E=h*f ?

[Capiert]

If E=m*(c^2)

 & light's_speed

 c=f*L

 where L is the wavelength Lambda

 then surely the energy must be related

 to the frequency_squared (f^2) instead,

 otherwise the units do NOT fit.

Please explain.

[Endy0816]

In the case of hf, the seconds cancel out.

f: 1/s

h: J*s

J*s * 1/s = J

 

J: kg * m^2 / s^2

 

Edited April 7, 2018 by Endy0816

[Strange]

1 hour ago, Capiert said:

If E=m*(c^2)

That doesn't apply because photons have zero mass. You need to use the full form of the equation for massless particles.

[swansont]

4 hours ago, Capiert said:

If E=m*(c^2)

 & light's_speed

 c=f*L

 where L is the wavelength Lambda

 then surely the energy must be related

 to the frequency_squared (f^2) instead,

 otherwise the units do NOT fit.

Please explain.

Explain how you think the units don't work.

E is energy and mc^2 has units of energy

c is a speed, and fL has units of speed

 

 

[Sensei]

On 7.04.2018 at 5:15 PM, swansont said:

Explain how you think the units don't work.

Because his understanding of physics is so weak.. ?

 

Edited April 9, 2018 by Sensei

[Capiert]

On 7 April 2018 at 5:15 PM, swansont said:

Explain how you think the units don't work.

E is energy and mc^2 has units of energy

c is a speed, and fL has units of speed.

Yes, the energy equivalent

 is KE, having speed_squared

 in its units.

Thus the (=my) logical conclusion

 would be to use the speed equivalent, also;

 which has both frequency f

 & wavelength L.

Using the f*L combination

 as speed,

 in a KE combination (equivalent)

 gives a frequency_squared,

 & wavelength_squared,

 substitution.

 

Is that unlogical?

 

On 7 April 2018 at 1:58 PM, Strange said:

That doesn't apply because photons have zero mass. You need to use the full form of the equation for massless particles.

Even if photons do NOT have a mass (which I doubt, because it's a number coeefficient for equations)

 I can still apply a (KE's) 'virtual' mass when equating energies.

[Strange]

6 minutes ago, Capiert said:

Is that unlogical?

It is just meaningless.

You can equate c to wavelength and frequency but there is no point. You just end up with c.

And this has absolutely nothing to do with photons or E=hf because:

1) E = mc² only applies to objects with mass. Photons have no mass. The full version of he equation can be used for photons

2) You format your posts like a spoilt child who just wants to annoy people.

12 minutes ago, Capiert said:

Even if photons do NOT have a mass (which I doubt, because it's a number coeefficient for equations)

 I can still apply a (KE's) 'virtual' mass when equating energies.

Photons do not have mass (your doubts are irrelevant as you don't know anything about the subject) so they don't have kinetic energy.

And that equation is not for kinetic energy; it is for mass-energy equivalence (E=mc² only applies to stationary objects; ie with no kinetic energy).

[Capiert]

27 minutes ago, Strange said:

It is just meaningless.

It makes sense to me,

 even if you can't understand.

Quote

You can equate c to wavelength and frequency but there is no point.

?

Quote

You just end up with c.

You just stated the point, there with that speed. (=Confirmation).

Quote

And this has absolutely nothing to do with photons

So do you mean photons do NOT travel at c? Ridiculous what you're saying.

Quote

or E=hf because:

1) E = mc² only applies to objects with mass. Photons have no mass.

How do you know photons have no mass?

Isn't that only an assumption?

Quote

The full version of (t)he equation can be used for photons

Maybe you can elaborate on that a bit,

 perhaps the need (why)?

Quote

2) You format your posts like a spoilt child who just wants to annoy people.

I suspect that is more your attitude than mine.

I thought this was a science website

 instead of a social discriminating website

 because you don't like children

 or writing styles.

Would you please stick to physics

 & refrain from your repetitive derogatory remarks.

If you don't like what I write,

 simply leave.

I'm sure you've got more important things to do.

(I know I'm not perfect,

 but I try to do my best.)

Edited April 29, 2018 by Capiert

[Strange]

6 minutes ago, Capiert said:

So do you mean photons do NOT travel at c?

Of course not. 

Just because an equation has a “c” in it doesn’t mean it magically applies to photons. E=mc2 only applies to objects with mass therefore not photons. 

7 minutes ago, Capiert said:

How do you know photons have no mass?

Theory and experimental evidence. 

8 minutes ago, Capiert said:

but I try to do my best

No you don’t. You could write like an adult but you choose not to. Purely to be annoying. 

