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. Can someone please help me with Lorentz Transformations?? Rate Topic: -----

#1 28 December 2011 - 07:58 PM I think out of the box 


Meson
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!


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#2 28 December 2011 - 09:54 PM Iggy 


Meson

View PostI think out of the box, on 28 December 2011 - 07:58 PM, said:

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:

t' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( t - v x/c^2 \right)
= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left( 0 - 0.6 \cdot 1/1^2 \right)
= -0.75

and according to her ruler it happens at:

x' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( x - v t \right)
= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left(1 - 0.6 \cdot 0 \right)
= 1.25
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#3 29 December 2011 - 05:41 AM I think out of the box 


Meson
THANK YOU!!!!!!!!!!!!!!!Posted ImagePosted ImagePosted ImagePosted Image can this be used with precession too?

View PostIggy, on 28 December 2011 - 09:54 PM, said:

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:

t' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( t - v x/c^2 \right)
= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left( 0 - 0.6 \cdot 1/1^2 \right)
= -0.75

and according to her ruler it happens at:

x' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( x - v t \right)
= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left(1 - 0.6 \cdot 0 \right)
= 1.25

This post has been edited by I think out of the box: 29 December 2011 - 05:58 AM

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#4 29 December 2011 - 05:26 PM Iggy 


Meson
No, they don't calculate precession.
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#5 29 December 2011 - 10:35 PM I think out of the box 


Meson
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!

View PostIggy, on 29 December 2011 - 05:26 PM, said:

No, they don't calculate precession.

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#6 30 December 2011 - 10:28 PM Iggy 


Meson

View PostI think out of the box, on 29 December 2011 - 10:35 PM, said:

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:

t' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( t - v x/c^2 \right)
= \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)
= -0.75 \ years

and according to her ruler it happens at:

x' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( x - v t \right)
= \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)
= 1.25 \ lightyears


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.

This post has been edited by Iggy: 30 December 2011 - 10:28 PM

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#7 1 January 2012 - 03:02 AM I think out of the box 


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

View PostIggy, on 30 December 2011 - 10:28 PM, said:

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:

t' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( t - v x/c^2 \right)
= \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)
= -0.75 \ years

and according to her ruler it happens at:

x' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( x - v t \right)
= \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)
= 1.25 \ lightyears


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.

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#8 21 January 2012 - 07:06 PM Widdekind 


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

p^{\mu}_i = p^{\mu}_f = p^{\mu}_1 + p^{\nu}_2 + ...

??

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

p^2 = p^{\mu}p_{\mu} (= m2 for one particle)

??
http://www.artofprob...p/LaTeX:Symbols
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