Can someone please help me with Lorentz Transformations?? I think it is this

 

x’ = gamma (x –vt) t’ = gamma(t- vx-c squared) gamma =1/sqrt 1-vsqaured/c squared

 

 

I am unsure if this is the actual formula as I have seen it posted here. Can you also be simple??

 

 

All I really need to see is an example using some simple "numbers" only. I am very thankful of you respond, thanks!

 

 

Can someone please help me with Lorentz Transformations?? I think it is this

 

x’ = gamma (x –vt) t’ = gamma(t- vx-c squared) gamma =1/sqrt 1-vsqaured/c squared

 

 

I am unsure if this is the actual formula as I have seen it posted here. Can you also be simple??

 

 

All I really need to see is an example using some simple "numbers" only. I am very thankful of you respond, thanks!

An event happens one lightyear to your right at t=0.

 

Sally is in your location moving to your right at .6c. As she passes, her clock also says t=0. According to her clock the event happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left( 0 - 0.6 \cdot 1/1^2 \right)[/latex]

[latex]= -0.75[/latex]

 

and according to her ruler it happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left(1 - 0.6 \cdot 0 \right)[/latex]

[latex]= 1.25[/latex]

THANK YOU!!!!!!!!!!!!!!! can this be used with precession too?

An event happens one lightyear to your right at t=0.

 

Sally is in your location moving to your right at .6c. As she passes, her clock also says t=0. According to her clock the event happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left( 0 - 0.6 \cdot 1/1^2 \right)[/latex]

[latex]= -0.75[/latex]

 

and according to her ruler it happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left(1 - 0.6 \cdot 0 \right)[/latex]

[latex]= 1.25[/latex]

Edited December 29, 201113 yr by I think out of the box

No, they don't calculate precession.

thanks, but I have one more question if you do not mind, how do scientist like you calculate fractions of the speed of light, and what does 1 squared mean??? I always thought that 1 squared was just 1, thanks!

No, they don't calculate precession.

thanks, but I have one more question if you do not mind, how do scientist like you calculate fractions of the speed of light, and what does 1 squared mean??? I always thought that 1 squared was just 1, thanks!

I'm not a scientist.

 

One squared means one times one and it does equal one.

 

When I said "moving to your right at .6c", that may have not been the best way to say that. I should have said "moving to your right at 0.6 lightyears per year". The speed of light is one lightyear per year. That may explain why I set v=0.6 and c=1.

 

The equations I wrote don't need anything to be expressed in fractions of the speed of light. You can use any units -- just as long as c and v are both expressed in the same units. I'll rewrite it and see if it makes more sense:

 

An event happens one lightyear to your right at t=0.

 

Sally is in your location moving to your right at 0.6 lighyears per year. As she passes, her clock also says t=0. According to her clock the event happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-(0.6 \ ly/yr)^2/(1 \ ly/yr)^2}} \right) \left( 0 \ yr - 0.6 \ ly/yr \cdot 1 \ ly/(1 \ ly/yr)^2 \right)[/latex]

[latex]= -0.75 \ years[/latex]

 

and according to her ruler it happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-(0.6 \ ly/yr)^2/(1 \ ly/yr)^2}} \right) \left(1 \ ly - 0.6 \ ly/yr \cdot 0 \ yr \right)[/latex]

[latex]= 1.25 \ lightyears[/latex]

 

You can do the same thing with any units.

 

If you do, however, want to know what fraction of the speed of light a velocity is... just divide the velocity by the speed of light. For example, 100 million meters per second is one third of the speed of light because 100 million divided by 300 million is one third. The speed of light in m/s is 300 million.

Edited December 30, 201113 yr by Iggy

YOUR NOT A SCIENTIST??????????? You really sound like one! GOOD JOB! Now I understand this better than ever..

I'm not a scientist.

 

One squared means one times one and it does equal one.

 

When I said "moving to your right at .6c", that may have not been the best way to say that. I should have said "moving to your right at 0.6 lightyears per year". The speed of light is one lightyear per year. That may explain why I set v=0.6 and c=1.

 

The equations I wrote don't need anything to be expressed in fractions of the speed of light. You can use any units -- just as long as c and v are both expressed in the same units. I'll rewrite it and see if it makes more sense:

 

An event happens one lightyear to your right at t=0.

 

Sally is in your location moving to your right at 0.6 lighyears per year. As she passes, her clock also says t=0. According to her clock the event happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-(0.6 \ ly/yr)^2/(1 \ ly/yr)^2}} \right) \left( 0 \ yr - 0.6 \ ly/yr \cdot 1 \ ly/(1 \ ly/yr)^2 \right)[/latex]

[latex]= -0.75 \ years[/latex]

 

and according to her ruler it happens at:

 

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

[latex]= \left( \frac{1}{ \sqrt{ 1-(0.6 \ ly/yr)^2/(1 \ ly/yr)^2}} \right) \left(1 \ ly - 0.6 \ ly/yr \cdot 0 \ yr \right)[/latex]

[latex]= 1.25 \ lightyears[/latex]

 

You can do the same thing with any units.

 

If you do, however, want to know what fraction of the speed of light a velocity is... just divide the velocity by the speed of light. For example, 100 million meters per second is one third of the speed of light because 100 million divided by 300 million is one third. The speed of light in m/s is 300 million.

In decays, wherein one particle "breaks apart" into two or more other particles, is relativistic four-momentum conserved through the decay, i.e.

 

[math]p^{\mu}_i = p^{\mu}_f = p^{\mu}_1 + p^{\nu}_2 + ...[/math]

??

 

And, if four-momentum is conserved, is not its "squared interval"

 

[math]p^2 = p^{\mu}p_{\mu}[/math] (= m

²

for one particle)

??