10 minutes ago, Capiert said:

Maybe you can elaborate on that a bit,

 perhaps the need (why)?

When you start writing properly. 

[Capiert]

24 minutes ago, Strange said:

Of course not. 

Just because an equation has a “c” in it doesn’t mean it magically applies to photons. E=mc2 only applies to objects with mass therefore not photons. 

I suspect you are confusing matter with mass.

Mass is a number, & matter is that stuff we can hold in our hand.

It's true we often don't bother distinguishing

 & take the shortcut (substitution).

Quote

Theory and experimental evidence. 

That doesn't tell me much.

Theory is the assumption.

What about the evidence?

Quote

No you don’t.

Yes I do my best.

Quote

You could write like an adult

What's that? a.dul(l).ed.

What are you you talking about?

Is it physics?

Quote

but you choose not to.

?

I write this way to reduce errors,

 also because this website's software

 produces errors.

I have not a better method

 to reduce them (errors).

If I make comprimises

 (I get confused)

 & you guys lock my threads faster.

Quote

Purely to be annoying.

That's NOT true!

That's only your opinion (=guess, or bias).

Quote

When you start writing properly. 

Newspaper columns

 are also narrow.

That I carriage return

  formally at the end of a phrase

 is not much different

 & has a natural pause.

What's your problem?

Edited April 29, 2018 by Capiert

[Phi for All]

5 hours ago, Capiert said:

It makes sense to me,

 even if you can't understand.

These words indicate you aren't doing science.

[Capiert]

1 hour ago, Phi for All said:

These words indicate you aren't doing science.

Science is independently reproducible results (experimentally),

if the method is stated (=given).

Interpretation is another thing.

No reference system (e.g. perspective) has priority. Einstein.

 

Btw DT entered politics because NOT enough "good" people were in the government,

 if the intention was to twist his meaning, or you unfortunately did NOT understand the paradox.

All paradoxes can be solved with simple common sense. Plato.

Who has it (that common sense)? Every healthy person.

Who uses it (=cs)? Not everyone.

Why? Education.

 

P.S. I don't know why your political quote is (being advertized) in physics.

People prefer to build walls, instead of bridges. Newton.

 

I think we should be working together,

 but I doubt that all can.

Edited April 30, 2018 by Capiert

[Capiert]

If you said light's momentum

 mom=k*f

 depended on frequency f

 (when the constant k=m*L

 is the mass m

 multiplied by the wavelength L),

 then I'd probably believe you

 because it is (only) has

 mass m

 multiplied by speed c=f*L.

But NOT energy,

 because (the) KE (equivalent)

 has speed_squared,

 where I'd expect c^2=(f*L)^2.

So I expect light's "energy" (e.g. KE equivalent)

 to depend on frequency_squared f^2;

 NOT simply f.

[Strange]

10 hours ago, Capiert said:

Theory is the assumption.

What about the evidence?

Theory is not a assumption. It is a well tested mathematical model. The results would be different if the photon had mass; we do not observe this.

https://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass

3 hours ago, Capiert said:

If you said light's momentum

 mom=k*f

 depended on frequency f

 (when the constant k=m*L

 is the mass m

 multiplied by the wavelength L),

momentum = h / L

https://en.wikipedia.org/wiki/Photon#Physical_properties

(This also includes the full form of the mass-energy equation that you can use for massless photons.)

3 hours ago, Capiert said:

So I expect light's "energy" (e.g. KE equivalent)

 to depend on frequency_squared f^2;

 NOT simply f.

What you expect is hardly relevant.

The equation E = hf is based on observation.

[swansont]

12 hours ago, Capiert said:

 I suspect that is more your attitude than mine.

I thought this was a science website

 instead of a social discriminating website

It's a science discussion site, which means a necessary component is effective communication. People who show up and have decided to redefine terminology are told not to do that, people who use all-caps for all of their writing are told not to do that, and people who use text-speak are told not to do that, for the same reason that you are being chastised for unconventional formatting. It's a distraction and impedes effective communication.

On 4/7/2018 at 11:15 AM, swansont said:

Explain how you think the units don't work.

E is energy and mc^2 has units of energy

c is a speed, and fL has units of speed

This is what I asked earlier, and after several followups, there hasn't really been any analysis of units.

I will ask again: how is it that the units "don't work"?

[Capiert]

On 7 April 2018 at 5:15 PM, swansont said:

Explain how you think the units don't work.

E is energy and mc^2 has units of energy

c is a speed, and fL has units of speed.

