The h Bar

I don't know the reasons. Either we've got busy enough to need a new mod or one of the mods has chosen to stand down. I assume there'll be an official announcement soon.

user116211

@JohnRennie ohh....sure.

user116211

Well, that is surprising; this would be my first election here at PSE.

John Rennie

Vote for me. I'll ban all discussion of thermodynamics! Hear that @KaumudiHarikumar :-)

user116211

@JohnRennie Seriously, you are in my list of nominees.

user116211

12:37 PM

@JohnRennie please, optics too.

John Rennie

@MAFIA36790 Nope, I'm too divisive a figure. I tend to make it obvious who I think the twits amongst us are. Moderators have to be more measured than I could manage.

user116211

Maybe we might be missing Kyle since he was also in my list for any future elections. He was active in meta as well as reviews.

Xasel

@JohnRennie:Well why you enemity with thermodynamics

user116211

@Xasel Who likes thermodynamics O.o

Xasel

I

John Rennie

12:41 PM

@Xasel I'm joking about thermodynamics. It's an exceedingly important part of physics. It's just that I found the thermodynamics courses deadly dull at university.

user116211

Let's see when we hear the official announcement.

ACuriousMind

@JohnRennie How did you get that information from a staff member? >100k rep privileges? :P

John Rennie

And the thermodynamics problems seemed to be just excuses for getting us to solve ridiculously complicated situations involving ideal gases, heart engines and walruses.

user116211

@JohnRennie nods, nods

ACuriousMind

...walruses?

Xasel

12:43 PM

@JohnRennie:so what shall we do to add poetic beauty to the subject so that it can allure/"bewitch" those on the quest of queching insatiable thirst of knowledge(9too much poetics)

ACuriousMind

What sort of thermodynamics exercise involves walruses?

John Rennie

@ACuriousMind I might have misremembered the walruses

user116211

@JohnRennie Global Warming homework problems ;P

John Rennie

@ACuriousMind I'd love to be able to boast about some exclusive access to the bowels of the SE, but it just came up in conversation.

Xasel

haha

ACuriousMind

12:45 PM

I don't think I want exclusive access to the bowels of anything

John Rennie

:-)

ACuriousMind

Couldn't you have chosen a...prettier metaphor? :)

Xasel

things happen when phsicst gets poetic

John Rennie

Anyhow having dropped my bombshell I now have to go score myself some lunch. See you all in a couple of hours.

@Xasel Walruses and bowels - not amongst the most poetic subjects even for a physicist :-)

user228700

@JohnRennie Noooo :P Y you do this?

Xasel

12:47 PM

@JohnRennie:it's all about perspective :)

John Rennie

@KaumudiHarikumar you're safe - there is zero chance of me being elected a modrator :-)

ACuriousMind

@JohnRennie At least you've managed to stay room owner longer than Chris, so maybe you're overestimating your divisiveness ;) (CW's account is gone entirely btw, for those that hadn't noticed)

0celo7

@JohnRennie why

@ACuriousMind He got banned?

ACuriousMind

@0celo7 The deletion was his own wish

John Rennie

@0celo7 principally because I'm not going to stand. That's going to make it kind of hard to elect me :-)

ACuriousMind

12:48 PM

He got chat banned for a day, though

0celo7

Holy [REDACTED], what happened?

user228700

I don't even understand all this about moderators. It was my understanding that all high rep. users became mods after a point. Somebody please explain?

ACuriousMind

@0celo7 He spoke his mind about a certain user, a flag was cast, he was banned. Appearently that upset him enough to stop participating in the site entirely, although of course there may be stuff I'm not privy to

Sir Cumference

@JohnRennie Yes, but many engineers still use pounds. In 1998 the U.S. spent $125 million on a Mars probe. Problem was, some of the engineers have their measurements in pounds, and the other engineers assumed it was in newtons. The discrepancy caused the probe to stray way off course, lost in space.

This is why I love this country...

0celo7

@ACuriousMind Do you know which user

Xasel

12:50 PM

I hate imperial unit :(

John Rennie

@0celo7 who do think?

0celo7

Looks like CW's a bad loser

@JohnRennie No idea.

I don't pay attention to drama around here.

Sir Cumference

@Xasel Preaching to the choir, buddy

0celo7

nor participate, etc.

John Rennie

@KaumudiHarikumar High rep users get some extra powers, but they are still only limited.

