.

 

However:

 

look at post #32.... hf cannot equal pc therefore something is wrong somewhere! The maths is all right, so the only logical explanation, backed my physics teacher who knows this is that the energy in each formula is different.

 

I do not know how to "try deriving it", I know what it's for and how to use, as well as logical processes such as that of post #32... are you asking me to tell you which energy is used??? (in which case I can try and find out)

[Johnny5]

Here is your post #32:

 

The e represented in [math]e = hf[/math] and the e in [math]e^2 = (pc)^2 + (m_0 c^2 )^2[/math] represent different types of energy.

 

Reason:

if it was the same type of energy' date=' which i thought it was then:

[math']e^2 = (pc)^2 + (m_0 c^2 )^2 = (hf)^2[/math]

and if m=0 then

[math]e^2 = (pc)^2 = (hf)^2[/math]

[math]e = pc = hf[/math]

but how can pc = hf when f is a variant depending on the photon's frequency and pc is a constant? The answer is that pc doesn't equal hf because they're referring to different types of energy... maybe that was obvious, i just realised it though!

 

Here you ask, "how can pc=hf when f is variant, and pc constant?"

 

First let me point out that momentum p is a vector, although the magnitude of that vector is a scalar. As for pc being constant... why do you assert that?

 

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

 

Where lambda is the wavelength of the object in question, and h is Planck's constant of nature. So P varies inversely with lambda.

 

Postulate:

 

[math] c = {f}{\lambda} [/math]

 

Hence

 

[math] p = \frac{hf}{c} [/math]

 

Hence:

 

 

[math] pc = hf [/math]

 

Regards

[5614]

Fair point!

 

OK, I will look into whether the [math]e[/math] in [math]e^2 = (pc)^2 + (m_0 c^2)^2[/math] is the same as the [math]e[/math] in [math]e=hf[/math]

[Johnny5]

Fair point!

 

OK' date=' I will look into whether the [math']e[/math] in [math]e^2 = (pc)^2 + (m_0 c^2)^2[/math] is the same as the [math]e[/math] in [math]e=hf[/math]

 

No you shouldn't just give in like that. I am asking you to explain yourself better.

 

Regards

 

PS: Think about the postulate I just pulled out of the air.

[5614]

well

is correct, at least I have learnt it is!

 

now the 2nd part looked a bit dodgy, but then

which all seems correct

 

and then you just take the original [math]p = \frac{hf}{c}[/math] multiply both sides by c and you get [math]pc = hf[/math] now probably im missing something obvious and I take your point "No you shouldn't just give in like that" but your postulate seems ok and therefore my post #32 is not valid evidence.

[Johnny5]

I have a bold suggestion.

 

Why don't you actually derive the formula for E^2 clearly, and then when you are done, explain the experimental origin of formula E=hf in some detail.

 

Regards

[swansont]

Being pretty confident' date=' isn't being certain. The simplest way to prove that the second postulate leads to a contradiction, is to prove that simultaneity is absolute not relative. The barn and ladder problem does this perfectly.

[/quote']

 

Simultaneity is not absolute, it is frame dependent. That's what the barn and the ladder example shows.

 

None of science is absolutely certain. You can't prove that things won't start falling up tomorrow. It's all about how confident we are in thephysical laws we discover.

[Johnny5]

Simultaneity is not absolute' date=' it is frame dependent. That's what the barn and the ladder example shows.

 

[/quote']

 

You have it backwards. The barn and ladder paradox was designed to show that simultaneity is absolute, not relative. It is mean't to show that simultaneity is NOT frame dependent.

 

Regards