Lorentz transform equation help, special relativity

[can't_think_of_a_name]

Consider the situation where we want to combine the space-time diagrams of Alice and Bob, where Bob is moving at a speed of 0.4c to the right (positive x direction). If we draw Alice’s x_A axis as horizontal and t_A axis as vertical, answer the following questions.(a) What is the equation, written in terms of x_A and t_A, that describes Bob's t_B axis (the world line where x_B = 0) on Alice’s diagram?

 

a) What is the equation, written in terms of x_A and t_A, that describes Bob’s t_B axis (the world line where x_B = 0) on Alice’s diagram?

 

I assume I want to change from Bob's  frame to  Alice's frame. Is this correct.

 

here are the relevant equation.

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

x' = y(x-vt)

How do I transform from the equations t' to t equation? I.O.W I want the non primed time equation.

 

If I made any mistakes in my assessment of the answer correct me. Thanks.

[joigus]

1 hour ago, can't_think_of_a_name said:

Consider the situation where we want to combine the space-time diagrams of Alice and Bob, where Bob is moving at a speed of 0.4c to the right (positive x direction). If we draw Alice’s x_A axis as horizontal and t_A axis as vertical, answer the following questions.(a) What is the equation, written in terms of x_A and t_A, that describes Bob's t_B axis (the world line where x_B = 0) on Alice’s diagram?

 

a) What is the equation, written in terms of x_A and t_A, that describes Bob’s t_B axis (the world line where x_B = 0) on Alice’s diagram?

 

I assume I want to change from Bob's  frame to  Alice's frame. Is this correct.

 

here are the relevant equation.

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

x' = y(x-vt)

How do I transform from the equations t' to t equation? I.O.W I want the non primed time equation.

 

If I made any mistakes in my assessment of the answer correct me. Thanks.

Where you write y, it should be (in standard notation),

\[\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}\]

Then you would have,

\[\sqrt{1-v^{2}/c^{2}}t'=t-vx/c^{2}\]

\[\sqrt{1-v^{2}/c^{2}}x'=x-vt\]

It's Gauss reduction of linear equations from here quite straightforwardly. If you need more clues, tell me.

Edit: Maybe Gauss elimination is a more familiar name...

Edited September 11, 2020 by joigus

[can't_think_of_a_name]

I don't know gauss elimination. What are the prerequisites to learn  gaus elimination? If I know the requirements I could just look up a video.

Edited September 12, 2020 by can't_think_of_a_name

[Endy0816]

2 hours ago, can't_think_of_a_name said:

I don't know gauss elimination. What are the prerequisites to learn  gaus elimination? If I know the requirements I could just look up a video.

This video covers it pretty well. You are likely already familiar with the matrices it involves.

https://www.youtube.com/watch?v=eDb6iugi6Uk

 

[studiot]

10 hours ago, can't_think_of_a_name said:

I don't know gauss elimination. What are the prerequisites to learn  gaus elimination?

It's just a formal procedure for solving simultaneous equations by substitution so that you always follow the same steps.
Substitution involves choice.

 

[joigus]

I'll give you a simpler numerical example. Suppose you've got,

\[2x-t=5\]

\[x-2t=3\]

and you want to solve in x and t. You can do the 1st equation minus twice the second:

\[-t+4t=5-6=-1\Rightarrow3t=-1\]

(you get only an equation in t)

And the second equation minus twice the first:

\[x-4x=3-10=-7\Rightarrow-3x=-7\] 

So the solution is,

\[t=-1/3\]

\[x=7/3\]

Now, you've got to do the same, but instead of 5 and 3 on the right-hand side; you've got t' and x' involved. But it's the same idea. You must eliminate t in one equation and x in the other.

Does that help?

[can't_think_of_a_name]

On 9/11/2020 at 9:42 PM, Endy0816 said:

This video covers it pretty well. You are likely already familiar with the matrices it involves.

https://www.youtube.com/watch?v=eDb6iugi6Uk

 

I don't know matrices but have been finding them quite easy to learn so far. What videos from the organic chemistry tutor should I watch to understand the video you linked.
So far I have watched

https://www.youtube.com/watch?v=yRwQ7A6jVLk

https://www.youtube.com/watch?v=vzt9c7iWPxs

Also what branch of math is matrices?

