Lorentz transformations losing on both sides

In summary, the conversation discusses two events in an inertial system K and the possibility of finding a frame K' in which the two events occur at the same place. The Lorentz transformation is used to determine if a frame K' exists, but it is found that the velocity needed for this transformation would be greater than the speed of light, making it impossible. The conversation also mentions an alternative method of checking if this is possible, by computing if the difference in space and time between the two events is smaller than the speed of light.
  • #1
Ascendant78
328
0

Homework Statement



Two events occur in an inertial system K as follows:
Event 1: x1 = a, t1 = 2a/c, y1 = 0, z1 = 0
Event 2: x2 = 2a, t2 = 3a/(2c), y2 = 0, z2 = 0

Is there a frame K' in which the two events described
occur at the same place? Explain.

Homework Equations



Lorentz transformation:
x' = (x-vt)/(sqrt(1-(v/c)^2)

The Attempt at a Solution



I set both equations up as x' (equal to each other), plugged in the values on each side for x1 and x2, and it resulted in me losing 'a' on both sides (cancels out). I'm assuming that this means there is no frame K' where they would occur in the same place, but I am a bit surprised at that answer and am not sure how to explain it. Can someone explain why I am losing 'a' on both sides of the equation and what that means when it occurs?
 
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  • #2
You are losing a on both sides because both t and both x are proportional to it. This is not the issue, the question is if you can find a v < c such that the equation you got is fulfilled. Can you?
 
  • #3
Orodruin said:
You are losing a on both sides because both t and both x are proportional to it. This is not the issue, the question is if you can find a v < c such that the equation you got is fulfilled. Can you?

Thanks for the information. As far as a v < c, the final answer I get is that v = -2c. Here is what I work with:

a - v(2a/c) = 2a - v(3a/(2c))

a - v(4a/(2c) = 2a - v(3a/(2c))

-a = va/(2c)

-2c = v

So, from what you're telling me, this basically means that it is impossible because the velocity would need to be 2x the speed of light in the -x direction?
 
  • #4
If |v| > c, then the gamma factor of the transformation is no longer real. There simply are no Lorentz transformations that have |v| ≥ c. But you are correct in that there is no such transformation and therefore no such frame.

A different way of looking at it would be to simply check whether or not you can travel from one event to the other with a speed smaller than c. The statements are equivalent. This is easiest done by computing if |Δx| < c |Δt|.
 
  • #5
Well thanks a lot for the information. I really appreciate it. Makes perfect sense.
 

FAQ: Lorentz transformations losing on both sides

How do Lorentz transformations lose on both sides?

Lorentz transformations describe the relationship between space and time in special relativity. When an object is moving at relativistic speeds, these transformations show that time and space measurements appear differently for observers in different reference frames. This is known as time dilation and length contraction. Therefore, it can be said that Lorentz transformations "lose" on both sides as they account for the difference in measurements between reference frames.

Are Lorentz transformations always symmetrical?

Yes, Lorentz transformations are always symmetrical. This means that the equations for transforming from one reference frame to another are the same regardless of the direction of the transformation. This symmetry is a fundamental principle of special relativity.

Can Lorentz transformations be applied to any reference frame?

Yes, Lorentz transformations can be applied to any reference frame, as long as the frame is in uniform motion relative to another frame. They can also be used to transform between frames in which one or both are accelerating, but the equations become more complex and may involve incorporating concepts from general relativity.

What happens to the speed of light in Lorentz transformations?

The speed of light is a constant in all reference frames, according to the principles of special relativity. Therefore, Lorentz transformations do not affect the speed of light. This is known as the "invariance of the speed of light."

Are Lorentz transformations only applicable in special relativity?

Yes, Lorentz transformations are only applicable in special relativity, which deals with the relationship between space and time for objects moving at constant speeds. In contrast, general relativity describes the effects of gravity on the curvature of spacetime and requires more complex equations to transform between reference frames.

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