The h Bar - 2019-05-04

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3:21 AM

-1

Q: Electric flux and electric field

Consider a point charge located at the vertex of a cone. Then field lines are perpendicular to the normal of the curve surface then there is no flux through the curve surface. But mathematically we get flux through the curve surface. How is that possible ?

user351417

@PhysicsMeta ::enormous sigh::

user351417

We have a physics question on the meta site.

user351417

Do people not realize that something's up once they start looking at the tag requirements?

3:58 AM

Taking over some back end for 34 hours. Will be slapping code all night 34 hours straight on relay with my boy on the other continent(europe_). No stopping us tonight.

Got tyga in my ears. Power coding like tomorrow does not exist

Hardcore !!

Tonight I take on the world

user351417

4:15 AM

@EmilioPisanty I just realized that the underwater railings near the edges are made out of k-nex. I used to love those things.

user351417

I can see those in E1 of the 2019 competition.

6:03 AM

slinging code . . . . Hard!

6:27 AM

Whole room is bumping . . going even harder. Just implemented some impossible stuff. Yaaaaaaaaaassssssssssssss

third coder just joined .

tonight we take over the world for a few hours

6:45 AM

Hey can anyone answer my question physics.stackexchange.com/questions/477550/…

Just the first doubt will do.

7:11 AM

@Sid Uhmm, it was closed as dupe. Check the original questions if they answer yours. If not, then edit it into your question, so visible as it is only possible, why not.

Constantine Black

7:46 AM

Markov chains for monte carlo with parallel processing(use of multiple CPU cores)? Anyone has a reference/example?
in Python3

1 hour later…

Novalium Company

8:59 AM

Can I find the energy of any wave using $E = hf$ ?

For example the energy of a wave on the beach?

nvm

I'm stupid

This formula is only for EM waves?

because of c = wavelength x frequency ?

@NovaliumCompany and de Broglie waves

Novalium Company

de broglie waves is basically any particle?

Electron?

Yes

Novalium Company

But I thought $E = mc^2$ is used for finding the energy of any particle?

$E^2 = p^2c^2 + m^2c^4$

That reduces to $E=mc^2$ only when $p=0$

i.e. $E=mc^2$ is the energy of a stationary particle.

Novalium Company

9:05 AM

So which formula is used to find the energy of a small molecule for example? Both work?

oh I'm stupid

@NovaliumCompany the equation I gave is the total energy. $E=mc^2$ gives you only the energy when stationary.

The equation comes from special relativity and applies to everything, both classical and quantum mechanics.

Novalium Company

$E = hf$ - This formula can be used to find the energy of EM and any particle (everything)?

The the de Broglie frequency is given by $E=h\nu$ and the wavelength by $\lambda=h/p$

Novalium Company

of what?

The particle. The photon, electron or whatever.

Novalium Company

9:10 AM

Where does this apply $E^2 = p^2c^2 + m^2c^4$ then?

I'm just trying to make some sense of this

user image

So using this $\lambda=h/p$, I can calculate the wavelength of my body?

where $p = mv$?

in my textboo it's given it's only significant for microscopic particles

so i guess no

@NovaliumCompany yes, though as @Scáthach says this doesn't apply to macroscopic bodies for reasons that are a bit complicated.

Novalium Company

alright then

thx :)

9:50 AM

@peterh and all others here, I modified my question and would be glad if someone could answer it,

1

Q: Derivation of Lagrange's Equation

So I was going through the derivation of the Lagrange's Equation given in Goldstein. And I have doubts about some of the steps. I would appreciate some clarifications on them. First: $$ \frac{d}{dt} \frac{\partial r_i }{\partial q_j} = \frac{\partial \dot{r_i}}{\partial q_j }$$ So in th...

classical-mechanics lagrangian-formalism velocity coordinate-systems

Constantine Black

10:01 AM

Hi. Should the order parameter of a system of hard disks be small at high or small densities?

@Sid Make it clear, at least in a comment (or in a short, well visible edit in the post), how it differs from the original questions. Avoid also to his another close reason, for example your question should not look homework-like. I.e. if your question is about some calculation, then 1) you should ask from some physical concept 2) you should show your own research.

@peterh I did all of those.

@Sid Unfortunately, the PSE is imho far more restrictive in the accepted questions, as it should be reasonable. Also you have a reopen vote related your own question, I can give you yet another, but a reopen requires 5.