14 hours ago, swansont said:

This is what I asked earlier, and after several followups, there hasn't really been any analysis of units.

I will ask again: how is it that the units "don't work"?

"Per second" [1/s] implies frequency

 & Joule has 2 of them.

Newton*second [N*s]

 has only 1.

 

Energy E has Joule(s) [J] units (in square_brackets)

 which is kilogram(s) [kg] (mass m)

 (multiplied by)

 meters_squared [m^2] (length_squared)

 per (=divided by)

 seconds_squared [s^2] (time_squared).

 

If wavelength

 (is length

 per cycle,

 then its) units

 are meter

 per cycle [m/c].

 

Frequency

 is cycles

 per second [c/s]=[cps]=[Hz] Herz.

 

Please notice (&/or forgive)

 the (slight) discrepancy

 between cycle (singular) [c]

 & cycles (plural) [c]'s

 (i(t')s awkward).

I'( wi)ll try to ignore the 's.

 

The wave's speed

 c=f*L

 (is frequency f

 multiplied

 wavelength L;

 & that) has units

 [cycles/second]*[meter/cycle]

 =[c/s]*[m/c]

Cancelling [cycle]'s

 we'( a)re left with [meter/second]'s=[m/s].

(As I said partial_plural is awkward

 & hopefully ignorable, or forgiveable.)

The point (there) is: the cycles

 in the

 top & bottom

 (=nummerator & denominator)

 either cancel

 or they don't;

 meaning they are there (implied)

 in(to) the equation

 or are not,

 but can be substituted in.

If we'( ha)ve substituted c=f*L

 for 1 of the (kinetic) energy's speeds,

 then we can (also) do that for the (=its) other speed,

 as well (just to be consistent).

Visually (for me), units of

 "per second_squared" [1/(s^2)]

 implies (to me) frequency_squared f^2

 when accompanied (=multiplied) with wavelength,

 & (=but) having no trace of [cycle]=[c] units

 because they (are) cancell(ed).

 

My beef (=complaint)

 is I'm getting indications

 of frequency_squared

 when I see units of

 "per second_squared"

 (in the energy unit: Joule)

 while trying to be consistent.

E.g.

If I have "per second" unit

 in a formula that has frequency f,

 that's obvious.

But instead that happens twice

 as "per second"*"per second"

 ="per second_squared",

 so I must conclude

 (to be consistent)

 that "frequency_squared"

 exists in that (energy) formula.

I know energy

 & momentum

 have different units.

& in momentum

 "per second"

 exists only once,

 so I must conclude

 momentum

 depends on a single frequency

 because it has only 1 "per second" unit.

I can NOT say that about energy

 because it has 2 "per second" units

 (i.e. multiplied together).

Thus I must conclude

 that energy depends

 on frequency_squared, instead.

E.g. Inspite of the nice or perfect constant h

 chosen (to fix the mess).

 

I hope that'( i)s a help.

Is that what you'( a)re looking for?

Edited May 1, 2018 by Capiert

[Strange]

4 hours ago, Capiert said:

Thus I must conclude

 that energy depends

 on frequency_squared, instead.

E.g. Inspite of the nice or perfect constant h

 chosen (to fix the mess).

But

that

is

not

what

is 

measured.

 

The

dimensions

of

h

are

chosen

to 

make

the 

relationship

E

=

h

f

work.

O

K

?

[swansont]

8 hours ago, Capiert said:

"Per second" [1/s] implies frequency

 & Joule has 2 of them.

Newton*second [N*s]

 has only 1.

 

 

Speed is meters per second. That's got 1/s and it's not a frequency. It's a rate.

So, no, 1/s does not automatically imply a frequency. 

Quote

 

Energy E has Joule(s) [J] units (in square_brackets)

 which is kilogram(s) [kg] (mass m)

 (multiplied by)

 meters_squared [m^2] (length_squared)

 per (=divided by)

 seconds_squared [s^2] (time_squared).

 

Yes.

Quote

 

If wavelength

 (is length

 per cycle,

 then its) units

 are meter

 per cycle [m/c].

 

Frequency

 is cycles

 per second [c/s]=[cps]=[Hz] Herz.

 

Cycles isn't a unit, as such.

 

Quote

 

The wave's speed

 c=f*L

 (is frequency f

 multiplied

 wavelength L;

 & that) has units

 [cycles/second]*[meter/cycle]

 =[c/s]*[m/c]

Cancelling [cycle]'s

 we'( a)re left with [meter/second]'s=[m/s].

(As I said partial_plural is awkward

 & hopefully ignorable, or forgiveable.)