ACuriousMind

12:51 PM

@KaumudiHarikumar You are right that many functions of moderation - like closing and reopening questions, editing, protecting question - are available to all high rep users, but a "moderator" is something more: They've got those diamonds next to their names on the site, and they are the people who deal with all the flags you raise that don't go into a high rep review queue

John Rennie

There's an enormous gulf between what I can do and what a moderator can do.

user116211

@KaumudiHarikumar stackexchange.com/about/moderators

user116211

36

Q: When do moderator elections take place?

I'm really confused on this. I have no idea on several things: When do elections begin? Is there a set time each year? Do elections actually take place each year? Can I vote? Where do I vote? Etc. Can someone please help me?

Xasel

maybe @ACuriousMind canstan for mod

user228700

Ohh, I seee. Thanks!

user116211

12:53 PM

@Xasel in my list too.

user228700

@JohnRennie Enormous? Okay...

John Rennie

High rep users can vote to close, delete, etc but it takes five votes to close a question. Modrators can singlehandedly close and delete questions.

user228700

Wow! Cool.

John Rennie

They can also suspend users for talking about thermodynamics and other antisocial behaviour.

Xasel

think of them as titans to greek gods

Sir Cumference

12:54 PM

@KaumudiHarikumar Not to mention, when a mod deletes a post, it is absolute. Only another mod can undelete it, not a high rep user.

Xasel

gree-->greek

0celo7

If high rep users were mods, Lumo would be a mod. And we'd all be dead, probably.

user116211

@Xasel You can edit within 2 min.

user228700

@JohnRennie Noo :P

Xasel

edit what @MAFIA36790

user228700

12:55 PM

@SirCumference Oh, I see...

John Rennie

@KaumudiHarikumar :-)

user116211

@0celo7 Not in my list; he doesn't participate in meta and review.

Sir Cumference

@JohnRennie Quite a heated subject

Xasel

or supergods to still mortal norse god

Sir Cumference

I'd say it takes too much energy to ban people though

ACuriousMind

12:56 PM

You can't just go aroujnd banning people at will as a mod, at some point someone will complain and you'll lose your diamond

0celo7

@ACuriousMind What if you delete everyone's account first?

Sir Cumference

Losing a diamond was what happened to Tildal, though I dunno the actual reason

0celo7

The music streaming service?

Sir Cumference

@0celo7 Stack Overflow staff bans you

John Rennie

Being a mod is a really hard job. It's about trying to keep this site running as smoothly as possible whilst ruffling as few feathers as possible. People underestimate just how much work our mods do.

3

ACuriousMind

12:58 PM

@0celo7 Mods cannot delete accounts, I think, that's a SE-employee level power.

Sir Cumference

They're the overlords in the end

0celo7

@ACuriousMind I need to become an SE employee

where do I sign up

Sir Cumference

It's a paid job 0celo

Though that is kind of obvious...

0celo7

@SirCumference I'm better qualified than @ACuriousMind

John Rennie

@0celo7 You're not a physicist

How can you be better qualified than ACM?

ACuriousMind

12:59 PM

@JohnRennie That. Being an invisible janitor most of the time strikes me as a thankless job.

3

Loong

@0celo7 stackoverflow.com/company/team

user116211

@JohnRennie Se employee doesn't need to be a physicist ;)

Sir Cumference

@0celo7 You'd ban people for disrespecting the holiness of GR

John Rennie

@ACuriousMind damn right

Sir Cumference

Or misspelling y'all

John Rennie

1:00 PM

Anyway I must go.

user228700

@JohnRennie Have a nice lunch!

0celo7

@JohnRennie He isn't either.

Secret

0

Q: A naive question over some General Relativity quantities

Really naively: what kind of information do those quantities give us? $$\text{Ricci Scalar} ~~~~~~~ R$$ $$\text{Ricci Tensor} ~~~~~~~ R_{\mu\nu}$$ $$\text{Riemann Tensor} ~~~~~~~ R_{\mu\nu\lambda\rho}$$ I mean: for the purposes of the description of some phenomenon or event in a gravitational...

general-relativity

OP should have googled

0celo7

People who ask questions that can easily be found on google are the worst

ACuriousMind

@0celo7 Aha.

0celo7

1:16 PM

@ACuriousMind Yes?

Bernard Meurer

So, it took me a couple of days...