@peterh Thanks

@Sid If your question doesn't get a reopen, you can also try physicsoverflow.com , which is far more broad in topicality. First make in your question very visible, how does it differ from the originals. If you are sure that the community vote had a false result, then you can ask a new question about the closure of your original question on the physics.meta.stackexchange.com .

@Sid Below your question, you can see a "reopen" link. Click that. Note, with your edit also a reopen vote was automatically started, and if you lose that, the reopen of your question becomes far harder.

10:08 AM

@peterh Ok Will do. Thank you very much

@Sid Qmechanic's answer to one of the duplicates contains an explicit expression for the total time derivative that answers your question already

@Sid A reopen vote can be started by: 1) if you edit your closed question 2) if anybody, included you, casts a reopen vote. Since you have more than 250 rep, also you have a reopen vote regarding your own questions. | To vote for the closure/reopen of the question of others, you would need 3000.

@Sid Mods can close and reopen any question with a single click, but they won't do it without a strong reason.

Abhas Kumar Sinha

10:53 AM

@Qmechanic hello

@AbhasKumarSinha : Hello.

Abhas Kumar Sinha

@Qmechanic Fixing atoms and molecules? XD

@Sid What's your problem with Lagr?

@Qmechanic What kind of mechanics do they use in Quantum Mechanics? Lagrangian or Hamiltonian?

@AbhasKumarSinha Both and neither. Canonical quantization approaches are more based on the Hamiltonian formalism, while path integrals usually stick more to the Lagrangian side of things.

There's different formulations of "quantum mechanics" just like Lagrangian or Hamiltonian mechanics are different formulations of "classical mechanics"

11:26 AM

@AbhasKumarSinha if you skim volume 3 they get the Schrodinger equation by differentiating $e^{icS}$ where $S$ is the action, $\frac{\partial S}{\partial t} = - H$ for $H$ the Hamiltonian (last chapter of mechanics) and $c$ is a constant to make the argument of $e^{icS}$ dimensionless and turns out to be 1 over $\hbar$ so you're using everything all over the place

Abhas Kumar Sinha

11:53 AM

@bolbteppa Oh okay

what was that relation between the lagrangian and hamiltonian Functions?

I forgot that,

probably some kind of like this : $\dot x \dfrac{\partial L}{\partial x} = H$

Or something like that...

@bolbteppa hello

Legendre Transformation

Abhas Kumar Sinha

what was the formula?

60

A: Physical meaning of Legendre transformation

Legendre transformations are commonly used in thermodynamics (to switch between different independent variables) and classical mechanics (to switch between the Lagrange and Hamilton formalisms). But you rightly ask: what exactly is a Legendre transformation? Where does it come from? What makes it...

Abhas Kumar Sinha

oh okay...

@bolbteppa While applying the Calculus of Variation in the action integral, can cause the path to change. So, how it doesn't affect the equation of motion (which is a different path) which is a relation of velocity, and position as a function of time

Can you rewrite that question as though we were talking about basic calculus and see if it helps

Abhas Kumar Sinha

12:07 PM

@bolbteppa Suppose there are 2 paths $A$ and $B$ from point $a$ to $b$ from time $t_1$ to time $t_2$. There is an equation of path let's denote it by $q$. I find the equation of KE and PE of an object travelling in that path and calculate the minima action integral by Calculus of Variation, no now we get second path $q$ which is minima of action integral, suppose. So it must be different from path $A$

So, equation we get after using the calculus of variation should be different from the different path?

So, how can both equation of path be correct?

I'm not sure what that means, basically you're just minimizing a function (the action integral) - it turns out the minimum of this specific function (the action integral) is a curve that must satisfy a differential equation

Abhas Kumar Sinha

from "calculate the minima action integral by Calculus of Variation", I mean that using the Euler Lagrangian equation.

@bolbteppa When I use Euler-Lagrangian equation, then, the path equation will be same to the original path?

If we're minimizing a function $f(x_1,x_2) = x_1^2 + x_2^2$ it turns out the minimum must satisfy the 'differential equation' $\frac{\partial f}{\partial x_i} = 0$, if we're minimizing the function $f(x_1,x_2) = \int [\frac{1}{2}m (\dot{x}_1^2 + \dot{x}_2^2) - \frac{1}{2}k(x_1^2 + x_2^2)]dt$ it turns out the minimum must satisfy the differential equation (EL equations)...

