Seahawks.NET AMAZON STOREFRONT

Vikings @ Rams

Discuss any and all NFL-related topics and matters of interest here. LANGUAGE RATING: PG-13
Re: Vikings @ Rams
Thu Sep 27, 2018 7:53 pm
  • getnasty wrote:meh, let me know when they win a playoff game.


    Great thread until this ignorance
    User avatar
    HawkerD
    NET Veteran
     
    Posts: 563
    Joined: Sun Oct 19, 2014 10:33 am
    Location: Covington WA


Re: Vikings @ Rams
Thu Sep 27, 2018 8:11 pm
  • Oddly that might have been the vikings best game this season.
    User avatar
    Uncle Si
    * NET Hottie *
     
    Posts: 13933
    Joined: Sat Mar 03, 2007 8:34 am


Re: Vikings @ Rams
Thu Sep 27, 2018 8:27 pm
  • Uncle Si wrote:Oddly that might have been the vikings best game this season.


    It was.
    User avatar
    hawknation2018
    NET Veteran
     
    Posts: 2877
    Joined: Mon Jan 01, 2018 1:04 pm


Re: Vikings @ Rams
Thu Sep 27, 2018 8:32 pm
  • Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.
    Sports Hernia wrote:Mayfield has grown on me a bit, but I disagree with him here.

    If an employer fires/terminates me, why do I owe them any sort of “loyalty”?

    &@$# that!

    That's what Sherman said
    User avatar
    5_Golden_Rings
    NET Veteran
     
    Posts: 1397
    Joined: Fri Sep 10, 2010 7:38 am


Re: Vikings @ Rams
Thu Sep 27, 2018 8:39 pm
  • Fade wrote:
    adeltaY wrote:What was that about Goff not being that good?


    The Same was said about Matt Ryan two years ago. It was Shanny.

    Goff is having the same benefit with McVay. The system & playcaller are making him appear better than he is.

    Play calling had nothing to do with the touch or accuracy. Goff did this in college, too: made yearly improvement.
    Sports Hernia wrote:Mayfield has grown on me a bit, but I disagree with him here.

    If an employer fires/terminates me, why do I owe them any sort of “loyalty”?

    &@$# that!

    That's what Sherman said
    User avatar
    5_Golden_Rings
    NET Veteran
     
    Posts: 1397
    Joined: Fri Sep 10, 2010 7:38 am


Re: Vikings @ Rams
Thu Sep 27, 2018 8:40 pm
  • For sure. Rams have too many potent weapons in too many places. They've even given porn stadium new respectability.
    SantaClaraHawk
    NET Starter
     
    Posts: 403
    Joined: Fri Sep 18, 2015 10:17 am


Re: Vikings @ Rams
Thu Sep 27, 2018 8:56 pm
  • 5_Golden_Rings wrote:Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.


    I’m with you on that one. After watching him at Cal, I thought Goff would be really good. His accuracy is off the charts.
    User avatar
    hawknation2018
    NET Veteran
     
    Posts: 2877
    Joined: Mon Jan 01, 2018 1:04 pm


Re: Vikings @ Rams
Thu Sep 27, 2018 8:58 pm
  • SantaClaraHawk wrote:porn stadium.


    I had no idea what this meant until I looked it up now. Whaaaaa?
    User avatar
    hawknation2018
    NET Veteran
     
    Posts: 2877
    Joined: Mon Jan 01, 2018 1:04 pm


Re: Vikings @ Rams
Thu Sep 27, 2018 8:59 pm
  • Wow. The Seahawks have to beat the Cardinals this Sunday, period.

    Because we're not beating the Rams without some serious flukey business.

    That offense is no joke.

    I miss 2013.
    Image

    "Shaquem Griffin tells ESPN after he got drafted by Seattle; 'I can't breathe.' That's the only time you'll hear him say he can't do something." - Dan Wetzel via Twitter.
    User avatar
    Aros
    [[ .NET Godfather ]]
     
    Posts: 12943
    Joined: Fri Feb 23, 2007 12:58 am
    Location: Just 4 miles from Richard Sherman!


Re: Vikings @ Rams
Thu Sep 27, 2018 9:16 pm
  • Yeeeaaahhh, I just wrote off all our games against the Rams for the next three years at least. That team is going to the Super Bowl this year.
    GO HAWKS!!!

    Visit my Seahawks blog at 17power.blogspot.com!
    User avatar
    MontanaHawk05
    * 17Power Blogger *
     
    Posts: 16158
    Joined: Fri May 01, 2009 8:46 am


Re: Vikings @ Rams
Thu Sep 27, 2018 10:06 pm
  • :179417:

    :179417:

    :179417:

    Well, that was both super stressful and fun.

    Now I’m starting to get what you all were talking about back when....

    And, Rams D has some serious issues. Can’t keep winning if they don’t get it together.
    When you get to Helllll, John. Tel em Daisy sentcha.
    User avatar
    RedAlice
    NET Veteran
     
    Posts: 3218
    Joined: Mon Dec 24, 2012 11:47 am
    Location: San Diego


Re: Vikings @ Rams
Thu Sep 27, 2018 10:17 pm
  • hawknation2018 wrote:
    SantaClaraHawk wrote:porn stadium.


    I had no idea what this meant until I looked it up now. Whaaaaa?


    LOL, it made the LA Times but nonetheless, I took it to the Shack;-)
    SantaClaraHawk
    NET Starter
     
    Posts: 403
    Joined: Fri Sep 18, 2015 10:17 am


Re: Vikings @ Rams
Thu Sep 27, 2018 10:20 pm
  • RedAlice wrote::179417:

    :179417:

    :179417:

    Well, that was both super stressful and fun.

    Now I’m starting to get what you all were talking about back when....

    And, Rams D has some serious issues. Can’t keep winning if they don’t get it together.


    RedAlice, as an expert especially on this team, would you mind expounding on the D's needs-improvement?
    SantaClaraHawk
    NET Starter
     
    Posts: 403
    Joined: Fri Sep 18, 2015 10:17 am


Re: Vikings @ Rams
Thu Sep 27, 2018 10:52 pm
  • SantaClaraHawk wrote:
    RedAlice wrote::179417:

    :179417:

    :179417:

    Well, that was both super stressful and fun.

    Now I’m starting to get what you all were talking about back when....

    And, Rams D has some serious issues. Can’t keep winning if they don’t get it together.


    RedAlice, as an expert especially on this team, would you mind expounding on the D's needs-improvement?


    I would love to!! But, I’ve had a bunch of cava celebrating the win - so I will get back to your request tomorrow.

    In the meantime, here is Suh in a live interview pretty much admitting our D needs some work.


    https://m.youtube.com/watch?v=NVlt-grFniA
    When you get to Helllll, John. Tel em Daisy sentcha.
    User avatar
    RedAlice
    NET Veteran
     
    Posts: 3218
    Joined: Mon Dec 24, 2012 11:47 am
    Location: San Diego


Re: Vikings @ Rams
Fri Sep 28, 2018 5:07 am
  • 5_Golden_Rings wrote:Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.


    you might be the smartest guy on the internet and I have seen some smart ones
    User avatar
    Smellyman
    NET Veteran
     
    Posts: 4721
    Joined: Tue Jan 08, 2013 6:58 pm
    Location: Taipei


Re: Vikings @ Rams
Fri Sep 28, 2018 5:59 am
  • Smellyman wrote:
    5_Golden_Rings wrote:Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.


    you might be the smartest guy on the internet and I have seen some smart ones


    Nick Foles had an awesome year too. Not saying he will or won't but certainly too early to count those chickens.
    Milehighhawk
    NET Veteran
     
    Posts: 732
    Joined: Sun Oct 07, 2012 2:33 pm


Re: Vikings @ Rams
Fri Sep 28, 2018 6:06 am
  • Rams - Chiefs superbowl. Chiefs pull off the upset 100-103. You heard it here first.
    I hate Tim Ruskell.
    User avatar
    Trrrroy
    NET Veteran
     
    Posts: 3263
    Joined: Mon Aug 17, 2009 9:24 am


Re: Vikings @ Rams
Fri Sep 28, 2018 7:28 am
  • My God I am not looking forward to next Sunday.
    HawkRiderFan
    NET Veteran
     
    Posts: 609
    Joined: Wed Jan 19, 2011 4:10 pm


Re: Vikings @ Rams
Fri Sep 28, 2018 8:31 am
  • Great game.

    With Zuerlein healthy we probably win 44-28 (FG instead of fake punt, FG made that Ficken missed, Vikings down 44-28 go for it after Suh's sack) - I say that knowing all teams deal with injuries, but kickers getting injured is a bit flukey..

    Goff won me over last night. I have been hard on him but he busted through my personal top 10 QB list with that performance. I'm not sure I've ever seen a QB be more accurate for an entire game, especially considering the talent the Vikings have on defense.

    Cooks/Woods/Kupp are very underrated. I think people still think of Woods in Buffalo, but with these three and McVay calling the plays, the Rams basically have three number one WRs - or "go to" WRs.

    Offense has been stunning, but everyone knows that..

    The defense....ugh.

    Talib will be out at least 8 weeks - and Peters had to be playing injured last night (he said he was fine but there's no way).

    We are seriously missing Mark Barron at ILB - Littleton is decent, but Ramik Wilson is not good. That will be a huge upgrade. I believe the Rams are playing Wilson over Barron on early downs and then Marqui Christian for Barron on passing downs - both guys are not playing well. His presence will make a huge difference.

    Pass rush has weirdly been a big issue. If I'm a Vikings fan, I'm pissed about Theilen being called down and the block in the back flag being picked up - those two combined for a 7 point swing (although even if they called the block in the back, Rams have the ball first and 10 at midfield - likely score anyways) - but the offensive holding on the Rams DL is next level. I know most fans complain about this, but I saw a bunch of neutral fans pointing it out yesterday too. I don't even blame d linemen for doing this - Donald is impossible to stop. But refs have to figure out a way to call it at least some of the time.

    Pass rush is the main culprit here IMO - the holding and lack of edge rush is biting us. But it took four weeks last year for the D to come together so I have hope here, and I still think Ebukam can develop into an average player. If an edge rusher doesn't emerge here, I expect to see one of either Suh or Donald split out to DE on obvious passing downs going forward.
    Ramfan128
    NET Veteran
     
    Posts: 762
    Joined: Mon Jan 13, 2014 11:46 am


Re: Vikings @ Rams
Fri Sep 28, 2018 8:33 am
  • Trrrroy wrote:Rams - Chiefs superbowl. Chiefs pull off the upset 100-103. You heard it here first.



    What's crazy is that we play them in Mexico on Monday Night in a few weeks..
    Ramfan128
    NET Veteran
     
    Posts: 762
    Joined: Mon Jan 13, 2014 11:46 am


Re: Vikings @ Rams
Fri Sep 28, 2018 8:34 am
  • Ramfan128 wrote:Great game.