The point (there) is: the cycles

 in the

 top & bottom

 (=nummerator & denominator)

 either cancel

 or they don't;

 meaning they are there (implied)

 in(to) the equation

 or are not,

 but can be substituted in.

 

Since cycles isn't a unit this doesn't matter, but as you say they cancel, so it still doesn't matter.

You end up with c as a speed, in m/s. No problem with the units.

 

Quote

 

If we'( ha)ve substituted c=f*L

 for 1 of the (kinetic) energy's speeds,

 then we can (also) do that for the (=its) other speed,

 as well (just to be consistent).

Visually (for me), units of

 "per second_squared" [1/(s^2)]

 implies (to me) frequency_squared f^2

 when accompanied (=multiplied) with wavelength,

 & (=but) having no trace of [cycle]=[c] units

 because they (are) cancell(ed).

 

First of all, why would you substitute c=fL into any equation that doesn't deal with light? That's what the equation applies to. There's no justification for randomly substituting into other equations.

Second of all, 1/s is not always a frequency, as I've already said. 

 

Quote

 

My beef (=complaint)

 is I'm getting indications

 of frequency_squared

 when I see units of

 "per second_squared"

 (in the energy unit: Joule)

 while trying to be consistent.

E.g.

If I have "per second" unit

 in a formula that has frequency f,

 that's obvious.

But instead that happens twice

 as "per second"*"per second"

 ="per second_squared",

 so I must conclude

 (to be consistent)

 that "frequency_squared"

 exists in that (energy) formula.

I know energy

 & momentum

 have different units.

& in momentum

 "per second"

 exists only once,

 so I must conclude

 momentum

 depends on a single frequency

 because it has only 1 "per second" unit.

I can NOT say that about energy

 because it has 2 "per second" units

 (i.e. multiplied together).

Thus I must conclude

 that energy depends

 on frequency_squared, instead.

E.g. Inspite of the nice or perfect constant h

 chosen (to fix the mess).

 

But since these aren't frequencies, this is an error on your part. And there's nothing wrong with the units. 

Quote

 

I hope that'( i)s a help.

Is that what you'( a)re looking for?

 

Something like that.

We've seen that:

A) There's nothing wrong with the units.

B) You have mistakenly associated 1/s with frequency. That's the error you need to fix. 1/s can be a frequency, but you can't say that it must be a frequency.

[Capiert]

On 1 May 2018 at 11:58 AM, swansont said:

Speed is meters per second. That's got 1/s and it's not a frequency. It's a rate.

So, no, 1/s does not automatically imply a frequency. 

Yes.

Cycles isn't a unit, as such.

Since cycles isn't a unit this doesn't matter, but as you say they cancel, so it still doesn't matter.

You end up with c as a speed, in m/s. No problem with the units.

 

First of all, why would you substitute c=fL into any equation that doesn't deal with light?

(Capiert:

E=m*(c^2). That's also a convertable "energy".

I'm irritated that the website's software

 does not work right here

 to allow the quote

 to split for commenting,

 although it allows at other places.)

Something like that.

We've seen that:

A) There's nothing wrong with the units.

B) You have mistakenly associated 1/s with frequency. That's the error you need to fix. 1/s can be a frequency, but you can't say that it must be a frequency.

Please identify which "per second" of the (kinetic) energy('s equivalent) is the frequency's;

 & which "per second" is for e.g. a speed

 from the h*f formula.

I can NOT distinguish it.

Can you?

What is the reason

 for the break in symmetry

 & the discontinuity?

Quote

But since these aren't frequencies, this is an error on your part. And there's nothing wrong with the units. 

Are you sure?

I see the units,

 as excellent clues (=hints),

 when consistency is disrupted.

Why should nature make exceptions

 (to her rules)?

Quote

That's what the equation applies to. There's no justification for randomly substituting into other equations.

Second of all, 1/s is not always a frequency, as I've already said. 

Please explain the asymmetry.

Edited May 2, 2018 by Capiert

[swansont]

1 hour ago, Capiert said:

Please identify which "per second" of the (kinetic) energy('s equivalent) is the frequency's;

Neither of them

1 hour ago, Capiert said:

 & which "per second" is for e.g. a speed

Rate. Both of them refer to a rate. The speed term is squared.

1 hour ago, Capiert said:

 from the h*f formula.

I can NOT distinguish it.

Can you?

The hf formula doesn't belong on the KE equation.

1 hour ago, Capiert said:

What is the reason

 for the break in symmetry

 & the discontinuity?

What symmetry? What discontinuity?