Abhas Kumar Sinha

Okay...

Got the logic... :O

12:24 PM

To minimize $x_1^2 + x_2^2$ you literally just take a partial derivative and set it to zero, $\frac{\partial}{\partial x_i}(x_1^2 + x_2^2) = 2 x_i = 0$, and then solve for the $x_i$ making this true, i.e. $x_i = 0$, but for $f(x_1,x_2) = \int [\frac{1}{2}m (\dot{x}_1^2 + \dot{x}_2^2) - \frac{1}{2}k(x_1^2 + x_2^2)]dt$.

If you blindly take the derivative $\frac{\partial}{\partial x_i} \int [\frac{1}{2}m (\dot{x}_1^2 + \dot{x}_2^2) - \frac{1}{2}k(x_1^2 + x_2^2)]dt = - \int k x_i dt$ this is clearly wrong because $\dot{x}_i$ involves $x$ so you need to peel off the derivatives, which is why you try to write the $\delta S$ as $\delta S = \int (...) \delta x_i dt$ which is as close as you can get to writing $\frac{\partial S}{\partial x_i} = 0$

2

A: Is anything expected to be found at CERN with 14 trillion electron volts that wasn't found at 13?

but it still seems to me that going from 13 to 14 is a very narrow window to find something new. It is normal for people when units go from GeV to TeV , just a little letter change, to see it as a small change, but the difference it makes to capabilities can be a step function, because 100...

ah ha!

ah ha!!!

> I was involved in the almost discovery of the Higgs at the time, with the ALEPH experiment:

found her.

(.... not that it makes all that much of a difference, tbh.)

and if $(...)\delta x_i$ is zero for all $\delta x_i$ then we can say $\delta S = (...) \delta x_i = 0$ i.e. $\frac{\partial S}{\partial x_i} = E.L. \ Eq's = 0$

Abhas Kumar Sinha

ah okay

doi.org/10.1103/PhysRevLett.33.1631 the next time someone starts bashing anna's (lack of) TeX usage, imma send them to check out how they had to render tables on a typewriter. Heck, that figure looks inked in.

Novalium Company

1:27 PM

If according to QM, particles dont have a defined position nor momentum, until observed, does that mean they didnt exist?

Also is it possible to imagine a particle?

a quantum particle

Novalium Company

1:57 PM

Also can somebody help me understand the relationship between these equation, how do they relate and to what.

E = hf

the de Broglie wavelength equation

Einsteins advanced E = mc2 but with the rest mass..

difference between momentum of a photon and electron...

(overall those baeic QM equations)

I'm afraid I'm not exactly sure what you want to know

user351417

@NovaliumCompany For a start, the mass-energy equivalence isn't 'basic quantum mechanics' : P

> so there are definitely substantial details left to explain there, which can hopefully be explained by a nuclear physicist.

@rob?

0

Q: Question on two-neutrino double electron capture

There was a fascinating paper in Nature recently, on the observation of two-neutrino double electron capture in xenon, with a half-life time of $1.8\times 10^{22}$ years. The process described in the article is $$^{124}\mathrm{Xe} + 2e^- \to {}^{124}\mathrm{Te} + 2 \nu_e.$$ According to Wikipe...

nuclear-physics radiation neutrinos

hi @NorbertSchuch

@EmilioPisanty Well, that was good timing. I'll look.

user351417

@ACuriousMind How did you find that web version of the results from last election which you sent me a while back? physics.stackexchange.com/election/3?tab=election doesn't include that...

2:09 PM

@Rishi physics.meta.stackexchange.com/q/9256/50583 has it

user351417

Thanks!

Novalium Company

3:17 PM

@ACuriousMind I think so :P

Can the photon energy formula E = hf be applied to electrons, small molecules or even humans, even though they have a very small wavelength?

And also whats the difference between the relativistic formula for energy vs the photon one?

take the relativistic relation $E^2 = p^2 c^2+m^2 c^4$ and suppose $m=0$ as for the photon

you get $E=pc$, which is the photon's energy-momentum relation

Novalium Company

ok? Im talking about which formula is used for what sized objects?

So pc = hf?

Sure. And $c=f\lambda$ so $p=hf/c = h/\lambda$

i.e. de Broglie wavelength

Novalium Company

3:32 PM

wow. nice.

