Go to the link below and click on page 49 and question 1.24.
https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

So assuming I am understand the question correctly Bob and Alice have the same location in both frames.

Event 0 everything is 0.

event 1
I assume both Alice and Bob are at the same location for every location. So I am just doing the calculations once for
Alice's frame for Alice and Bob's frame for Bob. I am not sure if the question asked it but I also did prime frame for  Alice and Bob.

gamma = 2
v = .866c
t = 1 sec
I go
d = vt = (1 s)(.866c) = .866LS (for lorentz transform)
Then t =  2 I could do the calculation or mulitply everything by 2. For t =  3 multiply by 3 etc
Now to find t' for event 1
t' = gamma (t -xv /c^2)
t' = 2(1 sec - (.866LS)(.866c) )
t' = .5 sec.
Now to get event 2 variables multiply time by 2 to get event 3 multiply 3.
Here is my basic formula

t_event_number = t_1_or_t'_1 (event_number)

t_event_number = t_1
t'_event_number = t_1'

Let me show an example
t_1' = .5 and t_1 = 1  
While, event number =  0 - infinity
Is this correct?

I remember reading that gut microbes could affect mental health.
I was thinking this could be similar to H pylori.
Has anyone tested the hallucinogenic gut microbe idea before? How are gut microbes mapped?

https://massivesci.com/notes/antidepressant-gut-health-bacteria/
https://www.inverse.com/mind-body/scientists-explain-link-between-depression-and-the-gut

By blinking I mean the screen seems like it is blinking.

The webcam is built into the pc.

I did a clean install in windows 10  then the webcam started acting up.

I did an avira virus scan and even an online set scan and malwarebytes and everything came back negative.

I also loaded some files from my usb and scanned my usb with malwarebytes and avira and nothing came back.

I even tried the method from the link below.

The 4th method didn't work because I couldn't create a new file.
 

 

 

Can someone help fix my webcam and make sure I don't have a virus etc?

The only things besides antiviruses's I downloaded were apps in firefox, which were privacy badger, https everywhere, Dencentraleyes and ublock origin.

 

Specs below

 

 

Thanks

5 hours ago, Ghideon said:

I agree; Over here we have at least one example of a concept store where there is no middle man at all. But maybe there is room for a developer to provide a platform shared by many if the middle man adds value. 

But if OP wishes to compete with platforms that connects individuals to individuals rather than companies to individuals then copycats could be an issue. We have at least two such apps around here. As an individual I prefer/need a man in the middle that handles the transactions, advertising, insurance etc but I could easily with to a copycat who provides a better experience.

@can't_think_of_a_name your questions are good but tricky to answer without more context. The business in the picture could use sensors etc to automate checkout and returns. In case individuals rent from each other there are different solutions available in different countries. The apps here rely heavily on local infrastructure.

Do you know the name of the store or apps so I could see what they are doing? Also sensors are expensive. I was thinking everyone who uses the app has a phone. So each phone would have a code that uses something like Bluetooth or NFC and send out a random generated code that isn't ever reused. During the rental of the item the code would be registered and the return would be registered. But this relies on good faith. Most people will follow good faith but a percentage won't. A friend suggested a rating system. I guess you could ban people below a certain star. But would that work?  

5 hours ago, Phi for All said:

Your problem isn't going to be copycats, imo. The people who own these products you want to rent out could easily arrange something like this, and cut you out as the middle man. The owners of the product don't have your administrative problems with confirming transactions and returning the goods, so they can do this more cheaply than you can. 

The motivations to rent as opposed to purchasing are tricky. And you have to be careful, since some of the same customers who prefer to rent a product do so because they're hard on products, and expect a replacement at no charge when they break it. 

Well I can think of a few ways around that. 1 only make money on the advertising on the platform. Can you make money off of that? 2 at a later date create an subscription fee. I assume you could create a subscription fee at a earlier date but would that work? 3 create a retail store front where everything is stored. I don't like 3. Is there anyway people would be okay with fees without cutting out the middle man as in me. Thoughts? 

Ignore the post right ahead of this.

So what your saying is the Lorentz transform doesn't change form a stationary frame to another stationary frame?

It seems like you have to start with L = L'/gamma in whatever reference frame you are in. Why can't you start with L' = L gamma?

Does the Galilean transform still work in special relativity in changing from Bob's stationary frame to Alice stationary frame? What is now confusing me is why the speeds in Bob's frame doesn't go from .983c to Alice frame -.0983. is this where I use Einstein velocity addition formula

I have come up with an idea. Without going into too much detail the idea is to rent products in your area. I would create a mobile app. People would rent the product. Each transaction I would make money off of. The problem I am having is how would I confirm the transaction went  through and confirm the product was returned?