But I have now realized that I can finally run the Intel C Compiler on this machine

You guys are all doomed

Aug 15 at 13:11, by Bernard Meurer

The only reason the human race still exists is the fact that I have an AMD processor

Brace yourselfs

0celo7

1:29 PM

What the hell

@BernardMeurer my birthday is coming up

You need to figure out what Reb is getting me

Bernard Meurer

@0celo7 Lol, tell her to text me

0celo7

@BernardMeurer I'm very afraid you'll find a way to contact her.

If you really wanted to, you could already.

Bernard Meurer

@0celo7 If I had no respect for you or your privacy I would've contacted her a long time ago lol

0celo7

Creep!

Bernard Meurer

:)

user218912

1:42 PM

@ACuriousMind what is the use/point of the newton-wigner localization/operator in qft?

user218912

my prof talked about it but didn't say what's used for.

ACuriousMind

@IceLord Well, those things are the closest to a "position operator" that you can get

user218912

@ACuriousMind what is the use of a position operator in qft

ACuriousMind

And so they are what you might use if you want to know whether a particle is inside a certain extent of space

user218912

since he said we're doing away with it

user218912

1:45 PM

oh

user218912

k

ACuriousMind

Well, we're doing away with it because it turns out even the N-W operators behave a bit strangely and don't really respect relativistic invariance

Qmechanic

@JohnRennie : Due to increased traffic, the phys mods have asked the SE team to increase the number of phys mods from 4 to 6, i.e. to organize an election for 2 new phys mods. [edited by DZ since this is a pinned post]

14

user218912

@ACuriousMind btw when would you want to do that?

user218912

can you give an example please?

ACuriousMind

1:46 PM

@IceLord Because you want to know whether your particle will hit a detector located at a certain spot or not, for instance

user218912

thanks!

hwlau

@ACuriousMind Isn't it just $\hat{x}$?

ACuriousMind

@hwlau Well, what does that mean in QFT? Remember, we made the fields $\phi(x)$ into operators, but not the coordinates $x$ themselves.

hwlau

@ACuriousMind Then it should be $\hat{\psi}(x)$

ok, I am not sure

ACuriousMind

@hwlau Well, that's a field operator, but what has that to do with a position operator?

Bass

1:52 PM

@ACuriousMind So they are usable as soon as we fix a specific Lorentz frame, but then we have lost Lorentz invariance, correct?

ACuriousMind

@Bass I think so, yes, but I never really worked with them either

It's best not to try and localize stuff ;)

Bass

OK I see.

Strange world. We started physics with "this thing is here, and later it will be there". Now the most fundamental theory we have cannot even localize stuff :)

hwlau

ok, i now know the reason for it

0celo7

2:14 PM

ACM will probably become a mod.

Sigh.

@ACuriousMind Don't do it

ACuriousMind

@0celo7 What are you on about? I haven't even said whether I might run or not.

Bernard Meurer

@Qmechanic We're doomed

@JohnRennie This thing is a tank

John Rennie

@BernardMeurer I've never tried using the Intel compiler because I've always been developing for Windows so I used MSVC. What's the Intel compiler like to use?

Secret

@Secret Is that your method? Asking to Google instead of asking to people who may be really into this field, knowing a LOT about, and knowing how to say it properly without confusion? Interesting. — Entropy 58 mins ago

Ok it seems I am too rude and my comment got deleted. Ok then next time I am not goign to remidn others to google things

Bernard Meurer

@JohnRennie It's a lot like GCC, or Clang, nothing special on the usage side of things IMO

The magic happens in the assembly it generates

It does some hardcore black magic and tends to create binaries that are significantly faster than GCC or Clang, even if you do GCC with -O2

John Rennie

2:31 PM

@BernardMeurer Cool :-) What debuggers will work with it's binaries? (Please don't say "printf statements" :-).

The single coolest thing about MSVC is that the debugger is ridiculously good.

Bernard Meurer

@JohnRennie I think only the IDB works with the ICC

IDB = Intel Debugger

John Rennie

That's a command line debugger isn't it?

Bernard Meurer

I don't know about windows dev at all, apart from the fact that supposedly VS is an amazing IDE

John Rennie

It's like the early 80s all over again :-)

Bernard Meurer

Yep, commandline

They see me ~~rolling~~ writing commands, they hatin'

0celo7

2:33 PM

@ACuriousMind ;_;

Xasel

guyas it seems to me that i have forgootent the basic algebra :(

0celo7

@BernardMeurer ???