But I dont understand if I can use both energy formulas on anything really?

Can I use the relativistic for microscopic and macroscopic objects and also can I use the photon one for micro and macro obj?

Okay hehe. You guys need to be careful. I got pinged when you guys were trying to ping the other @Sid

3:55 PM

@Sid Hey man

Novalium Company

4:44 PM

Can I use the relativistic formula for energy on microscopic and macroscopic objects and also can I use the photon one for micro and macro obj

5:02 PM

@NovaliumCompany $E=pc$ is only used for massless particles.

The total energy is more generally $E^2 = p^2 c^2 + m^2 c^4$. When we deal with low velocities (compared to $c$), using energy in special relativity approximately reduces to the same formulas seen in classical mechanics. To explain it further requires some calc 2 knowledge :/

Novalium Company

Oh, ok. So E = hf is for small massless particles (not electrons...) and $E^2 = p^2 c^2 + m^2 c^4$ is for bigger things such as rocks?

Im having hard time understanding which is for what

or maybe both formulas work for any sized ovj?

@NovaliumCompany OK, let's step back. As you mentioned, $E=mc^2$ is the rest energy, i.e. the energy that goes into resisting acceleration (equivalent to mass).

Novalium Company

No?

The idea that mass is equivalent to a form of energy is very important, since it means mass (rest energy) can be converted to other types of energy.

@NovaliumCompany No?

Novalium Company

I thought that formula was for just finding the potential energy any object has stored in it, no?

5:17 PM

@NovaliumCompany Special relativity doesn't usually deal with acceleration. So no forces, and therefore no potential energy

Novalium Company

oh oks

Instead, the idea is that $mc^2$ is a type of energy. But notice, we're just multiplying the mass by a constant $c^2$. So in essence, it's a straightforward conversion

Novalium Company

ok

We see that mass is equivalent to a form of energy, called "rest energy", that can be transferred to other types. Intuitively, it is the energy that goes into resisting acceleration

Novalium Company

got it

5:19 PM

The idea that mass can be converted to e.g. kinetic energy is huge, since it explains a lot of phenomena in the particle level. However, it only happens in noticeable amounts under extreme conditions, such as relativistic speeds.

Novalium Company

yep

In the macroscopic scale, we hardly notice it. Which is why we'd classically distinguish mass from a form of energy.

Novalium Company

we hardly notice what? The convertion of rest energy to kinetic?

OK, now back to the topic. $E_{total} = \sqrt{m^2 c^4 + p^2 c^2}$ is the total energy of the object, assuming there are no potential energy. In other words, it's simply rest energy plus kinetic energy (kinetic energy in SR is calculated as $E_{total} - E_{rest}$).

@NovaliumCompany Yep, or any kind of decrease in mass tbh. We classically have this notion of "mass is conserved", when in reality it's just a form of energy (which is still conserved relativistically) and can be converted to other types

Novalium Company

How can there be energy that resist energy? Its like its loosing energy to gain more :p

5:25 PM

@NovaliumCompany Energy that resists energy?

Novalium Company

rest energy?

I'm just saying that mass typically only converts to other types of energy in very small amounts

Rest energy is the energy that goes into resisting acceleration

Which lines up with your intuition of mass

Novalium Company

ahh a bit confusing

What is rest energy in the simplest terms?

Just take whatever intuition you have for mass, and multiply it by a constant $c^2$. That's rest energy :P

Novalium Company

got it

5:28 PM

The idea is that rest energy directly corresponds with mass, so they're simply the same concept. But the surprising thing is that rest energy is a type of energy, so it can be converted to other types (corresponding in a decrease in mass)

Novalium Company

got it

In everyday life, the conversion is too small to notice, so we simply assume "mass is conserved" as if it were not a type of energy.

Novalium Company

got it

In fact, remember when I said kinetic energy is just $KE=E_{total} - E_{rest}$?

Novalium Company

yep

5:31 PM

In classical mechanics, we mainly talk about changes in kinetic energy. If $E_{rest}$ is nearly the same before and after (i.e. we don't observe a noticeable change in rest energy), it'll just cancel out.

Novalium Company

yep

So for most non-relativistic purposes where rest energy hardly undergoes any conversion, it can be excluded from our calculations.