Also I know the basics of coding but never made a mobile app or anything really. I was thinking of starting off as an apple app  but eventually developing for android. Is there a way develop for both android and apple?

One problem is copycats could easily be created. Do you think the best way to avoid copycats is to get funding, assuming I could get? The app couldn't really be used until after covid is cured so it is far off.

Also how do you build an app that doesn't crash when it has too much traffic? Is it very hard to get funding? I am also planning on partnering with someone I assume I should pick someone in the same city. 
Hopefully I didn't give away the idea.

Does anyone know any good languages to create it in?

I just realized this could go in computer science area.
 

So what your saying is the Lorentz transform doesn't change from a stationary frame to another stationary frame?

It goes from stationary object to moving object,

but the moving object doesn't become stationary it stays moving. This always works from moving object to stationary object.

Does the Galilean transform still work in special relativity in changing  from Bob's stationary frame to Alice stationary frame ? What is now confusing me is why the speeds in Bob's frame doesn't go from .983c to Alice frame -.983c.  Is this where I use Einstein velocity addition formula?

I uploaded a picture just to make this clear for me. I think I got the correct answer.

But I have 1 question. L =  4.93 in Bob's frame. In Alice frame L' = 4.93. Therefore I apply Length contraction. L = L' / gamma.  This makes L = .82.

Why doesn't L' in Bob's frame become L = 29.58 in Alice's frame?
 

I know moving objects are smaller from a stationary observer.

But isn't stationary object bigger for a moving observer?
I apologize if this repetitive.

Ignore this picture below.

I mean d = L.

I Calculate the length for Alice in Alice's frame I get d = vt
Then I calculate d = vt  = (-.986c)(5 years) =  -4.93 LY. Alice sees the planets moving so I go L=L'/gamma. -4.93/5 = -0.986 LY

The part that was confusing me is the difference between d = vt and L=L'/y. I thought d = vt and L=L'/y are the same. How are they different?

Thanks

gamma should be 6 but ignore that just go with 5.

 No. If I am not mistaken there length needs to be the same to maintain symmetry for stationary Bob and stationary Alice. So how do I switch frames? Just to confirm L_stationary = L_moving / gamma. That doesn't switch frames? I assume the problem is the - sign.

Maybe I am just really confused but I have Bob's frame and Alice's frame. In Bob's frame I have Bob who is stationary and Alice who is moving. In Alice's frame I have Alice who is stationary and Bob who is moving in the opposite direction. Length contraction deals with either Bob's frame or Alice frame but it doesn't switch from Alice's stationary frame to Bob's stationary frame.

To answer your question exactly "Alice in Bob's frame" I just mean Alice is moving and Bob is stationary.

In my first post I just want to confirm length contraction in special relativity at a later date I will post my entire summary of special relativity. It is a work in progress.

I travel to a planet from earth it takes 5 years.
Bob is located on earth. Alice is moving at v = .986c on a spaceship to the other planet.
If I want to know the length for Bob in Bob's frame  d= vt = (.986c)(5 years) = 4.93 LY
If I want to know the length for Alice in Bob's frame. L = L'/ gamma =  L' = (L) (gamma) = L' = (4.93)(5 years) = 29.58 LY
 

If I want to switch velocity I use Einstein''s velocity addition formula and I get -.986c for planet earth and the other planet.
My main question is why time  for Alice in Alice's frame  = 5 years.
 If I am not mistaken.
Then I Calculate the length for Alice in Alice's frame I get d = vt
Then I calculate d = vt  = (-.986c)(5 years) =  -4.93 LY
Then I calculate  the length for Bob in Alice's frame .  L' = (L) (gamma = -29.58 LYLY.

So to summarize I have 2 frames Alice and Bob and 2 measurements in each frame a moving and stationary frame that I need to calculate.

If I made any mistakes please correct me if I didn't then tell me?

I also have one more question if I have c+t =g. If I want to move g to the left of the = sign. I don't care how pointless this is. I go c+t-g =  0?

Hopefully the last question about basic math.  Does √1 = 1^-1. Also I can go √1^2 = 1. Are there any other ways to remove a square root?

1 / Gamma = √ (1 - v^2) c^2/ c^2  is the same as

1 / Gamma = √ (c^2 - v^2) c^2/ c^2 

shouldn't this give  c / Gamma = √ 1 - v^2   if I do it my way. Or is that just not how algebra works?

I apologize for asking basic question I guess when I was originally in school either I didn't pay attention or wasn't taught properly. I assume the former but I can't be certain. I appreciate the help on basic questions.

why is this wrong?