Xasel

2x - x= 0

=> x = 0

0celo7

Is chamillionaire still alive

Bernard Meurer

@JohnRennie The other day my graphics driver failed and I restarted it with no image b/c of the command line

Xasel

2:34 PM

but if you add x to bth side

2 = 1

?

Bernard Meurer

It's amazing

@Xasel Wat?

John Rennie

@Xasel you're dividing by zero.

Bernard Meurer

$2=2$ I assure you

ACuriousMind

@BernardMeurer $2+2=5$. Everyone knows that.

Bernard Meurer

@ACuriousMind Only Radiohead fans know that :p

ACuriousMind

2:36 PM

@BernardMeurer Please report to room 101.

Xasel

huh..hmm...but what will you tell to a child who just started out algebra?w=how will you explain it to them

Bernard Meurer

@ACuriousMind :)

@ACuriousMind Do you want to come to lisbon and bring me a Flensburger chair?

I am in desperate need

@Xasel Tell them you can't divide by zero without breaking everything

John Rennie

@BernardMeurer ^^^

Secret

2x - x = 0 => x=0
2x - x + x = 0 + x
2x = x

0 = 0

ACuriousMind

@BernardMeurer No time to do so

Bernard Meurer

2:38 PM

@ACuriousMind Lies, everyone knows grad students don't do anything

Secret

@BernardMeurer Recall I have at least 5 years of semithesis that I can explain why you cannot divide by zero in nearly all conceivable algebraic structures

Bernard Meurer

@Secret So I'm right, you can't divide by zero without blowing shit up

Secret

Well the short answer is, because zero is such a 'good' absorber, any attempt to divide by zero you collapse the structure into the trivial structure {0}

Bernard Meurer

@JohnRennie You're a seasoned man

"Life During Wartime" Talking Heads LIVE

►

Tell me what is this guy on

Because David Byrne had something before this show, that's for sure

John Rennie

I never really got the Talking Heads. If you're male and in your fifties you're supposed to regard the Talking Heads and REM as gods. But somehow I just never quite got what they were trying to do.

Bernard Meurer

2:42 PM

I don't like REM much

I do absolutely wholeheartedly love Talking Heads

Secret

So mathematically yes, division by zero, loosely speaking, does create a balck hole and collapse reality

John Rennie

They've done some cracking good tracks. A Best of Talking Heads album is a good album.

Bernard Meurer

The Live one?

John Rennie

But I still struggle to raise much interest in them.

Bernard Meurer

I'm just wondering what drug does that to people because there's something wrong with that man lol

user218912

2:44 PM

@ACuriousMind why in $\langle \mathbf{k} | e^{-iHt} | \mathbf{x} = 0 \rangle$ in the answer the plane wave has an eigenvalue of the hamiltonian instead of the hamiltonian?

ACuriousMind

@IceLord I don't understand the question.

user218912

so it's $ = \frac{1}{2\pi^3}e^{\omega_k t}$

user218912

what did I say wrong?

user218912

$k$ is the basis state for spinless mesons.

user218912

and is an eigenstate of the hamiltonian

ACuriousMind

2:46 PM

1. You've given no context - which system are you looking at and what is $H$? 2. What do you mean by "the plane wave has an eigenvalue of the Hamiltonian instead of the Hamiltonian" - I just don't understand what you're trying to say with that

user218912

H is the relativistic hamiltonian.

user218912

and $\omega$ is the eigenvalue of it in the k basis.

ACuriousMind

Are we doing QFT or some rQM hack job?

Secret

Bernard Meurer: Here's some outline of my rambling manuscript about division by zero shown in my blog. Please note it is WIP and thus I cannot guarantee the validity for the moment being

ACuriousMind

Because I'm not good with those ad hoc rQM computations

0celo7

2:48 PM

This chat needs more balls

Secret

::Toss in a geodesic ball (whatever that means...)::

Bernard Meurer

@0celo7 I'll show you the balls ( ͡° ͜ʖ ͡°)

user218912

@ACuriousMind second one

0celo7

@ACuriousMind Interesting result: an isometry of a Riemannian manifold in the metric space sense is an isometry in the Riemannian sense, in particular it must be smooth.

ACuriousMind

Why are you telling me that?

user218912

2:53 PM

@ACuriousMind I'm just asking how do you compute $\langle \mathbf{k} | e^{-iHt} | \mathbf{x} = 0 \rangle$

2physics

Hi

user218912

how does the $\omega_k$ come in in the answer?