Edited September 13, 2020 by can't_think_of_a_name

[can't_think_of_a_name]

On 9/11/2020 at 5:56 PM, joigus said:

Where you write y, it should be (in standard notation),

 

γ=11−v2/c2−−−−−−−−√

 

Then you would have,

 

1−v2/c2−−−−−−−−√t′=t−vx/c2

 

 

1−v2/c2−−−−−−−−√x′=x−vt

 

It's Gauss reduction of linear equations from here quite straightforwardly. If you need more clues, tell me.

Edit: Maybe Gauss elimination is a more familiar name...

I drew the matrix as the image below. My biggest problem is I don't what to do next. I n the video it deals with a 4 by 3 matrix. I don't know how to apply Gauss elimination unless the matrix is 4 by 3. How do I apply it to different size matrices? I think I screwed up the matrix in this example.

I used this video. https://www.youtube.com/watch?v=eYSASx8_nyg

 

 

 

Edited September 15, 2020 by can't_think_of_a_name

[joigus]

6 hours ago, can't_think_of_a_name said:

I drew the matrix as the image below. My biggest problem is I don't what to do next. I n the video it deals with a 4 by 3 matrix. I don't know how to apply Gauss elimination unless the matrix is 4 by 3. How do I apply it to different size matrices? I think I screwed up the matrix in this example.

You don't need matrices. Look at my numerical example and try to understand what I did there.

In your OP case, it's v times the first equation plus the second equation, and vt -vt cancels and you're left with x in terms of t' and x'.

For the other one, it's c²/v times the 1st eq. minus the second. Now you've got cancellation x-x.Then there is some algebra left.

Gauss elimination with matrices is useful when you've got like 4x4 or even 3x3. But in a 2x2 system it's best to do it "by hand". 

[can't_think_of_a_name]

On 9/15/2020 at 2:16 AM, joigus said:

You don't need matrices. Look at my numerical example and try to understand what I did there.

In your OP case, it's v times the first equation plus the second equation, and vt -vt cancels and you're left with x in terms of t' and x'.

For the other one, it's c²/v times the 1st eq. minus the second. Now you've got cancellation x-x.Then there is some algebra left.

Gauss elimination with matrices is useful when you've got like 4x4 or even 3x3. But in a 2x2 system it's best to do it "by hand". 

I am having trouble solving the equation could you show the steps? Also do you have any tips when solving equations like these?

Edited September 18, 2020 by can't_think_of_a_name

[joigus]

OK. Let me know if you can follow the steps up to here:

\[v\left(t-vx/c^{2}\right)+x-vt=v\sqrt{1-v^{2}/c^{2}}t'+\sqrt{1-v^{2}/c^{2}}x'\]

\[\left(1-v^{2}/c^{2}\right)x=\sqrt{1-v^{2}/c^{2}}\left(vt'+x'\right)\]

\[x=\frac{\sqrt{1-v^{2}/c^{2}}}{1-v^{2}/c^{2}}\left(vt'+x'\right)\]

OK? If you have any difficulty with that, tell me. There's one minor step left. And then the other equation. But if you understand this one, it's almost done. You can PM me, if you want.

[joigus]

11 hours ago, can't_think_of_a_name said:

Also do you have any tips when solving equations like these?

Always break down problems into easy steps. Try a numerical example with say 3, -2, 7,... Get the idea. And then do it with a, b, c,...

Then go for powerful and general methods.

Once you get a solution, don't just be satisfied. Think: Does it make sense? If I make v=c, what do I get? What if I make v=0?

How does the equation of a light ray x=ct or x=-ct transform? Check.

Edited September 18, 2020 by joigus

[can't_think_of_a_name]

15 minutes ago, joigus said:

Always break down problems into easy steps. Try a numerical example with say 3, -2, 7,... Get the idea. And then do it with a, b, c,...

Then go for powerful and general methods.

Once you get a solution, don't just be satisfied. Think: Does it make sense? If I make v=c, what do I get? What if I make v=0?

How does the equation of a light ray x=ct or x=-ct transform? Check.

I think I may have gotten x = -ct or something really similar. I just got confused because I don't know the value of x and t. What are the values of x and t? 

[joigus]

1 minute ago, can't_think_of_a_name said:

I think I may have gotten x = -ct or something really similar. I just got confused because I don't know the value of x and t. What are the values of x and t? 

Arbitrary.

[can't_think_of_a_name]

1 minute ago, joigus said:

Arbitrary.

So I just plug in the values from the question?

Edited September 18, 2020 by can't_think_of_a_name

[joigus]

What question?

Oh, I see.