Novalium Company

yep

We also note that $E_{total} - E_{rest} = \sqrt{m^2 c^4 + p^2 c^2} - mc^2$ looks pretty ugly. But when velocities are very small compared to $c$, it approximately becomes $\frac{1}{2}mv^2$

To explain it precisely would require calc 2 knowledge though

Novalium Company

What has this have to do with where E = hf is used?

5:34 PM

Special relativity, like most types of modern physics, tries to generalize classical mechanics. We usually hope that in less extreme scenarios, our modern theories agree with classical mechanics precisely

@NovaliumCompany That's QM, not derivable from special relativity iirc.

Novalium Company

Im dont want to derive it

Im just asking if I plug the wavelength of a macro object into the formula, will I get its total energy?

It will work for light, but macroscopic objects are trickier

Novalium Company

they also have wavelengths with frequencies so why not :p

Using the de Broglie frequency of a macroscopic object is kind of unnecessary since their frequencies are so huge. It'll hardly give you a change in computations compared to regular classical mechanics, where we don't consider those

Novalium Company

Thank you

finally...

Ive been trying to understand this the whole day

saw endgame?

pretty nice

1 hour later…

Novalium Company

7:06 PM

I don't understand what exactly is quantized in QM? I know that the "amount" of each quanta step is Plank's Constant.

@EmilioPisanty It's the effect of pairing in nuclear structure. I'm pleased with the plot that I made there.

Novalium Company

There is a proton accelerator (the biggest in the world) that accelerates protons very very close to the speed of light and I was wondering, can we accelerate them above the speed of light? Would that mean that they'll travel back in time? Maybe we already have protons popping up in our present from our future?

@NovaliumCompany You can get as close as you want to $c$, but you can't exceed.

I like to tell beginning relativists about the electron accelerator at JLab, which two back-to-back linear accelerators arranged in a racetrack configuration.

The electron beam can make the loop at JLab as many as five and a half times.

The gamma factor $\gamma=1/sqrt{1-v^2/c^2}$ diverges to infinity as $v\to c$ from below (and is not well-defined for $v>c$)

The electrons are injected into the accelerator about about 50 MeV, which is already $\gamma = 100$.

7:20 PM

As such, the relativistic energy $E=\gamma mc^2$ diverges as $v\to c^+$. So if you could raise a proton to the speed of light, it'd have infinite energy.

Novalium Company

What's $\gamma$?

Each trip through a linac adds about 1 GeV to the beam energy.

It's a factor which shows up in a bunch of relativistic formulas.

Novalium Company

I need a simple answer please :D

In particular, the total energy $E$ is related to the rest energy $E_0=mc^2$ as $E=\gamma E_0$

7:21 PM

When I was at JLab, we ran experiments in three halls at once, and with "five-pass beam" there could be fifteen different electron energy bunches in the accelerator at the same time.

Novalium Company

@Semiclassical Are you talking to me? :P

So it tells you to what extend the total energy is boosted relative to its rest energy

But they don't run into each other, because they're all traveling essentially at $c$ --- at the same speed.

ya

@rob How much variation in speed would they typically have?

Novalium Company

I need a simple explanation with words, please.

7:22 PM

@NovaliumCompany Gamma is a number called the "Lorentz" factor, that shows up a lot in special relativity. It starts at 1, and gets higher and higher to infinity as you approach light speed

As an example of it showing up, remember $E_{total} = \sqrt{E_{rest} + p^2 c^2}$ from earlier?

Novalium Company

Yep

Another way you can calculate total energy is by $E_{total} = \gamma E_{rest}$

@Semiclassical Within the timing precision of the machine, the 50 MeV electrons and the 5 GeV electrons are all exactly at the speed of light.

Where $\gamma$ is the Lorentz factor, which depends on your speed

@rob nice

7:24 PM

Accelerator folks use other parameters to keep track of the beam quality.

(definitely a bit of FAPP in that description but that's how experiments go)

There's a lot of other uses of the Lorentz factor, and its formula is $\gamma = \frac{1}{\sqrt{1-( \frac{v}{c})^2 }$

Sigh yep I'm rusty

Novalium Company

Should I bother asking how you get to $E_{total} = \gamma E_{rest}$ ?

Wth is with LaTeX

For instance, electrostatic repulsion makes the electron bunches tend to spread out, so some of your focusing magnets have to delay the front of a bunch and accelerate the back of a bunch.

7:26 PM

@NovaliumCompany It's not an obvious result, but it's pretty beautiful. It'll require some background on the math of SR though

Novalium Company

@SirCumference SR?