    With Zuerlein healthy we probably win 44-28 (FG instead of fake punt, FG made that Ficken missed, Vikings down 44-28 go for it after Suh's sack) - I say that knowing all teams deal with injuries, but kickers getting injured is a bit flukey..

    Eh, calm down on that score prediction; the refs also illegitimately stole a TD back from the Vikes for you guys last night, lol.
    Image
    "VICTORYYYYYYY!" -Johnny Drama
    User avatar
    RolandDeschain
    * Spelling High Lord *
     
    Posts: 30061
    Joined: Fri May 01, 2009 8:39 am
    Location: Phoenix, AZ


Re: Vikings @ Rams
Fri Sep 28, 2018 8:55 am
  • RolandDeschain wrote:
    Ramfan128 wrote:Great game.

    With Zuerlein healthy we probably win 44-28 (FG instead of fake punt, FG made that Ficken missed, Vikings down 44-28 go for it after Suh's sack) - I say that knowing all teams deal with injuries, but kickers getting injured is a bit flukey..

    Eh, calm down on that score prediction; the refs also illegitimately stole a TD back from the Vikes for you guys last night, lol.


    Yeah, and wasn't Zuerlein hurt last year too? It's not a fluke for him if it happens another time.
    adeltaY
    NET Veteran
     
    Posts: 3281
    Joined: Tue Oct 11, 2016 8:22 pm
    Location: Portland, OR


Re: Vikings @ Rams
Fri Sep 28, 2018 10:01 am

Re: Vikings @ Rams
Fri Sep 28, 2018 10:50 am
  • RedAlice wrote::179417:

    :179417:

    :179417:

    Well, that was both super stressful and fun.

    Now I’m starting to get what you all were talking about back when....

    And, Rams D has some serious issues. Can’t keep winning if they don’t get it together.

    Nah. It’s tje defense faced another qb who I said was good that most everyone else here said sucked: Cousins. He is a quality QB, and he has two top of the line wide receivers. The Rams will not face many offenses that good. I expect at least one more shut out from them this season and a good chance of winning 13, 14, or 15 games. And yes, there’s enough talent and quality coaching that if everyone stays healthy and the Rams get lucky several times, 16 isn’t impossible.
    Sports Hernia wrote:Mayfield has grown on me a bit, but I disagree with him here.

    If an employer fires/terminates me, why do I owe them any sort of “loyalty”?

    &@$# that!

    That's what Sherman said
    User avatar
    5_Golden_Rings
    NET Veteran
     
    Posts: 1397
    Joined: Fri Sep 10, 2010 7:38 am


Re: Vikings @ Rams
Fri Sep 28, 2018 11:19 am
  • DomeHawk wrote:The best action was in the stands. http://www.tmz.com/2018/09/28/rams-viki ... tball-nfl/


    Really doesn't surprise me. Saw a Hawks fan get sucker punched by a Rams fan down in LA last year. Coward ran away up the stairs and peace'd out immediately. I'm not trying to generalize in regards to Rams fans, this type of stuff just happens semi-regularly at all types of games and sporting events. Pretty disgusting that people do this type of crap at games.

    Hope the Rams fan in the grey shirt has all applicable charges filed against him. Snatches a hat off a woman and then seemingly hits her in the face. Starts throwing fists at more fans and chucks a dude two rows down. That guy should be off the streets and in jail.
    User avatar
    JGfromtheNW
    NET Veteran
     
    Posts: 1842
    Joined: Mon Sep 24, 2012 9:37 am
    Location: Wenatchee


Re: Vikings @ Rams
Fri Sep 28, 2018 11:33 am
  • RolandDeschain wrote:
    Ramfan128 wrote:Great game.

    With Zuerlein healthy we probably win 44-28 (FG instead of fake punt, FG made that Ficken missed, Vikings down 44-28 go for it after Suh's sack) - I say that knowing all teams deal with injuries, but kickers getting injured is a bit flukey..

    Eh, calm down on that score prediction; the refs also illegitimately stole a TD back from the Vikes for you guys last night, lol.



    True.

    But then if we go down that road, the Vikings might have only had 10 points the whole game if they called every time Donald was held.
    Ramfan128
    NET Veteran
     
    Posts: 762
    Joined: Mon Jan 13, 2014 11:46 am


Re: Vikings @ Rams
Fri Sep 28, 2018 11:34 am
  • adeltaY wrote:
    RolandDeschain wrote:
    Ramfan128 wrote:Great game.

    With Zuerlein healthy we probably win 44-28 (FG instead of fake punt, FG made that Ficken missed, Vikings down 44-28 go for it after Suh's sack) - I say that knowing all teams deal with injuries, but kickers getting injured is a bit flukey..

    Eh, calm down on that score prediction; the refs also illegitimately stole a TD back from the Vikes for you guys last night, lol.


    Yeah, and wasn't Zuerlein hurt last year too? It's not a fluke for him if it happens another time.



    True too. Frustrating going from arguably the best kicker in the league to one of the worst.
    Ramfan128
    NET Veteran
     
    Posts: 762
    Joined: Mon Jan 13, 2014 11:46 am


Re: Vikings @ Rams
Fri Sep 28, 2018 12:37 pm
  • Smellyman wrote:
    5_Golden_Rings wrote:Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.


    you might be the smartest guy on the internet and I have seen some smart ones

    As an internet smart guy, here's some free tips on how to derive Einstein's E=mc^2 in three steps (because (1) I don't like being passively aggressively called stupid, and (2) I'm really really, shall we say, not "low" (I have all day to do nothing), so I'm going to be a smart ass while simultaneously doing a free public service to anyone who's nerdy enough to wonder about a logical justification for E = mc^2):



    Step 1:



    1. Assume the principle of relativity (the laws of physics are the same for all inertial reference frames)
    2. Assume isotropy of space and time (the direction you move in space or location in space doesn't change the laws of physics; the point in time you are in does not change the laws of physics).


    Now that we got that out of the way, we can infer that Newton's laws hold for small intervals of space and time in all inertial reference frames. Given that, we can imagine two reference frames, one S, and one S', moving with respect to each other at some speed v. The space and time coordinates (in 2 dimensions) for S will be x and t. The coordinates for S' will be x' and t'.

    Without making any assumptions about how time and space work, other than coordinate transformations between S and S' will be linear, we'll have this relationship (note we could use differentials if we wanted but why? It's linear):


    x' = Ax + Bt and t' = Cx and Dt

    where A, B, C, and D are four constants we're going to find out that will possibly depend on speed v.


    Now, consider the situation where the object S is watching move is at rest in S' (in other words, the object sits at the origin of S' while S' moves at speed v with respect to S).

    Then in that particular case (which exists for at least one inertial system), x' = 0.

    Thus, 0 = Ax + Bt, and therefore Ax = -Bt, and therefore x/t = v = -B/A. Which means B = -Av. Thus,

    Ax = x' - Bt
    Ax = x' - B(t' - Cx)/D
    Ax = x' - Bt'/D + BCx/D
    Ax - BCx/D = x' - Bt'/D
    ADx - BCx = Dx' - Bt'
    (AD - BC)x = Dx' - Bt'

    Now doing the same thing using this information with the equation relating time measurements in S to time measurements in S'.

    Dt = t' - Cx
    Dt = t' - C(x' - Bt)/A
    Dt = t' - Cx'/A + BCt/A
    Dt - BCt/A = t' - Cx'/A
    ADt - BCt = At' - Cx'
    (AD - BC)t = At' - Cx'

    Now go back to the first object moving from S'. It's moving in the other direction, so

    x/t = [(Dx' - Bt')/(AD-BC)]/[(At' - Cx')/(AD - BC)]
    x/t = (Dx' - Bt')/(At' - Cx')
    x/t = (Dv' - B')/(A' - Cv')

    x/t = (Dv' - B')/(A' - Cv')

    but since the first object is at rest in S, we can let x/t = 0 (just like we did with letting x' = 0 before), giving

    0 = (Dv' - B')/(A' - Cv')
    Dv' + B = 0
    v' = x'/t' = B/D

    Hence, B = -Dv, and then remembering that v = -B/A, we'll have:

    B = -D(-B/A)
    B = D(B/A)
    1 = D/A
    A = D

    So plug B = -Av and A = D back in and you'll see that we have:

    (AD - BC)x = Dx' - Bt'
    (A^2 + vAC)x = Ax' + Avt'
    (A^2 + vAC)x = A(x' + vt')


    Now, in real life a transformation like this requires that the same equations to work both ways except inverting the coordinates and reversing the sign (the x' to x, the t to t', and v to -v), and the only way for what I just wrote to hold for what we've derived so far is for (A^2 + vAC) to equal 1.

    So, set that equal to 1 and then solve for C:

    A^2 + vAC = 1
    vAC = 1 - A^2
    C = (1 - A^2)/(vA)



    Which means we can come to the following conclusions about the transformation equations that only depend on one of the constants, A:

    x' = Ax + Bt
    x' = Ax - Avt
    x' = A(x - vt)

    and (remembering to plug in the value for C):

    t' = Cx + At
    t' = At + Cx
    t' = At + x(1 - A^2)/(vA)
    t' = At - x(A^2-1)/(vA)
    t' = At -vx(A^2-1)/(v^2 A)
    t' = A(t - vx(A^2-1)/(v^2 A^2)
    t' = A(t - vx(A^2-1)/(v^2 A^2)




    So, the two transformation equations that depend only on one constant, A, are:


    x' = A(x - vt)

    and

    t' = A(t - vx(A^2-1)/(v^2 A^2))


    Since (A^2-1)/(v^2 A^2) is ugly and depends only on A and v, let it equal k. Then, we'll have to solve for k because the A is sitting there in the x' = A (x - vt) transformation equation. Doing that:

    (A^2 - 1)/(v^2 A^2) = k
    A^2 kv^2 = A^2 - 1
    A^2 - A^2 kv^2 = 1
    A^2 (1 - kv^2) = 1
    A^2 = 1/(1 - kv^2)
    A = 1/√(1 - kv^2)



    So we can plug that back into the two x' and t' transformation equations (and remembering what we defined k as for the time coordinate equation), giving:

    x' = [1/√(1 - kv^2)] (x - vt)

    and

    t' = [1/√(1 - kv^2)] (t - vxk)



    Now we've reached GENERALIZED TRANSFORMATION EQUATIONS. We are not yet at special relativity. These two equations are true for Galilean relativity, too. They depend on the value of k. Before we choose values for k, two things need to be made clear:

    (1) The only values of k that matter are 1, 0 and -1, because your choice of units can affect the value of k, and only those three numbers have true physical consequences.