1 hour ago, Capiert said:

Are you sure?

Yes.

1 hour ago, Capiert said:

I see the units,

 as excellent clues (=hints),

Hints, perhaps, but typically trying to discern the physics from the units is usually backwards, or leads you down the wrong path, other than checking for proper and consistent units.

Looking at units you might conclude that energy and torque are the same, but they aren't. That kind of analysis isn't very illuminating.

1 hour ago, Capiert said:

 when consistency is disrupted.

Why should nature make exceptions

 (to her rules)?

What rules are you referring to?

1 hour ago, Capiert said:

Please explain the asymmetry.

What asymmetry?

[Capiert]

3 hours ago, swansont said:

Neither of them

Rate. Both of them refer to a rate. The speed term is squared.

The hf formula doesn't belong on the KE equation.

But they are (related) equatable.

Quote

What symmetry? What discontinuity?

That from the the h*f (equation),

 1 of the units "per second"

 must represent frequency.

Why doesn't the other, also?

Quote

Yes.

Hints, perhaps, but typically trying to discern the physics from the units is usually backwards,

I'd say back engineering.

Quote

or leads you down the wrong path, 

I'd say: leads you back "up" to (pinpoint) the errors. =The errorful path.

Or in your version:

 get (down) to the bottom

 of the problem.

Quote

other than checking for proper and consistent units.

Bingo! You said it: consistent.

Quote

Looking at units you might conclude that energy and torque are the same, but they aren't.

We both know angle(s) plays an important role there,

 too often underestimated

 as not being its (=an) own parameter.

Instead you use vectors.

Quote

That kind of analysis isn't very illuminating.

What rules are you referring to?

c=f*L substitution. Why should that substitution be restricted to only once,

 when twice is possible?

You argue that "per second" does not have to be frequency,

 but we have recognized f in the h*f=KE equality (=substitution), KE~m*(v^2)/2,

 or E=m*(c^2).

Quote

What asymmetry?

That if we substitute c=f*L (once),

 what pervents its 2nd substitution

 in E=m*(c^2)?

 

Edited May 3, 2018 by Capiert

[swansont]

8 hours ago, Capiert said:

But they are (related) equatable.

Mathematically, yes. But that doesn't necessarily mean there is a reason to do so when you are looking at the physics.

8 hours ago, Capiert said:

That from the the h*f (equation),

 1 of the units "per second"

 must represent frequency.

Why doesn't the other, also?

Because speed isn't a frequency.

What does this have to do with symmetry?

8 hours ago, Capiert said:

I'd say back engineering.

I'd say: leads you back "up" to (pinpoint) the errors. =The errorful path.

And you would be wrong.

8 hours ago, Capiert said:

 Bingo! You said it: consistent.

Yes. If you are calculating an energy, you need to end up with an equation that has units of energy. We get people around here that try to write down equations which do not have consistent units.

They are a necessary condition for a correct equation. But they are not a sufficient condition for a correct equation. IOW, all you can tell from the units is if they are inconsistent, then you know there is something wrong with the equation.

 

8 hours ago, Capiert said:

 c=f*L substitution. Why should that substitution be restricted to only once,

 when twice is possible?

How is that a "rule"?

You use c= fL when you are talking about light, because that's what the equation is applied to.

8 hours ago, Capiert said:

You argue that "per second" does not have to be frequency,

 but we have recognized f in the h*f=KE equality (=substitution), KE~m*(v^2)/2,

 or E=m*(c^2).

What?

Where has anyone validly substituted hf into kinetic energy?  

 

8 hours ago, Capiert said:

That if we substitute c=f*L (once),

 what pervents its 2nd substitution

 in E=m*(c^2)?

 

Under what circumstances are you making this substitution?

[Strange]

9 hours ago, Capiert said:

but we have recognized f in the h*f=KE equality (=substitution), KE~m*(v^2)/2,

 or E=m*(c^2).

Although equating these doesn't really make any sense, you can equate the proper form of the equation for photons:

[math]E^{2}=p^{2} c^{2} + m^{2} c^{4}[/math]

A photon has no mass, so:

[math]E^{2}=p^{2} c^{2} [/math]

[math]E=p c[/math]

Now we can equate this: 

[math]E = p c = h f [/math]

Substituting for c = f * L (in your notation):

[math]E = p f \lambda = h f [/math]

Therefore:

[math] p  = \frac {h}{\lambda} [/math], which is correct.

One thing you have repeatedly avoided is the fact that E=hf is an observed property of photons. And the units of h are chosen to make there elationship work.

Edited May 3, 2018 by Strange