Special relativity

Novalium Company

oh

Don't derive it, pleaes ;D

Basically, $\gamma$ shows up the moment you start looking at how coordinate transformations work in SR

namely, with coordinate transformations that change the velocity of your reference frame

@NovaliumCompany Note that $E_{total} = \gamma E_{rest}$ and $E_{total} = \sqrt{m^2 c^4 + p^2 c^2}$ are both separate ways of finding the total energy, but both are equal. We can use the first if we know our particle's velocity, and we can use the latter if we know the momentum instead

Using them together allows us to relate a lot of quantities together, which is very useful

Novalium Company

7:28 PM

Got it

note that $1/\sqrt{1-v^2/c^2}$ shows up when doing stuff like length contraction / time dilation

Novalium Company

so what has that have to do with why we can't go up than c?

Think of v=c as the point where your formulas breakdown because you are trying to divide by zero.

@NovaliumCompany Let's say I'm moving at velocity $\vec{v}$ compared to my friend

Novalium Company

ye

7:31 PM

I want to know how he sees objects move. In classical mechanics, I'd subtract his velocity from everything else. That would tell me how he sees them move

It's a simple process. If I see a train moving rightward at 5 miles per hour, and I see my friend move rightward at 3 miles per hour, he sees the train move forward at 2 mph

Novalium Company

got it

next? :p

@NovaliumCompany He sees himself as stationary and me moving with speed $-\vec{v}$. Classically, this is simple and works very well. But it makes an incorrect assumption: it assumes that the only difference between our perceptions is the spatial motion of objects

Using that technique would imply that light can change speeds, which experiments have disproven

In special relativity, rather than simply adjusting how we perceive everything moving through space, we also consider that my friend and I are also observing time flow differently

Novalium Company

oks

Long story short, if my friend and I are moving at different velocities, I also need to account for how time moves differently for him. The greater our speeds differ, the more our flow of time differs

Novalium Company

yep

what has that have to do with why we can't go above the speed of light?

7:38 PM

Well, the Lorentz factor is based on how we convert between my perceptions and his. If I measure a time $\Delta t$ passing by, then the time he perceives passing is $\gamma \Delta t$

$\gamma$ depends on the difference in our speeds. When there is no difference, $\gamma = 1$. But as the difference goes closer to $c$, $\gamma$ goes higher and higher to infinity

It asymptotes

Novalium Company

Is that proven?

Yep, I tried writing down the Lorentz factor formula but Latex isn't working with me

The Lorentz factor or Lorentz term is the factor by which time, length, and relativistic mass change for an object while that object is moving. The expression appears in several equations in special relativity, and it arises in derivations of the Lorentz transformations. The name originates from its earlier appearance in Lorentzian electrodynamics – named after the Dutch physicist Hendrik Lorentz.Due to its ubiquity, it is generally denoted γ (the Greek lowercase letter gamma). Sometimes (especially in discussion of superluminal motion) the factor is written as Γ (Greek uppercase-gamma) rather...

Novalium Company

wikimedia.org/api/rest_v1/media/math/render/svg/…

:)

I can google don't worry :D

So already we're starting to see problems. Another interesting result is how we perceive energy

Novalium Company

wait

I have a few questions about this

7:41 PM

Yeah?

Novalium Company

Velocity is defined as the rate at which something passes through space for a given time. And the units we use to measure that distance are meters, kilometers... and that distance is defined based on the speed of light.

Something is weird here.

@NovaliumCompany So in classical mechanics, we assume all motion is relative. Special relativity says that all velocities are relative, except for one: the speed of light must always be constant

Novalium Company

The definition of speed of light depends on the speed of light wtf :D

@skillpatrol thanks for agreeing xD

jk

The formula converting between what I see and my friend sees is a bit complicated. For small differences in our speeds, its approximately just classical method of subtracting his velocity.

For higher speeds, it gets a bit different

Novalium Company

@SirCumference c = x m/s right?

7:46 PM

@NovaliumCompany Anyway, to answer your question, velocity and flow of time both get affected when I try to switch to my friend's perceptions

But they get affected in such a way that the speed of light remains constant

Novalium Company

I'm talking about something else

I'm talking about that the speed of light is based on the speed of light...

c = 299 792 458 m / s

nvm

I'm stupid

as always... :D

No worries, this stuff is tricky at first

I'd recommend learning a bit of linear algebra so you can understand the transformation from my perceptions to my friend's

It makes the subject a lot more geometric and intuitive

Novalium Company

It just seems weird to me that the speed of light is basically m/s, where the definiton of a meter is again based on the distance the light travels...