    (2) What is k in terms of physics? It is a function of the maximum speed limit, which is either finite or infinite. Using a lower case c to represent to the maximum speed limit, k = 1/c^2. Now again, either the maximum speed limit is infinity or it is finite (spoiler alert: the maximum speed limit is measured to be finite and is in fact the speed of light).

    So now let's look at what happens when we plug in values for k.



    If k = 0 (note if that were the case we'd have to use limit notation and let c approach infinity as k approaches zero so as to not divide by zero), here's what we have:


    x' = [1/√(1 - 0*v^2)] (x - vt)

    x' = x - vt

    and

    t' = [1/√(1 - 0*v^2)] (t - vx*0)

    t' = t

    And if you know anything about basic physics, those are the Galileo transformation equations, which hold true in pre-Einstein, Newtonian physics. In fact you can see this if you divide x' by t' to get v' (the speed the observer at rest in S' sees):

    x'/t' = (x - vt)/t'

    x'/t = (x - vt)/t

    x'/t = (x - vt)/t

    u' = u - v. This is the inverse Galilean velocity addition formula. It's old news (if you throw a ball 30 mph on the ground to my right, but I"m in a truck moving to your left at 60 mph according to the ground, I'll see the ball moving 30 mph to my left, i.e., at -30 mph). Moving on.




    If k = -1, we have:

    x' = [1/√(1 - (-1) v^2)] (x - vt)

    x' = [1/√(1 + v^2)] (x - vt)

    and

    t' = [1/√(1 - (-1)v^2)] (t - vx (-1))

    t' = [1/√(1 + v^2)] (t + vx)

    Now if you're really clever, you'll realize that leads to an impossibility: it allows for you to travel freely forwards and backwards through time, which we clearly cannot do. Furthermore, it leads to an indeterminate form (infinity over infinity) when you play with the transformations. I won't waste time with that.





    If k = 1, we have:


    x' = [1/√(1 - v^2)] (x - vt)

    and

    t' = [1/√(1 - v^2)] (t - vx)


    This is the Lorentz transformation in units in which the speed of light is one. Recalling that k = 1/c^2, if we change our units to the usual c = 3x10^8 meters/second, we get this:

    x' = [1/√(1 - v^2/c^2)] (x - vt)

    and

    t' = [1/√(1 - v^2/c^)] (t - vx/c^2)

    and here are the transformations from x to x' (swap the coordinates and change the sign on the v outside of the square root):

    x = [1/√(1 - v^2/c^2)] (x' + vt')

    and

    t = [1/√(1 - v^2/c^2)] (t' - vx/c^2)


    which you will see in any modern physics text book. For the sake of brevity, going forward I'm going to let the letter y = [1/√(1 - v^2/c^)] so I don't have to keep writing it, until it becomes necessary in step 2. This will make the time transformation from S to S' look like this (it will be needed for step 2):

    t = y (t' - vx'/c^2)






    That's step 1. Here's step 2:


    Consider the situation in which the moving clock is at rest with respect to the observer. This clock will measure what is called "proper time," and every measurement performed in any inertial reference frame will agree upon this value (which is why it is important in special relativity- it's invariant. Einstein actually wanted to call his theory the Theory of Invariants, but it was too late, sadly... ). So what happens when the observer S' is at rest with respect to his or her clock? x' = 0. So, look at the last time transformation I typed up and let x' = 0:

    t = y (t' - vx'/c^2)

    t = y (t' - v*0/c^2)

    t = yt'

    t' = t/y

    where y is that ugly square root, and in this case, t' is proper time (note that proper time will always be the shortest measured time interval by any observer for an event; when people saying "moving clocks run slow," they mean THEY'LL measure the moving clock having slowed time, so the moving clock observer himself will measure the smallest time interval, because to the moving clock observer, their clock is not moving. Everyone else watching the moving clock will conclude that the time interval is longer, because the moving clock will appear to be running slowly. Tricky, but you can figure it out).






    That's step 2. Here's step 3:


    Consider a random coordinate in our one spatial dimension simplification (we don't have to worry about all three spatial dimensions here because we're discussing two reference frames moving parallel with respect to each other with their respective distance axes aligned). This coordinate is a space AND time coordinate, and to make units match, time will be multiplied by the speed c (because units of time * units of distance/time = units of distance).

    Call it capital X. Then X = (x, ct).

    Now, because proper time is invariant, divide everything by proper time. Call this new thing capital V, and remember that proper time is t/y, where y is the ugly square root:

    V = (x/[t/y], ct/[t/y])

    simplify:

    V = (yu, yc)

    where u is coordinate velocity x/t and c is still the maximum speed limit (which by now I'm sure you've guessed has been measured to be the speed of light).


    Now we want to turn that into momentum. How do we do that in "old" physics? Multiply by mass. So let's do that. We'll use the symbol capital P for momentum, so mV = P

    P = (myu, myc)



    Now, because I'm sure you're getting bored, and because I don't want to do a massive ugly integral using trig substitution, I'm going to do a short circuit skipping step. What I'm going to do is ignore the spatial component of momentum and just look at the time component, and then take the time derivative to get a component of "force," and then use the work energy theorem to get "energy," although keep in mind this is cheating to skip space because this post is already huge.

    First, looking at the time "momentum" coordinate and taking the time derivative (note that m is constant with respect to time, so it can be pulled outside the derivative):

    F = dP/dt = m*d(yc)/dt



    Now, apply the work-energy theorem, which means if you integrate force over distance and choose your interval of integration correctly you'll get kinetic energy, except instead of dx for distance, because I'm working with time, I'm going to replace dx with cdt (this is the "cheating" I was speaking of).

    (Also note that m is constant with respect to distance, so it can be pulled outside the integral):

    W = F cdt = m [ d(yc)/dt] cdt

    note that c is constant with respect to time, so it can be pulled out of the differential d(yc) so that it can be written as cdy.

    W = m [cdy/dt]cdt

    Some more rearranging:

    W = m [c^2dy*dt/dt]

    and dt/dt = 1, so

    W = m [c^2dy]

    again, note that c is constant with respect to y, so it can be pulled out of the integral, too

    W = mc^2dy

    And as any first year calculus student knows,

    dy = y

    So, the indefinite integral is:

    W = ymc^2

    Now, choose the interval of integration to be from 0 to v, remembering that y is the nasty square root. This gives:


    W = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1 - 0^2/c^2)] mc^2 = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1)]mc^2

    W = [1/√(1 - v^2/c^2)] mc^2 - mc^2


    And THAT is the kinetic energy function for special relativity.


    Now, recall that total energy = kinetic energy + rest energy. That means kinetic energy = total energy - rest energy.

    Thus, total energy is [1/√(1 - v^2/c^2)] mc^2 and rest energy is mc^2.


    Proof this is true? Take the total energy equation found above and let v = 0:

    E = [1/√(1 - v^2/c^2)] mc^2

    E =[1/√(1 - 0^2/c^2)] mc^2

    E = [1/√(1 - 0)] mc^2

    E = [1/√(1)] mc^2

    E = mc^2





    And there you go. Now you can be rest assured beyond all doubt that I AM in fact the smartest person on the internet you've ever seen.
    Sports Hernia wrote:Mayfield has grown on me a bit, but I disagree with him here.

    If an employer fires/terminates me, why do I owe them any sort of “loyalty”?

    &@$# that!

    That's what Sherman said
    User avatar
    5_Golden_Rings
    NET Veteran
     
    Posts: 1397
    Joined: Fri Sep 10, 2010 7:38 am


Re: Vikings @ Rams
Fri Sep 28, 2018 12:47 pm
  • Image
    In 180 games, Walter Jones was called for 9 holding penalties in the course of 5,703 pass plays.
    First Round Inductee To Hall Of Fame 2014
    ESPN #1 Rated Seahawks Player of All Time
    User avatar
    KitsapGuy
    * NET Staff *
     
    Posts: 5174
    Joined: Fri Feb 23, 2007 12:09 pm
    Location: Kitsap County


Re: Vikings @ Rams
Fri Sep 28, 2018 3:04 pm
  • 5_Golden_Rings wrote:
    Smellyman wrote:
    5_Golden_Rings wrote:Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.


    you might be the smartest guy on the internet and I have seen some smart ones

    As an internet smart guy, here's some free tips on how to derive Einstein's E=mc^2 in three steps (because (1) I don't like being passively aggressively called stupid, and (2) I'm really really, shall we say, not "low" (I have all day to do nothing), so I'm going to be a smart ass while simultaneously doing a free public service to anyone who's nerdy enough to wonder about a logical justification for E = mc^2):



    Step 1:



    1. Assume the principle of relativity (the laws of physics are the same for all inertial reference frames)
    2. Assume isotropy of space and time (the direction you move in space or location in space doesn't change the laws of physics; the point in time you are in does not change the laws of physics).


    Now that we got that out of the way, we can infer that Newton's laws hold for small intervals of space and time in all inertial reference frames. Given that, we can imagine two reference frames, one S, and one S', moving with respect to each other at some speed v. The space and time coordinates (in 2 dimensions) for S will be x and t. The coordinates for S' will be x' and t'.

    Without making any assumptions about how time and space work, other than coordinate transformations between S and S' will be linear, we'll have this relationship (note we could use differentials if we wanted but why? It's linear):


    x' = Ax + Bt and t' = Cx and Dt

    where A, B, C, and D are four constants we're going to find out that will possibly depend on speed v.


    Now, consider the situation where the object S is watching move is at rest in S' (in other words, the object sits at the origin of S' while S' moves at speed v with respect to S).

    Then in that particular case (which exists for at least one inertial system), x' = 0.

    Thus, 0 = Ax + Bt, and therefore Ax = -Bt, and therefore x/t = v = -B/A. Which means B = -Av. Thus,

    Ax = x' - Bt
    Ax = x' - B(t' - Cx)/D
    Ax = x' - Bt'/D + BCx/D
    Ax - BCx/D = x' - Bt'/D
    ADx - BCx = Dx' - Bt'
    (AD - BC)x = Dx' - Bt'

    Now doing the same thing using this information with the equation relating time measurements in S to time measurements in S'.

    Dt = t' - Cx
    Dt = t' - C(x' - Bt)/A
    Dt = t' - Cx'/A + BCt/A
    Dt - BCt/A = t' - Cx'/A
    ADt - BCt = At' - Cx'
    (AD - BC)t = At' - Cx'

    Now go back to the first object moving from S'. It's moving in the other direction, so

    x/t = [(Dx' - Bt')/(AD-BC)]/[(At' - Cx')/(AD - BC)]
    x/t = (Dx' - Bt')/(At' - Cx')
    x/t = (Dv' - B')/(A' - Cv')

    x/t = (Dv' - B')/(A' - Cv')

    but since the first object is at rest in S, we can let x/t = 0 (just like we did with letting x' = 0 before), giving

    0 = (Dv' - B')/(A' - Cv')
    Dv' + B = 0
    v' = x'/t' = B/D

    Hence, B = -Dv, and then remembering that v = -B/A, we'll have:

    B = -D(-B/A)
    B = D(B/A)
    1 = D/A
    A = D

    So plug B = -Av and A = D back in and you'll see that we have:

    (AD - BC)x = Dx' - Bt'
    (A^2 + vAC)x = Ax' + Avt'
    (A^2 + vAC)x = A(x' + vt')


    Now, in real life a transformation like this requires that the same equations to work both ways except inverting the coordinates and reversing the sign (the x' to x, the t to t', and v to -v), and the only way for what I just wrote to hold for what we've derived so far is for (A^2 + vAC) to equal 1.

