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Course: Physics library > Unit 16
Lesson 3: Lorentz transformation- Introduction to the Lorentz transformation
- Evaluating a Lorentz transformation
- Algebraically manipulating Lorentz transformation
- Lorentz transformation derivation part 1
- Deriving Lorentz transformation part 2
- Lorentz transformation derivation part 3
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Algebraically manipulating Lorentz transformation
The forms in which we've introduced the Lorentz transformations are really nice in that they reveal the symmetry of the two axes of spacetime. But there are other representations: some are just more common in practical use, while others reveal other interesting similarities with classical mechanics!
Want to join the conversation?
Why is it so important to see the symmetry?(7 votes)
It's not important to see the symmetry, as long as you understand EXACTLY what spacetime is and how it behaves. In other words, seeing the symmetry just makes it easier to not only see why the spacetime diagram looks the way it does, but also to understand spacetime as a concept in the first place. I also think that it's easier to remember and makes more sense than the manipulated forms of the equation, which may seem arbitrary.(4 votes)
If v approaches c would the Lorentz FActor equal infinite?(3 votes)
As v approaches c we get something that approaches 1/0 (from the left side of zero), so in theory, yes, we get an infinity.
In the case of time, if we're not too rigorous, we can think of it as being T = t - xc/c /e
: . T = (t-x)/e, where e is some very small numer close to zero... so, T would essentially equal infinity.(3 votes)
Must a symmetry be the invariance of a particular laws form under some transformation(a change of something the law is described interms of, or can it be the invariance of something else? Like the speed of the light(the electromagnetic wave the laws form specifies.
Symmetries( when taken as invariances of laws forms),according to Noethers theorem imply and are implied by conservation laws. Each law that is invariant is linked to a conserved quantity and vise-versa.The symmetry (invariance ) of an X law undere transformationalchange Y correspons by Noethers theorem to the conservation of a quantity(WHAT QUANTITY??). I this quantity related to Y and or X and if so HOW is the conserved quantity related to the law that in invariant or the transformation under which it is so unchanged?
Given the symmetry of the transformation equations are x'=Y(x-Bct) and
ct'=Y(ct-Bx) show that space and time are really not separate but unify into one homogenized concept called "spacetime"(ST)what is the law in spacetime that is invariant? and under what transformations does it hold for? I think the law that is invariant is the speed of light(electromagnetic radiation which results from Maxwells laws of the electromagnetic field)[we could also say that what is invariant in the spacetime is its interval structure] and the transformations it is invariant under is the transformations of coordinates due to the "relative uniform motion" of the frames(motion between so called inertial frames).(2 votes)- Lets start off with non-Relativistic mechanics for simplicity.
Noethers theorem is based on the concept of something called action. Action is a function which is the time integral of the total kinetic energy minus the total potential energy of the system. (The laws of Newtonian mechanics can be derived using the principle of least action)
In the Wikipedia article on Noethers theorem there information about translational, rotational and time invarance as well as a generic derivation based on a single independent variable, the multi-variable version is much more complex.
Within Relativistic mechanics you don't have separate translational and time invarance they become a single invarance of space-time translation. What this translates to is something like conservation of liner momentum but it becomes the conservation of something called 4-momentium.(1 vote)
4:00was so amazing I can't even describe it. my question: Would it be accurate to say the Lorentz transformation is just a more informed and accurate method of Galilean transformations because we're following quantum rules?(1 vote)- It would literally be the most accurate statement.
Why? It was explained to me like this.
As physics gets older and older, our equations and predictions get more and more accurate. The discovery of special relativity, therefore, does not fundamentally change previous equations, it just adjusts them.
The best part? The some of equations we have now for the universe are not only incomplete, but show EXTREMELY SMALL inaccuracies. In other words, that equation itself might have another factor, or a slightly modified version. Good luck finding it!(2 votes)
- x' = gamma(x-beta*ct)
so if lorentz factor gamma is incredible large then x' will be incredibly bigger than x.
does this affect our reality? or what is the benefit of x' being incredibly larger than x.?(1 vote) - In what way is the function Y =[1/sqrt(1-v^2/c^2)] in the x scaling of the Galilean transformation seen as analogous to the projection operator functions cos Q evaluated at Q=tan-1 (v/c) and the Yv function analogous to the circular function sin, for projecting the x and t coordinates onto the x' tilted(at Q=tan-1(v/c) axis ? How do we see Y as a projection function analogous to the way we see cos as a projection function?(1 vote)
- At 5.03 where you say you are going to divide both sides by c to find t' why don't we divide Y by c also. Why don't we get (Y/c)[t-(v/c^2)x] instead of Y[t-(v^2/x^2)]?(1 vote)
- Any mistakes? Every time I look at it I notice something wrong.