2physics

I have a question, thankyou in advance for your help

ACuriousMind

@IceLord Well, if $\lvert k\rangle$ is an eigenstate of $H$ with eigenvalue $\omega_k$, then $\langle k \rvert \mathrm{e}^{-\mathrm{i}Ht} = \langle k\rvert \mathrm{e}^{\mathrm{i}\omega_k t}$.

2physics

what does it mean at the beginning of a book: "TEXT FLY WITHIN THE BOOK ONLY "

user218912

2:55 PM

thanks

user218912

@ACuriousMind wait is there a proof as to why that is true

ACuriousMind

what's there to prove?

user218912

why is that true

user218912

why does it work

0celo7

lol

user218912

2:59 PM

don't laugh

0celo7

you're in a QFT class asking about eigenvalues

ACuriousMind

Because for any function $f(H)$, $f(H) \lvert k\rangle = f(\omega_k)\lvert k\rangle$ by definition of what $f(H)$ means. If you define $\mathrm{e}^{\mathrm{i}Ht}$ by its power series instead, then expanding into the series and using $H^n\lvert k\rangle = \omega_k^n\lvert k\rangle$ yields the result.

user218912

okay

user218912

@0celo7 this was not in shankar

0celo7

@IceLord Such a lie lol

First chapter

ACuriousMind

3:00 PM

@0celo7 I took the question to be the same one as the one you asked recently: "What does $f(H)$ do?"

Balarka Sen

:P

Secret

I was almost going to ask how does the eigenvalue equation $H^n\lvert k\rangle = \omega_k^n\lvert k\rangle$ somehow become an exponential euqation. Looks like when going back to quanutm , I need to study more about exponential maps

I am still not very good at series manipulations. They are simply not as chunky as algebra stuff

0celo7

@ACuriousMind Yeah, that's basic material everyone knows.

ACuriousMind

@Secret No, you just need to think a bit harder :P You don't need any knowledge about "exponential maps" for this method, you just write $\mathrm{e}^{\mathrm{i}Ht} = \sum_n \frac{(\mathrm{i}Ht)^n}{n!}$, apply the eigenvalue equation for each summand and then write it back as an exponential.

0celo7

I was asking you about the functional analysis.

ACuriousMind

3:03 PM

lol

0celo7

brb gonna print a book on the work printer, hope I don't get fired

Secret

ACM: I know, this why I said I am bad at series. Even though I have learned them before in my class, I tend to end up having a fixation of solving problems by trying to use the operators directly without expanding into series

user218912

@0celo7 where in shankar does it say what acm said?

user218912

I read ch1 doood

0celo7

He discusses functions of operators.

I'll bet you a copy of Lee

Page 55

Equation 1.9.6

@IceLord Do you know why there's a minus on the left and a plus on the right?

2physics

3:17 PM

@0celo7 I personally prefer it to hear from a good teacher rather than reading a book to learn the stuff. and I think people here have a good knowledge and understanding of various physics issues which can teach us better than any mere text. that's why you better jump off the building now :D

0celo7

@2physics obe doesn't want to take a proper class on qm

2physics

or maybe drink some HF..

a proper class is nothing more than what you see here

0celo7

@ACuriousMind What the heck does $dp$ mean where $p\in M$

David Z

@Secret I wouldn't qualify your comment as rude. Though in the future, it couldn't hurt to be a bit more informative.

ACuriousMind

@0celo7 If you mean in an integral measure, that's just a notation for the volume form.

I.e. don't try to interpret it :P

Secret

3:21 PM

DavidZ: Noted, thanks

0celo7

@ACuriousMind they say it's an "infinitesimal movement"

this is in a legitimate mathematics text

I'm very confused

David Z

And for the record, we (current mods) kind of figured it was going to be another few weeks or so before election proceedings got started. That's why we didn't bother to make an announcement. No secrecy, just lack of motivation ;-)

Maybe the SE team had a more accelerated timeline in mind.

0celo7

I vote for @BalarkaSen

ACuriousMind

@0celo7 Well, I don't know any better what you're reading about than you do, and I can't tell from your scarce description

0celo7

@ACuriousMind That scare description is all I have!

I blame physicists

@ACuriousMind What's up with French and Russians having German names?

Lang, Berger, etc.

Is Berger pronounced the German way or a butchered French way?

