[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 109 KB, 1014x670, lensing.jpg [View same] [iqdb] [saucenao] [google]
8721096 No.8721096 [Reply] [Original]

Redpill me on gravitational lensing /sci/.

>> No.8721101

you can try it at home, just take a photo of your mom

>> No.8721133
File: 49 KB, 907x661, 100fam.jpg [View same] [iqdb] [saucenao] [google]
8721133

>>8721101
INCLUDE ME IN LE REDDIT SCREENCAP GUIS

>> No.8721139

>>8721101
OP BTFO

>> No.8721146

>>8721096
Is gravitational mirroring a thing also?

>> No.8721184

>>8721146
idk

>> No.8721196

>>8721101
wow

>> No.8721206
File: 2 KB, 125x125, 1482529099113s.jpg [View same] [iqdb] [saucenao] [google]
8721206

>>8721101
>>8721133
>>8721139
>>8721196
This thread is cancer

>> No.8721210

>>8721206
>So I'll bump it

Nice going, faggot. You're the real cancer.

>> No.8721211

>>8721210
bump

>> No.8721213

>>8721096
mass and energy bend spacetime (if you're interested read about einstein's GR)
when there is a large concentration of mass/energy (like a galaxy or a black hole) spacetime is bent to such an extent that it makes light move in an extremely curved path, just like a lens.
that's about it.

>> No.8721220

>>8721213
So basically wouldn't light be affected by gravity?

>> No.8721228

>>8721220
gravity = the curvature of spacetime.
we're basically talking about the same thing.

>> No.8721229
File: 391 KB, 3240x1200, icb0w[1].jpg [View same] [iqdb] [saucenao] [google]
8721229

>>8721096
Fun fact : If you're seeing gravitational lensing, you are looking directly at a blackhole.

>> No.8721231

>>8721228
>>8721228
But shouldn't light have mass to be affected by gravity?

>> No.8721238

>>8721231
No.

>> No.8721240

>>8721231
you don't have to have mass, you just need to have momentum, which light has.

>> No.8721242

>>8721240
So basically light acts like it has mass?

>> No.8721245

>>8721242
i wouldn't exactly put it like that, but yeah, that is the general result.

>> No.8721253

>>8721245
Ok, thanks anon.

>> No.8721255

WHO THE FUCK INVITED /POL/

>> No.8721257

>>8721255
some wagecuck brainlets

>> No.8721263

>>8721255
Dude, there's nothing wrong with people wanting to learn.

>> No.8721270

>>8721240
how can you have momentum with no mass?
p=mv
m=p/v
either p and v need to be 0
or m > 0

>> No.8721273

>>8721270
Light sort of acts like it has mass, but doesn't.

>> No.8721276

>>8721270
[math] p = \hbar / \lambda [/math]

>> No.8721280

>>8721270
E=mc^2 + p^2 c^4 actually
it's enough if you have energy
or
p = hv/c if you like where p is momentum

>> No.8721287

>>8721280
oops i fucked that up
it should have been
E^2 = (pc)^2 + (mc^2)^2

>> No.8721288

>>8721280
Isn't it E^2=(mc^2)^2+(pc)^2?

>> No.8721289

>>8721280
sorry anon i posted >>8721288 before the post >>8721287 appeared and then it appeared when i posted it.

>> No.8721293
File: 20 KB, 590x434, le cosmic shitposting face.jpg [View same] [iqdb] [saucenao] [google]
8721293

>>8721101

>> No.8721294

>>8721293
cuck

>> No.8721389

>>8721101
>>8721133
>>8721196
>>8721206
>>8721210
>>8721211
>>8721255
>>8721257
>>8721293
>>8721294
this

>> No.8721403

>>8721146
no.

>> No.8721446

>>8721276
what is this formula called?
Is this a different type of momentum than the type people commonly talk about, which is defined as p=mv? or does mv=ℏ/λ?

>> No.8721448

>>8721446
That's the de Brogile hypothesis.

>> No.8721460

>>8721446
momentum is classically taught as a 3 dimensional vector with 3 spacial components:
[math] p=m(v_x \hat{x} +v_y \hat{y} +v_z \hat{z} ) [/math]
But momentum is actually a 4-dimensional vector in space-time. The 3 spacial components, along with a time component which is equal to E/c (the energy of the particle divided by the speed of light). For photons, the mass is 0 so there is only a time component. The energy of a photon corresponds to the wavelength, that is where that formula comes from. So for photons, [math] p=E/c= \hbar / \lambda [/math] . For particles with no energy, [math] mv= \hbar / \lambda [/math] . But I'm not sure if particles without energy actually exist, so more generally you can say [math] (mv)^2 - (E/c)^2 = ( \hbar / \lambda )^2 [/math] (this is just pythagoreans theorem but the temporal component gets subtracted instead of added for some reason that nobody has really explained to me)