E^2=p^2*c^2+m^2*c^4
γ=cos(arctan(i*v/c))
p=m*v*γ=m*c*sin(arctan(i*v/c))
E^2=(m*c*i*sinh(arctanh(v/c)))^2*c^2+m^2*c^4=m^2*c^4*(i*sinh(arctanh(v/c)))^2+m^2*c^4=m^2*c^4(1-sinh(arctanh(v/c))^2)
sqrt(E^2)=sqrt(m^2*c^4(1-sinh(arctanh(v/c))^2))
E=m*c^2*cosh(arctanh(v/c))
E=m*c^2*γ
for completion sake, if v>c then
E^2=(m*c*i*sin(arctan(i*c/v)))^2*c^2+m^2*c^4
E=m*c^2*sin(arctan(i*v/c))
E=m*c^2*γv*i/c
E=m*cγ*v*i
E becomes negative when v>c when it is squared (E^2). There is a transformation that takes place when v/c > 1.
x/sqrt(x^2-y^2)+i*y/sqrt(x^2-y^2)
x/(i*sqrt(x^2-y^2))+i*y/(i*sqrt(x^2-y^2))
-i*x/sqrt(y^2-x^2)+y/sqrt(y^2-x^2)(1 vote) - If the primed frame is the one that is moving (usually by convention this is how it is I have found from textbooks and such) shouldn't x'=x1/gamma as "moving rods shrink" and we know that gamma is equal to or greater than 1? or in this video is S' frame stationary and S is the movign frame?(1 vote)
- Given that the symmetry of spacetime gives us the Lorentz transformations between x and x'; ct and ct' as x'=Y(x - B ct)
and ct'=Y(ct - B x)
and that this shows that space and time are not separate but are unified into one homogenized concept called spacetime(ST). But what is the law in the spacetime that is invariant, and under what transformations does this symmetry(invariance) hold(what change is this symmetry applicable for)?
I think the law that is invariant is the speed of light law of electromagnetic radiation (Maxwells electromagnetic field law) and the transformation it is invariant under is the transformations of coordinates between uniformly moving systems(of one inertial frame relative to another inertial frame).
What quantity is conserved between the systems(is it the interval ds^2(but is this physical or just mathematical)?
Why does ST possesses such Symmetry?(0 votes)
The Lorentz transformations are derived from the fact that the speed of light remains constant in all frames of reference. Regarding why the speed of light is constant in all reference frames... this is just how the universe behaves based on numerous experimental observations.(2 votes)
4:00Video transcript
- What I hope to do in
this video is get even more algebraically familiar with
the Lorentz transformation, so that we can recognize
it in its different forms and start to build our
intuition for how it behaves. So let's just write down
the Lorentz transformation, or at least the way
that I like to write it. So let's just remind ourselves, if I'm in my frame of reference, I am floating through
space, so I could say s for Sal's frame of reference. Some event in space time,
from my point of view, is going to have some x-coordinate, I will do that in green,
and it's going to have some ct coordinate, I
can do that in orange. So the Lorentz transformations
are going to go from coordinates in my frame of reference, space-time coordinates for an event, to my friend's frame of reference, so we can say that's the
s-prime frame of reference, and her frame of reference, the event, will have space-time coordinates x-prime, let me write it this way: x-prime, comma, ct-prime, I'm really having trouble
switching colors today, ct-prime, so let's just
write it down the way I've written it down
in the previous videos, and then I'll do a little bit
of algebraic manipulations, so we can recognize its different forms. So if we want to get x-prime, we see that it's going to be based on the Lorentz factor times x minus a scaled version of ct, and the scaling factor is beta, and we will redefine beta in a second, so beta times ct, where our Lorentz factor, let me write it over here, the Lorentz factor is
one over the square root of one minus v-squared over c-squared, or we could write it as
one over the square root of one minus beta-squared, where beta is equal to v over c. So there you go. That's how we get x-prime
and it's gonna be based on the Lorentz factors dependent on v and of course the rest
of this is going to be dependent on x and ct. And so how do we get ct-prime? Well ct-prime is going to be equal to, I'll just write it right over here, ct-prime is going to be equal to the Lorentz factor times, and once again this is going to be the nice symmetry we talked about, times ct, we'll do that in orange, so ct minus beta times x. And like I said before, I
like to write it this way. I find it easier to remember. I find it easier to