ACuriousMind

3:26 PM

@0celo7 I don't know in those specific cases, but most such German names either come from people that fled the Nazi regime or from German settlers (either during occupation or from earlier times)

Secret

0

Q: reflection in the micro world.

I have a hypothetical question in my mind. Suppose I have a single photon and an atom and I throw the photon towards the atom. So if the photon gets reflected, from where is it likely to do so?? From the nucleus or the orbiting electrons?

visible-light

Scattering problem?

John Rennie

I think that's a surprisingly deep question. The answer is that it's primarily the interaction with the electrons that does the scattering, but I'm not sure I know how to prove it.

That is, if you imagine removing all the electrons to just leave a grid of nuclei they would scatter much less strongly.

Secret

I felt like the scattering amplitude of the photon as it interacts with the electron density need to be calculated. However I am not really sure how this will be done, as for this scale QM is not enough and we need to go QED

ACuriousMind

@JohnRennie I'm not sure I understand the issue here: Unless the photon matches one of the energy level differences (of either the electron or the nucleus states), nothing will happen at all.

John Rennie

Rayleigh scattering

Rayleigh scattering (pronounced /ˈreɪli/ RAY-lee), named after the British physicist Lord Rayleigh (John William Strutt), is the (dominantly) elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation. Rayleigh scattering does not change the state of material, hence it is a parametric process. The particles may be individual atoms or molecules. It can occur when light travels through transparent solids and liquids, but is most prominently seen in gases. Rayleigh scattering results from the electric polarizability of the particles...

ACuriousMind

3:42 PM

@JohnRennie That's molecular, not atomic scattering

Bernard Meurer

@JohnRennie You just picked that because the guy is british

Secret

@ACuriousMind if the photon matches that/is resonant with any two levels of the electron and nuclear states, then I guess the photon will be absorbed instead and this result in the state to transit to an excited state

ACuriousMind

Molecules have additional d.o.f. (rotational/vibrational) that are much less selectively spaced and allow for greater occurence of scattering

John Rennie

Although the question does mention reflection so I wonder if they were thinking about metals ...

@ACuriousMind Monatomic gases, e.g. noble gases, would Rayleigh scatter as well.

ACuriousMind

It's meaningless to ask whether molecular scattering is "off the electrons" or "off the nuclei" because the molecule as a whole is involved

@JohnRennie Are you sure?

John Rennie

3:44 PM

@ACuriousMind Absolutely sure. A Rayleigh scatterer requires only a non-zero dimension.

The Wikipedia article that you didn't read says: The particles may be individual atoms or molecules

Honestly,you theoreticians :-)

ACuriousMind

Then I don't understand the mechanism behind this scattering at all.

John Rennie

Do I detect a HNQ effect here:

10

A: Why gases have weight?

Imagine a gas molecule in a closed box bouncing vertically between the top and boottom of the box. Let's suppose the mass of the gas molecule is $m$ and its speed at the top of the box is $v_t$. When the gas molecule moving upwards hits the top of the box and bounces back the change in momentum ...

Secret

@JohnRennie What is the mechanism of rayleigh scattering, as all I can see are just equations that describe how the intensity will be changed. Is it classical?

John Rennie

Yes, it's just Maxwell's equations. Actually I don't really know the details, but don't tell the theoretician that! :-)

Secret

ok

ACuriousMind

3:48 PM

Ohhhhh, it's not an absorb/reemit scattering at all, it's scattering off the induced dipole. Nothing quantum to see at all :/

This would be one case in which one of the more persistent anti-QM users on this site would be right with his "classical antenna theory" :P

John Rennie

:: John makes the sign of the cross in classic Hammer Horror style ::

ACuriousMind

::googles "Hammer Horror"::

John Rennie

Hammer Horror films!!

Hammer Film Productions

Hammer Films or Hammer Pictures is a British film production company based in London. Founded in 1934, the company is best known for a series of Gothic "Hammer Horror" films made from the mid-1950s until the 1970s. Hammer also produced science fiction, thrillers, film noir and comedies — and, in later years, television series. During its most successful years, Hammer dominated the horror film market, enjoying worldwide distribution and considerable financial success. This success was due, in part, to distribution partnerships with major United States studios, such as Warner Bros. During the late...

Awesomely bad films from the 70s.

Typically with Christopher Lee as Dracula and Peter Cushing as Van Helsing.

DanielSank

What's going on in here?

Oooohhhhhh, scattering!

@ACuriousMind As is so often the case.

Transcript for