@NovaliumCompany But finally to answer your question, let's say he has mass $m$. He sees himself as stationary, so he measures his mass as $mc^2$, i.e. just his rest mass. I observe him moving though, so I think he has energy $\gamma mc^2$ (remember, our formula from earlier).

Novalium Company

@SirCumference I understood what you were talking about, I've watched vids on time dilation, length contraction...

@SirCumference yep because $\gamma$ depends on his velocity?

7:50 PM

@NovaliumCompany Yep, exactly. Anything else that he saw as stationary also gets energy scaled by $\gamma$,

Novalium Company

got it

Now, let's say we have an object. Can we ever bring it to the speed of light, from my perspective? Well, the object has rest mass $mc^2$. From my perspective, it will have total energy $\gamma mc^2$; as it gets faster and faster, approaching the speed of light, the $\gamma$ term gets larger

$\gamma$ asymptotes at $c$. So if the object is to reach lightspeed from my perspective, I would need to see it gain infinite energy

Obviously adding infinite energy isn't practical :P

Novalium Company

When was $\gamma = 1$?

Zero velocity difference between the object and me

Novalium Company

got it

7:55 PM

So a lot of things get more extreme as you change to perspectives closer to light speed, e.g. time dilation, energy difference, etc. A lot of these depend on the Lorentz factor

At light speed, the Lorentz factor goes to infinity, so a lot of things also go to infinity and stop being possible. Beyond light speed, the calculations lead to ludicrous results

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So the explanation is that as something approaches the speed of light, it starts to gain infinite energy due to $\gamma mc^2$, which is impossible?

@NovaliumCompany Yep!

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You could have just said that at the beginning... xD

Well...yeah lol

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But thanks for the introduction to $\gamma$. It actually helped a lot!

7:58 PM

Np. Though again, I highly recommend learning linear algebra first

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You know what bothers me

Then a proper introduction to SR will become a lot more intuitive

Fewer (seemingly random) formulas, and more geometric beauty

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100 years ago, things in physics looked quite different and it seems as though we continue to think we've finally gotten to the "truth" and then a 100 years later, we find something else and everything changes... again.

So in my opinion SR might turn out to be false a few hundered years later.

We found out Newton's law weren't exactly what's going on, and what if our knowledge now is not what's going on. And when does this cycle end.

It's already based on some oversimplifications, e.g. it neglects gravity

GR is the more advanced and accurate theory, but even it is incomplete for extreme scenarios (such as dealing with black hole singularities, or dealing with the Big Bang)

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Nvm, in my opinion physics after a few hundred years will look completely different from what we think it is now.

And again, it's all based on our perception of reality and we have found out that our perception doesn't always tell the truth.

What what is true in general...

ahh just stop me here before I have an existential crisis again...

8:15 PM

@NovaliumCompany Should note before I forget, $E_{total} = \gamma mc^2$ assumes your object has mass. If you try to use it on a photon, then you'd simultaneously bring $m$ to zero but $\gamma$ to infinity (since its velocity is $c$), which causes problems (zero times infinity)

Using $E_{total} = \sqrt{m^2 c^4 + p^2 c^2}$ instead for that specific situation would avoid the problem

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Or simply $E = hf$?

Well yeah :P

Since we know light is actually a wave as well as a particle

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Yep :)

@SirCumference How do we know quarks and bosons... even exist (elementary particles in general)?

@NovaliumCompany Experiments maybe. Deriving them theoretically (if that's possible) is beyond what I know :)

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I suppose it's impossible to imagine those elementary particles because we cannot even imagine an electron nor a proton..?

Also, how do they send electrons through slits, I mean, don't the electrons bump into the air? It's done in a vacuum environment maybe?

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9:20 PM

For anything to have a de Broglie wavelength $\lambda = \frac{h}{mv}$, it must be moving?

(v must not equal 0)

9:36 PM

@NovaliumCompany yup, most particle experiments are in a vacuum

and yes it’s gotta have some momentum

9:59 PM

A possible, maybe too broad definition of unclear questions: You need more effort to understand it, than answer it.

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