Bob would be the primed system. Alice would be the un-primed system. So to see how Alice sees Bob, you must substitute x'=0 in the un-primed coordinates as a function of the primed ones.

x_A, t_A correspond to x, t

x_B, t_B correspond to x', t'

You have your notations mixed.

So you express x, t (x_A, t_A) as a function of x', t' (x_B, t_B) and substitute x=0 (x_B = 0) with v/c = 0.4

Does that help?

[can't_think_of_a_name]

15 hours ago, joigus said:

OK. Let me know if you can follow the steps up to here:

 

v(t−vx/c2)+x−vt=v1−v2/c2√ t′+1−v2/c√ x′

 

 

(1−v2/c2)x=1−v2/c2√ (vt′+x′)

 

 

x=1−v2/c2 √1−v2/c2(vt′+x′)

 

OK? If you have any difficulty with that, tell me. There's one minor step left. And then the other equation. But if you understand this one, it's almost done. You can PM me, if you want.

I can't understand in step 1 how you get (gamma)x' and (gamma)t'. Doesn't the x' and t' start on the left side. I can either divide x' and t'  from the left side or - from the left.
Where did I go wrong?

Formatting incorrect.

Edited September 19, 2020 by can't_think_of_a_name

[joigus]

2 hours ago, can't_think_of_a_name said:

I can't understand in step 1 how you get (gamma)x' and (gamma)t'. Doesn't the x' and t' start on the left side.

You mean (gamma)^-1x' and (gamma)^-1t'.

I may have swapped left and right. Equations are insensitive to swapping left and right. A=B is the same equation as B=A.

All the operations I told you already:

On 9/15/2020 at 9:16 AM, joigus said:

In your OP case, it's v times the first equation plus the second equation, and vt -vt cancels and you're left with x in terms of t' and x'.

For the other one, it's c²/v times the 1st eq. minus the second.

Couldn't you reproduce the partial steps up to,

\[x=\frac{\sqrt{1-v^{2}/c^{2}}}{1-v^{2}/c^{2}}\left(vt'+x'\right)\]

?

[can't_think_of_a_name]

Step 1 why is there only 1 "v" on the right?

Step 2 how do I remove the square root from "1-v^2 / c^2?" And how does "v + xt"  become positive?
 

Is there a latex guide to this site?

Do you know or have any idea where I can find examples of this difficulty of algebra to practice?

Edited September 19, 2020 by can't_think_of_a_name

[can't_think_of_a_name]

On 9/11/2020 at 5:56 PM, joigus said:

Where you write y, it should be (in standard notation),

 

γ=11−v2/c2−−−−−−−−√

 

Then you would have,

 

1−v2/c2−−−−−−−−√t′=t−vx/c2

 

 

1−v2/c2−−−−−−−−√x′=x−vt

 

It's Gauss reduction of linear equations from here quite straightforwardly. If you need more clues, tell me.

Edit: Maybe Gauss elimination is a more familiar name...

This post was made a while ago but I have quick followup question. Can gauss elimination or something similar be used in general relativity. I am not the greatest at solving equation like stated above. 

[joigus]

2 minutes ago, can't_think_of_a_name said:

This post was made a while ago but I have quick followup question. Can gauss elimination or something similar be used in general relativity. I am not the greatest at solving equation like stated above. 

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

"Linear" means that a combination of solutions S₁, S₂, like 2S₁ + S₂, or -S₁ + 3S₂, etc., is also a solution. That is, linear combinations of solutions are also solutions.

Also, the Gauss method is only valid for numeric equations, not for differential equations. GR is set in terms of differential equations (equations that involve the derivatives of an unknown function.)

I hope that helped.

[can't_think_of_a_name]

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

[joigus]

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

\[x^{2}-y^{2}=a\]

\[x+y=b\]

And assume,

\[b\neq0\]

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,

\[x^{2}-y^{2}=\left(x+y\right)\left(x-y\right)\]

and get to the linear system,

\[x-y=\frac{a}{b}\]

\[x+y=b\]

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

\[2x=b+\frac{a}{b}\]

\[2y=b-\frac{a}{b}\]

So that,

\[x=\frac{b^{2}+a}{2b}\]

\[y=\frac{b^{2}-a}{2b}\]

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

 

[can't_think_of_a_name]

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?

[joigus]

1 minute ago, can't_think_of_a_name said:

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

No. Computers "solve" equation by getting approximate solutions.

Humans who want to solve non-linear equations must be clever. That's what I mean.