    So, set that equal to 1 and then solve for C:

    A^2 + vAC = 1
    vAC = 1 - A^2
    C = (1 - A^2)/(vA)



    Which means we can come to the following conclusions about the transformation equations that only depend on one of the constants, A:

    x' = Ax + Bt
    x' = Ax - Avt
    x' = A(x - vt)

    and (remembering to plug in the value for C):

    t' = Cx + At
    t' = At + Cx
    t' = At + x(1 - A^2)/(vA)
    t' = At - x(A^2-1)/(vA)
    t' = At -vx(A^2-1)/(v^2 A)
    t' = A(t - vx(A^2-1)/(v^2 A^2)
    t' = A(t - vx(A^2-1)/(v^2 A^2)




    So, the two transformation equations that depend only on one constant, A, are:


    x' = A(x - vt)

    and

    t' = A(t - vx(A^2-1)/(v^2 A^2))


    Since (A^2-1)/(v^2 A^2) is ugly and depends only on A and v, let it equal k. Then, we'll have to solve for k because the A is sitting there in the x' = A (x - vt) transformation equation. Doing that:

    (A^2 - 1)/(v^2 A^2) = k
    A^2 kv^2 = A^2 - 1
    A^2 - A^2 kv^2 = 1
    A^2 (1 - kv^2) = 1
    A^2 = 1/(1 - kv^2)
    A = 1/√(1 - kv^2)



    So we can plug that back into the two x' and t' transformation equations (and remembering what we defined k as for the time coordinate equation), giving:

    x' = [1/√(1 - kv^2)] (x - vt)

    and

    t' = [1/√(1 - kv^2)] (t - vxk)



    Now we've reached GENERALIZED TRANSFORMATION EQUATIONS. We are not yet at special relativity. These two equations are true for Galilean relativity, too. They depend on the value of k. Before we choose values for k, two things need to be made clear:

    (1) The only values of k that matter are 1, 0 and -1, because your choice of units can affect the value of k, and only those three numbers have true physical consequences.

    (2) What is k in terms of physics? It is a function of the maximum speed limit, which is either finite or infinite. Using a lower case c to represent to the maximum speed limit, k = 1/c^2. Now again, either the maximum speed limit is infinity or it is finite (spoiler alert: the maximum speed limit is measured to be finite and is in fact the speed of light).

    So now let's look at what happens when we plug in values for k.



    If k = 0 (note if that were the case we'd have to use limit notation and let c approach infinity as k approaches zero so as to not divide by zero), here's what we have:


    x' = [1/√(1 - 0*v^2)] (x - vt)

    x' = x - vt

    and

    t' = [1/√(1 - 0*v^2)] (t - vx*0)

    t' = t

    And if you know anything about basic physics, those are the Galileo transformation equations, which hold true in pre-Einstein, Newtonian physics. In fact you can see this if you divide x' by t' to get v' (the speed the observer at rest in S' sees):

    x'/t' = (x - vt)/t'

    x'/t = (x - vt)/t

    x'/t = (x - vt)/t

    u' = u - v. This is the inverse Galilean velocity addition formula. It's old news (if you throw a ball 30 mph on the ground to my right, but I"m in a truck moving to your left at 60 mph according to the ground, I'll see the ball moving 30 mph to my left, i.e., at -30 mph). Moving on.




    If k = -1, we have:

    x' = [1/√(1 - (-1) v^2)] (x - vt)

    x' = [1/√(1 + v^2)] (x - vt)

    and

    t' = [1/√(1 - (-1)v^2)] (t - vx (-1))

    t' = [1/√(1 + v^2)] (t + vx)

    Now if you're really clever, you'll realize that leads to an impossibility: it allows for you to travel freely forwards and backwards through time, which we clearly cannot do. Furthermore, it leads to an indeterminate form (infinity over infinity) when you play with the transformations. I won't waste time with that.





    If k = 1, we have:


    x' = [1/√(1 - v^2)] (x - vt)

    and

    t' = [1/√(1 - v^2)] (t - vx)


    This is the Lorentz transformation in units in which the speed of light is one. Recalling that k = 1/c^2, if we change our units to the usual c = 3x10^8 meters/second, we get this:

    x' = [1/√(1 - v^2/c^2)] (x - vt)

    and

    t' = [1/√(1 - v^2/c^)] (t - vx/c^2)

    and here are the transformations from x to x' (swap the coordinates and change the sign on the v outside of the square root):

    x = [1/√(1 - v^2/c^2)] (x' + vt')

    and

    t = [1/√(1 - v^2/c^2)] (t' - vx/c^2)


    which you will see in any modern physics text book. For the sake of brevity, going forward I'm going to let the letter y = [1/√(1 - v^2/c^)] so I don't have to keep writing it, until it becomes necessary in step 2. This will make the time transformation from S to S' look like this (it will be needed for step 2):

    t = y (t' - vx'/c^2)






    That's step 1. Here's step 2:


    Consider the situation in which the moving clock is at rest with respect to the observer. This clock will measure what is called "proper time," and every measurement performed in any inertial reference frame will agree upon this value (which is why it is important in special relativity- it's invariant. Einstein actually wanted to call his theory the Theory of Invariants, but it was too late, sadly... ). So what happens when the observer S' is at rest with respect to his or her clock? x' = 0. So, look at the last time transformation I typed up and let x' = 0:

    t = y (t' - vx'/c^2)

    t = y (t' - v*0/c^2)

    t = yt'

    t' = t/y

    where y is that ugly square root, and in this case, t' is proper time (note that proper time will always be the shortest measured time interval by any observer for an event; when people saying "moving clocks run slow," they mean THEY'LL measure the moving clock having slowed time, so the moving clock observer himself will measure the smallest time interval, because to the moving clock observer, their clock is not moving. Everyone else watching the moving clock will conclude that the time interval is longer, because the moving clock will appear to be running slowly. Tricky, but you can figure it out).






    That's step 2. Here's step 3:


    Consider a random coordinate in our one spatial dimension simplification (we don't have to worry about all three spatial dimensions here because we're discussing two reference frames moving parallel with respect to each other with their respective distance axes aligned). This coordinate is a space AND time coordinate, and to make units match, time will be multiplied by the speed c (because units of time * units of distance/time = units of distance).

    Call it capital X. Then X = (x, ct).

    Now, because proper time is invariant, divide everything by proper time. Call this new thing capital V, and remember that proper time is t/y, where y is the ugly square root:

    V = (x/[t/y], ct/[t/y])

    simplify:

    V = (yu, yc)

    where u is coordinate velocity x/t and c is still the maximum speed limit (which by now I'm sure you've guessed has been measured to be the speed of light).


    Now we want to turn that into momentum. How do we do that in "old" physics? Multiply by mass. So let's do that. We'll use the symbol capital P for momentum, so mV = P

    P = (myu, myc)



    Now, because I'm sure you're getting bored, and because I don't want to do a massive ugly integral using trig substitution, I'm going to do a short circuit skipping step. What I'm going to do is ignore the spatial component of momentum and just look at the time component, and then take the time derivative to get a component of "force," and then use the work energy theorem to get "energy," although keep in mind this is cheating to skip space because this post is already huge.

    First, looking at the time "momentum" coordinate and taking the time derivative (note that m is constant with respect to time, so it can be pulled outside the derivative):

    F = dP/dt = m*d(yc)/dt



    Now, apply the work-energy theorem, which means if you integrate force over distance and choose your interval of integration correctly you'll get kinetic energy, except instead of dx for distance, because I'm working with time, I'm going to replace dx with cdt (this is the "cheating" I was speaking of).

    (Also note that m is constant with respect to distance, so it can be pulled outside the integral):

    W = F cdt = m [ d(yc)/dt] cdt

    note that c is constant with respect to time, so it can be pulled out of the differential d(yc) so that it can be written as cdy.

    W = m [cdy/dt]cdt

    Some more rearranging:

    W = m [c^2dy*dt/dt]

    and dt/dt = 1, so

    W = m [c^2dy]

    again, note that c is constant with respect to y, so it can be pulled out of the integral, too

    W = mc^2dy

    And as any first year calculus student knows,

    dy = y

    So, the indefinite integral is:

    W = ymc^2

    Now, choose the interval of integration to be from 0 to v, remembering that y is the nasty square root. This gives:


    W = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1 - 0^2/c^2)] mc^2 = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1)]mc^2

    W = [1/√(1 - v^2/c^2)] mc^2 - mc^2


    And THAT is the kinetic energy function for special relativity.


    Now, recall that total energy = kinetic energy + rest energy. That means kinetic energy = total energy - rest energy.

    Thus, total energy is [1/√(1 - v^2/c^2)] mc^2 and rest energy is mc^2.


    Proof this is true? Take the total energy equation found above and let v = 0:

    E = [1/√(1 - v^2/c^2)] mc^2

    E =[1/√(1 - 0^2/c^2)] mc^2

    E = [1/√(1 - 0)] mc^2

    E = [1/√(1)] mc^2

    E = mc^2





    And there you go. Now you can be rest assured beyond all doubt that I AM in fact the smartest person on the internet you've ever seen.


    So what is the speed of light? I learned that back in the 7th grade.
    R.I.P. Queen.
    User avatar
    Seahawkfan80
    NET Veteran
     
    Posts: 8310
    Joined: Sat Nov 05, 2011 12:20 pm
    Location: A little ways from Boise.


Re: Vikings @ Rams
Fri Sep 28, 2018 4:02 pm
  • Fade wrote:
    adeltaY wrote:What was that about Goff not being that good?


    The system & playcaller are making him appear better than he is.


    You are who you are. The system and play caller always makes the QB and vice versa, you can't be a great system and play caller without a good QB. Ultimately McVay and Goff aren't going anywhere soon, so the Rams seem to have a good offense for the foreseeable future. That's the beauty of having a head coach who is also the OC.
    User avatar
    94Smith
    NET Practice Squad
     
    Posts: 83
    Joined: Tue Sep 04, 2018 10:01 am


Re: Vikings @ Rams
Fri Sep 28, 2018 4:38 pm
  • 94Smith wrote:
    Fade wrote:
    adeltaY wrote:What was that about Goff not being that good?


