Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    A Lorentz boost in the x direction looks like \begin{bmatrix}\gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} where \beta=v/c and \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

    Everywhere I am reading it mentions boosts in single directions. Is boosting restricted to either x, y or z?

    For example to calculate the Lorentz boost for a direction I can take the Lorentz factor \gamma=\frac{1}{\sqrt{1-\frac{v_x^2+v_y^2+v_z^2}{c^2}}} and \beta_x=\frac{v_x}{c}, \beta_y=\frac{v_y}{c}, \beta_z=\frac{v_z}{c}

    How do I calculate a boost in x, y and z simultaneously?

    Lastly how does this differ from rotational? What exactly is meant by non rotating space?

    Thanks.
    Offline

    3
    ReputationRep:
    ngl, if I'd gone to my dynamics lectures this year I'd probably be able to help you. Soz
    • Thread Starter
    Offline

    2
    ReputationRep:
    Pessimisterious
    Offline

    12
    ReputationRep:
    (Original post by AishaGirl)
    A Lorentz boost in the x direction looks like \begin{bmatrix}\gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} where \beta=v/c and \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

    Everywhere I am reading it mentions boosts in single directions. Is boosting restricted to either x, y or z?

    For example to calculate the Lorentz boost for a direction I can take the Lorentz factor \gamma=\frac{1}{\sqrt{1-\frac{v_x^2+v_y^2+v_z^2}{c^2}}} and \beta_x=\frac{v_x}{c}, \beta_y=\frac{v_y}{c}, \beta_z=\frac{v_z}{c}

    How do I calculate a boost in x, y and z simultaneously?

    Lastly how does this differ from rotational? What exactly is meant by non rotating space?

    Thanks.
    I think the boosts are restricted to one direction for ease of understanding. In a lot of physics, we define things in their simplest terms, or simplest co-ordinate frames, knowing with confidence that the thing we're dealing with can be applied to anything more complex.

    In the case of Lorentz boosts, it obviously applies in any direction so doing it along the simple x-axis makes sense.

    Since all movements are in the x-y-z axes, all you'd need to do is a Lorentz boost on each axis separately.

    Alternatively, there's this which I found after a quick google search:


    (Image from here: https://en.wikipedia.org/wiki/Lorent...ransformations)

    If you look closely you should be able to see it's just the Lorentz boost on every axis at once.

    It's different from rotational space because a rotating frame adds another form of movement. So for example, on earth every movement you make in the northerly direction is incrementally altered to one side because of the ground very slightly rotating beneath your feet.

    Without going into any derivation at all, when something is accelerated in a rotating frame, the acceleration it feels isn't in a straight line, because the rotation affects it, leading to:

    \frac{d\vec{a}}{dt}|_{_{I}} = \frac{d\vec{a}}{dt}|__{{R}} + (\Omega \times \vec{a})

    where the left-hand derivative is the acceleration of the object in its own stationary ( 'inertial' )frame, the right-hand derivative is the acceleration of it in the rotating frame, and \Omega is the rate of rotation.

    There will be Lorentz boosts for this, but it'll need to factor in the situation I've alluded to in that little equation.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Pessimisterious)
    I think the boosts are restricted to one direction for ease of understanding. In a lot of physics, we define things in their simplest terms, or simplest co-ordinate frames, knowing with confidence that the thing we're dealing with can be applied to anything more complex.

    In the case of Lorentz boosts, it obviously applies in any direction so doing it along the simple x-axis makes sense.

    Since all movements are in the x-y-z axes, all you'd need to do is a Lorentz boost on each axis separately.

    Alternatively, there's this which I found after a quick google search:


    (Image from here: https://en.wikipedia.org/wiki/Lorent...ransformations)

    If you look closely you should be able to see it's just the Lorentz boost on every axis at once.

    It's different from rotational space because a rotating frame adds another form of movement. So for example, on earth every movement you make in the northerly direction is incrementally altered to one side because of the ground very slightly rotating beneath your feet.

    Without going into any derivation at all, when something is accelerated in a rotating frame, the acceleration it feels isn't in a straight line, because the rotation affects it, leading to:

    \frac{d\vec{a}}{dt}|_{_{I}} = \frac{d\vec{a}}{dt}|__{{R}} + (\Omega \times \vec{a})

    where the left-hand derivative is the acceleration of the object in its own stationary ( 'inertial' )frame, the right-hand derivative is the acceleration of it in the rotating frame, and \Omega is the rate of rotation.

    There will be Lorentz boosts for this, but it'll need to factor in the situation I've alluded to in that little equation.
    Yeah I figured that you can boost in 3Dimensions but I wanted to check because every single website I looked at showed boosts in single directions. So if observer O and observer O\prime are travelling in the x direction with the same constant velocity and observer O gets a boost in the x direction, observer O and O\prime will agree on y and z but not on x and t?

