How do Vectors transform from one inertial reference frame to another inertial reference frame in special relativity. A bound vector in an inertial reference frame (x,ct) has its line of action as one of the space axis in that frame and is described by x*i*,then what would it be in form of new base vectors (a) and (b) in a different inertial x‘,ct‘) moving with respect to the former inertial system with v*i* velocity.Let (i) and (j) be the two bounded unit vectors with the line of action as co-ordinate axis(x) and (ct) respectively and senses in the positive side of co-ordinates and similarly (a) and (b) are defined for co-ordinates (x‘) and (ct‘) respectively.

easternerjx 2022-09-17 Answered
How do Vectors transform from one inertial reference frame to another inertial reference frame in special relativity.
A bound vector in an inertial reference frame ( x, c t) has its line of action as one of the space axis in that frame and is described by x* i*,then what would it be in form of new base vectors ( a) and ( b) in a different inertial system ( x , c t ) moving with respect to the former inertial system with v*i* velocity.Let (i) and (j) be the two bounded unit vectors with the line of action as co-ordinate axis( x) and ( c t) respectively and senses in the positive side of co-ordinates and similarly ( a) and ( b) are defined for co-ordinates ( x ) and ( c t ) respectively.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Ruben Horn
Answered 2022-09-18 Author has 7 answers
Well, vectors (3D vectors) don't really transform linearly. Unlike Galilean transformations, you need now know anything "extra" when transforming a vector. Here, due to the "mixing" of space and time, you do. To transform displacement, you need to know time, and vice versa. Same with energy and momentum.
Four-vectors, on the other hand, transform linearly. These are four-dimensional vectors which transform linearly via the Lorentz matrix ( β = v c , γ = 1 1 β 2 ):
L = [ γ β γ 0 0 β γ γ 0 0 0 0 1 0 0 0 0 1 ]
For example, if you want to transform position and/or time, you use the four-position
X = [ c t x y z ]
this can be compactly written as ( c t , x )-this just means that you can expand the second "vector" term to get the next three four-vector components.
Anyway, the four vector transforms as:
X = L × X
(matrix product)
This, expanded, is:
[ c t x y z ] = [ γ β γ 0 0 β γ γ 0 0 0 0 1 0 0 0 0 1 ] [ c t x y z ] ,
which are your normal lorentz transformations. A property of four vectors is that if we are talking about the same four vector ( a , b ) in two frames, the value of a 2 | b | 2 is the same in both. For four-position, you get c 2 t 2 x 2 y 2 z 2 = c 2 t 2 x 2 y 2 z 2
Other four vectors are:
Four-velocity: ( γ c , γ u )
Four-momentum: ( E / c , p )
Four-current density: ( γ ρ , J )
Four-potential: ( ϕ c , A )

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-09-27
Einstein's Special Theory of relativity postulates that the speed of light is same for all frames.
Suppose a neutrino is there moving at the speed of light. Then will that neutrino also be flowing with same speed for all frames or is this a special property of EM waves?
asked 2022-05-18
Environmentally induced decoherence makes wave function collapse unnecessary. But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.) So, doesn't then the decoherence time depend on the motion of the prepared state relative to the environment? And, doesn't the ultimate environment, all particles in the universe, introduce a preferred frame into quantum mechanics in the sense that the decoherence time is relative to this frame? And would this be measureable, at least in principle
asked 2022-05-19
Is there a time + two spatial dimension representation of a Minkowski-space surface which could be constructed within our own (assumed Euclidean) 3D space such that geometric movement within the surface would intuitively demonstrate the “strange" effects of the Lorentz transformation (length contraction, time dilation)?
asked 2022-08-18
The kinetic energy is, in the usual form m 1 v 2 . As we've seen before, energy can be gained or lost, so if we want to, we can add an arbitrary quantity. Therefore, we can take the energy using 1 2 m v 2 . (...) But it is better to use the formula as m 1 v 2 , because it is valid for the conservation law.
asked 2022-09-23
Trying to derive the infinitesimal time dilation relation d t = γ d τ, where τ is the proper time, 𝑡 the coordinate time, and γ = ( 1 v ( t ) 2 / c 2 ) 1 / 2 the time dependent Lorentz factor. The derivation is trivial if one starts by considering the invariant interval d s 2 , but it should be possible to obtain the result considering only Lorentz transformations. So, in my approach I am using two different reference frames ( t , x ) will denote an intertial laboratory frame while ( t , x ) will be the set of all inertial frames momentarily coinciding with the observed particle, i.e. the rest frame of the particle. These frames are related by
t = γ ( t V x c ) , x = γ ( x V t ) ,
where V is some nonconstant (i.e. time dependent) parameter which is, hopefully, the velocity of the particle in the laboratory frame. Treating x, t and V as independent variables (for now) and taking the differential of the above relations,
d t = γ ( d t V d x c ) γ 3 c 2 ( x V t ) d V , and
d x = γ ( d x V d t ) γ 3 ( t V x c ) d V .
Imposing either the definition of the rest frame d x = 0 or (what should be equivalent) d x = V d t, the only way in which i obtain d t = γ d t is if d V = 0. So, the derivation breaks badly at some point or I must be wrong in using some of the above equations. Which one is it?
asked 2022-07-15
If an object has very energetic particles in it like that of the sun then wouldn't its mass be higher hence making its gravity greater than that of the still state ones ?
asked 2022-05-18
I am asked to show that the matrix
S = 1 i 4 σ μ ν ε μ ν
represents the infinitesimal Lorentz transformation
Λ μ ν = δ μ ν + ε μ ν ,
in the sense that
S 1 γ μ S = Λ μ ν γ ν .
I have already proven that S 1 = γ 0 S γ 0 , so I can begin with the left-hand side of this third equation:
S 1 γ μ S = γ 0 S γ 0 γ μ S = γ 0 ( 1 + i 4 σ μ ν ε μ ν ) γ 0 γ μ ( 1 i 4 σ μ ν ε μ ν ) = γ 0 γ 0 γ μ γ 0 γ 0 γ μ i 4 σ μ ν ε μ ν + γ 0 i 4 σ μ ν ε μ ν γ 0 γ μ γ 0 i 4 σ ρ τ ε ρ τ γ 0 γ μ i 4 σ μ ν ε μ ν = γ μ + 1 16 γ 0 σ ρ τ ε ρ τ γ 0 γ μ σ μ ν ε μ ν = γ μ + 1 16 ( σ ρ τ ε ρ τ σ μ ν ε μ ν ) γ μ .
Remember, we want this to be equal to
Λ μ ν γ ν = ( δ μ ν + ε μ ν ) γ ν = γ μ + ε μ ν γ ν .
The first term is there already, but I have no idea how that second term is going to work out to ε μ ν γ ν . Can anyone give me a hint, or tell me what I’ve done wrong?

New questions