Gamma = 1 / √ (1) - v^2 / c^2 =

1 / Gamma = √ 1 - v^2 c^2/ c^2 =

c / Gamma = √ 1 - v^2 +v^2 =

c + v / gamma = √ 1 =

switch the c + v to v + c =

v + c / gamma = √ 1=

v + c / gamma = √ 1 =

v + c gamma / gamma = √ 1 (gamma^2) =

v + c -c = √ gamma^2 =

v = √ gamma^2 - c

or

1 / Gamma - c^2 = √ 1 - v^2 / c^2 - c^2 =

1 / Gamma - c^2 = √ 1 - v^2 =

1 + v / Gamma - c = √ 1 - v^2 + v^2 =

1 +v / Gamma - c = √ 1 =switch the 1+ v to v + 1 =

v + 1 / gamma - c = √ 1=

v + 1 gamma / gamma - c = √ 1 (gamma) =

(v + 1) -c / - c = √ gamma -c^2 =

- (v+1) = √ gamma -c^2 =

-v -1 = gamma -c^2 =

-v = √ gamma^2 - c^2 + 1^2 =

- v / - = √ gamma^2 - c^2 + 1^2/ - =

v = √ gamma^2 - c^2 + 1^2 / - =

v = √ -gamma^2 + c^2 - 1^2

Okay I am not sure if this is good or bad answer. Worst case scenario the universe does repeat. I  Imagine I have pattern red green blue. This pattern repeats red green blue infinitely many times. I can either view the pattern as infinite or finite. Do I really need to count the additional red green blue or can I just view it as one pattern.

Just to make it clear
red green blue (happens once)
red green blue (happens infinite times)
I don't need any more information to know the pattern. Its like saying what is the number 1 is it 2/2/2/2 infinite times it doesn't add any information.
Of course you can view it as an overly complex infinite pattern  or you can view it as happening once. It reminds me of the glass half full half empty. It depends on your perceptive. I wasn't really going for the positive or negative outlook I was just saying both are correct but you can choose how to view it.

I apologize if this is a horrible answer. I am no way saying the universe repeats. Feel free to delete this answer if it sucks.

I meant v = c/ gamma

I know I got v = gamma/c. Like stated earlier why does  b= v/c give a different answer. I was solving a problem problem 3 A. I used V = c/gamma but they want b = c/v. This confuses me why does one works and  not the other? I guess I really didn't explain this well. In my course they never explain the difference. If the link doesn't work I will post the question.  

From here https://d3c33hcgiwev3.cloudfront.net/_d37cb29a797de375eb7866e695098a79_Wk7_problemsetsolutions.pdf?Expires=1603756800&Signature=SUmoIwTy2VAIs1CfSGO~F7C3BD7lIJKyMQJuedeO4RUqiCeo9HTLJk50r~oKjI6pAFoaSG5p-Pu3FRHuPiNizcGdd6EdCj4Eer1tU4BqbIBsdzW0WjLXxR8E~-5gZx3LDteO6L4ruR80eOQi1EzUdBj8Z9lnXJ6kQuj1fHpJGik_&Key-Pair-Id=APKAJLTNE6QMUY6HBC5A

6 hours ago, swansont said:

Could you be more specific? I don't see a use of the reciprocal for speed. I see a reciprocal for gamma mentioned, which they call alpha. Is that what you're referring to? The only mention of "reciprocal" is in this context.

I start with gamma.
The part that confuses me is that I want velocity = , from gamma . I get v =  c/gamma.(I could have made a math mistake.)  It never says I need B = v/c. I know B isn't the correct symbol I am just using it here.

Thanks for the prompt responses.

36 minutes ago, joigus said:

Not really. Not in general. You might get lucky and pull it off in particular examples. For example, consider the (non-linear) system:

 

x2−y2=a

 

 

x+y=b

 

And assume,

 

b≠0

 

The first equation is non-linear, because it involves powers of x and y different from 1. But you could use the second one to substitute x+y in,

 

x2−y2=(x+y)(x−y)

 

and get to the linear system,

 

x−y=ab

 

 

x+y=b

 

which can be solved by Gauss (by adding and subtracting both eqs.) to get,

 

2x=b+ab

 

 

2y=b−ab

 

So that,

 

x=b2+a2b

 

 

y=b2−a2b

 

That's the problem with non-linear equations. Each one is different.

 

Does this imply a computer can't solve non linear equations?

32 minutes ago, joigus said:

Absolutely not. Gauss elimination is for linear equations; GR is highly non-linear.

This math is beyond my understanding at this point but shouldn't this make it quite easy to solve to nonlinear equations?

https://www.quora.com/Is-there-an-algorithm-for-solving-nonlinear-systems-of-equations-akin-to-Gaussian-elimination