    The system & playcaller are making him appear better than he is.


    You are who you are. The system and play caller always makes the QB and vice versa, you can't be a great system and play caller without a good QB. Ultimately McVay and Goff aren't going anywhere soon, so the Rams seem to have a good offense for the foreseeable future. That's the beauty of having a head coach who is also the OC.


    Right and tell me how the system made Goff throw four absolute dimes down the field?
    adeltaY
    NET Veteran
     
    Posts: 3281
    Joined: Tue Oct 11, 2016 8:22 pm
    Location: Portland, OR


Re: Vikings @ Rams
Fri Sep 28, 2018 5:34 pm
  • Seahawkfan80 wrote:
    5_Golden_Rings wrote:
    Smellyman wrote:
    5_Golden_Rings wrote:Not to say I told you so on another quarterback, but I told you so on Goff. He’s going to be a quality qb for a long time.


    you might be the smartest guy on the internet and I have seen some smart ones

    As an internet smart guy, here's some free tips on how to derive Einstein's E=mc^2 in three steps (because (1) I don't like being passively aggressively called stupid, and (2) I'm really really, shall we say, not "low" (I have all day to do nothing), so I'm going to be a smart ass while simultaneously doing a free public service to anyone who's nerdy enough to wonder about a logical justification for E = mc^2):



    Step 1:



    1. Assume the principle of relativity (the laws of physics are the same for all inertial reference frames)
    2. Assume isotropy of space and time (the direction you move in space or location in space doesn't change the laws of physics; the point in time you are in does not change the laws of physics).


    Now that we got that out of the way, we can infer that Newton's laws hold for small intervals of space and time in all inertial reference frames. Given that, we can imagine two reference frames, one S, and one S', moving with respect to each other at some speed v. The space and time coordinates (in 2 dimensions) for S will be x and t. The coordinates for S' will be x' and t'.

    Without making any assumptions about how time and space work, other than coordinate transformations between S and S' will be linear, we'll have this relationship (note we could use differentials if we wanted but why? It's linear):


    x' = Ax + Bt and t' = Cx and Dt

    where A, B, C, and D are four constants we're going to find out that will possibly depend on speed v.


    Now, consider the situation where the object S is watching move is at rest in S' (in other words, the object sits at the origin of S' while S' moves at speed v with respect to S).

    Then in that particular case (which exists for at least one inertial system), x' = 0.

    Thus, 0 = Ax + Bt, and therefore Ax = -Bt, and therefore x/t = v = -B/A. Which means B = -Av. Thus,

    Ax = x' - Bt
    Ax = x' - B(t' - Cx)/D
    Ax = x' - Bt'/D + BCx/D
    Ax - BCx/D = x' - Bt'/D
    ADx - BCx = Dx' - Bt'
    (AD - BC)x = Dx' - Bt'

    Now doing the same thing using this information with the equation relating time measurements in S to time measurements in S'.

    Dt = t' - Cx
    Dt = t' - C(x' - Bt)/A
    Dt = t' - Cx'/A + BCt/A
    Dt - BCt/A = t' - Cx'/A
    ADt - BCt = At' - Cx'
    (AD - BC)t = At' - Cx'

    Now go back to the first object moving from S'. It's moving in the other direction, so

    x/t = [(Dx' - Bt')/(AD-BC)]/[(At' - Cx')/(AD - BC)]
    x/t = (Dx' - Bt')/(At' - Cx')
    x/t = (Dv' - B')/(A' - Cv')

    x/t = (Dv' - B')/(A' - Cv')

    but since the first object is at rest in S, we can let x/t = 0 (just like we did with letting x' = 0 before), giving

    0 = (Dv' - B')/(A' - Cv')
    Dv' + B = 0
    v' = x'/t' = B/D

    Hence, B = -Dv, and then remembering that v = -B/A, we'll have:

    B = -D(-B/A)
    B = D(B/A)
    1 = D/A
    A = D

    So plug B = -Av and A = D back in and you'll see that we have:

    (AD - BC)x = Dx' - Bt'
    (A^2 + vAC)x = Ax' + Avt'
    (A^2 + vAC)x = A(x' + vt')


    Now, in real life a transformation like this requires that the same equations to work both ways except inverting the coordinates and reversing the sign (the x' to x, the t to t', and v to -v), and the only way for what I just wrote to hold for what we've derived so far is for (A^2 + vAC) to equal 1.

    So, set that equal to 1 and then solve for C:

    A^2 + vAC = 1
    vAC = 1 - A^2
    C = (1 - A^2)/(vA)



    Which means we can come to the following conclusions about the transformation equations that only depend on one of the constants, A:

    x' = Ax + Bt
    x' = Ax - Avt
    x' = A(x - vt)

    and (remembering to plug in the value for C):

    t' = Cx + At
    t' = At + Cx
    t' = At + x(1 - A^2)/(vA)
    t' = At - x(A^2-1)/(vA)
    t' = At -vx(A^2-1)/(v^2 A)
    t' = A(t - vx(A^2-1)/(v^2 A^2)
    t' = A(t - vx(A^2-1)/(v^2 A^2)




    So, the two transformation equations that depend only on one constant, A, are:


    x' = A(x - vt)

    and

    t' = A(t - vx(A^2-1)/(v^2 A^2))


    Since (A^2-1)/(v^2 A^2) is ugly and depends only on A and v, let it equal k. Then, we'll have to solve for k because the A is sitting there in the x' = A (x - vt) transformation equation. Doing that:

    (A^2 - 1)/(v^2 A^2) = k
    A^2 kv^2 = A^2 - 1
    A^2 - A^2 kv^2 = 1
    A^2 (1 - kv^2) = 1
    A^2 = 1/(1 - kv^2)
    A = 1/√(1 - kv^2)



    So we can plug that back into the two x' and t' transformation equations (and remembering what we defined k as for the time coordinate equation), giving:

    x' = [1/√(1 - kv^2)] (x - vt)

    and

    t' = [1/√(1 - kv^2)] (t - vxk)



    Now we've reached GENERALIZED TRANSFORMATION EQUATIONS. We are not yet at special relativity. These two equations are true for Galilean relativity, too. They depend on the value of k. Before we choose values for k, two things need to be made clear:

    (1) The only values of k that matter are 1, 0 and -1, because your choice of units can affect the value of k, and only those three numbers have true physical consequences.

    (2) What is k in terms of physics? It is a function of the maximum speed limit, which is either finite or infinite. Using a lower case c to represent to the maximum speed limit, k = 1/c^2. Now again, either the maximum speed limit is infinity or it is finite (spoiler alert: the maximum speed limit is measured to be finite and is in fact the speed of light).

    So now let's look at what happens when we plug in values for k.



    If k = 0 (note if that were the case we'd have to use limit notation and let c approach infinity as k approaches zero so as to not divide by zero), here's what we have:


    x' = [1/√(1 - 0*v^2)] (x - vt)

    x' = x - vt

    and

    t' = [1/√(1 - 0*v^2)] (t - vx*0)

    t' = t

    And if you know anything about basic physics, those are the Galileo transformation equations, which hold true in pre-Einstein, Newtonian physics. In fact you can see this if you divide x' by t' to get v' (the speed the observer at rest in S' sees):

    x'/t' = (x - vt)/t'

    x'/t = (x - vt)/t

    x'/t = (x - vt)/t

    u' = u - v. This is the inverse Galilean velocity addition formula. It's old news (if you throw a ball 30 mph on the ground to my right, but I"m in a truck moving to your left at 60 mph according to the ground, I'll see the ball moving 30 mph to my left, i.e., at -30 mph). Moving on.




    If k = -1, we have:

    x' = [1/√(1 - (-1) v^2)] (x - vt)

    x' = [1/√(1 + v^2)] (x - vt)

    and

    t' = [1/√(1 - (-1)v^2)] (t - vx (-1))

    t' = [1/√(1 + v^2)] (t + vx)

    Now if you're really clever, you'll realize that leads to an impossibility: it allows for you to travel freely forwards and backwards through time, which we clearly cannot do. Furthermore, it leads to an indeterminate form (infinity over infinity) when you play with the transformations. I won't waste time with that.





    If k = 1, we have:


    x' = [1/√(1 - v^2)] (x - vt)

    and

    t' = [1/√(1 - v^2)] (t - vx)


    This is the Lorentz transformation in units in which the speed of light is one. Recalling that k = 1/c^2, if we change our units to the usual c = 3x10^8 meters/second, we get this:

    x' = [1/√(1 - v^2/c^2)] (x - vt)

    and

    t' = [1/√(1 - v^2/c^)] (t - vx/c^2)

    and here are the transformations from x to x' (swap the coordinates and change the sign on the v outside of the square root):

    x = [1/√(1 - v^2/c^2)] (x' + vt')

    and

    t = [1/√(1 - v^2/c^2)] (t' - vx/c^2)


    which you will see in any modern physics text book. For the sake of brevity, going forward I'm going to let the letter y = [1/√(1 - v^2/c^)] so I don't have to keep writing it, until it becomes necessary in step 2. This will make the time transformation from S to S' look like this (it will be needed for step 2):

    t = y (t' - vx'/c^2)






    That's step 1. Here's step 2:


    Consider the situation in which the moving clock is at rest with respect to the observer. This clock will measure what is called "proper time," and every measurement performed in any inertial reference frame will agree upon this value (which is why it is important in special relativity- it's invariant. Einstein actually wanted to call his theory the Theory of Invariants, but it was too late, sadly... ). So what happens when the observer S' is at rest with respect to his or her clock? x' = 0. So, look at the last time transformation I typed up and let x' = 0:

    t = y (t' - vx'/c^2)

    t = y (t' - v*0/c^2)

    t = yt'

    t' = t/y

    where y is that ugly square root, and in this case, t' is proper time (note that proper time will always be the shortest measured time interval by any observer for an event; when people saying "moving clocks run slow," they mean THEY'LL measure the moving clock having slowed time, so the moving clock observer himself will measure the smallest time interval, because to the moving clock observer, their clock is not moving. Everyone else watching the moving clock will conclude that the time interval is longer, because the moving clock will appear to be running slowly. Tricky, but you can figure it out).






    That's step 2. Here's step 3:


    Consider a random coordinate in our one spatial dimension simplification (we don't have to worry about all three spatial dimensions here because we're discussing two reference frames moving parallel with respect to each other with their respective distance axes aligned). This coordinate is a space AND time coordinate, and to make units match, time will be multiplied by the speed c (because units of time * units of distance/time = units of distance).

    Call it capital X. Then X = (x, ct).

    Now, because proper time is invariant, divide everything by proper time. Call this new thing capital V, and remember that proper time is t/y, where y is the ugly square root:

    V = (x/[t/y], ct/[t/y])

    simplify:

    V = (yu, yc)

    where u is coordinate velocity x/t and c is still the maximum speed limit (which by now I'm sure you've guessed has been measured to be the speed of light).


