Space-time interval Lorentz co/invariant

In summary, the student is trying to solve for ##\Delta s## by using Lorentz transforms, but he gets stuck on a problem with the expression for a boost in any direction. He tracks it down and finds that it reduces to the desired result when v=0.
  • #1
fluidistic
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Homework Statement


Hi guys!
I must show by brute force that ##\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2## is invariant under Lorentz transforms.


Homework Equations


Lorentz transforms:
##\Delta t' = \gamma \left ( \Delta t -\frac{\vec v \cdot \Delta \vec x}{c^2} \right )## and ##\vec \Delta x'=\gamma (\Delta \vec x - \vec v t)##.
I will use the convention ##\Delta \vec x= (\Delta x, \Delta y , \Delta z )## and ##\vec v = (v_x ,v_y, v_z)##

The Attempt at a Solution


Basically I must show that ##\Delta s'^2=\Delta s^2##. Where ##\Delta s'^2=-c^2\Delta t'^2+\Delta x'^2+\Delta y'^2+ \Delta z'^2##.
I have that ##\Delta x'^2+\Delta y' ^2+\Delta z'^2=\gamma ^2[\Delta x^2+\Delta y^2+\Delta z^2 +\Delta t^2 \underbrace{(v_x^2+v_y^2+v_z^2)}_{=\vec v^2}-2\Delta t (\Delta x v_x+\Delta yv_y +\Delta z v_z)]##
While ##\Delta t'^2=\gamma ^2 \{ \Delta t^2-\frac{2\Delta t}{c^2}(v_x\Delta x+v_y\Delta y +v_z \Delta z)+\frac{1}{c^4}[v_x^2 \Delta x^2+v_y^2\Delta y^2+v_z^2 \Delta z^2+2(v_x\Delta xv_y\Delta y+ v_y\Delta y v_z\Delta z+v_x\Delta x v_z\Delta z)] \}##.
So then I performed ##\Delta s'^2## and after all the simplifications I could do, I reached that ##\Delta s'^2=\gamma ^2 \left [ \Delta t^2 (\vec v^2-c^2) -\frac{\vec v \cdot \vec x}{c^2} +\Delta \vec x +\frac{2}{c^2} (v_x\Delta xv_y\Delta y+ v_y\Delta y v_z\Delta z+v_x\Delta x v_z\Delta z) \right ]## which does not seem equal to ##\Delta s##. I've redone the algebra 4 times (lost hours on this), I reach the same answer every time. So either the expression I found is right but I must simplify it even more and it indeed equals ##\Delta s##, or I'm wrong somewhere.
I'd appreciate some help... thank you!
 
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  • #2
fluidistic said:
##\Delta t' = \gamma \left ( \Delta t -\frac{\vec v \cdot \Delta \vec x}{c^2} \right )## and ##\vec \Delta x'=\gamma (\Delta \vec x - \vec v t)##.

For a boost in an arbitrary direction, I think your expression for ##\Delta t'## is ok. But the correct expression for ##\vec \Delta x'## is more complex than what you wrote. See for example

http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_any_direction

Are you sure you can't orient your axes so that the x-axis is parallel to the boost?
 
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  • #3
TSny said:
For a boost in an arbitrary direction, I think your expression for ##\Delta t'## is ok. But the correct expression for ##\vec \Delta x'## is more complex than what you wrote. See for example

http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_any_direction

Are you sure you can't orient your axes so that the x-axis is parallel to the boost?

I see, thank you very much.
Due to the complexity of the expression for a boost in any direction, I guess I can orient the axes so that it results to be in the x-direction. I will consider this case only. :biggrin:
 
  • #4
At first glance I get ##\Delta s'^2=\Delta t^2 (-c^2\gamma^2+v^2)+\gamma ^2 \left ( \Delta x^2+v^2 \Delta t^2 - \frac{v^2\Delta x^2}{c^2} \right )##.
I guess I'll have to redo the math.
Edit:Yeah I forgot 2 terms. I reach now ##\Delta s'^2=\Delta t^2[v^2+\gamma(v^2-c^2)]+\Delta x^2+\Delta y^2+\Delta z^2##. Almost the desired result but apperently I still have some job to do :)

Edit 2:This reduces to ##\Delta s'^2=\Delta t^2 (v^2-c^2)+\Delta \vec x##. I'm almost there. But I would reach the result only if v=0... which is not correct.
 
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  • #5
Everything multiplying ##\Delta t^2## should be proportional to ##\gamma^2##, so your expression in your first edit looks kinda odd.
 
  • #6
fluidistic said:
I reach now ##\Delta s'^2=\Delta t^2[v^2+\gamma(v^2-c^2)]+\Delta x^2+\Delta y^2+\Delta z^2##. Almost the desired result but apperently I still have some job to do :)

Trace back to see how you got the first term of ##v^2## in the first brackets. I don't think that should be there.
 
  • #7
TSny said:
Trace back to see how you got the first term of ##v^2## in the first brackets. I don't think that should be there.

I've tracked it down (already on my own), at ##\Delta x'=\gamma (\Delta x - v \Delta t)##. When I must take the square of this quantity, there's a term ##\gamma ^2 v^2\Delta t^2##. I couldn't cancel this term out though I must re-recheck the algebra.Edit: Nevermind, I made a mistake. I now reach the desired result! Thank you guys.
 
Last edited:

Related to Space-time interval Lorentz co/invariant

1. What is the space-time interval in the context of special relativity?

The space-time interval is a measure of the distance between two events in space and time, taking into account the effects of time dilation and length contraction. It is used in special relativity to describe the geometry of space and time.

2. What does it mean for the space-time interval to be Lorentz co/invariant?

A Lorentz co/invariant space-time interval means that it has the same value for all observers moving at different velocities. This is a fundamental principle of special relativity, which states that the laws of physics should be the same for all observers regardless of their relative motion.

3. How is the space-time interval calculated?

The space-time interval is calculated using the Minkowski metric, which takes into account the differences in space and time coordinates for different observers. It is given by the formula: Δs² = Δx² + Δy² + Δz² - (cΔt)² where c is the speed of light and Δs, Δx, Δy, Δz, and Δt represent the differences in space and time coordinates between two events.

4. What is the significance of a space-time interval being zero, positive, or negative?

If the space-time interval is zero, it means that the two events are occurring at the same time and in the same location for all observers. A positive space-time interval means that the events are separated by a distance in space and/or time, while a negative space-time interval indicates that the events are separated by a distance in space and/or time but in the opposite direction.

5. How does the concept of space-time interval relate to the concept of causality?

The space-time interval is closely related to the concept of causality, as it describes the order and relationship between events in space and time. In special relativity, causality is preserved for all observers, meaning that the cause of an event must always precede the effect in all reference frames. The space-time interval helps to determine this relationship between events and ensures that causality is maintained.

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