Lorentz Transformation/Time Dilation

In summary: Thanks for clearing that up!In summary, the equation for time dilation is t2-t1 = $(t2'-t1') which is the wrong result. The equation for length contraction is t1'=t2'=t.
  • #1
bon
559
0

Homework Statement



Ok so I know how to prove Time Dilation using the Lorentz transformation...

My problem is that I don't see why one couldn't use the Inverse Lorentz Transform and come to the opposite result as follows:

Time period in rest frame S: T = t2 - t1

ILT: t = $(t1' + B/c x') where $ is gamma

So: t2-t1 = $(t2'-t1') so T' = T/$ which is the wrong result...

I.e. how does one know that one must use the LT rather than ILT?

Thanks


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
  • #2
Why is that wrong? γ is greater than 1, so T' < T, as expected. The time elapsed in the moving frame is shorter than the time elapsed in the rest frame.
 
  • #3
No it should be the other way round. "Time Dilation" - time measured in any frame moving wrt rest frame should be dilated (i.e. longer time period) than in rest frame..
 
  • #4
I didn't look at it too closely, but I suspect that by using ILT you have implicitly inverted the rest frame and the moving frame.
Basically, you have relabelled T to T' and vice versa.
 
  • #5
Thanks, yes I agree. I just can't see where the inconsistency comes in in the maths... :S
 
  • #6
anyone?
 
  • #7
bon said:
No it should be the other way round. "Time Dilation" - time measured in any frame moving wrt rest frame should be dilated (i.e. longer time period) than in rest frame..
OK, so you're assuming the clock is in the rest frame and trying to calculate what someone in the moving frame would see?
 
  • #8
Your inconsistency stems from ignoring x' in the Lorentz transformation. It looks like you just erased it going from the ILT to your conclusion.
 
  • #9
but i though x1' = x2' since the two times were measured at the same position in S'?
 
  • #10
If the clock is at rest in S, it's moving in S', so x1' and x2' won't be equal.
 
  • #11
Ahh this is getting confusing now. Ok so its measured in S at a position x to have time interval t2-t1

Now what happens when the measurement is made in S' - what stays constant? My understanding was that the time interval is measured to be t2'-t1' made at a fixed position in S' - call this x'..

What have i don wrong?

thanks for your help
 
  • #12
Basically what i don't understand is the asymmetry:

using your above expression

time dilation: clock at rest in S. moving in S' therefore x1' and x2' aren't equal..
length contraction: clock at rest in S. moving in S'. Therefore t1' and t2' aren't equal..

but for length contraction t1' and t2' are equal - we use the inverse LT:

x1 = γ(x1'+Bct')
x2 = γ(x2' + Bct')

i.e. t1'=t2'=t
which gives the right result: x2-x1 = γ(x2'-x1')
 
  • #13
For time dilation, you're comparing the time elapsed between the same two events in different frames.

When measuring length, you're finding the spatial separation of, say, the two ends of a stick at the same instant. Since simultaneity is frame dependent, you're looking at different pairs of events in different frames.
 
  • #14
Thanks ok - so what went wrong in my last post? (no. 12)..

Thanks
 
  • #15
?
thanks
 
  • #16
Just when you said this: "length contraction: clock at rest in S. moving in S'. Therefore t1' and t2' aren't equal." When you measure the length in S', you have to have t1'=t2', right? You have to measure where the ends are at the same time (in S').
 
  • #17
but then by parity of argument, when we measure time in S', we have to measure at x1'=x2', right? You have to measure the time at the same point in space?
 
  • #18
No. The reason you have to measure where the endpoints are simultaneously is just an operational requirement. If the stick is moving, you can't measure where one end is at one time, measure where the other end is a second later, and subtract the two to get the length. You have to see where both ends are at the same time, and only then can you subtract those numbers to get the length.

With time dilation, you have two spacetime events, and different observers will say different amounts of time separate them, depending on their relative motion.

The symmetry you're looking for isn't there. Time dilation compares the same two events in different frames. Length contraction compares a pair of events in one frame to a different pair in another.
 

1. What is Lorentz Transformation?

Lorentz Transformation is a mathematical formula that describes the relationship between space and time in the theory of special relativity. It was developed by Dutch physicist Hendrik Lorentz and later refined by Albert Einstein.

2. How does Lorentz Transformation relate to time dilation?

Lorentz Transformation is responsible for the phenomenon of time dilation, which is the slowing down of time for an object in motion relative to an observer. This means that time will appear to pass at a different rate for the moving object compared to the observer.

3. Can you provide an example of time dilation?

One example of time dilation is the famous "twin paradox." In this scenario, one twin stays on Earth while the other travels at high speeds through space. When the traveling twin returns, they will have aged less than the twin who stayed on Earth due to the effects of time dilation.

4. How does Lorentz Transformation affect the speed of light?

Lorentz Transformation also explains the constant speed of light in a vacuum, regardless of the relative motion of the observer. This is a fundamental principle of special relativity and is known as the "Lorentz invariance."

5. Is Lorentz Transformation still relevant in modern physics?

Yes, Lorentz Transformation is still a fundamental part of modern physics and is used in many areas of research, including quantum mechanics and cosmology. It has been extensively tested and confirmed through various experiments and is an essential tool for understanding the behavior of matter and energy in the universe.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
853
  • Introductory Physics Homework Help
Replies
21
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
707
  • Introductory Physics Homework Help
Replies
3
Views
744
  • Introductory Physics Homework Help
Replies
4
Views
969
  • Introductory Physics Homework Help
Replies
1
Views
892
  • Special and General Relativity
2
Replies
54
Views
1K
  • Special and General Relativity
Replies
32
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
823
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top