    Now we want to turn that into momentum. How do we do that in "old" physics? Multiply by mass. So let's do that. We'll use the symbol capital P for momentum, so mV = P

    P = (myu, myc)



    Now, because I'm sure you're getting bored, and because I don't want to do a massive ugly integral using trig substitution, I'm going to do a short circuit skipping step. What I'm going to do is ignore the spatial component of momentum and just look at the time component, and then take the time derivative to get a component of "force," and then use the work energy theorem to get "energy," although keep in mind this is cheating to skip space because this post is already huge.

    First, looking at the time "momentum" coordinate and taking the time derivative (note that m is constant with respect to time, so it can be pulled outside the derivative):

    F = dP/dt = m*d(yc)/dt



    Now, apply the work-energy theorem, which means if you integrate force over distance and choose your interval of integration correctly you'll get kinetic energy, except instead of dx for distance, because I'm working with time, I'm going to replace dx with cdt (this is the "cheating" I was speaking of).

    (Also note that m is constant with respect to distance, so it can be pulled outside the integral):

    W = F cdt = m [ d(yc)/dt] cdt

    note that c is constant with respect to time, so it can be pulled out of the differential d(yc) so that it can be written as cdy.

    W = m [cdy/dt]cdt

    Some more rearranging:

    W = m [c^2dy*dt/dt]

    and dt/dt = 1, so

    W = m [c^2dy]

    again, note that c is constant with respect to y, so it can be pulled out of the integral, too

    W = mc^2dy

    And as any first year calculus student knows,

    dy = y

    So, the indefinite integral is:

    W = ymc^2

    Now, choose the interval of integration to be from 0 to v, remembering that y is the nasty square root. This gives:


    W = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1 - 0^2/c^2)] mc^2 = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1)]mc^2

    W = [1/√(1 - v^2/c^2)] mc^2 - mc^2


    And THAT is the kinetic energy function for special relativity.


    Now, recall that total energy = kinetic energy + rest energy. That means kinetic energy = total energy - rest energy.

    Thus, total energy is [1/√(1 - v^2/c^2)] mc^2 and rest energy is mc^2.


    Proof this is true? Take the total energy equation found above and let v = 0:

    E = [1/√(1 - v^2/c^2)] mc^2

    E =[1/√(1 - 0^2/c^2)] mc^2

    E = [1/√(1 - 0)] mc^2

    E = [1/√(1)] mc^2

    E = mc^2





    And there you go. Now you can be rest assured beyond all doubt that I AM in fact the smartest person on the internet you've ever seen.


    So what is the speed of light? I learned that back in the 7th grade.


    I get insecure when I’m blitzed lol.

    The speed of light is whatever you want it to be based on your choice of units. What does not change is the fine structure constant, and it’s the invariance and finiteness of the speed of light that matters, not the numerical value. But in “normal” units it’s about 3x10^8 m/s or 186,000 miles per second.
    Sports Hernia wrote:Mayfield has grown on me a bit, but I disagree with him here.

    If an employer fires/terminates me, why do I owe them any sort of “loyalty”?

    &@$# that!

    That's what Sherman said
    User avatar
    5_Golden_Rings
    NET Veteran
     
    Posts: 1397
    Joined: Fri Sep 10, 2010 7:38 am


Re: Vikings @ Rams
Fri Sep 28, 2018 6:28 pm
  • 5_Golden_Rings wrote:
    Seahawkfan80 wrote:
    5_Golden_Rings wrote:
    Smellyman wrote:
    you might be the smartest guy on the internet and I have seen some smart ones

    As an internet smart guy, here's some free tips on how to derive Einstein's E=mc^2 in three steps (because (1) I don't like being passively aggressively called stupid, and (2) I'm really really, shall we say, not "low" (I have all day to do nothing), so I'm going to be a smart ass while simultaneously doing a free public service to anyone who's nerdy enough to wonder about a logical justification for E = mc^2):



    Step 1:



    1. Assume the principle of relativity (the laws of physics are the same for all inertial reference frames)
    2. Assume isotropy of space and time (the direction you move in space or location in space doesn't change the laws of physics; the point in time you are in does not change the laws of physics).


    Now that we got that out of the way, we can infer that Newton's laws hold for small intervals of space and time in all inertial reference frames. Given that, we can imagine two reference frames, one S, and one S', moving with respect to each other at some speed v. The space and time coordinates (in 2 dimensions) for S will be x and t. The coordinates for S' will be x' and t'.

    Without making any assumptions about how time and space work, other than coordinate transformations between S and S' will be linear, we'll have this relationship (note we could use differentials if we wanted but why? It's linear):


    x' = Ax + Bt and t' = Cx and Dt

    where A, B, C, and D are four constants we're going to find out that will possibly depend on speed v.


    Now, consider the situation where the object S is watching move is at rest in S' (in other words, the object sits at the origin of S' while S' moves at speed v with respect to S).

    Then in that particular case (which exists for at least one inertial system), x' = 0.

    Thus, 0 = Ax + Bt, and therefore Ax = -Bt, and therefore x/t = v = -B/A. Which means B = -Av. Thus,

    Ax = x' - Bt
    Ax = x' - B(t' - Cx)/D
    Ax = x' - Bt'/D + BCx/D
    Ax - BCx/D = x' - Bt'/D
    ADx - BCx = Dx' - Bt'
    (AD - BC)x = Dx' - Bt'

    Now doing the same thing using this information with the equation relating time measurements in S to time measurements in S'.

    Dt = t' - Cx
    Dt = t' - C(x' - Bt)/A
    Dt = t' - Cx'/A + BCt/A
    Dt - BCt/A = t' - Cx'/A
    ADt - BCt = At' - Cx'
    (AD - BC)t = At' - Cx'

    Now go back to the first object moving from S'. It's moving in the other direction, so

    x/t = [(Dx' - Bt')/(AD-BC)]/[(At' - Cx')/(AD - BC)]
    x/t = (Dx' - Bt')/(At' - Cx')
    x/t = (Dv' - B')/(A' - Cv')

    x/t = (Dv' - B')/(A' - Cv')

    but since the first object is at rest in S, we can let x/t = 0 (just like we did with letting x' = 0 before), giving

    0 = (Dv' - B')/(A' - Cv')
    Dv' + B = 0
    v' = x'/t' = B/D

    Hence, B = -Dv, and then remembering that v = -B/A, we'll have:

    B = -D(-B/A)
    B = D(B/A)
    1 = D/A
    A = D

    So plug B = -Av and A = D back in and you'll see that we have:

    (AD - BC)x = Dx' - Bt'
    (A^2 + vAC)x = Ax' + Avt'
    (A^2 + vAC)x = A(x' + vt')


    Now, in real life a transformation like this requires that the same equations to work both ways except inverting the coordinates and reversing the sign (the x' to x, the t to t', and v to -v), and the only way for what I just wrote to hold for what we've derived so far is for (A^2 + vAC) to equal 1.

    So, set that equal to 1 and then solve for C:

    A^2 + vAC = 1
    vAC = 1 - A^2
    C = (1 - A^2)/(vA)



    Which means we can come to the following conclusions about the transformation equations that only depend on one of the constants, A:

    x' = Ax + Bt
    x' = Ax - Avt
    x' = A(x - vt)

    and (remembering to plug in the value for C):

    t' = Cx + At
    t' = At + Cx
    t' = At + x(1 - A^2)/(vA)
    t' = At - x(A^2-1)/(vA)
    t' = At -vx(A^2-1)/(v^2 A)
    t' = A(t - vx(A^2-1)/(v^2 A^2)
    t' = A(t - vx(A^2-1)/(v^2 A^2)




    So, the two transformation equations that depend only on one constant, A, are:


    x' = A(x - vt)

    and

    t' = A(t - vx(A^2-1)/(v^2 A^2))


    Since (A^2-1)/(v^2 A^2) is ugly and depends only on A and v, let it equal k. Then, we'll have to solve for k because the A is sitting there in the x' = A (x - vt) transformation equation. Doing that:

    (A^2 - 1)/(v^2 A^2) = k
    A^2 kv^2 = A^2 - 1
    A^2 - A^2 kv^2 = 1
    A^2 (1 - kv^2) = 1
    A^2 = 1/(1 - kv^2)
    A = 1/√(1 - kv^2)



    So we can plug that back into the two x' and t' transformation equations (and remembering what we defined k as for the time coordinate equation), giving:

    x' = [1/√(1 - kv^2)] (x - vt)

    and

    t' = [1/√(1 - kv^2)] (t - vxk)



    Now we've reached GENERALIZED TRANSFORMATION EQUATIONS. We are not yet at special relativity. These two equations are true for Galilean relativity, too. They depend on the value of k. Before we choose values for k, two things need to be made clear:

    (1) The only values of k that matter are 1, 0 and -1, because your choice of units can affect the value of k, and only those three numbers have true physical consequences.

    (2) What is k in terms of physics? It is a function of the maximum speed limit, which is either finite or infinite. Using a lower case c to represent to the maximum speed limit, k = 1/c^2. Now again, either the maximum speed limit is infinity or it is finite (spoiler alert: the maximum speed limit is measured to be finite and is in fact the speed of light).

    So now let's look at what happens when we plug in values for k.



    If k = 0 (note if that were the case we'd have to use limit notation and let c approach infinity as k approaches zero so as to not divide by zero), here's what we have:


    x' = [1/√(1 - 0*v^2)] (x - vt)

    x' = x - vt

    and

    t' = [1/√(1 - 0*v^2)] (t - vx*0)

    t' = t

    And if you know anything about basic physics, those are the Galileo transformation equations, which hold true in pre-Einstein, Newtonian physics. In fact you can see this if you divide x' by t' to get v' (the speed the observer at rest in S' sees):

    x'/t' = (x - vt)/t'

    x'/t = (x - vt)/t

    x'/t = (x - vt)/t

    u' = u - v. This is the inverse Galilean velocity addition formula. It's old news (if you throw a ball 30 mph on the ground to my right, but I"m in a truck moving to your left at 60 mph according to the ground, I'll see the ball moving 30 mph to my left, i.e., at -30 mph). Moving on.




    If k = -1, we have:

    x' = [1/√(1 - (-1) v^2)] (x - vt)

    x' = [1/√(1 + v^2)] (x - vt)

    and

    t' = [1/√(1 - (-1)v^2)] (t - vx (-1))

    t' = [1/√(1 + v^2)] (t + vx)

    Now if you're really clever, you'll realize that leads to an impossibility: it allows for you to travel freely forwards and backwards through time, which we clearly cannot do. Furthermore, it leads to an indeterminate form (infinity over infinity) when you play with the transformations. I won't waste time with that.