remember because it has this nice, beautiful symmetry to it. When I'm trying to solve for x-prime, it's x minus beta times ct. When I'm solving for ct-prime,
it's ct minus beta times x. And in both cases, I'm
scaling by the Lorentz factor. Well let's manipulate this a little bit, just to understand a little better, and reconcile with what you might see with other sources,
including, say, your textbook. Well we know that beta
is equal to v over c, and this is v over c, and so we can write this
as v over c times c, this c's going to cancel with that c. And so we could rewrite this as x-prime is equal to the Lorentz factor, gamma times x minus vt. V times t. Now this is really
interesting right over here, 'cause if you'd ignored gamma,
or if gamma was one here, this is essentially the
Galilean transformation. If it was just x minus vt, that was just our intuition about our everyday life, but Newtonian physics
would actually tell us, and so when you view it in this form, you really think, okay, you're
just going to scale that by this Lorentz transformation, which has this interesting behavior that if v is much, much smaller
than the speed of light, well then, this whole factor is going to be pretty close to one, and that's why the Galilean
transformation's worked for us, for kind of everyday velocities. But then if v starts to
approach the speed of light, this thing blows up and we
get a very different result than with our traditional
Galilean transformation. Well let's think about
what happens over here, and over here, instead
of staying in ct-prime, I'm gonna also divide both sides by c, so we're just solving for time as we normally associate it. Just the t-prime variables
as opposed to ct-prime. So let's also divide both sides by c, so you divide by c there, and
you can divide by c there, and you can divide by c there, those c's cancel, those c's cancel, and so we're left with t-prime is equal to the Lorentz factor, times t times t minus, now you're going to get v times x, and you're going to divide by
c twice, so over c-squared. So vx, v times x over c-squared. And this is actually a more
typical way, both of these, of seeing the Lorentz transformation. The reason why I don't
like this form as much, even though this does
have the neat, kind of, when you look at it, it looks
like you're just scaling up the Galilean transformation, is it you no longer
see the symmetry there, and you should see the symmetry there, because we're talking about space-time. We're not talking about this
independence of space and time. We saw how the angles in
the Minkowski diagram, how those were symmetrical,
the angles between the regular, or the unprimed axes, and the primed axes. And so what I don't like about this is you no longer see the symmetry, while you did see it the
first way that I wrote it. And frankly I find this
harder to remember. But let's just think
about what happens here when v is a very small
fraction of the speed of light. Well as we've already
said, our Lorentz factor is going to be pretty close to one, and if v is a very small fraction of c, well then this second
term right over here, is going to be pretty close to zero. And so if v is a small fraction of c, then this thing is going to be get pretty, just let me just write this down, so if v is much less than c, then this is going to reduce to t-prime being approximately equal to, because our Lorentz factor is going to be pretty close to one, this is going to be pretty close to zero. So this is going to be pretty close to t. Likewise, for v much lower than c over here, our Lorentz factor is going
to be pretty close to one, and so x-prime is going to be approximately equal to x minus vt. So for small v's, and small can even be
the speed of a bullet, or even the speed of the space shuttle, or things that are much, much smaller than the speed of light. Three times 10 to the
eighth meters per second, that's why the Galilean transformations are pretty good approximations. So hopefully this starts to give you a little bit of intuition. Start evaluating this,
evaluate this for v's in our everyday life,
and then see what happens when v starts to approach
.8 times the speed of light. .9 c, .99 c, think about what
happens to the Lorentz factor. And hopefully you'll
get an appreciation for how this whole thing behaves.