    Forgive me for being stupid but where is length contraction taken into account? They certainly would not both agree on the length of the object would they.
    • Thread Starter
    Offline

    2
    ReputationRep:
    double post
    Offline

    12
    ReputationRep:
    (Original post by AishaGirl)
    Yeah I figured that you can boost in 3Dimensions but I wanted to check because every single website I looked at showed boosts in single directions. So if observer O and observer O\prime are travelling in the x direction with the same constant velocity and observer O gets a boost in the x direction, observer O and O\prime will agree on y and z but not on x and t?

    Forgive me for being stupid but where is length contraction taken into account? They certainly would not both agree on the length of the object would they.
    They'll agree everywhere - that's the joy of the Lorentz transforms. I can't remember exactly how you apply it, but the whole point of the Lorentz transformation is that two objects at different velocities will have the same laws of physics relative to each other, with the only limit being the speed of light.

    Length contraction is a bit different. The two objects are related by that length contraction, so Object 1 sees Object 2 contracted in a certain way, and Object 2 sees the contraction in the same way happening on Object 1. Both objects experience the same thing, since their movement is relative to each other.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Pessimisterious)
    They'll agree everywhere - that's the joy of the Lorentz transforms. I can't remember exactly how you apply it, but the whole point of the Lorentz transformation is that two objects at different velocities will have the same laws of physics relative to each other, with the only limit being the speed of light.

    Length contraction is a bit different. The two objects are related by that length contraction, so Object 1 sees Object 2 contracted in a certain way, and Object 2 sees the contraction in the same way happening on Object 1. Both objects experience the same thing, since their movement is relative to each other.
    OK I think I understand it. Thanks.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Pessimisterious I'm so sorry to keep bothering you, this is the last time I'll ask specifically for your help. If you can just check this question for me that would be great. I didn't want to bother making another thread.

    Using the Lorentz transformations \Delta x and \Delta t, demonstrate (\Delta s)^2 is invariant.

    So I have the matrix to show a boost in the x direction given by \begin{bmatrix}\gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.

    So I can start off with:

    ct\prime=\gamma(ct)-\beta\gamma x

    x\prime=-\beta\gamma(ct)+\gamma x

    Now after squaring both equations:

    (ct\prime)^2=(\gamma ct-\gamma\beta x)^2

    =[(\gamma ct)^2+(\beta\gamma x)^2-2(\beta c\gamma^2 xt)]

    and:

    (x\prime)^2=[(\beta\gamma ct)^2+(\gamma x)^2-2(\beta c\gamma^2 xt)]

    The interval of frame S is defined as (\Delta s\prime)^2=(\Delta ct\prime)^2-(\Delta x\prime)^2 so this becomes

    (s\prime)^2=(ct)^2\gamma^2[1-\beta^2]+(x)^2\gamma^2[\beta^2-1]

    and finally using the definition of Lorentz factor I can reduce this to:

    (s\prime)^2=(ct)^2\gamma^2[1/\gamma^2]+(x)^2\gamma^2[-1/\gamma^2]

    =(ct)^2-(x)^2=(s)^2 and therefore (s\prime)^2=(s)^2

    There's probably a mistake in there somewhere. Is this how you would prove it?

    This will be the last thing I ask from you I promise
    Offline

    12
    ReputationRep:
    (Original post by AishaGirl)
    Pessimisterious I'm so sorry to keep bothering you, this is the last time I'll ask specifically for your help. If you can just check this question for me that would be great. I didn't want to bother making another thread.

    Using the Lorentz transformations \Delta x and \Delta t, demonstrate (\Delta s)^2 is invariant.

    So I have the matrix to show a boost in the x direction given by \begin{bmatrix}\gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.

    So I can start off with:

    ct\prime=\gamma(ct)-\beta\gamma x

    x\prime=-\beta\gamma(ct)+\gamma x

    Now after squaring both equations:

    (ct\prime)^2=(\gamma ct-\gamma\beta x)^2

    =[(\gamma ct)^2+(\beta\gamma x)^2-2(\beta c\gamma^2 xt)]

    and:

    (x\prime)^2=[(\beta\gamma ct)^2+(\gamma x)^2-2(\beta c\gamma^2 xt)]

    The interval of frame S is defined as (\Delta s\prime)^2=(\Delta ct\prime)^2-(\Delta x\prime)^2 so this becomes

    (s\prime)^2=(ct)^2\gamma^2[1-\beta^2]+(x)^2\gamma^2[\beta^2-1]

    and finally using the definition of Lorentz factor I can reduce this to:

    (s\prime)^2=(ct)^2\gamma^2[1/\gamma^2]+(x)^2\gamma^2[-1/\gamma^2]

    =(ct)^2-(x)^2=s^2 and therefore (s\prime)^2=s^2

    There's probably a mistake in there somewhere. Is this how you would prove it?

    This will be the last thing I ask from you I promise
    No problem, I don't mind being asked at all. In truth it's a subject I really didn't enjoy when covering it at uni so I'm happy that you're forcing me to go over it.

    Looking at your proof there, it seems about right to me.

    There's a more condensed version at the bottom an image I posted in one of your earlier threads:

 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.