    If k = 1, we have:


    x' = [1/√(1 - v^2)] (x - vt)

    and

    t' = [1/√(1 - v^2)] (t - vx)


    This is the Lorentz transformation in units in which the speed of light is one. Recalling that k = 1/c^2, if we change our units to the usual c = 3x10^8 meters/second, we get this:

    x' = [1/√(1 - v^2/c^2)] (x - vt)

    and

    t' = [1/√(1 - v^2/c^)] (t - vx/c^2)

    and here are the transformations from x to x' (swap the coordinates and change the sign on the v outside of the square root):

    x = [1/√(1 - v^2/c^2)] (x' + vt')

    and

    t = [1/√(1 - v^2/c^2)] (t' - vx/c^2)


    which you will see in any modern physics text book. For the sake of brevity, going forward I'm going to let the letter y = [1/√(1 - v^2/c^)] so I don't have to keep writing it, until it becomes necessary in step 2. This will make the time transformation from S to S' look like this (it will be needed for step 2):

    t = y (t' - vx'/c^2)






    That's step 1. Here's step 2:


    Consider the situation in which the moving clock is at rest with respect to the observer. This clock will measure what is called "proper time," and every measurement performed in any inertial reference frame will agree upon this value (which is why it is important in special relativity- it's invariant. Einstein actually wanted to call his theory the Theory of Invariants, but it was too late, sadly... ). So what happens when the observer S' is at rest with respect to his or her clock? x' = 0. So, look at the last time transformation I typed up and let x' = 0:

    t = y (t' - vx'/c^2)

    t = y (t' - v*0/c^2)

    t = yt'

    t' = t/y

    where y is that ugly square root, and in this case, t' is proper time (note that proper time will always be the shortest measured time interval by any observer for an event; when people saying "moving clocks run slow," they mean THEY'LL measure the moving clock having slowed time, so the moving clock observer himself will measure the smallest time interval, because to the moving clock observer, their clock is not moving. Everyone else watching the moving clock will conclude that the time interval is longer, because the moving clock will appear to be running slowly. Tricky, but you can figure it out).






    That's step 2. Here's step 3:


    Consider a random coordinate in our one spatial dimension simplification (we don't have to worry about all three spatial dimensions here because we're discussing two reference frames moving parallel with respect to each other with their respective distance axes aligned). This coordinate is a space AND time coordinate, and to make units match, time will be multiplied by the speed c (because units of time * units of distance/time = units of distance).

    Call it capital X. Then X = (x, ct).

    Now, because proper time is invariant, divide everything by proper time. Call this new thing capital V, and remember that proper time is t/y, where y is the ugly square root:

    V = (x/[t/y], ct/[t/y])

    simplify:

    V = (yu, yc)

    where u is coordinate velocity x/t and c is still the maximum speed limit (which by now I'm sure you've guessed has been measured to be the speed of light).


    Now we want to turn that into momentum. How do we do that in "old" physics? Multiply by mass. So let's do that. We'll use the symbol capital P for momentum, so mV = P

    P = (myu, myc)



    Now, because I'm sure you're getting bored, and because I don't want to do a massive ugly integral using trig substitution, I'm going to do a short circuit skipping step. What I'm going to do is ignore the spatial component of momentum and just look at the time component, and then take the time derivative to get a component of "force," and then use the work energy theorem to get "energy," although keep in mind this is cheating to skip space because this post is already huge.

    First, looking at the time "momentum" coordinate and taking the time derivative (note that m is constant with respect to time, so it can be pulled outside the derivative):

    F = dP/dt = m*d(yc)/dt



    Now, apply the work-energy theorem, which means if you integrate force over distance and choose your interval of integration correctly you'll get kinetic energy, except instead of dx for distance, because I'm working with time, I'm going to replace dx with cdt (this is the "cheating" I was speaking of).

    (Also note that m is constant with respect to distance, so it can be pulled outside the integral):

    W = F cdt = m [ d(yc)/dt] cdt

    note that c is constant with respect to time, so it can be pulled out of the differential d(yc) so that it can be written as cdy.

    W = m [cdy/dt]cdt

    Some more rearranging:

    W = m [c^2dy*dt/dt]

    and dt/dt = 1, so

    W = m [c^2dy]

    again, note that c is constant with respect to y, so it can be pulled out of the integral, too

    W = mc^2dy

    And as any first year calculus student knows,

    dy = y

    So, the indefinite integral is:

    W = ymc^2

    Now, choose the interval of integration to be from 0 to v, remembering that y is the nasty square root. This gives:


    W = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1 - 0^2/c^2)] mc^2 = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1)]mc^2

    W = [1/√(1 - v^2/c^2)] mc^2 - mc^2


    And THAT is the kinetic energy function for special relativity.


    Now, recall that total energy = kinetic energy + rest energy. That means kinetic energy = total energy - rest energy.

    Thus, total energy is [1/√(1 - v^2/c^2)] mc^2 and rest energy is mc^2.


    Proof this is true? Take the total energy equation found above and let v = 0:

    E = [1/√(1 - v^2/c^2)] mc^2

    E =[1/√(1 - 0^2/c^2)] mc^2

    E = [1/√(1 - 0)] mc^2

    E = [1/√(1)] mc^2

    E = mc^2





    And there you go. Now you can be rest assured beyond all doubt that I AM in fact the smartest person on the internet you've ever seen.


    So what is the speed of light? I learned that back in the 7th grade.


    I get insecure when I’m blitzed lol.

    The speed of light is whatever you want it to be based on your choice of units. What does not change is the fine structure constant, and it’s the invariance and finiteness of the speed of light that matters, not the numerical value. But in “normal” units it’s about 3x10^8 m/s or 186,000 miles per second.


    Yes. Thank you. I need to digest this post again. I REALLY Appreciate it. I was thinking about going back to my calculus books for idiots. I have note pads so I dont desecrate the books. I ran thru it kinda fast...should not have. I am a radar electronics Tech. But not that slow. thanks.
    R.I.P. Queen.
    User avatar
    Seahawkfan80
    NET Veteran
     
    Posts: 8310
    Joined: Sat Nov 05, 2011 12:20 pm
    Location: A little ways from Boise.


Re: Vikings @ Rams
Fri Sep 28, 2018 7:25 pm

Re: Vikings @ Rams
Fri Sep 28, 2018 8:31 pm
  • 5_Golden_Rings, you should teach. You have a great way of condensing this into a refresher.
    SantaClaraHawk
    NET Starter
     
    Posts: 403
    Joined: Fri Sep 18, 2015 10:17 am


Re: Vikings @ Rams
Fri Sep 28, 2018 10:02 pm
  • Seahawkfan80 wrote:
    5_Golden_Rings wrote:
    Seahawkfan80 wrote:
    5_Golden_Rings wrote:As an internet smart guy, here's some free tips on how to derive Einstein's E=mc^2 in three steps (because (1) I don't like being passively aggressively called stupid, and (2) I'm really really, shall we say, not "low" (I have all day to do nothing), so I'm going to be a smart ass while simultaneously doing a free public service to anyone who's nerdy enough to wonder about a logical justification for E = mc^2):



    Step 1:



    1. Assume the principle of relativity (the laws of physics are the same for all inertial reference frames)
    2. Assume isotropy of space and time (the direction you move in space or location in space doesn't change the laws of physics; the point in time you are in does not change the laws of physics).


    Now that we got that out of the way, we can infer that Newton's laws hold for small intervals of space and time in all inertial reference frames. Given that, we can imagine two reference frames, one S, and one S', moving with respect to each other at some speed v. The space and time coordinates (in 2 dimensions) for S will be x and t. The coordinates for S' will be x' and t'.

    Without making any assumptions about how time and space work, other than coordinate transformations between S and S' will be linear, we'll have this relationship (note we could use differentials if we wanted but why? It's linear):


    x' = Ax + Bt and t' = Cx and Dt

    where A, B, C, and D are four constants we're going to find out that will possibly depend on speed v.


    Now, consider the situation where the object S is watching move is at rest in S' (in other words, the object sits at the origin of S' while S' moves at speed v with respect to S).

    Then in that particular case (which exists for at least one inertial system), x' = 0.

    Thus, 0 = Ax + Bt, and therefore Ax = -Bt, and therefore x/t = v = -B/A. Which means B = -Av. Thus,

    Ax = x' - Bt
    Ax = x' - B(t' - Cx)/D
    Ax = x' - Bt'/D + BCx/D
    Ax - BCx/D = x' - Bt'/D
    ADx - BCx = Dx' - Bt'
    (AD - BC)x = Dx' - Bt'

    Now doing the same thing using this information with the equation relating time measurements in S to time measurements in S'.

    Dt = t' - Cx
    Dt = t' - C(x' - Bt)/A
    Dt = t' - Cx'/A + BCt/A
    Dt - BCt/A = t' - Cx'/A
    ADt - BCt = At' - Cx'
    (AD - BC)t = At' - Cx'

    Now go back to the first object moving from S'. It's moving in the other direction, so

    x/t = [(Dx' - Bt')/(AD-BC)]/[(At' - Cx')/(AD - BC)]
    x/t = (Dx' - Bt')/(At' - Cx')
    x/t = (Dv' - B')/(A' - Cv')

    x/t = (Dv' - B')/(A' - Cv')

    but since the first object is at rest in S, we can let x/t = 0 (just like we did with letting x' = 0 before), giving

    0 = (Dv' - B')/(A' - Cv')
    Dv' + B = 0
    v' = x'/t' = B/D

    Hence, B = -Dv, and then remembering that v = -B/A, we'll have:

    B = -D(-B/A)
    B = D(B/A)
    1 = D/A
    A = D

    So plug B = -Av and A = D back in and you'll see that we have:

    (AD - BC)x = Dx' - Bt'
    (A^2 + vAC)x = Ax' + Avt'
    (A^2 + vAC)x = A(x' + vt')


    Now, in real life a transformation like this requires that the same equations to work both ways except inverting the coordinates and reversing the sign (the x' to x, the t to t', and v to -v), and the only way for what I just wrote to hold for what we've derived so far is for (A^2 + vAC) to equal 1.

    So, set that equal to 1 and then solve for C:

    A^2 + vAC = 1
    vAC = 1 - A^2
    C = (1 - A^2)/(vA)



    Which means we can come to the following conclusions about the transformation equations that only depend on one of the constants, A:

    x' = Ax + Bt
    x' = Ax - Avt
    x' = A(x - vt)

    and (remembering to plug in the value for C):

    t' = Cx + At
    t' = At + Cx
    t' = At + x(1 - A^2)/(vA)
    t' = At - x(A^2-1)/(vA)
    t' = At -vx(A^2-1)/(v^2 A)
    t' = A(t - vx(A^2-1)/(v^2 A^2)
    t' = A(t - vx(A^2-1)/(v^2 A^2)




    So, the two transformation equations that depend only on one constant, A, are:


    x' = A(x - vt)

    and

    t' = A(t - vx(A^2-1)/(v^2 A^2))


    Since (A^2-1)/(v^2 A^2) is ugly and depends only on A and v, let it equal k. Then, we'll have to solve for k because the A is sitting there in the x' = A (x - vt) transformation equation. Doing that:

    (A^2 - 1)/(v^2 A^2) = k
    A^2 kv^2 = A^2 - 1
    A^2 - A^2 kv^2 = 1
    A^2 (1 - kv^2) = 1
    A^2 = 1/(1 - kv^2)
    A = 1/√(1 - kv^2)



    So we can plug that back into the two x' and t' transformation equations (and remembering what we defined k as for the time coordinate equation), giving:

    x' = [1/√(1 - kv^2)] (x - vt)

    and

    t' = [1/√(1 - kv^2)] (t - vxk)



    Now we've reached GENERALIZED TRANSFORMATION EQUATIONS. We are not yet at special relativity. These two equations are true for Galilean relativity, too. They depend on the value of k. Before we choose values for k, two things need to be made clear:

    (1) The only values of k that matter are 1, 0 and -1, because your choice of units can affect the value of k, and only those three numbers have true physical consequences.

    (2) What is k in terms of physics? It is a function of the maximum speed limit, which is either finite or infinite. Using a lower case c to represent to the maximum speed limit, k = 1/c^2. Now again, either the maximum speed limit is infinity or it is finite (spoiler alert: the maximum speed limit is measured to be finite and is in fact the speed of light).

    So now let's look at what happens when we plug in values for k.



    If k = 0 (note if that were the case we'd have to use limit notation and let c approach infinity as k approaches zero so as to not divide by zero), here's what we have:


    x' = [1/√(1 - 0*v^2)] (x - vt)

    x' = x - vt

    and

    t' = [1/√(1 - 0*v^2)] (t - vx*0)

    t' = t

    And if you know anything about basic physics, those are the Galileo transformation equations, which hold true in pre-Einstein, Newtonian physics. In fact you can see this if you divide x' by t' to get v' (the speed the observer at rest in S' sees):

    x'/t' = (x - vt)/t'

    x'/t = (x - vt)/t

    x'/t = (x - vt)/t

    u' = u - v. This is the inverse Galilean velocity addition formula. It's old news (if you throw a ball 30 mph on the ground to my right, but I"m in a truck moving to your left at 60 mph according to the ground, I'll see the ball moving 30 mph to my left, i.e., at -30 mph). Moving on.




    If k = -1, we have:

    x' = [1/√(1 - (-1) v^2)] (x - vt)

    x' = [1/√(1 + v^2)] (x - vt)

    and

    t' = [1/√(1 - (-1)v^2)] (t - vx (-1))

    t' = [1/√(1 + v^2)] (t + vx)

    Now if you're really clever, you'll realize that leads to an impossibility: it allows for you to travel freely forwards and backwards through time, which we clearly cannot do. Furthermore, it leads to an indeterminate form (infinity over infinity) when you play with the transformations. I won't waste time with that.





    If k = 1, we have:


    x' = [1/√(1 - v^2)] (x - vt)

    and

    t' = [1/√(1 - v^2)] (t - vx)


    This is the Lorentz transformation in units in which the speed of light is one. Recalling that k = 1/c^2, if we change our units to the usual c = 3x10^8 meters/second, we get this:

    x' = [1/√(1 - v^2/c^2)] (x - vt)

    and

    t' = [1/√(1 - v^2/c^)] (t - vx/c^2)

    and here are the transformations from x to x' (swap the coordinates and change the sign on the v outside of the square root):

    x = [1/√(1 - v^2/c^2)] (x' + vt')

    and

    t = [1/√(1 - v^2/c^2)] (t' - vx/c^2)


    which you will see in any modern physics text book. For the sake of brevity, going forward I'm going to let the letter y = [1/√(1 - v^2/c^)] so I don't have to keep writing it, until it becomes necessary in step 2. This will make the time transformation from S to S' look like this (it will be needed for step 2):

    t = y (t' - vx'/c^2)






    That's step 1. Here's step 2:


    Consider the situation in which the moving clock is at rest with respect to the observer. This clock will measure what is called "proper time," and every measurement performed in any inertial reference frame will agree upon this value (which is why it is important in special relativity- it's invariant. Einstein actually wanted to call his theory the Theory of Invariants, but it was too late, sadly... ). So what happens when the observer S' is at rest with respect to his or her clock? x' = 0. So, look at the last time transformation I typed up and let x' = 0:

    t = y (t' - vx'/c^2)

    t = y (t' - v*0/c^2)

    t = yt'

    t' = t/y

    where y is that ugly square root, and in this case, t' is proper time (note that proper time will always be the shortest measured time interval by any observer for an event; when people saying "moving clocks run slow," they mean THEY'LL measure the moving clock having slowed time, so the moving clock observer himself will measure the smallest time interval, because to the moving clock observer, their clock is not moving. Everyone else watching the moving clock will conclude that the time interval is longer, because the moving clock will appear to be running slowly. Tricky, but you can figure it out).






    That's step 2. Here's step 3:


    Consider a random coordinate in our one spatial dimension simplification (we don't have to worry about all three spatial dimensions here because we're discussing two reference frames moving parallel with respect to each other with their respective distance axes aligned). This coordinate is a space AND time coordinate, and to make units match, time will be multiplied by the speed c (because units of time * units of distance/time = units of distance).

    Call it capital X. Then X = (x, ct).

    Now, because proper time is invariant, divide everything by proper time. Call this new thing capital V, and remember that proper time is t/y, where y is the ugly square root:

    V = (x/[t/y], ct/[t/y])

    simplify:

    V = (yu, yc)

    where u is coordinate velocity x/t and c is still the maximum speed limit (which by now I'm sure you've guessed has been measured to be the speed of light).


    Now we want to turn that into momentum. How do we do that in "old" physics? Multiply by mass. So let's do that. We'll use the symbol capital P for momentum, so mV = P

    P = (myu, myc)



    Now, because I'm sure you're getting bored, and because I don't want to do a massive ugly integral using trig substitution, I'm going to do a short circuit skipping step. What I'm going to do is ignore the spatial component of momentum and just look at the time component, and then take the time derivative to get a component of "force," and then use the work energy theorem to get "energy," although keep in mind this is cheating to skip space because this post is already huge.

    First, looking at the time "momentum" coordinate and taking the time derivative (note that m is constant with respect to time, so it can be pulled outside the derivative):

    F = dP/dt = m*d(yc)/dt



    Now, apply the work-energy theorem, which means if you integrate force over distance and choose your interval of integration correctly you'll get kinetic energy, except instead of dx for distance, because I'm working with time, I'm going to replace dx with cdt (this is the "cheating" I was speaking of).

    (Also note that m is constant with respect to distance, so it can be pulled outside the integral):

    W = F cdt = m [ d(yc)/dt] cdt

    note that c is constant with respect to time, so it can be pulled out of the differential d(yc) so that it can be written as cdy.

    W = m [cdy/dt]cdt

    Some more rearranging:

    W = m [c^2dy*dt/dt]

    and dt/dt = 1, so

    W = m [c^2dy]

    again, note that c is constant with respect to y, so it can be pulled out of the integral, too

    W = mc^2dy

    And as any first year calculus student knows,

    dy = y

    So, the indefinite integral is:

    W = ymc^2

    Now, choose the interval of integration to be from 0 to v, remembering that y is the nasty square root. This gives:


    W = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1 - 0^2/c^2)] mc^2 = [1/√(1 - v^2/c^2)] mc^2 - [1/√(1)]mc^2

    W = [1/√(1 - v^2/c^2)] mc^2 - mc^2


    And THAT is the kinetic energy function for special relativity.


    Now, recall that total energy = kinetic energy + rest energy. That means kinetic energy = total energy - rest energy.

    Thus, total energy is [1/√(1 - v^2/c^2)] mc^2 and rest energy is mc^2.


    Proof this is true? Take the total energy equation found above and let v = 0:

    E = [1/√(1 - v^2/c^2)] mc^2

    E =[1/√(1 - 0^2/c^2)] mc^2

    E = [1/√(1 - 0)] mc^2

    E = [1/√(1)] mc^2

    E = mc^2





    And there you go. Now you can be rest assured beyond all doubt that I AM in fact the smartest person on the internet you've ever seen.


    So what is the speed of light? I learned that back in the 7th grade.


    I get insecure when I’m blitzed lol.

    The speed of light is whatever you want it to be based on your choice of units. What does not change is the fine structure constant, and it’s the invariance and finiteness of the speed of light that matters, not the numerical value. But in “normal” units it’s about 3x10^8 m/s or 186,000 miles per second.


    Yes. Thank you. I need to digest this post again. I REALLY Appreciate it. I was thinking about going back to my calculus books for idiots. I have note pads so I dont desecrate the books. I ran thru it kinda fast...should not have. I am a radar electronics Tech. But not that slow. thanks.

    Sounds like a cool job. I wish I could take the time to study electronics and electromagnetism in general beyond the theoretical stuff taught in EM. I want to make things. I made a current creating device once, but never beynd that. Just never have the time to learn it.
    Sports Hernia wrote:Mayfield has grown on me a bit, but I disagree with him here.

    If an employer fires/terminates me, why do I owe them any sort of “loyalty”?

    &@$# that!

    That's what Sherman said
    User avatar
    5_Golden_Rings
    NET Veteran
     
    Posts: 1397
    Joined: Fri Sep 10, 2010 7:38 am


Re: Vikings @ Rams
Mon Oct 01, 2018 3:55 am
  • They still gave up 31 points, nobody seems to be talking about that.
    R.I.P. THE EDGAR, YOU WILL BE MISSED......
    User avatar
    SoulfishHawk
    NET Pro Bowler
     
    Posts: 10821
    Joined: Thu Sep 06, 2012 9:59 am
    Location: Sammamish, WA


Re: Vikings @ Rams
Mon Oct 01, 2018 4:13 am
  • SoulfishHawk wrote:They still gave up 31 points, nobody seems to be talking about that.


    I’m always excited to see what Soul has to say.

    He’s the best Rams hater I’ve ever seen.

    I just want to capture him and put him in a jar in my kitchen. It’s the most cutest adorablest thing ever!!!! It’s making this season so much better every word he types!!!!!!!!!!!!!!!!

    :179417: :179417: :179417:
    When you get to Helllll, John. Tel em Daisy sentcha.
    User avatar
    RedAlice
    NET Veteran
     
    Posts: 3218
    Joined: Mon Dec 24, 2012 11:47 am
    Location: San Diego


Previous


It is currently Fri Dec 14, 2018 8:56 am

Please REGISTER to become a member

Return to [ NFL NATION ]




Information
  • Who is online